Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Which way do black hole jets spin? The centers of black holes and quasars often have jets coming out the two poles of an accretion disk, say north and south. Is it known if the two jets spin in the same direction or opposite directions to each other?
| The Blandford-Znajek process and variants are leading contenders for explaining black hole jets. The driving energy for the jets is extracted from the black hole spin and transferred to the outgoing plasma by twisted magnetic fields. Along with energy, the jets must contain the lost angular momentum from the black hole, which then ultimately comes from material being accreted from the disk.
Thus I would expect that the angular momentum vector in the jets (in both directions) is in the same sense as the black hole angular momentum. i.e. the material spirals away from the black hole in the same sense in both directions. e.g. If the black hole spins clockwise looking down on its "north rotation pole", then the angular momentum vector points south. Material travelling towards you and away from you will also spin clockwise.
See also models for astrophysical relativistic jets from compact objects and
Why are polar jets emitted along the axis of rotation?
| {
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A question about the coupling between string and gauge field $A_{\mu}$ I have a question about deriving the coupling term of string and the gauge field on brane. According to David Tong's lecture note p184/(191 in acrobat), the coupling is given by
$$ S_{\mathrm{end-point}}=\int_{\partial M} d \tau A_{a}(X) \frac{d X^a}{d \tau} \tag{1} $$
It is said that the coupling is obtained by exponentiating the vertex operator, "as described at the beginning of Section 7",
$$
V_{\mathrm{photon}} \sim \int_{\partial M} d \tau \zeta_a \partial^{\tau} X^a e^{ i p \cdot x} \tag{2}
$$
My question is about the logic of exponentiating the vertex operator.
In the beginning of section 7 of the lecture note, in order to obtain the coupling between string and the gauge fields, the Polyakov action is extended in curved space
$$ S= \frac{1}{4 \pi \alpha' } \int d^2 \sigma \sqrt{g} g^{\alpha \beta} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} G_{\mu\nu}(X) \tag{7.1} $$
The coupling comes from the bending of spacetime, e.g.
$$G_{\mu\nu} (X) = \delta_{\mu\nu} + h_{\mu\nu} (X) $$
$$ Z= \int \mathcal{D} X \mathcal{D} g e^{-S_{\mathrm{Poly}} -V} = \int \mathcal{D} X \mathcal{D} g e^{-S_{\mathrm{Poly}} } (1-V +\frac{1}{2} V^2 + \dots ) $$
$$ V= \frac{1}{4 \pi \alpha'} \int d^2 \sigma \sqrt{g} g^{\alpha \beta} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} h_{\mu\nu}(X) \tag{7.2} $$
In order to obtain Eq. (1) from (2), where is the bending of metric?
| There is no 2-D metrics here, because we are working with the boundary $\partial M$
You could imagine a standard action $S_0 = \int_{\partial M} d\tau A_a \frac{d X^a}{d \tau}$, where $A_a$ is constant.
With a small perturbation, we will have : $A_a(X) = A_a + \epsilon_a(X)$, and we have an action $S = \int_{\partial M} d\tau A_a(X) \frac{d X^a}{d \tau}$
A partition function would be $Z = \int dX e^{-S_0 - V} = \int dX e^{-S_0}(1-V +\frac{1}{2} V^2 + \dots )$, with $V = \int_{\partial M} \epsilon_a(X)\frac{d X^a}{d \tau}$
$A_a(X)$ are coherent states of photons, as $G_{\mu\nu}(X)$ are coherent states of gravitons.
In some sense, we may consider that the photon vertex operator corresponds to a very special (non-coherent) case where $\epsilon_a(X) = \zeta_a e^{ip.x}$, in the same way as the graviton vertex operator is a very special (non-coherent) case where $h_{\mu\nu}(X) = \zeta_{\mu\nu}e^{ip.x}$
| {
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What is the difference between the Balmer series of hydrogen and deuterium? In my quantum mechanics textbook, it claims that the Balmer series between hydrogen and deuterium is different. However, I was under the impression that the Balmer series
$$H_\alpha, H_\beta, H_\gamma$$ is related by the equation $$\lambda=C\frac{n^2}{n^2-4}$$
where
$$
C=3646 \mathring{\text{A}}
$$
$$n=3,4,5$$
Is there a equation relating the mass and the Balmer series?
Any hint would be appreciated
| $$
\frac{1}{\lambda} = \frac{4}{C_M}\left(\frac{1}{4} - \frac{1}{n^2}\right) = R_M\left(\frac{1}{4} - \frac{1}{n^2}\right),
$$
where $R_M = 4/C_M$ is the Rydberg constant for the particular atom:
$$
R_M = R_\infty\left(1+\frac{m_\text{e}}{M}\right)^{-1},
$$
with $m_\text{e}$ the electron mass, $M$ the mass of the atomic nucleus and
$$
R_\infty = 1.0973\,731\,568\,539\times 10^7\;\text{m}^{-1}.
$$
For Hydrogen I get,
$$
\begin{align}
R_H &= R_\infty\left(1 + \frac{5.4857990943\times 10^{-4}\;\text{u}}{1.007276466812\;\text{u}}\right)^{-1},\\
&= 1.0967\,758\,341\times 10^7\;\text{m}^{-1},\\[2mm]
C_H &= 4/R_H = 364.70534\;\text{nm},
\end{align}
$$
and for Deuterium I get,
$$
\begin{align}
R_D &= R_\infty\left(1 + \frac{5.4857990943\times 10^{-4}\;\text{u}}{2.013553212724\;\text{u}}\right)^{-1},\\
&= 1.0970\,742\,659\times 10^7\;\text{m}^{-1},\\[2mm]
C_D &= 4/R_D = 364.60613\;\text{nm}.
\end{align}
$$
Note: I haven't verified these numbers.
| {
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Feynman's subscript notation Consider this vector calculus identity:
$$
\mathbf{A} \times \left( \nabla \times \mathbf{B} \right) = \nabla_\mathbf{B} \left( \mathbf{A \cdot B} \right) - \left( \mathbf{A} \cdot \nabla \right) \mathbf{B}
$$
According to Wikipedia, the notation $\nabla_\mathbf{B}$ means that the subscripted gradient operates on only the factor $\mathbf{B}$. Can somebody explain the term $\nabla_\mathbf{B} \left( \mathbf{A \cdot B} \right)$ in detail, give a concrete example, or an expression in components because I do not understand it at all. I encountered this identity in electromagnetism.
| First of all, I always tell every student who asks me about such expressions to stop remembering all these infinite rules and learn to do that with Kronecker delta and Levi-Civita symbols. For example your initial expression looks like this:
$$\mathbf{A}\times[\nabla\times\mathbf{B}]=\varepsilon_{ijk}A_j\varepsilon_{klm}\partial_lB_m = \varepsilon_{ijk}\varepsilon_{klm}A_j\partial_lB_m$$
As you can see, when you've written it like that, you don't need to care about non-commutativity of the cross-product any more. So you can take these Levi-Civitas and do the transformation:
$$\varepsilon_{ijk}\varepsilon_{klm} = \varepsilon_{kij}\varepsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
First I just made a cyclic permutation of the indexes, and the second equality is the only equality you need to remember (which is really easy: same indices go with "+", and swapped indices go with "-"). Inserting that, one gets:
$$\varepsilon_{ijk}\varepsilon_{klm}A_j\partial_lB_m = \delta_{il}\delta_{jm}A_j\partial_lB_m - \delta_{im}\delta_{jl}A_j\partial_lB_m=A_j\partial_i B_j-A_j\partial_jB_i$$
Here I've used the property of Kronecker delta and actually came to the result as in your expression. Lets expand the sums to clarify what those terms actually mean:
$$A_j\partial_i B_j-A_j\partial_jB_i =\left(\begin{array}{c}A_x\frac{\partial B_x}{\partial x}+A_y\frac{\partial B_y}{\partial x}+A_z\frac{\partial B_z}{\partial x}\\A_x\frac{\partial B_x}{\partial y}+A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_z}{\partial y}\\A_x\frac{\partial B_x}{\partial z}+A_y\frac{\partial B_y}{\partial z}+A_z\frac{\partial B_z}{\partial z}\end{array}\right)-\left(\begin{array}{c}A_x\frac{\partial B_x}{\partial x}+A_y\frac{\partial B_x}{\partial y}+A_z\frac{\partial B_x}{\partial z}\\A_x\frac{\partial B_y}{\partial x}+A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_y}{\partial z}\\A_x\frac{\partial B_z}{\partial x}+A_y\frac{\partial B_z}{\partial y}+A_z\frac{\partial B_z}{\partial z}\end{array}\right)$$
| {
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Surface gravity of Kerr black hole I'm going through Kerr metric, and following the 'Relativist's toolkit' derivation of the surface gravity, I've come to a part that I don't understand.
Firstly, the metric is given by
$$\mathrm{d}s^2=\left(\frac{\Sigma}{\rho^2}\sin^2\theta\omega^2-\frac{\rho^2\Delta}{\Sigma}\right)\mathrm{d}t^2-2\frac{\Sigma}{\rho^2}\sin^2\theta\omega \mathrm{d}\phi \mathrm{d}t+\frac{\Sigma}{\rho^2}\sin^2\theta \mathrm{d}\phi^2+\frac{\rho^2}{\Delta}\mathrm{d}r^2+\rho^2 \mathrm{d}\theta^2$$
With
$$\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2,$$
$$\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}$$
The Killing vector that is null at the event horizon is
$$\chi^\mu=\partial_t+\Omega_H\partial_\phi$$
where $\Omega_H$ is angular velocity at the horizon.
Now I got the same norm of the Killing vector
$$\chi^\mu\chi_\mu=g_{\mu\nu}\chi^\mu\chi^\nu=\frac{\Sigma}{\rho^2}\sin^2\theta(\Omega_H-\omega)^2-\frac{\rho^2\Delta}{\Sigma}$$
And now I should use this equation
$$\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu$$
And I need to look at the horizon. Now, on the horizon $\omega=\Omega_H$ so my first term in the norm is zero, but, on the horizon $\Delta=0$ too, so how are they deriving that side, and how did they get
$$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta$$
if the $\Delta=0$ on the horizon? Since $\rho$ and $\Sigma$ both depend on $r$, and even if I evaluate them at $r_+=M+\sqrt{M^2-a^2}$ they don't cancel each other.
How do they get to the end result of $\kappa$?
| The calculation can be done in this coordinate system just fine, even though it doesn't extend across the horizon. Surface gravities are very commonly computed in coordinate systems which go bad at the horizon. For example, the surface gravity of Schwarzschild
$ds^2 = -f dt^2 + f^{-1} dr^2 + r^2 d\Omega_2^2, \qquad f= 1- \frac{r_+}{r},$
is easily found to be $\kappa = \frac{f'}{2} = \frac{1}{2r_+}$.
I think your problem is that you are evaluating quantities at the horizon before taking derivatives. It's important to first take derivatives, and then to evaluate at the horizon.
| {
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Why doesn't Bernoulli's Principle apply to Current and Resistors in a circuit? Bernoulli's principle makes sense when you apply it to fluids. If you decrease the diameter of a pipe then the velocity of the fluid increases because it needs to keep the same rate of fluid moving through the pipe.
So my question is:
If Voltage == Diameter of the pipe
and
Current == Rate of which the fluid is moving
Why do resistors work?
Shouldn't the resistor only actually work within itself but then return the current to it's actual rate once you have passed it?
Or have I taken the analogy of wires being like pipes of water to far?
| So you can apply Bernoulli's theorem but not according to your analogy take velocity of fluid as electric field at that position now electric flux is conserved since no charge enclosed in it so as diameter is large electric field will be less.
| {
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Spinor and Scalar Bose-Einstein condensate I read about an order paramater that describes a Bose-Einstein condensate. But I don't understand, the classification into "scalar" condensate and "spinor" one. Is it linked with spin of atoms that take part in condensation? Or is it a property of the condensate?
Does anyone have a good explanation?
| You call the condensate 'scalar' when the atoms are spin-0. When instead atoms have a non trivial spin you talk about 'spinor condensates'.
| {
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Adiabatic filling of a container Suppose a thermally insulated container is filled with atmospheric air until the pressure reaches 5000 psi. This could represent the filling of a diving cylinder, before thermal dissipation becomes significant.
Initially, then, some mass of air has atmospheric temperature $T_1$ and $p_1$. When forced in to the tank adiabatically, we expect its final temperature to be
\begin{align}
T_2 &= T_1\left(\frac{p_2}{p_1}\right)^{1-1/\gamma} \\
&= (300\,\mathrm{K})\left(\frac{5000\,\mathrm{psi}}{14.5\,\mathrm{psi}}\right)^{1-1/1.4} \\
&\approx 1590\,\mathrm{K}
\end{align}
This final temperature seems far too hot. A diving cylinder would likely melt at this temperature.
More likely, though, something is wrong with my reasoning. Is this not the correct use of the temperature-pressure relations, assuming an ideal gas? If not, what other information is needed to predict the final temperature of the container?
| I think you forget that when you fill the tank, you add mass to it. Your relations assumes that the mass in the system is constant.
| {
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Period $T$ of oscillation with cubic force function How would I find the period of an oscillator with the following force equation?
$$F(x)=-cx^3$$
I've already found the potential energy equation by integrating over distance:
$$U(x)={cx^4 \over 4}.$$
Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), but I'm stuck on how to approach the problem. I can set up a differential equation:
$$m{d^2x(t) \over dt^2}=-cx^3,$$
$$d^2x(t)=-{cx^3 \over m}dt^2.$$
But I am not sure how to solve this. Wolfram Alpha gives a particularly nasty solution involving the hypergeometric function, so I don't think the solution involves differential equations. But I don't have any other leads.
How would I find the period $T$ of this oscillator?
| Starting from
$$ \frac{1}{2} \left( v(x)^2 - v_0^2 \right)= - \frac{c}{m} x^4 $$
with initial velocity $v_0$ when $x=0$, the time relationship is
$$ t = \int_0^x \frac{1}{v(x)}\,{\rm d} x $$
I use in intermediate variable $\xi$ for distance $x = \sqrt[4]{\frac{2 m v_0^2}{c}}\, \xi $ .
I integrate the energy relationship to get
$$ t = \int_0^x \frac{1}{\sqrt{v_0^2 - \frac{c x^4}{2 m} }} \,{\rm d} x=\sqrt[4]{ \frac{2 m}{c v_0^2} } \int_0^\xi \frac{1}{\sqrt{1-\xi^4}}\,{\rm d} \xi $$
$$ t = \sqrt[4]{ \frac{2 m}{c v_0^2} }\; {\rm EllipticF}( \sin^{-1}\xi, -1) $$
Note that the elliptic integral has a taylor expansion of
$${\rm EllipticF}(x,m) \approx x + \frac{m}{6} x^3 - \frac{m}{30} x^5 + \ldots $$
which makes the above solution approximately (for small displacements)
$$ t = \frac{ \tau}{2\pi} \; \sin^{-1}\xi $$
with period $\tau = 2 \pi \sqrt[4]{ \frac{2 m}{c v_0^2} }$ and final solution
$$ x = \sqrt[4]{\frac{2 m v_0^2}{c}}\,\sin \left( \frac{2 \pi t}{\tau} \right) $$
| {
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when we rub objects together, what determines which material will pick up electrons? For example We know glass when rubbed by silk will become positively charged while the silk will be charged negative.
What exactly makes glass appropriate for losing electrons in that experiment? (
| Each object has a certain value that defines whether it will lose electrons when rubbing against an object with a higher value, and vice-versa. Glass can loose electrons much more easily than other objects, making it a useful tool for teaching about static electricity. Electrons are held in place by the positive balance of protons.
"different materials have different affinities for electrons. By rubbing a variety of materials against each other and testing their resulting interaction with objects of known charge, the tested materials can be ordered according to their affinity for electrons. Such an ordering of substances is known as a triboelectric series. One such ordering for several materials is shown in the table at the right. Materials shown highest on the table tend to have a greater affinity for electrons than those below it. Subsequently, when any two materials in the table are rubbed together, the one that is higher can be expected to pull electrons from the material that is lower. As such, the materials highest on the table will have the greatest tendency to acquire the negative charge. Those below it would become positively charged."
