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How fast does an electron move? I've been reading this website: www.physics.wayne.edu/~apetrov/PHY2140/Lecture8.pdf to learn how fast an electron moves in a circuit.
On page #8, #9 and #10 It says to take the Cross-sectional Area of the wire, The current, The density, The Charge and the electrons^3
Area- 3.14x10^-6 ( 2mm thick wire = 3.14 × (0.001 m)^2 = 3.14×10^−6 m^2 = 3.14 mm^2)
Current- 10 I
Density of copper- 8.95 g/cm^3
charge of 1 electron- 1.6x10^-19
electrons^3- 8.48x10^22 = ( 6.02*10^23 mole * 8.95 g/cm^3 * (63.5 g/mole)^-1 )
Total: 10 / 8.48x10^22 m^3 * 1.6x10^19 * 3.14x10^-6 m^2 = 2.48x10^-6 m/s
But they say that with 2.48x10^-6 m/s It'll take the electrons 68 minutes to travel 1 meter, How is that possible?
When I calculated that equation I end up with 5.9245283e+35, Then when I try to calculate again to get 68 minutes to travel 1 meter I can never get it right.
I'm not the best at math, The m's confused me. What am I missing ?
| The calculation on page 9 of the PDF is incorrect because of the author's confusion mixing meters and centimeters combined with a numerical error. (Sad!) The correct calculation of the drift velocity for the input values given is
$$v_d=\frac{I}{nqA}=\frac{10.0\,\text{C/s}}{(8.48\times 10^{28}/\text{m}^3)(1.6\times 10^{-19}\,\text{C})(3.00\times 10^{-6}\,\text{m}^2)}=2.46\times 10^{-4}\,\text{m/s}$$
so the time to go one meter is
$$\frac{1\,\text{m}}{2.46\times 10^{-4}\,\text{m/s}}=4065\,\text{s}=67.75\,\text{min}.$$
You might want to find a better site for studying this subject.
| {
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Difference between left- and right-handed, helicity and chirality What is the difference? I know there is the (almost) same question What's the difference between helicity and chirality? but when a particle is given as left-handed. Is it helicity or chirality?
| When we consider spinors of the Lorentz group $SO(3,1)$, recall that the universal covering of $SO(3,1)^+$ (the component of the Lorentz group connected to the identity) is isomorphic to $SL(2,\mathbb C)$.
Two-component spinors are elements of two-dimensional irreducible modules of $SL(2,\mathbb C)$. However, noting that the complexification of the Lie algebra of $SL(2,\mathbb C)$ is $A_1 \oplus A_1$, there are two inequivalent such modules.
These modules have weights $(1,0)$ and $(0,1)$, or in physics language, spins $(\frac12,0)$ and $(0,\frac12)$ respectively. Objects with indices corresponding to each have different transformation properties, namely, for the former,
$$\psi_\alpha \to M^\beta_\alpha \psi_\beta$$
for some $M\in SL(2,\mathbb C)$ whereas for the latter,
$$\psi_{\dot\alpha} \to \overline M^{\dot\beta}_{\dot\alpha} \psi_{\dot\beta}.$$
Typically, we refer to the undotted indices as left-handed and the dotted indices as right-handed. Note that in some cases in lower dimension, they are not distinct (which is simpler as one does not need Van der Waerden notation to distinguish them.)
It should be noted they transform in the same way under rotations, but they transform oppositely under boosts, motivating the nomenclature. Normally you may have been introduced to spinors first through the Dirac spinor, which lies in the $(\frac12,0) \oplus (0,\frac12)$ representation, being comprised of two chiral spinors.
| {
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Heat reduction based on compression So I was thinking about air conditioner today and how we run air across compressed freon to cool down air but why do we need freon at all why not store just compressed air. My guess is because its inefficient.
My question given a 1 cubic meter tank of air, if the room temperature at 1 atmosphere is 80 degrees. At what atmosphere would you have to have the tank so that the initial release of the air would be the temperature 70 degrees?
|
Compression brings the molecules closer together thus cooling it down.
This isn't correct. The temperature of a gas isn't related to how close together the molecules are, but their speed.
By compressing the gas, they are closer together, but the work done in compression has sped them up as well. If you wait a while and let the gas cool, they'll still be closer together, but will be moving at the original speed.
My question is given air pressure at sea level. 80 degrees at 14.70 psi, how much pressure would you have to add to cool the air to 70 degrees if it was stored in a tank for 1 cubic meter?
There is no such pressure. Applying pressure will increase the temperature in the short term, not lower it. Any temperature change by doing this is only temporary. The gas will then exchange heat with the environment and move to ambient temperature.
| {
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How does phase transition occur in finite sized ising model? I was simulating the square lattice Ising model via Metropolis Algorithm and found that at 0 magnetic field, there is spontaneous magnetisation below some temperature.
I have used Periodic Boundary Condition in a 100x100 lattice.
Is this an instance of a Phase Transition?
I have heard that phase transitions occur in thermodynamic limit. So how does this spontaneous magnetisation occur?
If this is not a phase transition, is it an artifice of the metropolis algorithm and relates to non convergence of this algorithm?
If this is a phase transition how does spontaneous magnetisation occur at all since the probabilities carry the symmetry of the hamiltonian and the partition function is finite allowing the microstates to have boltzmann distribution for all magnetisation values?
| In a strict mathematical sense, you will not observe a phase transition in a finite volume, for the reason you mention. If you measure thermodynamic quantities and their derivatives, when you expect a completely sharp transition, you will instead see a smooth curve that approximates the "correct" behavior if the volume gets larger and larger. There is a theory of finite-size scaling that addresses this quantitatively, and in fact explains how these finite-size effects can be exploited to measure critical exponents effectively.
In practice, on a 100x100 lattice you should have no problem at all to detect a phase transition. If you use a good algorithm (like a cluster algorithm that flips many spins at once) you will find that the susceptibility obtains a maximum $\chi_\text{max}(L)$ at some temperature $T_c(L)$ that slightly depends on the number of spins $L$. By measuring these quantities for different box sizes $L$ you can obtain estimates for the actual critical temperature $T_c$ and the critical exponents $\nu$ and $\gamma$.
| {
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$SU(2)$ and $SO(3)$ WZW models It seems that the $SU(2)_1$ and $SO(3)_1$ Wess-Zumino-Witten models are quite different despite the Lie algebras being identical. The $SO(3)_1$ model has central charge 3/2 and is equivalent to 3 free Majorana fermions. The $SU(2)_1$ model has central charge 1, and can be expressed in terms of a compactified free boson (see for instance section 15.6 in Di Francesco et al's CFT textbook).
So unless I'm misunderstanding something, through ordinary bosonization the $SU(2)_1$ model should be equivalent to 2 Majorana fermions and thus equivalent to the $SO(2)_1$ model rather than $SO(3)_1$.
This situation seems very strange to me. Can someone point out where the global difference between $SO(3)$ and $SU(2)$ leads to a loss of a Majorana fermion?
Note that in this related question the brief answers claim the $SU(2)_1$ and $SO(2)_1$ WZW models are not equivalent, but frankly I don't see why that is the case. So perhaps my confusion with that question is related to this one.
| The $G$-WZW model depends not only on the group $G$, but also on a number $k$ called the level. The symmetry algebra is an affine Lie algebra, and it also depends on $k$. Both $SU(2)$ and $SO(3)$ have the same affine Lie algebra, and the central charge is
$$ c = \frac{3k}{k+2}
$$
where $k\in \mathbb{N}$ for $SU(2)$ and $k\in 2\mathbb{N}$ for $SO(3)$.
It seems you are considering the $SU(2)$ WZW model at level $k=1$ (so $c=1$), and the $SO(3)$ WZW model at level $k=2$ (so $c=\frac32$). Their symmetry algebras differ because their levels differ.
Even at the same level, the $SO(3)$ and $SU(2)$ WZW models would differ. They would have the same symmetry algebra, but different spectrums. (Diagonal for $SU(2)$, non-diagonal for $SO(3)$.)
(I am currently trying to improve the Wikipedia page on WZW models. Help and suggestions are welcome.)
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Solution to Maxwell-Lorentz equations I am trying, without success, to find an example (preferably simple) of solution for the Maxwell-Lorentz equations, i.e., the coupled system of Maxwell equations + dynamics of a charged particle given by Lorentz force. Say we have a (for simplicity, non-relativistic) particle of mass m, charge q, position $\vec x$ and velocity $\vec v$, then the Lorentz force will give
$$m \vec x''(t) = q ( \vec E (\vec x(t),t) + \vec v(t) \times \vec B (\vec x(t),t ))$$
Is there any system for which we can exhibit at some instant $t_0$ the 'state' of the system $( \vec E(\vec r,t_0),\vec B(\vec r,t_0) ,\vec x(t_0),\vec v(t_0))$?
Standard textbooks seems not to consider solutions of coupled Maxwell-Lorentz equations, the only one I didn't check is Jackson's, because I don't have a copy with me.
| The Maxwell-Lorentz equations for point-like charged particles are meaningless. This is well-known since the beginnig of the 20th century. Older textbooks (like that of Becker) written between the two world wars discuss it in all details. The devil lies in the self-interaction. A hand-made correction, excluding from the Lorentz force the field produced by the particle itself has still survived. Ignoring all magnetic forces this approach leads to the Coulomb Hamiltonian used also in the non-relativistic quantum mechanics, where the Coulomb terms i=j are just omitted.
Actually, one should not even teach the electrodynamics of point-like classical particles, since it is basically wrong. Has neither Lagrangian nor Hamiltonian formulation. A consistent formulation of the electrodynamics of charged particles may be formulated only in the frame of the field theory followed by a quantization.
I recommend You the recent pedagogical arXiv preprint ( a future chapter of a textbook in preparation):
A field-theoretical approach to non-relativistic QED.
by Ladislaus Alexander Bányai and Mircea Bundaru,
arXiv:1907.13053v1 [quant-ph] 30 Jul 2019
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Units in general relativity My question is pretty straight-forward: what are the units of the tensors in General Relativity? This should sound easy, but I always studied those in natural units ($c=1$) so I can't figure it out. In particular, what are the units of
*
*$G_{\mu\nu}$
*$g_{\mu\nu}$
*$R^\rho_{\mu\sigma\nu}$
*$R_{\mu\nu}$
*$R$
*$T_{\mu\nu}$
*$\Gamma^\lambda_{\mu\nu}$
?
| If you choose coordinates with the units of length, such as $(ct, x, y, z)$, then the metric tensor and its inverse are dimensionless, the Christoffel symbols have the dimensions of inverse length, and the curvature tensors are inverse length squared. In these coordinates the energy-momentum tensor has the dimensions of energy density.
| {
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Bose-Einstein condensation: Bogoliubov Approximation I'm trying to understand the Bogoliubov approximation from "Statistical Mechanics" by Pathria and Beale. First of all they say
Since $a_0^{\dagger}a_0=n_0=O(N)$ and $(a_0a_0^{\dagger}-a_0^{\dagger}a_0)=1<<N$, it follows that $a_0a_0^{\dagger}=(n_0+1)\simeq a_0^{\dagger}a_0$
and this part is clear. It's not clear the following logic step
The operators $a_0$ and $a_0^{\dagger}$ may, therefore, be treated as c-numbers, each equal to $n_0^{1/2}\simeq N^{1/2}$
Can someone explain me why we can treat these operators as c-number?
| The first statement reformulated gives you $[a_0, a^\dagger_0] \simeq 0$ (there is an assumption that operators involving $a^\dagger_0a_0$ are replaced by their expectation values). Then, the approximation is that since the operators commute, they can be replaced by classical objects.
So $a_0$ is a number equal to $n^{1/2}_0$ and its conjugate is $a^*_0 = \left(n^{1/2}_0\right)^*$.
EDIT : Here is a short justification of the replacement of operators by numbers. You can replace an operator by a number when states are always eigenvectors of the operator in question. In this case, we suppose that the number of particles is well defined, so we can replace $N$ by a number. Then, within the approximation that $a$ and $a^\dagger$ commute, they also commute with $N$. Since states are eigenvectors of $N$, they’re also eigenvectors of $a$ and $a^\dagger$, so we replace them by numbers.
| {
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Conservation of linear momentum with mass defect Suppose we have an insulating container, like a perfect black body, which absorbs all the radiation coming from a radioactive element placed in the center (or some equivalent process like matter annihilation). Assuming a spherically symmetric emission. It will transform all radiation into thermal energy until reaching thermal equilibrium.
If the $\Delta$m, due to mass defect, doesn't "pop out" from the body, like the example of the rocket, how is momentum preserved?
Does the body accelerate????
$\sum_{i} F_i = \frac{dp}{dt}=\frac{d(vm)}{dt}$
| With a simple calculation, it can be shown that there is no acceleration.
Let's assume $dm/dt$ is constant we call $\delta$. We then have
\begin{equation}
\frac{dv}{dt}=-\delta v
\end{equation}
Then:
\begin{equation}
\frac{dv}{v} = -\delta dt
\end{equation}
By integrating:
\begin{equation}
\ln (v) = -\delta t + A
\end{equation}
Where $A$ is an integration constant, dependent on initial conditions as we shall see.
\begin{equation}
v=B e^{-\delta t}
\end{equation}
Where $B=exp(A)$. But at $t=0$ we have $v=B$. But $v=0$ at $t=0$, then $B=0$ ($A\rightarrow - \infty$). So now:
\begin{equation}
v=0 ; \forall t\geq 0
\end{equation}
Thus:
\begin{equation}
a=\frac{dv}{dt}=0
\end{equation}
No acceleration!
(The case for $dm/dt$ not constant which is realistic is not solvable for most realistic states. Should try for $dm/dt=Cexp(-ct)$ like a radioactive material, but in my knowledge this isn't solvable analytically, I'm not that good.)
| {
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Why are the left- and right-hand sides of a differential equation with two separated variables equal to a constant? While deriving the Time Independent Schrodinger Equation, my book mentioned this line.
So time and position of a particle are two independent variables. If they are equal to one another for all values of $t$ & $r$, then why should they be equal to a constant?
Can't we have other solutions to this other than treating both the sides as a constant?
|
Can't we have other solutions to this other than treating both the
sides as a constant?
No. The TDSE (and many other similar PDEs, like the TISE, the Newtonian wave equation and Fourier's equation) has a numeric separation constant. They are known as eigenvalue problems.
| {
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Differences between charge quantity and electric charge As a senior middle school from China mainland, I am teaching physics about electric field. I work with my workmates, and we got a problem now. We cannot get an agreement. There are three viewpoints.
The first is that: electric charge is physical attribute and a physical quantity. It means electric charge is a physical quantity. The unit of electrical charge is the coulomb (symbolized C).
The second is that: electric charge is physical attribute. Charge quantity is a physical quantity. The unit of charge quantity is the coulomb (symbolized C). Electric charge has no unit.
The third is that: electric charge is physical attribute and a physical quantity. Charge quantity is a physical quantity too. The unit of electrical charge is the coulomb (symbolized C). The unit of charge quantity is the coulomb (symbolized C) too.
| Charge is a fundamental and inherent physical property of matter and we know that for sure because we can measure it. It is defined:
*
*by a magnitude, which helps us understand how strongly matter experiences or produces electric, magnetic or electromagnetic fields
*by a unit of measurement, which is a definite magnitude decided by a convention ($C$, $e$, $\frac{[Current Intensity]}{[Time]}$, ...)
