Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Weight in Interplanetary Space How is weight zero in interplanetary
space? The Moon is orbiting the Earth because of the gravitational pull of earth. Then gravity must exist in interplanetary space too. So any body in space must also have an acceleration due to gravity ($g$) but $g$ must actually be 0 for weight to be zero.
Can anyone please help me with this?
| Weight depends on the reference frame.
In the reference frame of the Earth, the gravitational acceleration onboard the International Space Station is about 8.5 m/s$^2$ (about 90% of $g$).
In the reference frame of the ISS, the gravitational acceleration onboard the ISS is between -0.0001 and +0.0001 m/s$^2$, depending on whether you are closer to the floor or to the ceiling.
So the weight of an astronaut depends on which reference frame you want to express it in.
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Determining the reverse breakdown voltage of a Zener diode graphically I have been measuring the reverse bias IV characteristics of a Zener Diode BZX (2.7 V) at three different temperatures, -196.2 celcius (liquid nitrogen temperature), 22.0 celcius (room temperature) and 99.0 celcius. Now I need to determine the breakdown voltages at said temperatures. Below is a plot of my results, which I processed on python using matplotlib.
I understand it is quite easy to roughly determine the breakdown voltage by eye but I was wondering if there was a more sophisticated and accurate way of determining the breakdown voltage. One thought I had was to fit an exponential of the form $y = -e^{-kx}$ but then I'm not sure which parameter of the exponential would correspond to the breakdown voltage. Perhaps I could then (numerically) take the second derivative of the data, set it to equal 0, which would then give me the voltage at which the slope is steepest (and thus corresponds to the breakdown region?) Is there another method I'm missing which could be more accurate?
Also, as an aside, my next step is to calculate the temperature coefficient for the zener diode. Is this simply a matter of calculating the following: $$T_{c} = \frac{V_2 - V_1}{T_2 - T_1}$$ the numerator is the difference between the breakdown voltage measured at temperature 2 and temperature 1, and the denominator is the change in temperature between temperature 2 and temperature 1?
| Function at extremum point (minimum or maximum) does not change, so you need to evaluate differential :
$$ \frac {d}{dx} y = \frac {d}{dx} \left(-e^{-kx}\right) = k e^{-k x} \tag 1$$
Now expand (1) into Taylor series, take couple of terms and set it to zero :
$$ k - k^2 x = 0 \tag 2$$
Solving (2) for $x$, gives approximate breaking voltage :
$$ V = \frac 1k~~~~~~\text{for}~k \ne 0 \tag 3$$
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Order Parameter, Phase transition
*
*Is order parameter for a phase transition is unique?
*Is it always true that in one phase order parameter have zero value and in another phase it has non zero value?
*Is there any standard rules for chosing order parameters?
*Can a system undergoing phase transition can have more than one order parameter?
| Order parameter is defined as a quantity that is zero in one phase and not zero in the other - this answers the second question, but also the first one, since any quantity satisfying this condition can be an order parameter. In particular, if $M$ is an order parameter, then any function of $M$ that leaves it zero in one phase and non-zero in the other is also an order parameter.
To my knowledge there are no standard rules, since the order parameter is often a physical property, specific to the system considered (e.g., magnetization for ferromagnet, density for liquid-gas transition, etc.)
As systems may have very complex phase diagrams, there may be multiple order parameters, controlling where on the phase diagram the system finds itself. See an example in the figure below (taken from here)
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What gives us the equations of motion in GR? Maybe stupid question, but to my understanding, the Einstein equation tells us the differential equation governing the Geometry of spacetime. That's all good and fine , but suppose I had an actual particle with some forces acting on it, how would I find it's trajectory? Would Newton's second law still hold?
Absurd example: Hypothetically suppose that I wanted to study block on a ramp with friction but with the system near a black hole. If I were to do this, then how exactly can I fit in the fact that the block is effected by friction into the idea that the block follows Geodesics in space time..?
| Yes, Newton's first law sort of holds. You have to generalize it to curved geometries. You can think of Newton's first law as the statement that "In the absence of external forces, a particle's velocity gets parallel transported along its trajectory".
In the absence of curvature, the parallel transport is the trivial one : A straight line trajectory (assuming a nice co-ordinate system).
In the presence of curvature, there is no nice co-ordinate system which will give you the trivial straight line parallel transport. Instead, you have to use the geodesic equation, which parallel transport a vector in a general geometry.
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Experimentally Measuring the Velocity of Water coming out of an Orifice I plan on doing an investigation into Torricelli's Law, where I will be looking at one of the following:
*
*How the cross-sectional area of an orifice affects the velocity of water coming out of it (constant height).
*How the height of an orifice affects the velocity of water coming out of it (constant orifice area).
However, I was unsure about how to accurately measure the velocity of water coming out of the orifice. Videos on YouTube only suggest one method, which is using the horizontal and vertical displacements of the water stream to calculate velocity. However, when I've done this experimentally I've found an about $15\%$ error compared to expected values. The process is also not very exact per se, i.e. it is hard to judge the exact marking of a ruler that the stream lands on.
Therefore, I was wondering if there were any accurate means to measure the velocity of water coming out of an orifice, using equipment typically found in a school laboratory.
| Water is incompressible. So you can put a bucket under your stream and let it run for some time. That way you measure the volume of water coming out per time [units m$^3/$s]. Divide by the cross-sectional area of the orifice [m$^2$] to get the water speed [m$/$s].
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Can we determine what the wavefunction (states) of a particle is before we decide which measurement to make? If we are measuring spin-up or spin-down then we write the wavefunction (I think) as
$$ \require{physics} \psi = \tfrac{1}{\sqrt{2}} |{\uparrow}\rangle+\tfrac{1}{\sqrt{2}} |{\downarrow}\rangle $$
But if we emit a particle, only to make a measurement after a long period of time and distance, and we haven't decided what to measure yet, does the particle still have a wave function? We might measure up/down, left/right or some other quantum parameter.
Would it include all possible measurements each with 0 amplitude (since they are infinite as many). And after we decide to measure up or down you have a wavefunction coalescing into two distinct states? Kind of like a sub-collapsing before the actual collapsing that happens during the measurement.
*
*What can we know apriori before the measurement, and how is that encoded into the wavefunction?
*What about two entangled particles? Is there a relationship in their wavefunctions before measurements? Is a relationship like $\int \langle \psi_1 | \psi_2 \rangle = 0 $ where the two wavefunctions are "orthogonal" to each other, or something like that?
| Note that your spin 1/2 wave function:
$$ \require{physics} \psi = \tfrac{1}{\sqrt{2}} |{\uparrow}\rangle+\tfrac{1}{\sqrt{2}} |{\downarrow}\rangle = |\rightarrow\rangle $$
where:
$$|\rightarrow\rangle \equiv |S=\tfrac 1 2, S_x=\tfrac 1 2 \rangle $$
That is, it is an eigenstate of the $\hat S_x$ operator.
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Why is an equilateral triangle not a 2d unit cell? An equilateral triangle obeys the crystallographic restriction theorem, but it is not a part of 2d crystal structure. What symmetry does it lack? Why can't it be a Bravais lattice?
| It is impossible to make a Bravais lattice of triangular cells (equilateral or not). It is the essence of the Bravais lattice concept that it is based on translational symmetry (in 2D, along two independent directions). In order to build a Bravais lattice with an equilateral triangular cell, one would need to ensure that, by rigidly displacing a triangle along two of its sides, it is possible to have a tessellation (i.e., a covering) of the plane without overlapping and without holes. This is not possible with triangles. One needs parallelograms or other 2D figures that do not need rotations after displacement to tassell the plane.
What could be misleading is that, in 2D, there is the so-called triangular lattice. This is a misname. The elementary cell is a rhombus with $60^{\circ} $ acute angles. The Wigner-Seitz cell is instead a regular hexagon made of six equilateral triangles. However, no cell is a triangle. If we see it as a crystalline structure, it can be thought as a triangular Bravais lattice with a basis made by two equilateral triangles sharing one side. But that it is not a Bravais lattice of triangles.
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Balmer proportionality How did Johannes Balmer arrive at
$$
\lambda \propto \frac{n^2}{n^2-4}, \quad (n=3,4,\dots),
$$
and then how did Rydberg mathematically derive
$$
\frac{1}{\lambda}=R\left(\frac{1}{n^2_1}-\frac{1}{n^2_2}\right)?
$$
I know $n$ stands for the shells but in the textbook, it doesn't define what $n$ is at first. Was this because Balmer did not know what shells were at that time?
| They looked at the patterns in the spectra and found formulae that matched them. Physics isn't derived from mathematics: the phenomena drive the mathematical models.
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Difference between Symmetric and Phase difference? For a simple coupled oscillator system such as the one here, with equal spring constants and equal masses (with a displacement from equilibrium of $x_1$ and $x_2$), it follows that:
$(\ddot{x}_1+\ddot{x}_2)\propto x_1+x_2$
$(\ddot{x}_1-\ddot{x}_2)\propto x_1-x_2$
Wikipedia refers to this as the 'normal modes' of the system and the former as antisymmetric and the latter as symmetric. Does this imply that these proportionalities demonstrate the masses are in phase and in antiphase?
(Note for answering: this is my first attempt at understanding coupled oscillators; I am not familiar with modes in a generic sense either. If it helps, I visualised these equations as the dot-products of the displacements and acceleration vectors of $x_1$ and $x_2$ with the eigenvectors generated with their coupled differential equations, as a way of projecting the transformation of $(x_1,x_2)$ to $(x_1,x_2)''$ onto components along lines of invariance [if that makes sense])
| IMHO, the best approach to mass-spring systems is via matrix formulation,
$\underline{\underline{M}}\underline{\ddot{x}} + \underline{\underline{K}}\underline{x} = \underline{F} $,
where the vector $\underline{x}(t)$ collects the degrees of freedom of the problem, and finding the eigensolution of the homogenous system,
$\left[s^2 \underline{\underline{M}} + \underline{\underline{K}} \right]\underline{\hat{x}} = \underline{0} $.
In your example,
$\underline{\underline{M}} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}$
$\underline{\underline{K}} = \begin{bmatrix} 2 k & -k \\ -k & 2 k \end{bmatrix}$
and the determinant of the matrix $s^2\underline{\underline{M}} +\underline{\underline{K}} $ reads
$\det (s^2\underline{\underline{M}} +\underline{\underline{K}} ) = (s^2 m + 2k)^2 - k^2 = m^2 s^4 + 4 km s^2 + 3 k^2$
the eigenvalues reads
$s_{1,2}^2 = - \left[ 2 \pm
1 \right]\dfrac{k}{m} $
and thus
$s_1^2 = - \dfrac{k}{m} $$\quad \rightarrow \quad$$s_{1\pm} = \pm j \sqrt{\dfrac{k}{m}}$
$s_2^2 = - 3 \dfrac{k}{m} $$\quad \rightarrow \quad$$s_{2\pm} = \pm j \sqrt{3\dfrac{k}{m}}$,
while the corresponding eigenvectors are:
$\underline{\hat{x}}_{1} =\begin{bmatrix} 1 \\ 1 \end{bmatrix}$,$\quad$$\underline{\hat{x}}_{2} =\begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
The first eigensolution is the mode with the two masses oscillating "in phase" with the same amplitude and pulsation $\omega_1 = \sqrt{\frac{k}{m}}$.
The second eigensolution is the mode with the two masses oscillating "in anti-phase" with the same amplitude and pulsation $\omega_2 = \sqrt{3\frac{k}{m}}$.
We can use these eigensolutions to build the general solution of the homogeneous system as
$\underline{x}(t) = A_1 \underline{\hat{x}}_1 e^{j( \omega_1 t +\phi_1)} + A_2 \underline{\hat{x}}_2 e^{j( \omega_2 t+\phi_2)}$
$x_1(t) = A_1 \hat{x}_{11} e^{j( \omega_1 t +\phi_1)} + A_2 \hat{x}_{21} e^{j( \omega_2 t+\phi_2)}$
$x_2(t) = A_1 \hat{x}_{12} e^{j( \omega_1 t +\phi_1)} + A_2 \hat{x}_{22} e^{j( \omega_2 t+\phi_2)}$
| {
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Should a system be “uniform” to qualify for it to be in steady state? I am wondering about whether a system need to be “uniform” to qualify for it to be in steady state or can not uniform systems also act as a steady system.
Can someone please clarify this.
Thanks in advance!
Any help would be appreciated.
| Steady state implies only that local thermodynamic state variables do not change in time. They can be non-constant function of position in space, e.g. there can be a temperature gradient and corresponding heat flux. If temperature function $T(x)$ does not change in time, it is a steady state.
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What is the use of the low pressure helium lamp in this plasma jet experiment? I would like to ask about this experimental set up. What is the use and significance of the reference cell (Low pressure helium lamp)? Why is it necessary on this set up?
Link to paper is as follows.
https://iopscience.iop.org/article/10.1088/1361-6595/ac4e21
| This is part of a laser stabilization or wavelength detection setup.
As long as the laser is on one of the Helium 2P-2S transitions, the beam gets absorbed in the helium lamp and the power on Photo Diode#2 is at a minimum. (Wavelengths: 1082.91, 1083.03 and 1083.03 nm)
With this, you can detect if your laser is at resonance. With a little bit of feedback-loop electronics you can even enable a frequency-lock of the laser.
The paper explicitly mentions this on page 3
The second beam goes
through a low-pressure helium discharge to provide the precise
wavelength positions of the He(2^3PJ –2^3S1) transitions.
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Why does the kinetic energy of particles increase on heating? One of the obvious facts in thermodynamics is, when we provide heat to a system, some of the heat energy get stored into the internal energy of the system. It is thought that it was stored either in form of potential energy or kinetic energy, but here we give more emphasis to the kinetic energy term, because it is in the core of the theoretical framework of kinetic theory of gases. It is very easy to experimentally confirm this fact, but what is theoretical reasoning behind it?
More specifically, why and how does the kinetic energy of the particles of a system increase when heat is provided to the system?
| When we add heat to matter its energy increases, both kinetic and potential. How much of the energy we add goes to kinetic versus potential depends on the details of molecular interactions. In the ideal gas state there is no interaction, therefore all heat goes into kinetic energy. When molecules interact, some heat goes into increasing the potential energy as well, which is to say, molecules can spend more time in configurations that have high potential energy. This is what happens when a liquid, whose molecules are mostly trapped in regions of negative potential due to attraction, turns into a gas, whose molecules are mostly at nearly zero (and thus higher) potential.
We talk more about kinetic than potential energy because it is mathematically simpler to deal with. The total kinetic energy is the sum of kinetic energies of all particles, a simple additive formula. Potential energy on the other hand is much harder to handle because it depends on the arrangement of molecules in space. Move one molecule and the potential energy of all molecules changes in a way that cannot be calculated by a simple sum. The ideal gas is so simple to describe mathematically precisely because it has no potential energy.
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Derivation of path equation, how to derive $$ r''= - \frac{L^2}{m^2}u^2 \frac{d^2u}{d\theta^2 } $$ So specifically I don't know how to go from $r'$ to $r''$. If we start with $r'$, where the substitution $r=u^{-1}$ is used.
$$ r'= - \frac{L}{m} \frac{du}{d\theta } $$
$$ (1): r''= - \frac{L^2}{m^2}u^2 \frac{d^2u}{d\theta^2 } $$
I start with differentiating $r'$ w.r.t $t$:
$$\ddot{r}=\frac{d}{dt}\dot{r}=\frac{d}{dt}\left(-u^{-2}\right)\dot{u} -\ u^{-2}\frac{d}{dt}\dot{u}$$
Which I find
$$ =u^{-2}{\dot{\theta}}^2\left(2u^{-1}\left(\frac{du}{d\theta}\right)^2-\frac{d^2u}{d\theta^2}\right)$$
And I don't know how to carry on to find equation $(1)$
| I assume that with $L$ you mean the orbital angular momentum. Hence, in polar coordinates, $L=mr^2\dot{\theta}=mr^2\frac{d\theta}{dt}$. Moreover, for a function $f(\theta(t))$ we have
$$ \frac{d}{dt}f(\theta(t))=\frac{df}{d\theta}\frac{d\theta}{dt} $$
Hence:
$$ \frac{d}{dt}\dot{r}=-\frac{L}{m}\frac{d}{dt}\frac{du}{d\theta}=-\frac{L}{m}\frac{d^2u}{d\theta^2}\frac{d\theta}{dt}=-\frac{L}{m}\frac{L}{mr^2}\frac{d^2u}{d\theta^2}=-\frac{L^2}{m^2r^2}\frac{d^2u}{d\theta^2}$$
which is what you want if you define $u=1/r$.
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Propagator for Dirac field The Feynman propagator for Dirac field is defined as:
$$S_{\alpha\beta} = \langle 0 | T \psi_\alpha(x)\bar{\psi}_\beta(y) | 0 \rangle.$$
It should, however, be a transition amplitude so that a particle at location $\mathbf{y}$ is generated at time $y_0$, moves to $\mathbf{x}$ and is destroyed at time $x_0$.