Links:
http://www.physicsclassroom.com/class/estatics/u8l2a.cfm
| {
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Distribution of point charges on a line of finite length How will $N$ freely moving charges confined to a line with length $L$ be distributed? What are their equilibrium positions?
| This is a more down-to-earth answer as opposed to the fancy mathematics in the other one. This problem is easily solved numerically. The equations are easily stated: inverse-square forces to the right from the particles to the left and to the left from the particles to the right. Thus, for a system of $n+2$ charges where the first and last are fixed at $x=0$ and $x=L$ by external forces (to prevent the whole system from flying apart), the positions $x_1,\ldots,x_n$ of the middle particles obey
$$
\frac1{x_i^2}+\sum_{j=1}^{i-1}\frac1{(x_i-x_j)^2}-\sum_{j=i+1}^n\frac1{(x_i-x_j)^2}-\frac1{(x_i-L)^2}=0,\ \ i=1,\ldots,n.
$$
This can be solved numerically to give the positions. This is a plot of the particle positions for $n$ from 1 through 50:
Mathematica code available on request.
The distribution is relatively close to uniform but it is not so exactly. This is evident from the slight bunching up at the ends, and can be seen more clearly in the probability density for the positions at large $n$:
I should note that this sort of distribution is more or less what you'd expect from the zeros of a large-order orthogonal polynomial, though the Jacobi case is even more bunched up at the edges:
(Though actually the two histograms are so different, and the Coulomb one is so uniform, I'm starting to doubt whether there does exist a Calogero electrostatic model for an inverse-distance interaction potential.)
| {
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$\pi/2$ phase shift in RC circuit I just recently learnt Alternating Current and RCL circuits and I also learnt about phasers and various graphs. Now I also know that current lags in an LR circuit. My intuitive understanding of this is according to Lenz Law, the inductor will oppose the current and hence slow it down. So the current lags. But I wanted to know if there is any similar logic for a RC circuit. The capacitor doesn't seem to make the current lead by a phase difference of $\pi/2$. I have already studied the mathematical proof and everything works, but I wanted to have an intuitive understanding of both.
| When a resistance $R$ is put in series with some other impedance $Z$, by Ohms law you have $U=(R+Z)\cdot I$, or
$$I=U\cdot \tfrac{1}{R+Z}=U\cdot \tfrac{1}{R+Z}\tfrac{\overline{R+Z}}{\overline{R+Z}}=U\cdot \tfrac{(R+\mathbb{Re}(Z))-i\ \mathbb{Im}(Z)}{|R+Z|^2}.$$
If $Z$ has an imaginary part, then $I$ and $U$ differ by some phase. And yes, this is the case wether $Z$ comes from an inductance $Z_L$ or a capacitance $Z_C$.
On a macroscopic level, the inductance of an AC system is given by the impedance $Z_L=i\omega L$, where $i$ is the imaginary unit and both the frequency $\omega$ as well the inductance $L$ is some real, so $Z_L$ is purely imaginary and multiplication by $i$ corresponds to a rotation in the complex plane by $90°$ or $\tfrac{\pi}{2}$.
A capacitance gives the impedance $Z_C=\frac{1}{i\omega C}$, which you can re-write as $Z_C=-i\omega\left(\frac{1}{\omega^2 C}\right)$. So for a given input voltage frequency $\omega$, and a given circuit element with inductance $L$, you can design a capacitor circuit element with $Z_C=-Z_L$. Same goes in reverse- So these two passive impedances work similarly, and there is a phase shift too, albeit in the other rotating direction.
The behaviour of these circuit elements can be understood by considering the voltage law for circuits, which expresses energy conservation. If energy is stored by one circuit element (voltage drop at a resistor), it must go missing somewhere else. The different distribution of energy of an circuit with or without complex element make a difference in the current flow. The inductance voltage is, more microsopically motivated, given via $U_L=L\frac{\mathrm d}{\mathrm dt}I$ (relates to Lenz law), and described the reactive behaviour of the circuit element to current. The capacitance stores charges from the current and the relation is $U_C=\frac{1}{C}\int^t I \mathrm dt$. These two equations are the origin of $Z_L=i\omega L$ and $Z_C=\frac{1}{i\omega C}$ if you consider in a sinodial like current.
| {
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On Einstein's equivalence principles There are two foundative Equivalence Principles in General relativity:
Weak Equivalence Principle (WEP): the dynamics of a test particle in a gravitational field is independent of its mass and internal composition.
(WEP is equivalent to say that the ratio between the gravitational mass $m_g$ and the inertial mass $m_i$ has a universal value $k$. This value can be considered as the scalar $1$).
Einstein equivalence principle (EEP): A frame linearly accelerated relative to an inertial frame in special relativity is LOCALLY identical to a frame at rest in a gravitational field.
Most textbooks say that "obviously" (EEP) implies (WEP) but the converse is not true.
I don't understand why the implication (EEP) $\Rightarrow$ (WEP) is true, and moreover I'd like a counterexample showing that (WEP) $\not\Rightarrow$ (EEP).
| I agree that in the textbook you considered, explanation is not clear. First, please check my answer here:
graviton and principle of equivalence
Personally, I prefer to unify WEP and EEP, but let's distinguish them as in your textbook. The example you provided for WEP is definitely mechanical, so not-gravitational experiment, hence the implication (EEP) ⇒ (WEP) indeed is obvious. As counter example I would suggest an experiment with photon. For instance the Doppler effect, being a consequence of special relativity is a part of general relativity. Being an experiment on photons, it does not tell you anything about mass of particles.
| {
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Open problem? Square of the wave function $\Psi(x)_{x_o} = \delta(x-x_0)$ of a particle localized at a point $x_0$? Does anybody know the status of the problem to define the wave function (non-relativistic Quantum Mechanics) of a particle localized at a definite point?
Landau-Lifshitz says in chapter 1 that this function is $\Psi(x)_{x_o} = \delta(x-x_0)$ and gives an explanation that it produces the correct probability density when it is used to span some other arbitrary wave function $\Psi(x)$. The problem is of course that the wave function given above squares to a non integrable function. As far as I know this problem is unsolved. My question is if anybody knows the status quo of this problem. I am sorry if this question may be duplicated, I could not find it amongst the answered questions.
| Most scientists agree there are still some interpretational issues in QM, so it is hard to make unequivocal statements. IMO, point-like single quantum pure states (i.e., Hilbert space "rays") are not physically measurable or even physically realizable. They are idealizations of zero entropy and thus not realizable by the Nernst statement of the third law of thermodynamics. No physical object property can manifest existence in a spacetime interval of zero extent, e.g., no object can manifest a real-numbered position, not even its CG. This means every realizable quantum state is a mixed state of multiple concurrently existing incoherent pure states. (They are incoherent superpositions, otherwise they would be expressible as a coherent sum of pure states, i.e., a single component, zero entropy, pure state.) This, however, casts doubt on the probability interpretation of QM. The appropriate vertical metric that should be applied to the magnitude squared of a realizable quantum mixed state distribution must be a physical, or ontic, metric, not probability. So OP, real-numbered (point-like) eigenvalues are not physically realizable, all manifest physical property values are distributions, not real numbers.
| {
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Entropy of the Sun
*
*Is it possible to measure or calculate the total entropy of the Sun?
*Assuming it changes over time, what are its current first and second derivatives w.r.t. time?
*What is our prediction on its asymptotic behavior (barring possible collisions with other bodies)?
| The entropy of the sun is roughly $10^{35} \mathrm{J} /\mathrm{K}$ (see http://dx.doi.org/10.1103/PhysRevD.7.2333 where the entropy of the sun is given as $10^{42}\mathrm{erg}°\mathrm{K}^{-1}$)
It can be calculated using Boltzmanns law ( $S = k_B \ln{W} )$ where $W$ is the disorder parameter of the sun (depends on the number of atoms in the sun).
I did not get you second question.
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How does humidity affect the path of a bullet? Background
Last night, I was reading the FM 23-10 (The U.S. army official field manual for sniper training), and I've noticed that they're potentially teaching snipers incorrect information.
Generally speaking, when we say "impact goes up" it means that the bullet was either somehow made faster or its path was easier, therefore the curve in its ballistic trajectory is smoother. Thus, it will hit higher. When we say impact goes down, we mean the opposite.
For example, atmospheric heat will, loosely speaking, make the air "thinner" and therefore the impact will be higher. Cold weather will do the opposite. This part is correct.
What about humidity?
The FM 23-10 says:
The sniper can encounter problems if drastic humidity changes occur in
his area of operation. Remember, if humidity goes up, impact goes
down; if humidity goes down, impact goes up.
They're basically saying that when humidity goes up, then the bullet's travel will be more difficult-> steeper trajectory curve -> lower point of impact.
However, as far as I know, dry air is denser than humid air because air has higher molecular mass than water vapour. In humid air water vapour replaces other gases, thus bringing the whole density down. So, the point of impact should be higher with higher humidity.
So my question is:
All other factors being equal, does humid air pose less resistance to the bullet making the point of impact higher than in dry air?
| Today I have noticed that our humidity is extremely high compared to yesterday. The temp is about approx 5degrees F warmer today than yesterday. I just shot 30 rounds and noticed without any bias on this subject (no previous opinion) that my rounds were landing .125" higher than yesterdays 30 rounds. Hence why I am researching this subject of humidity and the impact on subsonic rounds. So for what it is worth, my experience has been humidity higher, POI higher. Then again, it could just be me.
| {
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How does a pressure suit work? I recently learnt that a suit called pressure suit is worn by fighter plane pilots to prevent red-outs and black-outs. And it seems to be work by -
"..applying pressure to selective portions of the body."
How do these suits work; i.e. by what means, selective portions of the body are pressurised?
Do astronauts wear these while takeoffs, and also F1 drivers?
| If the body is subjected to high accelerations directed towards the feet the heart is unable to pump blood to the brain and unconsciouness results. There are no end of TV programmes showing such experiments in centrifuges, for example a quick Youtube search found this.
The type of pressure suit you mention compresses the legs and forces the blood in them upwards. This makes it easy for the heart to maintain the circulation and allows the pilot to remain conscious to higher forces.
| {
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Why does a laser beam diverge? I was wondering why a laser beam diverges. If all the photons are in the same direction, I would imagine that it would stay that way over a long distance. I am aware that a perfectly collimated beam with no divergence cannot be created due to diffraction, but I am looking for an explanation based on photons rather than wave physics.
| Electromagnetic waves are diffracted, so a plane wave can only exist at a single location along the axis of propagation (in a uniform homogeneous medium) . In a semiconductor laser, the end mirrors might be planar crystal faces; but they aren't always; for example they aren't in VCSELs; where Bragg mirrors are often used.
Smaller source diameters lead to larger diffraction angles, which depend on the ratio of source diameter and wavelength, so semiconductor lasers can have very large beam angles.
A cavity resonator with parallel end mirrors is unstable, so is a poor choice for a laser. In practice, there is a physical "gain medium" that the waves are propagating in in the resonator, and inhomogeneities in that medium will render the effective cavity not parallel; particularly in semiconductor lasers, where impurity doping, will render the refractive index non uniform.
| {
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Does a different opposing force affect work? Suppose a man exerts $10~N$ as he lifts a $1~kg$ box a distance of $2~m$ against Earth's gravity.
To determine work we can use the following equation:
$$
W = F \cdot d \\
W = (10~N) \cdot (2~m) = 20~J
$$
The work in this case is $20~J$.
Would work be the same if the man performed this task on the moon rather than the Earth?
Mathematically, the equation shows that if the man exerts the same force ($10~N$) over the same distance ($2~m$), then the work will remain the same ($20~J$) -- but I'm having trouble conceptualizing this.
$$
MoonGravity < EarthGravity
$$
The force opposing our movement when lifting the box on the moon would be less than that on Earth since the gravity on the moon is far smaller. Conceptually, it seems that it'll be "easier" to lift the box on the moon, and thus should take less work.
Where is the fault in my logic?
| the fault is in the assumption that work done on the box only goes towards gravitational potential energy. take note that if the applied force exceeds the weight of the box at any time, the box will accelerate and also attain kinetic energy.
assuming $g_{earth}$ = $10N/kg$, then the box never accelerates, and the full $ 20J$ work done goes into increases the box's potential energy.
assuming $g_{moon}$ = $1.6N/kg$, then by $W_{PE}\approx mgh$, $W_{PE}= 3.2J$. the "missing" $16.8J$ goes towards kinetic energy (this box is moving very fast by the time it reaches the top).
I'll leave it to you to confirm that the KE is indeed 16.8J. (hint $v^2=2as$, where $a=\frac{(10N-1.6N)}{1kg} $and $s=2m$
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Determing time to complete known distance with constant acceleration Answer: $a=v^2/d$ is the formula I needed.
This is a problem in a programming assignment, I haven't taken physics.
I have a starting speed (0 m/s), an final speed (208.33m/s) and the distance it took to reach that speed (200km). I need to get the time it will take to travel any arbitrary distance lower than 200km.
From what I remember in highschool, I used basic calculus to get $a/2(t^2)=distance$ or for what I have $t=\sqrt(400000(m/s)/a)$. what I don't know is the acceleration, usually to calculate acceleration I'd need time. I'm not sure where to go from here.
| According to the question while the car travels from 0 to 200 m/s with constant acceleration it covers a total distance of 200km.
Applying the formula $$2as=v^2-u^2$$ where v is the final speed,u is the initial speed and s is the distance. Here u=0m/s,v=208.3m/s and s=200km.Putting the values we get $$a=0.1085034 m/s$$
Finally using $$s=ut+1/2at^2$$ where s is the distance covered, u is the initial velocity,a is the acceleration and t is the time, we get the distance(in m) covered at any time t(in seconds).
Here u=0m/s. We get $s=1/2at^2$. Given any s<200km you can now find the time it takes to cover that distance.
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Is it possible to find the ground state of generalized Ising models? Is there a general solver (or a theoretical algorithm) for obtaining the ground state configuration of the extended Ising model, which involves an arbitrary lattice, arbitrary coordination number (i.e. $n$-body interactions for arbitrary $n$), arbitrary Hamiltonian, and arbitrary spin of each particle (-1,0,1 and so forth)? Except exact enumeration, what could we do better?
I know such a requirement is too stringent. But is there some research or established knowledge that aims to accomplish this?
| As in your question the stress was on the word general, I have some bad news:
an efficient "general solver (or a theoretical algorithm) for (...) extended Ising models, which involves an arbitrary lattice" does not exists.
Of course, one can invent algorithms that, in principle, could find the ground state. The most trivial would be checking the energy of all configurations (I guess this is what you referred to by exact enumeration). However, the time used by this algorithm would scale exponentially with the number of sites. One may ask - as you did - whether one can do something substantially better, e.g., finding an algorithm where the "used time" scales polynomially with the number of sites. Unfortunately, there is no such algorithm (if we assume that NP$\ne$P). Already for the ordinary Ising Hamiltonian
$$H= \sum_{ab} J_{ab}S_aS_b $$
with $J_{ab}\in\{+1,-1,0\}$ and with connectivity on an $L\times L\times 2$ cubic lattice, it was shown that finding the ground state is an NP hard problem. The proof of this can be found here:
F. Barahona, On the computational complexity of Ising spin glass models, J. Phys. A: Math. Gen., 15 3241 (1982).
Of course, if you don't consider the general case, but restrict your attention to a set of "easier" lattices or graphs (e.g. to planar graphs), then there could be a polynomial time algorithms (depending on the structure of the specific restricted cases).
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Deriving the condtion for spontaneity using gibbs free energy While deriving the condition for spontaneity, $\Delta$G$\leqslant$0,
we start by saying that
$\Delta S_{tot}$ $\ge$0 $\Rightarrow$ $\Delta S_{sys}$ + $\Delta S_{surr}$ $\ge0$
If $Q$ is the heat transferred to the system from the surroundings, then $−Q$ is the heat lost by the surroundings, so that $\Delta S_{ext}$ = - ${Q \over T}$, corresponds to the entropy change of the surroundings.
$\Delta S_{int}$ - ${Q \over T}$ $\ge 0 $,
But, isn't $\Delta S_{int}$ = ${Q \over T}$ , since $Q$ is the heat gained by the system?