I'm not exactly sure what you mean by physical attribute or physical quantity but in that case, I believe charge by definition is closer to your third viewpoint
| {
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Rotational invariance of the conductivity tensor (Classical Hall Effect) In classical Hall effect, the conductivity tensor is given as
$\sigma = \frac{\sigma_{DC}}{1+\omega_B^2 \tau^2} \begin{pmatrix}
1 & -\omega_B \tau \\ \omega_B \tau & 1
\end{pmatrix}$
where the author suggests that since it is rotationally invariant, it must be in the form of
$\sigma = \frac{\sigma_{DC}}{1+\omega_B^2 \tau^2} \begin{pmatrix}
1 & -\omega_B \tau \\ \omega_B \tau & 1
\end{pmatrix} = \begin{pmatrix}
\sigma_{xx} & \sigma_{xy} \\ -\sigma_{xy} & \sigma_{xx}
\end{pmatrix}$.
My understanding of rotational invariance is that of
$\forall A \in SO(2), \quad A^T \sigma A = \sigma$ , i.e. $[\sigma, A]=0$, which would result in that form as outlined in https://math.stackexchange.com/questions/173639/properties-for-a-matrix-being-invariant-under-rotation.
However, I want to know what it really means physically; $\sigma$ is a conductivity, hence a tensor, which used to be a scalar when magnetic field is absent. So what does it mean that the conductivity is rotationally invariant?
| A generic $n\times n$ matrix $\sigma$ can be decomposed as
$$\sigma = \sigma_0 + \sigma_H + \sigma_S$$
where $\sigma_0$ is proportional to the identity matrix, $\sigma_H$ is antisymmetric, and $\sigma_S$ is symmetric and trace-free. Explicitly for $2\times 2$ matrices,
$$\pmatrix{a&b\\c&d}=\frac{a+b}{2} \pmatrix{1&0\\0&1} + \frac{b-c}{2}\pmatrix{0&1\\-1&0} + \Delta \pmatrix{\cos(\varphi)&\sin(\varphi)\\ \sin(\varphi)&-\cos(\varphi)} $$
where $\Delta \equiv \sqrt{\left(\frac{a-d}{2}\right)^2+\left(\frac{b+c}{2}\right)^2} $ and $\tan(\varphi)=\frac{b+c}{a-d}$.
Writing $\mathbf E = \sigma \mathbf J$, we can examine the contributions to $\mathbf E$ from the various parts of $\sigma$. The scalar part contributes a scalar multiple of $\mathbf J$, while the antisymmetric part contributes a component which is orthogonal to $\mathbf J$.
The symmetric part requires additional interpretation. It defines two special directions in the plane, offset by $\varphi$ from the chosen reference axes. $\mathbf J$ is resolved into its components along these special axes, scaled by $\Delta$ in one direction and $-\Delta$ in the other, and finally added back together. You can see this in the following GIF:
Rotational symmetry precludes the existence of these two special directions, which means that the only contribution can come from the scalar and antisymmetric parts.
In slightly more mathematical language, the scalar part of $\sigma$ transforms (unsurprisingly) like a scalar, while the antisymmetric part transforms as a pseudoscalar in $2$ dimensions. Under proper rotations, both of these parts are invariant. The symmetric trace-free part transforms non-trivially under rotations, and so imposing rotational symmetry rules it out.
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Centre of mass of solid hemisphere I am trying to find the centre of mass of a solid hemisphere using theta as a variable. But I am getting the wrong answer. Can you please tell me where I am wrong. My working is shown in the image attached....
| The thickness of the thin disc cannot be $ R \ d \theta $. Otherwise a disc at $ \theta = 0 $ would be counted as having the same thickness as a disc (sweeping the same angle $ d\theta $) at angle $ \theta = \pi / 2 $, when clearly the disc is thicker at $ \theta = 0 $. So your expression for $ dm $ is missing a factor. I suggest drawing a bigger diagram with a bigger $ d \theta $.
| {
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As $SL(2,\mathbb{C})$ is a double cover of the Lorentz group, is $SL(2,\mathbb{Z})$ a discrete subgroup of the Lorentz group? The group $SL(2,\mathbb{C})$, the group of $2 \times 2$ complex matrices with determinent $1$, is a double cover of the Lorentz group. (These transformations can be understood as Mobius transformations on the Riemann sphere, which correspond to the action of Lorentz transformations on a sphere of light rays shooting out from the origin.) The Modular group $SL(2,\mathbb{Z})$ of $2 \times 2$ matrices with integer entries and determinent $1$, is a subgroup of $SL(2,\mathbb{C})$. Does this group therefore correspond to a discrete subgroup of the Lorentz group? What is its significance? How can it be thought of?
| As emphasized in the other answers, $SL(2,\mathbb{Z})$ is a discrete subgroup of the double cover of the Lorentz group (actually of its connected component), so $SL(2,\mathbb{Z})/\{\pm 1\}$ is a discrete subgroup of the Lorentz group itself. The purpose of this new answer is to say something about the geometric significance of this discrete subgroup.
To get some geometric insight, represent a vector with components $(t,x,y,z)$ as a $2\times 2$ matrix
$$
X :=
\left(\begin{matrix}
t+z & x+iy \\
x-iy & t-z
\end{matrix}\right)
$$
with $X^\dagger=X$ and $\text{det}(X)=t^2-x^2-y^2-z^2$, so that a Lorentz transformation can be expressed as
$$
X \mapsto M X M^\dagger
$$
with $M\in SL(2,\mathbb{C})$. The fact that $\pm M$ both give the same transformation of $X$ corresponds to the fact that $SL(2,\mathbb{C})$ is the double cover of the Lorentz group.
One thing to notice immediately is that the $y$ component of $X$ is invariant under every $M\in SL(2,\mathbb{Z})$. This follows from the fact that the real and imaginary parts of $X$ are separately self-contained under such a transformation, and from the fact that the reflection $y\to -y$ is excluded from the connected component. Therefore, after quotienting by $\{\pm 1\}$, the set of trasnformations with $M\in SL(2,\mathbb{Z})$ implements a subgroup of the Lorentz group of the three-dimensional spacetime with coordinates $t,x,z$.
Now, recall that the modular group $SL(2,\mathbb{Z})$ is generated by the two transformations
$$
\left(\begin{matrix}
0 & 1 \\
-1 & 0
\end{matrix}\right)
\hskip1cm
\text{and}
\hskip1cm
\left(\begin{matrix}
1 & 1 \\
0 & 1
\end{matrix}\right).
$$
The first of these implements the Lorentz transformation
$$
\left[\begin{matrix}
t \\ x \\ z
\end{matrix}\right]
\mapsto
\left[\begin{matrix}
t \\ -x \\ -z
\end{matrix}\right],
\tag{1}
$$
and the second one implements the Lorentz transformation
$$
\left[\begin{matrix}
t \\ x \\ z
\end{matrix}\right]
\mapsto
\left[\begin{matrix}
(3t+2x-z)/2 \\ t+x-z \\ (t+2x+z)/2
\end{matrix}\right].
\tag{2}
$$
Notice that the coefficients in this transformation are not all integers, although they are rational. As a check, we can confirm directly that this really is a Lorentz transformation:
$$
\left(\frac{3t+2x-z}{2}\right)^2
- (t+x-z)^2 -
\left(\frac{t+2x+z}{2}\right)^2
= t^2-x^2-z^2.
$$
The geometric significance of (1) is obvious, but (2) is more difficult to visualize.
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Should the 4D normalization constant $8\pi$ in Einstein field equations (EFE) be changed to $(n-2)S_{n-2}$ in other spacetime dimensions? Should the 4D normalization constant $8\pi$ in Einstein field equations (EFE) be changed to $(n-2)S_{n-2}$, where $S_{n-2}$ denotes the area of a $n-2$-sphere, in higher dimensions? The reason is that the factor $8\pi$ essentially comes from the Poisson equation and the Green function of Laplacian in 3D.
Or is this just a matter of convention?
| It is, indeed, just a matter of convention since Einstein equations in higher dimension are just mathematical, they have no currently known physical application. This means you could even drop any reference to units in them. The typical choice is to place $\kappa$ in front of the stress-energy tensor as a catch-all coupling constant as Cham mentioned in the comments. As a result, the equations are
$$R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu}\,.$$
But the truth is most mathematicians researching Einstein equations in $D$ dimensions are either way interested in ($\Lambda$-)vacuum space-times, which leads to equations without reference to $\kappa$
$$R_{\mu\nu} = \frac{2 \Lambda}{D-2} g_{\mu\nu} \,.$$
If you find solutions such as black hole space-times, where the "matter" is essentially a boundary condition at the singularity, then in does appear in the metric as a parameter, but you just define it to get the simplest form of the metric and usually do not care about "physical" numerical factors.
But yes, if you wanted to have a mass $M$ (defined by a $D$-volume integral of $T^{00}$) exerting a gravitational acceleration $\approx G_{\rm D} M/r^{D-2}$ in the Newtonian limit, then $\kappa = (D-2)S_{D-2} G_D$ is the right convention.
| {
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Does light have mass? Why? I've been wondering whether light has mass. Yet given the wave-particle duality of light, the statement seems to be affirmative. With that, how to calculate it?
| I would avoid mass concept of photon at all, because it doesn't have rest mass. Relativistic mass is very slippery thing and is not unambiguously defined. Someone just put Lorentz factor $$ {\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}} $$ besides rest mass (multiply by $m_0$) and has called it "relativistic mass". Now we know for sure that photon has NO rest mass, so you can't substitute something for $m_0$ in photon case. Thus photon doesn't have relativistic mass too !
However, it has momentum:
$$p = mc$$
using Einstein famous relation between mass and energy $E=mc^2$ and photon energy $E=h\nu $, we get momentum as:
$$ p = \frac{h\nu}{c} $$
Still a bit strange that some object without a rest mass can have momentum, but it is so !There are numerous experiments which proves light pressure. To me photon and fields in general are very strange things.
| {
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When is total pressure not conserved in a system? I came across a problem that involved two compartments that are separated by a movable, adiabatic wall. As the wall moves, the pressure is not conserved- rather total pressure decreases- assuming this is an ideal gas. How is this possible? Doesn't the second law of thermodynamics tell us that total pressure should be conserved?
| When you talk about conserving pressure it sounds like you think pressures are additive. You might be thinking in terms of adding the pressures before and after the wall moves. The initial pressures were 1 and 4 and you added them to make 5. After the wall moves you add 1.75 and 1.75 and get 3.5 and wonder why they don't add up to 5 as well. Is that correct?
If that's what you are thinking you are treating pressure as an extensive thermodynamic property, like mass, which it is not. Pressure, like temperature, is an intensive property.
Let's say you have a room filled with air at one atmosphere pressure. If you divide the room in half with a wall, will the pressure on each side of the be one half of an atmosphere? It's the same with the temperature of the air in the room. If the air temperature is 20 C in a room, will it be 10 C if the room is divided by wall? Of course the answers are no because pressure and temperature are intensive properties. Mass and energy, however, are extensive properties. If you divide the room in half, each will have half the mass and half the internal energy.
If that's what you were thinking, I hope this helped. If not, I will delete the answer.
| {
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Why do things cool down? What I've heard from books and other materials is that heat is nothing but the sum of the movement of molecules. So, as you all know, one common myth breaker was "Unlike in movies, you don't get frozen right away when you get thrown into space".
But the thing that bugs me is that things in the universe eventually cool down. How is that possible when there are no other things around to which the molecules can transfer their heat?
| This can be understood easily: If your temperature is higher than the surrounding temperature heat will flow out to the surrounding. It is analogous to electric current which moves from a higher potential to a lower potential. Similarly heat current flows from high heat potential(high temperature) to a lower heat potential(low temperature)
| {
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Is the Earth a gyro? Due to rotation and low friction, can the Earth be considered a gyroscope? If so, any interesting implications to this? Thanks
| The gyroscopic motion of the Earth is the reason we have seasons. For half the year the northern hemisphere is closer to the sun than the southern hemisphere, and for the other half of the year the opposite is true. This is because the axial direction of the Earth stays fixed as it rotates around the sun, a direct result of the gyroscopic motion of the Earth (resulting from the Earth's rotation).
| {
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How exactly is white light a combination of several wavelengths? I have read that light is an electromagnetic wave. Every ray of light has a specific wavelength. The colour perceived by any observer is dependent upon the wavelength of the incident light.
What I don't understand is that how do electromagnetic waves of different wavelengths combine to form a single wave of another wavelength? Simply put, I have the following two related questions to ask:
*
*When we look upon a, say, completely white object, what is the composition of the individual rays that strike our eyes? Are those rays waves formed by the addition of waves corresponding to individual wavelengths that constitute white colour? I get that white light is composed of all wavelengths of visible light, but how are those wavelengths combined into a single unit which we call white light?
*If it is so, then how are prisms able to disperse light into its constituent colours?
Also, as a side question, how does all of this relate to light being composed of photons?
| David White's comment is correct, and I think the existing answers are confusing the point. The poster asks:
How are those wavelengths combined into a single unit which we call
white light?
They aren't. There is no "unit" called white light. Our eyes have receptors for light of three wavelength ranges (graph here), and a collection of incident photons that excites each of them roughly equally is what we sense as "white." It's similar to blue and yellow light giving the sensation of green. There is no "wave interference" that makes a blue wavelength and yellow wavelength combine to green; it's solely an aspect of human perception.
While it is physically possible to make "white" light with electromagnetic wave packets that are very short, and so spread out in frequency throughout the visible range, this is neither necessary nor typical.
| {
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Will a can filled with vacuum move when we let in air?
This picture is from L.C.Epstein's book Thinking Physics. The upper can is filled with compressed air, and, when an opening is made on the right, the air comes out and the can shoots left. The question is what happens to the lower can, filled with vacuum, when we similarly make an opening. Does it move left - right - not at all?
Epstein says that the lower can doesn't move at all, "except for a
momentarily slight oscillation about the center of mass". I'm not sure I understand this. The explanation is that the air incoming into the bottle provides force on the left inner wall to compensate for the lack of force on the opening, and this balances the force on the left outer wall from the outer air. Which seems convincing, but opens a path to more questions:
*
*Shouldn't the can still start moving from the moment we make the opening and until the air pressure inside the can is equalized with the outside air?
*If that in fact happens, why would it stop and return ("a momentary slight oscillation about the center of mass") and not simply continue moving right with the constant velocity it's acquired?
| If the pressure difference between inside and outside is the same (but opposite) then the force on the can in both situations is equal and opposite. If the geometry is the same then the pressure difference evolves in the same way but with opposite sign. The motion of the can, opposite, is only different due to difference in the amount of gas resistance.
So I don't fully agree with Epstein on this.
| {
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In de Sitter space, does the cosmic horizon change its shape for fast-moving observer? If an observer moves at a speed close to the speed of light, will the horizon deviate from spherical shape?
