Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Is it really true that valence band is completely filled at zero temperature? Is it really true that valence band is completely filled at null temperature?
Indeed, I would think that if we apply an electric field, this would give some energy to the electrons from the valence band, so would they be prevented to leave the valence band to go to the conduction band, thanks to the energy from the electric field?
I don't see why would the electrons know if the energy that they receive is due to temperature or from electric field source?
| If a perfect semiconductor/insulator is at zero temperature, than indeed all its electrons are in the valence band, and exciting them to the conduction band requires, as a minimum, the gap energy, $E_g$.
If a uniform electric field is applied, then, obviously, there will be degeneracy between the valence band and the conduction band states, separated by distance of order
$$eEd=E_g$$
(More precisely, we have to recalculate the states of the crystal, taking into account the electric field, since the translation symmetry is broken now in one direction.) This degeneracy results in electrons tunneling from the valence band to the conduction band, which is a kind of dielectric breakdown, known as Zener effect.
However, in practice this effect is observed only at very large electric fields or in artificially created materials with small band gap (superlattices). The reason for that is that the tunneling is possible only when the tunneling length is comparable to the coherence lengths, which is usually rather short in real materials.
Note also, that in a real material the imperfections (dislocations, impurities, etc.) will typically result in both valence and conduction band having tails extending well into the gap, so there might be always a residual concentration of electrons and holes, even at zero temperature - but too low to result in detectable electric conductance.
| {
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'Drummy' sound when striking with a hammer In building there is a common test for masonsry structures that involves striking the structure with a hammer and listening to the resulting sound. If the sound is ringing the structure is fine but if the sound is 'drummy' (maybe like a dull thud) then there is an issue with the structure.
I'm presuming this somehow relates to vibrations in the structure. But what actually is the difference between the ringing sound and the 'drummy' sound in terms of the vibrations within the structure?
| in the ringing sound, the structure is vibrating as if it were a solid, seamless metal bar like in a xylophone or a wind chime, which indicates that all the individual bricks in it are tightly and firmly cemented together into a single mass, which is a good thing.
if the sound instead is dull and "drummy", it means there are gaps and breaks or "debonds" between the individual bricks which prevent the entire structure from responding as a single big chunk of stuff. instead it responds like a big pile of smaller chunks that are poorly-connected together- which is a very bad thing for a masonry structure.
| {
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Photoelectric current vs anode potential Attached is the graph of photoelectric current vs Anode potential as given in my book for same intensity and different frequencies of incident light for the same metal(hence same work function).
In my opinion this graph isn't the graph for the conditions as mentioned rather, it should be for light with same number of photons incident per unit time per unit area(n), with different frequencies.
$I=nh \nu$, where I is the intensity. As $\nu$ is different, $I$ is also different.
In my opinion, the graph for
photoelectric current vs Anode potential for same intensity and different frequencies of incident light for the same metal should be as attached in the free hand drawn diagram as drawn by me(I couldn't find any reference regarding the diagram I got).
Could someone please explain whether my book or I or neither is correct?
All my other reference books also seem to suggest the same as given in my book.
P.S. I found few other similar questions asked here before but none of which could answer my query.
| Yes, you are right.
What you drew can be confirmed from here:
Stopping Potential vs Frequency
| {
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Is 'Curl of magnetic field in electrostatic is zero' only empirical? I was looking up on the uniqueness of the displacement current.
About the uniqueness of the displacement current this question was exactly what I was looking for, but all the answers seem to go with 'empirically, when the electric field is constant and current density is zero, the curl of the magnetic field is also zero.'
Are there any more explanation other than 'That's what we have found experimentally'? or are we solely relying on the fact the Ampere-Maxwell equation matches the experiments.
| No, it's not purely empirical. If the curl of an electrostatic magnetic field was going to be nonzero, it would have to be determined by some new term on the right-hand side of that Maxwell's equation. There is nothing available to serve that purpose that would preserve linearity and have the right symmetry properties under C, P, and T. For instance, $\operatorname{curl}\textbf{B}=\alpha \textbf{B}$, where $\alpha$ is a constant, would violate parity.
At a deeper level, there is very little wiggle room in writing down Maxwell's equations because they're the wave equations of an antisymmetric tensor field. If you try to write down something relativistically valid, you don't have a lot of other options. The main options are either to add magnetic monopoles or to do various trivial redefinitions like flipping the direction of the thing we call the magnetic field.
| {
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Heat produced if earth stops rotating In my textbook there is a question that is as follows:-
If earth stops rotating about it's own axis,the increase in its temprature will be(Here R=radius of earth,ω=angular velocity of earth,J=mechanical equivalent of heat,C=average specific heat capacity of earth)
Here i have doubt why heat is produced if earth stops rotating?and please explain how to solve it.
| I agree with the comments of @G.Smith and @Pounik. I'm thinking that one possible explanation as to why the temperature of the earth might increase is due to the inertia of the molten core. The crust would stop but the molten core would continue to rotate. This would result in friction between the molten core and the crust which, in turn, might raise the temperature at the interface. But I must admit, this is purely speculation on my part.
Anyways, as possible food for thought, I hope this helps.
| {
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What would be the potential due to a charge $Q$ at $Q$ itself? We know that the potential due to a point charge $Q$ at any distance is given by $V=\frac{kQ}{r}$, but what would the potential be at the charge itself?
| Potentials and charge distributions are not mathematical, they are physical observation that lead to mathematical models, which have to describe the data.
The classical Maxwell equation solutions fit a multitude of observations with great accuracy and also are predictive for new systems. BUT note the term classical. It is in contrast with quantum mechanical. When distances approach zero one is in the quantum mechanical regime, the one that is given by the Heisenberg uncertainty principle. $Δx*Δp>h/{2π}$ .
The classical potential function of 1/r cannot be tested for small distances because a test charge that would measure the potential can never reach the other charge. This is one of the reasons that quantum mechanics was invented. Together with the black body radiation, the periodic table, and the spectrum of light from atoms it was necessary to forbid two point charges to fall on each other, because atoms would not exist, the electrons would fall to the nucleus and neutralize it.
In quantum mechanics, the potential of the charged nucleus enters into the Schrodinger ( more correctly the Dirac) equation and the solutions for the electron are quantized, and there is always a ground state which allows to form stable states. ( see the hydrogen atom).
So it is the quantum mechanics from which all classical physics emerges that provides the mathematical models for r=0 .
The same problem exists with the 1/r of the gravitational potential, that is why quantization of gravity is a holy grail of theoretical physics. An effective quantization is used at the moment in cosmological models like the Big Bang, and they are looking for definitive models.
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What is a free parameter? Soft question here, but I was wondering just what exactly free parameters are? I have a murky understanding on the concept but I would much appreciate someone shedding some light on the matter. Is Newton's gravitational constant $G$ a free parameter of that theory? What exactly are the requirements for a parameter to be considered a free parameter?
| Free parameters are not predefined but must be estimated by theory or experimentally. Or it can be a parameter used in fitting a dataset with an expression. The free parameters are varied to get a good fit to the data.
G in Newton's gravitational equation is a free parameter and has been measured but not with high accuracy.
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Derivation of density of states (free electrons) I am reading Condensed matter physics from M.Marder.
This is the derivation for the density of states for free electrons.
$\begin{aligned} D(\mathcal{E}) &=\int[d \vec{k}] \delta\left(\mathcal{E}-\mathcal{E}_{\vec{k}}^{0}\right) \\ &=4 \pi \frac{2}{(2 \pi)^{3}} \int_{0}^{\infty} d k k^{2} \delta\left(\mathcal{E}-\mathcal{E}_{\vec{k}}^{0}\right) \\ &=\frac{1}{\pi^{2}} \int_{0}^{\infty} \frac{d \mathcal{E}^{0}}{\left|d \mathcal{E}^{0} / d k\right|} \frac{2 m \mathcal{E}^{0}}{\hbar^{2}} \delta\left(\mathcal{E}-\mathcal{E}^{0}\right) \\ &=\frac{m}{\hbar^{3} \pi^{2}} \sqrt{2 m \mathcal{E}} \end{aligned}$,
where
$\int[d \vec{k}] \equiv \frac{2}{V} \sum_{\vec{k}}=\int d \vec{k} D_{\vec{k}}=\frac{2}{(2 \pi)^{3}} \int d \vec{k}$,
The factor of 2 accounts for electron spin. V is the volume.
and $\mathcal{E}_{\vec{k}}^{0}=\frac{\hbar^{2} k^{2}}{2 m}$.
In step one he says that he changes the integral to polar coordinates because $\mathcal{E}^0$ depends upon the magnitude and not the direction of $\mathbf{k}$. So, where does the $4\pi$ come from? Shouldn't it be $2\pi$?
In second step he writes $k$ in terms of $\mathcal{E}^0$. In the end I will have
$\begin{equation}\frac{m}{\hbar^3 \pi^2} \int_0^\infty d \mathcal{E}^0 \sqrt{2 m \mathcal{E}^0} \delta(\mathcal{E}-\mathcal{E}^0)
\end{equation}$
I think he used the property
$\int_{-\infty}^{+\infty} f(x) \delta(x-a) dx = f(a)$
but the integral goes from 0 to infinity. Can you explain to me what is going on here?
| In step one, the angular part of the integral is $4\pi$ because the $\vec{k}$ integral is an integral in 3 dimensions, so you're integrating over a sphere:
$$\int d\vec{k} = \underbrace{\int_{0}^{2\pi} \,d\phi \int_{0}^\pi \sin\theta \,d\theta }_{=\,4\pi} \int_0^\infty k^2 \,dk \,.$$
In step two, the lower bound of the integral can be extended to $-\infty$, since the integrand is zero for $\mathcal{E}^0<0$, i.e.:
$$\int_0^\infty \sqrt{2m\mathcal{E}^0} \,d\mathcal{E}^0 = \int_{-\infty}^\infty \sqrt{2m\mathcal{E}^0} \,d\mathcal{E}^0 \,.$$
| {
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How to calculate inertia tensor of composite shape with angle? I have I have some objects assembled like this :
The inertia tensor would be :
$$I=I_1+I_2+I_3-m_1 \,\tilde{r}_{01}\,\tilde{r}_{01}-m_3\,\,\tilde{r}_{03}\,\tilde{r}_{03}$$
Where :
$$\tilde{r}_{01}=\begin{bmatrix}
0 & -z & 0 \\
z & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}$$
and
$$\tilde{r}_{03}=\begin{bmatrix}
0 & +z & 0 \\
-z & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}$$
According to this.
Detailed notations here.
But what is the inertia if -
*
*I rotate the planar object on the very top(everything with it) around $Z$-axis
*Just rotate the rod and sphere (except the planar object), just like a pendulum. Let angle with $X$-axis is $\alpha$ and angle with $Y$-axis is $\beta$
*Do both together
| The inertia tensor obeys the congruent transformation from the local coordinates to the world coordinates.
If you have a 3×3 rotation matrix $\mathbf{R}$ then you have
$$ \mathbf{I}_{\rm world} = \mathbf{R} \, \mathbf{I}_{\rm body} \mathbf{R}^\top $$
So the combined inertia would be
$$ \mathbf{I} = \mathbf{R} \, \left(\mathbf{I}_1 +\mathbf{I}_2 + \mathbf{I}_3 \right) \mathbf{R}^\top - m_1 \overline{\boldsymbol{r}}_1 \overline{\boldsymbol{r}}_1 - m_3 \overline{\boldsymbol{r}}_3 \overline{\boldsymbol{r}}_3$$
You must make sure the vectors $\boldsymbol{r}_n$ point to the center of mass in the world coordinate system after the body is rotated.
See also this answer to a similar question.
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Is the Olympic running race fair? I noticed that the 200-meter sprints are conducted on curved tracks. (See this video: world championship semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(Source)
Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.
(The extra centrifugal force is roughly 2 Newtons from my calculation).
I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?
| Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).
| {
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The right hand rule confusion?
I have a question regarding this problem. By using the right hand rule, I thought the answer would be A, but the answer key says it's B.
Doesn't the current come from the + side, so you wrap your fingers towards yourself(?) so that the thumb points to the left?
| Electric and magnetic fields are best understood using the terminology "electric field ${\bf E}$" and "magnetic field $\bf B$". In the case of a magnetic field this is better terminology than "lines of force" because the force produced by a magnetic field on a small object such as a current-carrying wire or a moving charge is perpendicular to the field not along it. And when a small magnet such as a compass is placed in a magnetic field, the effect of the magnetic field is to align the compass without producing any net force on it in any direction (it just produces a torque until the small magnet is aligned with the field). Therefore I will avoid the term "line of force" in my answer. I will call them "field lines" or "lines of magnetic field".
In the case of a straight current-carrying wire, the magnetic field loops around the wire in a right-handed direction. That is, if you make your right thumb point along a current then the curled fingers of your right hand show the magnetic field direction. If you do this for each small section of a current-carrying loop, then you find that the field through the middle of the loop has to be in the direction given by your right thumb when your right hand fingers curl around the loop. The current in the circuit shown in the question goes from $+$ to $y$ to $x$ to $-$ so we deduce that the magnetic field lines (lines of $\bf B$) go from $y$ to $x$ inside the coil, and then loop around outside till they come back to $y$.
The terminology "north pole, south pole" is also poor terminology because in fact there is no pole for the field ${\bf B}$, it goes in loops. But the convention is that the end where the field lines point away outside the magnet is the "north" so in this example $x$ is the north pole and $y$ the south pole.
| {
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Work done by friction in a complicated path
A block of mass $M$ is taken from point $A$ to point $B$ in a complex path by a force $F$ which is always tangential to the path. We also have coefficient of friction as $K$. What will be the work done by force $F$ when it reaches point $B$ from point $A$? Given that the vertical displacement from $A$ to $B$ is $h$ and the horizontal displacement from $A$ to $B$ is $l$.
In this question, I tried solving the problem using conservation of energy, we know that the total energy will remain constant. So with that, we will have,
$$\Delta U_{gravity}+W_{friction}+W_{F} = O$$ But how do you calculate the work done by friction in this case?
Moreover, in the answer, the work done by friction is only dependent on l!!
EDIT:
1.The body is moved very slowly.
| In fact the answer is a general result for all particles taken slowly from one point to another when it is on any inclined surface.
The Normal reaction is given as:-$$N=mg\cos\theta$$ This relation is valid as the body is hauled up slowly so the acceleration perpendicular to the surface tends to zero.
$\theta$ is the variable angle made by the the tangent at any point on the surface with the horizontal.$$f=kmg\cos\theta$$
Work done by friction in moving the particle by a distance $ds$ along the path:-$$dW_{friction}=kmg\cos\theta ds$$
Now $$ds\cos\theta=dl$$
So $$dW_{friction}=kmgdl$$
$$W_{friction}=\int kmgdl$$
$$W_{friction}=kmgl$$
We conclude that if the object takes any random path the work done would only depend on horizontal distance between initial and final point.
EDIT 1
As you have asked about the work done by $F$ in the comment section.You can simply use work energy theorem to solve it.
$$W_{total}=\Delta K$$
$$\Delta K=0$$
$$W_{friction}+W_{gravity}+W_{F}=0$$
$$W_{F}=kmgl+mgh=mg(h+kl)$$
| {
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Modern form of Brown-Henneaux formula Almost every paper mentioning Brown and Henneaux's matching of asymptotic symmetries of AdS$_3$ with the Virasoro algebra of a $1{+}1$-dimensional CFT summarizes their results in the formula
$$c=\frac{3R}{2G},$$ whereby the central charge $c$ is expressed in terms of the AdS radius $R$ and the gravitational constant $G$.