Consequently, wouldn't we always get $\Delta S_{int}$ - ${Q \over T}$ = ${Q \over T}$- ${Q \over T}$=0 ?
| There is an asymmetry between the system and the surroundings. In fact, at constant temperature $T$, we are in a canonical or grand-canonical situation (depending on cases), where the "surrondings" is a heat reservoir, with Energy, Number of Particles,etc... much greater than the system. When we modify some external data (ex : volume), the set (sytem + heat reservoir) is evolving until it reaches a thermodynamic equilibrium, and this is done by a positive variation of the total entropy (sytem + heat reservoir) and a diminution of the (Helmholtz, Gibbs) free energy of the system.
The heat reservoir is so big, than we may consider it approximatively in thermodynamic equilibrium, during this evolution, with a temperature T,so, if we consider the heat gained by the heat reservoir $Q_R$, the variation of entropy of the heat reservoir is approximately $\Delta S_R = \frac{Q_R}{T}$.
On the other hand, during the evolution, the system itself cannot be considered in thermal equilibrium, so the notion of temperature is not well defined, so we cannot express a variation of entropy of the system using variation of heat and temperature like $\Delta S = \int dS$ with $dS = \frac{\delta Q}{T}$.
For instance, for simplicity, consider Helmotz free energy, and consider that no work is exchanged between the system and the reservoir. The conservation of energy is written :
$$\Delta U + \Delta U_R=0 \tag{1}$$
where $U$ and $U_R$ are the internal energies of the system and the heat reservoir.
We have :
$$ \Delta U_R = \Delta Q_R = T \Delta S_R\tag{2}$$
The first equality comes from the hypothesis of no work exchange, and the second equality comes from the approximated thermodynamic equilibrium of the heat reservoir.
The total entropy (system + heat reservoir) is increasing during the evolution, so we have :
$$\Delta S + \Delta {S_R} \ge 0 \tag{3}$$
Now, from $(1)$ and $(2)$, we get : $\Delta {S_R} = - \frac{\Delta U}{T}$, so, finally : $\frac{1}{T} (T\Delta S - \Delta U) >0$, and noting the Helmoltz free energy $F = U- TS$, we may write :
$$\Delta F \le 0 \tag{4}$$
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Is it possible for the entropy in an isolated system to decrease? As far as I can tell, the concept of entropy is a purely statistical one. In my engineering thermodynamics course we were told that the second law of Thermodynamics states that "the entropy of an isolated system never decreases". However, this doesn't make much sense to me.
By counter-example: Consider a gas-filled isolated system where the gas has maximum entropy (it is at equilibrium). Since the molecular motion is considered to be random, at some point in the future there will be a pressure gradient formed by pure chance. At this point in time, entropy has decreased.
According to Wikipedia, the second law purely states that systems tend toward thermodynamic equilibrium which makes sense. I then ask a) is the second law as we were taught it wrong (in general), and b) what is the use of entropy (as a mathematical value) if it's effectively an arbitrary definition (i.e. what implications can we draw from knowing the change in entropy of a system)?
Thanks in advance for your help.
| If you consider the case of different configurations of the gas in question will get a number of microstates. With them you can define an equivalence relation that form a macrostate. All this microstates share the general characteristics of the system (as the gas pressure exerted on the walls of the container).
These macrostates can be assigned a probability of occurrence based on the microstates that comprise it. We determine that there is some physical quantity that allows us to define nature's preference for certain states over others and called entropy.
In this way we are not talking about whether the system may or may not be in a particular state, but what is the probability of that happening. Importantly entropy allows us to know what will be the state of the system may not need to develop mechanical equations for each of the particles. Additionally does not tell us how long it will happen, only that a state is more likely to occur will have a higher entropy value.
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Circumference of a circular path Suppose a car moves with a constant speed of $20 \text{m/s}$ a quarter of a circle, and completes the quarter in $5$ seconds. One way to calculate the circumference is simply $20 \cdot 5 \cdot 4 = 400 \text{m}$. However, I know that $a=\frac{v^2}{R}$ and the circumference is $2 \pi R$. The acceleration is defined as $\frac{\Delta v}{\Delta t}$ and thus from Pythagorean theorem we get $a = \frac{\sqrt{20^2+20^2}}{5}=4 \sqrt{2} ~~ \text{m/s}^2$. Now to find the circumference we plug our result into $2 \pi \frac{v^2}{a}$, thus the circumference: $2 \pi \frac{20^2}{4 \sqrt{2}} \approx 444.29 \text{m}$ which is clearly wrong. So why the second way gave me a wrong result?
| Acceleration is the derivative of velocity with respect to time, which is not the same as just dividing the difference in velocity by the difference in time.
$$a=\frac{dv}{dt}\neq \frac{\Delta v}{\Delta t}$$
$dv$ and $dt$ can be thought of as infinitely small $\Delta v$ and $\Delta t$. So if your $\Delta v$ and $\Delta t$ are quite small, the result of your calculation will be pretty close to the actual acceleration. But in your case, $\Delta v$ and $\Delta t$ are pretty big. Therefore, your calculation is way of. We can see that our result becomes more accurate if we consider an eighth of a circle:
$$\frac{\Delta v}{\Delta t}=\frac{ \sqrt{(v_{x0}-v_{x1})^2 + (v_{y0}-v_{y1})^2} }{5/2}$$
Now we plug in $v_{x0}=20$, $v_{y0}=0$ and $v_{x1}=v_{y1}=20\cdot\sqrt{\frac{1}{2}}$.
$$\frac{\Delta v}{\Delta t}=2\frac{ \sqrt{20^2(1-\sqrt{\frac{1}{2}})^2 + 20^2(0-\sqrt{\frac{1}{2}})^2} }{5}=40\frac{ \sqrt{1-2\sqrt{\frac{1}{2}} + \frac{1}{2} + \frac{1}{2}} }{5}=8\sqrt{2-\sqrt{2}} \approx a$$
And we use
$$a=\frac{v^2}{R} \Rightarrow R =\frac{v^2}{a}$$
$$C=2\pi R = 2\pi \frac{v^2}{a} \approx 2\pi \frac{20^2}{8\sqrt{2-\sqrt{2}}} \approx 410.46\space m$$
As you can see, this is much closer to the real value of the circumference. This is because we chose a smaller part of the circle, and because of that a smaller $\Delta v$ and $\Delta t$. If you pick smaller and smaller parts of the circle, your value will get closer and closer to the real value. And in the limit of choosing a part of the circle with length 0, it will be the exact value. This is exactly what the '$d$' in $\frac{dv}{dt}$ means.
You can also try choosing a larger part of the circle, and you ranswer will come out a lot worse. In the worst case, if you use the entire circle, you will find that $\Delta v=0$, and therefore $a=0$, and conclude that the circle has an infinite circumference.
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Do orbitals overlap? Yes, as the title states: Do orbitals overlap ?
I mean, if I take a look at this figure...
I see the distribution in different orbitals. So if for example I take the S orbitals, they are all just a sphere. So wont the 2S orbital overlap with the 1S overlap, making the electrons in each orbital "meet" at some point?
Or have I misunderstood something?
| An orbital is essentially a wave function from which a probability distribution of the location of an electron upon measurement can be inferred. What is depicted will be something like the region within which the probability is 50% (shaped in a way that depends on a decomposition of the state function in a radial part and an angular part).
If you mean to ask if these regions overlap, yes, they certainly do. If you mean to ask if the regions of space where the probabilities are non-zero overlap, they even more certainly do, as the probability is non-zero almost everywhere (i.e. zero on set of volume 0).
If you mean to ask if they (or rather the spaces they span) in the state space overlap, then no: see Programming Enthusiast's answer.
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Why does voltage remains same over Parallel Circuit Why does voltage remains same over parallel circuit. If a resistor is connected in the circuit some of the charge should be transformed into heat and make a lack of charge after the resistor (in my sense). So, what's the reason in it?
| Two points:
1) The voltage across the two resistors is the same. The definition of a perfect conductor is that there is no voltage drop along it's course. Of course nothing is perfect but copper is a very good conductor and it would be difficult to measure the voltage drop across relative short segments. In a parallel circuit the conductors on either side of the multiple resistive elements maintain nearly identical voltages. It is the current that gets divided. As others have pointed out, the charge is a conserved quantity (since it's just the numbers of electrons) so it gets "divided" between the various resistive elements and then gets recombined back at the other side so I_out is the same as I_in at the current source.
In the diagram offered by another poster, Waffle's Crazy Peanut, there is a "wire" from the start of R1 to the start of R2. In circuit theory that section of "wire" has no resistance and it's equivalent to both resistors being connected to the same point which implies that they have the same voltage at their positive sides.
2) The heat is coming from the drop in energy as the electrons cross the resistors. Since the amount of drop is the same for each passing electron, the amount of heat is proportional to the current (or number of electrons) that pass. Electrical engineers sometimes talk about across quantities and through quantities. Voltage is an across quantity while charge is a through quantity. You raise the issue of heat which is a measure of power generation. Heat from each resistor in a parallel circuit will be calculated as the product of the voltage (the same for each) and current (different and inversely proportional to the resistances.)
You should think of voltage as pressure and the charge as the fluid being pushed forward. If you turn off the faucet attached to a garden hose (decreasing pressure or by analogy the voltage) , the water doesn't decrease in amount; rather only the flow of charge (current) decreases.
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Vanishing of the Ricci tensor in higher spacetime dimensions I understand how, if the Riemann tensor is 0 in all its components, since we construct the Ricci tensor by contracting the Riemann, Ricci tensor would be 0 in all components as well.
I've read that vanishing of the Ricci tensor in 3 spacetime dimensions implies the vanishing of the Riemann curvature tensor, but that in higher dimensions that does not hold.
Can somebody explain why is that so? Is it because we have more independent components of the Riemann tensor in 4 spacetime dimension, than in 3 (20 vs 6)?
Also if the number of independent components of Riemann tensor in $n$ spacetime dimensions is
$$N(n)=\frac{n^2(n^2-1)}{12}$$
and since we know that the Riemann tensor has 256 components, does that limit the spacetime dimensions of it's usage? Or does that mean, that for example in 10 spacetime dimensions, there won't be any independent components of Riemann tensor?
| So, let's take your formula, and set $n=3$. This gives you $N = \frac{9\cdot 8}{12} = 6$. Well, how many independent components of the Ricci tensor do you have? Well, since it's a 3x3 symmetric tensor, you've got six independent components. Therefore, there is no room in the Riemann tensor to have additional components. Since, for $n > 3$, you will always have fewer components of the ricci tensor than the riemann tensor (figure out what the fomula for independent components of a symmetric $n\times n$ matrix), for higher dimensions, there will always be additional extra components.
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Why 3 dipole terms in a multipole expansion? As can be seen on this page http://en.wikipedia.org/wiki/Multipole_expansion when we take a multipole expansion without assuming azimuthal symmetry we end up with $2l+1$ coefficients for the $l^{th}$ moment in the expansion. So the dipole moment has 3 terms, the quadrupole has 5 and so on. This is different to the case of azimuthal symmetry as each we need only one co-efficient for each term.
Interpreting 3 coefficients for the dipole moment isn't too bad. I'm guessing it represents the dipole moments along the 3 Cartesian axes? And how do we interpret having to have $2l+1$ coefficients for each term?
| The spherical multipole expansion arises from the solution of the Laplace equation in spherical coordinates. We try to solve by separation of variables and a eigenvalue equation appears .
The number 3 in the second term is caused by the degeneracy of certain eigenvalue (the are 3 linearly independent solutions for the same eigenvalue). The same argument is valid for higher terms.
A happy coincidence allow us to identify the potential of the dipole (a pair of charges of opposite sign with a little separation) in the second term but for higher terms it isn't so easy.
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Why is moment dependent on the distance from the point of rotation to the force? The formula for moment is:
$$M = Fd$$
Where F is the force applied on the object and d is the perpendicular distance from the point of rotation to the line of action of the force.
Why? Intuitively, it makes sense that moment is dependent on force since the force "increases the intensity". But why distance? Why does the distance from the line of action of the force to the point of intensity affect the moment?
I am NOT looking for a derivation of the above formula from the cross product formula, I am looking for intuition. I understand how when I am turning a wrench, if the wrench is shorter its harder to turn it but I don't understand WHY.
Thanks.
|
I understand how when I am turning a wrench, if the wrench is shorter its harder to turn it but I don't understand WHY.
Suppose a bolt can be unscrewed with one turn, and the process consumes $E$ Joules. Then since $w=F d$, we have
$$E=2\pi rF.$$
Thus
$$F=\frac{E}{2\pi r}.$$
That's why it's harder to unscrew a bolt using a short wrench. You need to push harder. Does this answer your question?
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Quantum randomness and brownian motion in biological systems, e.g., fertilization I am looking for examples of physical indeterminacy impacting the macroscopic world. By physical indeterminacy, I mean physical sources of randomness such as quantum indeterminacy or brownian motion.
One example is of particular interest here: whether such randomness influences which sperm cell fertilizes a particular ovum, or whether such biological systems are too large to be affected by random perturbations, either at the atomic level (quantum) or molecular level (brownian motion).
| Some kinds of mutation provide an example of this kind of indeterminacy. UV light can be bad for our health. One of the reasons is that, when we are exposed to sunlight, UVB photons are absorbed by double bonds in pyrimidines, which break open, become reactive, and dimerize (photo-dimerization). This damages the DNA in the same way that it would damage a zipper if 2 adjacent teeth were fused together. The cell responds to this kind of damage with a repair process that may lead to mutation. Eventually the effect of the mutation will be manifested macroscopically. The causal sequence is complex, but it starts with a photon, which has an indeterminate path due to quantum indeterminacy. It is indeterminate whether a photon will hit a particular site in the genome.
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Why do same/opposite electric charges repel/attract each other, respectively? I know plus pushes another plus away, but why, really, do they do that? On the other hand, molecules of the same type are attracted to each other. I find that weird.
I do know some stuff about four universal forces. But why in general the general "rule" is that opposite charges pull each other?
Yes, I do realize this could be connected to very basic stuff that science is still trying to figure out, and can be traced to the Higgs, but still, there must be something to tell.
| It's all about the energy, in the sense that everything is positive or negative energy. Opposite charges attract each other in order to complement the lack or surplus of energy.
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Is the uniqueness theorem correct in superconductivity? There is an uniqueness theorem in electromagnetism. It says that the solution of Maxwell's Equations is determined uniquely by boundary conditions.
We can treat superconductivity as a completely diamagnetic material with magnetic susceptibility $\chi=-1$.
I think the uniqueness theorem is still correct in this case, but I am confused very much.
If we think about a superconducting ring, and we know the magnetic field at infinite distance, the boundary condition at infinity, we can not be sure about the magnetic field. We don't know whether the ring have flux.
If we cool down the material before adding the magnetic field, the ring will have a zero flux.
If we add the magnetic field before cooling down the ring, the ring will have a non-zero flux.
Does it mean that the uniqueness theorem is not correct? The solution is not determined by state of the system only, but is determined by history and state of the system.
Could you answer my question? Thank you very much!
| I think you have missed a step in the uniqueness theorem. After all, the fields in a ferromagnet are governed by non-superconducting E&M, but internal magnetization depends on the history and state of a system.
The uniqueness theorem applies to regions of space where the boundary fields and all charges and currents are specified. However in a material it's possible to have quite complex bound currents --- especially in a superconductor.
| {
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Are quantum decoherence and Everettian approaches to the measurement problem necessarily distinct? As I understand it, there is a large contingent of physicists who believe that the measurement problem is "solved" by decoherence, without, for example, needing to postulate the existence of "many worlds." Yet at the same time my understanding is that in the decoherence picture there is only unitary evolution of the wave function, and that while the appearance of collapse is explained, the global superposition of states still in fact exists, and whether or not multiple states within the universal wave function observe the same appearance of collapse (but to different eigenvalues) is a question that is left completely unaddressed.
Therefore my reading of the decoherence picture is that it is virtually identical to an Everettian approach, except that it purposefully ignores an obvious interpretational consequence of its description. Is this true, or do decoherence-based approaches somehow argue that there really is only a unique observer within the universal wave function that observes a collapse to unique eigenvalues, and that there is some form of symmetry breaking that allows this to happen at the expense of all the other potentially conscious components of the universal wave function?
| Decoherence and collapse/MWI are actually complementary, since they address completely different aspects of the classical limit. Decoherence explains what happens with interference terms on macroscopic systems, but it doesn't address the problem of individual measurements. The classical probability distribution of a mixed state describe what happens in a statistically large sample of measurements. Collapse and MWI both address what happens on a specific instance of measurement.