But how can a matrix be a transition amplitude?
I tried to find something about this on the internet, but I found nothing.
|
It should, however, be a transition amplitude so that a pond at location $\mathbf{y}$ is generated at time $y_0$, moves to $\mathbf{x}$ and is destroyed at time $x_0$.
...pond? Anyway, the amplitude does not literally have this position-based interpretation. This is more of a rule-of-thumb type of interpretation that people use, and it is related to the way that such terms appear in Feynman diagrams - which also are suggestive, but not known to be literal interpretations of what is going on in the interaction region.
But how can a matrix be a transition amplitude?
The answer here is that with spin, you need a "matrix" of transition amplitudes. If we have spin 1/2, then there are 4 different amplitudes which each have their own respective probabilities:
*
*When starting with a spin $+1/2$ particle of momentum $p$, the probability to transition to a spin $+1/2$ particle of momentum $p'$
*When starting with a spin $+1/2$ particle of momentum $p$, the probability to transition to a spin $-1/2$ particle of momentum $p'$
*When starting with a spin $-1/2$ particle of momentum $p$, the probability to transition to a spin $+1/2$ particle of momentum $p'$
*When starting with a spin $-1/2$ particle of momentum $p$, the probability to transition to a spin $-1/2$ particle of momentum $p'$
Of course, these amplitudes will depend on momentum, unless you want the probability to get a particle at any momentum, in which case you integrate $\int d^3p$. On the other hand, often we don't care which spin, and so we just sum over all outgoing spin combinations and average over the incoming ones, which tend to be a 50/50 split.
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Experiment preparation - Box to reduce background photons noise I am planning to perform experiment with a photomultiplier tube (PMT) in my lab. The system is not large, with a length scale of about 10 cm. I am looking to reduce the background photons noise as much as I can, and for this I thought to perform the experiment inside a box of the above length scale. Does anyone here have recommendations for a box that can do that job? I was looking at Amazon/Ebay/Alli express, but there are so many options, so I thought asking another people would be beneficial.
*My budget are flexible
| Use a "Pelican box" or similar black and waterproof cases. Works perfectly.
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Does quantum mechanics require classical mechanics for its own formulation? Is true that quantum mechanics requires classical mechanics (as a limiting case) for its own formulation?
| There are different theories that we call classical theories. There is Newtonian mechanics with all its later developments. Then there is classical field theory, such as Maxwell's equations for electromagnetism. And then there is thermodynamics, with statistical physics. Quantum mechanics is built on concepts from all these classical theories. So, without the different classical theories that inform quantum mechanics, we would not have been able to formulate quantum mechanics.
BTW, that is also why the idea that classical theories can be considered as a certain limiting case of quantum mechanics is somewhat misleading. Which classical theory should we end up with in this limit?
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Why not define tensors under Galilean or Poincare transformations? I have seen vectors (and tensors, in general) defined under rotations,
$$V^i=R^i_{~j}V^j$$
and under Lorentz transformations,
$$V^{\prime\mu}=\Lambda^\mu_{~~\nu}V^\nu$$
where $R,\Lambda$ are the matrices representing rotations and Lorentz transformations.
My definition of tensors is old-fashioned i.e., in terms of how a bunch of numbers transforms under rotation or Lorentz transformations.
Why doesn't one talk about vectors (and tensors, in general) under the Galilean or Poincare transformations (or groups)? Wouldn't that be more general? Why doesn't one define tensors under translations, spatial, or spacetime? Please do not use more sophisticated definitions of tensors as it will be very painful for me to follow.
|
Why doesn't one talk about vectors under the Galilean or Poincare transformations (or groups)? Wouldn't that be more general?
The tuple $(V^1,\ldots,V^n)$ is interpreted as the representation of some vector $V$ w.r.t. a basis. That being said, if we fix two bases, then the transformation of vector components is obviously linear (e.g. a rotation or a Lorentz transformation as you said). But Galilean transformations and Poincaré transformations are affine transformations.
The generalization you are looking for is the general linear group, the transformations between two arbitary bases. The point is that euclidean spaces and Minkowski space do not only have an affine structure, but also come with a bilinear form on the translation space. Each chart has an orthonormal basis associated to it and the transformation of vector components from one coordinate system to another is precisely the differential of the coordinate transformation (i.e. a Lorentz transformation in the case of Minkowski space).
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In firing a single photon at the center divider of a double slit, does it ALWAYS go through the slits? If we think of a single photon approaching the slits as a wave function, and we fire the photons at the midpoint of the two slits, one at a time, then I would think the probability function is highest at this midpoint. That would mean that, depending on the separation of the two slits, some of the photons would impact the wall separating the slits. If I fire a photon at a thin piece of material, why would it not at least occasionally 'hit' the material, (with a possible photoelectric effect). How is it that ALL the photons avoid this 'wall' and pass through one slit or the other?
| Don't try to analyze a diffraction problem using particles: you'll just confuse the issues.
Thinking of waves, well, what happens when waves impinge on opaque material? You see this every day. Some of the energy is reflected, some is absorbed. There's nothing particularly profound beyond this that happens in the double slit experiment. For narrow slits, you may need to account for polarization, but classical electrodynamics is fine for that, no quanta needed.
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Why do the units of entropy include energy and temperature? I am trying to learn what entropy actually is, and I read this answer about how entropy is the information needed to specify a full quantum state. However, if entropy is just information, why does it have units related to energy and temperature? So I was wondering if someone could explain how information can have these units.
| It's a quirk of scientific history that energy is quantified with one set of units (joules, ergs, calories) and temperature with another (kelvins, degrees Rankine). In Thermodynamics and an Introduction to Thermostatistics, Callen notes that energy and temperature have the same dimensions: $\mathrm{mass}\cdot\left(\frac{\mathrm{length}}{\mathrm{time}}\right)^2$. It would be reasonable to take entropy as dimensionless and unitless, as you note, being a representation of certain information—namely, the number of microstates consistent with a given macrostate. In this framework, the conversion coefficient of Boltzmann's constant (which would otherwise give entropy its conventional units of $\mathrm{J}/\mathrm{K}$ in SI, for example) is not necessary, and temperature and energy share units.
However, perhaps it's just as well that temperature has different units than energy. For one, it lets us state the temperature within several orders of magnitude of 1, which wouldn't typically be possible otherwise. In addition, look at all the pedagogical challenges associated with internal energy, thermal energy, heat, and work all having the same units (and thus being conflated more or less by new practitioners, especially for the case of thermal energy and heat). If temperature were measured in units of joules, for instance, we could expect still more confusion. But it would nevertheless be a valid (and arguably more logical) choice.
| {
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Are there materials which allow light beams to interact within them? For example: if you point two perpendicular lasers in air, nothing will happen. I'm wondering if there's any material in which these lasers would combine, bounce off each other, basically anything that would result in a different outcome than just passing through each other.
Thank you for any help.
| Nonlinear optics is the study of materials which, if the light is intense enough, do not react linearly to the electromagnetic fields that propagate through them. When this happens, the superposition principle breaks, and beams of light propagating through the material can modify each other's phase, propagation direction, and frequency, and even combine to make new beams of light at new frequencies and propagation directions. The Wikipedia page linked above contains a long list of examples of processes within this class.
For this to happen, the material typically needs to satisfy certain constraints, the most notable of which is certain types of crystal symmetry depending on the order of the process involved. Moreover, for nonlinear processes to happen at any measurable rate, you need the light to be intense enough, with the threshold intensity determined by the nonlinear susceptibility of the material.
| {
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Why is exhaust stroke isochoric in otto cycle? Please help me clarify this simple concept. In otto cycle, after the power stroke, the gas expands adiabatically and during the exhaust stroke why is the volume of the gas constant? Doesn't the volume of the gas change after the exhaust valve is open? Is the volume considered constant because the temperature of the gas drops quickly at the end of the power stroke? If so, how does the temperature of the gas drop so quickly in such small time interval? I am confused, please help me. Thank you!
|
Doesn't the volume of the gas change after the exhaust valve is open?
Yes it does, for the real Otto cycle which is an irreversible open system. But it does not for the reversible closed system model of the cycle.
The ideal reversible model of the Otto cycle does not include the intake or exhaust strokes because it is a closed thermodynamic system model of the cycle. Instead of thermal energy being lost along with mass in the exhaust stroke for an open system, it is shown as heat rejected at constant volume for the reversible closed system model. See the PV diagram of the ideal reversible cycle.
For reference I have included arrows to represent the intake and exhaust strokes of a real open system cycle which are not normally shown for the reversible closed system model. This has no impact on the net work done, since there is no net work done for the combination of the exhaust and intake strokes both of which occur at the same constant external pressure and the same increase and decrease in volume. All the thermal energy lost during the real exhaust stroke is accounted for as heat rejected at constant volume for the reversible cycle model.
Hope this helps.
| {
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How are electrons really moving in an atom? Niels Bohr proposed the solar system model of the atom, which is the most recognizable, and assumed electrons revolve in circular paths called orbits/energy levels. However, we know that the Bohr model is incorrect due to later observations, like the Stark and Zeeman effect.
Now the quantum model of an atom is the best model which we have so far which tells us that electrons are located in regions of probability called orbitals.
Are electrons really in motion or what is the nature of their motion inside the atom? Is the motion of an electron predictable or observable by some means or is it really haphazard motion, which is unpredictable?
EDIT:
Thank you everyone for kind attention, all answers really helped me understand and clear my confusion.
| There is no such thing as a "trajectory" of a particle in quantum mechanics. If a particle followed a single trajectory -- that is, for every time $t$, you could associate a unique position $x(t)$ where it was located -- then you could also assign a unique momentum $p(t)=m\dot{x}(t)$. This would contradict the Heisenberg Uncertainty Principle, which says you cannot know the position and momentum to arbitrary precision simultaneously.
The orbitals you plotted represent the probability of finding an electron at a given location in the atom for states of the atom with definite energy. That is all the information that exists about the position of the atom; in quantum mechanics, there is no "extra information" about where the particle is beyond what you see in the wave function, until you make a measurement of position.
| {
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Orbital motions Suppose two satellites are moving in circular orbits around the earth and then crash into one another.
My Question is this, how do we know that the bodies move in an elliptical orbit after the crash? Why not just remain in a circular orbit with a smaller radius (r)? Why not instead go into a hyperbolic or another orbit?
| As an intuitive answer to "why can't it result in a smaller circular orbit":
The collision occurs at radius $r$. Therefore the resulting trajectory must pass through a point $r$ distance from the central body.
A hypothetical smaller circular orbit would have some radius $r'$. But a circular orbit by definition has constant radius. A body moving in that orbit is always at $r'$ distance, never at $r$. So for an instantaneous collision to produce a radius $r'$ circular orbit, the body would be required to teleport from distance $r$ to distance $r'$. That's clearly impossible.
So that means that if the body is to end up in a circular orbit at $r'$, there must be some intermediate non-circular trajectory taking the body from radius $r$ to radius $r'$, before it can take up its final circular orbit. But if the impact puts it on such a trajectory, why does it later change trajectory again (from a non-circular orbit into a circular one) when it reaches radius $r'$?
For this to happen would require another force to cause the change in trajectory; another impact, or an engine firing, etc. But that would mean the lower circular orbit at radius $r'$ isn't the result of the initial collision, but the result of both events (taking place at different times and places). If the impact is the only event we're analysing, then the "intermediate" trajectory is the final one.
Basically: an object can never transition from a circular orbit at one radius to a different circular orbit at another radius (whether lower or higher) as the result of a single event. And the reason is quite straightforward: the object is simply not located on the second circle, so it can't possibly be in a circular orbit on that circle!
| {
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Does horizontal acceleration affect gravity? If we apply 1G horizontally in some object, will this constant force equal to G affect the time of falling? If the force does not affect gravity, why gravity is prioritized over this force if both are equal?
Edited: For the ones who didn't understand what i mean well, i mean why we can't switch them and say gravity is the horizontal force, and the force we applied is gravity? why the trajectory does not change as shown in this image?:
| The vertical component of net acceleration which is equal to $g$ directed vertically downward is prioritized because this component can only decide the time of flight ,as soon as the distance of projectile from the ground becomes zero then we consider as it has stopped going forward, which mainly is ruled by this component that when is the projectile going to complete its trajectory.
(Furthermore after the collision bouncing may happen depending on the coefficient of restitution between projectile and the ground, but this is not the case we are talking about, here we are considering only a single parabolic trajectory motion)
Else the horizontal component of net acceleration will only decide the range of that projectile.
| {
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Evolution of a general state - Quantum Quench Suppose we have a state prepared in ground state of some Hamiltonian. Say
$$H \vert\psi_{0}(g)\rangle = E_{0}\vert\psi_{0}(g)\rangle$$
If I evolve this state with a different Hamiltonian
$$ \exp({-iH_{2}t})\vert\psi_{0}\rangle = \sum_{n} \exp(-iE_{n}^{'}t) \langle\phi_{n}\vert\psi_{0}\rangle \vert\phi_{n}\rangle $$
where $H_{2}\vert\phi_{n}\rangle = E_{n}^{'}\vert \phi_{n} \rangle$
(Q1) My doubt is whether the state evolved by different Hamiltonian can be written in the basis of new Hamiltonian for a very short time $t<<1$.
(Q2) This is called quantum quenching. Obviously at $t=0$, the state of the system is still the ground state of old Hamiltonian. The state starts changing once the evolution is started, how long should i wait for the state to evolve, so that the state actually evolved to new eigenbasis for new hamiltonian? Maybe this is a stupid question to ask! correct me if I am wrong!
| You can always write any state in terms of any complete basis, including the eigenbasis of any given Hamiltonian.
For your second question none of the terms in the expansion
$$
e^{-\imath H_2 t} |\psi\rangle = \sum_n e^{-\imath E_n t} \langle \phi_n|\psi\rangle |\phi_n\rangle
$$
ever decay, so the initial state will never decay into an eigenstate of the new Hamiltonian. You need some form of non-unitary dynamics, say by interacting with an external bath, to bring the state into an eigenstate of the new Hamiltonian
| {
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What is the state of an entangled photon after its twin is absorbed? Let's two photons are entangled in polarization after a laser beam passes through a Betha Barium Borate crystal. They take different paths and one of them (1) is absorbed in a black sheet. What is the state of the leftover photon (2)? Is it in superposition of polarization h/v or it must flip spontaneously in a certain polarization? What if the black sheet atoms absorb photons only with a certain polarization (say h)? Will the absorbed photon (1) take h polarization in the process of absorption and hence the second twin flip to v?
| What if the black sheet atoms absorb photons only with a certain polarisation?
https://apps.dtic.mil/sti/pdfs/AD1096363.pdf
They have used holes, rather than a black sheet, so this experiment may not be exactly what you want.
But,
https://uwaterloo.ca/institute-for-quantum-computing/sites/default/files/uploads/files/01-2gr-activity-answers.pdf
I found the above activities very helpful!
| {
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Why there is no $s$-channel for fermion-fermion scattering? I'm learning the Lagrangian for Yukawa theory, where $L_{int} = \phi\bar{\psi}\psi$. For the fermion-fermion scattering, we can draw the Feynman diagrams as
My question is why we can't have $s$-channel here? If it exists, it still seems like we can have a diagram with the appropriate vertices. Also, does the direction of the arrow matter for that internal propagator?
| For $s$-channel, the fermions need to annihilate, so:
$$ e^+ e^- \rightarrow e^+ + e^- $$
is $s$-channel. There is still a $t$-channel, of course, but no $u$-channel as the electron and positron are not identical particles.
| {
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Electrons after Bremsstrahlung radiation Where do electrons go after they have been braked in the Bremsstrahlung effect? In particular, those who have lost almost all their kinetic energy, could they be captured or interact with the nuclei?
| Bremsstrahlung is a name for a general phenomenon, which may happen in many different situations. E.g., in X-ray tube the electrons simply are collected by the anode.
The scenarios described in the OP are more characteristic of Beta-decay - in this case the answer is yes, in some situations electrons are captured.
| {
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Quickway to calculate density matrix knowing $[S_{x,y,z}]$ Consider the spin $\frac{1}{2}$ system. A general quantum state could be written as $|\alpha^{(i)}\rangle=c_i|+\rangle+d_i|-\rangle$ and thus we have the density operator $$\rho=\sum_i w_i|\alpha^{(i)}\rangle\langle\alpha^{(i)}|.$$Suppose we know $[S_x]=\text{Tr}(\rho S_x)$, $[S_y]$ and $[S_z]$. I am wondering if there is a quick way of calculating $\rho$ using the fact that the trace of any operator is independent of the basis we choose to evaluate it?
Edit: I should add the following constraints
*
*$\sum_i w_i=1$
*$c_i^2+d_i^2=1$; $c,d\in\mathbb{C}$
| Yes. Note that the set of Pauli matrices and the identity operator on $\mathbb C^2$, $\{\sigma_0:=\mathbb I_2,\sigma_1,\sigma_2\sigma_3\}$, is an orthonormal basis in the space of complex $2\times 2$ matrices $M_2(\mathbb C)$ equipped with an inner product $\langle \cdot,\cdot\rangle: M_2(\mathbb C) \times M_2(\mathbb C) \longrightarrow \mathbb C$ defined by $\langle A,B\rangle := \frac{1}{2}\mathrm{Tr}\, A^\dagger B$.