However, Brown and Henneaux's original paper doesn't contain this formula. Instead, it derives an asymptotic algebra of the form
\begin{equation}
\{J[L_n],J[L_m]\}^\star = (n-m) J[L_{n+m}] + 2\pi i \sigma R n (n^2 - 1) \delta_{n,-m}
\end{equation}
for charges $J[L_n]$ (on page 222), which looks very close to the standard Virasoro algebra
\begin{equation}
[L_n,L_m] = (n-m) L_{n+m} + \frac{c}{12} n (n^2 - 1) \delta_{n,-m} \ .
\end{equation}
But I don't see how this form would relate the central charge in any way to $G$ nor figure out where the factor $3/2$ comes from. Can anyone help me figure out the missing steps?
| Computing the Poisson bracket gives
$$I=i\{L_{m}^{(+)},L_{n}^{(+)}\}=\frac{il}{\kappa}\int_{0}^{2\pi}d\phi e^{imx^{+}}\left(e^{inx^{+}}\partial_{+}L_{+}+2L_{+}\partial_{+}e^{inx^{+}}-\frac{1}{2}\partial_{+}^{3}e^{inx^{+}} \right),$$
where $\kappa=8\pi G$ and $L_{m}^{(\pm)} = \frac{l}{\kappa}\int_{0}^{2\pi}d\phi\ L_{\pm}\left(x^{\pm}\right)e^{imx^{\pm}}$. Integrating by-parts on $\partial_{+}$ gives us
$$I=\frac{il}{\kappa}\int_{0}^{2\pi}d\phi\ \left(\underbrace{-i(n+m)e^{i(n+m)x^{+}}L_{+} + 2inL_{+}e^{i(n+m)x^{+}}}_{=-i(m-n)L_{+}e^{i(m+n)x^{+}}} - \frac{1}{2}(-i)n^{3}e^{i(n+m)x^{+}} \right).$$
Now, the integral representation of the discrete-delta funciton is given by: $\delta(n) = \frac{1}{2\pi}\int_{0}^{2\pi}e^{int}dt$ see here. Thus, we have
$$I = (m-n)L^{(+)}_{(m+n)} + m^{3}\delta_{(m+n),0}\frac{l}{8G},$$
where the central charge is $m^{3}\delta_{(m+n),0}\frac{l}{8G} = \frac{c}{12}m^{3}\delta_{(m+n),0}$ with the Brown-Henneaux central charge $c=\frac{3l}{2G}$
| {
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What is the time derivative of resistance? Is there a unit for $\frac{\Omega}{sec}$? I have tried looking it up, but I can’t find anything
| Contrary to what the other answers say, resistors do change their value with time, even the most accurate ones, even at perfectly constant temperature. This is due to various phenomena, e.g. release of internal stresses, contamination from impurities etc. For instance, National Metrology Institutes keep historical records of the drift of their standard resistors, which is typically quite predictable and can be used to interpolate resistance values between calibrations.
The resistance drift of a resistor is usually specified in relative terms, that is, by the relative drift coefficient
$d = \dfrac{1}{R}\dfrac{\mathrm{d} R}{\mathrm{d} t},$
the SI unit of which is $\mathrm{s}^{-1}$. When the drift coefficient is constant, you can predict the value of the resistance $R(t)$ at time $t$ from its value at time $t_0$, with the equation
$R(t) = R(t_0)[1+d(t-t_0)].$
It is also worth noting that the unit $\Omega/\mathrm{s}$, which you mentioned in your question, really is an SI unit, it's just a unit without a special name (e.g. farad, symbol F, is just a special name for $\mathrm{s}^4\mathrm{A}^2\mathrm{m}^{−2}\mathrm{kg}^{−1}$).
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How do we calculate the force applied by a ball on a wall which bounces back? If we have a ball which we throw toward a wall which touches the wall and bounces back then how will you calculate the force applied by the wall on the ball because the the contact time of the ball and the wall is infinitely small so force must be infinitely large?
In a similar case
If a ball is dropped from a height of 80cm (10kg ball),
What amount of force will the ground apply on the ball(suppose ball comes to halt after touching the ground)?
In both the cases the duration of change of momentum is infinitely small so should the force be infinitely large ??
Note: Pls try and ignore any mistakes in the question because I am new to stack exchange
| For something like a tennis ball, one or two assumptions are in order:
1) Assume some amount of "flatness" of the ball when it is most compressed against the wall.
2) Estimate the distance from the center of the ball to the wall at this point. Call this distance "d".
Note that before the collision, the distance from the center of the ball to the surface of the ball is the ball's radius ("r"). Therefore, the center of the ball is one radius away from the wall at the instant when it first touches the wall. One instant later, the center of the ball is a distance "d" from the wall.
Since the velocity of the ball just as it first touches the wall is known ($v_i$), the velocity of the ball at maximum compression against the wall is known ($v_f=0$), and the distance moved during the compression against the wall is known ("r-d"), you can calculate the average acceleration from the formula $v_f^2 = v_i^2 + 2a \Delta x$, where $\Delta x = r - d$. Knowing the average acceleration, you can calculate the force on the ball, which is also the force on the wall due to Newton's 3rd law, from the equation $F=ma$.
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The potential at a point According to my book, 'The potential at a point is said to be 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.' But I wonder how it is possible. As the charge is being brought from infinity, the work done = force * infinity, thus, the work done would be infinity indeed. Please help me out.
| Work isn’t $Fd$ when the force changes with position. It’s an integral, and the integral is finite even over an infinite distance because the force goes to zero sufficiently rapidly at large distances.
The integral is a standard homework problem so I am not going to write it or evaluate it.
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How to interpret the wave function for non point-like objects The accepted interpretation of a single-particle wave function is that it represents (among other things) the probability of finding the particle at any point. The wave function is normalised so that the probability sums to 1 over space.
In principle, how might the accepted interpretation address the hypothetical case of a non-point-like object with no component parts?
For those of you who might not see the point of the question, consider the 2D 'particle in a box' scenario, in which a point particle is constrained to lie on some portion of the X-axis by a pair of potential barriers. Now replace the point particle by a square particle the length of whose sides is greater than half the width of the box. The particle can move, but there are now values of x for which the particle will always be found, so it is no longer possible to normalise the wave function, and it is no longer possible to associate the particle unambiguously with a given value of x.
| The wave function will necessarily include all degrees of freedom the particle has: position, orientation, deformations, etc.
| {
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Physics of air flow in Kipchoge's sub-2:00 marathon Yesterday Eliud Kipchoge became the first human to run a marathon distance in under two hours. Part of what allowed him to do it seems to have been that he had pacers running along with him to break the wind. These pacers ran in a strange formation like a "Y:"
Kipchoge is the white circle. Is there any explanation of how this was arrived at? Was it purely empirical? Is there some physical way to understand why this would be a good formation, from the point of view of fluid dynamics? The inversion of the wedge is very counterintuitive to me.
| Maybe this will help: a picture of the flow around 8 cylinders. This is a viscous 2D flow of an incompressible fluid. The color in the figures corresponds to the magnitude of the flow velocity. A numerical solution is obtained by integrating the Navier-Stokes equations using FEM - see code on community.wolfram.com/groups/-/m/t/1433064
One cylinder is highlighted here (corresponds to the champion position). The drag coefficient of this cylinder is negative!
| {
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Are Imaginary Numbers Really “Imaginary?” I find the naming convention of “Imaginary” misleading, as it does give a sense that the quantity is merely an abstract construct used to mitigate the difficulties of performing some mathematical operations.
My question is, other than the wavefunction for example, where else do complex numbers have physical significance?
Also, since only $|\Psi|^2$ has physical meaning i.e. probability amplitude, does this really mean that the “Imaginary” part is really significant, or just another purely mathematical construct?
| It is possible to do all computations that traditionally involve complex numbers, without reference to imaginary numbers at all. An equation written in complex notation can always be separated into two equations in real notation. Example:
$$Z_1^2 +Z_2^2 = C$$
is the same as $$x_1^2 -y_1^2 + 2i(x_1y_1) + x_2^2-y_2^2 + 2ix_2y_2= C_r +iC_i$$ which is exactly the same as the pair of equations $$x_1^2 -y_1^2 +x_2^2 -y_2^2= C_r$$ and $$2x_1y_1+2x_2y_2 = C_i$$
However, complex notation using i, and especially i in an exponent, greatly simplifies the notation and provides a useful way to visualize the meaning of the computation.
| {
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What is the difference in the way a single rope and a double rope would behave under dynamic load? Suppose I have a single rope attached to a fixed point via a load cell, which gives a number (in kN) based on the load it's experiencing. I take a weight (x) and attach it to the rope at a fixed point and raise it up to a given height and then drop it. The rope has elastic properties and has some elongation.
Now, suppose I repeat the process, but this time I use two ropes hanging in parallel (actually, it's the same rope with a mid-line knot which prevents the tension/elongation in one half being transferred to the other half).
What I need to understand is whether the number on the load cell would be different between the two scenarios, and if so, why?
| Is this a question about rock climbing? If so, then the answer is never to use twin ropes, because they're a total pain, and their theoretical advantages never show up in actual climbing. (And note that (1) twin ropes have different physical specs, and (2) the twin ropes will usually not share the load at all equally.)
But anyway, as a pure physics question, the answer to this is not one number. The tension is a complicated function of time in both cases. The masses will move as in simple harmonic motion when the rope is under tension, but this can be interspersed with periods of free fall when the rope is slack.
You could change the question to ask about the maximum tension in the rope. Then conservation of energy tells us (for small damping) that $T\propto \sqrt{k}$, where $T$ is the maximum total tension and $k$ is the equivalent spring constant of the ropes. (To see this, observe that the work done by the rope is fixed, and equals $(1/2)kx^2$. The maximum tension is $kx$, which is therefore proportional to $\sqrt{k}$.) Using two ropes in parallel doubles $k$, and therefore increases the maximum total tension $T$ by $\sqrt{2}$.
| {
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Do Hermite polynomials imply a weight for quantum harmonic oscillator wavefunctions? I know that solutions of quantum harmonic oscillator can be expressed in the form of Hermite polynomials.
But I recently came to know that Hermite polynomials are actually orthogonal polynomials having $e^{-x^2}$ weight function.
So, here is my question (out of curiosity):
Is there a connection between the exponential weight function and the harmonic oscillator solutions? Perhaps the oscillator problem can somehow be transformed into a different space having $e^{-x^2}$ weight function with simpler solutions.
Thanks for any help.
Sorry for missing out some technical details as I am a beginner in Physics.
| Absolutely. The Hermite polynomials
$$
H_n(x) = (-1)^n e^{x^2} \partial_x^n e^{-x^2} = \left(2x - \partial_x\right)^n \cdot 1
$$
are orthogonalized by
$$
\int_{-\infty}^\infty H_m(x) H_n(x)\, e^{-x^2} \,dx = \sqrt{\pi}\, 2^n n! ~ \delta_{nm} ~,
$$
whereas the (nonpolynomial) Hermite functions
$$
\psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{-\frac{x^2}{2}} H_n(x) = (-1)^n \left (2^n n! \sqrt{\pi} \right)^{-\frac12} e^{\frac{x^2}{2}} \partial_x ^n e^{-x^2}
$$
are orthogonalized trivially,
$$\int_{-\infty}^\infty \!\psi_n(x) ~ \psi_m(x) \,dx = \delta_{nm} ~,
$$
so they obey the quantum oscillator eigenvalue equation,
$$
\partial_x^2 ~ \psi_n(x) + \left(2n + 1 - x^2\right) \psi_n(x) = 0.$$
It is these functions that are the eigenfunctions of the Fourier transform,
as the oscillator Hamiltonian is invariant under Fourier transformation.
The Hermite polynomials happened to have been invented first and be more mathematically meaningful by virtue of being an Appel sequence,
$$
\partial_x H_n(x) = 2nH_{n-1}(x),
$$
since
$$
H_n(x) = 2^n ~e^{-\partial_x^2 /4} ~ x^n ~.
$$
| {
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How unique is the length scale picked out by intelligent life? Our human bodies pick out a length scale (let’s say 1m). How unique is this scale and why did it arise?
In other words, how much smaller could humans, or multicellular lifeforms in general, be while sticking with approximately the same architecture of life? I know that the weight to muscle ratio is one argument against scaling things up massively, but what about smaller? Would such arguments still be available we just lower the gravitational field strength? If so, might there be intelligent aliens kilometers big?
The question is vague, I’m not looking for one particular answer.
| This answer is extremely approximative:
The scale of life should be somewhere between the smallest size physically meaningfull: the Planck lenght $\ell_{\textsf{P}} \approx 1,6 \times 10^{-35}~\mathrm{m}$, and the largest size set by causality in our universe: the Hubble lenght $\ell_{\textsf{H}} \approx 2 \times 10^{26}~\mathrm{m}$. The most likely scale of life should be around the geometric average of these extreme scales:
\begin{equation}
\ell_{\text{Life}} \sim \sqrt{\ell_{\textsf{P}} \, \ell_{\textsf{H}}} \approx 6 \times 10^{-5}~\mathrm{m} = 0.06~\mathrm{mm}.
\end{equation}
This is indeed about the size of a human cell.
| {
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Doubt in understanding the potential energy of dipole in external electric field? When a dipole is placed in external electric field it experiences a torque $$\vec \tau = \vec p \ \times \vec E $$
whose magnitude is
$$||\vec \tau|| = ||\vec p|| \cdot ||\vec E|| \cdot sin\theta$$
On calculating the potential energy:
$$W=-\Delta U$$
$$\therefore W = -\int pEsin\theta\cdot d\theta $$
Which gives:
$$W = pEcos\theta$$
But its wrong and is given as $-pEcos\theta$ in many text books.
Please Explain.
| Let me start from the beginning
Wnet = $\Delta$K.E
Wconservative+Wnon-conser.+Wexternal = $\Delta$K.E
Wnon-conser. = Zero [No non-conservative force is acting ]
$\Delta$K.E = zero [Assumption here is that the dipole is rotated very slowly therefore change amounts to zero]
Therefore the equation reduces to
-Wconservative = Wexternal
-d(Wconservative) = $\Gamma$*d$\theta$
-$\int$d(Wconservative) = $\int$ $\Gamma$*d$\theta$
-Wconservative = $\int$ p*E sin$\theta$ d$\theta$
-Wconservative = p*E $\int$ sin$\theta$ d$\theta$
-Wconservative = p*E(-cos$\theta$)
$\Delta$U = -p*E cos$\theta$ [ Work done by conservative force is negative of change in potential energy of system.]
It would make sense to integrate having definite limits then comparing the equations on both sides to understand what the potential of the dipole at and angle would be, I haven't done that in hopes that you can carry it from here(that is if you want it as such). Either way the result would be the same.
| {
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Expand superspace function into component form In 2D (1,1) superconformal field theory, the invariant "distance" between two points $Z_1=(z_1,\theta_1)$ and $Z_1=(z_1,\theta_1)$ in superspace is $$Z_{12}=z_1-z_2-\theta_1\theta_2.$$
My question is how to compute the quantities when the $\theta$ appears in denominator or in square root, for example, how to expand $\frac{1}{Z_{12}}$ into component form?