Individual measurements are fundamentally non-unitary transformations relative to the observer. They make losses of quantum state and information that are irrecoverable even in principle. We've know this for almost a century, but there are still a majority of physicists that are assuming that measurements can be explained as a complex interaction of purely unitary operations.
| {
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Do objects rotate around the torque vector or its center? If a sphere has a torque vector coming out of it at point A, would the sphere rotate about its center or the axis of the torque vector?
| If a body moves only because of the influence of a torque, then it will rotate about the center of mass.
There is no location for torques, only directions. You you take the equations of motion as seen here (https://physics.stackexchange.com/a/80449/392) you will see that the location of the torque does not enter into the equations. Only the location of the forces.
As a result the acceleration of the center of mass is zero, and only angular velocity will exist. The body will rotate about its center of mass.
Note that these two statements are equivalent:
*
*A pure force thorugh the center of mass (with no net torque about the center of mass) will purely translate a rigid body (any point on the body).
*A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of mass.
Consider a motionless rigid body with a pure instantenous torque $\vec{\tau}$ applied on it. The motion of any point A not on the center of mass is
$$ 0 = m \vec{a}_A - m \vec{c}\times \vec{\alpha}
\\ \vec{\tau} = I_c \vec{\alpha} - m \vec{c} \times \vec{c} \times \vec{\alpha} + m \vec{c} \times \vec{a}_A $$
where $\vec{c}$ the position vector of the center of gravity relative to point A. The soltution to the above is
$$ \vec{a}_A = \vec{c} \times \vec{\alpha}
\\ \vec{\tau} = I_c \vec{\alpha}- m \vec{c} \times \vec{c} \times \vec{\alpha}+ m \vec{c}
\times \vec{c} \times \vec{\alpha} = I_c \vec{\alpha} $$
$$ \vec{\alpha} = I_c^{-1} \vec{\tau} \\ \vec{a}_A = \vec{c} \times I_c^{-1} \vec{\tau} $$
From the above is it obvious that the only point A not moving is at $\vec{c}=0$ and all points parallel to $ \vec{\alpha}$ through the center of mass.
| {
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Why can't batteries be charged quickly? Charging of laptops, cell phones take so much time. Why can't we make such batteries easily/commercially which are charged more quickly? What's the thing behind this limiting?
| Heat. Batteries have internal resistance and so produced heat when current flows through them (Joule heating). Also, the heat generated increases by the square of that current. E.g, doubling the charging current causes the heat produced to be increased 4 times.
Ultracapacitors are a different technology that can be used like batteries--they have very very low internal resistance and so can be charged or recharged at currents hundreds of times what you see in regular batteries like laptop ones. The tradeoff is that capacitors have really low capacity in comparison (they run out too quickly and must be recharged again).
| {
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Why does Newton's third law exist even in non-inertial reference frames? While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?
|
While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?
Interesting interpretation. I would put it exactly the other way around: in a noninertial frame, the first and second laws hold, but the third law doesn't.
Let's say we're in a rotating frame, and in that frame, a baseball experiences a centrifugal force. There is no third-law partner for this force: the baseball doesn't create a force back on any other object. This is because the centrifugal force is not an interaction between two objects, so we can't have the third-law pattern of A on B, B on A.
On the other hand, the first and second laws certainly apply to the baseball, provided that we include the centrifugal and Coriolis forces as forces. These fictitious forces also obey the law of vector addition, which is a fundamental law of Newtonian mechanics, although not traditionally considered one of Newton's laws.
I suppose the opposite interpretation, as given in the question, occurs if you refuse to consider fictitious forces as forces. Then they don't violate Newton's third law, because they're not forces. (Dogs can't violate the law against murder, because the law only applies to people, and dog's are not considered people.) The first and second laws are then violated, because we refuse to put in the inertial forces that would have been needed in order to make them work.
| {
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What is the opposite of the Planck length? What "large size" unit of length could be considered at the opposite end of spectrum from Planck's length?
Is there a table of smallest and largest value for various physical quantities that can be defined from well-known constants?
Edit
I was teaching the exponential function and scientific notation to kids and I was looking for example of physical quantities that occur on vastly different scales. Length is the easiest and there are some demos as in The Scale of Universe. As the size of universe seems to be a function of time I wondered about other large lengths.
| Graham's number is the only practicable answer I can think of to answer your question.
| {
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do lenses with curved focal planes exist I know about spherical, aspherical, cylindrical lenses, but are there lenses that could have a curved focal plane? for example, the retina of the eye is curved, and still we see sharp images of distant objects.
In a recent problem I studied, I wanted to have a lens that would transform a flat object (a book cover, for example) into a curved image. That is, each point of my book cover would be imaged correctly (i.e. the image would be sharp enough) by my lens - only, on a curved screen, like a spherical screen for example.
Would there be a specific form of lens that could do exactly this, the same way that aspheric lenses do "perfect" images of distant objects on flat screens, better than any spherical lens?
I am asking this because, in my application, the screen will need to be strongly curved, and I still want the best possible stigmatism in these conditions, without reducing the aperture too much...
| Well ALL lenses have curved focal surfaces; there are NONE that have perfectly flat sharp focal planes over any finite angular field of view.
It is however theoretically possible to design a lens with a flat focal surface, but only for imaging one specific object plane onto one specific image plane.
The problem is a factor called the "Petzval curvature." which is a curved surface on which the image will be formed IF it is free from Astigmatism. Ordinary, uncorrected Astigmatism, will make the image field even more curved, than the Petzval surface. It is a common trick to deliberately incorporate a certain amount of overcorrected astigmatism, in order to produce a flatter best image plane; but the images will be somewhat fuzzy.
The Petzval curvature (sum) is dependent ONLY on the curvatures of the surfaces, and the refractive index of the media. It is unaffected by thicknesses or the order of the surfaces.
| {
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How can doped semiconductor be neutral? I have studied about the two types of doping which result in p and n type semiconductors. I also came to know that they are neutral. But, how can it be?
Is it that the positive charge(holes) in p-type and negative charge in n-type are negligibly small to affect the overall neutrality of the substance? But, in that case with very few holes or negative charges how can semiconductors work the way it actually does?
So, I think there is another possible explanation and I would like to know the correct one.
| An atom contains not only the electrons but also the nucleus which consists of an equal number of protons. Hence an atom is neutral.
The reason why your doped semiconductor carries a neutral charge is because it has equal number of electrons as there are protons, be it boron doped or phosphorous doped.
While the whole crystal remains neutral, by doping you are vastly increasing the conductivity of the semiconductor. And hence, the specialty of semiconductors and doping.
| {
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Expansion of the Universe: Conversion of gravitational potential energy to kinetic energy? Suppose there is an object floating in space which over time begins to fall toward the source of a gravitational field. As it falls, its motion happens to be such that it gets locked in orbit around the source with a greater velocity than it had before it 'began to fall'. So it's gravitational potential has been converted to kinetic energy. According to relativity, this increased speed should increase the gravitational potential of the object (and therefore the object + the original source of the field), correct?
Does this mean that as a result of this, the expansion of the Universe should slow down slightly (because there is now a slightly greater gravitational potential in the Universe)? And if so, would that imply that the conversion from gravitational potential to kinetic energy is in a sense a conversion between the bulk kinetic energy of the expanding Universe and the local kinetic energy of a test mass?
| Energy is not conserved in GR, in the sense that there is no global, tensorial measure of energy that is conserved and can be defined in all spacetimes. See this question: Total energy of the Universe
However, if the system described in your example is sufficiently isolated, then we can describe that region of space as asymptotically flat, and we have various conserved measures of energy such as the ADM energy. This energy remains constant during the process you describe, so there is no effect on the broader cosmos.
| {
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Why are lab Magnets painted red? Wy are lab magnets painted red in color?
I tried searching everywhere but couldn't get a satisfactory answer.
| Isn't it just a convention ? And are all the laboratory magnets around the worle painted red ?
I would agree with you that mostly they are painted red as i once bought a lot of laboratory bar magnets made of AlNiCo all of them were painted red while same was not true when I bought some higher power loadstone magnets they were simply in their original state (no paints and blackish) electromagnets are not painted at all as well. Some extremely good permanent magnets I don't remember what they were made of, but they were round and very powerful were finished in silver.
Maybe it is a convention which makes nearly all the permanent lab magnets red, but I can assure you that not all of them are red.
| {
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2 dimensional Coulomb's law equation We can notice that in the Coulomb's law equation,
$$\begin{equation}\tag{1}F=\frac{1}{4\pi\epsilon}\cdot\frac{q_1q_2}{r^2}\end{equation} $$
$4\pi r^2$ factor in the denominator expresses directly the surface of a virtual sphere with radius $r$. Actually we can look at this equation as it was for $3$ dimensional objects. If we suppose want to consider for $2$ dimensional objects, can we modify the equation as,
$$\begin{equation}\tag{2}F=\frac{1}{2\pi\epsilon}\cdot\frac{q_1q_2}{r}\end{equation}$$
Here we can think of $2\pi r$ as area of virtual circle. I don't really know whether it works or not. So, can we have equation (2) as the modified equation for electrostatic force between two $2$ dimensional uniformly charged objects?
| What is 2D charge? It is not a good definition, and you should be more carefully. The force will be proportional to the charge or the square of charge? 2D charge in the vortex is the topological charge? The thing in 2D makes different.
| {
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Can a very thin sheet of any material float inside a liquid? When a body is immersed in a liquid,buoyancy is the net force of all the forces acting on it.
Now the forces are equivalent to those which will act on the same volume of liquid.Rightly so,but considering the chaotic motion of the molecules of the liquid,for every force acting on that particular volume of liquid,there will be another force in the opposite direction but of different magnitude.The magnitude differs because of the difference in height.
Now let's consider a very very thin sheet immersed in the liquid.The opposite forces now will be of the same magnitude as difference in height is negligible and hence will cancel each other!so the only force acting will be its weight.
So the thin sheet will never be balanced!So does that imply that no thin sheet will ever float inside a liquid??
| If the density of the material of your sheet is higher than the density of the liquid, then, the situation will be unstable, and it will collapse immediately by gravitational forces.
The process you are describing is basically just diffusion. If your material is also a fluid, and basically dissolved in the fluid, then the sheet, however thin, will always diffuse away.
If your material is a separate liquid, then surface tension will probably immediately break up the sheet. (If there is no surface tension, then it is dissolve and diffusion will take over).
If your material is a solid, then force by the random motion of water molecules (I don't want to call this diffusion, but it is probably somehow related) will tend to deform the solid sheet. You will then come into the field of solid mechanics. I would think that elasticity and bulk parameters come into play, which for infinitely thin sheets can not withstand the forces by water.
| {
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Is there a heat transfer equation that takes conduction, convection, radiation and dT/dt into account? I see equations that take 2 or 3 of the listed parameters into account but I haven't been able to find one that is that complete. I'm seeking to solve this equation using matlab for a simulation project to study how the heat dissipates on a microheater given an electrical power or current as input and see how quickly the temperature rises, and so forth. But first, I need the proper equation.
| The general form of the heat equation would be
$$ \frac{\partial T}{\partial t} + (\vec{u} \cdot \nabla) T =a \nabla^2T + S$$
The first term is the time derivative, the second term convection, the third term diffusion of heat and the last term is the source term, which can be anything.
However, normally I would expect radiation to come into the equation via the boundary conditions as some kind of heat flux, and not as a source term.
Radiation is often described by the Stefan-Bolzman law, e.g.
$$ \text{heat flux} \propto T^4 $$
which is a boundary condition.
In terms of modelling, one normally linearized this term, e.g $4T_0^3 (T-T_0) $.
| {
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Have we found a Higgsino? In supersymmetry, for each particle (boson/fermion), there is a symmetric particle which is a fermion/boson.
The MSSM predicts five Higgs bosons: two neutral scalar ones (H and h), a pseudo-scalar (A) and two charged scalars.
Does this mean that there are five higgsinos?
My question arises because I read:
What is the difference between the SM Higgs and the supersymmetric H and h?
Nothing.
So I'm confused, is it saying that H and h are the supersymmetrical parters of the Higss boson?
| The Higgs field of the standard model is a complex SU(2) doublet. This means it has two complex-number-valued components which interact with the weak-force gauge bosons. Two complex-number-valued degrees of freedom, equals four real-valued degrees of freedom. Three of these degrees of freedom are absorbed into W+,W-,Z bosons (giving them each a third, spin-0 polarization, and thus making them massive rather than massless), and the fourth real degree of freedom is "the Higgs boson". It is the left-over part of the Higgs field.
In the minimal supersymmetric standard model, there are two Higgs superfields (they differ from each other by having opposite "hypercharge"). One will give mass to up-type quarks, the other to down-type quarks and to the charged leptons. In the standard model, it's the same unique Higgs field which gives mass to all these fermions, but supersymmetry forbids certain couplings, so the MSSM is more complicated.
Each Higgs superfield has a bosonic part that is a scalar field with two complex-number-valued components, as above, and a fermionic part consisting of two two-component spinors. This fermionic part is the higgsino.
If we add up everything, the MSSM then has two higgsinos and eight real degrees of freedom in the bosonic part of its Higgs superfields. As before, three of the real degrees of freedom are absorbed by W+,W-,Z bosons, but now there are five real degrees of freedom left over, and these are the five Higgs bosons of the MSSM. They are conventionally denoted h, H, A, H+, H-, and have various charges and symmetry properties. h and H are neutral and "CP-even" like the lone Higgs boson of the nonsupersymmetric standard model. That is the sense in which they are the same as it.
You can read about all this in "A Supersymmetric Primer".
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A question on Bernoulli's principle Which is more appropriate regarding Bernoulli's principle
*
*fast moving air causes low pressure or
*lower pressure causes fast moving air.
| Definitely the second, assuming that what you meant to say was "Relatively low pressure causes fast moving air." Bernhard is absolutely correct on this point, in that really it's the pressure gradient, not the pressure itself which causes local accelerations in the flow. The language I am using is very deliberate, suggesting that the pressure gradient causes the flow accelerations and not the other way round. After all, Newton's Second Law tells us what an object will do when a net force is applied to it, not what forces will be generated when it just begins to move apropos of nothing. Forces cause accelerations, and so analogously the pressure gradients are what cause the local flow accelerations (although the phenomena are dynamically coupled).
| {
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Explicit time dependence of the Lagrangian and Energy Conservation Why is energy (or in more general terms,the Hamiltonian) not conserved when the Lagrangian has an explicit time dependence?
I know that we can derive the identity:
$\frac{d \mathcal{H}}{d t} = - {\partial \mathcal{L}\over \partial t}$
but is there a more physical and intuitive explanation for the conservation?
| I) If a system (and hence the Lagrangian) depends explicitly on time, it can often be physically interpreted as that the system is not an isolated system. In other words, that it interacts with the environment, and hence there is no reason that the energy of the system should be conserved.
Example: A non-relativistic 1D point particle with Lagrangian
$$\tag{1} L~=~\frac{1}{2}m \dot{q}^2 -V(q,t),$$
where the potential
$$\tag{2} V(q,t)~=~V_0(t) - F(t) q +\frac{1}{2}k(t)q^2$$
is quadratic in $q$ and the three coefficients $V_0(t)$, $- F(t)$ and $k(t)$ depend explicitly on time. We may interpret:
*
*$V_0(t)$ as a fluctuating (choice of) zero-point energy. This is a total derivative term in the Lagrangian, and does not affect the equations of motion.
*$F(t)$ as an external force.
*$k(t)$ as a changing spring constant.
II) On the other hand, if the Lagrangian does not depend explicitly on time, so that time translation is a symmetry of the Lagrangian, then Noether's (first) theorem yields that the corresponding Noether charge (=the energy) is conserved.
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How important is mathematical proof in physics? How important are proofs in physics? If something is mathematically proven to follow from something we know is true, does it still require experimental verification? Are there examples of things that have been mathematically proven to some reasonable degree of rigor (eg satisfy a mathematician) that turned out to be false based on experiment?
| Mathematical proofs relate to exactly how a MODEL will behave. They don't have much to do with how the real world behaves. If the mathematics is carried out correctly, then one has "proven" how the model will behave.