So every complex $2\times 2$ matrix can be expanded in this orthonormal basis; in particular every density matrix on $\mathbb C^2$:
$$\rho = \sum\limits_{k=0}^3 \langle \sigma_k, \rho\rangle\,\sigma_k \quad .$$
Thus, given the inner product of a density matrix with the Pauli matrices $\sigma_k$ for $k=1,2,3$ (which are trivially related to the $S_k$), we can obtain the coefficients in the above basis expansion and hence can reconstruct $\rho$ immediately. We further have that $$\langle \rho, \sigma_0\rangle = \frac{1}{2}\mathrm{Tr}\, \rho \,\mathbb I_2 = \frac{1}{2}\quad ,$$ by the trace normalization of $\rho$.
Finally, we can ask the following: Given some operator $\tilde \rho$ expanded in this basis, can we deduce from the coefficients of this basis expansion whether $\tilde \rho$ is a density matrix? This is discussed here.
| {
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Do we need a quantum gravity theory to model an hydrogen atom on earth? The hydrogen atom is a quantum mechanical system. However, it is also attracted by the gravitational pull of the earth. Therefore, do we need quantum gravity to model its behavior correctly? Conversely, can we study hydrogen atoms on earth to obtain new information about quantum gravity?
| No, we do not need quantum gravity to model a hydrogen atom, at least not for any practical purpose. We are in fact able to model hydrogen atoms extremely well with existing theories, so well that quantum electrodynamics is among the best verified theories in all of science.
Gravity is an extremely weak force, and its contributions to the behavior of an individual atom are so incredibly tiny that they produce no practically measurable effects. The error introduced by just ignoring gravity completely is well below any experimental uncertainty. Moreover, as @Andrew said in his answer, if we did want to include gravity in our model of the atom we could use a classical or semi-classical approximation which would reduce the error even further.
| {
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Why is the stress on a body not a vector? In my textbook, Physics, Part II—Textbook for Class XI, there's a line which talks about why stress is not a vector:
Stress is not a vector quantity since, unlike a force, stress cannot be assigned a specific direction. Force acting on the portion of a body on a specified side of a section has a definite direction.
It does not elaborate why stress cannot be assigned a specific direction. I know that the stress on a body is the restoring force (applicable when the body is deformed) per unit area. Has it got something to do with the fact that area itself is a vector? Moreover, we often say that tensile (or compressive) stress is applied perpendicularly to the surface. That's specifying direction, isn't it?
An intuitive explanation (instead of a rigorous mathematical one) is highly appreciated.
| We can put a wire under tensile stress by pulling each end with a force of equal magnitude. If the wire has an East-West alignment we need to pull its eastern end to the East and its western end to the West (even though one of these forces may be inconspicuously applied by a fixed anchorage to which one end of the wire is attached). You need both these forces to put the wire under uniform stress. So the stress doesn't point to the East or to the West. It has an alignment rather than a direction. That doesn't stop us from calculating the tensile stress by dividing the magnitude of either force by the area over which it acts.
Stress, then, is not a vector quantity. [Instead it can be represented by a mathematical object called a second rank tensor. Tensors enable us to represent not just tensile stress but other sorts of stress (such as shear stress), or combinations of different sorts of stress.]
| {
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Analysis of the reflection in metal In order to solve the reflexion in a material with a complex index the solution I've found on textbooks is to define $\hat{n}\cos{\phi}:=a+bi$ where $\hat{n}=n_r+n_ii$ is the complex index and then solve the system
$$a^2-b^2=n_r^2 - n_i^2 -n^2\sin^2(\phi)$$
$$ab=n_rn_i$$
Then, with the complex expression for $\hat{n}\cos{\phi}$ you could determine the $r_\parallel$ and $r_\perp$ coefficients.
However, to solve this, I've attempted simplifying the problem by solving for the complex expression for $\cos \phi '$.
To do this, by using Snells law and Pythagoras law I get the expression'
$$\cos \phi '= \sqrt{1-(\frac{n}{\hat{n}}\sin \phi )^2 }$$
Now, defining $\sqrt{z}=\pm \sqrt{|z|}e^{i\phi/2}$ I can get the complex expression for the cosine. Each of the two solutions for the square mathematically represents an offset of $\pi$ in the complex phase.
The problem with this is that I don't know when I should choose the positive or the negative solutions for the complex module of the cosine.
| Why are using a law of refraction (snell's law) for reflection in metals?
You can not compute reflections in metal without taking into consideration of wavelength.
Refractive index of metals varies largely for different wavelengths. Different metal absorbs different wavelength that also give metals its color.
The refractive index for plastic tends to decrease with increasing wavelength wheras RI of metals increase with increasing wavelength.
| {
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What would a standing wave of light look like? I want to know what a standing wave of light would like and what properties it might have that are interesting.
| The resonant cavity of a laser is a standing wave. It doesn't really look like anything in particular because the standing waves are not travelling to your eyes. However, you can let some of the wave escape the cavity to do all sorts of interesting things.
The most interesting property is the ability to control cats due to the tight collimation:
Other less interesting properties include that the light is coherent and monochromatic. It can also be very intense and easily focused. It can produce more heat at a target than in the cavity, thus allowing to thermally ablate materials.
| {
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Some basic questions about applying operator in quantum mechanics Given a momentum operator
$$
\def\bra #1{\langle#1|}
\def\ket #1{|#1\rangle}
\def\braket #1{\langle#1\rangle}
$$
$$
P \equiv - i \hbar \frac{\partial}{\partial x},
$$
I want to calculate $\bra{a} P \ket{a}$. Here are the steps provided by my professor.
$$\begin{aligned}
\bra{a} P \ket{a}
&= \int dx \int dx' \braket{a | x} \bra{x} P \ket{x'} \braket{x' | a}
&& \left( \text{insert two sets of bases} \int dx \ket{x} \bra{x} = I \right) \\
&= \int dx \int dx' \braket{a | x}
\bra{x}
\left(- i \hbar \frac{\partial}{\partial x'}\right)
\ket{x'}
\braket{x' | a}
&& \left( \text{pull in } P = - i \hbar \frac{\partial}{\partial x'} \right) \\
&= \int dx \int dx' - i \hbar \braket{a | x}
\braket{x | x'}
\left(\frac{\partial}{\partial x'}\right)
\braket{x' | a}
&& \left( \text{move } \ket{x'} \text{ to the left} \right) \\
&= \int dx \int dx' - i \hbar \braket{a | x}
\delta(x - x')
\left(\frac{\partial}{\partial x'}\right)
\braket{x' | a} \\
&= \int dx - i \hbar \braket{a | x}
\left(\frac{\partial}{\partial x}\right)
\braket{x | a}
&& \left( \int f(x') \delta(x - x')\, dx' = f(x)\right) \\
&= -i \hbar \int dx
\Psi_a^*(x)
\left(\frac{\partial}{\partial x}\right)
\Psi_a(x)
&& \left( \braket{x | a} = \Psi_a(x) \right) \\
\end{aligned}$$
I have a few questions.
*
*Is $\ket{x'}$ a function? Can I treat it as $f(x')$.
*If $x$ and $x'$ mean "position", why they can become a series of the basis? $\int dx\, \ket{x} \bra{x} = I$
*Is the equation below correct?
$$
\left(- i \hbar \frac{\partial}{\partial x'}\right)
\ket{x'}
\braket{x' | a}
= - i \hbar \left(
\frac{\partial}{\partial x'}
\ket{x'}
\braket{x' | a}
\right)
$$
*Why $\ket{x'}$ can be moved to the left at the third step?
| Your basic definition is simply wrong/ambiguous. Use @Emilio’s informal rule above, or the correct definition of Sakurai and Napolitano, (1.248), namely,
$$
\def\bra #1{\langle#1|}
\def\ket #1{|#1\rangle}
\def\braket #1{\langle#1\rangle}
$$
$$
P \equiv - i \hbar\int \!\! dx~\ket{x} \frac{\partial}{\partial x}\bra{x},
$$
sometimes encoded as
$$
P_x \equiv - i \hbar \frac{\partial}{\partial x},
$$
as long as you know what you are doing, which your instructor muffed, to your detriment.
From the correct definition, note
$$
\bra{y} P\ket{z}= \int\! dx \braket {y|x} (-i\hbar \partial_x) \braket{x|z}= -i\hbar \partial_y \delta(y-z),
$$
and the rest will follow correctly.
$$
\bra{a} P \ket{a}= \int\! dy d z ~\braket{a | y} \bra{ y} P \ket{ z} \braket{ z | a}\\
=\int\! dxdy d z ~\braket{a | y} \braket {y|x} (-i\hbar \partial_x) \braket{x|z} \braket{ z | a}\\
=\int\! dx ~ \braket {a|x} (-i\hbar \partial_x) \braket{ x | a}\\
=-i\hbar \int\!\! dx ~~ \Psi_a^*(x) \partial_x \Psi_a(x)~.
$$
You can, of course, go from the first to the penultimate line directly. I left them in to humor your previous wrong derivation.
*
*Yes.
*No.
2&4. are meaningless.
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Why do we choose the Dirac delta function as the eigenstate of position operator? When we try to find the eigenstates of the position operator, we get that the product of (x-y) and the eigenstate must be zero. It is obvious then that for x different than y, the eigenstate must be zero.
Now for x equal to y, how do we know that the eigenstate is infinite so that we get the Dirac delta function? What if we choose any other form of the eigenstate for x=y - it would still be zero and satisfy our math, right?
| If $\psi$ were $0$ except for some finite value at $x=x_0$, then the integral of $|\psi|^2$ — that is, the probability of finding the particle anywhere at all — would be $0$. Yet experimenters keep finding particles in various places.
| {
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Is it possible to realize a probabilistic Maxwell's demon using Tesla valve? Imagine a container, with balls of diameter 1x and 10x, moving randomly in all directions. These balls are mixed, so it is a low order system.
Now this container is connected to another container via a slightly modified Tesla valve which has the passage of width 8x. So this passage allows only the balls of diameter 1x and filters the 10x balls. The image below is a section of the Tesla valve, many such can be connected in series to increase the efficiency.
In the allowed direction, the probability of a ball making it through the valve is high. But the probability of a ball travelling in opposite direction is very small, as it has to take the correct path all the time, and even if it takes a single wrong path, it will end up moving in the opposite direction.
Now after a period of time one container has mostly 10x balls and another container has only 1x balls. It is as if though they are sorted automatically, hence a higher order system came out of a lower order system.
Does this work? Is second law of thermodynamics not work in this scenario? What's happening here?
| Maxwell's demon highlights the importance of information and ignorance in thermodynamics. By "information" I mean the type of knowledge that is accessible to the observer and by "ignorance" the type of knowledge that remains inaccessible. In the statistical mechanical view of a molecular system the information available to us is macroscopic: the volume of the container, the number of molecules it contains and its energy. On the other hand we are ignorant regarding all microscopic details: where each molecule is and how fast it is moving. Maxwell's demon demonstrates the implications of the fact that some information is unavailable to us: if we could gain access to it we would be able to do things that we otherwise can't.
To get back to the original question, sorting that is based on macroscopic properties (i.e., on knowledge that is accessible to the sorter) is possible and violates no laws (I sort my socks after laundry each time). What is impossible is to sort things based on properties that we cannot observe. When we put it this way there is nothing controversial about Maxwell's demon.
| {
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Maximum Range for Projectile Motion This is a question concerning a trick I observed while solving for the angle responsible for maximum range of a projectile.
What I have observed is :
If you draw two lines, one opposite to the acceleration faced by the body(i) , another in the direction we want to find the projectile(ii), the angle for maximum range is always half the angle between the two lines .
In other words , the body should be thrown along the angle bisector of the two lines mentioned above to get the maximum range .
We note that this trick works for all the normal projectile cases because we know the angle should be 45° .
For an example , let's say a body is projected and faces acceleration due to gravity , 'g' in both vertical and horizontal directions. Line (i) will be at 135° from the horizontal opposite to the resultant acceleration √2g and line (ii) will be along 0° from the horizontal. So 67.5° is the angle for maximum range which is also the result derived after differentiating the range w.r.t the angle .
For projections along/down inclined planes as well , the trick seems to work. If x° is the angle of the inclined plane,line (i) is along y-axis and line (ii) ,x° from the horizontal. The angle for maximum range is thus (90°-x)/2 = 45°-x/2 from the inclined plane .
For down the incline , line (i) is as usual 90° from the horizontal and rotating line(i) through 90°+x° gives us line(ii) . The angle for maximum range is thus 45°+x/2 from the inclined plane .
I would like to know if there is some solid mathematical proof behind this trick . Why does it seem to work everytime ? Are there any cases where it doesn't?
I am sorry if I am missing something trivial .
| Lets call the dimension along which range is to be maximised $x$ and the perpendicular dimension $y$. Let $g_x$ and $g_y$ be acceleration components and $u_x$ and $u_y$ be initial velocity components. Then, the time of flight is given by $t = -\frac{2 u_y}{g_y}$ and the range is given by $R = u_xt + \frac{1}{2}g_x t^2$.
Substituting $t$ in $R$, we get
$R = -2 u_x u_y\frac{1}{g_y} + 2u_y^2 \frac{g_x}{g_y^2}$ ---- (1)
Let $\theta$ be the angle between $x$ and $g$ and $\alpha$ be the angle between $x$ and $u$
Then,
$g_x = g \cos{\theta}$
$g_y = g \sin{\theta}$
$u_x = u \cos{\alpha}$
$u_y = u \sin{\alpha}$
Putting this in (1),
$R = -u^2 (2 \sin(\alpha) \cos(\alpha)) \frac{1}{g_y} - u^2 (2 \sin^2(\alpha))\frac{g_x}{g_y^2}
= -\frac{u^2}{g_y^2} (\sin(2\alpha) g_y + (\cos(2\alpha)-1) g_x)
= -\frac{u^2}{g_y^2} (\sin(2\alpha) g \sin{\theta} + \cos(2\alpha) \cos {\theta} - g \cos{\theta}) = -\frac{u^2 g}{g^2 \sin^2{\theta}} (\sin(2\alpha) \sin{\theta} + \cos(2\alpha) \cos {\theta} - \cos{\theta}) = \frac{u^2}{g \sin^2{\theta}} (\cos{\theta} - \cos({2\alpha - \theta}) ) $
$\theta, g$ and $u$ are constants. Hence, $\alpha$ which maximises $R$ should minimize $\cos({2\alpha - \theta})$ which is when $\cos({2\alpha - \theta}) = -1
\implies 2\alpha - \theta = (2m+1)\pi \implies \alpha = m\pi + \frac{\pi}{2} + \frac{\theta}{2}$
Observe that $\theta \in [-\pi,0]$ for time $t$ to be positive because $g \sin{\theta}$ needs to be negative.
If we choose $\theta \in [-\pi, 0]$, then $\alpha = \frac{\pi}{2} + \frac{\theta}{2}$
| {
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Is household alluminium foil good for observing photoelectric effect? I have built an electroscope(very primitive, just a wire and some aluminum foil). I'm thinking about using it to observe the photoelectric effect. But I don't have any high frequency wave producing equipment.
So, before I try to obtain something for that purpose, I want to know if aluminum foil used in houses is suitable for observing the photoelectric effect.
Is it covered with some kind of metal oxide or something?
Is the 'work function' of house aluminum foil similar to the pure aluminum metal?
| Aluminum foil is covered with a layer of aluminum oxide as soon as it is exposed to air. This tendency for metals to oxidize and also to adsorb water molecules makes work functions difficult to repeatably measure when tested in air.
In addition, certain rare-earth oxides as used to coat vacuum tube emitters dramatically reduce the work function of the metal they have been applied to.
Meaningful measurements of work functions therefore require atomistically clean surfaces, which you almost never experience in contact with air.
| {
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Are the field lines around a magnet (and the repulsion felt between magnets) EM radiation? If not, is it those field lines which turn into EM radiation when a charge is accelerated?
| No, because it is a static situation.
An accelerated charge generates EM radiation and if it is strong enough, could make iron filings on a piece of paper oscillate. So yes, theoretically, moving field lines could show EM radiation.
We can make an analogy with gravitational field and how strong is the force downward depending on the steepness of points in the dunes of a desert.
If instead of sand there is water in the sea, the outcome is waves, not static dunes. And the water waves are indeed called gravity waves.
| {
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What does a dot over a spinor index signify? My questions should be rather simple. I was trying to get through one of my professor’s papers, and I saw the following notation, first with regards to Dirac and Weyl spinors, but the notation continues throughout
$$
\psi=\begin{pmatrix}\lambda_\alpha\\\bar\lambda^{\dot\alpha}\end{pmatrix}
$$
In the $\bar\lambda^{\dot\alpha}$ term, what does the dot over the $\alpha$ signify?
| Starting from the defining representation $A(\vec{\alpha},\vec{u})$ of ${\rm SL}(2,\mathbf{C})$, also $A^\ast$, $(A^T)^{-1}$ and $(A^\dagger)^{-1}$ furnish representations of ${\rm SL}(2, \mathbf{C})$. The representations $A$ and $A^\ast$ are inequivalent and likewise the representations $(A^T)^{-1}$ and$(A^\dagger)^{-1}$. On the other hand, the representation $A$ and its contragredient representation $(A^T)^{-1}$ are equivalent and, analogously, the complex representation $A^\ast$ and its contragredient $(A^\dagger)^{-1}$.