And how to expand
\begin{equation}
\sqrt{1-\frac{Z_{12}Z_{34}}{Z_{13}Z_{24}}}
\end{equation}
into component form?
| TL;DR: It is defined by expanding in the finite Taylor series of the $\theta$s.
More details:
*
*Recall that a supernumber $z=z_B+z_S$ has a body $z_B\in\mathbb{C}$, which is a complex number; and a soul $z_S$, which belongs to the ideal generated by Grassmann-odd generators.
*For any analytic function $f$, the supernumber $f(z)$ is defined via its formal Taylor series around the body $z_B$.
In particular, for the reciprocal function $f(z)=z^{-1}$, this prescription only make sense if the body $z_B\neq 0$ is non-zero.
*Point 2 can be generalized, so that if supernumbers $z,w$ share the same body $z_B=w_B$, then $f(w)$ may be defined as a formal Taylor series around the other supernumber $z$.
In particular for a superfield $\Phi(\theta)=\phi_0+{\cal O}(\theta)$ and its lowest component field $\phi_0$, they share the the same body $\Phi(\theta)_B=\phi_{0B}$, so that $f(\Phi(\theta))$ may be defined as a Taylor series (which happens to be finite!) around $\phi_0$.
*This generalizes in a straightforward way to several superfields.
| {
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Is dark matter inside galaxies different from dark matter in intergalactic space? I just read a text about astronomy and when talking about dark matter the author says:
[...], the dark matter responsible for the orbits of the stars in the Milky Way is probably different from the dark matter responsible for the orbit of the Milky Way within the local super-cluster of galaxies.
Is this true? How would it be different? And why?
For context, this is the whole paragraph:
Since the 1930's astronomers have measured the orbits of galaxies in clusters of galaxies, clusters of galaxies in clusters of clusters, and so forth. They have found similar anomalies in the angular velocity of galaxies at these larger scales. Again the anomalous high angular velocities of the galaxies and clusters of galaxies may be explained by postulating a mysterious dark matter that fills the universe. It is doubtful that one can explain the anomalous angular velocities at different scales by the same type of dark matter. Thus, the dark matter responsible for the orbits of the stars in the Milky Way is probably different from the dark matter responsible for the orbit of the Milky Way within the local super-cluster of galaxies.
And the article can be found here:
https://mathblog.com/wp-content/uploads/2009/12/Keplers-New-Astronomy.pdf
| The standard model of cosmology (for now) is called Lambda-Cold-Dark-Matter. It has only one kind of dark matter, and it agrees well with the observational data.
Other types of dark matter, such as “warm” or “hot” rather than “cold”, have been considered, and some people have considered models in which more than one kind of dark matter exist. These more complicated models do not appear to be necessary to explain what we observe.
The precise nature of dark matter doesn’t matter in cosmology. What matters is the relationship between its energy density and its pressure, because in General Relativity both energy density and pressure cause gravity. What is important is how relativistic dark matter is, because that determines its pressure. Cold dark matter is non-relativistic and has negligible pressure. All of its gravity comes from its energy density.
| {
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Confusion about Ohm's law So does ohms law say if the resistance is increased the voltage will also increase but not the current? And in non ohmic conductors the current increases with the voltage even though the resistance is also increasing? (meaning it shouldn't but still is, defying the law thereby)
| Ohms law relates three variables.
If you change one of those, then the other two have to change in a coordinated way: if R goes up, the V has to go up, or I has to go down, or some combination.
How do you figure out what happens? The resistor has given you one relation, but that’s not enough. There needs to be some other constraint.
That constraint comes from the characteristics of what the resistor is connected to:
*
*if it’s connected to an ideal battery, the voltage can’t change, so the current will.
*if connected to an ideal current source, the current can’t change, so the voltage will
*and in others, the voltage and resistance of the connected circuit will determine the mix of V and I changes
| {
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Would an infinitely (or very long) diffraction grating produce a diffraction pattern? Take a typical science lab diffraction grating which is producing a pattern on a screen.
Let's consider the location of say the first maximum on the right side.
Let's draw straight lines from the slits to the first maximum and think in terms of ''rays'' travelling from the slits to the maximum on the screen.
These rays all converge at the maximum and although they are close to being parallel we know that they are not perfectly parallel (because they converge).
Therefore the differences between the distances that neighboring rays have to travel will not be exactly the same.
i.e. If one ray has to travel one lambda more than its neighboring ray, then the following ray will have to travel slightly more or less than its neighboring ray.
If there are many slits very close together I can see that if the variance is small compared to the wavelength then we can ignore these differences and assume that each ray travels exactly one wavelength more/less than its neighboring ray.
But if the grating is long enough, these differences will eventually build up until they are significant enough to cause destructive interference.
My understanding is that a very long grating will have destructive interference at all locations and will not produce a pattern.
Is this true ?
What are peoples thoughts on this ?
|
Would a very long diffraction grating produce a diffraction pattern?
Yes. You might need to set the screen further back to ensure that the far-field condition is still satisfied, but given that the screen is sufficiently far away (for the length of the grating), arbitrarily large gratings can be used.
(Alternatively, you can drop the idealized introductory-textbook configuration and use a realistic design, which includes things like a curved screen and focusing optics before the grating.)
Would an infinitely long diffraction grating produce a diffraction pattern?
This is equivalent to asking "can one get infinitely far away from an infinitely long grating?". This kind of ill-defined double limit is not worth considering.
| {
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Why does heating an atom make it emit certain wavelengths? We're going over quantum basics in chemistry right now and I'm very confused.
Electrons can only accept in discreet quanta to move up an energy level, right? And they reflect other forms of light that don't supply energy in their specific quanta, right? And flame is just infrared electromagnetic radiation, right?
Then when heated by flame, why do the electrons move up to a next energy level? Shouldn't they need a certain wavelength to move up - one that isn't provided by the flame, given the emission for the atom is in the visible light range?
| There's not really any such thing as heating an individual atom. When you heat a gas like in a candle flame, the heat is the random motion of all the different atoms. The randomness is what makes us call it heat. If there's only one atom, there's no way to say if its motion is random or not.
And they reflect other forms of light that don't supply energy in their specific quanta, right?
Or they just transmit it (act transparent to it).
Then when heated by flame, why do the electrons move up to a next energy level? Shouldn't they need a certain wavelength to move up - one that isn't provided by the flame, given the emission for the atom is in the visible light range?
The atoms are being excited by collisions with other atoms. They're not being excited by absorbing light.
| {
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Measuring mass density of electrons in the universe Pardon me if the question is too naive, but only baryon and dark matter masses are considered while making measurements of the mass density in the universe.
What I am interested to know is how small is the total mass of electrons, relative to the other two that it was safely ignored from calculations in the Standard Model of cosmology?
| If we assume the universe is electrically neutral (it might not be completely neutral, but it is neutral to a large degree) then there are roughly the same number of protons as there are electrons. Because the mass of protons (and neutrons) is about 1836 times that of electrons, it's safe to assume that the contribution from electrons is negligible.
| {
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Small amplitude approximation in waves on string In transverse waves on a string, we neglect the longitudinal displacement of the particles, the reason given in most books is that the slope and the amplitude of the string is very small.
Can someone please prove it mathematically that it's reasonable to neglect the longitudinal motion of the element of the string under these conditions?
|
( The angle $\phi$ is exaggerated in figure)
A very small displacement, d, would create a very small angle, ϕ, with the strings original position. The transverse displacement, T, would be proportional to d∗cos(ϕ) and the longitudinal displacement,L, would be proportional to d∗sin(ϕ). So as ϕ tends to zero the transverse displacement would tend to d and the longitudinal displacement would tend to zero.
| {
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Why does it seem everything I push moves at a constant velocity? I am aware that a constant force causes a constant acceleration but friction can counteract this. However, if I push something across a table, for example, it seems no matter how hard I push, the object travels at a constant velocity, even if I apply more force than the kinetic friction. The object seems to always travel at the same velocity as my hand, does this mean I am not actually applying a constant force?
|
The object seems to always travel at the same velocity as my hand,
does this mean I am not actually applying a constant force?
Yes it does mean you are not applying a constant force. The force is due to an interaction at the surfaces of your hand and the object, and that interaction depends on how closely in contact and compressed those two surfaces are. The compression depends on the very precise difference in position between your hand and the object.
Let's assume your hand and the object each have constant speeds. Of course the speeds will change later, but this assumption is realistic for thinking about a very short time period.
If the object is moving slower than your hand, then it is getting closer, and if touching both things compress. As this happens the force very rapidly increases.
If the object is moving slower, then the opposite happens, and the force very rapidly decreases.
The result of this is that the force rapidly stabilizes to whatever level is necessary to move the object at the same speed as your hand.
| {
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Confusion regarding the Fermi-Dirac distribution and related formulas I'm studying semiconductor physics and having a problem with some of the terms.
The definition we were given in class for the Fermi-Dirac distribution is:
$$f_{FD}(E)
= \frac{1}{1+e^\frac{E-E_f}{kT}}
$$
From this formula it appears that $E_f$ is a constant independent of temperature, otherwise, it would have been written as a function of $T$.
But then, there are the formulas for the intrinsic fermi levels:
$$E_i-E_V
= \frac{E_g}{2}+\frac{3}{4}kT\ln\left(\frac{m_p^*}{m_n^*}\right)
$$
$$E_f-E_i
= \frac{kT}{2}\ln\left(\frac{n}{p}\right)
= kT\ln\left(\frac{n_i}{p}\right)
= kT\ln\left(\frac{n}{n_i}\right)
$$
Here $E_f$ changes with temperature? Is that the same $E_f$ from the Fermi-Dirac distribution function?
| It depends whether you are keeping the number of particles fixed or not. Usually we write
$$
f(\mu, T,E)= \frac 1{1+\exp\{(E-\mu)/kT\}},
$$
where $\mu$ is the chemical potential. Then the total number of particles is
$$
N=\int dE g(E) f(\mu,T,E).
$$
Here $g(E)$ is the energy density of states.
If we want $N$ to stay fixed as we vary $T$, then $\mu$ will have to depend on $T$. It may be however that our device is set up (by connecting to a reservoir of particles) so that it is $\mu$ that remains fixed and then $N$ will depend on $T$. The Fermi energy $E_F$ is really the value of $\mu$ at $T=0$, but people are rather casual about distinguishing between $\mu$ and $E_F$.
| {
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If a particle and its antiparticle annihilate upon contact, how do they form bound states? I am reading about some old discoveries in particle physics and early collider experiments from Perkin's Introduction to High Energy Physics. However, I didn't get the answers to all my questions.
If two beams of electrons and positrons are collided head-on, the collision can produce various quark-antiquark meson states called quarkonium. For example, $e^-e^+$ annihilations have produced various $c\bar{c}$ and $b\bar{b}$ meson bound states came to be collectively known as charmonium and bottomonium respectively. The most stable charmonimum is $J/\psi$ and bottomonium is $\Upsilon$. Toponium states do not exist since top quarks decay too fast to form mesons. The also exists bound states of $e^+e^-$ and $\mu^+\mu^-$, respectively called positronium and muonium.
If particles and antiparticles annihilate each other how can there be a bound state of them in the first place?
| A bound state is possible for awhile because the particle and antiparticle are separated in space and not significantly “in contact”. You can think of them as orbiting each other; their kinetic energy and angular momentum keep them apart. But, just as in a hydrogen atom, they are really described by wavefunctions. Eventually they annihilate because their wave functions do overlap a bit and there is a small probability that the particle and antiparticle are at the same point at the same time. So the bound state has a probability to decay and has a finite average lifetime.
| {
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Does the wavelength of a particle depend on the relative motion of the particle and the observer? The de Broglie wave equation states:
$$\lambda = \frac{h}{p},$$
where $\lambda$ is the wavelength of the “particle”, $h$ is Plank's constant, and $p$ is the momentum of the particle.
Momentum is usually written $\,p=mv$, where $m$ is the mass and $v$ is the velocity of the particle. But presumably $v$ is the relative velocity between the observer and the particle.
So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?
Or, perhaps more accurately, when a particle is incoming to another particle, in as much as an interaction between the particles depends on their relative speed, or the energy of impact, it thus also has something to do with their relative wavelengths.
Is that a conclusion, or simply a restatement of the premise, using different words that mean the same thing?
| Your understanding is correct: the de Broeglie wavelength of a particle as measured by an observer depends on the relative motion of the observer and particle. It doesn't make a lot of sense to state that an interaction depends on the relative wavelengths of two particles, because an observer on either particle will perceive the wavelength of his particle as infinite.
However, an external observer watching the two particles will see each as having its own wavelength, and from the wavelengths (and masses) can deduce their momenta and kinetic energies as measured from any moving frame. The kinetic energy in the rest frame of the center of mass of the two particles will be the same as calculated from any other frame.
Note that the above is not quantum mechanically precise.
| {
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Equilibrium in a spoon hanging from a toothpick While searching about equilibrium I came to this YouTube video which was quite astonishing but the video maker didn't explained the physics behind it. My query is that why the spoons aren't falling? The obvious answer that someone might give would be that they are in equilibrium, but why are they in equilibrium? I mean the mass is decreasing on one side, shouldn't it be unstable then? Also if the answers explaining this phenomenon in terms of forces will be much appreciated.
| The short answer is the location of the toothpick on the rim of the glass happens to be the location of the center of mass of the system which, in turn, happens to not be located at a point on the system itself. Since it’s the location of the center of mass, balance is achieved.
Hope this helps
| {
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Velocity not affecting heat produced by impact
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 $\text{m s}^{–1}$. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
The answer given in the book I am working from is that heat produced due to impact will be same in both the cases (when the elevator is moving with uniform velocity and when it is stationery) but I am getting different answers.
Why is the amount of heat produced the same irregardless of velocity? I don't follow the reasoning.
| Since all frames of reference are equally valid, in a uniformly descending, non accelerating elevator, all physical occurrences would be indistinguishable from an elevator considered at rest. So in both cases the bolt would release the same energy on impact with the floor. If you were in the moving elevator, you would see the bolt start it's fall at 0 velocity, fall 3 m at about 9.8 meters per second squared, and hit the floor. You would see the same in an elevator considered at rest. An external observer on the ground would see the bolt fall for the same amount of time in both cases, since no relativistic speeds are involved.
| {
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Heisenberg vs. Bandwidth Principle Is it true that the Heisenberg Uncertainty Principle is "obvious" because of the Bandwidth Theorem (Fourier Transforms)?
| 'Obvious' is a subjective term- my wife might not agree with my obvious excellent taste in ties.
However, the Heisenberg uncertainty principle is easy to understand if you are familiar with the properties of Fourier transforms. To localise a wave packet more tightly you need an increasing spread of frequency components, so there will always be a trade-off between being able to specify a location and being able to specify a frequency- it is strictly impossible to specify them both exactly for a given wave packet.
| {
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Derivatives of polar coordiantes? I'm a undergraduate and I was reading about the polar coordinate system specifically this paper. I don't understand the term: $$\frac{de_r}{d\theta} = e_\theta \text{, and } \frac{de_\theta}{d\theta} = -e_r$$
I don't see how you can have a the derivative of $de_r$ over $d\theta$ since they are not the same variable, if that makes any sence.
| $\hat{e}_r$ is a radial unit vector. It obviously points in different directions for different values of $\theta$. So it has a derivative expressing that rate of change.
| {
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What does the magnitude of centripetal acceleration actually represents? A fan moving with an angular velocity of 5π rad/second has a centripetal acceleration of 175.185 m/s^2.The Centripetal acceleration doesnt express the change in velocity as it is constant. Then what does 175.185 m/s^2 actually represents?
| In addition to the previous answers:
The precise value, $175.185 \text{ m/s} ^2$ is the linear acceleration of any point on the fan blade exactly $0.71$ metres from the axis of rotation, if any such point exists.