The reason for experimentation, is to find out if the completely fictional MODEL that somebody simply made up, behaves in any way the same, as observations suggest the real world behaves. If it doesn't, that is not an indication of a mathematical error, it simply means the fictional model is not a good description of the real world, and must be changed.
| {
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Catapult vs. Trebuchet I have been looking at trebuchet designs lately, and I have noticed that most, if not all, have a sling attached to them. Without such a sling, the machine would be a catapult. In terms of the speed and energy of a launched projectile, what is the general difference between a catapult and trebuchet? Would trebuchet projectiles have extra centripetal acceleration due to the sling?
| Well ultimately the range depends on launch velocity. This is a combination of acceleration (applied force) and time or distance (during acceleration.)
The trebuchet seems to give a much longer "casting stroke" , so reaches a greater launch velocity. Very clever gadgets.
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Lorentz transformation of the Spinor Field I'm reading chapter 3 of Peskin and Schroeder and am stuck on page 43 of P&S. They have defined the Lorentz generators in the spinor representation as:
\begin{equation}
S^{\mu \nu} = \frac{i}{4}[\gamma^\mu,\gamma^\nu]
\end{equation}
such that a finite transformation is given by:
\begin{equation}
\Lambda_{1/2}=e^{-\frac{i}{2} \omega_{\mu \nu} S^{\mu \nu}}
\end{equation}
where $\gamma^\mu$ are the gamma matrices and $\omega_{\mu \nu}$ are the elements of a real and antisymmetric matrix. According to P&S on page 43 (between equation (3.32) and (3.33), they say that the adjoint of a Dirac spinor transforms as follows:
\begin{equation}
\psi^\dagger \rightarrow \psi^\dagger \left(1+\frac{i}{2} \omega_{\mu \nu}(S^{\mu \nu})^\dagger \right)
\end{equation}
However, I would expect the transformation to be:
\begin{equation}
\begin{aligned}
\psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\&
= \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\&
= \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right)
\end{aligned}
\end{equation}
where in the last line I made use of the fact that $\omega$ is a real and antisymmetric matrix:
\begin{equation}
(\omega_{\mu \nu})^\dagger = (\omega_{\mu \nu})^T = \omega_{\nu \mu} = - \omega_{\mu \nu}
\end{equation}
This implies that according to my calculations, equation (3.33) of P&S should actually be:
\begin{equation}
\overline{\psi} \rightarrow \overline{\psi} \Lambda_{1/2}
\end{equation}
This equation must be wrong because it means that $\overline{\psi} \psi$ does not transform as a scalar and therefore the Dirac Lagrangian is not correct. However, I do not know where my mistake is and was hoping someone could help me out?
| The mistake you are making is in "daggering" the object $\omega_{\mu\nu}$. For each $\mu, \nu = 0,\dots 3$, the symbol $\omega_{\mu\nu}$ is a real number, so its dagger (which is really just complex conjugation in this case) does nothing; $(\omega_{\mu\nu})^\dagger = \omega_{\mu\nu}$.
When we say that $\omega_{\mu\nu}$ is an antisymmetric real matrix, we really mean that the matrix with these numbers as components is such a matrix, not that $\omega_{\mu\nu}$ is a matrix for each $\mu$ and $\nu$.
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"url": "https://physics.stackexchange.com/questions/82704",
"timestamp": "2023-03-29T00:00:00",
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Could velocity be taken as fundamental instead of time? In physics time and length are taken as fundamental in the SI system and, as it seems, in the thinking of physicists. Could one instead take velocity, with c as its unit, together with length as fundamental and then understand time by dimensional analysis in terms of l/v (length divided by velocity)? If not, why not?
| From the point of view of dimensional analysis alone, this can be done. It would not give mathematically inconsistent results. Mathematically, any quantities $A$, $B$, $C$ can be chosen as 'fundamental', as long as no relationship of the form $f(A, B, C) = 0$ exists between them.
However, from the point of view of physics itself, such an approach needs to be questioned. In physics, we consider certain concepts to be primitive - meaning that they cannot be explained in terms of simpler ideas. Space and time are amongst such ideas. Other concepts are derived - meaning that they can be explained or defined in terms of existing ideas.
If we consider time and length as fundamental quantities, we should also be able to provide a simple prescription for measuring them. This can be easily done for time and length - for eg. - as the no. of oscillations of a particular pendulum, or as the no. of multiples of some unit length. In the case of velocity however, it would be quite difficult, or perhaps impossible to provide such a prescription which does not involve using the ideas of space or time. Hence, there does not appear to be a strong case for considering velocity as 'fundamental'. It does appear that certain quantities are really more fundamental than others, it does not appear to be just a conventional element.
| {
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What is the depth in meters of the pond? A small spherical gas bubble of diameter $d= 4$ μm forms at the bottom of a pond. When the bubble rises to the surface its diameter is $n=1.1$ times bigger. What is the depth in meters of the pond?
Note: water's surface tension and density are $σ= 73 \times 10^{-3} \mbox{ N}$ and $ρ= 10^3 \mbox{ kg/m}^3$, respectively. The gas expansion is assumed to be isothermal.
My attempts:
I used the equation of pressure:
$P_1V_1=P_2V_2$
where $P_1$ is the pressure at the top and $V_1$ is the volume of bubble at the top and $P_2$ is the pressure at the bottom, and $V_2$ is the volume of the bubble at the bottom
Because the bubble at the bottom, it received the hydrostatic pressure, so the equation became:
$P_1V_1$= $(P_0+\rho g d) V_2$
Since $V_1$ and $V_2$ is sphere, we can use the sphere volume. And $P_1$ is same with atmospheric pressure= $10^5 \mbox{ Pa}$
$10^5 \cdot (\frac{4}{3} \pi r_1^3) = (10^5 + 10^3 \cdot 9.8 \cdot d) \cdot (\frac{4}{3} \pi r_2^3)$
Cancel out the $\frac{4}{3}\pi$ and we get:
$10^5 \cdot (r_1)^3 = (r_2)^3 \cdot (10^5 + 9800d)$
Substitute $r_1 = 4 \times 1.1= 4.4 \mbox{ μm}= 4.4\times10^{-6} \mbox{ m}$ and $r_2 = 4 \times 10^{-6} \mbox{ m}$
$10^5 \cdot (4.4 \times 10^{-6})^3 = (4 \times 10^{-6})^3 \cdot (10^5 + 9800d)$
$10^{-13} \cdot (4.4)^3 = 64 \times 10^{-18} \times 10^5 + 64 \times 10^{-18} \times 9800d$
$85.184 \times 10^{-13} = 64 \times 10^{-13} + 627200 \times 10^{-18} d$
$21.184 \times 10^{-13} = 627200 \times 10^{-18} d$
$d= 3.37 \mbox{ m}$
So, the depth of the pond is $3.37 \mbox{ m}$. My question: what is the useful of $σ = 73 \times 10^{-3} \mbox{ N}$ ?
I really confused about it. Thanks
| The surface tension of the bubble would help you to find out how much work the bubble is doing while it increases its volume.
| {
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Do atoms get created or are they recycled? Basically, are the atoms that make up my body right now something that has existed since the big bang?
| Watch out, though, because particles can be indistinguishable. Roughly, you can't write a label on an atom, even in principle, so to say that your hydrogen atoms are this old and my hydrogen atoms are that old is not very well defined. What is closer to reality is that there are this many atoms in this state, and that many atoms in that state, and which is which is none of your business :)
This is easier to understand if you think of elementary particles (quarks, electrons, etc.) as being excited states of a single "thing" called a field. There is only one electron field, and, depending on what quantum state that field is in, there exist a number of electrons, all of which are identical. You can't claim ownership of a particular electron any more than you can claim ownership of the letters in your name.
| {
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Dimensional regularization - integral How can I derive the following formula?
$$\int d^{d+1} k \frac{e^{i K X}}{K^2} = \frac{\Gamma (d-1)}{(4\pi)^{d/2} \Gamma (d/2) |X|^{d-1}}, \quad K^2 = k_0^2 + \vec k^2, KX = k_0 \tau + \vec k \vec x$$
What I tried so far:
*
*Integrate over $k_0$ while promoting $k_0$ to complex variable, so I can use residue theorem. Then I move to n-spherical coordinates. I obtain:
$$ \propto {\rm sign} (\tau) \int d \Omega \int dr \, r^{d-3} e^{-r |\tau| + ir |\vec x| \cos \phi_1}$$
And I'm stuck.
*I change variables to n-spherical ones and I end up with this:
$$\int d \Omega \int dr \; r^{d-2} e^{ir |\vec x| \cos \phi_1}$$
And I'm stuck as well.
| The second route should work, look up the form of $d \Omega$ and note that it has a $d \cos \phi$ term that appears in it. From there, a quick change of variables and some research on gamma functions and solid angle integrals and you should be home free.
| {
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Is the uncertainty principle valid? The uncertainty principle says that the product of the uncertainties in position and momentum can be no smaller than a simple fraction of Planck's constant $h$.
Several articles lately suggest this is not true.
Today in Physicsworld.com
Looking at the position and momentum of spin-polarized neutrons, they found that, as Ozawa predicted, error and disturbance still involve a trade-off but with a product that can be smaller than Heisenberg's limit.
and earlier work in Toronto, Canada
Aephraim Steinberg and colleagues at the University of Toronto conducted an optical test of Ozawa's relationship, which also seemed to confirm his prediction. Ozawa has since collaborated with researchers at Tohoku University in another optical study, with the same result.
Can we safely throw out the Heisenberg uncertainty principle and start to view physics with more certainty?
| "Can we safely throw out the Heisenberg uncertainty principle and start to view physics with more certainty?"
No, you definitely can not throw out the HUP. If you have two operators $A$ and $B$, and define $\Delta A = A - \langle A \rangle $, it is easy to prove that (see, e.g., Sakurai's QM book, section 1.4)
\begin{equation}
\langle (\Delta A)^2\rangle \langle (\Delta B)^2\rangle \geq
\frac{1}{4} | \langle [A,B] \rangle |^2,
\end{equation}
which is just a more general form of the HUP. If this inequality was wrong, it would mean that the whole formulation of QM in terms of vectors and operators in a Hilbert space is also wrong, which seems extremely unlikely.
The uncertainty in position and momentum is part of the quantum mechanical description of matter.
The articles in your links are probably talking about the so-called Heisenberg's error-disturbance uncertainty relation (that applies to measurements and is not a property of matter), which is not the same as the fundamental uncertainty relation in the equation above. Also note that the uncertainties in position and momentum are still a fraction of $h$ in all the works cited in those links; the fraction is simply smaller than Heisenberg initially thought.
| {
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Does antimatter curve spacetime in the opposite direction as matter? According to the Dirac equation, antimatter is the negative energy solution to the following relation:
$$E^2 = p^2 c^2 + m^2 c^4.$$
And according to general relativity, the Einstein tensor (which roughly represents the curvature of spacetime) is linearly dependent on (and I assume would then have the same mathematical sign as) the stress-energy tensor:
$$G_{\mu \nu} = \frac{8 \pi G}{c^4}T_{\mu \nu}.$$
For antimatter, the sign of the stress-energy tensor would change, as the sign of the energy changes. Would this change the sign of the Einstein tensor, causing spacetime to be curved in the opposite direction as it would be curved if normal matter with positive energy were in its place? Or does adding in the cosmological constant change things here?
| Here's a naive argument for expecting antimatter and matter to both have "attractive" properties under gravity. General relativity describes gravity in terms of a valence 2 tensor $g_{\mu\nu}$. Upon quantization one would therefore expect a spin $2$ particle. The propagator might look something like
$$(g_{\mu\rho}g_{\nu\sigma}+g_{\mu\sigma}g_{\nu\rho})\frac{i}{q^2+i\epsilon}$$
which in the nonrelativistic limit yields a universally attractive potential by comparing low energy scattering with the QM Born approximation. For more details see Peskin and Schroeder page 126.
| {
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Are there models/simulations of antigravitational antimatter-galaxies? In the comments to another question's answer, I started wondering:
Assuming antimatter possessed negative gravitational mass§ (which is not proven impossible to date, though deemed unlikely), basically entire galaxies consisting of antimatter could form. Due to the repulsion of ordinary matter, they would rarely collide with usual ("pro-matter-")galaxies% and actually drift apart, maybe even partly contributing to inflation. But I don't want to speculate too much, so my questions is
Have entire galaxies consisting of negative-gravitational-mass antimatter been considered in any sound theory, model or simulation?
§ I.e. while such antimatter gravitationally repulsed conventional matter, it would still attract other antimatter.
% Unless there exists e.g. a (currently unknown) force that attracts antiparticles to their exact counterparts
| There are strong constraints on antigravitating antimatter, because it could, in principle, be used, to create a perpetual motion machine.
1) Use energy $E$ to create a particle/antiparticle pair at height $h_{i}$
2) Raise the particle/antiparticle pair to a height $h_{f}$. This takes zero work, because the antiparticle will be pushed up in the potential and the particle will be pushed down, and the forces will be equal, since they are a particle/antiparticle pair.
3) annhilate the particle/antiparticle pair, getting the energy $E$ back.
4) direct the resultant $N$ photons downward to the start point. During the trip down, they will be blueshifted by a factor $\sqrt{\frac{h_{i}\left(c^{2}h_{f}-2GM\right)}{h_{f}\left(c^{2}h_{i}-2GM\right)}}$
5) Now, you have photons that have $N\hbar f_{0}\left(\sqrt{\frac{h_{i}\left(c^{2}h_{f}-2GM\right)}{h_{f}\left(c^{2}h_{i}-2GM\right)}}-1\right)$ more energy than when you started, and you can use the remnant to restart the cycle.
| {
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Normal force of ball sliding on concave surface Imagine a ball is sliding along a surface shaped like $y=x^2$. Like ,
but please ignore the fact that the center of the ball is on the surface instead of the edge.
When the ball is stationary, I can calculate the normal force fairly easily. $F_\text{net}$
in the direction normal to the surface is 0. The normal force is normal to the surface, so it is the force that is exactly strong enough to cancel the normal component of the force of gravity.
The normal force comes out to $<\frac{-a}{(1 + a^2)}, \frac{1}{(1 + a^2)}>$, where $a=f_{\text{surface}}'(x)$ (or in this specific case: $a=2x$)
I initially expected this to be true even after the ball begins sliding down the surface, but I found that it was not. Intuitively, I understand that a velocity with a non-zero component in the normal direction affects the strength of the normal force. This seems like the same mechanism by which the normal force is able to stop a falling object as it hits the ground, for example.
In the falling ball example, the relationship between velocity and normal force seems a bit sticky mathematically. Assuming neither the ground nor the object compress, the normal force would have to be infinite initially, right? It would only be infinite for an infinitesimal amount of time (so the integral would still be finite). However, in that case, the $v(t)$ curve is discontinuous, while it seems like $v(t)$ would be continuous for a ball sliding on $y=x^2$, so hopefully the relationship is less obnoxious in the continuous case.
Is there a mathematical formulation of this relation between normal force and velocity? Can I add a velocity parameter to my equation for $F_\text{net}(x)$ (making it $F_\text{net}(x, v)$) to take this relationship into account?
Note: I also see a conservation of energy solution that sidesteps the normal force issue altogether, but I'd like to understand the forces aspect.
| When a particle follows a path then at any instant there is tangential vector $\vec{e}$ and a normal vector $\vec{n}$ allowing for the velocity to be defined as $$\vec{v} = v \vec{e}$$ and the acceleration $$ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n} $$ where $\rho$ is the radius of curvature of the path.
For a path defined by a function $f(x)$ with derivatives $f'(x)$ and $f''(x)$ the radius of curvature is
$$ \rho(x) = \frac{\left(1+{f'(x)}^2\right)^\frac{3}{2}}{f''(x)} $$
and the direction vectors are
$$ \begin{aligned}
\vec{e}(x) &= \begin{pmatrix}
\frac{1}{\sqrt{1+{f'(x)}^2}} \\
\frac{f'(x)}{\sqrt{1+{f'(x)}^2}}
\end{pmatrix} & \vec{n}(x) &= \begin{pmatrix}
- \frac{f'(x)}{\sqrt{1+{f'(x)}^2}} \\
\frac{1}{\sqrt{1+{f'(x)}^2}}
\end{pmatrix} \end{aligned} $$
The last part for you, is the fact that gravity only affects the tangential acceleration $\dot{v} = \vec{g} \cdot \vec{e}$ (with $\cdot$ the vector dot product)
$$ \dot{v} = -\frac{g f'(x)}{\sqrt{1+{f'(x)}^2}} $$
with total acceleration
$$ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n}
\\\vec{a} = \begin{pmatrix} - \frac{g f'}{1+f'^2} - \frac{v^2 f' f''}{\left(1+f'^2\right)^2} \\
- \frac{g f'^2}{1+f'^2} + \frac{v^2 f''}{\left(1+f'^2\right)^2} \end{pmatrix} $$
The total force applied is $\vec{F} = m \vec{a}$ and the reaction force is the component normal to motion or $$\boxed{R = \vec{n} \cdot \vec{F} = \frac{m v^2}{\rho} = \frac{m v^2 f''}{\left( 1 + f'^2 \right)^\frac{3}{2}}}$$
note: this area of study is called differential geometry (Frenet formulas). Look it up.
| {
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How does energy transfer between B and E in an EM standing wave? I'm trying to understand how an electric field induces a magnetic field and vice versa, its associated energy, as well as relating it to my understanding of waves on a string.