In index notation (van der Waerden notation), a Weyl spinor $\chi$ transforming with respect to the ${\rm SL}(2, \mathbf{C})$ representation $A$ is written with lower indices, transforming as $\chi^\prime_\alpha = A_\alpha ^{\, \, \beta} \chi_\beta$. A Weyl spinor $\bar{\varphi}$ transforming with respect to the representation $A^\ast$ is written with lower dotted indices, transforming as $\bar{\varphi}_\dot{\alpha}= A^{\ast \, \, \dot{\beta}}_{\, \dot{\alpha}} \bar{\varphi}_\dot{\beta} $. Spinors transforming with respect to the contragredient representation $(A^T)^{-1}$ are denoted by upper indices, $\chi^{\prime \, \alpha}= A^{-1 \, \, \alpha}_{\, \, \, \, \beta} \chi^\beta$. Analogously, spinors transforming with respect to $(A^\dagger)^{-1}$ are written with upper dotted indices, $\bar{\varphi}^{\prime \, \dot{\alpha}}=(A^\ast)^{-1 \, \, \dot{\alpha}}_{\, \, \, \, \dot{\beta}} \bar{\varphi}^\dot{\beta}$.
Employing the fact that $A$ and $(A^T)^{-1}$ are equivalent representations, spinors with upper and lower indices can be related by $\chi^\alpha = \varepsilon^{\alpha \beta} \chi_\beta$ and $\chi_\alpha = \varepsilon_{\alpha \beta} \, \chi^\beta$, where the antisymmetric invariant $\varepsilon$ tensor is defined by $\varepsilon^{1 2} =\varepsilon_{2 1}=1$ (all other elements are determined by antisymmetry). Analogously, one has the relation $\bar{\varphi}^\dot{\alpha}= \varepsilon^{\dot{\alpha} \dot{\beta}} \bar{\varphi}_\dot{\beta} $.
A Dirac spinor consists of two Weyl spinors $\chi_\alpha$ and $\bar{\varphi}^\dot{\alpha}$.
| {
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The relation of time-evolution operators from Schwartz's textbook In the section 7.2.2 of Schwartz's QFT textbook, it says:
define the generation definition of time-evolution operators:
$$U_{21}\equiv U(t_2,t_1)=T{\exp[-i\int^{t_2}_{t_1} dt'V_I(t')]}\tag{7.46}$$
where $V_I$ is the interaction part of Hamiltonian in the interaction picture.
Then it has:
$$U_{21}U_{12}=1\tag{7.47}.$$
I don't understand how this relation is arrived. I tried to expand the definition directly, but to the second order of integral, it seems it cannot be canceled:
$$\int^{t_2}_{t_1} dt' dt'' V_I(t')V_I(t'')-\int^{t_2}_{t_1} dt' dt'' T[V_I(t')V_I(t'')]$$
which is obviously not zero.
| He must be tacitly using anti-time ordering when $t_2<t_1$. Then the cancellation works.
He should really use the path-ordering symbol "P" rather than "T".
| {
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What does it mean to say the universe is not locally real? Pardon me if this is a naive question.
What is difference between saying space-time is not locally real, and saying it is not real?
The proposal that the universe is not locally real seems to imply that it is is non-locally real. But what does this mean?
There is a good discussion of a related question elsewhere on the forum, but I find the answers confusing. Physicists use words differently from philosophers, and on this topic I find the differences difficult to resolve.This may be because I'm being a bit dense, but I wonder whether the issue can be explained in a more general philosophical way, or at least in a way I can grasp.
| Real, but not locally real, means that one can overcome the Bell argument (that there is nothing real being measured in a quantum measurement) by allowing information to be conveyed faster that the speed of light.
| {
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Peskin and Schroeder eqn 9.14 I am not familiar with functional integral, and in the text like $$\int D\phi D\pi \exp [i\int^T_od^4x(\pi\dot{\phi}-\frac{1}{2}\pi^2-\frac{1}{2}(\nabla \phi)^2-V(\phi))].\tag{9.14}$$ I try to compile it but get:
$$\int D\phi\left[ \prod_{x,t}\int d\pi\ \exp [i\epsilon^4(\pi\dot{\phi}-\frac{1}{2}\pi^2)] \right] \exp([i\int^T_od^4x(-\frac{1}{2}(\nabla \phi)^2-V(\phi))]$$
$$=\int D\phi \left[\prod_{x,t}\sqrt{\frac{2\pi}{i\epsilon^4}} \exp [i\epsilon^4\frac{\dot{\phi}^2}{2}] \right] \exp([i\int^T_od^4x(-\frac{1}{2}(\nabla \phi)^2-V(\phi))]$$
$$=\int D\phi \left[\prod_{x,t}\sqrt{\frac{2\pi}{i\epsilon^4}}\right] \exp[i\int^T_od^4x(\frac{1}{2}\dot{\phi}^2-\frac{1}{2}(\nabla \phi)^2-V(\phi))].$$
I know things were terrible wrong, but what is it?
| OP's infinitely many Gaussian integrations of the momentum field are essentially the correct method. The infinite product in OP's last line is usually tucked away by normalizing the path integral appropriately.
| {
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Third principle of thermodynamics and the unattainability of absolute zero Consider a $S-T$ diagram (entropy-temperature) and consider cooling a substance by doing a series of succesive isothermal and reversible adiabatic processes between two volumes $V_{1}$ and $V_{2}$.
Now when cooling the substance from $T_{1}$ to $T_{2}$ in the reversible adiabatic process we can write:
$$S(0, V_{1})+\int_{0}^{T_{1}}\frac{C_{V}}{T}dT = S(0, V_{2})+\int_{0}^{T_{2}}\frac{C_{V}}{T}dT$$
letting $T_{2}=0$ will lead to:
$$\underbrace{\int_{0}^{T_{1}}\frac{C_{V}}{T}dT}_{>0} = \underbrace{S(0, V_{2})-S(0, V_{1})}_{=0}$$
a contradiction showing that the third principle of thermodynamics implies that absolute zero cannot be achived.
Is this reasoning correct?
| Your reasoning proves that we cannot reach absolute zero by reversible adiabatic process from a non zero temperature. I prefer to reach this conclusion as follows:
Your first equation applies to any reversible adiabatic process between states $(T_1,V_1)$ and $(T_2,V_2)$. If we set one temperature to absolute zero the other must also be zero because both states must have the same entropy –zero. In other words, we cannot connect $T=0$ to any finite temperature via an isentropic path.
| {
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How can rigid body have more than 1 speed? I have a doubt that circular ring is a rigid object and we know every point on the rigid obj must have same velocity or speed. But in case of pure rolling of sphere or others we can find topmost point has velocity 2v lowermost have 0 how's that possible..i am not able to visualise it.
| . . . . we know every point on the rigid object must have same [linear] velocity . . . is not true; it is the angular velocity which is the same.
Perhaps the simplest rigid body to think about in the context of rotation is a door.
When a door is opened it is clearly the case that all parts of a door do not travel at the same velocity.
The motion of each particle which make up a rotating wheel which is not slipping relative to the ground can be thought of as the sum of the constant rotational motion about the centre of mass of the wheel and the constant horizontal translational motion as shown below.
Note that when these two motions are added together the linear velocity of the particles which make up the wheel are not the same.
| {
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Need help about motion So,In my book it is written that -
Translational motion is the motion when all the particles of a moving body move the same distance in the same same in the same direction.It is of two types-1)Rectilinear and 2)Curvilinear
Now, my question is that to have translational motion doesn't need to have uniform velocity,right? Because when a body with fixed mass moves then all of its particles move the same distance in the same time in the same direction. So can't we just say translational motion is the motion when a body with fixed mass moves from one point to another? doesn't every body experience translational motion when it moves?either in a linear path or curved?
|
... all the particles of a moving body move the same distance in the same same in the same direction
So if two points of the body $A$ and $B$ move from positions $A_0$ and $B_0$ to $A_1$ and $B_1$ respectively, this says that
$\vec{A_1}-\vec{A_0} = \vec{B_1}-\vec{B_0}$
which can be rearranged to give
$\vec{B_1}-\vec{A_1} = \vec{B_0}-\vec{A_0}$
In other words, the position of $B$ relative to $A$ at the end of the motion is the same as at the start of the motion. Since this applies to any two points of the body, then translational motion simply means that the body is in the same orientation in space at the end of the motion as it was at the start of the motion - so if you compare the orientations of the body at the start and the end of the motion, there is no rotation. Apart from this, there is no restriction on the path that the body takes or the speed or orientation of the body during the motion.
I think your textbook is giving a rather obscure definition of a very simple concept.
| {
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Why is Newton's second law seemingly not applicable to a ball rolling down incline plane? A homework problem asked us to find the acceleration of a ball (pure) rolling down an incline plane without friction. I thought it was simply $a=g \ \sin(\alpha)$ where $a$ is the acceleration of the CM and $\alpha$ is the angle of the incline to the horizontal.
The solution says, however, that the acceleration (like in this related thread: Acceleration of ball rolling down incline) is $(3/5) \ g \ \sin( \alpha)$.
Since we are only interested in the acceleration of the Center of Mass and the only force is the parallel component of gravity (since there is no friction and the normal force compensates the perpendicular component), shouldn't Newton's second law gives us the simple $g \ \sin(\alpha)$ answer?
Where am I going wrong? I asked my tutor but he said, he doesn't really know, and only that Newtons second law like that would leave out the fact its rotational motion.
I thought Newton's law would suffice to describe the translational motion of the CM like that but apparently not, so why is this wrong?
| Your error is here:
... since there is no friction ...
Pure rolling is impossible without friction. No friction would imply no rolling (just sliding smoothly without rotation). Pure rolling implies a static friction force that is exactly large enough to prevent the contact point from moving.
You can solve your task by starting with considering Newton's 2nd law in its rotational version:
$$\sum \tau = I \alpha.$$
The key is to link the linear properties (which is what you are seeking, linear/translation acceleration) with the rotational properties, and for that you might as well need a geometric bond, such as $a=r\alpha$.
| {
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Remote mass detection and location Could a collection of gravitational sensors (eg. MEMS accelerometers), placed in a 3d grid (eg. 10 x 10 x 10 sensors forming a 1 x 1 x 1 m cube) detect the the presence and position of a remote mass concentration, eg. an airplane in the sky?
What accuracy would be needed to do so?
| Gravitational sensors in orbiting satellites have been used to map density variations in the rocks under the ground. (See here for a recent example.)
At shorter range, gravimetric gradiometers are being researched for detecting underground tunnels, objects behind walls, and stealthed aircraft and submarines. (See here for a list of recent projects.)
You would probably not have much success with a 1x1x1 metre grid; you need to take measurements across a wide range of angles to localise sources. So you either have to move your sensor around (the usual technique), or have a widely-dispersed array of sensors.
| {
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How to prove $\mathrm{Tr}[(\partial_\mu U)U^\dagger]=0$? I am studying ChPT by referring to "A Primer for Chiral Perturbation Theory" by Stefan Scherer.
I'm having a problem with the consideration of terms that appear in the Lagrangian.
The textbook says only $\mathrm{Tr}(\partial^\mu U\partial_\mu U^\dagger)$ is important and other terms such as $\mathrm{Tr}[(\partial_\mu \partial^\mu U) U^\dagger]$ are irrelevant because $\mathrm{Tr}[(\partial_\mu U)U^\dagger]=0$.
Proving $\mathrm{Tr}[(\partial_\mu U)U^\dagger]=0$ is also an exercise and my question.
Below is the corresponding part from the textbook.
Do you have any ideas?
| Following the detailed hint,
$$
\begin{align}
(\partial_\mu U) U^\dagger
&= \left( \frac{i \partial_\mu \phi}{F_0} + \frac{(i \partial_\mu \phi) i \phi + i \phi (i \partial_\mu \phi)}{2! F_0^2} + \cdots \right) U^\dagger \\
&= \frac{i \partial_\mu \phi \ U^\dagger}{F_0} + \frac{(i \partial_\mu \phi \ U^\dagger) i \phi + i \phi (i \partial_\mu \phi \ U^\dagger)}{2! F_0^2} + \cdots
\end{align}
$$
I used $[\phi, U^\dagger] = 0$ in the last line above to get $i \partial_\mu \phi$ and $U^\dagger$ together in each term. Now taking the trace will allow to move each $(i \partial_\mu \phi \ U^\dagger)$ factor to the front of every term,
$$
\begin{align}
\text{Tr}((\partial_\mu U) U^\dagger)
&= \text{Tr}\left(
\frac{i \partial_\mu \phi}{F_0} U^\dagger + 2 \frac{i \partial_\mu \phi}{F_0} U^\dagger \frac{i \phi}{2! F_0} + \cdots
\right) \\
&= \text{Tr}\left(
\frac{i \partial_\mu \phi}{F_0} U^\dagger \left(1 + \frac{i \phi}{1! F_0} + \cdots \right)
\right) \\
&= \text{Tr}\left( \frac{i \partial_\mu \phi}{F_0} U^\dagger U \right) \\
&= \frac{i \partial_\mu \phi_a}{F_0} \text{Tr} (\Lambda_a) \\
&= 0
\end{align}
$$
| {
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The motion of a rigid body Consider a rigid body with $n$ forces acting on it. What I intend to know is how to determine the motion of the body, more specifically:
*
*How to determine a point through which the axis of rotation of the body passes
*How to determine the direction in which the axis of rotation points
*How to determine the angular velocity of the body
I admit that answering these questions might be time consuming, so it would be helpful if you could provide some reading material which can answer these questions. Also, if you do intend to answer my question, please do provide some mathematical backing to it.
| *
*axis of rotation is passing through a point , with respect to which angular velocity of all the point on the body is same.
*this can be determined by right hand palm rule (used in cross product rule to determine the direction)->your thumb should point in direction of axis of rotation,four fingers in direction of point which is rotating and the palm will face in direction of velocity of that point, you will be given point , velocity and then you can finger the direction of thumb.
3)for this i will recommend you to watch "Rotational dynamics by Mohit Tyagi" on youtube(if you dont know hindi ,then you can see subtitle and understand the concept).
| {
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The conservation of energy in both the cases: Will it be different or the same? On an inclined plane of same height $h$ and every factor the same but the only factor changes is the presence of friction. In the first case the friction is present in the top half and in the second case the friction is present in the second half of the inclined plane...
Which one has more kinetic energy when the ball reaches the end of the incline
| The energy lost to friction will be $F_k d$ where $F_k$ is the magnitude of the kinetic friction force and $d$ is the distance over which the friction acts.
From the description the $d$ is equal in both cases. So which one is greater depends only on the force.
In the usual case $F_k=\mu_k N$ where $\mu_k$ is the coefficient of kinetic friction and $N$ is the magnitude of the normal force. Since the normal force is the same in both cases and since the coefficient is also the same in both cases then the force is the same in both cases. So then the energy lost to friction is the same and the KE is the same in both cases.
Note: $F_k=\mu_k N$ is a very simplistic formula that approximates real friction. In reality friction can be a function of velocity too, and temperature, and other factors. So if those other factors are important then $F_k$ may not be equal in the two cases, like the simplified model suggests
| {
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Magnetic force calculation for parallel wires using Maxwell stress tensor. Issue with shear forces I am trying to calculate the forces in between permanent magnets and ferromagnetic surfaces with the Maxwell stress tensor using image theory and the Biot-Savart law. However I discovered a weird behavior regarding shear forces where I somehow must use the Maxwell stress tensor wrong.
I can break down the problem to the force calculation in between two parallel wires with similar direction of current.
The calculation of the flux density is quite easy based on amperes law. For two parallel wires aligned in the y axis the flux density component in the centerline of the permanent magnets ($y =50$ in the graphic) must be zero.
In theory I can now find the magnetic forces in 2D by utilizing the Maxwell stress tensor:
$$
F_x = \int_{-\infty}^\infty \frac{1}{\mu_0}( B_x(x,0)^2 - \frac{1}{2}(B_x(x,0)^2+B_y(x,0)^2) + B_x(x,0)\cdot B_y(x,0) dx
$$
$$
F_y = \int_{-\infty}^\infty \frac{1}{\mu_0}(B_x(x,0)\cdot B_y(x,0) + B_y(x,0)^2 - \frac{1}{2}(B_x(x,0)^2+B_y(x,0)^2) dx
$$
Note: in the graphic the $0$ is at the $y$ value $50$ I just kept the zero in the equation to explain it here.
This works fine for $F_y$. The result matches the Lorentz force equation well.