The formula for centripetal acceleration is:$$a=\omega ^2r=\frac{v^2}{r}$$where
$a$ = the linear acceleration. in $\text{ metre/second} ^2$, of a point moving in a circle;
$\omega$ = the angular velocity in radians per second;
$v$ = the linear velocity of the point, in metres/second;
$r$ = the radius of the circle the point is moving in
Since $a$ and $\omega$ are given, the value of $r$ can be calculated.
If we assume a $1$ gram mass ($.001$ kg) is attached to the fan blade at this particular radius, Newton's Second Law requires a force be applied to the mass:$$F = ma =0.1758 \text{ Newtons}$$
If the blade is not strong enough to supply this tension force, the blade will fail and the mass will fly off in a straight line.
| {
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Equations of motion in curved spacetime I'm trying to understand how to use covariant actions to derive equations of motion. A simple example would be the free scalar field
$$
S = \int\;d^4x\; \sqrt{-g} \left( -\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2 \right)
$$
Now from classical mechanics we know that the equations of motion pop out when we set the variation of the action to zero. How would we do that in this formulation?
The $\nabla_\mu$ s represent covariant derivatives!
What I have tried:
Using the $\partial_\mu \rightarrow\nabla_\mu $ perscription the lagrangian of the free scalar field will be $$\mathcal{L} = \left( -\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2 \right)$$
Hence the equations of motion will come from the modified E-L equation:
$$
\frac{\partial\mathcal{L}}{\partial\phi}=\nabla_\mu\frac{\partial\mathcal{L}}{\partial(\nabla_\mu\phi)}
$$
My question is basically how do we arrive to this equation from varying the action?
| From Ref. 1, one sees that the Euler-Lagrange equations generalize to $$\frac{\partial \mathcal{L}}{\partial\phi}=\nabla_\mu\left(\frac{\partial\mathcal{L}}{\partial(\nabla_\mu\phi)}\right)$$
In general, when one does these things, one must make sure of two things: that the index placement is the same, and that the indexes are NOT the same in the derivative and the Lagrangian. Use the metric to raise and lower indexes as needed, and a simple relabeling will take care of the the second. Thus, I change your Lagrangian as such: $$\mathcal{L}=-\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2\to -\frac{1}{2}\nabla_\alpha\phi\,g^{\alpha\beta}\nabla_\beta\phi-\frac{1}{2}m^2\phi^2$$ We also note that for an object with one index, $\frac{\partial V_\alpha}{\partial V_\beta}=\delta^\beta_\alpha$.
Thus, your scalar Lagrangian becomes
$$\begin{align}
-m^2\phi&=-\frac{1}{2}\nabla_\mu\left(\frac{\partial}{\partial(\nabla_\mu\phi)}\left(\nabla_\alpha\phi\,g^{\alpha\beta}\nabla_\beta\phi\right)\right)\\
& =-\frac{1}{2}\nabla_\mu\left(g^{\alpha\beta}\delta^\mu_\alpha\nabla_\beta\phi+g^{\alpha\beta}\nabla_\alpha\phi\delta_\beta^\mu\right) \\
&=-\frac{1}{2}\nabla_\mu\left(\nabla^\mu\phi+\nabla^\mu\phi\right)
\end{align}$$
So, finally, we have $$\nabla^\mu\nabla_\mu\phi-m^2\phi=0$$
which is just the flat spacetime version generalized by the usual prescription.
Ref. 1: Explicit gauge covariant Euler–Lagrange equation
| {
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Double slit experiment with electron one at a time I understand how a wave create the interference pattern, but what is the mechanism for a single electron after it pass the slit and scatter and land at different location on the screen to produce the same interference pattern.
Does the electron pass through one slit or it split into two before the slit and pass through both the slit?
Thanks.
| All matter has both wave and particle properties, for small particles it is possible to observe the wave properties. For massive particles it is impossible to observe the wave properties, the theoretical wavelength is too short. The reason particles have wave properties is because all matter interacts with each other thru the EM field, i.e when a particle is colliding with another the electrons are the first to "see" or "sense" each other (there is something called virtual photons). Because the interaction is in the EM it has to have a wavelength/wavefunction associated with it. The term interference is historical, the slit experiment for photons showed a pattern similar to what water waves do. Even single photons "interfere" but it is better to think of an allowed path explanation (Feynman) where the brights spots occur. The electron will choose one slit at random, with this new path it will look to interact with the screen at a point that works with its wave function, this causes the INTENSITY pattern (note: I did not want to use the word "interference"!):
| {
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WKB treatment for unstable particle I was wondering if the WKB treatment of particles entering a potential (there is some reflection (R) and transmission (T) coefficient such that R+T =1) works only for stable particles. Essentially the WKB gives me a wave function of a state and the probability it will be reflected/transmitted across a potential.
What is I have an unstable intermediate particle, do I always need to use the stable final states?
Thanks!
| This is not a complete answer, but is too long for a comment.
Interesting question. I guess spontaneous fission would be an example, since it takes place by tunneling through a barrier. (The coordinate is some measure of the shape of the nucleus, leading to a saddle point and then scission.) People certainly do use WKB concepts to discuss fission, and the products are often unstable nuclei. So I think the answer is that it's at least sometimes OK.
My gut feeling would be that you'd run into problems if the half-life of the product was too short. The fission example would seem to indicate that it can be OK even if the half-life of the product is short compared to the rate of transmission. Perhaps there would be a problem if the half-life of the product was short compared to one over the frequency of assaults on the barrier (although such a thing is not actually very well-defined in examples like fission and alpha decay, since preformation of the product would violate the exclusion principle).
| {
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Why gapped systems are called incompressible? I study quantum Hall systems and I haven't studied Fermi liquid theory yet. But I understand the concept of having gap or being gapless. But why do we use the term incompressibility to correspond the presence of gap in the bulk state? Is there any relation to physical compressibility here?
| The electronic compressibility $\kappa$ is defined as
$$\kappa =\frac{\partial \rho}{\partial \mu} $$
where $\rho$ is the electron density, and $\mu$ the chemical potential. The region where the $\rho(\mu)$ is constant indeed indicates that there is an energy gap. This can simply be understood by the fact that in the energy gap there is a region where there are no electronic states that can contribute to the electronic density $\rho$, and thus it remains constant. The region of $\mu$ values over which $\rho$ remains constant determines the energy gap.
The criterion often used for insulating behavior is that there is a certain region where $\kappa$ vanishes. However, keep in mind that this criterion for insulating behavior applies to standard insulators, as well as electron-driven insulators (e.g. Mott insulator). Nonetheless, there are insulators where $\kappa$ does not vanishes, these are disorder-induced Anderson insulator. So keep in mind that this is not a universal characterization for insulating behavior!
| {
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How do I experimentally measure the surface area of a rock? I hope this is the right place to ask this question.
Suppose I found a small irregular shaped rock, and I wish to find the surface area of the rock experimentally. Unlike for volume, where I can simply use Archimedes principle, I cannot think of a way to find the surface area. I would prefer an accuracy to at least one hundredth of the stone size.
How can I find the surface area experimentally?
| If you have access to a planimeter, then you might try the method used in this research paper on the strength of cements used on teeth.
In order to compare the strength of the cement, the authors needed to separate the effect due to the cement from effect due to the varying surface areas of the real teeth used in the tests.
For each tooth used, the authors laid aluminum foil over the teeth and used a burnishing tool to make the foil follow the contour of the surface of each tooth. Overlapping areas were then cut away and the foil was removed from the the tooth then pressed flat. A tracing of the outline of each piece of foil was made, and the area measured using a planimeter.
I happen to have bought a planimeter of the exact same model as used in the referenced paper, and actually found that paper while searching the internet for information about the planimeter I had just bought at a fleamarket.
| {
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Velocity in SHM
By book defines SHM as
Simple harmonic motion is defined as the projection on any diameter of a graph point moving in a circle with uniform speed.
but in the next line it says -
Moving back and forth along the line AB, the mass point is continually changing speed vx' Starting from rest at the end points A or B, the speed increases until it reaches C.
Can somebody explain - If the original particle is moving with a constant speed then how come it's mass point have different velocity ?
Also for a simple harmonic motion resorting force should be proportional to displacement (towards mean position) .In this diagram i see no force acting on mass point?
| SHM is governed by the following equation:- $$\ddot x+\omega^2 x=0$$
Here $\omega$ is a positive constant which is also known as angular frequency of simple harmonic motion and $x$ is the displacement from the mean position.
The diagram given in the post gives an analogue of simple harmonic motion which simplifies the problems we deal with but the above definition clearly convinces us that the net force is always directed towards the equilibrium position and is proportional to its distance from the mean position.
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Is nuclear force a kind of strong interaction? I'm trying to understand the role of Yukawa potential, and it seems to describe the nuclear force. But at the top of the article it says:
This article is about the force that holds nucleons together in a nucleus. For the force that holds quarks together in a nucleon, see Strong interaction.
Not to be confused with weak nuclear force.
So it seems that it's neither strong interaction nor weak interaction? So what is it? Or is it actually electromagnetic interaction in the case of nuclear? I have a feeling that strong interaction is more likely to be correct.
| Behind the scenes, it is the strong force. But on the nuclear scale, it looks quite different, and is mediated by pions (composed of two quarks) rather than gluons. Sometimes the resulting force is called the residual strong force.
| {
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Why don't people just put a water sprinkler on their roof for cooling? Some air conditioners work with evaporative processes. So I'm not sure why people wouldn't just use a sprinkler that turns on for a minute every hour or half hour or whatever in order to wet the roof. The water will not only absorb the heat from the sun, but when it evaporates won't it also take away the heat from the house, thus cooling it down.
I'm not suggesting it's going to be as good as an air conditioner since it's outside, and only cooling the roof. But surely it will use very little power and water for a significant reduction in heat in the house, wouldn't it? Are there factors I'm missing?
| This is a waste of water and will catastrophically exacerbate the effects of heat and drought when applied at large scale. Paint your roof white, insulate your house (roof, walls, floor, double glazing) and use solar power for your airco is the way to go.
| {
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Is there any relationship between S-matrix elements and the path integral? Reading Peskin&Schroeder I've made the following curious observation: Comparing S-matrix elements to the definition of the path-integral they look remarkably similar:
$$_{out}\langle \mathbf{p}_1 \mathbf{p}_1| \mathbf{k}_A \mathbf{k}_B\rangle_{in}= \lim_{T\rightarrow \infty} \langle \mathbf{p}_1 \mathbf{p}_1| e^{-iH(2T)}|\mathbf{k}_A \mathbf{k}_B\rangle=\langle \mathbf{p}_1 \mathbf{p}_1| S|\mathbf{k}_A \mathbf{k}_B\rangle \tag{4.70 +4.71}$$
compared to:
$$\lim_{T\rightarrow \infty} \langle \phi_b(x)| e^{-iH(2T)}|\phi_a(x)\rangle = \int {\cal D}\phi \exp\left[i\int_{-T}^{T} d^4x \cal{L}\right] \tag{9.14}\equiv Z$$
They are just the time evolution operator sandwiched between appropriate quantum states. I even guess that multi particle states like $|\mathbf{k}_A\mathbf{k}_B\rangle$ can be developed in field states $|\phi_a(x)\rangle$ with some a-priori unknown coefficients. So the transformation from one to the other does not look easy at all or is even impossible. But nevertheless, as the definitions look so similar, I ask the question: Are S-matrix elements related to the path integral or is it just a silly question, i.e. my observation is an accidental coincidence ?
| Yes, you can write down S-matrix elements directly in terms of path integrals. This was figured out by L. Fadeev and is explained in his 1975 Les Houches lecture notes. A review of his work is also in Bailin and Love's gauge field theory textbook.
References:
Ludwig Faddeev, Introduction to Functional Methods, p. 1-39 in Methods in Field Theory, North Holland, 1976 (and references therein).
| {
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Mass Dimension 6 QED-Lagrangian Consider the QED Lagrangian
$$\mathcal{L}_{\text{QED}}=-\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \bar{\psi}(i D_{\mu} \gamma^\mu -m) \psi.$$
I need to extend the Lagrangian up to mass dimension 6, of course respecting all the symmetries/invariances of the theory.
My professor told me, that one can ignore pseudo-scalar terms such as $\bar{\psi}\gamma_5\psi$, since the theory has to be parity invariant.
But what about the product of the two pseudoscalars $$\Delta \mathcal{L} = \bar{\psi}\gamma_5\psi \bar{\psi}\gamma_5\psi.$$ Why can this not be a term in my Lagrangian? Is there a problem with one of the invariances?
| We are talking about chiral invariance, right?
First of all, the mass term
$$
m\bar{\psi} \psi
$$
breaks the chiral symmetry. So if your professor demands chiral invariance, then we are dealing with massless QED.
For massless QED, you can add a chiral symmetric mass dimension 6 term like (NJL 4-fermion interaction)
$$
\Delta \mathcal{L} = g (\bar{\psi}\psi \bar{\psi}\psi - \bar{\psi}\gamma_5\psi \bar{\psi}\gamma_5\psi).
$$
Note that
*
*The individual pseudoscalar-pseudoscalar-interaction term (second
term) is not chiral symmetric (no problem with local gauge $U(1)$ invariance and Lorentz invariance though). However, the aggregation of the scalar and pseudoscalar terms does respect the chiral symmetry.
*The mass dimension 6 4-fermion interactions are non-renormalizable. Hence a specific regularization regime is part and parcel of the model.
On the other hand, if you forgo chiral symmetry, then a "complex" mass term is perfectly legit:
$$
m\bar{\psi} e^{\theta i\gamma_5} \psi = m\cos\theta \bar{\psi} \psi + m\sin\theta \bar{\psi} i\gamma_5\psi.
$$
See details here: Why is the Higgs $CP$ even?
Since you are considering mass dimension 6 terms, to be complete, don't miss out on mass dimension 5 terms like
$$
i\bar{\psi}\gamma^\mu \gamma^\nu F_{\mu\nu} \psi,
$$
and mass dimension 6 terms like
$$
i\bar{\psi}\gamma^\mu \gamma^\nu \gamma^\rho F_{\mu\nu} D_\rho\psi.
$$
| {
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How and why Liquid helium climbs the walls of the container it is kept in? Liquid helium (He-II) shows a strange phenomenon, where it flows on its own, forming films across the surfaces of the container it is kept in. How and why this happens and how is it possible?
| Liquid Helium as He-IV and He-III behave in this manner. The property of a material to have zero viscosity is known as superfluidity. This is only possible at cryogenic temperatures at the nano-kelvin level.
You may know viscosity as a measure of the 'thickness' of the fluid or a measure of how easily it can flow. When the viscosity drops to zero, there are no shear forces within the liquid and the liquid will try to spread out as much as possible. This will continue till the till the film becomes one molecule thick. The specifics of why this happens is explained by the Bose-Einstein Condensate statistics.
| {
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Local phase invariance of our lagrangian So in lectures we have just started looking at classical field theory. We were introduced to symmetries and we were told in general our lagrangian wont be invariance wrt to local phase changes but if we require it to be we can introduce the covariant derivative. This introduces a gauge field that acts to restore local phase invariance.