If no, will it be the same horizon as for stationary observer (at the same position)?
| No, unless the observer is accelerating. A linearly accelerating observer will perceive a non-spherical Rindler horizon at minimum distance $c^2/a$, where $a$ is proper acceleration. This Rindler horizon will be closer than the cosmic horizon as long as the observer's proper acceleration is greater than $c^2/R_u$, where $R_u$ is the radius of the observable Universe.
| {
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Deriving the Heat capacity from Fermi-Dirac statistics I was watching the lectures on Solid state physics by Steve Simon (Oxford). He was explaining how to find Heat capacity of metal due to electrons from Fermi-Dirac statistics. You can write the total number of electrons as $$N =g(E_f) \int_0^\infty \frac{E^\frac{1}{2}}{1+e^{\beta(E-u)}}dE.$$
Here, $g(E_f)$ is the density of states. Now, from this formula, if you know $N$ and temperature, you should be able to figure out $u$ (chemical potential). Then, you can find average energy by averaging the integral above with $E$ multiplied. From there, you can get specific heat by differentiating with respect to temperature. He didn't do it this way as it involves a lot of algebra according to him. I tried solving the integral but couldn't do it. Can anyone solve it or provide some reference?
| There is the handwaving argument that the width of the edge is about $kT$ and that increasing the temperature will cause electrons to occupy states that are something like $kT$ higher in energy.
So the electron energy increases with $T^2$. The electronic heat capacity $c_v$ is the derivative, the cause of the linear term in the low-temperature specific heat of metals.
If you want to try calculate it is probably necessary to assume a constant DoS and a value of $kT$ much smaller than $E_F$.
| {
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Measuring acceleration due to gravity in the lab I am measuring the acceleration due to gravity in the lab with an electromagnet apparatus.
My textbook says to take the average time for a number of falls (keeping the height constant of course).
But I recall many moons ago being told to take the shortest value, not the average.
The explanation being that the electromagnet can retain some magnetism for a short time after being switched off. So the ball can take longer than it should but not shorter.
Any thoughts on this?
Should I go with the shortest value or the average?
| Generally there are two types of errors in an experiment, random errors and systematic errors. In this case there is a random error due to your limited ability to record the time of fall precisely. There may or may not be a systematic error due to the fact the electromagnet does not release the ball the instant you press the switch.
Random errors show up in your measurements because they are random. That is, when you measure the same thing many times you get results that are scattered. We generally assume the errors follow a normal distribution, so then we can calculate a standard deviation $\sigma$, and the final standard error from doing $N$ measurements is $\sigma/N$.
Systematic errors are much harder to find because they can't (or at least can't easily) be spotted from a statistical analysis of your results. If your electromagnet took e.g. 0.1 seconds to release the ball after you pressed the switch you wouldn't easily be able to spot this. So as a general rule we do the best we can i.e. calculate the random error. That's why you are being told to take the average value of the time (and the standard deviation?). If there is a systematic error due to the magnet you would have to find that by other means. For example you could do the experiment for a range of different heights.
| {
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In the Stern-Gerlach experiment, why is there a nonzero force even though the atoms were electrically neutral I know that the magnetic moment of a particle is given by:
$\vec{\mu} = \frac{gq}{2mc}\vec{S}$
I know that in the Stern-Gerlach experiment, neutral silver atoms were used. Additionally, the deflection in this experiment was due to the force $F = \nabla (\vec{\mu} \cdot \vec{B})$.
How is a nonzero force experienced, given that $\vec{\mu}$ is dependent on charge $q$, which is zero for silver atoms.
| The Stern-Gerlach experiment is famous because it verified quantization of angular momentum in quantum mechanics. However, your question is really a question about classical electromagnetism. For example, you can replace the silver atom with a loop of wire carrying a current, and the question is the same: why does the loop experience a nonzero force, when its charge is zero?
The answer is that when you have a mixture of particles with different charge-to-mass ratios, the relation between the magnetic moment and the angular momentum doesn't have to hold. It holds only for each type of particle, not for the aggregate.
| {
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Motivation for introducing quantum field theory in particle physics Why is it so that because particles can be destroyed and recreated we introduce QFT? I read at the begining of some textbook that this is so. My main problem is not the rest of the book but the first motivation for introducing QFT for modeling.
My thinking...till now every quantum operator id est observable was attached to a particle in question but when you have variable number of particles you can not do this so you imagine that there is a more fundamental thing which we observe and one observable is also the number of particles.
Also, somewhere I read that because we are now in relativistic regime we have to define observables that are spacelike separated to commute. And because of that also we have to define observables as functions of spacetime points.I dont see it.
| Quantum Mechanics is about mechanics, and Quantum Field Theory is about fields. Given that all the forces in nature are described by fields, this would mean that QFT is the more fundamental theory. In fact, we can describe QM as a zero dimensional QFT. Zero dimensional as particles are zero dimensional.
It turns out that QFT requires particle creation and annhilation, hence it's also called the many particle theory.
QFT is usually motivated in most textbooks as the unification of the relativity principle from Einsteinian mechanics and QM. Hence it is a partial unification of dream theory that physicists are busy searching for, that is a full quantisation of General Relativity and which would require quantising the metric.
| {
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Standard Model Lagrangian and Euler-Lagrange Equations First off, note that I only know physics through quantum mechanics, so forgive me if this is a foolish question. I've seen the Standard Model's Lagrangian density written out in full. My question is, could one send this through the Euler-Lagrange equations to get a system of equations describing the evolution of the fields, or does quantum field theory have different ways to extract information from it?
| That is exactly what you do.
Note that you don't need quantum field theory till the very end.
You've got your nice Lagrangian density $\mathcal{L}$, and apply the classical field theory toolkit of symmetries (global, local), gauge covariant derivatives and Euler-Lagrange equations.
You then get the equation of motion describing your (classical) field.
The quantumness of the problem arises when you want to write you classical field $\psi(x)$ in terms of creation and annihilation operators, $\psi(x) \propto a + a^\dagger$, because you want to quantise the theory. I.e. write in terms of discrete excitations, be it waves on a string or particles.
Note: for the Dirac equation, for example, you'll find that you need at least a $(4\times1)$ object (called a spinor) to have a non-zero solution. Though you don't need to quantise the theory to get to this result, quantum mechanics allows you to interpret this as the presence of spin.
| {
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Why do thermal cameras work at infrared radiation and not at visible one? From the Wien's law I have computed the Spectral Irradiance of a black body at 1000°C.
From different thermal cameras manufacturers web sites, a lot of Thermal Cameras work in the range 6-14 $\mu m$. These cameras are able to measure usually from 200 - 1400 °C.
From the following plot the maximum of the radiation is in the rage 2-3 $\mu m$ (in the visible band) not at 6-14 $\mu m$. So my question is:
Why do thermal cameras work at infrared radiation and not at visible one?
I try to plot the Wien's law for black body at 200°C but the maximum is around 6 $\mu m$ not at $\frac{6+14}{2}=10$.
| The simple answer is that visible light does not correlate with temperature as well as infra-red light does.
The EM waves received by the camera are a mixture of radiated waves and reflected waves from ambient sources. For example, consider your laptop on your desk in the daylight- it will be emitting black-body radiation and reflecting black-body radiation from the sun. Since the black-body spectrum from the sun is dominated by light in the visible range, the overall radiation from your lap-top is predominantly visible, and its intensity is more influenced by the colour of your laptop than by its temperature.
If you want to take an image that correlates better with temperature you have to filter out the EM radiation that is most strongly represented by reflected visible light, ie the radiation that represents the colour of objects rather than their temperature.
If you want to consider a much more extreme example of the same principle, imagine a radio transmitter, a camera and a radio. Both the camera and the radio receive EM radiation from the transmitter. The camera is sensitive to the visible radiation given off by the transmitter, and is useless for detecting the radio signal. The radio is very good at picking up the radio signal, but cannot take a photograph. A device that created an 'image' that was a mix of the visible and radio radiation strength might be an interesting toy to play with!
| {
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How is pressure related to the energy cost Consider the following problem found in this webpage.
Consider a spherical bubble of radius $R$, of a certain fluid of density $\rho$, trapped inside of some other
fluid. The bubble is stabilized by the presence of surface tension. Namely, suppose that the bubble has a
nearly, but not perfectly, spherical surface, which we describe by a function $\zeta (\theta, \phi)$, denoting the difference
$\zeta = r - R$ between the actual radius $r$ and the original radius $R$. One can then write the energy cost of
this deformation as:
$$ E = \alpha \int d\theta d \phi \sin\theta(R+\zeta)^2\sqrt{1+ \left(
\frac{1}{R+\zeta} \frac{\partial \zeta}{\partial \theta} \right)^2
+ \left( \frac{1}{(R+\zeta)\sin \theta} \frac{\partial \zeta}{\partial \phi}\right)^2} .$$
The problem is then to argue that the pressure (difference from equilibrium pressure) at the surface of the bubble is:
$$ P = \frac{2 \alpha \zeta}{R^2} + \frac{\alpha}{R^2} \nabla^2 \zeta $$
where $\nabla^2$ is spherical Laplacian.
I am not sure even in general case if one is given energy cost $E$ then how one would obtain pressure $P$. Is there a general definition or procedure that one would do?
| I think one can get the expression for $P$ without that for $E$. Let $p_o$ be the pressure due to the fluid outside and $p_i$ be the pressure due to the fluid in the bubble. Since the bubble is initially spherical,
\begin{equation}\tag{e1}\label{e1}
p_o - p_i = \frac{2\alpha}{R}.
\end{equation}
When the bubble is deformed its radius is given by $r = R + \zeta(\theta, \phi)$ so that the equation of its surface if $f(r, \theta, \phi) = 0$ where $f = r - R - \zeta(\theta, \phi)$. If the pressure inside the bubble is $p_f$,
\begin{equation}\tag{e2}\label{e2}
p_o - p_f = \alpha\Delta f,
\end{equation}
where $\Delta$ is the Laplace operator in $r, \theta, \phi$. Note that $R$ is a constant, so that
\begin{equation}\tag{e3}\label{e3}
\Delta f = \frac{2}{r} - \frac{\nabla^2\zeta}{r^2},
\end{equation}
where $\nabla^2$ is the Laplacian in $\theta, \phi$ alone. Now,
\begin{equation}
r = R\left(1 + \frac{\zeta(\theta,\phi)}{R}\right).
\end{equation}
If $\zeta(\theta,\phi) \ll R$, we can approximate
\begin{equation}\tag{e4}\label{e4}
\frac{1}{r} = \frac{1}{R} - \frac{\zeta}{R^2}.
\end{equation}
Similarly,
\begin{equation}\tag{e5}\label{e5}
\frac{1}{r^2} = \frac{1}{R^2} - \frac{2\zeta}{R^3}.
\end{equation}
Substitution equations (e5) and (e4) in (e3) we get
\begin{equation}\tag{e6}\label{e6}
\Delta f = \frac{2}{R} - 2\frac{\zeta}{R^2} - \frac{\nabla^2\zeta}{R^2},
\end{equation}
where we have ignored the term $\zeta\nabla^2\zeta$ it being of a higher order in $\zeta$. From equation (e2) and (e6),
\begin{equation}\tag{e7}\label{e7}
p_o - p_f = \frac{2\alpha}{R} - \frac{2\alpha\zeta}{R^2} - \frac{\alpha\nabla^2\zeta}{R^2}
\end{equation}
Subtracting (e7) from (e1) we get
\begin{equation}\tag{e8}\label{e8}
p_f - p_i = \frac{2\alpha\zeta}{R^2} + \frac{\alpha\nabla^2\zeta}{R^2}.
\end{equation}
$p_f - p_i$ is the difference in the deformed bubble from its equilibrium pressure.
The integral in the expression for $E$ is just the area of a deformed sphere. You may want to refer to one of my questions for more details.
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About the pressure of a confined gas According to fluid mechanics, we have Pascal's principle $P_2 = P_1 + \rho gh$. So, the pressure of a confined gas is different depending on the depth.
However, in thermodynamics, we have another formula $PV = nRT$. They use a confined gas to use this formula. Here, what is the $P$? Does $P$ refer to the pressure exerted onto the piston by a gas or the average pressure?
Or, since the density $\rho$ of the confined gas is so low that we can neglect the pressure difference?
Also, on the microscopic level, they say the reason why the (ideal) gas exerts on the walls is that every molecule collides with the walls. In this case, how can we explain the difference of the pressure depending on the depth of the wall?
| The ideal gas law assumes constant pressure and temperature throughout the volume of the gas. In case these quantities were vary in the volume, you can use a local form of the law, for instance: $$P=\rho \dfrac{R}{M}T$$
where $P$, $\rho$ and $T$ are the local pressure, density and temperature, respectively, and $M$ is the molar mass of the gas.
As you said, variation of a gas pressure with depth is usually neglected due to the low density of gases, unless for very large values of $h$, like the ones encountered in atmospheric fluid dynamics.
On the microscopic scale: In statistical mechanics, the average kinetic energy of gas molecules depends only on absolute temperature, thus, molecules will have the same energy anywhere in the container if temperature is uniform. However, the gas density is higher at the bottom of the container, and there will be more collisions per unit area per unit time, which translates to a higher pressure.
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Is there a way to inspect thin metal weld (2 mm - 3mm) by ultrasonic testing? I am working with an ultrasonic device to inspect welds. So far, I have learned that the minimum thickness of the metal sheet for this inspection is 6mm - 8mm. But the product of mine has 3mm thick welds:
I have basic knowledge working with angle beam transducer to inspect thick metal welds. How can I inpsect thin metal welds by using conventional ultrasonic device?
| If the thickness is less than 3-4 of $\lambda$ bulk waves cannot propagate (limits to min. 5-6mm). You need "normal" or so called Lamb waves, you can achieve them with a wedge sensor (inclined angle to the surface, goes definitely below 3 mm) or submerging methods (which is not really very practically).
see e.g. Comparison of Ultrasonic Non-Contact Air-Coupled Techniques
Lamb Wave interaction
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Differentiability of wave function at boundary in infinite square well I was told in class that a wave function should have the following properties:
*
*Finite and single-valued
*Continuous
*Differentiable
*Square integrable
But if we consider the wave function in an infinite square well, the wave function isn't differentiable at the boundaries since $\Psi (x)$ is:
$$\Psi (x) =\begin{cases}
\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L}), & 0<x<L \\
0, & \text{elsewhere}
\end{cases}$$
This violates one of the properties of wave functions. So how is this an acceptable wave function?
| If you consider the differentiability of the wavefunction at the boundary from inside an infinite square well, you find: $\frac{d\Psi(x)}{dx}$ = $\sqrt{\frac{2}{L}}\frac{\pi}{L}$ as $x\rightarrow0^{+}$, and $\frac{d\Psi(x)}{dx}$ = $-\sqrt{\frac{2}{L}}\frac{\pi}{L}$ as $x\rightarrow L^{-}$. Outside the well, you obtain: $\frac{d\Psi(x)}{dx} = 0$ at $x=0$ and $x=L$. The derivatives do not match, hence the function is not differentiable at either boundary of the well.
This is easily seen by graphing the hybrid wavefunction (for say the ground state, $n=1$) and noting the sharp points.
The infinite square well is one of the first problems taught in undergraduate QM. It is an attempt at taking a free particle and trapping or localizing it over a finite domain, in order to obtain a normalizable wavefunction and demonstrate novel effects like energy quantization.
The problem is, you physically cannot produce a free particle over some finite domain, and then on either side arbitrarily "clamp it down" by imposing an infinite potential to instantaneously drive the wavefunction to zero. This is why you end up violating basic properties of the wavefunction, as you have mentioned.
You could, for instance, use a Fourier transform to build up a wave packet from a sum of plane waves of varying amplitude and momentum, approximating a sinusoidal solution over $0<x<L$, however, the probability of finding the particle outside of this domain would not sharply change to zero, as suggested by the infinite square well problem.