For $F_x$ however the results and the equation itself makes no sense at all.
I would be expecting zero force in the direction of $x$.
However if we insert $B_x(x,0) = 0$ (based in symmetry) into the equation for $B_x$ we get
$$
F_x = \int_{-\infty}^\infty \frac{1}{\mu_0}( 0 - \frac{1}{2}(0^2+B_y(x,0)^2) + 0 dx = \int_{-\infty}^\infty \frac{1}{\mu_0}(\frac{1}{2}(B_y(x,0)^2) dx \neq 0
$$
Because $B_y$ is squared the part to integrate will always be positive which is resulting in a net force in the x direction which makes no sense at all.
Can anybody explain to me where my mistake in using the Maxwell stress tensor is ?
| You've applied the stress tensor incorrectly. If we want to find the force in the $i$-direction on a collection of charges & currents using the stress tensor, then it is
$$
F_i = \oint T_{ij} n_j da,
$$
where $\hat{n}$ is the normal to the surface (and I'm using Einstein summation). In your case, for the lower half-space you have
$$
\hat{n} = \hat{y} \quad \Rightarrow \quad n_x = n_z = 0, \quad n_y = 1
$$
and so
$$
F_x = \oint T_{xy} \, da \qquad F_y = \oint T_{yy} \, da.
$$
Note that the $i$ index has to match on both sides of the equation, and that the $j$ index is set equal to $y$ by virtue of the fact that $\hat{n}$ only points in the $y$-direction.
You were calculating instead
$$
F_x = \oint (T_{xx} + T_{xy}) \, da \qquad F_y = \oint (T_{xy} + T_{yy}) \, da;
$$
both expressions were technically incorrect, but you got the right result for $F_y$ because $T_{xy} = 0$ in this case.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Product of Pauli-matrix exponentials Given Pauli matrices $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ and $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, can one write $e^{\alpha \sigma_z} e^{\beta \sigma_x}$ in terms of $e^{\beta \sigma_x} e^{\alpha \sigma_z}$ ($\alpha$, $\beta$ are some complex numbers). For example, is it possible to find some $M$ such that
\begin{equation}
e^{\alpha \sigma_z} e^{\beta \sigma_x} = e^{\beta \sigma_x} e^{\alpha \sigma_z}M?
\end{equation}
| Use the standard formula,
$$
e^{\alpha \sigma_z}= I\cosh \alpha + \sigma_z \sinh\alpha = , \\
e^{\beta \sigma_x}= I\cosh \beta + \sigma_x \sinh\beta ,
$$
to compute the quadruple product suggested in the comments,
$$
M=e^{-\alpha \sigma_z}e^{-\beta \sigma_x}e^{\alpha \sigma_z}e^{\beta \sigma_x}\\ =
e^{-ia\sigma_z}e^{-ib\sigma_x}e^{ia \sigma_z} e^{ib\sigma_x}\\
= \Biggl (I\cos a\cos b+i\left (-\sigma_y\sin a \sin b -\sigma_z \sin a \cos b-\sigma_x\sin b \cos a \right ) \Biggr)\\ \times \Biggl ( I\cos a\cos b +i\left (-\sigma_y \sin a \sin b + \sigma_z \sin a \cos b + \sigma_x \sin b \cos a \right ) \Biggr)\\
= I(1-2\sin^2a ~\sin^2 b)\\ +i\Bigl (\sigma_x 2\sin^2 a\sin b\cos b-\sigma_y 2\sin a \cos a\sin b\cos b -\sigma_z 2\sin^2b\sin a \cos a \Bigr ).
$$
Can you take it from here? You might be amused taking $a=\pi$, and seeing it geometrically, or else $b=\pi$ .
Note the all-important unitarity constraint/check: the sum of the squares of the coefficients of I and the σ s, respectively amounts to 1, as it should!
From the coefficient of the identity matrix, you find its arccos, θ, and you just confirmed that the coefficient of the $i\hat n\cdot \vec \sigma$, for normalized unit $\hat n$, messier to write down, must be $\sin\theta$! So,
$$
M=e^{i\theta \hat n\cdot \vec \sigma}.
$$
For imaginary hyperbolic angles, so real trigonometric angles, this amounts to a quadruple group element product for SU(2), dubbed the group (not algebra) commutator.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Do objects always speed up as they fall? In the thermodynamics video I watched, it was stated that objects at higher elevations have more energy. And that objects must speed up as they fall. My question is, "what happens if an object reaches it's terminal velocity? It's still falling but no longer speeds up" Doesn't this make the statement that "objects must speed up as they fall", incorrect?
| Force is spatial rate of energy transference. When it is applied on an object, the object gains momentum or loss depending upon state of motion of an object and direction of applied force. It is never become constant instantaneously. When it becomes constant that means an object reach at equilibirium and rate of gain in momentum equals rate of loss in momentum.
Just like thermal equilibirium in which rate of absorption equals to rate of emission there is mechanical equilibirium in which rate of gain of momentum equals to loss of momentum. At mechanical equilibirium, an object attains constant speed. For mathematical expression visit link below.
https://physics.stackexchange.com/a/741397/344834
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I use the Schwarzschild metric to calculate space curvature and time curvature seperately? I want to understand the math behind the idea that around Earth time dilation accounts for 99.99% of gravity, while around a black hole it only accounts for 50% of gravity while space curvature accounts for the other 50%. I think it can be shown with the first two terms of the Schwarzschild metric, but I can't figure out how. Can you give some guidance on how I use the SM to show this discrepancy? Or to put it in a different way; why — using the SM — does space curvature grow at such a rapid rate while time dilation does not when near a black hole?
| I quite dislike this particular turn of phrase - there is no meaningful sense in which curvature can be split into temporal and spatial parts. We can talk about pure spatial curvature in the sense that we can foliate a $d$-dimensional spacetime into spacelike leaves $\Sigma_t$, each of which constitutes a $(d-1)$-dimensional Riemannian manfold. Each leaf inherits a Riemannian metric $\gamma$ from the pseudo-Riemannian metric $g$ of the full spacetime, and so we can talk about its inherited Riemann/Ricci tensors.
However, this is not really an intrinsic property of the full spacetime - a different foliation will result in different leaves with different curvatures. More to the point, even in those situations where we can write the full spacetime as $\mathcal M = \mathbb R \times \Sigma$, a 1-dimensional manifold has no intrinsic curvature and therefore it doesn't make sense to talk about the "curvature of time."
What one presumably means when talking about this is the following. If we consider the radial geodesic equation in Schwarzschild coordinates and assume only radial motion, we obtain the following equation (the dot denotes differentiation with respect to the proper time):
$$\ddot r + \Gamma^r_{\beta \nu} \dot x^\beta \dot x^\nu = 0 $$
$$\implies \ddot r = -\Gamma^r_{tt} (c\dot t) ^2 - \Gamma^r_{r r} \dot r^2 - 2 \Gamma^r_{t r} (c\dot t) \dot r$$
Recall the formula for the Christoffel symbols:
$$\Gamma ^\mu_{\beta\nu} = \frac{1}{2} g^{\mu\rho} \big( \partial_\beta g_{\rho \nu} + \partial_\nu g_{\beta \rho} - \partial_\rho g_{\beta \nu}\big)$$
$$\implies \matrix{\Gamma^r_{tt} = \frac{1}{2} g^{rr} \big(\partial_t g_{rr} + \partial_t g_{rr} - \partial_r g_{tt}\big) = -\frac{1}{2}\big(1+\frac{2\Phi}{c^2}\big) \big(2\frac{\phi'}{c^2}\big)\\
\Gamma^r_{rr} = \frac{1}{2} g^{rr}\big(\partial_r g_{rr} + \partial_r g_{rr} - \partial_r g_{rr}\big)= \frac{1}{2}\big(1+\frac{2\Phi}{c^2}\big) \frac{-2\Phi'/c^2}{\big(1+\frac{2\Phi}{c^2}\big)^2}\\
\Gamma^r_{tr} = \frac{1}{2}g^{rr}\big(\partial_t g_{rr} +\partial_r g_{tr} - \partial_r g_{tr}\big) = 0 }$$
where $\Phi(r) = -GM/r$ is the Newtonian gravitational potential. After some simplification, the radial geodesic equation becomes
$$\ddot r = -\left(1+\frac{2\Phi}{c^2}\right)\frac{\Phi'}{c^2}\left(c^2\dot t^2 + \frac{\dot r^2}{\left(1+\frac{2\Phi}{c^2}\right)}\right)$$
$$= -\Phi'\left(1+\frac{2\Phi}{c^2}\right)\left(\dot t^2 + \frac{\dot r^2/c^2}{1+\frac{2\Phi}{c^2}}\right)$$
From the line element,
$$\mathrm d\tau^2 = \left(1+\frac{2\Phi}{c^2}\right) \mathrm dt^2 - \frac{1}{\left(1+\frac{2\Phi}{c^2}\right)} \mathrm dr^2$$
$$\implies \dot t^2 = \frac{1 + \frac{\dot r^2}{(1+2\Phi/c^2)}}{1+\frac{2\Phi}{c^2}}$$
and so finally
$$\ddot r = -\Phi'\left( 1 + \frac{2\dot r^2/c^2}{1+\frac{2\Phi}{c^2}}\right)$$
I assume that the author of your statement is referring to the relative magnitudes of the terms in this expression.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/745960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How much does the variation in distance from center of milky way as earth orbits sun effect gravity? As the earth rotates around the Sun our distance from the center of the galaxy is varying - what sort of variations does this cause in the force of gravity here on earth?
[Edit]
Given that Earth orbits around the Sun at a radius of 150e+6 km while the Sun orbits within the Milky Way at a distance somewhere between 2.283e+17 km and 2.685e+17 km of the center (with a motion that weaves up and down in relation to the galaxies flattened plane).
| Easy to estimate: the gravitational force goes like $1/(\mathrm{distance})^2$. With the distances of Earth to Sun and Sun to the Galactic center respectively,
\begin{equation}
\Delta F \sim \left( \frac{150\times 10^6 \mathrm{km}}{2\times 10^{17}\mathrm{km}} \right)^2 \sim 10^{-20} ,
\end{equation}
i.e. by nothing.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Convert reflection spectrum to optical density I have the spectrum 380-780 nm at 10 nm intervals of a nearly grey patch. I need to input its optical density into a calibrating programme. This is what a densitometer would do - but I don't have one and I don't know what the densitometer would be measuring. Is there a formula for making the translation? And I have 104 patches to convert. Thank you
| Here are a couple related questions: How do I convert light with a given spectrum to an RGB color? See this.
How do I convert RGB to grayscale? See this.
The idea is that lighter grayscale colors are those that appear brighter to the eye. Light appears brighter if it is more intense, and if its spectrum in in the region to which the eye is more sensitive.
You have a grey patch. Presumably you have illuminated it with light that is white - It covers the full visible spectrum. It isn't necessarily uniform at all wavelengths, but it emits enough to measure the entire range.
A colored object would reflect a different amount at each wavelength. You have a grey patch that presumably reflects roughly the same at each wavelength. You don't believe it is absolutely uniform, or you wouldn't need to measure the spectrum. So you find the fraction of light reflected at each wavelength. This determines the color of the grey patch. It is the color reflected white light would have.
The gray scale value is a weighted average of the reflectance over the visible spectrum, weighted by the sensitivity of the eye.
The last step is to convert reflectance to optical density.
$$OD = log_{10}(I_0/I) = log_{10}(reflectance)$$
| {
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Is Intensive property/Intensive Property an Extensive Property? We know that Extensive property/Extensive Property is Intensive is most of the cases, but is Intensive/Intensive an Extensive property ? if so, is there any examples
| No, intensive divided by intensive is always intensive. Here is the math.
A function $f(x)$ is homogeneous with degree $\alpha$ if
$$f(\lambda x) = \lambda^\alpha f(x)$$
for all $\lambda>0$. Intensive properties are homogeneous with degree $\alpha =0$, extensive properties are homogeneous with degree $\alpha=1$.
*
*Ratio of extensive properties: Define $h(x) = f(x)/g(x)$ where $f$ and $g$ are extensive:
$$
h(\lambda x) = \frac{f(\lambda x)}{g(\lambda x)}
= \frac{\lambda f(x)}{\lambda g(x)}
= h(x)
= \text{intensive}
$$
*Ratio of intensive properties: Define $h(x) = f(x)/g(x)$ where $f$ and $g$ are intensive:
$$
h(\lambda x) = \frac{f(\lambda x)}{g(\lambda x)}
= \frac{f(x)}{g(x)}
= h(x)
= \text{intensive}
$$
In both cases the ratio is intensive. Actually, we have proven a more general result: the ratio of homogeneous properties with the same degree of homogeneity is always intensive.
| {
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Cause non-linear relation between intensity and photo-current I wanted to investigate and validate the relationship between intensity and photo-current with a simple experimental setup. I expected a linear relationship based on the photo-diode's datasheet (I've added the link to the datasheet below).
The experimental setup consisted of the following parts: A 0.95 mW 630 nm laser, a polaroid to control the intensity of the laser, a 50/50 beam-splitter, which reflects a part of the beam towards a powermeter and transmits the other part towards the photo diode. We also used a lens to converge the laserbeam on the light sensitive area of the photo diode (picture below).
Based on our measurement data, we have obtained the following quadratic relationship, which does not make any sense to me, and I really don't know what could've gone wrong. Could it be the polaroid altering the relationship between the two quantities?
https://www.conrad.nl/nl/p/osram-fotodiode-to-39-820-nm-55-bpw-21-152977.html?WT.mc_id=affiliates:tradetracker:feed:152977&utm_medium=affiliate&utm_source=tradetracker&utm_campaign=316050&utm_content=ShopForward%20NL%20Shopping
| As suggested I've used a ND filter instead of a polaroid to control the laser's intensity. This gave me the expected results, as shown in the figure below. As also said before: the beamsplitter affects the polarized light, in a for me, unknown way but it seems to change the relation between the two quantities (as shown in the figure below). If anyone has a detailed explanation, I'm very curious as to why it wouldn't work with a polaroid!
| {
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Are the equations of the Poynting vector and energy density of an electromagnetic wave only for the real waves? So, my book says that the Poynting Vector associated to an electromagnetic wave in matter with permeability $\mu$ is $\mathbf{S} = \frac{1}{\mu} \mathbf{E} \times \mathbf{B}$. The thing is, I am unsure whether I can compute this vector with the complex fields "associated" to my original waves, and then get the real part, or if this formula is only valid when using the original real fields.
My exercise in particular gives me the wave $E(y,z,t)=E_0cos(ay+bz-wt) \hat{x}$, which has associated complex wave $$E(y,z,t)=E_0e^{i(ay+bz-wt)} \hat{x}$$
The magnetic field (which I got from Maxwell's third equation) is:
$$B(y,z,t)=\dfrac{E_0}{w}(b\hat{y}-a\hat{z}) e^{i(ay+bz-wt)} $$
Now, if I compute the poynting vector with the first method(complex waves and then extract real part) I get: $\mathbf{S}=\dfrac{E_0^2}{\mu w}(b\hat{z}+a\hat{y})cos(2(ay+bz-wt))$
Whereas if I use directly the real fields I get: $\mathbf{S}=\dfrac{E_0^2}{\mu w}(b\hat{z}+a\hat{y})cos^2(ay+bz-wt)$.
Obviously they cannot both be right, since one has $cos(2\alpha)$ and the other $cos^2(\alpha)$.I also have a similar confusion with the formula for the energy density
$\frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2$. Should I use the modules of the complex waves, or the modules of the real waves? Thanks in advance.
| Complex number expressions for fields in EM theory are just a different mathematical representation of fields to make solving differential equations on paper easier. The actual physical field can be described by either the real or the imaginary component of those complex expressions, depending solely on our choice; usualy the real component is meant to represent the actual field.
The Poynting energy expressions are derived and meant to be used with the actual physical fields. It is not correct to put auxiliary complex representations of fields into them, you would get wrong results.
| {
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What is the relevance of the Lorentz factor in general relativity? The Lorentz factor is ubiquitous in Special Relativity and is used to express "how much the measurements of time, length, and other physical properties change for an object while that object is moving."
However, in my fairly introductory study of General Relativity, mention of the Lorentz factor disappears. Does the Lorentz factor have any relevance in General Relativity and if so, what is its intuitive meaning?
| The Lorentz factor turns up in GR when physical observables are involved, particularly under transformations between different (local) frames. Sadly there is usually little attention given to such observables in textbooks and courses, which is why you haven't seen the Lorentz factor there. The formula $\gamma = -\langle\mathbf u,\mathbf v\rangle := -g_{\mu\nu}u^\mu v^\nu$ between two 4-velocities at the same point (as mentioned in robphy's answer), does appear in Carroll §2.5 and Hartle §5.6, but not many others. Their "relative 3-velocity" $\gamma^{-1}\mathbf v - \mathbf u$, as one might term it, does appear in Misner, Thorne & Wheeler §2.8 eqn 2.35. You will find the Lorentz factor a lot in the "spacetime splitting" literature by Bini, Jantzen, and others in this field, but it gets fairly technical.
| {
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Does a double star system have more mass than its constituents? According to Einstein, energy is equal to mass. Consider a planet that is in gravitational attraction to two stars. Normally I would say that the gravitational attraction is proportional to the masses of the two stars. But if they are orbiting each other, they possess energy.