My question is what is the physical motivation for wanting our lagrangian to be invariant wrt to local phase changes? What are some cases where we want local $U(1)$ invariance (& why) and some cases where we dont (again & why)?
| If you use Noether's theorem, there is a locally conserved current associated with local $U(1)$ symmetry. We can then identify this current as electric current. Without the symmetry, we wouldn't have a locally conserved current to use for electric current, so it's pretty important. As an additional note, we can find Lagrangians that have $SU(2)$ or $SU(3)$ local gauge invariance. When we do this, we find that the Lagrangian for $SU(3)$ describes the behavior of the strong interaction between quarks.
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Help understanding unit: Micromoles per square meter per second The context is light, illumnination, photons.
The units seem to suggest something different from the definitions I have found:
$$\frac{µmol}{m^2 s}$$
This, to me, says I have one millionth of a mole ($6.022×10^{17}$) in photons landing on a one meter square area every second.
However, the definitions I have found pretty much state something similar to the above and then add that the quantity is divided by a mole ($6.022×10^{23}$).
Frankly, not sure what that means. Is it meant to represent what fraction of a micro-mole lands on a one square meter per second? I realize mol is unit-less, which might make it confusing.
In other words, you count the photons you have in one second, divide that by Avogadro's number (or is it Avogadro's number divided by 1E6?) and that's your number.
| 1 mol is defined as Avogadro's number $6.022\cdot10^{23}$ and you could count anything using moles. We could count some events happening in a fixed time, for example water molecules flowing through a pipe. If we divide the total count of those molecules by elapsed time we get rate of flow which in this case would measured in $\frac{mol}{s}$.
If we wanted to change the size of that pipe, rate of flow per area might be handy, which would thus be measured in $\frac{mol}{s\cdot m^2}$. This is just one example but there are many more. Also 1 micromole is $10^{-6}$ mol.
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A problem on thermal stress Here's the question:
There are 2 rods placed between rigid supports. Where $Y_{i}, \alpha_{i}$ and $A_{i}$ are the Young's Modulus, Coefficient of linear expansion and Cross sectional area of the rods. When the system is heated to temperature $\theta_{2} $ from $\theta_1$, find the relation between $A_1$ and $A_2$ such that there lengths remain constant.
Now I know that if the length does not change I could consider that a thermal stress of $Y\alpha\Delta\theta $ is developed in the rods. The problem is the area that I have to consider to equate the force. My teacher tells me that I have to consider this equation:
$$A_{1}Y_{1}\alpha_{1}\Delta\theta=A_{2}Y_{2}\alpha_{2}\Delta\theta$$But I don't understand this equation. According to me, it must be $A_2$ on both the sides instead of $A_1$ and $A_2$. As the stress developed in the rods could only be transferred through the common area (I think...). Could someone explain me what's wrong?
| The reason you take the complete area on $A_{1}$ is because its an approximation: no matter how locally the stress on its surface is being applied, the whole object feels the stress as if it were applied evenly across its cross-section. Otherwise there would be local deformation.
This isn't true in reality though.
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Why doesn't the block fall?
I came upon this question as I was going through the concepts of tension. Well according to Newton's third law- every action has an equal and opposite reaction. Here my question is that if the tension at point B balances the tension at point A then which force balances mg as it can't be balanced by the reaction force of mg which attracts the earth towards the mass as it is not in contact with the mass. Then why doesn't the mass fall?
| It is the normal force ( https://en.wikipedia.org/wiki/Normal_force ) of the Earth against the supports that hold the platform up. Normal force is what keeps our feet from sinking into the Earth, due to our weight, it keeps a book on a table from falling through the table. It will keep the supports from falling into the Earth, the supports hold the platform, which holds the string and mass.
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Creating Em waves Is it necessary for creation of em waves be from same source.
If a positive charge particle and negative charge particle is accelerating with same acceleration will the em waves created be identical in all aspects?
|
If a positive charge particle and negative charge particle is accelerating with same acceleration will the em waves created be identical in all aspects?
Just for the record, your statement is true only for a charge and its anti-particle. To show that this is the case, one has to imagine that the acceleration of an electron requires less energy than the acceleration of the heavier proton. So when the electron and the proton are accelerated at the same speed, they release different amounts of energy.
dominecf commented: “There is no term in Maxwell equations that takes into account the particle's properties other than charge.” But there is the photon spin, which for reasons of symmetry needs a closer look.
Maxwell equations take into account the empirical fact that the direction of propagation, the electric and the magnetic field are direction-dependent. If for example the thumb points into the direction of propagation and the second finger in the direction of the electric field at one moment, you have to take the right hand to point with the third finger into the direction of the magnetic field at that moment. This dependency is measurable for radio waves.
Radio waves are generated from accelerated electrons. Until now radio waves from protons or positrons are not generated nor measured. It is a speculation that these particles during acceleration emit left handed EM waves.
Just for the record 2. R.W. Bird wrote, “If you had identical positive and negative charges oscillating side by side, the two waves would be out of phase. They cancel each other and you have no wave from a net zero charge.” is misleading. Once emited, EM waves are not canceling out like water waves. For water, the energy gets dissipated into heat. For EM waves that is not the case. Observable is the undisturbed crossing of both Eam waves.
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Speed of light in different mediums with different frequencies As I know the speed of the light doesn't change while it travels through a vacuum. But while it travels through a prism, it shows different deviation angles to different frequencies.
So according to the $$ n = \frac {sin (A+D)/2}{sin (A/2)} $$ formula $n$ is different with the frequency of the light, when $n$ is different the velocity should be different through the medium.
Why does the velocity change with the frequency of the light while it travels through a medium but not in vacuum?
| Light’s velocity in a medium changes compared with its velocity in vacuum because the electrons in the atoms of the medium experience forces due to the light passing through, and, as they are accelerated, radiate light of their own which superposes with the incoming light. (Showing that this additional radiated light changes the phase velocity of the incoming light requires some math.)
The velocity shift is frequency-dependent because the electrons respond differently to different incoming light frequencies. They are basically driven harmonic oscillators. They have a natural frequency determined by the atomic or molecular physics of the medium, and when the driving frequency of the incoming light is close to their natural frequency they accelerate more easily. This is just like how a child on a swing moves more easily when you push the swing at the frequency it “wants” to swing at.
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What's 'force per second'? For example, if a force of 10 N per second (10 N/s) is applied to an object, does this have a name or a definition? I'm not referring to impulse - which is Ns.
An airplane's engine thrust is simply given as a force, but this must be a force applied by the engines each second (N/s)?
Thanks!
| It does not make sense to discuss a force with units of $\rm N/s$ since those are not the units of force. The physical quantity that has these units is called yank and refers to the derivative of force with respect to time.
An airplane's engine thrust is simply given as a force, but this must be a force applied by the engines each second (N/s)?
To answer this, I ask why you think the force should be applied every second, rather than some other time interval, such as every fortnight or Planck time? The problem here is that, in order for a finite force to have an effect, it must act over some non-zero duration. We see this in the equation for impulse.
$$\vec J = \int_{t_1}^{t_2}\vec F\ \mathrm{d}t$$
Since force is inherently related to a rate of change, by Newton's Second Law, you are naturally inclined to think that its units should have seconds in the denominator. However, the actually already do, since the newton is defined by
$$\rm N = kg\cdot m/s^2 = \frac{kg\cdot m/s}{s}$$
Note here that $\rm kg\cdot m/s$ is the unit of momentum and that net force is the time derivative of momentum.
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What vector field property means “is the curl of another vector field?” I'm an undergraduate mathematics educator and I teach a lot of multivariable calculus. I posed this question on MSE over four years ago and I haven't gotten any definitive answers (despite 12 upvotes and a bounty posted). It could be there's no answer, but someone suggested I ask on this forum.
I know that a vector field $\mathbf{F}$ is called irrotational if $\nabla \times \mathbf{F} = \mathbf{0}$ and conservative if there exists a function $g$ such that $\nabla g = \mathbf{F}$. Under suitable smoothness conditions on the component functions (so that Clairaut's theorem holds), conservative vector fields are irrotational, and under suitable topological conditions on the domain of $\mathbf{F}$, irrotational vector fields are conservative.
Moving up one degree, $\mathbf{F}$ is called incompressible if $\nabla \cdot \mathbf{F} = 0$. If there exists a vector field $\mathbf{G}$ such that $\mathbf{F} = \nabla \times \mathbf{G}$, then (again, under suitable smoothness conditions), $\mathbf{F}$ is incompressible. And again, under suitable topological conditions (the second cohomology group of the domain must be trivial), if $\mathbf{F}$ is incompressible, there exists a vector field $\mathbf{G}$ such that $\nabla \times\mathbf{G} = \mathbf{F}$.
It seems to me there ought to be a word to describe vector fields as shorthand for “is the curl of something” or “has a vector potential.” But a google search didn't turn anything up, and my colleagues couldn't think of a word either. Maybe I'm revealing the gap in my physics background. Does anybody know of such a word?
TL;DR: gradient is to conservative as curl is to ___?
| In the general case -- i.e in any number of dimensions, the analogue of $\nabla\times(\nabla \phi)=0$ and $\nabla\cdot(\nabla\times {\bf A})=0$ is $d^2=0$ where $d$ is the exterior derivative anding on $p$-forms. This means that if $\omega=d\eta$ then $d\omega=0$. A p-form $\omega$ such that $d\omega=0$ is said to be closed. If $\omega= d\eta$ then $\omega$ is said to be exact.
You mentioned cohomology, so probably you know what I have just written, and are instead asking what are "closed" and "exact" are called ordinary vector calculus. The answer is that I think that there is no standard name for this situation because most vector calculus is done in contractable spaces where closed $\Rightarrow$ exact. Certainly I have never seen a name.
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Is there some truth to the often told story that the running of couplings is the result of screening through virtual particles? It's a well established fact that coupling parameters changes with the energy scale at which we probe a given process:
A popular way to explain this phenomenon goes as follows. Particles are surrounded by clouds of virtual particles which screen (or anti-screen) the bare charge, analogous to what happens in a dielectric material.
As we probe the process at higher energies, we start to see more and more of these virtual electrons and thus the net charge that we observe becomes weaker.
Virtual particles are, of course, a controversial topic. But I was wondering if there is some truth to this story?
| To give some context for what I have in mind, here's the best answer I came up with myself. Corrections, comments and better answers would be much appreciated.
The proper mathematical context to discuss this question are the renormalization group equations. For example, in $\phi^4$-theory we have:
$$ \lambda_R(s_1) = \lambda_R(s_0) +C \ln \left( \frac{s_1 }{s_0}\right) \lambda_R^2(s_0) + \ldots $$
The idea is that we choose some reference scale $s_0$. By definition, the process is at this scale completely described by the simple single-vertex diagram with coupling $\lambda_R(s_0)$ at the vertex. But as soon as we probe the process at a different scale $s_1$, we must take corrections into account which are described by the renormalization group equation. Formulated differently, if we consider a perturbative expansion in the renormalized coupling at reference scale $s_0$, $\lambda_R(s_0)$, we only describe the process completely by the simple single vertex diagram at scale $s_0$. At any other scale $s_1$, we must take corrections due to additional diagrams into account. In this sense, the effects of virtual particles (which correspond to lines in the loop) become important at higher energy scales.
In particular, higher order correction become more and more important as we move farther away from $s_0$. This is a result of the logarithmic dependence on $s_1/s_0$. In this sense, the effects of virtual particles become even more important at higher energy scales. And, as a result, the charge becomes, at least in this case, more anti-screened.
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Dependence Of A Cross-Section I am trying to understand the dependence of a differential cross-section on $$\sigma(\theta) = \left(\frac{s}{ \sin \theta}\right)\left|\frac{ds}{d\theta}\right|,$$ where $s$ is the impact parameter and $\theta$ is an angle that is between the scattered and incident directions.
Any explanation as to what is going on would be a big help.
| I can't give a complete answer but I'm sure someone will. For a low energy projectile, lower than the first excited state of the target, where the projectile is neutrons, this is just the elastic or inelastic scattering cross-section. The neutron scatters like a billiard ball off the stationary nucleus. Applying conservation of momentum and conservation of energy you can calculate the cross-section. As the energy increases more channels open. You can excite the nucleus into a higher energy level which will decrease the energy of the outgoing neutron. Other scattering mechanisms become possible with increase in energy.
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Is the relation between change in potential energy and work by internal conservative force can be used even in presence of non conservative forces? We know that work done by internal conservative forces is the negative of change in potential energy of the system stored in conservative force field. But does this logic still hold when there are non-conservative forces like friction or resistance?
Do the non-conservative forces only withdraw from the kinetic energy part and not affect the potential energy in any way? Consider for an example a system of two charges having some mass kept at a finite distance and both are free to move over a rough surface and released.
| Non-conservative forces change the total mechanical energy of the system, since
$$W_\text{nc}=\Delta E=\Delta K+\Delta U$$
assuming all conservative forces are internal to the system.
However, nothing from this tells us how the kinetic and potential energies change. More information is needed. For example, with a mass sliding on a horizontal surface with friction, the non-conservative work only changes the kinetic energy as the mass slows down. However, if you set up the mass to slide down an incline with friction at a constant speed then only the potential energy is changing. In each case we have the same non-conservative force, but the changes in kinetic and potential energies are different.
Of course one could argue that in the incline case that the potential energy is being converted into kinetic energy that is instantly removed from the system by friction, but at that point it's just a difference in interpretation that yields the same result.
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Because things smell, is everything evaporating? Everything, in theory, can have a smell, but that is not the whole point of this question.
My main query is, since things do smell, does that mean that everything is slowly evaporating (or, sublimating, I suppose)?
For example, if we perceive metal to have a scent, this must be because some of the molecules of metal are in the air or are at least affecting the gasses between the metal and our noses.
What is happening in terms of physics that causes almost anything to be perceivable by a (biological or otherwise) molecule receptor?
(There are a few similar questions closed as off-topic, but I would like to argue that this is a valid physics question.)
| Technically, everything IS evaporating, it's just a question if our olfactory abilities are advanced enough to detect it.
But the metal smell coming off metal is actually the result of a reductive chemical reaction between skin lipid peroxides and the metal itself. These usually produce an array of molecules that do evaporate readily (unlike metal), and give each metal a distinctive smell, but the main ingredient of this smell is Oct-1-en-3-one.
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How does a time varying magnetic field confined in a cylindrical region produces induced electric field even outside the cylindrical region? We know from Faraday's Law of Electromagnetic Induction that due to Time Varying Magnetic Field (TVMF), a non conservative electric field will be induced.
Now if we consider a cylindrical region in which the magnetic field is varying with time then outside the region of cylinder there is no magnetic field and no variation of it either, so how does the electric field gets induced there?
I understand that in the interiority of the cylindrical region, the electric field will be induced as there is TVMF, but how it gets induced outside it?
Isn't it true that Induced electric field should only generate at the location of time variation of magnetic field?
Isn't it strange TVMF can produce Induced Electric Field even at a distant location?
Is it a law of nature kind of thing that TVMF can produce induce electric field at a distant location?
| Outside the solenoid, I assume this is the system that you have mind, is a vector potential. It falls off as 1/r and is directed tangentially. When the current through the solenoid is varied, dA/dt acts as an electric field.
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QED electron self-energy in 1PI effective action The electron self-energy at one-loop is given by the one-particle irreducible graph
I know how to calculate it using the Feynman rules but I was wondering how this diagram appears in the QED effective action (which it in principal should as a 1PI graph).