Using a standing wave as an example, I came up with the equations
$\vec{E}=A\sin(\omega t)\sin(kx)\hat{y}$
$\vec{B}=\frac{Ak}{w}\cos(\omega t)\cos(kx)\hat{z}$
I checked them against Maxwell's equations, and they're self-consistent. At time 0, this reduces to:
$\vec{B}=\frac{Ak}{w}\cos(kx)\hat{z}$
Since the electric field is 0, based on the Poynting vector, there's no energy transfer at this time. At this time, at a node where $\vec{B}=0$, there's neither electric field nor magnetic field. If there's no energy transfer, and no energy stored in either field, then how can an electric field exist at this point at some time later? How is the energy stored, or transferred from elsewhere?
| Ah, but there is energy stored in the field: recall that the energy density is given by
$$
u(t,x)=\frac12\left(\varepsilon_0\vec{E}\cdot\vec{E}+\frac{1}{\mu_0}\vec{B}\cdot\vec{B}\right)
$$
which is clearly non-zero at $t=0$. Surely there are zeroes in the energy at $t=0$, but for $t>0$, the energy is transferred from the nearby non-zero fields! In fact, that's exactly what the Poynting vector tells you: how much energy is transfered per unit area in a particular direction.
| {
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Does the moon affect the Earth's climate? So, this morning I was talking to a friend about astronomical observations, and he told me that lately there has only been good weather when there was a full moon in the sky, which was a shame.
I jokingly said: 'maybe there's a correlation!', but then I started thinking: wait, if the moon can affect the oceans, why shouldn't it also make an impact on the atmosphere, which is just another fluid. So... are there atmospheric tides? Does the moon affect the weather or the climate in a significant way?
| The moon DOES affect weather but not significantly.
Just like it affects ocean tides, it affects the atmosphere in a similiar way. When the moon is full or new for example, it creates a "bulge" in an ocean, which is why we have tides. A similar thing happens with the atmosphere; it attracts the atmosphere to itself.
When the Sun, moon and Earth line up, and the moon is at its perigee (closest to the Earth), you can expect lower temperatures. This is due to the combined gravitational pull of the moon and the Sun.
Nobody is completely sure how the moon affects weather exactly, but it doesn't affect it significantly.
| {
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B3LYP vs PBE functionals for conjugated organic systems Two of the most popular (exchange and correlation) functionals for density functional theory are B3LYP and PBE. Out of the people I've worked with / learned from, mostly the computational chemists were using B3LYP, and the physicists were using PBE.
Now, different functionals have different strengths and weaknesses, but these two functionals seem to have a lot of overlap in their applicability. Right now I'm interested in ground-state properties of molecules with extended pi-frameworks in the gas-phase. For example, I'd like to compute the geometries, ground-state electron densities, and molecular polarizabilities of the acenes.
Does anyone have concrete information or references on the performance of B3LYP or PBE in this situation, or in general?
| For static polarizability calculations it seems that both B3LYP and PBE functionals do a pretty good job; for benzene and napthalene I am getting numbers within a few percent of experimental values. What is much more important is the basis set. In particular, it's absolutely necessary to include diffuse functions. For benzene it's so extreme that AUG-cc-pVDZ is closer to basis-set convergence than cc-pVTZ, cc-pVQZ, or even cc-pV5Z.
| {
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Is false vacuum bubble nucleation possible in our universe?
*
*Is it possible that a false vacuum bubble to nucleate into our universe rather than a true vacuum one?
*If yes, it will expand at speed of light within our spacetime or what?
| If there is another metastable point below the energy our universe lives in but above the true vacuum level then yes (a nice description on page 3 here). The information within it would travel at the speed of light and any photons that may be within but I am unsure if it might experience an inflationary period or not.
| {
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Classical Wave Equation - Approximations I don't understand the derivation of the wave equation given below -
$$T \sin (\theta _1) - T \sin (\theta ) = T\tan (\theta _1 )-T\tan (\theta ) = T \left. \left(\frac{\partial f}{\partial z} \right|_{z + \Delta z} - \left. \frac{\partial f}{\partial z}\right| _z \right) = T \frac{\partial ^2 f}{\partial z^2} \Delta z$$
I understand that the small angle approximation was used, but I'm at a loss for figuring out we turned $\tan$ into a derivative, and then after made it become a second derivative.
The derivative of $\tan \theta$ is of course $\sec \theta$ which is equal to $\frac{1}{cos \theta}$, which was taken with respect to $\theta$, maybe there's a way to use the chain rule to find $\partial _z f$?
| As we can see from the previous answers, we have the following approximations:
$tan(\theta) = \frac{opposite}{adjacent} = \frac {rise}{run} = gradient = \frac {∂y}{∂x}$
and we can say, still through approximation:
slope(z+∆z)=slope(z)+slope_variation_tax*∆z, which mathematically means:
$$\frac{\partial f(z+∆z)}{\partial z} = \frac{\partial f(z)}{\partial z}+\frac{\partial }{\partial z}(\frac{\partial f(z)}{\partial z}) \partial z=\frac{\partial f(z)}{\partial z}+\frac{\partial ^2 f}{\partial z^2} \partial z$$
Then you have, after substitution:
$$T\frac{\partial f(z+∆z)}{\partial z}-T\frac{\partial f(z)}{\partial z}=T\frac{\partial f(z)}{\partial z}+T\frac{\partial ^2 f}{\partial z^2} \partial z-T\frac{\partial f(z)}{\partial z}=T\frac{\partial ^2 f}{\partial z^2} \partial z$$
And that's the last member of your equation.
Hope I helped ;)
| {
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How much extra distance to an event horizon? How much extra distance would I have to travel through space to get from Earth to a stellar mass event horizon?
(compared to the same point in space without a black hole)
| I suspect you're not asking the question you're really interested in, because the answer to your question is really boring. If you jump into a black hole you'll see the event horizon retreating before you, and you'll never cross it. The distance you've travelled is an ambiguous quantity since of course in your frame you're stationary and have travelled no distance at all. The time you take to pass the distance $r=r_s$ then hit the singularity is finite, and for stellar mass black holes very short.
A far more interesting question is if you hover outside the horizon and let down a tape measure how long would it have to be to reach the horizon i.e. what do you get by integrating $dr$ in a radial direction towards the event horizon? The Schwarzschild metric is:
$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 d\Omega^2 $$
Lets suppose we hover at a distance $r_1$ from the singularity and let down a tape measure to measure the distance to some point at a radial distance of $r_2$. Because $dt$ and $d\Omega$ are constant the equation for the line element simplifies to:
$$ ds = \frac{dr}{\left(1-\frac{r_s}{r}\right)^{1/2}} $$
To get the length of the tape we just need to integrate this expression from $r_1$ to $r_2$:
$$\begin{align}
s &= \int_{r_2}^{r_1} \frac{dr}{\left(1-\frac{r_s}{r}\right)^{1/2}} \\
&= \int_{r_2}^{r_1} \frac{r^{1/2}dr}{\left(r-r_s\right)^{1/2}}
\end{align}$$
To integrate this we cheat and look up the answer in a GR book, the result being:
$$ s = \left[ z \sqrt{z ^2 - r_s} + r_s \ln \left( z + \sqrt{z ^2 - r_s} \right) \right]_{z_2}^{z_1} $$
where we've used the substitution $r = z^2$ to make the integral manageable.
To make this concrete lets take a black hole with the mass of the Sun, so $r_s$ = 2954m, and we'll start from 5km out i.e. $r_1 = 5000$. Let's graph the length of the tape as a function of $r_2$:
The magenta line is the Newtonian result, i.e. if space was flat, and the blue line is what we actually measure. The tape measure distance from $r = 5km to the event horizon is about 4,780m compared to the Newtonian calculation of 2046m.
So the affect of the curvature is to make the distance measured radially greater than $r_1 - r_2$. However I must emphasise that this is not what you'd measure if I threw you into the black hole. This is the distance measured by an observer hovering far from the event horizon.
| {
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Chinook Helicopter Torque The Chinook Helicopter has 2 rotors to counteract the torque generated by spinning the blade.
Theoretically, could you use a smaller "back" rotor that is farther away from the main rotor to achieve the same result, ie no twisting?
| The separation between the rotors does not actually matter. What matters is that the torque exerted by each of the motors on the respective rotor be the same and in opposite directions. Those torques then add vectorially and cancel out to give zero net torque on the helicopter.
It's not clear to me exactly what you mean by "small". It is indeed possible to have a rotor with shorter blades, but it will need to be driven faster to provide the same torque. This will typically be associated with a smaller power consumption and less lift coming from that rotor.
In general, though, it is the symmetry between the rotors which makes the configuration work. While in practice they won't be exactly the same, they will be very similar to each other, and straying far from this configuration will be very impractical unless you have some stringent requirements on your 'secondary' motor, as well as radical new engines for it to match.
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One way insulation? I know from basic physics lessons that a box painted black will absorb heat better than a box covered in tin foil. However a box covered in tin foil will lose heat slower than a black box.
So what is the best way to conserve the temperature of a box? (aiming for 0 degrees Celsius inside the box when it's -60 outside).
I mean would painting the outside of the box black, and having tin foil on the inside work? So the box can absorb heat better (black paint) and the tin foil making it harder for heat to escape?
| If your box (at 0-20°C) starts out hotter than the environment (at -60°C) then your best strategy is to prevent any heat flowing out of the box into the environment i.e. insulate the box.
Using foil will reduce radiative energy transfer, however in most cases the cooling is dominated by convection rather than radiation and foil is a rather good conductor of heat. You can demonstrate this by wrapping yourself (at about 37°C) in foil and standing in a -60°C wind (though I wouldn't do this experiment for very long). Mind you, painting yourself black would also do little to keep you warm when you're standing in a -60°C wind.
However, suppose it's a clear winter day and the Sun is shining brightly. In that case painting the box black would help because it would increase the absorption of energy from the Sun.
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Electric flux for a rectangular surface? I have the following homework problem:
A line of charge $\lambda$ is located on the z-axis. Determine the
electric flux for a rectangular surface with corners at coordinates:
$(0, R, 0)$, $(w, R, 0)$, $(0,R, L)$, and $(w, R, L)$.
This is what I have come up with so far:
The line of charge, is located on the $z$-axis. We can recall that $\Phi = \int_{S}\vec{E}~\mathrm{d}A$. We initially note that this is parallel to the $xz$-plane, ergo we will integrate in respect to $x$ and $z$. Due to the fact that this is an infinite line of charge, there is no change in the field as we vary the distance along the $z$-axis. Our integral is $$\int_0^L\int_0^w\vec{E}~\mathrm{d}x\mathrm{d}z$$ We can recall that $$E=\frac{\lambda}{2\pi\varepsilon_0r}$$ We can see that $r=\sqrt{x^2+R^2}$ by the Pythagorean Theorem. By substitution we have the following integral: $$\int_0^L\int_0^w\frac{\lambda}{2\pi\varepsilon_0\sqrt{x^2+R^2}}\mathrm{d}x\mathrm{d}z = \frac{\lambda L}{2\pi\varepsilon_0}\int_0^w\frac{1}{\sqrt{x^2+R^2}}\mathrm{d}x$$
When I solve this I get:
$$\Phi=\frac{\lambda L}{2\pi\varepsilon_0}\sinh^{-1}\left(\frac{w}{R}\right)$$
I am not sure where I am going wrong. I may be doing something conceptually incorrect.
| I think the integral would simply be $Φ=(λ/L2πε0)\sin^{−1}(w/R)$ and not $\sinh^{-1}$.
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What is fundamental difference between wave and its 180 flip phase? I'm studying property of sound wave and I was wondering what is difference between two waves (one is original and one is 180 flip phase of original) ?
Amplitude and frequency remains same and also wavelength is same, so are they same?? I could not detect any difference from hearing two sounds. If different what is different and can human/computer detect it? What is basically difference between two waves?
| If you play them simultaneously then they will destructively interfere provided they arrive at the listener at the same time (no additional phase difference). This is the basis of noise cancellation.
| {
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Uncertainty principle in atomic clocks? How does the uncertainty principle limit the accuracy of atomic clocks. I know line width and measurement time are important but not exactly why?
| A caesium clock generates a microwave signal that it tunes to match the absorption peak in the Caesium spectrum. A counter counts every oscillation of the generated microwave, and every 9,192,631,770 counts = 1 second. That's how the clock counts the seconds and keeps time.
But if the peak in the caesium spectrum is broad it's difficult to tune the microwave generator to exactly the peak maximum. That means there is a potential error in the microwave frequency and hence in the measured time. This will make the clock run slow or fast.
So the narrowness of the peak is very important to keeping accurate time. When I last looked the limitation on the peak width was Doppler broadening. This happens because some of the caesium atoms are moving towards the microwave generator and some caesium atoms are moving away. This motion shifts the position of the absorption peak, and the end result is that the peak broadens.
You can design the clock to reduce the speed the caesium atoms move, and in principle you can eliminate Doppler broadening. However the uncertainty principle sets a lower bound for the peak width. For an electronic transition like the one in Caesium the uncertainty principle tells us that:
$$ \Delta E \Delta \tau >= \frac{\hbar}{2} $$
where $\Delta \tau$ is the lifetime of the excited state i.e. the average time taken for the excited state to emit a photon and relax to the lower energy state. This $\Delta \tau$ is a characteristic of the caesium atom and is effectively a constant beyond our control. This matters because the energy $E$ is related to frequency by $E = h\nu$, so we end up with an uncertainty in the frequency given by:
$$ h\Delta \nu >= \frac{\hbar}{2} \frac{1}{\Delta \tau} $$
This tells us we will never be able to tune the microwave frequency to better than $\Delta \nu$, so there is a fundamental source of error that we cannot eliminate.
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Why are high voltage lines “high voltage?” If I have two spheres of the same size and one sphere has a small amount of charge compared to the other that has a lot more charge, then clearly the sphere with the larger charge has a larger voltage (relative to the ground). My question is do high voltage power lines have a lot more charge that is placed on them? Is that what gives them the high voltage? I think I have a grasp of the step up stations that use transformers to kick up the voltage from power plants.
This question seems almost silly to me but I have been struggling with this for a long time. I’ve done several searches online and I am not able to find answers. If there is a link that someone can provide, I much appreciate it.
| For an analogy, you should consider voltage more like a "pressure" than an amount of charge. It just so happens that in your sphere example, with a fixed number of charges (electrons), you get this nice correlation between charge and voltage.
But in a conductor, the pressure can come from far away so to speak - from the step-up transformer in your example.
Put another way, the voltage is defined as the potential to conduct current through a resistance. This property increases when you "push" more in the other end. In the spheres, the push comes from the increased number of charges in a confined space, and in the high-power wires, the push comes from/through the transformer.
The actual reason why a domestic current-distribution network uses high voltages as opposed to low voltages is another question.
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Why don't metals bond when touched together? It is my understanding that metals are a crystal lattice of ions, held together by delocalized electrons, which move freely through the lattice (and conduct electricity, heat, etc.).
If two pieces of the same metal are touched together, why don't they bond?
It seems to me the delocalized electrons would move from one metal to the other, and extend the bond, holding the two pieces together. If the electrons don't move freely from one piece to the other, why would this not happen when a current is applied (through the two pieces)?
| I believe this is essentially what happens in gilding, owing to the special properties of gold (malleability and lack of corrosion).
Extremely flat surfaces can get stuck together due to Van der Waals forces as well as air pressure. I once accidentially stuck two quartz optical windows together, and had a hell of a time separating them.