Finally, other issues arise as a result of this problem. Namely, the relationship $E=\frac{p^2}{2m}$ does not hold, where $p=\hbar k$, $k$ the wavenumber associated with the momentum, and you end up obtaining a continuous not discrete distribution of momenta, even though the energy is quantized. Carefully reading the Wikipedia article https://en.wikipedia.org/wiki/Particle_in_a_box helps elucidate these issues.
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Two-coupled oscillator: Doubt in finding normal modes and natural frequency I want to find the natural frequency of a two coupled oscillator system like this-
My book does it this way but I don't really get it.
The equations of motion for the pendula are-
$$I\frac{d^2\theta_1}{dt^2}=−M_\text{eff}\ gL\sin \theta_1− \kappa l^2(\sin \theta_1−\sin \theta_2)$$
$$I\frac{d^2\theta_2}{dt^2}=−M_\text{eff}\ gL\sin \theta_2+ \kappa l^2(\sin \theta_1−\sin\theta_2)$$
To find the natural frequencies of the system, we take the sum and subtraction of the equations and we obtain (Using small angle approximation):
$$I\left(\frac{d^2\theta_1}{dt^2}+\frac{d^2\theta_2}{dt^2}\right)=−M_\text{eff}\ gL(\theta_1+\theta_2)$$
and
$$I\left(\frac{d^2\theta_1}{dt^2}-\frac{d^2\theta_2}{dt^2}\right)=−M_\text{eff}\ gL(\theta_1−\theta_2)−2\kappa l^2(\theta_1−\theta_2)$$
The two equations above are uncoupled and represent the two normal modes of the coupled system. The $\theta_1+\theta_2$ mode or ‘+’ mode represent the in-phase motion of the pendula where both the pendula are moving with same phase (same direction). The $\theta_1−\theta_2$ mode or ‘−’ mode represent the out-of-phase motion of the pendula where the pendula are moving with opposite phase (opposite direction).
I have marked the portions I don't understand in bold above.
Doubts:
*
*What is meant by uncoupled?
*Why does the two equations represent the normal modes?
*Why does $\theta_1+\theta_2$ represent in-phase and $\theta_1-\theta_2$ represent out of phase motion?
| Consider $\theta_{\rm sum} = \theta_1 + \theta_2$ as one variable, and $\theta_{\rm diff} = \theta_1 - \theta_2$ as a second variable.
The two equations become
$$ \begin{aligned}
I \frac{{\rm d}^2}{{\rm d}t^2} \theta_{\rm sum} & = − \left(M_\text{eff}\ g L\right) \theta_{\rm sum} \\
I \frac{{\rm d}^2}{{\rm d}t^2} \theta_{\rm diff} & = − \left( M_\text{eff}\ gL−2\kappa l^2\right) \theta_{\rm diff}
\end{aligned} $$
Now it is clear they are two decoupled equations. Each differential equation is only in terms of one unknown.
*
*Now if they two pendulums where out of phase by the same amount $\theta_{\rm sum} = \theta + (-\theta) =0 $ and $\theta_{\rm diff} = \theta - (-\theta) = 2 \theta$. So the second equation describes the out of phase vibration.
*Conversely, if they are in phase $\theta_{\rm sum} = 2 \theta$ and $\theta_{\rm diff} = 0$, which means the first equation describes the in-phase motion.
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Damped Oscillation and Period In my school experiment, I wanted to measure the gravitational constant ($9.81 \ m/s^2$) by using a pendulum. If we take into account the damped oscillation (i.e. friction forces), does that affect the period?
As far as I know, the position $x$ of a particle undergoing Simple Harmonic Motion can be expressed as a function of time: $$x(t) = e^{-t} \sin t. $$
But the answer in the following link says that the friction force might be affected by factors like the velocity, or perhaps the position of the pendulum bob:
Does damping force affect period of oscillation?
In addition, since I am conducting an experiment to find the $g$ value, I am concerned if the changing period affects the value of $g$.
Could anybody clarify?
| The free vibration frequency is affected by damping.
If you assume the damping force is proportional to velocity, the math is well known. See http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html for example, particularly the difference between the damped frequency $\omega_1$ and the undamped $\omega_0$ in the last equation.
The answer in the link you posted is hand-waving, not rigorous math, but it gives the right physical idea of why the frequency changes.
You can find the amount of damping by measuring how much the amplitude of oscillation decreases from one cycle to the next.
For a pendulum swinging in air, the damping force is not proportional to the velocity (it is more likely to be proportional to the square of the velocity) but for small amounts of damping, you can still fit a curve of the form $x = A e^{-\gamma t}\cos \omega_1 t$ to the measured displacements and then find the undamped frequency using $\omega_0^2 = \omega_1^2 + \gamma^2$ as in the hyperphysics link.
For a pendulum, the change in frequency is likely to be small, but it is still worth checking if the correction to your experimental data is significant or not.
| {
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Does a maximum thermometer assume a room temperature? Consider a liquid-in-gas maximum thermometer. That is usually conventional thermometer designed for measuring body temperature.
Wikipedia says:
One special kind of mercury-in-glass thermometer, called a maximum thermometer, works by having a constriction in the neck close to the bulb. As the temperature rises, the mercury is pushed up through the constriction by the force of expansion. When the temperature falls, the column of mercury breaks at the constriction and cannot return to the bulb, thus remaining stationary in the tube. The observer can then read the maximum temperature over the set period of time. To reset the thermometer it must be swung sharply. This design is used in the traditional type of medical thermometer.
If I understand it correctly, the the liquid in the bulb has body temperature, but the liquid in the pipe cools quickly down to the room temperature. If it did not, the bar in the pipe would partially go down when the thermometer would be taken out of the body…
But if I understand correctly, the thermometer assumes some range of room temperature in order to show the value within some defined error. So, measuring body temperature outdoor in cold winter or hot summer would give bad results. Am I correct?
| You are in principle right, that the mercury (or whichever liquid you use) in the "tube" part of the thermometer could expand or constrict independently from the temperature the thermometer bulb is at. However, the expansion/constriction of the mercury is volumetric and the volume of liquid in the bulb is generally much larger than the volume of liquid in the tube. Therefore the majority of the thermal expansion is due to the temperature of the bulb, and possible other expansion/constriction in the tube would not contribute significantly to the temperature reading. This is true for any liquid-filled thermometer regardless if it's a maximum thermometer or not, which is why you can use an old mercury thermometer to measure, e.g., the temperature of the contents of a lab beaker with the stem sticking up in the free air.
| {
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On work done by internal forces which is coming out to be not equal to zero 1) Let us consider a block which explodes due to some internal mechanism into two smaller fragments of equal masses.The system was initially at rest and now is having some finite kinetic energy(due to momentum conservation).We can hence comment that the work has been done by the internal force by the Work-energy Theorem since there are no other forces acting on the system.But this seems to contradict the fact that work done by internal forces is always 0.Where am I going wrong? I have researched similar questions on stack and other site but to no avail.
Also,textbooks for some reason do not consider a lot of theory on this matter for some reason which adds to my woes.
2) I have another question that in a two mass spring block system does the spring do any work?It should be 0 according to me as it is an internal force when solving from COM frame but is this also true from a ground frame?While writing the work energy theorem on this system, would the spring work show up even in the form of potntial energy?
| The answer to your first question is that work done by internal forces only sums to zero in the case of rigid bodies, so the principle does not apply to an exploding body.
The same is true for the two blocks linked by a spring- it is not a rigid body.
The two cases are analogous. In the case of the exploding block, potential energy stored in the explosive was converted into the KE of the two moving parts. Likewise the spring can store PE which is converted into the KE of the blocks.
| {
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Why doesn't hot charcoal glow blue? I was learning about black body radiation and the explanations given by Max Planck and Albert Einstein when a thought crossed my mind.
When we heat an iron piece, its color changes gradually from red, orange, yellow to bluish white. Yet such a change is not visible in a glowing piece of charcoal obtained from wood. Why is it that, wood charcoal is not able to glow in colors of higher frequencies?
| Iron is supposed to glow blue-white at around 1600 degrees C. page 7.
Would carbon have that color at that temperature?
Spectral lines of iron
Spectral ines of carbon
Maybe not. Maybe those bright yellow lines would add too much. Carbon might have to be hotter.
Do we have examples of carbon heated very hot? How about a carbon arc lamp? Pass a whole lot of electricity between two pieces of carbon. Some carbon vaporizes, and the hot carbon vapor gets very bright.
Does this look bluish-white to you? Maybe it's possible, and maybe the charcoal has to be vaporized to do it. Can you find a way to get charcoal to do that by burning it with pure oxygen? I don't know. Can you get that color with solid charcoal, or would the carbon vaporize first? Carbon's sublimation point is 3642 C, so maybe.
Sorry I can't give a more definite answer.
(Here's a tip if you want to look for yourself. If you do an image search that includes "white-hot" or "blue-white-hot", be sure to set your censor. For reasons I didn't think of ahead of time, these searches produce a lot of porn.)
| {
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The potential at a point According to my book, 'The potential at a point is said to be 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.' But I wonder how it is possible. As the charge is being brought from infinity, the work done = force * infinity, thus, the work done would be infinity indeed. Please help me out.
| Force is dependent of distance $r$, so thing move like this
$$\newcommand{\newln}{\\&\quad\quad{}} \begin{align}&\int^{r_b}_{r_a}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_a-U_b) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_r-U_\infty) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}} =-U_r ~~~~~~~ [U_\infty = 0]\newln \Rightarrow \int^{r}_{\infty}k\cdot\frac{q.q_o}{r^2}d\mathrm{\mathbf{r}}=-U_r ~~~~~~~ [\textrm{Coulomb's Law}]\newln \Rightarrow kq\cdot q_o\int^{r}_{\infty}\frac{1}{r^2}d\mathbf{r}=-U_r\newln \Rightarrow kq\cdot q_o\left[\frac{-1}{r} \right]^r_\infty=-Ur\newln\Rightarrow \frac{-kq.q_o}{r}=-U_r\newln \Rightarrow U_r=\frac{kq.q_o}{r}
\end{align}
$$,
And potential energy is $(-) work done by conservative force.
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Physical reason for $T^2=a^3$ when $T$ is in years and $a$ is in AU Kepler's third law states $$T^2\propto a^3$$
When $T$ is in years and $a$ is in AU, the proportionality constant becomes $1$. This can't be a coincidence; I would like to know the physical reason for it.
| It's not really a physical reason but you are correct in suspecting that this shouldn't be a coincidence. The fact that the proportionality constant becomes unity follows from the definition of a year and an astronomical unit. A year is defined precisely as the period of time it takes for the Earth to complete a revolution around the Sun. An astronomical unit is defined precisely as the distance from the Earth to the Sun. Since the orbit of the Earth is nearly circular, one can take the semi-major axis to be approximately the same as the radius of the Earth's orbit, which is the same as the distance from the Earth to the Sun in this approximation. Thus, if $T^2=ka^3$ then in the units of a year and $\text{AU}$ for $T$ and $a$ respectively, for the case of the Earth, by definition, both $T$ and $a$ are $1$. This determines $k$ to be unity.
Edit
Since other responses to the post invoke the precise form of the gravitational force, I would like to point out that while Kepler's law obviously follows from the inverse-square nature of the gravitational force, the fact that the proportionality constant for the relation $T^2\propto a^3$ is unity is simply a matter of definition--explaining it doesn't even need to invoke the nature of the force. If Kepler's law had been $T^2\propto a^5$ (i.e., if the force hadn't had the inverse-square nature) then also the proportionality constant would have been unity given the definition of the said units.
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Understanding simple LC circuits I'm trying understand the physics of simple inductor-capacitor circuits such that there is just an inductor L and a cacpacitor C and a switch.
Imagine first that the capacitor is fully charged and the switch is then closed.
I do not understand why the current increases from an initial low value as the charge difference between the plates DECREASES because this is in direct contradiction to how a capacitor discharges in isolation.
I know the solution lies in the inductor being present but I can't seem to follow the physics of cause and effect to understand it properly.
Any illumination would be appreciated.
| Actually, LC Circuit is cause of LC oscillation,
When you apply kvl you get.
$Q/C=Ldi/dt$, and $dQ/dt$ so after double differentiating we get
$Q/C=Ld^2Q/dt^2$, which look quite same as shm equation $a=-w^2x$,so we get $w=1/(LC)^1/2$, so energy oscillate,
At any instant instant energy will. Be equal. To. Field. =energy stored in capacitor +energy stored in inductor. Like we have in simple pendulum shm, I think if you know how actually inductor work you can easily answer your own question now.
| {
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Is tension always the same as centripetal force? For example, if a ball is attached to a string and released from a vertical height and then pivots around a point to initiate circular motion, tension is equal to centripetal motion.
If, on the other hand, a ball hands from a string and it’s hit in such a way that it travels in a vertical circle. Tension is not just equal to centripetal force.
When is tension equal to centripetal force, and when is it another value?
The scenarios above are taken from previous problems I’ve seen in class. I’m not sure if I’ve explained them as clearly as needed, but I think the general idea should be understood.
| "Centripetal" is Latin for "towards the center." A centripetal force is not a particular type of force like a frictional force or a magnetic force. It's just a force that makes an object go in a circle. The word "centripetal" describes the direction of the force, not the type of force.
When a car drives around in circles on level ground, the centripetal force is a frictional force. When we whirl a ball around on a string, the centripetal force is a normal force of the string on the hook it's tied to, and the magnitude of this normal force is equal to the tension in the string. (Tension is not a type of force.)
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Can someone explain what is the force the ball will exert? If a ball is falling under free fall then the force exerted by the ball on the ground would be $mg$. But that's not the case in real life ball would hit with more force. But when i draw free body diagram there is only one force that is acting on it $mg$
Can someone explain what is the force the ball will exert ?
|
the force exerted by the ball on the ground would be $mg$
Why do you say that?
The force that the ball exerts on the ground depends on the modulus of elasticity (stiffness) of the ground and can be very high. The normal reaction of the ground stops the ball over a certain distance. If the modulus of elasticity of the ground is high, the stopping distance is short and the force of reaction is high.
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Why Don’t We Use Radios for Propulsion? We’ve all heard the idea of laser propulsion before, but why can’t a high powered RF source be used to propel a body? A simple high current coil with an AC signal applied to it to it could produce 100s of watts if not kilowatts of EM power. So why not use these instead of our underpowered 10W Red lasers?
What makes RF propulsion energy inefficient?
| In principle the thrust-to-radiated-power ratio of all electromagentic radiation drives is identical.1 Frequency doesn't enter directly into the issue.
So it is tempting to say "use the band with the highest ratio of radiated-power-to-supplied-power". Good idea.
But then there is the question of collimation and the influence of diffraction.
However you produce it, your beam is going to try to spread out. Applying acollimating system is generally a good idea, but diffraction limits the degree of success you can have based on the size of the collimator (big is good) and the frequency/wavelength of the EM signal (high-frequency/short-wavelength is good). Qualitatively you should be thinking of the opening angle $\theta$ (expressed in radians) as being given by
$$ \theta \propto \frac{\lambda}{D} $$
where $D$ is the size of the aperture and $\lambda$ is the wavelength the constant of proportionality is geometry dependent.
The main problem with using radio-band (aside from the very low thrust of photon drive in general) is that without a large and therefore massive collimator you will lose non-trivial thrust to diffraction.