With
\begin{equation} \Gamma = S + i \,{\rm Str}(\ln(S^{(2)})) \end{equation} I do not see how I get a loop with two different particles in it.
So how does this diagram appear in the QED effective action?
| *
*The 1-loop quantum correction to the 1PI effective action is given by a functional superdeterminant$^1$
$$ \begin{align}\Gamma_{\text{1-loop}}[\phi_{\rm cl}]
~=~&\frac{i\hbar}{2} \ln{\rm sDet}\left(S^{-1}_2 \frac{\delta^2 S[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)\cr
~=~&\frac{i\hbar}{2} {\rm sTr}\ln\left(S^{-1}_2 \frac{\delta^2 S[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)\cr
~=~&\frac{i\hbar}{2} {\rm sTr}\ln\left({\bf 1}+S^{-1}_2 \frac{\delta^2 S_{\neq 2}[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)\cr
~=~&-\frac{i\hbar}{2}\sum_{n\in\mathbb{N}}\frac{1}{n} {\rm sTr}\left(-S^{-1}_2 \frac{\delta^2 S_{\neq 2}[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)^n,\end{align}$$
cf. e.g. my Phys.SE answer here.
*OP's amputated 1-loop 1PI QED self-energy diagram with 1 fermion propagator and 1 gauge field propagator is included in the term with $n=2$ containing 2 propagators $G^{k\ell}_0=-(S^{-1})^{k\ell}$. The 2 propagators do not need to correspond to the same field.
--
$^1$For notation, see my Phys.SE answer here.
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How is it possible that entropy of the universe has increased since big bang but temperature of the universe has decreased? How is it possible that entropy of the universe has increased since the big bang, but the temperature of the universe has decreased?
I know that the increasing temperature of the system tends to increase in the entropy of the system thing tends to go higher entropic states.
Here System = Universe.
| The universe is expanding, so its volume is constantly increasing, stretching out radiation wavelength and cooling down and decreasing the density of matter, which increases entropy, also entropy always increases in an isolated system. Entropy increasing/decreasing with temperature is only valid when kept in the same volume.
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Why is electric field Fourier transformed to produce image in radio astronomy? In radio interferometry, to get an image, the correlation of electric field observed in different antennas are Fourier transformed. This gives the "brightness function in sky coordinates".
Why do we need to Fourier transform? Does not the electric field in a direction directly give information about how the object looks?
I am looking for an explanation that does not involve much mathematics.
| Think about using a photo camera. If you simply take the photo sensor without any lens in front of it you are not getting an image on your sensor. You need to put the objective in front of the sensor to form an image. The electro-magnetic wave in the lens plane and the sensor plane are directly connected to each other by a Fourier transform. A thin lens projects the Fourier transform of the wavefield into its focus plane.
With radio waves it will be quite hard for you to build a working lens in front of your antenna array. But you have a distinct advantage over visible light, i.e. the waves are of long wavelength and slow frequency - therefore you can detect the phase of the waves, a thing that is impossible with visible light. In the consequence you can measure the full information about the radio waves and perform the function of the objective lens in your computer to obtain the image.
So in imaging you always need to perform a Fourier transform, depending on the problem it is either done optically (with a lens) or digitally on a computer.
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Meaning of "harmonic" I'm trying to understand the meaning of the term "harmonic". IE, appearing in following sentence of Fluctuation-dissipation relations for stochastic gradient descent
The second relation (FDR2) further helps us determine the properties
of the loss function landscape, including the strength of its Hessian
and the degree of anharmonicity, i.e., the deviation from the
idealized harmonic limit of a quadratic loss surface and a constant
noise matrix.
What is the meaning of "harmonic limit" here?
Also, is it appropriate to term "harmonic approximation" to refer to a method which uses Gaussian-like approximation? (ie, assumes that cumulants of rank-3 and above are zero)
| If you make the equivalence between the loss surface (or loss function) and a physical potential then the "harmonic limit" here is the one of a Brownian evolution into a harmonic (or quadratic) potential. That means the stochastic gradient views as a stochastic process is the same than the evolution of a particle driven by a thermal noise into a quadratic potential.
As the solution of the Fokker-Planck equation associated with such evolution are actually Gaussian, it could be thought as a Gaussian-like approximation. But the approximation here is really on the loss function.
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Constant $g$ acceleration from astronaut's frame of reference When a spaceship is experiencing a constant acceleration of $10m/s^2$, the astronauts will be moving at nearly the speed of light after about a year in the earth's reference frame. This means the spaceship's energy will start to diverge as a function of the speed $v$ so there will be a huge amount of energy necessary to increase the speed of the ship any further. This way, the speed of light can never be crossed.
All of this is clear to me, but all of this is also formulated in earth's reference frame. But from the astronaut's reference frame: the spaceship is simply accelerating at $10m/s^2$ and so the force on the spaceship is constant. Then why would we need huge amounts of energy to accelerate the spaceship?
For example, I read somewhere that the amount of energy that would be needed to accelerate a large spaceship to half the speed of light is more than 2000 times the current world annual energy consumption. How does this make sense in the astronaut's (non-inertial) frame?
| From the rocket's frame of reference, the rocket is at rest and the Earth is travelling faster and faster, approaching c. In both frames of reference, the relative velocity approaches c so the energy needed diverges.
I'm not sure whether this answers your question.
| {
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Can the forces change with frame of reference? Consider a ball kept on man's head (mass $M$) on the Earth. Now supposing I throw the ball from height $h$ of tall building then why does he gets more hurt? Isn't the force still mg?
I would like to know what happens in ideal case (no air resistance)
and then in real case (with air resistance)
| The force is still $mg$, but note - it is the force that is applied on the ball, not on the man's head!
As the ball falls from above, it picks up speed due to its constant acceleration $g$, which respectively increases its momentum, $p = mv$. Bigger momentum means a bigger force, so that's why the poor person's head suffers more when you drop the ball from a certain height, than when you drop it from a height that's substantially smaller.
P.S - this is true for a scenario with air resistance, and for a scenario without air resistance.
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Why in mercury barometer pressure inside the glass tube at a point is same as the pressure outside the glass tube at the same height? I am quite familiar with the pascal’s law but i still think that being on the same points still they should have different pressures as above on one point is atmosphere and on the other is only the column of mercury can anyone give physical explanation for this and also if the pressure inside the torricelli vacuum is only due to mercury vapours which is quite low than why the vacuum does not get crushed under the atmospheric pressure?
| The pressure inside the tube at height is not equal to the pressure outside. It is a vacuum, or partial vacuum created by the weight of the column of mercury in the tube. Higher atmospheric pressure at the open pool of mercury at the base of the tube will push the column of mercury higher in the tube. Lower atmospheric pressure at the base pool will let the pool level rise, making the column of mercury drop. So the pressures equal at the surface of the mercury pool at the bottom, not at height inside the tube. see https://en.wikipedia.org/wiki/Barometer#/media/File:MercuryBarometer.svg
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Solving a differential equation using factorisation In Griffiths Introduction to Quantum Mechanics, when discussing ladder operators in Chapter 2, he write Schrodinger's equation as,
$$
\frac{1}{2m}\left[\left(\frac{\hbar}{\mathrm i}\,\frac{\mathrm d}{\mathrm dx}\right)^2+\left(m\omega x\right)^2\right]\psi=E\psi\tag{2.40}
$$
Then he says,
The idea is to factor the term in the square brackets. If these were numbers, it would be easy:
$$u^2+v^2=(u-iv)(u+iv).$$
Here, however, it's not quite so simple, because $u$ and $v$ are operators, and operators do not, in general, commute. Still this does invite us to look at the expressions,
$$
a_\pm=\frac{1}{\sqrt{2m}}\left(\frac{\hbar}{\mathrm i}\frac{\mathrm d}{\mathrm dx}\pm m\omega x\right).\tag{2.41}
$$
What is behind this idea to factorize? Equation 2.40 doesn't seem like it could be solved by factorising the term in the brackets. After all, the R.H.S of 2.40 is not zero, then why would one factorise to solve?
| The basic idea behind the factorization is to replace a 2nd-order differential equation by a pair of first order ones. It was made popular in physics by the work of Hull and Infeld:
Infeld, Leopold, and T. E. Hull. "The factorization method." Reviews of modern Physics 23.1 (1951): 21
although in fact earlier examples, such as the factorization of the harmonic oscillator into creation and destruction operators, were known before the aforementioned paper. The review paper is apparently openly accessible if you follow the GoogleScholar link
The method is actually quite general and Infeld and Hull do a good job of showing how one can factorize some 2nd-order differential equations.
It is at the core of supersymmetric quantum mechanics, where one finds operators $\hat A^\dagger$ and $\hat A$ so that $\hat A^\dagger \hat A$ and $\hat A\hat A^\dagger$ give two different Hamiltonians, connected by a superpotential. Suppose
\begin{align}
\hat H\psi(x)=\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V_-(x)\right)\psi_0(x)=0\, , \qquad V_-(x)=V(x)-E_0
\end{align}
One can easily show that the Hamiltonian can be rewritten in the form
\begin{align}
\hat H_-=\frac{\hbar^2}{2m}\left(\frac{d^2}{dx^2}+\frac{\psi_0^{''}(x)}{\psi_0(x)}\right)=\hat A^\dagger\hat A\, ,
\end{align}
where
\begin{align}
\hat A=\frac{\hbar}{\sqrt{2m}}\left(\frac{d}{dx}-\frac{\psi^{'}_0(x)}{\psi_0(x)}\right)\, ,\qquad
\hat A^\dagger=\frac{\hbar}{\sqrt{2m}}\left(-\frac{d}{dx}-\frac{\psi^{'}_0(x)}{\psi_0(x)}\right)\, .
\end{align}
The interesting "supersymmetric" bit is that $\hat A\hat A^\dagger$ is the Hamiltonian for a different potential.
More details can be found on this in the review paper by Fred Cooper et al (e-copy here):
Cooper, Fred, Avinash Khare, and Uday Sukhatme. "Supersymmetry and quantum mechanics." Physics Reports 251.5-6 (1995): 267-385.
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Effective power of a lens-mirror combination In case of thin lenses in contact the effective power of the combination is given as: $P=P_1+P_2+P_3$.....
For a lens - mirror in contact, like a lens with silvered surface (where lens and mirror are seperated by virtually $0$ distance) can we say that the effective power is the sum of power of mirror and lens.If so how?
| For a silvered lens, the power formula would be $P_{eqv.}=P_{lens}+P_{mirror}+P_{lens}$.
(the lens power is doubled as rays travel through it twice).
eg. for a biconvex lens ($\eta,R$) with one side mirrored:
1. $f_{lens}=\frac{R}{2(\eta-1)}$
2. $f_{mirror}=R/2$
3. $P_{eqv.}=\frac{2(2\eta-1)}{R}$
4. in other words $P_{eqv.}=(2+\frac{1}{k})P_{lens}$ where $k=f_m/f_l$
The power formula doesn't seem to care what the underlying optical instrument is as long as they are thin and close enough for linearity to hold. When mirrors are involved appropriate multiplicities account for multiple passage of light rays.
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Tension in a massless string being pulled at its ends with unequal forces There is a question in my textbook. If a massless inextensible string is pulled on with a force of $10 N$, at both ends, what is the tension in the string?
It’s a very common question. The answer is $10 N$, cf. e.g. this & this Phys.SE posts. It can be proved using Newton’s $2nd$ and $3rd$ law. If we think of the string as a series of links in a chain, for example, or if we think about the adjacent molecules in the string, then we can prove using Newton’s $2nd$ and $3rd$ law that tension in the string is $10N$ at each point along its length.
But what if we pulled on the ends of the string with forces of unequal magnitudes? This question occurred to me and I kind of got confused. My intuition says that the string has a net force acting on it, and hence it would accelerate. But because the string is massless, Newton’s 2nd law did not help me understand this situation. My question is,
If we pull on the ends of a string that is massless and inextensible, with forces of $60N$ and $70N$ respectively, what would be the tension in the string?
Will it be $60N$? Will it be $70N$?
I gave it some thought, and I thought this situation is similar to an atwood machine, two masses $6kg$ and $7kg$ respectively, hanging from a pulley. The pulley is massless and frictionless. The string is massless and inextensible. Because of gravity, one end of the string is being pulled on with $60N$, and the other end is being pulled on with $70N$, isn’t this situation similar? If I work out the tension in the string using $T$ = $\frac{2m_1m_2g}{m_1 + m_2}$, it gives $T$ = $64.6N$.
So can I say that If we pull on the ends of a string that is massless and inextensible, with forces of $60N$ and $70N$ respectively, tension in the string would be neither $60N$, nor $70N$, but somewhere in between ($64.6N$)?
| The arrangement you describe is impossible. The tension of the string will be 70N. Whatever was trying to restrain the end of the string with a force of 60N will be subject to a force of 70N by the string. As a result it will accelerate subject to a net force of 10N. The reaction on the string will be 70N.
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How does a Wehnelt electrode both extract and focus electrons? Now that I can try to pronounce it I'd like to understand how a Wehnelt lens, grid, cap, etc. extracts electrons from a cathode and simultaneously focuses them.
Naively I'd think that it would have a positive potential with respect to the filament in order to produce a field strong enough to overcome at least part of the work function and convince the electrons to leave the cathode material and enter free space, but it also appears to have a repulsive force on them in order to produce a converging beam, at least as shown in the diagram below, which may be incomplete.
How does this work?
Source
|
How does a Wehnelt electrode both extract and focus electrons?
It doesn't. Since it is biased negative with respect to the cathode it would suppress electron emission rather than extract them if it were the only electrode.
Instead, the anode further down stream is positive with respect to the cathode and the opening in the Wehnelt is sufficiently large that the anode still produces a net attractive field at the cathode's surface. Once they leave the cathode the Wehnelt's circular aperture's effect is to repel the electrons, forcing them back towards the axis and thereby producing an image of the cathode downstream, or at least a crossover.
See slide 14 of Presentation on Electron Sources Chapter 5 which shows the same drawing as a Quora answer to What is an electron gun?:
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What are laws in physics? Many times a precise definition of something in physics is not available but yet there exist some rough definitions that guide us through. I need the same rough (if not precise) definition of physical laws. The reason being that we would consider $\mathbf a= \frac {d \mathbf v}{dt}$ as a mere definition but $\mathbf F =m \mathbf a$ as a law (as it is called as Newton's second law of motion). But to me it seem to be more inclined towards being as definitions rather than laws. I mean contrasting Newton's laws of motion to the law of conservation of energy (momentum, etc) it seem to be an attempt to define forces ( though I might be wrong at this (but I don't know reason for such)).
*
*So what is the definition of a physical law that can encapsulated it's entirety?
| As pointed out by Safesphere, the physical laws that rule the world are expressed at a higher and more general level of (mathematical) abstraction which underlie Newton's laws. We can then derive Newton's laws from them as needed to solve everyday problems in real-world dynamics.
These more general relationships are generically called symmetries; if you know what they are for a given system then you can in a sense solve the problem once by working with those symmetry equations and apply the general form of the solution to any of an entire class of related problems, making them easier to solve. This is a very powerful and general technique which is why physicists use it.
As pointed out by dmckee, this question has been asked and answered before but the reason I am answering it here is in doing so I get the chance to test my own understanding of the topic. If I get it wrong, then the experts here will point out my errors and I then learn something new.
Please write back if you are interested in examples of symmetries and I'll furnish you with some.