Is it correct to say that this star system therefore has a stronger gravitational pull that is greater than just the two added masses of the stars?
| The sum of the kinetic and potential energy of a bound system is negative. This must be the case, because you would have to inject more energy into the system to separate the components to infinity.
Therefore the "gravitational mass" of the binary - what you would measure with another orbiting test mass at greater distance - would be less than you would expect from the sum of all the masses of the components of the system measured when they are far apart and stationary.
Whether this is an important effect (i.e. the relative size of the correction) can be judged from the ratio of the absolute value of binding energy to the rest mass energy:
$$\alpha \simeq \frac{GM_1M_2}{2Rc^2(M_1+M_2)}\ ,$$
where $R$ is the separation and the result is exact for a circular orbit.
The effect can be important in binaries featuring compact stars (e.g. neutron stars) that have stellar masses and where $R$ can be quite small - the ratio above is a few per cent for a pair of neutron stars separated by 30 km.
As an aside, this consideration also applies to single stars, where the sum of their internal kinetic energy and gravitational potential energy is also negative. Again, this is important in white dwarf and neutron star physics.
| {
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"timestamp": "2023-03-29T00:00:00",
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In the context of the holographic principle, is the bulk-boundary correspondence due to entanglement? According to the holographic principle, a "bulk" region of D dimensions corresponds to a "boundary" region of D-1 dimensions. In this context, the laws of physics of the bulk can be "encoded" on the boundary, so there is a correspondence between the bulk and the boundary.
My question is:
Does this correspondence arise because there is an entanglement between the bulk region and the boundary? Could they become unentangled, so that the fundamental laws of physics in the boundary could become radically different compared to those in the bulk?
| The idea of holography is that the bulk degrees of freedom and the boundary degrees of freedom are the same. They can't be entangled with themselves.
In a simple holographic universe consisting of a single qubit, there is a description of an arbitrary state $α|0\rangle + β|1\rangle$ as a configuration of the bulk, and another description of it as a configuration of the boundary, and you can choose to use either description. There are no states like $(|00\rangle + |11\rangle)/\sqrt2$ because there are no states $|00\rangle$ or $|11\rangle$, only $|0\rangle$ and $|1\rangle$.
Even if the bulk and boundary states were distinct, entanglement couldn't explain their evolving in lockstep, because entangled subsystems don't evolve in lockstep; they just have correlated initial states. You could argue that identical subsystems evolving under identical laws with no nonunitary collapse would remain identical forever, but that would apply just as well to unentangled initial conditions ($|ψ\rangle\otimes|ψ\rangle$).
| {
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How to justify this small angle approximation $\dot{\theta}^2=0$? Suppose the equation of motion for some oscillating system takes the following form:
$$\ddot{\theta}+\dot{\theta}^2\sin\theta+k^2\theta\cos\theta=0$$
Applying small angle approximation to $\theta$ gives $\sin\theta\approx\theta$ and $\cos\theta\approx1$,
$$\ddot{\theta}+\dot{\theta}^2\theta+k^2\theta=0$$
But I am wondering if it is alright to set $\dot{\theta}^2=0$ so that the equation simplifies to
$$\ddot{\theta}+k^2\theta=0$$
If so, how can this approximation be justified?
| By drawing a phase plot you can more visually see where the approximations are valid. Define
\begin{align}
x&\equiv \theta\\
y&\equiv\dot\theta
\end{align}
This transforms your initial equation into the following first order differential equations:
\begin{align}
\pmatrix{\dot x\\\dot y}=\pmatrix{y\\-y^2\sin x-k^2x\cos x}
\end{align}
We can plot the vector on the right as a velocity field to see the dynamics. You are probably familiar with this. I use the following Mathematica code to do this:
range = 3;
expressions = {
{{y, -k^2 x}, {y, -k^2 Sin[x]}},
{{y, -y^2 x - k^2 x}, {y, -y^2 Sin[x] - k^2 x Cos[x]}}
};
Grid@Table[
StreamPlot[
expressions[[i, j]] /. k -> 1, {x, -range, range}, {y, -range,
range}, PlotLabel -> MatrixForm@{expressions[[i, j]]}],
{i, 1, 2}, {j, 1, 2}
]
I added the small angle approximation of the harmonic oscillator as well for comparison.
My interpretation: as long as $x$ and $y$ are both small the approximation is valid. The region of validity is fairly small though and outside the unit circle the approximation becomes pretty bad. The valid region will probably change for different $k$.
| {
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Higher-order perturbation in Kondo problem In chapter 6 of the book about condensed matter physics written by Gerald Mahan, the self-energy of the conduction electron is calculated to the third order in $J$ (the Kondo coupling) to show that the imaginary part of the self-energy diverges logarithmically as the energy scale (or temperature) decreases.
What I am wondering is whether this 'the third order expansion in $J$' can be derived from the Anderson's impurity model.
Kondo problem, modelled by a Hamiltonian
$$
H_{K}= \sum_k \varepsilon_k {c^\dagger _k} {c_k}+J\boldsymbol{S}\cdot\boldsymbol{s}(0),
$$
where $\boldsymbol{s}(0)$ is the spin-1/2 operator of the conduction electron at the origin and $\boldsymbol{S}$ is the spin of the localized electron $f$, is derivable from the Andersons's impurity model:
$$
H_{A}=\sum_k \varepsilon_k {c^\dagger _k} {c_k}+ V\sum_k ({f^\dagger} c_k+{c^\dagger_k} f)+Un_{f,\uparrow} n_{f,\downarrow}
$$
through Schrieffer-Wolff transformation. In this derivation, $J$ in $H_K$ is related to $V$ and $U$ in $H_A$ by $J \propto V^2/U$.
However, in the transformation, higher-order corrections proportional to, for example, $V^{6}/U^3$ are neglected and only the first order correction (proportional to $J \propto V^2/U$) from the high energy excitaion is kept.
Therefore, it seems to me that if $V^{6}/U^3$ correction in Schrieffer-Wolff transformation is kept, the result of the third order expansion (derived in the book by Mahan) will look different.
If this is the case, why are such higher order corrections usually neglected in Schrieffer-Wolff transformation to derive Kondo problem even if we need to calculate the perturbation series beyond the first order in $J$ to solve Kondo problem?
| Yes, indeed. See, e.g., Perturbative approach to the nonequilibrium Kondo effect in a quantum dot (also published as a PRL paper), where the perturbation theory is carried out up to the fourth order in $U$.
Note also that Kondo effect in Kondo and Anderson models was solved exactly by Bethe ansatz respectively by Andrei&Destri and Wiegman&Tsvelik (Not sure about the references here, but likely [35-38] in this preprint.)
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What does squaring a vector mean? So,in the 3rd equation of motion,i.e $v²=u²+2as$,what does the square on the final velocity and initial velocity actually mean?And how does it make an effect on their direction?
suppose upwards vector is positive and downwards vector is negative.And,we are throwing a ball from a height of h from ground.In this case,u=-4m/s and there is some final velocity downwards
While using $v²=u²+2as$ here, $u$ is -4m/s and so $u²$ will be 16(m/s)^2. Doesnt this change its direction?
| This equation should be written:
$$
v_{x}^2 = v_{0x}^2 + 2 a_x (x - x_0)\,,
$$
where $v_x = v_x(t) = \vec{v}(t) \cdot \hat{i}$ and $x = x(t) = \vec{r}(t) \cdot \hat{i}$, meaning that it is an equation for the components of the vectors, under the assumption of constant acceleration vector, $\vec{a}$, and a choice of coordinate system.
To see that this is true, recall that the basic equations of motion for a constant acceleration vector are:
\begin{align*}
\vec{r}(t) &= \vec{r}(0) + \vec{v}(t) \,t + \frac{1}{2} \vec{a} t^2\\
\vec{v}(t) &= \vec{v}(0) + \vec{a} t
\end{align*}
which are found by integrating $\frac{{\rm d}^2 \vec{r}}{{\rm d} t^2} = \vec{a}$ twice. What is true for these vector equations is also true for any component of the vectors, under any choice of coordinate system. So, given such a choice, we have the following equations for the $x$ components of the vectors:
\begin{align*}
x &= x_0 + v_x t + \frac{1}{2} a_x t^2\\
v_x &= v_{0x} + a_x t
\end{align*}
where I've suppressed the time dependence of $x$ and $v_x$ and written $x_0 = x(0)$ and $v_{0x} = v_x(0)$.
We can do algebra on these two component equations to eliminate time and obtain an equation that relates the $x$ components of velocity and position. Namely, multiplying the first equation by $2 a_x$, we obtain
$$
2 a_x (x - x_0) = 2 a_x v_{0x} t + a_x^2 t^2
$$
and then squaring $(v_x - v_{0x})$ in the second equation and inserting the second equation again:
\begin{align*}
(v_x - v_{0x})^2 &= a_x^2 t^2 \\
v_x^2 - 2 v_x v_{0x} + v_{0x}^2 &= a_x^2 t^2\\
v_x^2 + v_{0x}^2 - 2 (v_{0x} + a_x t) v_{0x} &= a_x^2 t^2\\
v_x^2 - v_{0x}^2 = 2 a_x v_{0x} t + a_x^2 t^2
\end{align*}
and finally inserting our first result into this second result we obtain:
$$
v_x^2 - v_{0x}^2 = 2 a_x (x - x_0)\,.
$$
So the equation you are presenting is actually an equation for the components of the position, velocity, and acceleration vectors along any chosen direction (again, under the assumption that the acceleration vector is constant). The vector equation version presented above in gandalf61's answer is obtained by summing the $x$, $y$, and $z$ versions of this equation together, but this component equation cannot be determined from that vector equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Setting of the canonical ensemble for actual interacting systems The canonical ensemble formalism is usually derived by considering a given small system (the system under study) weakly coupled to a huge system (the thermal bath), so that the microcanonical ensemble computations can be applied to the pair of system, regardless of the details of the interaction between the system and the bath.
However, the applicability of the canonical ensemble to a system does rely (?) on the existence of an actual interaction between the system and a bath, so that studying, e.g. the Ising model with say, periodic boundary condition with hamiltonian
$$ H = -J \sum_{<i,j>}\sigma_i \sigma_j $$
in the canonical ensemble actually means, from a physical point of view studying a larger system with fixed hamiltonian
$$H = -J \sum_{<i,j>}\sigma_i \sigma_j + H_\text{interaction} + H_\text{bath} $$
where $H_\text{interaction}$ has an explicit expression, acts on the physical sites of the system, etc.
It is not immediately obvious what are the most general forms of $H_\text{interaction}$ and $H_\text{bath}$ that are acceptable (i.e. such that the canonical ensemble predictions are indeed recovered), nor that the actual form of the interaction with the bath has no physical effect: for instance, $H_\text{interaction}$ can be small and still induce correlations (?) between sites of the Ising lattice.
So my question is what does it mean concretely to couple a system to a heat bath, so that the canonical ensemble formalism makes sense for a given hamiltonian system ? What is the most general form of the coupling to a thermal bath that preserves the physics that we get from canonical ensemble computations ?
| A rather general consideration usually given in statistical mechanics books is that the energy of the system of interest is proportional to its volume, whereas the interaction energy (with the bath) is proportional to the size of the contact surface. This means that for a sufficiently big system (i.e., in thermodynamic limit) the contact interaction can be neglected, in the same way we neglect the residual interactions within the system itself - although these are necessary for establishing the thermodynamic equilibrium, they are negligible ehen describing the properties of this equilibrium.
| {
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How did Enrico Fermi compute when the Chicago Pile-1 nuclear reactor would become critical? I'm trying to understand the first nuclear reactor, the Chicago Pile-1, specifically the math Fermi did to figure out when the reactor would go critical. There's a nice report available from Fermi, where he tracks the value of $R_{eff}^2/A$, where $R_{eff}$ is the effective radius of the pile, and A is the measured neutron intensity at the center of the pile, some screenshots below.
Fermi then claims that when $R_{eff}^2/A$ reaches 0, the pile will become critical. This is where I get lost - I read through the report and some other sources, but I don's see where this math is coming from -> why should the pile become critical when $R_{eff}^2/A=0$?
| This kind of plot is typically called a "1/M Plot", where M is the reactor multiplication. To measure the multiplication, you need a neutron source and a detector. As you add more and more material to the reactor, the detector signal increases. When the reactor goes critical (or more precisely supercritical), the detector signal (M) will grow to infinity. If you plot the inverse of the signal (1/M), it will become zero when the reactor goes critical.
If you plot 1/M as a function of the reactor size (or control rod position, or number of rods, etc.), you can do a curve fit to estimate where the point of criticality is. In Fermi's plots, they were adding material (layers) and the plot shows the reactor when critical at layer 57.
| {
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Is $\phi^5$ a descendant in $\phi^4$-theory (at the conformal Wilson-Fisher fixed point)? I'm wondering if $\phi^5$ is a descendant in $\phi^4$-theory in $d = 4 - \epsilon$ at the conformal Wilson-Fisher fixed point, where the coupling constant is $\lambda$. The e.o.m. tells us that $\phi^3$ is a total derivative (and therefore also a descendant) of $\phi$ (let us not care about the numerical factors, and suppress the spacetime indices on the derivatives that is summed over)
\begin{equation}
\partial^2\phi \sim \lambda\phi^3 \ .
\end{equation}
Now regarding $\phi^5$, we find from $\partial^2\phi^3$ and the e.o.m. that $\phi(\partial\phi)^2$ is related to $\phi^5$
\begin{equation}
\partial^2\phi^3 \sim \phi(\partial\phi)^2 + \lambda\phi^5 \quad\text{or}\quad \phi(\partial\phi)^2 \sim \lambda\phi^5 + \partial^2\phi^3 \ .
\end{equation}
Since $\partial^2\phi^3$ is a descendant (which in turn is a descendant of $\phi$ due to the e.o.m.), we should only treat either $\phi^5$ or $\phi(\partial\phi)^2$ as a primary.
However, if we now study $\partial^4\phi$ using the equation above, we find
\begin{equation}
\partial^4\phi \sim \lambda^2\phi^5 + \lambda\partial^2\phi^3 \quad\text{or}\quad \lambda^2\phi^5 \sim \lambda\partial^2\phi^3 + \partial^4\phi \ .
\end{equation}
From this equation it looks like $\phi^5$ is a descendant of $\phi$. Is this correct?
If so, we can use the reasoning for $\phi^6$ (by studying $\partial^2\phi^4$ and $\partial^4\phi^2$) and find that $\phi^6$ is a descendant.
If my reasoning is incorrect, which ones are the primaries with scaling dimension $\Delta = 5 + \mathcal{O}(\epsilon)$?
| Your last equation does not follow from the others.
The combinations of fields and derivatives which give primary operators in Wilson Fisher are almost the same as those for the free theory. Due to multiplet recombination, there is one exception at spin zero ($\phi^3$) and one exception at each odd spin $\geq 5$. If $\phi^5$ were to be eaten, the free theory would need to have a primary at the unitarity bound of dimension $5 - n$ which is killed by $n$ derivatives and not by any smaller number of derivatives. There is no such operator so $\phi^5$ stays primary.
| {
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What exactly does a Lorentz transformation provide?
The primed reference frame is moving relative to the unprimed frame. So if we were to take the lorentz transformation of point P from the unprimed to primed, would it be the point A or B that it returns ? Assuming that the Lorentz transformation is passive i.e. we are talking about the same event in two frames, my first guess was B, because only at B will the primed reference frame see P to occur. A on the other hand is what the unprimed reference frame sees when P occurs. Is this true ?
PA is parallel to the x axis and PB is parallel to the x' axis.
| See below for an update to address the "passive transformation".
The [active] Lorentz-boost-transformation of event P is neither your events A nor B,
but an event between your A and B
that is on the hyperbola centered at the origin that passes through P.
The boosted event is joined to the original event by a line
that is not parallel to either axis of these two frames.
From my spacetime diagrammer https://www.desmos.com/calculator/emqe6uyzha ,
the boosted events lie on a common hyperbola (which are events that "equidistant in future time from the meeting event" according to the Minkowski metric).
The spatial-axis of a frame is parallel to the tangent to that hyperbola (shown as a dotted line).
It may be instructive to consider the Euclidean version of the construction (move the slider to $E=-1$):
The rotated point is joined to the original point by a line
that is not parallel to either axis of these two coordinate systems.
UPDATE:
Concerning the passive transformation...
the events are left unchanged in spacetime.
But now the passive Lorentz boost provides the coordinates
of an event in another inertial frame.
In this case, your event B is the event simultaneous with P, according to the moving inertial observer.
The diagram below is for $v=(3/5)c$.
So, the transformation would assign the event $P$ with red-coordinates $(t=1,x=0)$
with blue coordinates $(t=1.25, x=-0.75)$.