1 Identically tiny. You get about 3 micro-newtons per killowatt.
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Justification for excluding gravitational energy from the stress-energy tensor I did general relativity years ago at Uni and was just revising with the aid of Dirac''s brilliant book; the beauty of this book is that it is so thin and concise. On reading this book I find that I have a few questions regarding energy.
One thing I had not appreciated before was that the energy in the energy tensor only includes all energy excluding gravitational energy. Is this true? What is the evidence for this position? How could we know that this energy term actually excludes gravitational energy?
The only argument that I can see is that the energy in the energy tensor is not fully conserved, so you could infer that there is a missing energy term and that that energy is gravitational energy. But if you take the missing quantity and call it the gravitational energy, this quantity turns out not to be a tensor.
Hence, its form will in general always look different in at least some different coordinate systems regardless of whichever quantities you use to write it out in. This latter point might only have mathematical consequences, but does it have physical consequences?
| There are various definitions of the total global mass-energy contained in a spacetime: the ADM mass, Bondi mass, etc. In present understanding, these require specific conditions such as symmetry in time, or asymptotic flatness.
Then there are an awful lot of quasi-local proposals. The idea is that, since you can't define the gravitational energy at a point, draw a small box around a region of spacetime, and define the gravitational energy inside it. For instance, Hawking's proposal examines how light rays exiting the box diverge; the motivation is mass-energy curves spacetime which affects the light. The proposal by Epp (& Mann & McGrath), studies the acceleration of the walls of the box. Another proposal by Bartnik is more mathematically motivated, I'm told. Most definitions agree for simple cases, apparently.
| {
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Is there a limit of electrons a single hydrogen atom can have? Is there a limit of electrons a single hydrogen atom can have? If so what is it? why? Is the the answer to why scalable to helium?
| By definition, "hydrogen atom" refers to the neutral system with one proton and one electron, so it cannot hold any extra electrons.
However, protons can hold more than one electron, in which case the system is termed a hydrogen anion. This is a stable, bound system, and the reaction
$$
\mathrm{H}+e^- \to \mathrm{H}^-
\tag 1
$$
releases about $0.75\:\rm eV$ (an energy known as the electron affinity of the hydrogen atom), plus whatever kinetic energy the electron came in with, through the emission of a photon.
(As an aside, the hydrogen anion, and particularly the reaction $(1)$ above together with its converse in the form of photodetachment, is incredibly important ─ this is the reason why the Sun's spectrum is continuous.)
Free atoms of most elements tend to have positive electron affinities, which means that their singly-charged negative anions are stable systems, and they release energy when they capture their first extra electron. There's a few exceptions, though, starting with helium: atoms which have stable closed shells can 'reject' that extra electron, as it's forbidden from sitting in the closed valence shells and it's forced to sit at higher-energy shells that are too far uphill in energy to be stable.
If you want to up the game and go to a second extra electron, though, to get to $\rm H^{2-}$, the game runs out, and indeed it runs out for every element ─ all the second electron affinities are negative. That is, it takes work to cram a second extra electron in, and the resulting dianion will at best be in a metastable state that's ready and jumping to give that energy back out by dissociating into the single anion and a free electron. It's just too hard to try and hold two extra electrons (and their resulting mutual electrostatic repulsion) within the confines of an atomic system.
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Why impose invariance of the Lagrangian under infinitesimal coordinate transformations? I am reading Cubic order spin effects in the dynamics and gravitational wave energy flux of compact object binaries by Sylvain Marsat.
In section 2B the author imposes the invariance of the Lagrangian under infinitesimal coordinate transformations: which is the physical menaing of this?
| Rather than asking why we should impose this invariance, I think it would make more sense to ask why we should relax it. The Lagrangian is a relativistic scalar, which means that it has to be invariant under any coordinate transformation. An infinitesimal change of coordinates is just one type of coordinate transformation.
The physical interpretation is that coordinates don't in general have a physical interpretation. Coordinates are just names. A change of coordinates is just a renaming of points in spacetime. If we're going to derive equations of motion from a Lagrangian, we want those equations of motion to be true regardless of how we rename points in spacetime.
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Interpretation of the photon scattering rate? The photon scattering rate $\Gamma$ describes the rate at which photons scatter off an atom$^1$. In a two-level system, the ansatz for the photon scattering rate often is given by
\begin{equation}
\Gamma = \rho_{22}\gamma
\end{equation}
where $\rho_{22}$ is the probability to find the atom in the excited state and $\gamma$ is the rate of spontaneous decay. However, I don't see the connection between the ansatz above and what the photon scattering rate is physically meant to be.
$^1$In my imagination, the photon scattering rate is the absorption rate for photons at a certain frequency $\omega$. Hence $\Gamma(\omega)$ shows the saturation broadened Lorentzian absorption line of the atom, centered around a resonance frequency.
| If $\rho_{22}$ is the population of the excited state then for sure $\rho_{22} \gamma$ is the rate at which the atom is emitting photons. I guess the question arises from not realising that $\rho_{22}$ is itself dependent on the conditions the atom is under. For example, if the atom has been left alone for a while then you will have $\rho_{22} = 0$. If the atom is in an electric discharge then you will have $\rho_{22} > 0$. If the atom is located in a beam of light then $\rho_{22}$ will have some value which depends on the intensity and frequency of the light. It seems that the question has the latter scenario in mind. In that case you will have
$$
\rho_{22} = \rho_{22}(I, \omega)
$$
and a typical dependence on intensity and frequency (for a two-level model of the atom, in steady state) is
$$
\rho_{22} = \frac{(1/2)\gamma^2 I/I_s}{(\omega-\omega_0)^2 + \gamma^2/4 + \gamma^2 I/I_s}
$$
where $I_s$ is the saturation intensity. This leads to the Lorentzian function for $\Gamma(\omega)$ which the question asks about.
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Prove equivalence of the definition for coherent states Quote:
The coherent states $|z\rangle$ is defined as $$|z\rangle
=e^{-|z^2|/2}\sum_{n=0}^\infty\frac{z^n}{\sqrt{n!}}|n\rangle
=e^{-|z^2|/2} e^{a^\dagger z}|n\rangle,$$
which was, very understandable and mathematically easy.
However, I encountered an article that using translation operator to define the coherent states, See reference here page 28. Quote:
$$|\tilde{x}_0\rangle \equiv T_{x_0}|0\rangle=e^{-\frac{i}{\hbar}\hat{p}x_0}|0\rangle.\tag{4.14}$$
But this has suddenly be somewhat confusing, the former stated that coherent states is a superposition of infinite states of probability amplitude distribution of Gaussian wave pack. The later sates that coherent states is a spacial translation of $|0\rangle$ eigenstates.
Further, the first definition involve only $\hat{a}^\dagger$ while the latter involved $\hat{p}$, a linear combination of $\hat{a}$ and $\hat{a}^\dagger$.
Could you explain to me what's going on? Especially, what's the inituition between define coherent states from $T_{x_0}$? How to prove they are same or not the same?
| Note that later in the text on p. 34 the formula (4.54) explains exactly how Zwiebach's coherent states $|\tilde{x}_0\rangle$ are related to the standard coherent states.
The intuition is that actual position-eigenstates does not belong to the Hilbert/Fock space, so instead we roughly speaking consider Gaussian wavepackets, centered around $x_0$.
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Is this question physically possible? Question: Two smooth vertical walls stand on a smooth horizontal plane and intersect at right angles. A small smooth sphere of mass m is moving with velocity (4i + 3j) m/s when it hits one of the walls. It rebounds from the wall with velocity (i + 3j) m/s and goes on to hit the second wall. Given that the coefficient of restitution between the sphere and each wall is the same, find the total kinetic energy lost by the sphere in both collisions
Problem I have: When I worked on this I get an impulse from the first collision of -3mi (which wouldn't be perpendicular to the first wall). So surely this problem doesn't "work" (unless I have made a mistake)
| Yes the impulse isn't perpendicular to the wall so the restitutive coefficient will be both vertical and horizontal to the wall.
The fact that it says that the coefficient of restitution is the same for each wall implies that it is invariant to incident angle (not in any way realistic, but hey that's what logically must be concluded) and that these components will be the same.
Thus you can figure out the wall angle by balancing the horizontal and vertical components to equal. Then go on to the other wall.
More likely the examiner made a mistake and one of the i velocity component should have changed sign.
| {
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Dimensional regularization of a divergent integral Suppose there is an integral in four dimension Euclidean space
\begin{equation}
I_{d=4}=\int_0^\infty d^4x\frac{1}{|x|^2},~
\end{equation}
which is divergent. $|x|$ is the length of the vector. Can one use dimensional regularization to compute this integral by using $d^4x \to d^dx$,with $d=4-\epsilon$ ?
Or more abstractly my question is that If I want to compute an integral $I_{d=4}$, but it divergent for example at range $2<d<5$, can we use dimensional regularization by writing $d=4+\epsilon$. Then at the end of calculation let $\epsilon\to0$ ?
| The method of dimensional regularization in QFT comes with a few definitions which are crucial to evaluating integrals of this type. Following Zinn-Justin, they are the properties of these integrals under the following:
*
*Translations:
$$
\int d^d p \, F(p + q) = \int d^d p \, F(p)
$$
*Dilatations:
$$
\int d^d p \, F(\lambda p) = |\lambda|^{-d} \int d^d p \, F(p)
$$
*Factorizations:
$$
\int d^d p \, d^{d'}q \, F(p) G(q) = \left( \int d^d p \, F(p) \right) \left( \int d^{d'}q \, G(q) \right)
$$
From these properties, you can already address some of the integrals you have mentioned. In particular, the first two properties immediately imply the "identity"
$$
\int \frac{d^d p}{(2 \pi)^d} \frac{1}{(p + q)^{2\alpha}} = 0,
$$
for all $d$ and $\alpha$.
In the comments, you have also mentioned the integral
$$
\int_{\mathbb{C}} \frac{d^2 z}{(z - z_i)(\bar{z} - \bar{z}_j)}.
$$
You can consider applying dimensional regularization to this integral, either by introducing multiple copies of $\mathbb{C}$ or writing it as an integral over $\mathbb{R}^2$ and then generalizing to an integration over $\mathbb{R}^d$. You'll find that if $z_j = z_i$, the integral is zero in dimensional regularization, but if $z_j \neq z_i$, I see no reason why it should vanish.
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Should the ground state electron density of an atom go to zero at the origin? I have heard from my professor that the particle density of electrons (in the ground state) of an atom should vanish near the nucleus. Hydrogen is an obvious counter-example. So I am trying to work out what he could have meant? Which quantum phenomenon is he thinking of?
|
Hydrogen is an obvious counter-example.
Indeed it is. The claim as stated is false.
So I am trying to work out what he could have meant?
You'll have to ask him. There's no way for us to read his mind.
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Charm prediction: Required electroweak or just weak? Everywhere I look it up the prediction of the charm quark is predicted by the electroweak interaction, which of course recovers the weak interaction with the spontaneous symmetry break and the Brout-Englert-Higgs mechanism.
My question is, did the new features of the electroweak theory allow for the prediction of the charm quark? Or it could have been predicted with the old features of the weak theory?
As far as I am concerned what the electroweak unification gave rise to was the existence of neutral currents, i.e. interactions mediated by the $Z^0$ boson. But the prediction of the charm was made to avoid neutral current interactions in semileptonic decays with $\Delta S \neq \Delta Q$ as well as $\Delta S = 2$, which hadn't been observed. These neutral currents could have been avoided in weak theory because neutral currents hadn't been observed, or in electroweak because decays with $\Delta S \neq \Delta Q$ hadn't been observed.
So again, was electroweak theory necessary for the prediction of the existence charm quark?
Thank you.
| You might wish to move or repost your question to HSM where such issues are discussed.
Indeed, the Weinberg-Salam model did not figure at all in the historic GIM paper, 1970, not even as a reference. What is required for the suppression of FCNC is a "conventional mixing" of charged currents, reasonably well understood at the time.
(Now, Glashow and Bjorken in 1964 had speculated on a 4th quark, but without making it necessary, as a component of an explanation of a physical "fact on the ground" like GIM.)
Subsequently, Gaillard, Lee, and Rosner, 1975 discussed all components going into their anticipation of the mass of the charmed quark,
Gaillard & Lee 1974, but that was much later, after the discovery of neutral currents and hence the triumph of WS: so the latter prediction does reference Weinberg and Salam, but this is only because by that time 't Hooft had proven renormalizability of the model, so of course loop corrections should be mindful of and grateful for that...
| {
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Does the wavelength of a particle depend on the relative motion of the particle and the observer? The de Broglie wave equation states:
$$\lambda = \frac{h}{p},$$
where $\lambda$ is the wavelength of the “particle”, $h$ is Plank's constant, and $p$ is the momentum of the particle.
Momentum is usually written $\,p=mv$, where $m$ is the mass and $v$ is the velocity of the particle. But presumably $v$ is the relative velocity between the observer and the particle.
So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?
Or, perhaps more accurately, when a particle is incoming to another particle, in as much as an interaction between the particles depends on their relative speed, or the energy of impact, it thus also has something to do with their relative wavelengths.
Is that a conclusion, or simply a restatement of the premise, using different words that mean the same thing?
| Yes, the observed wavelength of a particle depends upon the relative motion between the observer and the particle. This is called the Doppler Effect.
| {
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Problem Regarding Buoyancy
A spherical marble of radius $1\,$cm is stuck in a circular hole slightly smaller than its own radius (for calculation purposes , both are equal) at the bottom of a filled bucket of height $10\,$cm. Find the force on the marble due to the water.
I have always been troubled by problems like this. Does the marble not displace a certain volume of fluid? Should a buoyant force not act on it? However, in this problem, the answer happens to equal the product of the pressure, and the projection area....
And, when I came across this similar problem :-
A steel ball is floating in a trough of mercury. If we fill the empty part of the trough with water, what happens to the steel ball?
The answer to this one is that the steel ball rises.
Here, instead of multiplying the pressure and area of projection, and arguing that a net downward force acts, we argue that the steel ball displaces water, and causes an upward buoyant force to act.
My question is, when does one know which force to apply?
| It all depends on the fluid contact.
Buoyancy comes about due to hydrostatic pressure differences on a submerged or floating object.
For submerged or floating objects, the fluid pressure acts on the submerged volume. Because fluid pressure increases with depth due to hydrostatics; when you submerge an object, the pressure at the top is less than the pressure at the bottom. This causes the net upwards force on the object that we call buoyancy. As long as the fluid is below the object, it will have buoyant force.
When your marble is in the bottom of the bucket with only fluid above it, it is not the same as being submerged. There is no higher pressure fluid below it, only a high pressure fluid above it, so the net force due to fluid acts downwards, not upwards as it does when the top and bottom faces of the object have pressure acting on them.
TL;DR: You need to see if the fluid is actually surrounding the ball as in the second case, or if it's just acting on top of the ball, as in the first case.
| {
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Where do we see centrifugal acceleration? A body in circular motion always possesses centripetal acceleration which is felt by a person sitting at the center of mass. It will not be felt by a person viewing the motion from the ground frame. Then where do we feel the centrifugal acceleration? We cannot be anywhere on the body itself because at every point we will perceive the motion to be circular by considering our point as the center.
| If you are sitting in any kind of vehicle which is moving along a curved path, then you feel the seat exert a centripetal force on you. If you are swinging a mass around in a circle on the end of a rope, then the force which the rope exerts on you might be considered centrifugal.