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Viewing LEDs at great distance Say I put a satellite in orbit with several LEDs on the outside: red, yellow, blue, white and green. Could I see them from the ground with consumer-grade binoculars or telescopes?
If I can see them, could I differentiate between the colors? (e.g. would the color affect how well I could see them?)
Say I have 20/20 vision with no color blindness and I travel to an ideal dark observation point on the earth.
| It really depends on the resolution of your binoculars...
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Entropy of mixing Consider the case of a wall dividing a box into sections 1 and 2, each of volume $V_0$.
Let be $X$ is an ideal gases. Section 1 contains $N$ particles of $X$ and section 2 contains $N$ particles of $X$.
The entropy of mixing (removing the divider) is $0$ in the above case. I understand that an argument for this is that it is a reversible process as we can just as well place the divider back and the system will be in an identical state to its initial condition (assume indistinguishable particles).
My question is that if entropy is a measure of the number of states a particle can take on, a state being a specification of all the values of the variables needed to completely describe the particles, i.e its position and momentum.
If we remove the divider, each particle can now take up twice as many positions and so should have twice as many available states to be in, this surely increases the entropy?
I can't seem to resolve the two different ideas.
Can someone clarify on this?
| You are surprised that $S(T, 2V, 2N) = 2S(T, V, N)$ but this formula can be obtained with statistical physics considering a mix of two identical gas
When you remove the barriers and the two gas are identicals you do not increase the number of unknowns configurations of the system, therefore, entropy does not increase. (This is a really intuitive view of the problem I don't know if you should consider it like that)
But when you remove the barriers with two differents gas, it is obvious that there is a change of the unknowns configuration of the system
Considering the last statement and the demonstrations for both discernable and non-discernable mix on the French Wikipedia you should be able to understand it
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Where does the power delivered to car's wheels go? Okay, so power is work/time. Most cases, when power is provided to something, energy is gained as kinetic energy or lost to friction.
But in a car, the engine puts power ( torque x rpm/5252) to wheels, but very little ends up in the wheels, assuming friction keeps them from spinning.
So where does the power go? Do the wheels thru the friction forces cause the energy to go to the car? Essentially the wheels do work on the car which transforms the rotational energy to kinetic energy of the car right?
| The engine applies torque to the wheels. The wheels turns and apply friction force to the road. By Newton's third law the road applies force to the wheels which make the car moving. Engine power goes to kinetic energy of the car, dissipated heat to friction, air resistance force, battery charging and air conditioning
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Why is frequency a fundamental property of waves? We are taught that although wavelength can change from one medium to other, frequency of a wave doesn't whenever Velocity varies in different media.
But why at a deep level, is frequency so fundamental? Why can't both frequency and wavelength change? Or why isn't wavelength fundamental? Or velocity fundamental?
| If a chunk of matter absorbs 100 wave crests and emits 101 wave crests, then we would say that the chunk of matter created some wave crests by itself, at least one. Right?
If 100 crests go in and then 100 crests come out at slower rate, then it is possible that the waves that came out are the same as the waves that went in.
Like if we send a continuous sound wave through an air mass whose temperature is dropping, then 100 waves that went in during 1 seconds time, those same waves may come out during 2 seconds, so this changing medium halves the frequency, and doubles the duration. (And contains at least 50000 wave crests after 1000 seconds)
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I can't understand one of deduction in Simple Harnomic Motion, can anyone help? source:http://farside.ph.utexas.edu/teaching/336k/lectures/node18.html#e4.8
in order $x=0$ to be a stable equilibrium point we require both
$$f(0)=0$$
and
$$\frac {df(0)}{dx}<0$$
Now, our particle obeys Newton's second law of motion,
$$m \frac{d^2x}{dt^2}=f(x)$$
Let us assume that it always stays fairly close to its equilibrium point. In this case, to a good approximation, we can represent $f(x)$ via a truncated Taylor expansion about this point. In other words,
$$f(x)\simeq f(0)+\frac{df(0)}{dx}x+O(x^2)$$
the article say that the above expression can be written
$$f(x)\simeq -m\omega_0^2x$$
where $\frac{df(0)}{dx}=-m\omega_0^2$
my qustion is that i can't understand what the $\omega_0$ stand for?
why $\frac{df(0)}{dx}=-m\omega_0^2$?
| Physically, $\omega_0$ is the angular frequency of the simple harmonic motion. It is related to the period by $T = 2\pi / \omega_0$. The definition of $\omega_0$ is (in the notation of your link)
\begin{align}
\omega_0 = \sqrt{-\frac{1}{m}\frac{df}{dx}\bigg|_0}.
\end{align}
The reason we define $\omega_0$ this way is because if we set
\begin{align}
m \frac{d^2x}{dt^2} = f(x) \approx \frac{df}{dx}\bigg|_0 x = -m\omega_0^2 x,
\end{align}
we get the differential equation of motion for a simple harmonic oscillator with period $T=2\pi /\omega_0$,
\begin{align}
\frac{d^2x}{dt^2} + \omega_0^2 x = 0.
\end{align}
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Brownian Motion I’m currently interested in learning some topics about the Brownian motion and the random walk (in general, from a pure statistical and probabilistic way).
For that, I would like to ask you if you can recommend some books so I can learn the basic stuff and prepare myself for the real deal.
As I learn these topics, I would like to complement my knowledge about this subject by using programs to make simulations, any idea where I can learn about this?
If you have any ideas of where should I begin, please let me know and tell me the book/author. Thank you!
| While you can construct mathematical models as simple as integrating white noise which evolve trajectories that have the appearance of Brownian motion, physical Brownian motion occurs due to a more complex, stochastic dynamic equilibrium between macroscopic particles and atoms/molecules. So if you are interested in real Brownian motion you need to consider the macroscopic properties such as mass, size, viscosity, geometry, etc. together with the atomic properties of the suspending fluid such as temperature, pressure, atomic weight, etc. to formulate the equation of dynamic equilibrium. That's what Einstein first figured out in his paper in Annalen der Physik which became his PhD thesis.
My recommendation is that you start there.
Beyond that more recent, interesting work models Brownian motion rather using Fractional Calculus (as opposed to the integral calculus Einstein used). You can google Bruce J West for work in this area.
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Why do the $L_z$ and $L^2$ operators share eigenfunctions, but the $L_x$ and $L_y$ operators don't? In my lecture notes the following was written:
I would understand in the case of an applied field if there was some symmetry breaking feature which would allow for a preferred axis or something which could explain why the $L_z$ operator and $L^2$ operator share eigenfunctions as mentioned in the above notes.
I would've thought in the case of no external field, there is no reason to assume the x y and z axis's have any distinguishing factor between them.
Why does this happen?
| The reason is quite simple:
The operators $\hat{L}^2$ and $\hat{L}_z$ have common eigen-functions
because they commute with each other, i.e. $[\hat{L}^2,\hat{L_z}]=0$.
The operators $\hat{L}_x$ and $\hat{L}_y$ don't have common eigen-functions
because they don't commute with each other, i.e. $[\hat{L}_x,\hat{L}_y]\ne 0$.
So the reason is just the algebraic relations between the angular momentum operators.
An external field is not relevant for the reasoning.
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Why does the potential energy of the system decrease when two charge balls are connected using a connecting wire I am confused because I've seen in textbooks and online solutions to questions that when connected,the potential energy of a system of two charged spheres decreases.But according to law of conservation of energy shouldn't it remain the same?
| Simple answer is that there is no possibility for the system to gain energy, in any form whatsoever. So it can do two things either loose no energy, or release some energy of heat, assuming ideal behaviour. To prove that it would loose energy if there is any charge transfer consider the following
Initial charge on sphere 1 is $q_1$ and on sphere 2 is $q_2$. The potential energy initially would be
$$k\frac{q_1^2}{2r_1}+k\frac{q_1^2}{2r_2}$$
The factor of half here is due to the assumption that the balls don't interact when left free.
Now after you connect the wires, the charges re-distribute in such a way that they are at the same potentials. Thus we may say
$$\frac{q_1}{r_1}=\frac{q_2}{r_2}$$
If you now write down the total energy after the redistribution (you will have to use conservation of charge as well), you will find out that unless the charges were already in the above ratio, there's always a lots of energy.
Also there is no violation of conservation of energy as the deficit energy has been lost as heat.
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Echoes and Pitch Change I encountered something strange regarding echoes and pitch.
One Fourth of July , I saw rockets reach a height its hard for me to gauge. I'd call it two to three dozen meters.
I had noticed that shortly after a rocket went off, Pop!, I'd hear a Pang! echoed from the wall of a nearby building. The Pang came about a half-second after the Pop. The Pang was little shorter in duration from the Pop and had a higher pitch than the Pop.
That night I was talking to my professor. He pointed out that reflection doesn't change pitch at an interface and said I must be mistaken.
Is it possible I wasn't? Perhaps there is some physical mechanism that would have filtered out the pitches?
Perhaps longer wave lengths were more dampened by the aluminum siding leaving higher intensities for the higher pitches? It was probably a hot day. Perhaps temperature differences could effect the pitches along the direct path differently from those from an echoed path so that there was a different balances of pitches when the echo arrived? A non-physical explanation, my left ear was facing the building the Pops were in the direction of my right ear. Maybe I've lost high pitch hearing in my right.
If this helps, the building was about 5 to 8 m high and Between 10 and 15 m long. The siding had 3cm depth by 3cm width ridges running the height of the side of the building facing the fire works show, periodically.
I'm about 168cm tall. I believe was two times the distance from the fireworks as I was from the building. I was roughly 6m from the building. I was located roughly in the blue location marked ME and the poppers were around SRC in red.
| I think this is most likely due to the doppler effect. the explosion of the rocket occurs while it is moving away from you which pitch-shifts the tone of the "pop" down. Meanwhile, the sound reflection off the sides of the nearby building (which is tall) will come sideways from the exploding rocket, and hence will be doppler-shifted less. This means the reflected sound will be higher in pitch.
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Is it possible to conserve the total kinetic energy of a system, but not its momentum? It is possible to conserve momentum without conserving kinetic energy, as in inelastic
collisions. Is it possible to conserve the total kinetic energy of a system, but not its momentum? How?
To clarify, I am not necessarily talking about an isolated system. Is there any scenario which we could devise in which momentum is not conserved but kinetic energy is?
| If non-isolated system are of interest, then what you’re looking for is an external force that does no work.
*
*The central force in a circular orbit: the satellites energy in unchanged, but its momentum is continuously changing.
*An electron moving across a constant magnetic field: ditto
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Work done by battery on a capacitor Work done by battery on a capacitor is QV/2, where V is the final potential across the capacitor plates an Q is the charge. I know that the Q charge which gets stored on the capacitor comes from the connecting wires. However, since
Positive charge on one plate is reducing (Assuming conventional flow) and increasing on the other it is convenient to assume that the charge is going directly from one plate to the other as it makes calculating work done on the charges more easier.
If we say some Q charge left one plate (let us call these charge carriers set A) and some charge Q ended up on the other plate (let this be B).Now, since it is not necessary that A went through the battery let us call the charge Q that went through the battery C. If battery didn't do work on A or B why do we say that the energy stored in the capacitor comes from the WORK done by the battery in transferring charges from one plate to the other.
According to my professor this work done by the battery can be assumed to be on the same charges since electric field is conservative and only depends on the initial and final states of the system.
I have assumed charges A,B and C to have the same magnitude. I know that distinguishing between charges is pointless but I have done so in order to make my question more clear.
| If the net charge on a plate is $0$, it takes no effort to move initial charge to that plate, so work done is $0$.
After you moved some charge to the plate, this excess charge on the plate opposes further new charge to be moved, so you must do positive work to overcome this opposition.
In a circuit, battery manages to move charge from one plate to the other. This means the battery is doing positive work against the electric field from existing net charge on the plate.
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Reference request for a mathematical motivation for the Born rule I was reading the popular science book The Hidden Reality by Brian Greene. My question is about a part in the notes at the end of the book. It is chapter 8, note 9. Brian Greene describes a mathematical motivation for the probabilistic interpretation (i.e. the Born rule). He does this by defining a frequency operator and then taking the limit by increasing the number of states to infinity.
Greene also states that the majority of the terms in the expansion of $|\phi \rangle ^{\otimes n}$ must be considered as 'non-existent'. However, the challenge is understanding what this 'means', physically speaking (if it already means something at all).
I'm interested in more reading material about these topics. I'm interested in more advanced explanations in the literature (possibly by the authors mentioned by Greene in the note). However, I could not find material for this. Perhaps someone who has read the book or is knowledgeable about this topic could suggest some resources.
| I have found an interesting discussion about this question in the book (in french) "Mécanique quantique, Bases et applications" by Constantin Piron.
He proves a Gleason-like theorem (A.2 Théorème fondamental, p.172) stating something like: if you would like to associate to each state (= vector) and proposition (= closed subspace of the Hilbert space) a real number between $0$ and $1$ behaving like a probability (there are mathematical hypotheses corresponding to this) then the numbers are given by the usual formula.
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What would the water pressure be at this point in a static incompressible fluid? I have a bowl full of water. I invert a glass and place it upside down in the water, leaving a small pocket of air.
My question: what would the water pressure at positions P1 and P2 be? I know P1 = P2 because there's no hydrostatic effect as the water is level, but what is this value?
*This may seem rather trivial but I can't decide between two lines of reasoning. From where my confusion arises:
The atmospheric pressure acts to push the water level in the bowl down and up into the glass. The small pocket of air in the glass has the opposite effect, acting to push the water level in the glass down and increase the level in the bowl. Since the air pressures are equal, the water is at the same level in the bowl as in the glass.
1) Now, since the fluid is in equilibrium, P0 must equal the pressure at P1 for no flow to occur. Likewise, then P2 must equal P0 in the glass. Then P1 = P2 = P0
2) Or the water pressure at P1 and P2 completely zero? And this would be because the pressure from the air in the glass counters the pressure from the air outside acting on the bowl?
Thank you in advance, sorry for the length of this.
| Let L be the total length of the glass (assuming constant cross sectional area), z be the depth that the lower lip of the glass is inserted below the water surface, and h be the height that water rises inside the glass above the lower lip. Then the pressure of the air trapped in the glass is given by:$$p_a=p_{atm}+\rho g z-\rho g h$$This trapped air pressure is also given by the ideal gas law: $$p_a=p_{atm}\frac{L}{L-h}$$So, from these two equations,$$p_{atm}+\rho g z-\rho g h=p_{atm}\frac{L}{L-h}$$This equation is non-linear in h. But, if we assume that h << L, we can linearize the equation to obtain:$$p_{atm}+\rho g z-\rho g h=p_{atm}+p_{atm}\frac{h}{L}$$The solution to this equation for h is $$h=\frac{z}{\left[1+\frac{p_{atm}}{\rho g L}\right]}$$which is significantly less than z but greater than zero.
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Why is the normal contact force horizontal on an inclined ladder?
There is only one force acting on the ladder which is its weight and it acts vertically downwards. Then why does the normal contact force from the vertical wall act horizontally on the ladder? There must be a horizontal force acting on the wall to exert a horizontal force on the ladder. What causes the horizontal force on the wall and what is it called?
| Normal forces act perpendicular to the surface in contact. The force acting in the ladder is actually somewhere in between the horizontal and vertical forces shown. Those are just the components of the normal force.
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Correct way of viewing a (1,1)-tensor returning a vector I am currently watching this excellent video series building up to general relativity. We have finally started looking at tensors and a question came up from the audience which (to my understanding) was asking why tensors are defined as multi-linear maps from sets of vectors and covectors to the real numbers yet (1,1)-tensors are often viewed as linear maps from vectors to vectors.