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Why force $F$ is $ma$ but not $md$ or $mv$? How can I observe and understand "force" in real life? As a layman, i can calculate approx "displacement" just by observing the moving object. And accurately by using a simple "scale". Similarly, again, I can calculate angle from origin by using displacement in $x$ and $y$ dimensions. Similarly I can use a stopwatch and scale to understand velocity.
But when I read about force, first of all it confuses with the english word "force", we use in real life. To some extent I am sure, it has nothing to do with that. So exactly what is it in the real sense. How can a layman, see force as, just like he can see displacement or angle? Or force is just a quantity defined by physicists to simplify the combinations of $ma$, they might be facing every time. And thus came up with term "force" ( which is similar in spelling to english word in oxford dictionary). And lastly, why not force has just been called as something proportional to mass and it's displacement or velocity. Why something at the level of change of velocity has been used to define it.
| A force is something that can modify the state of motion of a system.
If a system has a constant velocity (it means it doesn't accelerate : $\vec{a}=\vec{0}$), then no force is applied to the system ($\vec{F}=m\vec{a}=\vec{0}$).
If the system has a velocity that changes (it accelerates or turns) then the acceleration is not null. We can quantify this change, namely measure the force.
This explanation is correct to some extend. If you want more information, I can develop my answer.
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Atwood machine problem
Sorry for the bad drawing, but I hope that this will help you get a hold of the problem.
Consider an Atwood Machine with a total of two blocks, a mass less pulley, ideal string. One block rests on the floor, while the other one is at a height (H). Now, the string near the block that rests on the surface is slack. So, the other block falls freely, and later induces a jerk in the other block. How do I calculate the initial velocity of the two block just after the string is taut.
My approach was:
1) Calculate the velocity of the block in motion (initially) at the point when the is ALMOST taut.
2) Now conserve mechanical energy b/w the point where string was almost taut but impulse wasnt generated, and the point where impulse was generated, and the second block had JUST started motion.
But, my book says, I should conserve Linear Momentum between the same two points. I think this is wrong, because, The string that holds the pulley in place, will have an impulsive tension as soon as the impulse is generated in the string that joins the two masses.
According to you, what is right, and why?
| The answer is that both mechanical energy conservation and linear momentum conservation are not valid.
Linear momentum can't be conserved because there will be an impulse exerted on the system by the pulley. This impulse will be in the upward direction because notice how the tensions in the string on the pulley are downward, so the pulley will apply a force upwards to counteract this impulse.
Mechanical energy can't be conserved because we don't know whether all the impulses are conservative or not.
What you can do is consider an impulse $J$ on both the blocks upwards, and use change in momentum equals the impulse applied.
Thus you have
$$J=m_1 v_{\mathrm{final}}$$
for the first block, and
$$J=m_2 v_{\mathrm{final}}- m_2 v_{\mathrm{initial}}$$
for the second free falling block.
$v_{\mathrm{final}}$ will be the same for both blocks because they are constrained by the string. With these two equations, you can find $v_{\mathrm{final}}$
| {
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Difference between primitive unit cell and the associated basis? As I understand it, the basis is the group of atoms whilst the primitive unit cell is the unit space that fits the total space without any gaps, and only containing one lattice point? How do the two relate to each other? Thanks.
| I'm going to maximally disagree with P3trus and say that they're exactly the same thing: "A crystal can be categorized by its lattice and the atoms that lie in a primitive cell (the basis)" (source).
| {
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Is there a theoretical maximum for refractive index? May there be materials yet to be discovered which may have a higher refractive index than today's known materials (for wavelengths within the visible range)?
Is there a theoretical limit for the refractive index of a material?
| Since the refractive index is given by $\displaystyle{n_{12}=\frac{\sin \theta_1}{\sin \theta_2}}$, theoretically there is no limit at all on the value of refractive index. You could say that it must be positive, but then check this out: http://en.wikipedia.org/wiki/Negative_refraction
| {
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Is any apparent horizon a minimal surface? I faced "any apparent horizon is a minimal surface", but I don't know how I can relate a physical concept (apparent horizon) to pure mathematical concept (minimal surface). How can I prove it?
| I found the answer:
The apparent horizon $\mathcal{H}$ is defined as the outer boundary of the region of $\Sigma$ (a hypersurface of spacetime with induced metric $h_{ab}$ and extrinsic curvature $K_{ab}$) which contains trapped or marginally trapped surfaces. $\mathcal{H}$ itself must be a marginally trapped surface, and thus it satisfies
$$k+K^{ab}(h_{ab}-n_an_b)=0$$
where $k$ is the trace of the extrinsic curvature of $H$ as a submanifold of $\Sigma$ and $n^a$ is the unit outward normal to $H$ on $\Sigma$.
Now, in the case of time symmetric data (the extrinsic curvature $K_{ab}$ of $\Sigma$ has been set to zero) by above equation one can conclude $k=0$. Then minimal surface and apparent horizon coincide in the case of time symmetric data.
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Does changing the electric / magnetic field cause self-reinforcing induction of the other? I understand that changing electric field produces magnetic field and changing magnetic field produces electric field.
Are these produced magnetic and electric field produced due to one defined to be constant or variable?
If these are defined to be variable then do they continue to produce one another? By this I mean if changing electric field produces changing magnetic field, does this changing magnetic field produce a new electric field or the same one again?
| Instead of thinking about one field changing in response to the change of the other, it is more correct to say that
whenever the magnetic field is changing, so is the electric field, and vice versa.
The way these fields change is governed by Maxwell's equations. This way, we do not arrive at the confusion OP has.
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Symmetries of a Uniform Magnetic Field Simple question. A system with a uniform electric field everywhere in space has translational invariance in the directions perpendicular to the electric field but no translational invariance parallel to it. This system also has rotational invariance in the plane perpendicular to the electric field.
What about a uniform magnetic field? Judging from the Hamiltonian $(\vec{p}-q\vec{A})^2/2m$, it would seem that the symmetries depend on our choice of gauge. Choosing a different $\vec{A}$ breaks different symmetries. Is there a most symmetric choice for $\vec{A}$?
| Your hamiltonian is still gauge invariant because the canonical momentum, that is the generator of translations, is $m \dot{\vec{x}} - q \vec A$. In other words, your Hamiltonian is still $H = \frac{1}{2m}\Pi^2 - q \Phi$, where $\vec \Pi = m \vec{ \dot{x}}$ is the mechanical momentum.
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Detecting a photon without changing it: Does it break conservation laws? This is about an article published on ScienceMag: Nondestructive Detection of an Optical Photon. I don't have access to full text, but you can see a brief transcription in this link.
Basically, it says that a photon causes a phase shift in another system. This phase shift can be detected, and it does not change photon properties, such as frequency (pulse shape) and polarization.
How can that be true? I thought that for a photon to cause any change on a system, it must lose some energy, which is transferred to the detector. What am I missing?
| There exists elastic scattering where only directions are changed , momentum is conserved and energy is conserved.
There exists elastic scattering of photons:
In Thomson scattering a photon interacts with electrons.
Thomson scattering is the elastic scattering of electromagnetic radiation by a free charged particle, as described by classical electromagnetism. It is just the low-energy limit of Compton scattering: the particle kinetic energy and photon frequency are the same before and after the scattering. This limit is valid as long as the photon energy is much less than the mass energy of the particle: nu very much smaller than mc^2/h .
In Rayleigh scattering a photon penetrates into a medium composed of particles whose sizes are much smaller than the wavelength of the incident photon. In this scattering process, the energy (and therefore the wavelength) of the incident photon is conserved and only its direction is changed. In this case, the scattering intensity is proportional to the fourth power of the reciprocal wavelength of the incident photon.
Without having read the experiment I would consider it a utilization of elastic scattering of photons, which should involve phases since directions of individual photons change in elastic scattering.
| {
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Photons from stars--how do they fill in such large angular distances? It would seem that far-away stars are at such a distance that I should be able to take a step to the side and not have the star's photons hit my eye. How do stars release so many photons to fill in such great angular distances?
| The only stars you can reliably see are ones that are spewing enough photons at your eyeballs to appear stable.
Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough.
So your question is basically embroiled in observer bias; it assumes that the stars you see are all the stars there are, and it assumes that you could see a single photon if it hit your eye.
| {
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Predominant light frequency over the day? I'm not well-versed in physics, so I hope you'll forgive me if this question is significantly off the mark.
I'm interested in the predominant frequencies of light over the course of one day. My understanding is that as the sun rises and falls, the atmosphere absorbs differing amounts of light, and that is responsible for the changing colour of the daylight, for example becoming more red as the sun sets.
I guess that there is some curve from (all but) absolute darkness, through to broadband white light during the middle of the day, back to darkness again over the course of twenty-four hours?
If so, is it possible to approximate the makeup of the light at a given point during the day? And if so, how? I appreciate that there are probably significant complicating factors, so any rough approximation would be fine.
Again, I'm coming from a layman position here, so I apologise if I've drastically misunderstood something.
| I found a good paper that can help you. However, due to copyright issues I cannot put the spectra here.
Try to get this article:
"The Distribution of Energy in the Visible Spectrum of Daylight". A. H. TAYLOR and G. P. KERR. J. Opt. Soc. Am. 31 no. 1, pp. 3-8 (1941)
. Also available here (pdf).
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Effective mass of a black hole? Suppose a black hole forms from a given mass of particles such as the core of a star going supernova. The black hole formed will have an effective mass due to the curvature of space time induced. Such a mass is presumably the inferred mass deduced from the effect of the black hole on the motion of nearby bodies. How does this effective mass compare with the original mass of particles all of which ended up "in the black hole"? Is the effective mass less greater or the same as the formation mass?
| Answer: slightly less mass.
In general relativity, mass/energy of any system is conserved from the point of view of a distant observer (I forget the name of this theorem). We use this mass when we talk about how heavy a black hole is because we are (very) far away.
If you start with particles that are stationary, the total mass of the system is slightly less than the mass of the particles (because of gravitational binding energy). When they fall toward each-other to make a hole, additional (a few % typically) energy is released as gravitational waves (and light rays, neutrinos, etc). Due to conservation of energy, this means less mass for the final hole. In fact, asymmetric radiation of gravitational waves also can create a recoil effect that ejects the hole formed (although this case is for two black holes merging):
http://www.youtube.com/watch?v=dNXWvJoN_sw
Note that mass/energy never leaves the hole itself, the expelled stuff originates from spacetime regions before/outside of the hole.
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Faradays law for a falling magnet If experimentally it can be proven that the velocity of a falling magnet through a coil is proportional to the emf induced, has it also been proven that dB/dt is proportional to the emf induced and how so ?
| For slowly moving magnets, you can measure the strength of the magnetic field using another magnet mounted on a torsion balance. The degree of deflection would be proportional to the field strength. For fast moving magnets, you run into difficulties with your apparatus "lagging behind" and it is more properly the case that we measure $dB/dt$ by measuring things like EMF. The reason we do this is because mechanics and electrodynamics seem to be self-consistent. In other words, we believe it to be the case because the theory makes the right predictions. The proof, as you would say, is in the pudding.
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Is there a way for an astronaut to rotate? We know that if an imaginary astronaut is in the intergalactic (no external forces) and has an initial velocity zero, then he has is no way to change the position of his center of mass. The law of momentum conservation says:
$$ 0=\overrightarrow{F}_{ext}=\frac{d\overrightarrow{p}}{dt}=m\frac{d\overrightarrow{v}_{c.m.}}{dt}$$
But I don't see an immediate proof, that the astronaut can't change his orientation in the space. The proof is immediate for a rigid body (from the law of conservation of angular momentum). But the astronaut is not a rigid body.
The question is: can the astronaut after a certain sequence of motions come back to the initial position but be oriented differently (change "his angle")? If yes, then how?
| There's another way to do this also, more akin to how spacecraft actually do it:
Take a weight on a string, hold it up and spin it. You'll turn in the opposite direction. When you stop it you also stop turning.
Of course this will produce an off-axis force that will be a real pain to deal with. Real spacecraft do it by means of a set of internal wheels so they can turn on any axis.
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Why is it easier to walk diagonally upstairs Try it yourself on a set of wide steps! Work is given by
$$\int_C \mathbf{F} \cdot d\mathbf{x}$$ where $C$ is a path integral. In this case I think $\mathbf{F}$ is a rotational vector field because the stairs are essentially a set of discontinuities. This would mean that the integral is path dependent.
Here is why this seems like a paradox: if you where to climb the same height over a smooth hill, then the integral is path independent since $\mathbf{F}$ is "smooth" and irrotational. This means that the amount of work done never changes given the path you take, hence if you take a longer (diagonal) path you must exert less force over that distance than if you where to take a straight path. In the case of walking up stairs, you must always cross the same number of stairs no matter the path you take. All other distance you make is along the horizontal parts of the tops of stairs and hence requires much less force. This would seem to suggest to me walking up stairs always requires the same force to scale the vertical parts of each stair.
Am I hallucinating that walking diagonally up stairs is easier?
| It's because of how the body is oriented when going up diagonally which allows for the lateral thigh muscles to contribute to the effort, versus only the frontal muscles when going straight up. Since the effort is spread over more muscle areas, it feels less strenuous.
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A naive question about the Second Quantization? Let's consider a single-particle(boson or fermion) with $n$ states $\phi_1,\cdots,\phi_n$(normalized orthogonal basis of the single-particle Hilbert space), and let $h$ be the single-particle Hamiltonian. As we all know, the second quantization Hamiltonian $H=\sum\left \langle \phi_i \mid h \phi_j \right \rangle c_i^\dagger c_j$ of $h$ should not depend on the single-particle basis we choose(where $c_i,c_i^\dagger$ are the bosonic or fermionic operators.), and this can be easily proved as follows:
Choose a new basis, say $(\widetilde{\phi}_1,\cdots,\widetilde{\phi}_n)=(\phi_1,\cdots,\phi_n)U$, where $U$ is a $n\times n$ unitary matrix. Further, from the math viewpoint, an inner product can has two alternative definitions, say $\left \langle \lambda_1\psi_1\mid \lambda_2 \psi_2 \right \rangle=\lambda_1^*\lambda_2 \left \langle \psi_1\mid \psi_2 \right \rangle(1)$ or $\lambda_1\lambda_2^* \left \langle \psi_1\mid \psi_2 \right \rangle(2)$.
Now, if we think $(\widetilde{c}_1^\dagger,\cdots,\widetilde{c}_n^\dagger)=(c_1^\dagger,\cdots,c_n^\dagger)U$ combined with the definition (1) for inner product, then it's easy to show that $\sum\left \langle \widetilde{\phi}_i \mid h \widetilde{\phi}_j \right \rangle \widetilde{c}_i^\dagger \widetilde{c}_j=\sum\left\langle \phi_i \mid h \phi_j \right \rangle c_i^\dagger c_j$; On the other hand, if we think $(\widetilde{c}_1,\cdots,\widetilde{c}_n)=(c_1,\cdots,c_n)U$ combined with the definition (2) for inner product, one can also show that $\sum\left \langle \widetilde{\phi}_i \mid h \widetilde{\phi}_j \right \rangle \widetilde{c}_i^\dagger \widetilde{c}_j=\sum\left\langle \phi_i \mid h \phi_j \right \rangle c_i^\dagger c_j$.
Which combination of transformation for operators and definition for inner product is more reasonable? I myself prefer to the former one.
| The entirety of the modern quantum mechanics literature uses inner products that are linear in the second argument, and antilinear in the first one. Mathematicians often use the other convention, but I've never seen it used in physics. This is of course pure convention, but you will find grief, at least when you try to publish, if you go against the flock on this one.
| {
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Physics of the inverted bottle dispenser
When you invert a water-bottle in a container, the water rises and then stops at a particular level --- as soon as it touches the hole of the inverted bottle. This will happen no matter how long your water-bottle is. I understand this happens, because once the water level touches the hole, air from outside cant go inside and therefore there is nothing to displace the water that falls out of the container.
Now according to the laws of pressure ---- the pressure at the water level must be same everywhere --- whether it's inside the water bottle or outside. And that must be equal to the atmospheric pressure. Therefore the pressure of the water column + air column inside the inverted bottle must be equal to the atmospheric pressure.