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Why do we need short wavelength to measure the position of a particle accurately? I am reading "A brief history of time" by Stephen Hawking. It is explaining the uncertainty principle:
... However, one will not be able to determine the position of the particle more accurately than the distance between the wave crest of light, so one needs to use light of a short wavelength in order to measure the position of the particle precisely.
Why is that? Why can't we determine the position of a particle accurately if the wavelength is long?
| It is analogous to the separation between the physical marks on a measuring tape or ruler. If you have a tape that is only marked to the nearest inch, you will not be able to make as precise a measurement as you might with a tape marked at 1/16th of an inch intervals.
| {
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Why does a chain or rope move the way it does when suspended and rotated on a vertical axis? I have always been interested in why objects like chains, ropes, etc. move the way they do when "rotated" around a vertical axis while being held only where it is suspended. It forms a shape if you will, resemblant of a "C" or a wave depending on the length. I am curious to know what laws and effects of physics are at play.
| If i understand correctly you spin the chain on an vertical axis with angular speed $w$. The 3-dimensional shape of the spinning chain is rather complex, but a 2-dimensional projection of this shape on a vertical plane resembles a standing wave. Where the nodes are the points where the chain is kept in the middle. If you increase the length of the chain you will have to adjust the angular speed, but more nodes will appear. And i guess that these nodes have the same relation with the length of the chain and $w$ as a simple 2-dimensional standing wave
| {
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Question about description of Gibbs free energy When introduced to the gibbs free energy, it was derived as follows:
First law: $dU=dq+dw$
Second law: $dS>dq/T$ for a spontaneous change.
Note $dq$ and $dw$ are inexact differentials.
Subsituting $dq=dU-dw$, into the second law gives us:
$TdS>dU-dw$
using $dw=-P_{ext}dV$
$Tds>dU+P_{ext}dV$
or,
$dU+P_{ext}dV - TdS<0$
Now, keeping pressure and temperature constant, we can say that:
$dU+P_{ext}dV - TdS<0$
= $d(U+P_{ext}V - TS)<0$
= $dG<0$, where $G$ is the gibbs free energy.
Here is my problem.
A few lectures later when we were being introduced to the idea of chemical potential, the gibbs free energy was re written as a function of pressure and temperature in the following way.
$dG=Vdp-SdT$, this expression was derived using the result above. My question is that if pressure and temperature were constant in the above expression, isnt $dp$ and $dT$ always 0? If so, how is this a valid expression of $G$?
| G is not defined by the equations you wrote. For a pure substance or a mixture of constant chemical composition, it is defined by $$G=U+PV-TS$$And this equation applies only to thermodynamic equilibrium states. So, $$dG=dU+PdV+VdP-TdS-SdT$$But since $dU=TdS-PdV$ we are left with $$dG=-SdT+VdP$$
| {
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Would looking out from inside or near a black hole be unimaginably bright? I have the following assumption based on limited knowledge.
A black hole appears dark to us because any light that would be emitted from it, and any light that passes nearby, is caught by the gravity well and either captured or bent from its original course.
If this is true, from any point within the gravity well or towards the center of a black hole, looking outward there would be immense amounts of light and radiation coming in.
Edit for specificity:
The idea is not necessarily that a human or object would be looking out while at a point, but theoretically a fixed point that has light and matter passing by. Of perhaps if we somehow invented something capable of not exceeding but matching the gravitational force and could stay still. Or a sensor reading at a certain point, pointing a camera or infinite lumen sensor outward.
| The light is not exactly bent from it's original course, the blackhole curves spacetime and the light follows that curvature. The thing is that the time component of "spacetime", as well, is bent near a black hole, so that as the light approaches the event horizon, the time it takes (from an outside perspective) for the light to get closer and closer to the event horizon gets longer and longer such that eventually, for the light to proceed an inch closer to the horizon, in the necessary time elapsed the universe would have died. So there is, seemingly, a paradox here, in the "time" it takes you to "perceive" anything at the event horizon (assuming you somehow can survive the gravitational tides) the universe would have ended. Of course your experience is separate and equal from an outside perspective and time would feel normal to you, but the question is, if the universe in which you are contained is destroyed in a different perspective, what happens to you from your own perspective? There really is no currently conceivable way to know the answer to your question.
| {
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What is dislocation loop? What is dislocation loop? Is it something different from dislocations? If so can dislocations at higher temperature can combine to form dislocation loop?
| a dislocation loop is a linear dislocation in which its two ends meet together- that is, it's still a dislocation in the crystal lattice but by closing in on itself it has special propagation properties, about which entire chapters of upper-division materials science textbooks have been written.
Years ago, a team of Japanese materials scientists and electron microscopists made a movie using transmission electron microscopy which actually shows dislocation loop propagation as a function of applied stresses. If I can find a copy on-line I will edit & link it to this answer.
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Hermiticity of spin-orbit coupling in real space In the Kane-Mele model, the spin-orbit coupling is defined in real space as
$$\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}$$
where the sum is over next-nearest-neighbor sites on a honeycomb lattice, and $\nu_{ij} = - \nu_{ij} = \pm 1$ depends on the orientation of the next-nearest-neighbor bonds (I don't believe the details of how $\nu_{ij}$ is calculated is relevant for Hermiticity). I am having difficulty understanding how this term is Hermitian. Taking the Hermitian conjugate seemingly gives
$$\left(\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta}\right)^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} (-i) t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha} c_{j \beta}^\dagger = \sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{j \beta}^\dagger c_{i \alpha}
\\
= -\sum_{\langle \langle i j \rangle \rangle \alpha \beta} i t_2 \nu_{ij} s^z_{\alpha \beta} c_{i \alpha}^\dagger c_{j \beta} $$
where in the final line we have relabeled indices and used that $\nu_{ij} = - \nu_{ji}$, $s^z_{\alpha \beta} = s^z_{\beta \alpha}$. I must be missing something obvious, but this seems to show that the term is anti-Hermitian, instead of Hermitian. What am I missing here?
| From the first to second expression on the first line, the Conjugation operation should transpose the matrix $v_{ij}$ to $v_{ji}$, giving the missing minus sign.
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Why don't we lose generality when we assume $z^0 > x^0 > y^0$ in the proof of Wick's theorem? In proof's of Wick's theorem it's typically stated (see e.g. page 87 in Coleman's notes) that it's sufficient to consider just one possible time-ordering. For example for the product $T(\phi(z)\phi(x)\phi(y))$ this would mean that we assume $z^0 > x^0 > y^0$.
Why is this possible without losing generality?
| Inside the brackets of the time ordering operator all fields commute. Thus the times are irrelevant as the fields are always ordered so that the assumption is satisfied
| {
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What is the $R$-symmetry group for ${\cal N}=6$ supergravity in $D=4$ dimensions? What is the $R$-symmetry group for ${\cal N}=6$ supergravity in $D=4$ dimensions?
| This is an old post, but I would like to clarify this for others who, like me, stumble upon this question now. As Qmechanic noted, the $R$-symmetry group in $D = 4$ is $U(\mathcal{N})$, which is Table 12.1 on page 240 of Ref. [1]. However, in supergravity theories, the kinetic terms for the scalars are non-trivial and determine the scalar manifold (see section 12.5 of Ref. [1]). Scalar manifolds of all supergravities with more than 8 real-component supersymmetries are symmetric spaces, i.e. they are spaces $G/H$, with $G$ non-compact (the isometry group) and $H$ its maximal compact subgroup (the isotropy group). An overview of the scalar manifolds for various supergravity theories is given in Table 12.3, page 250 of Ref. [1]. Ash's comment on $\mathcal{N} = 8, D = 4$ supergravity actually refers to the isotropy group being $\text{SU}(8)$, as given in that table, probably explaining the confusion. The $R$-symmetry group is always a factor of the isotropy group.
References:
*
*D.Z. Freedman & A. Van Proeyen, Supergravity, 2012
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How do I experimentally measure the surface area of a rock? I hope this is the right place to ask this question.
Suppose I found a small irregular shaped rock, and I wish to find the surface area of the rock experimentally. Unlike for volume, where I can simply use Archimedes principle, I cannot think of a way to find the surface area. I would prefer an accuracy to at least one hundredth of the stone size.
How can I find the surface area experimentally?
| The task isn't well-defined.
Do you include cracks? If yes, you'll see finer and finer cracks adding to the surface, and in the end you'll be at atomic level and find it hard to even define what's part of the rock and what isn't.
If you don't include cracks: What's your rule for distinguishing a mere unevenness from a crack?
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When the direction of a movement changes, is the object at rest at some time? The question I asked was disputed amongst XVIIe century physicists (at least before the invention of calculus).
Reference: Spinoza, Principles of Descartes' philosophy ( Part II: Descartes' Physics, Proposition XIX). Here, Spinoza, following Descartes, denies that a body, the direction of which is changing, is at rest for some instant.
https://archive.org/details/principlesdescar00spin/page/86
How is it solved by modern physics?
If the object is at rest at some instant, one cannot understand how the movement starts again ( due to the inertia principle).
If the object is not at rest at some instant, it seems necessary that there is some instant at which it goes in both directions ( for example, some moment at which a ball bouncing on the ground is both falling and going back up).
In which false assumptions does this dilemma originate according to modern physics?
| I assume you are talking about an object that reverses direction through a collision. Take the case of a ball bouncing off the ground. As it hits the ground, the ball deforms and the ground compresses a tiny amount. The resistance of the ball to deformation and the resistance of the ground to compression decelerate the ball to the point at which its centre of mass is stationary, and thereafter accelerate the ball upwards.
The 'dilemma' is the mistaken result of assuming that no force acts on the object at the point at which its motion is stopped.
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What determines whether we use a vector or scalar potential? I understand that electrostatic potential is scalar because the curl of the field is zero, and this implies the electrostatic field is the gradient of the scalar potential to satisfy this. Similarly the divergence of a magnetostatic field is zero so a magnetostatic field is the curl of the vector potential.
But what actually determines when you would use which potential? Is it purely to do with these definitions in electro and magneto statics, or is it something else?
| From a more general point of view, the scalar potential and the vector potential are parts of the electromagnetic four-potential.
The four-potential is defined up to a gauge transformation. To answer your question, one uses the gauge that is the most convenient for the specific problem. For example, in the case of electrostatics one can find such a gauge that the vector potential vanishes and use the scalar potential. In a general case, this is not possible, but one can always use such a gauge that the scalar potential vanishes (Weyl gauge). Therefore, the vector potential is always sufficient for any electromagnetic problem, not just for magnetostatics (for example, one can use it for electrostatics as well), but it is not always convenient.
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Why coupled oscillators tend to seek integer frequency ratios? In this document, the author writes (page 225)
Coupled oscillators have a tendency to seek frequency ratios which can be expressed
as rational numbers with small numerators and denominators. For example,
Mercury rotates on its axis exactly three times for every two rotations around
the sun, so that one Mercurial day lasts two Mercurial years. In a similar way, the
orbital times of Jupiter and the minor planet Pallas around the sun are locked in a
ratio of 18 to 7 (Gauss calculated in 1812 that this would be true, and observation
has confirmed it). This is also why the moon rotates once around its axis for each
rotation around the earth so that it always shows us the same face.
Is that true? Can we prove mathematically that Coupled oscillators love rational frequency ratios?
Oh, it appears that planetary motion is not an oscillator. But anyway, I just want some reference to verify whether this is true, preferably with mathematical derivations.
| They may in special cases, but this is by no means a rule, and most certainly doesn't hold for all cases. As I'm a firm believer in concrete examples, I'll illustrate this with the following simple case of two blocks, each of mass $m_1=m_2=m$, connected to walls and each other by Hookean springs (which are harmonic oscillators) of spring constant $k$, as shown in the following image
$$$$
Now, the Lagrangian for this system is, of course, $$L=\frac{1}{2}m\left(\dot{x}_1^2+\dot{x}_2^2\right)-\frac{1}{2}kx_1^2-\frac{1}{2}k\left(x_2-x_1\right)^2-\frac{1}{2}kx_2^2$$
which returns the equations of motion $$\begin{align}
\ddot{x}_1 &=-\frac{2k}{m}x_1+\frac{k}{m}x_2 \\
\ddot{x}_2 &=\frac{k}{m}x_1-\frac{2k}{m}x_2\\
\end{align}$$
One can solve these equations with a bit of linear algebra, and one gets, with vanishing initial velocity,
$$\begin{align}
x_1(t) &= \frac{1}{2}\left(x_1(0)+x_2(0)\right)\cos\left(\sqrt{\frac{k}{m}}\,t\right) +\frac{1}{2}\left(x_1(0)-x_2(0)\right)\cos\left(\sqrt{\frac{3k}{m}}\,t\right) \\
x_2(t) &=\frac{1}{2}\left(x_1(0)+x_2(0)\right)\cos\left(\sqrt{\frac{k}{m}}\,t\right) -\frac{1}{2}\left(x_1(0)-x_2(0)\right)\cos\left(\sqrt{\frac{3k}{m}}\,t\right)
\end{align}$$
Now, even this simple case isn't periodic except in cases where $x_1(0)=x_2(0)$ or $x_1(0)=-x_2(0)$, as the terms in the cosines differ by a factor of $\sqrt{3}$. This nonperiodicity as can be seen in these plots, where I have set $k=1.2321$, $m=0.771203$, $x_1(0)=e+0.231$, and $x_2(0)=1$.
There are many, many more examples I could show. In general, the motion of coupled oscillators is complex, is not necessarily periodic, and does not tend toward rational frequencies. Now, are there cases where the motion is periodic and they are in sync? Of course, but they are not the general rule. For this example, set $x_2(0)=-x_1(0)$, and they are in a ratio of $1:1$, as shown in the below plots (everything is the same, except $x_1(0)=-1$).
To sum it up, while there are some cases where oscillators tend towards being in a rational sync, this is not always true.
| {
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Are the boundary conditions purely a consequence of Maxwell's equations? The boundary conditions, namely
were all these, realized only by looking at Maxwell's equations? Or is there a physical reasoning behind them? For example, Why does the component of the electric field parallel to the surface of interface remain unaltered? I also read that the reason light bends when it passes through another medium is because only the normal component gets altered and the horizontal component remains the same(whereas the velocity gets altered because of the other electrons in the material that are driven by the source and produce a separate wave with a different phase and the superposition of these two waves seem to alter the speed of light in a medium $^\dagger$).
My question in short is, What would be my answer,if someone asked me, to explain the boundary conditions, without equations*.
*If that's purely based on equations, please ignore "without equations", but there's got to be something that's physically occurring which led us to create a model, right?.
$^\dagger$Is that right?
| In short: Yes.
Those equations can be understood as boundary conditions, but they can also be understood as being valid between any two points inside the domain. This is because they result from integrating the Maxwell equations over an arbitrary test volume, and hence link the quantities at either end of this test volume.
| {
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Where does $\pi$ come from in the Heisenberg equation? In class today we were taught about Heisenberg’s equation, $$\Delta x\Delta p\ge\frac{h}{4\pi}. $$
Experience tells me that any time an equation involves pi, circles aren’t far behind. Obviously this is true in geometry, but even pure number theory equations, such as $\Sigma_{n=1}^{\infty} \frac1{n^2}=\frac{\pi^2}6$, you can always find a way to construct the problem such that circles are involved and the solution, including pi, naturally jumps out.