He says that the two ways of viewing the (1,1)-tensor contain the same information, but I don't understand how. If anyone could provide another explanation for why this is true I would really appreciate it.
| A $(n,m)$ tensor eats $n$ vectors and $m$ co-vectors and spits out a real number. A $(1,1)$ tensor eats one vector and one co-vector and spits out a real number. Let's call our $(1,1)$ tensor $f(.,.)\colon V \times V^{*} \mapsto \mathbb{R} $. The first argument is a vector and the second argument a co-vector. So for generic vectors $a \in V, b \in V^{*}$, $f(a,b) \in \mathbb{R}$.
What about $f(a,.)$? Clearly, it eats a co-vector (in the empty argument) and spits out a real number. But this is precisely the definition of a vector. So, $f(a,.)$ is a vector. Therefore $f(.,.)$ ate a vector(i.e $a$) and produced a vector $f(a,.)$.
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Why don't we define potential due to a magnetic field? We define electric potential and gravitational potential and use them quite often to solve problems and explain stuff. But I have never encountered magnetic potential, neither during my study (I am a high-schooler), nor during any discussion on physics.
So, does magnetic potential even exist? According to me, it should because a magnetic field is a conservative one and so we can associate a potential with it? Also if it's defined, then why don't we encounter it as often as we do the others (electric potential, gravitational potential, etc.)?
I have encountered magnetic potential energy only in the cases where a dipole is subjected to a magnetic field. Is magnetic potential limited only to this scenario, or is there a general expression for the magnetic potential?
I should have said this earlier, but don't restrain your answers due to my scope. You can use vector calculus as I am quite familiar with it. Also, this question is meant for everyone, so even the answers which are out of my scope are appreciated.
| There is the vector potential $\bf A$ for which ${\bf \nabla} \times {\bf A} = {\bf B}$. So $B_z = dA_x/dy - dA_y/dx$ and similar for other components.
The classical field theory of electromagnetism is based on the four potential $A^\mu = (\phi,{\bf A})$ . $\phi$ is the Coulomb potential .
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Adding Angular Momenta Operators in QM Consider $j,m$ to be the angular momentum magnitude and $z$-projection eigenvalues corresponding to a total angular momentum operator $\hat{J}$, composed of angular momentum $\hat{J}_1$ and $\hat{J}_2$ with eigenvalues $j_1,m_1$ and $j_2,m_2$. We want to know what values $j$ and $m$ can take on in terms of $j_1,m_1,j_2,m_2$. It is commonly stated that
$$ \hat{J}_z = \hat{J}_{1z} + \hat{J}_{2z}, $$
from which one can immediately derive $m = m_1+m_2$.
What is the explanation for the simple addition of the $z$ operators? If there is some vectorial model explanation, then is it also true that $\hat{J}_x = \hat{J}_{1x} + \hat{J}_{2x}$, for example? Is there some other way to prove this?
Further, if we are looking at a vectorial model, why isn't it true that the magnitudes are the same, i.e. that $j = j_1+j_2$?
| I am gonna give a much shorter answer than @Cyro.
*
*Yes it is also true for the other components of angular momentum.
*It is simply due to vector addition. The total angular momentum of the system (were it classical) would be $\mathbf{J} = \sum_i\mathbf{J}_i$. For quantum, it's the same thing, but you just quantise the observable into an operator.
*The "catch" is that while $J_{\mathrm{tot}}$ and $J_z$ commute, the $x,y,z$ components of the angular momentum do not commute among themselves. The choice of the $z$ axis is conventional in this context. So you only know the value of the total angular momentum $\sqrt{j(j+1)}$ and the value of its $z$ projection $m_z = m_{z_1} + m_{z_2}$.
In other words, $m_z = m_{z_1} + m_{z_2}$ but $m_x \neq m_{x_1} + m_{x_2}$.
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Doubt regarding derivation of escape velocity In my text book the derivation goes like this:
The minimum speed required to project a body from the surface of the Earth so that it never returns to the surface of the Earth is called escape speed. If a velocity greater than the escape velocity is imparted, the body will escape and leave the surface. If a velocity lesser than the escape velocity is given, it will fall back to the surface or be in an orbit. A body thrown with escape speed goes out of the gravitational pull of the Earth. Work done to displace the body from the surface of the Earth ($r=R_e$) to infinity ($r=\infty$) is given by:
$$\int dW=\int^{\infty}_{R_e}\frac{GM_e m}{r^2}dr$$
or
$$W=GM_e m\int^{\infty}_{R_e}\frac{1}{r^2}dr=-GM_e m\frac{1}{r}\Biggr| ^{\infty}_{R_e}$$
$$=-GM_e m\left(\frac{1}{\infty}-\frac{1}{R_e}\right)\Rightarrow W=\frac{GM_e m}{R_e}$$
Let $v_e$ be the escape speed of the body of mass m, then kinetic energy of the body is given by:
$$\frac{1}{2}mv^2=\frac{GM_e m}{R_e}\Rightarrow v_e=\sqrt{2gR_e}=11.2 \:\text{kms}^{-1}$$
But isn’t work done $Fdx=Fdx\cos z$. The direction of force and displacement is anti parallel but there is no -ve sign in the derivation. Have I misunderstood something?
|
Work done to displace the body from the surface of the Earth ($r=R_e$) to infinity ($r=\infty$) is given by:
$$\int dW=\int^{\infty}_{R_e}\frac{GM_e m}{r^2}dr$$
That's rather sloppy, and it's incorrect. Unfortunately, once a mistake gets into one physics text from India to mistake gets propagated to many physics texts from India. I've found this sloppy mistake in three Indian physics textbooks so far.
It's obvious that the work has to be negative. The initial velocity is nonzero and hence the initial kinetic energy is positive. The final velocity is identically zero and hence so is the final kinetic energy. This means the change in kinetic energy is negative. Since work equals the change in kinetic energy in the case of a conservative force, the work done by gravity must also be negative. When computed correctly, this negative amount of work is exactly what is needed to derive escape velocity as $v_e = \sqrt{2 g R_e}$.
| {
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Precise definition of the vertex factor Just a short question about the vertex factor in QFT. When I have an interaction Lagrangian
$$\mathcal{L}_{\mathrm{int}}=-\frac{\lambda}{3!}\phi^3$$
with a real scalar field $\phi$, is the vertex factor given by $-i\lambda$ or $-i\frac{\lambda}{3!}$?
Because as far as I learnt, the vertex factor is $-i\frac{\lambda}{3!}$ and $3!$ is the symmety factor of the diagram. But I saw in many books that they claim that $-i\lambda$ is the vertex factor of this interaction.....
| It is conventional to write interactions normalized by the number of permutations of identical fields. So, there will be a $\frac{1}{n!}$ factor for each interaction with $n$ identical fields. This factor is then canceled by the $n!$ ways of permuting the $n$ identical lines coming out of the same internal vertex.
The diagram is therefore associated with just the prefactor, e.g. $\lambda$, from the interaction.
If the diagram presents a symmetry, you have also to divide by the geometrical symmetry factor.
| {
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What is gravitational radiation? What is gravitational radiation (in association with gravitational waves)? Is it a form of energy/mass? Or is it just another word for gravitational waves?
| Strictly speaking gravitational waves are a subset of gravitational radiation. Gravitational radiation could in principle be radiated as solitons, and while these can be constructed from gravitational waves by Fourier synthesis we wouldn't normally describe them as a gravitational wave.
However this is a somewhat trifling objection and for all practical purposes you can use the two terms interchangeably.
| {
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Is the energy of an electromagnetic radiation the sum of the energy of each photon? In A.P. French's Special relativity the author said,
We suppose that an amount
$E$ of radiant energy (a burst of photons) is emitted from one end
of a box of mass $M$ and length $L$ that is isolated from its surroundings
and is initially stationary. The radiation
carries momentum $E$/c.
The mass and length of the box are irrelevant here.
He said the momentum of the radiation is $E_{radiation}/c$. We know that the momentum of a photon with energy $E_{photon}$ is $p_{photon}=E_{photon}/c$.
So is $E_{radiation}$ the sum of the energy of each photon, $E_{photon}$?
| Is the purchasing power of my checking account equal to the sum of the purchasing power of each dollar in my checking account? Well, in a sense yes. But in another sense, not really, because there is no such thing as an individual dollar in my checking account. If I have a five dollar balance, there is no meaningful way to point to individual dollars that make up that balance.
Likewise with photons.
| {
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Why is $|\alpha\rangle$ not eigenstate of $a^{\dagger}$ for $\alpha^*$ I know that even if we have:
$$a |\alpha \rangle = \alpha |\alpha\rangle$$
We don't have:
$$a^{\dagger} |\alpha \rangle = \alpha^* |\alpha\rangle$$
Actually as explained in the second answer here Eigenvalue for the creation operator for a coherent state $a^{\dagger}$ doesn't have eigenvalues.
But what disturbs me is the "matrix" vision.
If I choose a basis composed of $|\alpha\rangle$ and then some other states such that I have an orthogonal basis in the end, I can write down the (infinite) matrix $a$ in this basis.
The first diagonal coefficient will be $\alpha$. Then to go to $a^{\dagger}$ I conjugate transpose the matrix which will give me $\alpha^*$ on the first diagonal element.
Then through this vision it seems that $|\alpha\rangle$ is eigenstate of $a^{\dagger}$ with $\alpha^*$ as eigenvalue.
Where is the mistake in this reasoning ? Is it because we work in infinite dimension space so the matrix vision can induce errors ?
| Your mistake - while $\alpha$ will be on the diagonal (and $\alpha^*$ on the diagonal of the hermitian conjugate) it is not guaranteed that all the other terms in the row (and the column in the Hermitian conjugate representation of $a^{\dagger}$) will be zero. Therefore it will not be an eigenstate. It will have the property $\langle \alpha | a^{\dagger} |\alpha\rangle = \alpha^*$ as this is the value on the diagonal.
Note, for example, that the coherent states do not form a normal orthogonal basis, but an over-complete basis. Such that $\langle \alpha | \beta \rangle \propto \exp(|\alpha-\beta|^2)$ if I remember correctly.
| {
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Usually, how much does a phonon travel without scattering? Phonons propagate without problems in a lattice, until they scatter on something, like a defect, an electron, or another phonon. But in a typical solid at room temperature, how much (or how long) is the mean free path of a phonon? I know that depending on the temperature one type of scattering can be more probable than the other, but do mean free paths of these different types of scattering tend to be very different from each other? (With different orders of magnitude?)
| Phonons are lattice vibrations. The distance between two consecutive phonons is of the order of $1/N$ where $N$ is the number of atoms in the lattice. At room temperature a phonon travels approx 10 to 100 lattice constants before scattering.
In this article
they say that a phonon travels $< 1 \mu m$ before scattering, that is approx 100 lattice constants (the lattice constant is approx 1 nm)
| {
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Why is the sunset not bluer My question is a duplicate of this; Clarification on Rayleigh scattering causing various sky colors.
The accepted answer from the link above says that at sunset the scattering occurs farther away and does not reach the observer, which is unsatisfactory and vague to me. (not even sure if it's the right and correct answer)
It doesn't quite make sense to me that at noon, we see scattered blue light whereas, at sunset, we see the sunlight itself minus the scattered blue light. At noon or at sunset, we are not seeing the sun, but the scattered sunlight indirectly. The logic should apply both cases equally and should imply that at sunset we see bluer sky than at noon because of more scattering.
Please let me know if this duplicate question was unnecessary because the accepted answer from the link above was enough to answer the question.
| When you look at the sun overhead (not advisable) you see the white light from the sun with a little bit of the blue scattered horizontally. The blue sky is blue light scattered from sunbeams going elsewhere. When the sun is near your horizon, its light passes through a much greater distance of dense atmosphere, and most of the blue is lost from the beam to give other people a blue sky.
| {
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On the derivation of the north-south aberration angle In A.P. French's Special relativity, page $39$, the author said the north-south aberration angle $\alpha$ below, in figure (b), is equal to $v\sin(\theta_{0})/{c}$, where $\theta_{0}$ is the angle when the earth is stationary (no aberration).
How did $v\sin(\theta_{0})/c$ came about?
This value seems rather suspicious, especially because of the $\sin(\theta_{0})$. According to this Wikipedia article, the angle $\theta$ above (named $\phi$ in the article) satisfies $$\tan(\theta)=\frac { \sin(\theta_{0})}{v/c + \cos (\theta_{0})},$$
thus,
$$\tan(\alpha)=\tan(\theta_{0}-\theta)=\frac{\frac{\sin(\theta_{0})}{\cos(\theta_{0})}-\frac { \sin(\theta_{0})}{v/c + \cos (\theta_{0})}}{1+\frac{\sin(\theta_{0})}{\cos(\theta_{0})}\times \frac { \sin(\theta_{0})}{v/c + \cos (\theta_{0})}}.$$
I highly doubt that taking the $\tan^{-1}$ of the above would result in $v\sin(\theta_{0})/c$.
Is there a simpler way to compute the aberration angle to check whether it conforms with $v\sin(\theta_{0})/c$?
| Use 4-vectors (with $c\equiv 1$):
$$ k_{\mu} = (\omega, k_x, k_y, k_z) = k(1, -\cos{\theta_0}, -\sin{\theta_0}, 0) $$
is the wave-vector in the stationary frame. The angle of arrival is given by
$$\theta = \tan^{-1}{\frac{k_y}{k_x}}=\tan^{-1}{\frac{-\sin\theta_0}{-\cos\theta_0}}=\theta_0$$
Boost by $v$:
$$ k'_{\mu} = (\gamma(\omega-vk_x), \gamma(k_x-\omega v), k_y, k_z) $$
$$ k'_{\mu} = k(\gamma(1+v\cos\theta_0), -\gamma(\cos\theta_0+v), -\sin\theta_0, 0) $$
so:
$$\theta' = \tan^{-1}{\frac{k'_y}{k'_x}}=\tan^{-1}{\frac{\sin\theta_0}{\gamma(\cos\theta_0+v)}}$$
which reduced to the wiki expression at solar system velocities ($\gamma < 1+10^{-8}$).
Note: I once did this with aerospace engineers trying to point a star tracker in deep space. They were taking a 3+1 approach and were in knots. They down right thought it was physicists' voodoo.
So now that we have a rigorous expression, we need to consider the fact that $v \ll c$ and that $\alpha \ll 1$. That means two things (with $\gamma \approx 1$):
$$ \tan{\theta} = \frac{\sin\theta_0}{\cos\theta_0 +v}=\frac{\tan\theta_0}{1+v/\cos{\theta_0}}\approx \tan\theta_0[1-v/\cos\theta_0]$$
Meanwhile, a Taylor expansion about $\theta_0$ gives:
$$ \tan\theta \equiv \tan{(\theta_0-\alpha)} \approx \tan\theta_0 - \alpha\cdot[\tan x]'_{|x=\theta_0}=\tan\theta_0 - \alpha/\cos^2{\theta_0}$$
$$ \tan\theta = \tan\theta_0[1 - \frac{\alpha}{\cos\theta_0\sin\theta_0}]$$
Setting:
$$ \frac v{\cos\theta_0} = \frac{\alpha}{\cos\theta_0\sin\theta_0}$$
and reintroducing $c$ gives:
$$\alpha = \frac v c \sin\theta_0$$
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