The passive transformation identifies event B as the event the moving inertial observer says is simultaneous with P
and calculates the blue time-coordinate of P to be equal to the blue time-coordinate of B as $t=1.25$.
(The transformation will also identify an event (not shown in the image) on the moving x-axis (at blue time t=0) that is "at the same place as P" according to blue. The spatial-coordinate of that event is $x=-0.75$.)
| {
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4-Vector Potential, transformation under Lorenz Gauge I am given an initial vector potential let's say:
\begin{equation}
\vec{A} = \begin{pmatrix}
g(t,x)\\
f(t,y)\\
0\\
g(t,x)\\
\end{pmatrix}
\end{equation}
| i think you are misunderstanding slightly, to obtain your potentials, you must have a specific chosen gauge that these potentials satisfy already.
Suppose my vector potential satisfies a specific gauge already,
$$\nabla \cdot \vec{A} = G(x,y,z,t)$$
I now perform a gauge transformation,
$\vec{A} \rightarrow \vec{A}'$ such that:
$\vec{A}' = \vec{A} +\nabla f$
$\phi' = \phi -\frac{\partial f}{\partial t}$
These leave the field invariant
Suppose that these transformations also satisfy:
$\nabla \cdot \vec{A}' +\frac{1}{c^2}\frac{\partial\phi'}{\partial t}=0$
substitute the definition of the transformed potentials $\vec{A}' \phi'$ into our expression:
$\nabla \cdot [\vec{A} +\nabla f] +\frac{1}{c^2}\frac{\partial[\phi -\frac{\partial f}{\partial t}]}{\partial t}=0$
$\nabla \cdot \vec{A} + \nabla^2f + \frac{1}{c^2}\frac{\partial \phi}{\partial t} -\frac{1}{c^2} \frac{\partial^2f}{\partial t^2} = 0$
$ \nabla^2f -\frac{1}{c^2} \frac{\partial^2f}{\partial t^2} = -[\nabla \cdot \vec{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}]$
Here we have an equation for f, that would need to be used in order to transform our potentials in the lorentz gauge.
Plugging in our known value for the original gauge that my untransformed potentials satify ( something which you havent mentioned yet), and time derivative for the scalar potential we obtain our f.
From our obtained value of f, we then transform the potentials that we have defined earlier.
$\vec{A}' = \vec{A} +\nabla f$
Our new potential $\vec{A}'$ thus satisfies the lorentz gauge whilst leaving the field B invariant.
| {
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Making sense of band theory for nonmetals For the case of metals, we observed that metals often have nearest-neighbours in excess of the maximum valency (for example, 8 for Li, which has only one valence electron) and that metals display great conductivity. On this basis, it was suggested that electrons are delocalised over the solid, and we used LCAO (linear combination of AO) over all the valence orbitals to form a band - the 2s band in this case.
The above is from section two in MIT's solid state lecture notes here. They talk about the above two points suggesting an atom in a metal lattice can interact with a large number of others. This then leads to thinking in terms of the orbitals of the "system" or the entire "macromolecule"
However, for non-metallic solids, like diamond for example, bonds are viewed as sp3 localised bonds and the number of nearest neighbours equals how many covalent bonds it can form, and it has poor conductivity as well. The poor conductivity is often explained by saying the band gap($E_g$) to the conduction band is large. But I don't understand the meaning of forming a band in this context, when the assumption we used of delocalised electrons over the entire solid doesn't hold.
Also, the description for the formation of the valence and conduction band for diamond is given in terms of sp3 orbitals combining to the form the bands. Why is hybdridisation (from VBT/VSEPR) being brought in to the picture here?
| Existence of band structure is the result of Bloch theorem, which predicts that electron spectrum in a periodic potential consists of continuous patches, states being label by a continuous number/vector $\mathbf{k}$ and a discrete band index $n$. Thus, the existence of bans is the consequence of crystal structure.
Excess of valency comes into play when distinguishing metals and semiconductors - in metals it results in the last band typically not completely filled, which means that there are mobile electrons present. On the other hand, in diamond structure, characteristic of Carbon, Silicon, and Germanium, all the bonds are saturated and we can distinguish last fully filled band (valence band) and the first empty band (conduction band). Similar situation takes place for close elements, as in GaAs or AlGaAs - which are combinations of elements of valencies three and five.
| {
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Why does magnetic force only act on moving charges? I don't understand why the magnetic force only acts on moving charges. When I have a permanent magnet and place another magnet inside its field, they clearly act as forces onto one another with them both being stationary. Also, I am clearly misunderstanding something.
| A magnetic field does have an influence on a stationary electric charge.
"when i have a permanent magnet and place another magnet inside its field they clearly act a forces onto one other with them both being stationary"
is exactly the case that is also true for an electron.
An electron is both an electric charge and a magnetic dipole. Any description of internal currents in the electron to create the magnetic dipole is superfluous. It is quite sufficient to consider the magnetic field of the electron to be just as fundamental as its electric field.
For a detailed description of why an electron moving through a magnetic field is deflected laterally (Lorentz force, Hall effect(s) ), see here.
| {
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What is happening at the particle level in the Bernoulli Principle? One might think that increasing the speed of particles would increase pressure -- if I understand what the Principle states, it is very counterintuitive.
My guess is, the pressure has something to do with particles moving perpendicular to the fast motion and maybe the particles spend less time hitting the sides of the pipe but there are also more particles per unit time.
I have some other ideas but the above guess is the kind of explanation I am looking for.
| The particles, on average, keep their energy, but the motion becomes more organized. They have more velocity along the streamlines, and less toward the sides.
| {
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Why do we drop the renormalization term in momentum Klein-Gordon Field Theory? I'm following Peskin & Schroeder's book on QFT. I managed to prove expression (2.33) which gives us the 3-momentum operator for the Klein-Gordon Theory:
$$\mathbf{P}=\int \frac{d^3p}{(2\pi)^3}\mathbf{p}a_\mathbf{p}^\dagger a_\mathbf{p} + renorm.$$
However, the book drops the renormalization term that is proportional to $\delta^3(\mathbf{0})$. I understand that, in the case of energy we cannot measure changes in energy, so that term is irrelevant. But why is this so in the case for momentum? Is it because we can choose a frame that moves alongside that infinite momentum term?
That solution doesn't really make sense, since it would imply infinite velocity (or mass)
| That term has nothing to do with renormalization, it’s just a result of ordering ambiguity in transitioning from classical physics to quantum physics.
| {
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Why do some objects tend to sink after some time in water even if they float at the start? I have observed this phenomenon in swimming pools: I have seen many dead insects floating on the surface, but after some time some they tend to sink down without any external influence. Why does this exactly happen? It even happens with paper: When it is fully immersed in water, after some time, it overcomes surface tension and buoyant force and sinks down. Do paper/insects gain more density?
| In order for an object to float, its density has to be less than or equal to the density of the liquid.
The objects you describe sink after a while because a sufficient fraction of the volume of air originally in each object, due to its porosity, is replaced by water which has a higher density than air. The end result is the overall density of the object becomes greater than the density of water after water replaces air in the object.
Hope this helps.
| {
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What if the universe was not uniform...? In this popular science article they say that if our universe resulted to be non-uniform (that is highly anisotropic and inhomogeneous) then the fundamental laws of physics could change from plate to place in the entire universe.
But what theoretical basis does this claim have? I mean, I know that, as far as we know, the universe appears to be isotropic and homogeneous, but even then, let's assume for a moment that this was not the case. How does an anisotropic-inhomogeneous universe lead to this? If our spacetime was highly anisotropic and inhomogeneous would this be possible?
| It sort of happens by definition. The wordings are a bit circular in this sense. Isotropy speaks to a symmetry in the universe which we can use to predict that things operate the same in all directions. If it is anisotropic, then there is no such symmetry. "Anisotropic" is less of a cause of the fundamental laws changing and more of a description of it happening.
Of course, such anisotropy could also occur within the structure of deeper fundamental laws of the universe, more fundamental than the ones we have gleaned through science, but the question becomes how could we determine what the laws are at extremely far distances, as we simply cannot travel there.
| {
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What is sound and how is it produced? I've been using the term "sound" all my life, but I really have no clue as to what sound exactly is or how it is created. What is sound? How is it produced? Can it be measured?
|
What it sound?
Sound is nothing but a mechanical wave. The human ear can detect mechanical waves from 20 Hz to 20,000 Hz.
How is it produced?
It is produced when energy transformation occurs. For eg. beating a drum - mechanical energy.
Can it be measured?
Yes, it can be measured by the frequency of the wave.
| {
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How to show that the Coriolis effect is irrelevant for the whirl/vortex in the sink/bathtub? There is a common myth that water flowing out from a sink should rotate in direction governed by on which hemisphere we are; this is shown false in many household experiments, but how to show it theoretically?
| The calculation of the Coriolis force is dependent on latitude:
$F = m a$
where $a = 2 \Omega sin(lat)$, with $\Omega$ being the Earth's angular velocity
$m$ is the mass of the object in question
The Earth's angular velocity is (about) $7.29 \times 10^{-5}$ rad/sec
So, for a sink with a couple gallons of water in it at 45 degrees north... the Coriolis force is about $7.57 \times 2 \times 7.29 \times 10^{-5} = 1.10 \times 10^{-3}$ N.
Interesting reading about this over this way, and (of course) here.
| {
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Mnemonics to remember various properties of materials I'm trying to figure out how to remember that
*
*hardness: how resistant it is to deformation
*toughness: how resistant it is to brittle failures
*stress: force on a surface area
*strength: ability to withstand stress without failure
*strain: measurement of deformation of a material
Does anyone know of a mnemonic or easy way? I only know these from googling them, and I'm finding it tricky to remember them.
| I would suggest learning about some of the formulas relating them - that way you're not just memorizing things but actually have some grasp of what goes into them. In particular, I only really know about stress and strain, and it's because I think of them as being the analogue in linear elasticity theory of "force" and "displacement" in Hooke's law.
| {
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Is Newton's Law of Gravity consistent with General Relativity? By 'Newton's Law of Gravity', I am referring to
The magnitude of the force of gravity is proportional to the product of the mass of the two objects and inversely proportional to their distance squared.
Does this law of attraction still hold under General Relativity's Tensor Equations?
I don't really know enough about mathematics to be able to solve any of Einstein's field equations, but does Newton's basic law of the magnitude of attraction still hold?
If they are only approximations, what causes them to differ?
| To be direct: the answer is no. There's no spatial curvature for Newtonian gravity, when it is rendered in geometric form as a curved space-time geometry. All the curvature is in time.
One of the most prominent ways this stands out is in the relation between radial distances versus circumferences. For Newtonian gravity, the circumferences $C_1$ and $C_2$ of two coplanar circular orbits will differ by $2π$ times their respective radii; or more precisely: the closest distance between the respective orbits will be $|r_1 - r_2|$, where $r_1 = C_1/(2π)$ and $r_2 = C_2/(2π)$.
For General Relativity, approximating the exterior of the gravitating body by the Schwarzschild metric, the closest distance between the orbits will be
$$\left|\sqrt{r_1(r_1 - r_0)} - \sqrt{r_2(r_2 - r_0)} + r_0 \log{\frac{\sqrt{r_1} + \sqrt{r_1 - r_0}}{\sqrt{r_2} + \sqrt{r_2 - r_0}}}\right|,$$
where $r_0 = (2GM)/c^2$, with $G$ being Newton's constant and $M$ the mass of the gravitating body, where $r_1$ and $r_2$ are as above. By assumption, we're talking about the exterior of the body, so $r_1 > r_0$ and $r_2 > r_0$. In the limit as $c → ∞$, $r_0 → 0$, and the distance approaches the limit
$$\left|\sqrt{{r_1}^2} - \sqrt{{r_2}^2}\right| = \left|r_1 - r_2\right|,$$
where $r_1 > 0$ and $r_2 > 0$.
The spatial curvature will also show up as a precessing of a fixed axis - such as that of a gyroscope - when taken around the gravitating body in an orbit. That test can be done ... and has been done, on the space shuttle.
The $r$ coordinate in the Schwarzschild metric, in General Relativity, is actually not radial distance at all, but is the "circumference radius" - that is: the circumference divided by $2π$. The two do not coincide in relativity, because of the spatial curvature.
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Book about classical mechanics I am looking for a book about "advanced" classical mechanics. By advanced I mean a book considering directly Lagrangian and Hamiltonian formulation, and also providing a firm basis in the geometrical consideration related to these to formalism (like tangent bundle, cotangent bundle, 1-form, 2-form, etc.).
I have this book from Saletan and Jose, but I would like to go into more details about the [symplectic] geometrical and mathematical foundations of classical mechanics.
Additional note: A chapter about relativistic Hamiltonian dynamics would be a good thing.
| As already mentioned the standard introductory books in hamiltonian geometrical (point-) mechanics are Foundations of Mechanics by Abraham and Marsden and Arnolds Mathematical Methods of Classical Mechanics. Another standard book is: "Classical Mathematical Physics" by Walter Thirring.
You may also have a look at "Symmetry in mechanics: a gentle, modern introduction" by Stefanie Frank Singer, which bridges the gap between standard college courses on classical point mechanics and books like those mentioned above.
Another interesting book for geometrical hamilton mechanics is "Introduction to Symmetry and Mechanics" by J. Marsden, which gives a nice introductory overview to that topic.
For more examples from a geometrical point of view you may also consult "Global aspects of classical integrable systems" by Cushman and Bates.
One sould also mention "Mathematical Aspects of Classical and Celestical Mechanics" by Arnold, Kozlov and Neishtadt.
For a less advanced (and less rigorous approach) with very much examples you may have a look at the german book "Klassische Mechanik" by F. Kuypers.
| {
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What will happen if we add salt to boiling water? I would like to have a good understanding of what is happening when you add salt to boiling water.
My understanding is that the boiling point will be higher, thus lengthening the process (obtaining boiling water), but at the same time, the dissolved salt reduce the polarization effect of the water molecules on the heat capacity, thus shortening the process.
Is this competition between these two effects real ? Is it something else ?
| I think the dominant effect might actually be the fact that the salt you add might not be at boiling temperature. But this is just based on the fact that the boiling-point elevation due to salt in water is actually quite low for typical amounts of salt used in cooking, say. I'm not too familiar with the second effect you mention though.
| {
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Intuitively, why is a reversible process one in which the system is always at equilibrium? A process is reversible if and only if it's always at equilibrium during the process. Why?
I have heard several specific example of this, such as adding weight gradually to a piston to compress the air inside reversibly, by why should it be true in general?
EDIT:
Here is something that would firmly convince me of this:
Suppose I have a reversible process that is not always in equilibrium. Describe a mechanism for exploiting this process to create a perpetual motion machine.
| As I am new to Stack Exchange, I happened to see the question only now and mine is a belated answer. If Mark is still interested in an answer, then:
It is a good question gaining added strength with the EDIT.
For thermodynamic analysys, a process must connect two equilibrium states A, B of a system. Take a state C of the system on the way from A to B. We can now consider the process A to C or C to B. To qualify for such a consideration, C must satisfy the condition that it is an equilibrium state of the system. C being chosen arbitrarily, it follows that every state of the sysyem along the path A to B must be an equilibrium state.
Coming to the EDIT:
To simplify the discussion, let us assume the syatem to be an adiabatic system. Suppose A is an equilibrium state and B is not. Then, let the system go from A to B reversibly (if possible), then the entropy of the system (and of universe in this case) decreases. Therefore, the process B to A becomes a spontaneous process. We can then employ the spontaneous process B to A to do work for us. Therefore, we let the process go reversibly from A to B, then let it go from B to A spontaneously, doing work for us. We can repeat this process endlessly and perpetually extracting work (energy) from nothing! We thus achieve perpetual motion if an adiabatic system goes from an equilibrium state A to a non equilibrium state B reversibly.
Radhakrishnamurty Padyala
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Would it help if you jump inside a free falling elevator? Imagine you're trapped inside a free falling elevator. Would you decrease your impact impulse by jumping during the fall? When?
|
Would you decrease your impact impulse by jumping during the fall?
Yes
When?
Soon enough that it's before impact, late enough that you don't hit the ceiling of the elavator.
Beyond that I don't think it matters much.
Would it help if you jump inside a free falling elevator?
Probably not. Indeed I expect it would make things worse.
The consequences of a fall cannot be adequately explained by rigid body models. What matters is the peak force delivered to critical parts of your body.
If you don't jump then the elavator and you experience the crash as a system. Some elevators have explicit buffers at the bottom of the shaft. Even if they don't it is likely that there will be some compression of the elavator car and of whatever happens to be at the bottom of the shaft which will limit the peak force delivered to your body.
If you jump at the "right time" then you reduce your "impact impulse" but the elevator and you crash at different times. So the elavator can no longer help in spreading the impact over time.
If you jump two early then you hit the ceiling of the elevator. This gives you the worst of both worlds. You experience the original "impact implulse" and the elevator can no longer help in spreading it out over time.
| {
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Finding the volume of this irregular shape I have I have an approximately basketball-sized non-hollow piece of aluminum sitting in my house that is of irregular shape. I need to find the volume of it for a very legitimate yet irrelevant reason.