The natural question, then, is: what do circles have to do with Heisenberg? Why is Planck’s constant divided by a multiple of pi, and why specifically $4\pi$?
| There are two different conventions for the constant used in the uncertainty principle, which are written as $h$ and $\hbar$. They are related by
$$\hbar = \frac{h}{2\pi}$$
and the reason for the $2\pi$ is because both appear in the expression for the energy of a photon:
$$E = hf = \hbar \omega$$
and the relation between frequency $f$ and angular frequency $\omega$ is,
$$\omega = 2\pi f$$
from whence the relation to circles is more obvious. Indeed, I'm more familiar with writing the HUP as
$$\Delta x \Delta p \ge \frac{1}{2}\hbar$$
with no $\pi$ in it and $\hbar$ as the basic constant. And in theoretical papers on quantum theory, you'll much less often encounter $h$, and you amy not even see $\hbar$ because Planck units are used where $\hbar = 1$!
| {
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Mean free path equation derivation I was reading about mean free path equation derivation online and stumbled upon this:
We will derive the equation using the following assumptions: let’s assume that the molecule is spherical, and the collision occurs when one molecule hits another, and only the molecule we are going to study will be in motion and the rest of the molecules will be stationary.
Let’s consider our single molecule to have a diameter $d$ and all the other molecules to be points. This does not change our criteria for collision, as our single molecule moves through the gas, it sweeps out a short cylinder of cross section area $πd^2$ between successive collisions...
which got me confused. Wouldn't the cross section area be equal to $πr^2 = πd^2/{4}$?
To be clear, I read that here (under the section 'Derivation of Mean Free Path').
| The cross-sectional area is indeed $A_c = \pi d^2$ but the explanation on this page is not precise in this regard. You basically consider a single particle of diameter $d$ and its collisions with other particle of the same size.
Simple considerations in 2D
The chance for it to collide with another particle is proportional to the center of that other particle being in a circle with radius $2 r = d$ around the center of the particle under consideration.
Just think of it that way: What is the neighbourhood of a particle that another particle with the same radius would have to be in for a collision? Obviously they would collide if the two radii would touch or the distance would be smaller. Thus $\pi (2r)^2 = \pi d^2$ is the area the center of the second sphere must lie in for a collision between the two particles.
More general considerations in 3D
More generally in three-dimensional space one has to consider the area perpendicular to the relative velocity of the two particles. And integrate over this cross-sectional area
$$ d A_c = r \, dr \, d \phi $$
depending on the position of the two particles. Introducing an angle $\psi$ in between the line connecting the two centers and the relative velocity can be calculated according to
$$ r = d \, \sin \psi \hspace{2cm} dr = d \, \cos \psi d \psi$$
Now integrating over all potential angles in direction of the relative velocity $0 \leq \phi \leq 2 \pi$ and $0 \leq \psi \leq \frac{\pi}{2}$ considering the identity $\sin \psi \, \cos \psi = \frac{\sin ( 2 \psi)}{2}$ we yield
$$ A_c = \int\limits_{\phi = 0}^{2 \pi} d \phi \int\limits_{\psi = 0}^{\frac{\pi}{2}} \frac{d^2 \, \sin ( 2 \psi)}{2} d \psi = \pi d^2 $$
| {
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Commutator of $L^{2}$ and $L_{z}$ I'm trying to work through a proof of why $[L^{2},L_{z}]=0$, and am getting lost on this step:
We can use the commutation relation $[\hat{L}_{z},\hat{L}_{x}]=i\hbar\hat{L}_{y}$ to rewrite the term as:
$$
\hat{L}_{x}\hat{L}_{x}\hat{L}_{z} = \hat{L}_{x}\hat{L}_{z}\hat{L}_{x} - i\hbar\hat{L}_{x}\hat{L}_{y}
$$
I understand that they are trying to get the first term on the RHS to cancel with the next term in the commutator, and that the Levi-Cevita symbol dictates the negative sign for the second term, I'm just unsure how they come about from the commutation relationship. Any help would be appreciated.
| Given
$$
[ L_z, L_x ] = i \hbar L_y
$$
just multiply both sides by $L_x$:
$$
L_x \bigg( L_z L_x - L_x L_z \bigg) = i \hbar L_x L_y
$$
which gives
$$
L_x L_x L_z = L_x L_z L_x - i \hbar L_x L_y
$$
| {
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How do you tell if a wave is reflected or refracted at an interface? My physics textbook says that a wave traveling through medium $1$ will enter medium $2$ if medium $1$ has a higher index of refraction. Otherwise, the wave will be reflected. This makes absolutely no sense to me, since this would mean all sunlight traveling through the vacuum of space would just bounce off Earth's atmosphere since air has a higher index of refraction than a vacuum.
There has to be something else that goes on for determining if a wave will be reflected or refracted at an interface, but my textbook doesn't elaborate any further past the statement already mentioned. Can someone tell me what the actual criteria is?
Edit
Nevermind, I misread the passage. Turns out the terms "reflect" and "invert" are not nearly as interchangeable as I thought they were.
| It's good that it makes absolutely no sense to you, because that is absolutely wrong, and if your book truly says that then it should be cast into the nearest volcano as soon as possible.
In general, light incident upon an interface will be neither completely reflected nor completely transmitted. The Fresnel equations are used to determine the reflection and transmission coefficients $R$ and $T$, which tell you what fraction of the incident light is reflected and transmitted, respectively.
Unfortunately, the Fresnel equations are rather complicated, and depend both on the refractive indices of the two media as well as the polarization of the incoming wave. If we make a few simplifying assumptions for normal materials and visible light, we have the following:
For $p$-polarized waves, in which the electric field oscillates in the same plane as the interface, the reflection coefficient is
$$R_p = \left|\frac{n_1\sqrt{1-\left(\frac{n_1}{n_2}\sin(\theta_i)\right)^2}-n_2\cos(\theta_i)}{n_1 \sqrt{1-\left(\frac{n_1}{n_2}\sin(\theta_i)\right)^2}+n_2\cos(\theta_i)}\right|$$
For $s$-polarized waves, in which the electric field oscillates in the direction normal to the interface, the reflection coefficient is
$$R_s = \left|\frac{n_1\cos(\theta_i) - n_2\sqrt{1-\left(\frac{n_1}{n_2}\sin(\theta_i)\right)^2}}{n_1\cos(\theta_i) + n_2\sqrt{1-\left(\frac{n_1}{n_2}\sin(\theta_i)\right)^2}}\right|$$
In both cases, the amount of light (specifically, the intensity of the beam as a fraction of the incident intensity) transmitted is $T_{s/p} = 1-R_{s/p}$.
| {
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Can the work by kinetic friction on an object be zero? We know that friction is of two types - static and kinetic. Static friction acts when there is no relative motion between the surfaces in contact. Kinetic friction takes place when surfaces rub against each other. I was wondering whether the work done by the kinetic friction can be positive, negative or zero.
*
*Positive work - When an object is placed on a rapidly moving belt, it moves along with the belt but with slipping (relative motion between the surfaces exist) when there is no enough friction to prevent slipping. Here the work done by the kinetic friction is positive, as the direction of frictional force and the displacement is same.
*Negative work - Work done by kinetic friction, when an object moving on a rough surface slows down, is negative as the direction of friction and displacement are opposite to each other.
I'm unable to think of any circumstances when the work done by kinetic friction is zero because of the following reasons:
*
*Work done on an object is zero if displacement is zero. In our case, if displacement is zero, the frictional force acting on the object is static and not kinetic in nature.
*Work done is also zero when the force and displacement are perpendicular to each other. The only example I am aware of is circular motion. As the point at which the wheel touches the ground is at rest. The nature of friction is again static.
So, can the work by kinetic friction on an object be zero?
Please note: I read the answers for the following two related questions. There is no clear explanation on the two aspects of friction (static and kinetic) in those answers. Simply they don't have enough details.
*
*Work done by Friction. Can it be positive or zero?
*Positive work done by friction
| Hold a piece of wood against a sanding belt. In your frame, the block is not moving, but
*
*kinetic friction is exerting a force: you have to hold the block still
*energy is transferred: the block gets hot, and pieces are pulled off it
| {
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Conservation of momentum in photon-atom collision Is this explanation correct:
When a photon with the appropriate energy hits an atom, the electron will make a transition from the ground state to a excited state. This will make the potential energy of the atom higher. Also, momentum is conserved, and the velocity of the atom will change in the direction from the incoming photon. When the electron returns back to it’s ground state, a photon is emitted in a random direction. Again, momentum is conserved and the velocity of the atom changes in the opposite direction from the emitted photon.
I am a bit confused because if that's the case, the atom will gain kinetic energy. Doesn’t that violate the conservation of energy? Since the energy is already conserved by absorbing and emitting a photon with the same wavelength.
|
I am a bit confused because if that's the case, the atom will gain kinetic energy.
In the way you've set it up (atom is initially at rest in our reference frame), then that is correct.
Doesn’t that violate the conservation of energy?
That depends on the energy inputs and outputs. We haven't described them completely yet.
Since the energy is already conserved by absorbing and emitting a photon with the same wavelength.
That's not correct here. Because of the recoil, the emitted photon will be less energetic. If you instead chose the frame where the center of momentum were at rest (same as the excited atom at rest), then the photons would have the same energy. But in that frame, the atom has identical kinetic energy before and after the interaction.
| {
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Will August be always summer in the northern hemisphere? Is the Earth orbit precessing, or are there other effects which will create a shift between our calendar (day counting), and the Earth's orbit?
I imagine these effects to be small, but I'm asking for long timescales.
[Edit] To formulate my question better, let me be more precise. Assume that our calendar never changes, meaning we keep counting days always in the same way (second...days defined by an atomic clock, 365 days = 1 year, usual leap years, etc), and consider what will happen in -I don't know- 100k-1M year? Or, if this timescale is wrong, what should it be to see an effect of a shift between seasons and months, as we are used to?
|
A sidereal year is the time taken by the Earth to orbit the Sun once with respect to the fixed stars. ...
The sidereal year differs from the tropical year, "the period of time required for the ecliptic longitude of the Sun to increase 360 degrees", due to the precession of the equinoxes. The sidereal year is 20 min 24.5 s longer than the mean tropical year
Our Gregorian calendar is based on the tropical year, so it has the seasonal drift "built in" to it. (This has nothing to do with leap days. They are needed because the tropical year is not an integer number of days.)
| {
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Status of Space-Time Many physicists conjecture that space-time is not fundamental.
Is this the orthodox view in physics these days?
Follow ups - If a philosopher argues that space-time is reducible, are any physicists likely to argue? Are there many or any theories (for instance versions of string theory) that actually require a fundamental space-time?
This not asking how space-time can be emergent. The question is asking whether the view that space-time is emergent is considered orthodox, or to what extent it is endorsed by physicists.
| It is the prevailing view among high-energy physicists that spacetime should be emergent from some more basic structure because of the following basic argument: if you want to probe regions of spacetime that are arbitrarily small then according to the principles of quantum mechanics (and special relativity), it corresponds to concentrating arbitrarily high energy in a small enough region. However, since gravity exists, such an attempt would turn the intended region to be probed into a black hole. And thus, you cannot probe it. So, this shows an in-principle impediment to probing small enough regions of spacetime. This motivates one to think that the notion of spacetime itself should somehow breakdown at such high-energy scales and should be replaced by some other construct. However, this cannot be considered an "orthodox" view because we do not yet have an actual framework in which spacetime can be fully replaced by some more fundamental entity.
The closest we have come to having a framework for talking about emergent spacetime in a robust way is via the $\text{AdS-CFT}$ correspondence in which one space dimension of a theory with gravity can be seen as emergent because such a theory is dual to a theory without gravity with one lower number of space dimensions. While space and time are supposed to be on the same essential footing due to Lorentz symmetry, we don't yet have a working framework like the one provided by $\text{AdS-CFT}$ in which time is also emergent.
| {
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Unitarity/Hermiticity condition for $osp(m,n|\mathbb{C})$ superalgebra According to Dictionary on Lie Superalgebras (page 82), the compact form of $OSP(m,n|\mathbb C)$ Lie superalgebra must satisfy $M^{\text st}H\,M=1$ and $M^{\ddagger}M=1$ (is this the unitarity condition?), this means that the conditions for the corresponding superalgebra are $E^{\text st}H+HE=0$ and $E^{\ddagger}+E=0$ (right? it could be $E^{\ddagger}-E=0$?) due to the exponential map. So $E$ must be anti-Hermitian (right?, see Reduced matrix elements of the orthosymplectic Lie superalgebra (page 32) in which it looks that this condition needs to be applied "by blocks"). Here, $$H=\begin{pmatrix}
\mathbb{I}_m & 0\\
0 & \mathbb{J}_{2p}
\end{pmatrix},\quad \mathbb{J}_{2p}=\begin{pmatrix}
0 & \mathbb{I}_p\\
-\mathbb{I}_p & 0
\end{pmatrix}\tag{1}$$
and $E=\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}$ is an even supermatrix, so $A$ and $D$ are bosonic, and $B$ and $C$ are fermionic. The supertranspose operation is
$$E^{\text st}=\begin{pmatrix}
A^{\text t} & C^{\text t}\\
-B^{\text t} & D^{\text t}
\end{pmatrix}\tag{2}$$
as defined in page 84 of (1). By imposing
$$E^{\text st}H+HE=0 \tag{3}$$
I get the usual conditions for the $SO(m)\times Sp(n)$ bosonic subalgebra, and others for the fermionic part.
My problem is with the operation $\ddagger$. According to (1) (page 84) (see also Graded Lie algebras: Generalization of Hermitian representations ),
$$E^{\ddagger}=(E^{\text st})^\#$$
The $\#$ operation is a "superconjugation" or superstar. It is not clearly expressed in (1) so I went to CURRENT SUPERALGEBRAS AND UNITARY REPRESENTATIONS (page 18) in which $\#$ is defined as
$$E^{\#}=\begin{pmatrix}
A^{*} & -iC^{*}\\
-iB^{*} & D^{*}
\end{pmatrix}\tag{4}$$
involving usual conjugation and a transpose!(does this agree with the definitions in (1) and (5)). So it looks like $\#$ is already something like $\ddagger$. With this, a unitary representation of $\mathscr{gl}(m,n|\mathbb C)$, according to (4) must satisfy $E^{\#}+E=0$ from which $C=iB^{*}$. this operation is given also in Cornwell's Group theory in physics, vol. 3 (page 11) but it is not so clear for me.
At the end, $-E=E^{\ddagger}:=(E^{\text st})^\#=-E^{\text st}\Rightarrow E=E^{\text st}?$ and from the condition of $OSP$, $E=-HEH^{-1}$? So this condition with $E^{\text st}H+HE=0$ allows to get the unitary $osp$ superalgebra?
Notice also Superstrings on AdS4 × CP3 as a Coset Sigma-model (page 5) in which eq. (2.4) corresponds to the hermiticity (unitarity) condition.
| Apparently $E^{\#}\equiv E^{\ddagger}$ and $*\equiv\dagger$, so the condition $$E^{\#}=-E\tag{1}$$
for unitarity, is the same as $$E^{\ddagger}=-E \tag{2}$$
That's why you have a transpose in the definition of $E^{\#}$! SO, actually $C=-iB^{\dagger}$. And, with $E^{\text{st}}H+HE=0$, you get the generators of $uosp(m,n|\mathbb C)$.