What I dont understand is, no matter how long a bottle you take, the water level will always stop at the hole. So that means that no matter how long a bottle you take, the pressure of the water column + air column inside the water bottle will be equal to the atmospheric pressure. How could this be possible?
Also I'd like to let you know that, if you pierce the upper part of the bottle with a small pin, then the water level rises and overflows out of the container. I'm assuming air from outside rushes in and pushes the water out.
| Your mistake is to assume that the water will stop "no matter how long a bottle you take". It will not - you just need a longer bottle than you expect. To be precise, you need a column of water 10 meters high to counteract atmospheric pressure.
| {
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Quantizing the Dirac Field: which commutation relations are more fundamental? When quantizing a system, what is the more (physically) fundamental commutation relation, $[q,p]$ or $[a,a^\dagger]$? (or are they completely equivalent?)
For instance, in Peskin & Schroeder's QFT, section 3.5, when trying to quantize the Dirac field, they first say what commutation relation they expect to get for $[\Psi(\vec{x}),\Psi^\dagger(\vec{y})]$ (where $i\Psi^\dagger$ is the conjugate momentum to $\Psi$), in analogy to the Klein-Gordon field, then they postulate a commutation relation between $[a^r_{\vec{p}},a^{s\dagger}_{\vec{q}}]$ etc., and then verify that they indeed get what they expected for $[\Psi(\vec{x}),\Psi^\dagger(\vec{y})]$.
Why did we need to postulate the value of $[a^r_{\vec{p}},a^{s\dagger}_{\vec{q}}]$? Couldn't we have just computed it off of $[\Psi(\vec{x}),\Psi^\dagger(\vec{y})]$? (which, by expecting to get it, we could have just as well already postulated it).
I suppose that would entail explicitly writing something like:
$$ a_{\vec{p}}^r = \frac{1}{\sqrt{2E_{\vec{p}}}} u^{r\dagger}(\vec{p})\int_{\mathbb{R}^3}d^3\vec{x}\,e^{-i\vec{p}\cdot\vec{x}}\Psi(\vec{x})$$
and a similar expression for $b$.
| Firstly, note that they postulate those commutation relations in the beginning of section 3.5 in order to show that they are wrong, which they demonstrate in the ensuing pages. The ultimate point is to show that one needs to impose anti-commutation relations on fermionic fields.
In fact, the correct relations are postulated in equation 3.96;
\begin{align}
\{\psi_a(\mathbf x), \psi_b^\dagger(\mathbf y)\} &= \delta^{(3)}(\mathbf x - \mathbf y) \delta_{ab}
\end{align}
You could then ask, are these equivalent to the anti-commutation relations of the mode operators that they write in (3.97)? Namely,
\begin{align}
\{a^r_\mathbf p, {a^s_\mathbf q}^\dagger\} = \{b^r_\mathbf p, {b^s_\mathbf q}^\dagger\} = (2\pi)^3\delta^{(3)}(\mathbf p - \mathbf q)\delta^{rs}
\end{align}
And the answer is yes.
To show that the second set implies the first, write the fields in their integral mode expansions, compute the anti-commutator of these integral expressions, and apply the anti-commutators between modes. To show that the first set implies the second, invert the integral expressions for the fields in terms of the modes to obtain integral expressions for the modes in terms of the fields, and do the analogous thing.
Main Point. The commutators/anti-commutators between fields are equivalent to the commutators/anti-commutators between modes.
| {
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Exciting Surface Plasmon-Polaritons with Grating Coupling I'm very new the topic of SPPs and have been trying to understand this particular method of exciting surface plasmons using a 1D periodic grating of grooves, with distance $a$ between each groove. If the light incident on the grating is at an angle $\theta$ from the normal and has wavevector ${\bf k}$, then apparently if this condition is met:
$\beta = k \sin\theta \pm\nu g$
where $\beta$ is corresponding SPP wavevector, $g$ is the lattice constant $2 \pi/a$ and $\nu={1,2,3,...}$ then SPP excitation is possible.
I haven't ever really had a formal course in optics, so my question is where this condition comes from. It seems like Fraunhofer diffraction, but only for the light being diffracted at a $90^\circ$ angle to the normal. Most books don't state how they get this result, they just say it's because of the grating "roughness" which really confuses me.
Any help would be greatly appreciated.
| Yes light can change propagation direction when scattering off of a grating into pre-defined directions that depend on the angle of incidence and the grating period (as well as the refractive medium the grating resides in). This is just Bragg's Law. The change in direction (or momentum) is determined by the above parameters. You can think of the change in momentum as a "kick" if you wish, though I've never heard it explained that way before. Anyway, the diffracted light can couple into a Surface Plasmon Polariton (SPP) grating mode, also called an SPP Bloch mode. I believe, as you mention, the diffracted light needs to be propagating parallel to the surface. This is similar to the TIR condition being met in prism-coupled SPP. You need to couple through a higher index material through the (thin) metal and into a lower index material to see the SPP resonance in the prism-coupling approach. This condition is relaxed when using grating coupling.
| {
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Bulk modulus of Liquid helium and first sound Does anyone know where to find the bulk modulus of liquid helium ?
I've been looking all over the internet but everywhere I get N/A. Any tips ?
I'd need it to estimate the speed of first sound in liquid helium which is given by:
$c_1=\sqrt{\left.\frac{\partial p}{\partial\rho}\right|_\sigma}$
| The best way is to use the inverse value, the compressibility, this one is more easilly found.
| {
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Calculate average speed with unknown variable accelaration I am in the middle of a vehicle tracking project where I have to calculate the distance traveled by the vehicle in a given amount of time.
Data I am getting:
Speed : 30.2 km/hr 12.7 km/hr 15 km/hr 21.8 km/hr
Time : 11:00:00 11:00:22 11:00:45 11:01:10
That is I am getting the speed of the vehicle every 20-25 seconds. So what is the best way to calculate the distance traveled by the vehicle during this whole duration? Is taking the median of two speeds the best way to calculate the average speed here?
| Well, this is an easy problem in kinematics. "The rule" for solving it is drawing a graphic of the (differences of) time versus the speed. You will get some points. Join these points to obtain a trapezoid histogram. So, what do you think it is the area beneath this curve? It has the dimensions of a length, so... Now, you want an average speed, that is a constant speed for the whole path. In the previous graphic, a constant speed would be a horizontal line. You can solve this problem drawing a horizontal line that creates a rectangle in your graphic, whose area is the same as that of the trapezoid.
These are just a guidelines. The analytical part follows straightforward from these hints..!
| {
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Why do some beverages stay cold longer than others? For my daughter's science experiment, she placed six beverages (cola, diet cola, milk, chocolate milk, apple juice, and water) in the exact same amount in the exact same type and size of plastic cups, and placed all of the cups in a refrigerator at the same time and allowed them to remain in the refrigerator for the same period of time (4 hours). When the beverages were taken out and set on the same counter, in a room at approx. 70 degrees, a cooking thermometer was then used to measure the temperature of each of the beverages at different time intervals (the intervals were the same for all beverages)---i.e., the temperature of each was measured at 15 minutes, 45 minutes, 1 hour, and 90 minutes. The results were that all but two of the beverages held the same temperatures throughout the measuring period. The two beverages with different (colder) temperatures were diet cola and (white) milk. Of course, my daughter is supposed to cite to references that support (and explain) the basis of the test results/conclusions, and we are finding nothing that addresses this. "Viscosity" is the closest principal that we have found, but we cannot find anything that provides the viscosity levels for the separate beverages (hence, no conclusion can be drawn). Any ideas on where to find research references related to this? thanks,
| It is mostly because different liquids have different Thermal conductivity and Heat capacity.
http://en.wikipedia.org/wiki/Heat_capacity
http://en.wikipedia.org/wiki/Thermal_conductivity
Of course viscosity of Liquid is effective but it is a secondary factor.
| {
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Dirac Lagrangian density in curved spacetime I'm trying to derive this form of the Dirac Lagrangian density in curved space-time:
$$ \mathcal{L}~=~\det\left(e\right)\bar{\Psi}\Bigg (\frac{i}{2}\gamma^{a}\partial_{a}-m+\gamma^{a}\gamma^{5}B_{a}\Bigg)\Psi $$
starting from this form:
$$ \mathcal{L}~=~\sqrt{-g}\Bigg (\frac{i}{2} \bar{\Psi} \gamma^{a} \stackrel{\leftrightarrow}{D}_{a} \Psi-\bar{\Psi}m\Psi \Bigg) $$
I can get the first term, however, in the last term I get a factor of $ \quad - \frac{1}{2} \quad$ out the front, i.e.
$$ \mathcal{L}~=~\det(e)\bar{\Psi}\Bigg (\frac{i}{2}\gamma^{a}\stackrel{\leftrightarrow}{\partial_{a}}-m-\frac{1}{2}B_{a}\gamma^{a}\gamma^{5}\Bigg)\Psi $$
I've tried deriving it in several different ways and always end up with the above expression. The problem is, in all of the papers that I've read on the subject, it is given in the form as presented at the top of this page.
If it helps, here is a link to my derivation in full: http://we.tl/mUgiw0BkMh
Would really appreciate any help in solving this problem.
| What does B correspond to here? Is it related to the spin connection? The axial current?
Also the link you propose does not work anymore.
If you want a rather nice demonstration of the Dirac equation in curved space, you can try "Nonlinear spinor equation and asymmetric connection in general relativity" by Hehl and Datta.
| {
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What is the area in Faraday's law if we have only a piece of metal moving in a magnetic field? If a piece of metal of length $l$ is moving with a speed $v$ in a region where there is a uniform magnetic field $B$ perpendicular to it, there will be a potential difference across its terminals equal to $lvB$ which is known as motional EMF. This can be shown and understood in terms of magnetic and electric forces on the free charges in the metal.
How can one calculate such EMF from Faraday's law,
$\displaystyle\mathcal{E} = \left|\frac{d\Phi_B}{dt}\right|$?
(where $\Phi_B$ is the magnetic flux $\int \bf{B}\cdot d\bf{a}$)
(If $B$ is not changing, then the change in the magnetic flux must be due to change in an area, but the area of what? What are the boundaries of this area?)
| I think your question answers itself, indeed the area of what? Faraday's law is for a closed loop of wire, thus Faraday's law is inappropriate and we should look for an alternative, as you have done by considering the Lorentz force. If the metal were a closed loop of circumference $l$ then Faraday's law would be valid. The forces on the electrons from the top and the bottom would both act in the same direction but in opposite orientations around the loop and thus no EMF around. Also the force on the protons would be counter to that on the electrons and thus no overall force on the loop, thus we conclude that Faraday's Law saying that there would be no net EMF is correct.
| {
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All objects radiate energy, but we cannot see all objects in the dark. Why? We claim that all objects radiate energy by virtue of their temperature and yet we
cannot see all objects in the dark. Why not?
| Cold bodies radiate mostly in the infrared zone (invisible to the human eye), but as the temperature increases the body will emmit higher frequencies with more intensity. So room temperature obects will not be seen due to black body radiation.
As you can see, hot bodies are visible because they emmit visible light mostly.
| {
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Why are roofs blown away by wind? Whenever there are high winds, such as in storms, thin metal roofs on sheds as well as concave roofs on huts are sometimes blown away.
One explanation provided to me said that the higher velocity of the air outside causes the air pressure above the roof to decrease and when it has decreased to a certain extent such that the air pressure above the roof is lesser than the air pressure beneath the roof and due to some kind of osmosis, the air particles move from the area of higher pressure (beneath the roof) to the area of low pressure. In this process, the roof is blown away.
Another explanation, specifically about the thin metal roofs, said that it was blown away due to the lift caused by the air and this is the same kind of lift you get when you blow on paper.
Both these explanations puzzle me.
What really bothers me is the basis of the first one, how can an increase in velocity cause pressure to drop? I can't seem to correlate that with the Force per unit area definition of pressure.
Please, oh great physicists of the internet, help me and every other ordinary person to understand how and why roofs get blown away.
| Bernoulli to the rescue!
Does this answer the question?
Keep in mind, hurricane speeds are often twice small aircraft stall speeds, and typical aircraft wing loading is in the range of $50 kg/m^2$, so a roof could see 4 times that.
Roofing material would have to be really heavy not to be lifted by that.
| {
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"timestamp": "2023-03-29T00:00:00",
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Heating metal with laser? Can a laser be used to heat metal's to the point they glow red?
How are laser cutters able to cut metal very easily?
How much heat can a laser generate?
| *
*Can a laser be used to heat metal's to the point they glow red?
Yes
*
*How much heat can a laser generate?
How big can it be? What laser are you talking about?
*
*How are laser cutters able to cut metal very easily?
Now that is a good question.
Some laser cutters will melt the object they are cutting to cut them. More sophisticated laser cutters, and more energy efficient, use ultra short pulses to create ablation. Ultra short pulsed laser ablation, typically converts a very specific amount of material into a plasma. This will allow scanning devices to verify the results before new pulse(s) are sent to cut more.
| {
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What is quark transverse momentum? When you google my question you get something on the order of 400 000 results but none of them explains how it is defined (No I didn't check them all).
I know what the words quarks, transverse and momentum mean. But what is quark transverse momentum? Transverse with respect to what?
| The only way we know that quarks exist is by a series of deep inelastic scatterings with leptons and with protons. This is a reconstructed event at LEP
quark antiquark at 12:00 o'clock and 4:00 o'clock gluon the third one.
The lepton colliders have the advantage that most of the energy taking part in the collision can be detected in four pi detectors.
The transverse momentum, i.e. the momentum perpendicular to the beam is useful, as explained in this answer to a similar question, in the case of hadron colliders , or lepton hadron target experiments because the projections to transverse direction to the beam , due to momentum conservation contain all the energy of the deep inelastic interaction that has gone in the transverse direction. In the beam direction the debris of soft interactions and other possible beam tracks can very easily destroy signals and correlations.
To get the idea about the complexity of hadron hadron events here is a Higgs candidate from the ATLAS detector :
Higgs boson decaying into two τs, each of which subsequently decay into either an electron (blue line) or a muon (red line).
If you go to the link you will see that most of the event displays show the transverse projection, because it is the cleaner from noise, from soft interactions of the three quark to three quark collisions and from extraneous beam particles colliding softly.
So the quark transverse momentum in LHC data will be the momentum of the quark jet perpendicular to the beam. Incidentally, the first indication of the quark content distribution in the protons was given by the existence of high p_t events in deep inelastic scattering of leptons on targets, and this discovery eventually lead to the establishment of QCD as the strong interaction force .
| {
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How does air escape from a pneumatic tire? Obviously, it is caused by the difference in pressure between the inside of the tire and its surrounding environment; but how specifically is the air escaping?
| Air escapes via multiple mechanisms.
Diffusion through the material: Most materials have some permeability to air. Air molecules can fit between the rubber molecules. The N2 and O2 molecules are so few and far between that they don't interact. The average force on a molecule is zero as it randomly walks through the rubber, but the concentration difference means that by chance more will leave than enter. Air escapes ~ pressure difference, so double the pressure difference and twice the mass of air will leave per day.
Diffusion through a tiny (nano-sized) pore: Similar mechanism, but the loss rate is higher as there is more freedom for the air to escape.
Diffusion through a small hole (~100nm to a few 100 um): Air molecules now interact since the mean free path is smaller than the hole size. Air acts as a fluid and flows out smoothly. Air escapes ~ pressure difference.
Larger hole (above a few hundred um): Air flows out like a fluid, but this time it is turbulent and makes an audible hissing sound. There is usually no smooth transition to turbulence as you increase the hole size, it happens suddenly the exact transition is somewhat unpredictable. Air escapes ~ sqrt(pressure difference).
| {
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As the universe ages, will we see more stars or less? After a very long time will we see more stars (due to the fact that more light is get to us) or less stars (as the universe expends and light have to pass larger distance)?
In general, can stellar objects go outside of the scope of the observable universe or is it only growing with time? Calculations are always better than just discussion.
Does it have any connection to dark matter?
| Star formation will slowly start to decrease in galaxies as the universe ages because of the conversion of gas, such as hydrogen, into heavier elements, such as carbon and iron. Essentially the universe is slowly running out of fuel. Stars have a hard time fusing heavier elements.
Eventually there will be no stars left. What will remain are black dwarves, black holes, neutron stars, etc. These will slowly decay over time into a cosmic soup of fundamental sub-atomic particles.
The expansion of space will mostly make it harder to see galaxies not stars.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/90024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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