What is the best way I can do this? In fact, what are all the ways I could feasibly do this without going to a lab? (I don't live near any labs)
| Eureka! As Archimedes said, according to legend.
In principle, "TheMachineCharmer's" answer is feasible, but I would recommend recording the change in the volume of water instead (if you need an accurate measurement), because (1) it could be difficult to measure the volume of the spilled water, and (2) it is also a little less accurate to do so. (Some water will be left on the sides of the first container, and inserting the object into the filled container, catching all of the spillover while making sure the water level doesn't drop below the brim, could be difficult.)
If you only need a rough idea, the other way is fine.
If you need a more accurate measurement, try one of these:
*
*Get the volume of a container by filling it with water (e.g. from a graduated vessel). Empty the container, then place the object in it. Fill it with water again, measuring how much water you added. Subtract this number from the volume of the container to find the volume of your object. (The order in which you do these doesn't matter, of course.)
*This one could be a little harder, because you need a large graduated vessel. Fill the empty vessel with roughly enough water to submerge the object. Put the object in and record the change in the water volume.
| {
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Why don't spinning tops fall over? One topic which was covered in university, but which I never understood, is how a spinning top "magically" resists the force of gravity. The conservation of energy explanations make sense, but I don't believe that they provide as much insight as a mechanical explanation would.
The hyperphysics link Cedric provided looks similar to a diagram that I saw in my physics textbook. This diagram illustrates precession nicely, but doesn't explain why the top doesn't fall. Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground. However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral. Another reason I am not satisfied with this explanation is that the calculation is apparently limited to situations where: "the spin angular velocity $\omega$ is much greater than the precession angular velocity $\omega_P$". The calculation gives no explanation of why this is not the case.
| when the mass is spinning it has an angular momentum pointing in a direction perpendicular to the plane it's spinning on.
The angular momentum has to be conservate: i.e. has to keep pointing in the plane-perpendicular direction. As cedric said, the gravity, works for the axis of the spinng mass to fall horizontally on the plane: if this happens also the angular momentum as to torque! and this is not convenient from an energy conservation point of view..
Then u can consider that the magnitude of the angular momentum is proportional to the spinning speed: so as the spinning velocity gets higher it gets, for lack of a better word, "easier" for the top to resist the gravity..
If u try to spin a top on an inclined plane you will need to spin it faster to obtain the same "resistance to gravity"!
| {
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How do subgrid-scale parametrisations in climate models work? Global Circulation Models typically have grids of 100-300km on a side. There are obviously lots of atmospheric processes that happen at smaller scales than this. Convection, cloud formation, the effect of mountains...
How are these processes built in to the model?
| In general small scale motions (like convection and formation of local eddies) in geophysical fluid dynamics are treated as turbulence, that is a regime characterized by chaotic motions and rapid, quasi random variations of pressure, temperature and velocity. Those random processes cannot be neglected in boundary layers (layers of flow close to bounding surfaces atmosphere-soil, atmpsphere-ocean) because they averagely involve a flux of momentum and heat from the atmosphere to the soil or to the ocean (or viceversa). In geophysics most notable boundary layers are:
--> The atmospheric boundary layer, that is the bottom part of the atmosphere, about 1000 m thick, in contact with soil and sea surface.
--> The oceanic boundary layer, that is the top layer of the sea, about 10-100 m thick, close to the boundary with the atmosphere.
Hence a climatic large scale model ivolving boundary layers (for example models describing wind driven oceanic circulation) must take into account turbulence average effects: indeed parameterizing small scale phenomena means "taking their average effects into account".
Conversely, far from the boundary layers, turbulence can be neglected. For example global circulation models describing motions of high atmosphere usually neglect turbulence.
The simplest way to take into account the average effects of turbulence is to introduce in the equations of dynamics terms that represent the average friction, for example that described by drag equation:
Fd = - ρ Cd |U| u
where ρ is fluid density, |U| is velocity scale, u is velocity. Fd is called drag force, and is a force by surface unit. It represents the average friction exerted by the atmosphere on the surface. Cd is called "drag coefficient", and can be estimated through experimental observations. Its value can be different in different situations. Drag equation is an empirical relation, and can be deduced by purely dimensional consideretions, like Reynolds number. In particular we can find, using Buckingham theorem, that Cd depends only on Reynolds number.
| {
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Planet orbits: what's the difference between gravity and centripetal force? My physics teacher says that centripetal force is caused by gravity. I'm not entirely sure how this works? How can force cause another in space (ie where there's nothing).
My astronomy teacher says that gravity is (note: not like) a 3D blanket and when you put mass on it, the mass causes a dip/dent in the blanket and so if you put another object with less mass it will roll down the dip onto the bigger mass. Is this true and is this what causes the centripetal force.
| cetripital force only exists when you have prescribed motion (due to constraints). Think of a roller coaster car riding on a rail. To keep the car on the rail and tangetial to its direction a force and moment need to be applied to the car. When the path is circular we call the cetripetal force. In fact, with any path, instanteously it is said to be following a circle and therefore there is always an instanteneous centripetal force (unless in free fall).
Planets and things in orbit do not have a prescribed motion, but are following the free fall path whichever way they need to go. Gauss called this the principle of least action.
I hope this helps.
| {
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How can I measure the mass of the Earth at home?
*
*How can I measure the mass of the Earth at home?
*How was the mass of the Earth first measured?
| You can make estimates of the Earth's mass $M_\mathrm E$ by estimating its average density $\rho$ and using the formula $M_\mathrm E = \rho \cdot V$, where of course $V= \frac{4}{3} \pi R^3$, so you have to know the Earth's radius $R$. This method is rather a gamble because you don't know the Earth's average density (suppose the Earth would be hollow?). Newton made such an estimate himself.
Newton's law of gravitation $F_\text{grav} = G \cdot \frac{M_\mathrm E \cdot m}{R^2} = m \cdot g$ describes the attraction between a mass $m$ on the Earth's surface and the Earth. While $m$, $g$, so $F_\text{grav}$ and $R$ can be determined, there are two unknowns: $M_\mathrm E$ and $G$.
So one can only determine the mass of the Earth together with this constant $G$: the gravitational constant.
You cannot measure the mass of the Earth at home unless you have access to a Cavendish torsion balance to measure the gravitational constant. The Cavendish experiment is described in this Wikipedia article. Approx. 100 years after Newton the first experiment was done to measure $G$.
The value of $G \cdot M_\mathrm E$ can be measured very accurately ($3.9860042\times10^{14}\ \mathrm{m^3/s^2}$).
The value of $G$ can be measured less accurately ($6.674\times10^{-11}\ \mathrm{m^3/(kg\ s^2}$) so that's the bottleneck for the determination of the Earth's mass.
The Earth's mass is not constant; it is estimated that every day 40 tons of dust from meteorites hit it. This is totally negligible for a mass of $5.9736\times10^{24}\ \mathrm{kg}$.
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Advantages of high-energy heavy-ion collisions over proton-proton collisions? Some high-energy experiments (RHIC, LHC) use ion-ion collisions instead of proton-proton collisions. Although the total center-of-mass energy is indeed higher than p-p collisions, it might happen that the total energy per nucleon is actually lower. What are the advantages of using ion-ion collisions (e.g. gold-gold or lead-lead) instead of proton-proton collisions, considering the same accelerator?
| As a clarification the energy per nucleon is always lower: for example, currently in the LHC the proton top energy is 3.5 TeV. Now the Pb energy is 3.5 TeV times Z so the energy per nucleon is 3.5*Z/A and A is greater than Z for every nucleus (except the proton where it is equal to one).
But the goal of ion-ion collision is not to increase the total energy or the energy per nucleon: it is to obtain a different type of collision.
It should be noted than in a proton-proton collision, the energy involved in the real collision process is variable: each quark and gluon carry a fraction of the energy of the proton, and hard collision involve a collision between a quark/gluon of one proton against a quark/gluon of the other.
In the case of ion-ion collision you have the same process: the energy is shared by the protons/neutrons and they can have different energies.
The goal of such collision is also to obtain a volume (bigger than in a p-p collision) with a very high energy density. In such a volume, a "state of matter" called quark-gluon-plasma is believed to be possibly created. The study of this QGP is one of the main goal of the ALICE experiment at the LHC.
A few references:
*
*The ALICE experiment at CERN
*Live control screen of the LHC now running as a Pb-Pb collider at 574 TeV in the center of mass
| {
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What software programs are used to draw physics diagrams, and what are their relative merits? Undoubtedly, people use a variety of programs to draw diagrams for physics, but I am not familiar with many of them. I usually hand-draw things in GIMP which is powerful in some regards, but it is time consuming to do things like draw circles or arrows because I make them from more primitive tools. It is also difficult to be precise.
I know some people use LaTeX, but I am not quite sure how versatile or easy it is. The only other tools I know are Microsoft Paint and the tools built into Microsoft Office.
So, which tools are commonly used by physicists? What are their good and bad points (features, ease of use, portability, etc.)?
I am looking for a tool with high flexibility and minimal learning curve/development time. While I would like to hand-draw and drag-and-drop pre-made shapes, I also want to specify the exact locations of curves and shapes with equations when I need better precision. Moreover, minimal programming functionality would be nice additional feature (i.e. the ability to run through a loop that draws a series of lines with a varying parameter).
Please recommend few pieces of softwares if they are good for different situations.
| I use TKPAINT which still works very well.
http://www.netanya.ac.il/~samy/tkpaint.html
First, one has to download ActiveTcl for Windows or its Tcl counterparts for Linux or whatever you use. It can draw filled or empty disks, ellipses, squares, rectangles, splines, rotate them, quickly copy them, move them, texts with many fonts, colors, grid, and it may be exported as EPS - encapsulated postscript as well - which is a standard way to embed similar diagrams in TeX papers on the arXiv and beyond.
I've used it in many papers when I was writing them.
Cheers
LM
| {
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Would a magnet attract a paperclip indefinitely? Let's say we have a magnet stuck to a metal bar, suspended above the ground. If I attach a paperclip to the magnet, where is the energy to hold the paperclip coming from (against the force of gravity), and for how long will the paperclip remain there - will it remain there forever?
| When the magnet attracts and moves the paperclip, moving it in the gravitational field of the earth, the energy comes from the potential energy that [in that case] we can associate to the magnetic field.
By attracting the paperclip you increase the gravitational potential energy of the paperclip but you reduce the one it has due to the magnetic field.
You do not need energy to hold the paperclip. As an analogy: when you hold something in your hands, you feel like you are "working" even in a static case; but in mechanics, work = exchange of energy is defined as $W = \int \vec{F} d\vec{s}$.
When the situation is stationary, no energy exchange is involved, no force does no work at all, so it can stay forever.
| {
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Distance travelled in free-fall When an object is in free fall, we have:
$$a(t) = g - \frac{c}{m}v(t)^2$$
where $g$ is acceleration due to gravity, $m$ is the mass of the object, and $c$ is the coefficient of air resistance.
How does one get the distance traveled after t seconds? I tried integrating it, giving
$$v(t) = gt - \frac{c}{3m}s(t)^3$$
$$s(t) = \sqrt[3]{\frac{3m}{c}(gt - v(t))}$$
Which is a function defined in terms of its derivative. How would I find a better definition of $s(t)$?
| When accelereration is given as a function of velocity only then the following apply
$$ x(u)=\int\frac{u}{a(u)}\,\mathrm{d}u+K_{1} $$ and $$ t(u)=\int\frac{1}{a(u)}\,\mathrm{d}u+K_{2} $$
the rest is solved similarly to David Z (above).
| {
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Notation of plane waves Consider a monochromatic plane wave (I am using bold to represent vectors)
$$ \mathbf{E}(\mathbf{r},t) = \mathbf{E}_0(\mathbf{r})e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)}, $$
$$ \mathbf{B}(\mathbf{r},t) = \mathbf{B}_0(\mathbf{r})e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)}. $$
There are a few ways to simplify this notation. We can use the complex field
$$ \tilde{\mathbf{E}}(\mathbf{r},t) = \tilde{\mathbf{E}}_0 e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)} $$
to represent both the electric and magnetic field, where the real part is the electric and the imaginary part is proportional to the magnetic. Often it is useful to just deal with the complex amplitude ($\tilde{\mathbf{E}}_0$) when adding or manipulating fields.
However, when you want to coherently add two waves with the same frequency but different propagation directions, you need to take the spatial variation into account, although you can still leave off the time variation. So you are dealing with this quantity:
$$ \tilde{\mathbf{E}}_0 e^{i\mathbf{k} \cdot \mathbf{r}} $$
My question is, what is this quantity called? I've been thinking time-averaged complex field, but then again, it's not really time-averaged, is it? Time-independent? Also, what is its notation? $\langle\tilde{\mathbf{E}}\rangle$?
| Stationary field or monochromatic field. Yes, basically that is the field including the $e^{i\omega t}$ term, but even when it is omitted one still knows what is meant.
| {
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How are neutrons produced from cosmic ray particles? What are the details of how neutrons are produced as a result of cosmic ray particles hitting our planet's atmosphere?
For instance, what is the pathway that creates the highest number of neutrons from cosmic ray particles?
The article "Single event upset" states:
An SEU happens due to cosmic particles which collide with atoms in the atmosphere, creating cascades or showers of neutrons and protons. At deep sub-micrometre geometries, this affects semiconductor devices at sea level.
But it does not describe the mechanism.
The Wikipedia article on the neutron states something similar:
Cosmic radiation interacting with the Earth's atmosphere continuously generates neutrons that can be detected at the surface.
What are the processes, in detail?
| Spallation of air atoms' nuclei is the easiest way - there are neutrons kicked out of the nucleus directly or emmited by radioactive elements, which were activated by cosmic rays.
Also neutrons can be produced upon hadronization of quark-gluon-plasma or from electron capturing in air nuclei.
| {
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Why does kinetic energy increase quadratically, not linearly, with speed? As Wikipedia says:
[...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.
Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ \mathrm{m/s}$ to $1\ \mathrm{m/s}$?
My intuition is wrong here, please help it out!
| In comes down to definitions.
Momentum is defined as $p = mv$. Momentum grows linearly with velocity making momentum a quantity that is intuitive to understand (the more momentum the harder an object is to stop). Kinetic energy is a less intuitive quantity associated with an object in motion. KE is assigned such that the instantaneous change in the KE yields the momentum of that object at any given time:
$\frac{dKE}{dv} = p$
A separate question one might ask is why do we care about this quantity? The answer is that in a system with no friction, the sum of the kinetic and potential energies of an object is conserved:
$\frac{d(KE + PE)}{dt} = 0 $
| {
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Material resistency to lasers beam Keeping the average power constant, why some materials are more eager to be damaged by pulsed laser with respect of C.W. lasers, or viceversa?
When i talk about pulsed lasers i think for examples of duty cycles in the order of $10^5$.
For example optical elements (such as a vortex phase plate for donut-shaping the beam) have different tolleration regimes regarding the incident power not simple dependent on the average power but also on the peak power for pulsed beam.
| Not being an exhausting list. Some phenomena that allows for this to happens is non-linear effects. A short pulse allow a higher intensity for a small period to time which enhance non-linear effects such as two-photon absorption and self-focusing. Two photon absorption will make a material absorb light which it wouldn't if light was low intensity. Self-focusing would make the light focus more which would decrease area which the heat can be diffused, or increase intensity for two or more photon absorption. The self-defocusing is possible too which would allow for decrease damage in the material if pulsed.
| {
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Chosing a reference frame in which the Earth is at rest and doesn't rotate We may choose a non-rotating earth as our reference frame and ask ourselves: how about the planetary and stellar motions. A star at a distance of 10 million light years would turn around the earth in 24h with a velocity of 10^18 m/s.
A friend once told me that actually articles have been published delving into this problem, e.g. to prove that fictitious forces emerge from the choice of such a bizar reference frame that ensure that the earth is still (somewhat) flattened at it's poles.
Questions:
1) Does anybody know of such a publication?
2) I know that even such speeds of 10^18 m/s are not in contradiction with relativity because a limiting velocity only exists for exchange of information, which apparantly does not occur.
Still: could anybody explain why such bizar velocities are allowed?
| Planetary motions in such a frame would be depicted properly by the deferent an epicycle model of Ptolemy:
http://en.wikipedia.org/wiki/Deferent_and_epicycle
So, the question is actually answered by Ptolemy. For more reference, look up details in the Almagest! :-)
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Evolution in the interpretation of the Dirac equation As I understand, Dirac equation was first interpreted as a wave equation following the ideas of non relativistic quantum mechanics, but this lead to different problems.
The equation was then reinterpreted as a field equation and it is now a crucial part of quantum field theory.
My question is: could you provide me a reference (paper, book) that explains this evolution, including the different historical steps, etc. ?
I have a good knowledge of QM and I studying field theories, but I would like to have a clearer view on this historical evolution.
| In my opinion, a good introductory text for Relativistic Quantum Mechanics is Quantum Field theory- by Itzykson and Zubor. Another text is Relativistic quantum mechanics of string fields by - Greiner and Muller
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