In order to "unify" definitions, this is what happens
$$E^{\ddagger}:=(E^{\text{st}})^{\#}:=\begin{pmatrix}
A^t & C^t\\
-B^t & D^t
\end{pmatrix}^{\#}=\begin{pmatrix}
A^{t*} & -iC^{t*}\\
-iB^{t*} & D^{t*}
\end{pmatrix}=-E \tag{3}$$
| {
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Interacting lagrangian with multiple terms We know that for a $\phi^4$ interacting theory with $$
\mathcal{L}_{int} = -\frac{\lambda}{4!}\phi^4
$$
the interaction $\phi \phi \rightarrow \phi \phi$ gives a vertex with a factor $-\frac{i\lambda}{4!}$ which stems from Wicks theorem.
If we now use the same procedure for an interacting lagrangian of the form $$
\mathcal{L}_{int} = -\frac{g}{6}\phi^3-g\Phi^*\Phi\phi
$$
for example, what would be the vertex? When we go through Wicks theorem, do we sum the terms or just treat them seperately?
| Your second Lagrangian would generate of two vertices:
A cubic self interaction for $\phi$ (in your notation $\phi \phi \rightarrow \phi$, say) and an inter-field interaction between $\Phi$ and $\phi$ (in your notation $\Phi^{*}\Phi \rightarrow \phi$).
Both vertices have three legs; in the first they're all $\phi$ fields and in the second there is a $\Phi$ anti-particle, a $\Phi$ particle and a $\phi$. In your example, both couplings are equal, $g$.
| {
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How can the mechanism of electrons in an atom be explained? I am a high school student who takes both Physics and Chemistry.
Recently I learnt about the quantum mechanical point of view of looking at electrons or nuclei. I also learnt that the wave functions can be obtained by solving the Schrodinger's equation with various conditions specific to the problem (such as the particle in a box).
My shallow understanding of quantum mechanics is that we can only know the probability of an electron existing at a certain position and time, and the actual position can be determined when the 'observation' takes place.
The chemical bondings and chemical reactions are the results of electric interactions between nuclei and electrons. The Coulomb force is a function of the distance between two charges, so it is important that the exact locations of electrons should be known. But taking into consideration quantum mechanics, we don't even know where the electrons are, and we built up a subject called Chemistry, and most importantly, CHEMISTRY STILL WORKS VERY WELL.
So, what is going on?
| curios,
It is important to make a clear distinction between what is known or knowable and what it exists.
Quantum mechanics does not say that the electrons do not have precise positions or precise momenta. It tells you that:
*
*You cannot prepare a state where both the position and momentum of an electron are known with arbitrary accuracy (Heisenberg uncertainty principle).
*If the electron is not in a position eigenstate you can only predict the result of a position measurement probabilistically.
In other words quantum mechanics limits what can be known. It says nothing about what can exist.
Your observation that the classical Hamiltonian, based on Coulomb's law, works very well seems indeed to contradict the idea that particles do not have precise positions. It's not a rock-solid argument, but seems a little bit mysterious.
| {
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Hamiltonian for 2-qubit system What is the general form of a Hamiltonian for a 2 qubit system?
The Hamiltonian for D-Wave system is:
$H = K_1\sigma_x^1 + K_2\sigma_x^2 + H_1\sigma_z^1 + H_2\sigma_z^2 + J_{12}\sigma_z^1\sigma_z^2 $
https://arxiv.org/abs/1512.01141
Is this true for all superconducting qubit system? If not, then what's the Hamiltonian? For instance, What's the Hamiltonian of the qubit in the IBM quantum computer?
| The hamiltonian of a two qubit system can be any $4\times4$ hermitian matrix. In general, we can write this
\begin{align}
H = \sum_{i,j=0}^3 h_{ij} \,\sigma^1_i\sigma^2_j
\end{align}
where the coefficients $h_{ij}$ are real, and the $\sigma_i$ are Pauli matrices, with $\sigma_0$ the identity.
What the hamiltonian of a given quantum computer is will presumably depend on what you are trying to compute.
| {
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Why is torque a cross product? If I'm not wrong, torque is perpendicular to both the radius and force i. e. It is along the axis of rotation. Questions that arise are- why do we consider the length between the axis/point of rotation while calculating torque? More importantly why is torque a cross product?
| (a) "why do we consider the length between the axis/point of rotation while calculating torque?"
We can calculate torque about any point, O, that we choose; it doesn't have to be a physical axis of rotation. But it's often more useful to calculate torque about a possible physical rotation axis, for example when thinking about what torque we need to apply on a nut with a spanner (wrench) in order to undo it. As for why length (or, specifically, perpendicular distance from O to the line of action of the force) comes into the definition, just think about trying to undo that nut!
(b) "why is torque a cross product?"
In vector notation we define the torque about O due to a force $\vec{F}$ acting at a point displaced by $\vec{r}$ from O to be $\vec{r} \times \vec{F}$. The magnitude, $|\vec{r} \times \vec{F}|$of this torque vector is exactly equivalent to the "force $\times$ perpendicular distance" definition that I quoted from in (a). The direction of $\vec{r} \times \vec{F}$ is at right angles to the plane containing $\vec r$ and $\vec F$ and therefore tells you the alignment of the (possibly imaginary) axis about which the torque, acting on a nut, would turn it! [In fact. with the usual 'right handed' convention for defining cross product, if you point the thumb of your right hand in the direction of $\vec{r} \times \vec{F}$, the fingers of that hand will tend to curl round in the sense that the nut will turn!]
| {
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How do physicists compare the relative strengths of the four forces? Since the four forces are different, with different force carriers, how are they (seemingly) directly compared? I often read that the weak force, for example, is many orders of magnitude stronger than gravity, and that electromagnetism is several orders of magnitude stronger than the weak.....
I do notice that different sources give different ratios when comparing the four forces' relative strengths, though....
| In particle physics, the strength of fundamental forces is specified by either masses of force-carrier particles (e.g. masses of $Z$, $W^{\pm}$ bosons for weak interaction) and/or by a numerical "coupling" coefficient (e.g. electron charge in electromagnetic interaction).
One of the ways to measure the relative strength is by comparing the ratios of particle productions. If during a collision of two hydrogen atoms $K$ particles are produced through the electromagnetic interaction, $L$ particles through the strong interaction, and $M$ through the weak - $K : L : M$ ratio could be used to quantatively compare their strength. This is generally going to be dependent on various factors like angles of detectors, center-of-mass energy of collision, but can be nevertheless used make qualitative statements about the strengths of the forces.
| {
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Could approaching an unknown object in space electrocute you? Say you're out for a spacewalk on the ISS one day, and you see an unknown object floating next to you (ignoring the probability that it somehow matched your orbit perfectly and didn't just plunge straight into Earth's atmosphere). Let's say a bowling-ball sized metallic sphere.
Given the relative nature of voltage/potential difference, assuming this thing hasn't come into contact with anything else recently and ignoring any other ways a mysterious floating object could kill you, is it possible that the potential difference between you and the object could be so great that contact with it would instantly/almost instantly kill you? Would the situation be any different in an atmosphere, or with a bigger object?
| Putting your hands on opposite terminals of a large, high voltage capacitor will do it. They are dangerous.
| {
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Different apparent brightness of a distant star from a moving frame - an apparent paradox Consider a stationary star at a distance $L$ from the Earth.
A spaceship at Earth moving with velocity $v$ towards the star will find the star to be at a distance of $\dfrac{L}\gamma$ in its own frame.
But it won't actually see the star at that distance since the speed of light is finite.
The light that it will get at that instant will be from a star that was at distance $\dfrac{Lc}{\gamma(c-v)}$ which is $>L$.
At $v=0.6c$, $\dfrac{Lc}{\gamma(c-v)}=2L$.
Therefore, the spaceship will see the star $0.25$ times as bright as an observer on the Earth.
But this is unexpected since both spaceship and the Earth are bathed in the same starlight and if anything, the observer on the Earth sees the spaceship gather more photons due to its motion.
How to resolve this apparent paradox?
| The star will look dimmer because to the moving observer the star is aging slower, so to this observer, the star emits less energy per unit of time. Also, the total number of photons each observe between two events will be the same, but they will disagree on how much time has passed between the two events and about their synchronization
| {
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Might a cast iron pan set on top of a microwave oven affect the operation? Our microwave seems to take longer to get the job done these days. I notice that someone is storing a large, heavy cast iron pan on the top of the microwave. Is there any way at all that it is possible that the iron pan is interfering with the microwave?
I could try with/without the pan but it isn't that dramatic.
| Elaborating a bit on terri's answer... If (big, hypothetical if), some object like the cast iron pan were reducing the energy going into the food, then it should be heating up. But, a big hunk of metal in firm contact with the Faraday cage that is your oven's enclosure should at most act as part of the cage.
Metal objects will in general, respond to microwaves in either of two ways (depending on how things are arranged): (1) Eddy currents will be induced in the metal; it will absorb energy and heat up. (2) It will reflect the microwave energy.
As microwave ovens age, the power they deliver does drop for a few reasons. One of them is as follows: There is a "window" between the oven's main compartment and the emitter; this window looks opaque but is transparent to microwaves. It can accumulate contamination from spilled/spattered food which will reduce the amount of energy that can pass through, reducing the amount of energy that goes into your food.
| {
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Misconception in partial derivatives of Lorentz transformation Let us consider a Lorentz transformation of four vectors from frame S to S' where S' is moving with relative velocity $\textbf{v}$ with respect to S. The boost is given by
$$t'=\gamma(t-vx), \quad x'=\gamma(x-vt), \quad y'=y, \quad z'=z.$$
The inverse transformation is given by
$$t=\gamma(t'+vx'), \quad x=\gamma(x'+vt'), \quad y=y', \quad z=z'.$$
Now here comes the crucial part. Notice that
$$\frac{\partial x}{\partial x'} = \gamma, \quad \frac{\partial x'}{\partial x} = \gamma. \tag{1}$$
I have thought about this for a while, but more thoughts always lead me to the same conclusion that this is true.
However, then we have a problem:
$$\frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} = \gamma \frac{\partial}{\partial x} \tag{2}.$$
This seems fine. Continue:
$$\frac{\partial}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} = \gamma \frac{\partial}{\partial x'} \tag{3}.$$
This also seems fine. Continue:
$$ \frac{\partial}{\partial x'}=\gamma \frac{\partial}{\partial x} = \gamma \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} = \gamma^2 \frac{\partial}{\partial x'}$$
where we have reached a contradiction since $\gamma^2 \neq 1$.
Where's have I messed up in (1), (2), (3)?
| You forgot that $x'$ is not only function of $x$, but also of $t$: $x' = x'(x, t)$. Similarly, $x=x(x', t')$. Hence:
$$ \frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} + \frac{\partial t}{\partial x'} \frac{\partial}{\partial t}$$
and
$$ \frac{\partial}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} + \frac{\partial t'}{\partial x} \frac{\partial}{\partial t'}$$
| {
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Phase transition on magnetic materials Is ferromagnetic to paramagnetic phase transition a reversible process?
If I start with a ferromagnetic material with a spontaneous magnetization below the Curie temperature, and then I start to heat it, it will become paramagnetic above the critical temperature. If I then start to drop the temperature slowly to below the Curie temperature then will I achieve the ferromagnetic behaviour with same spontaneous magnetization as before?
| Spontaneous symmetry breaking
Unless there is a preferred direction for magnetization, specified, e.g., by the external magnetic field or crystal symmetries, there is no reason for the magnetization to point in the same direction as before. The direction chosen by the magnetization when entering the ferromagnetic phase is an example of spontaneous symmetry breaking.
Domains
Also, as @user137289 has correctly pointed out, unless the crystal is cooled all the way to zero temperature, the magnetization is not homogeneous, but would rather consist of many homogeneously magnetized domains, which are differently oriented in respect to each other.
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If a basis set is complete, are the elements in it mutually orthonormal? If a basis set is complete, are the elements in it mutually orthonormal? For example, we can express the field operator in the basis of the creation and annihilation operators.This basis is complete, are the elements in it mutually orthonormal?
| The answer is NO
Given any complete basis, one can construct another by taking independent combinations.
Assuming the original basis is complete and orthogonal, and contains $V_1$ and $V_2$, (which are thus orthogonal).
Replacing $V_2$ by $V_2'=V_1+V_2$ does not change the completude of the basis. But now two vectors of the new basis, namelt $V_1$ and $V'_2$ are not orthogonal.
If the questio is, whether it is always possible to find an orthogonal basis is one thing. But there is no need for a basis to be orthogonal to be complete.
| {
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Is time slower on the far galaxies? Since far galaxies move away faster, what would be the speed of their time relative to us?
If there is a difference:
*
*What determines whose time would be faster?
*(If I haven't understood it wrong) To resolve the twin paradox, acceleration is required. Is the expansion of the universe creates acceleration? If not how can we explain this difference?
I think following question is the same:
Imagine we have a some kind of machine that can bend the space-time, and create a gravitational pull in front of us. So we gain speed without feeling any acceleration. If one of the twin in the twin paradox would use such a machine to gain speed; what would happen?
I don't have a physics or math background, and I hope my question makes sense.
| When you talk about farther galaxies moving faster relative to us, you are presumably talking about the effects of the expansion of the Universe. When you refer to the twin paradox, you are presumably talking about the special relativistic effects of relative motion. These are not the same thing. To see this, note that in the first sense, some of those far-away galaxies are receding faster than the speed of light, so if there were time dilation in the second sense, their clocks would be running backward in our frame.
Conclusion: The question is very hard to make sense of. The recession of the galaxies certainly does not cause time dilation in the "twin paradox" sense that you seem to be thinking of.
| {
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Acoustic Standing Wave and Shape of Pipe Does the shape of the container matter when constructing an acoustic standing wave? Is there an advantage of using a cylindrical container over a rectangular prism?
| The shape effect on fundamental pitch is slight, but the effect on the overtone series is significant. Most wind instruments have cylindrical shapes for this reason, and also because square cross-section pipes with bends in them are more difficult to fabricate. This is a topic about which a lot has been written in the field of musical instrument acoustics/physics of musical instruments- far too much to cover here.
| {
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How come Goldstone boson, PQWQ axion, be able to have mass at all? Quote:
Goldstone's theorem: For every spontaneously broken continuous symmetry, there is a massless particle created by the symmetry current.
However, under $U(1)_{PQ}$ symmetry, I read that PQWW axion can obtain mass from $G\tilde{G}$.
Both these sentence make sense, but, being a Goldstone boson, there seemed to be a contraction around PQWW axion.
I read that this might be connect with the approximation from t' Hooft's determinental interaction. Could you explain to me what the determinental interaction was?
Does that mean Goldstone's theorem only works for the first order of t' Hooft determinental interaction?
Does PQWW boson has mass or not?
Could you explain to me why a Goldstone could every obtain a mass?
| Two conditions for pseudo goldstone boson:
*
*The broken symmetry is not gauged, otherwise the (pseudo)
goldstone boson will be eaten by the gauge field.
*Something else (chiral anomaly in this case) breaks the symmetry
even before the spontaneous symmetry breaking is taking place.
To visualize, think about a tilted Mexican hat potential. The lateral move in the tilted Mexican hat groove would be uphills (non-zero mass), as opposed to being flat (massless) for the usual Mexican hat potential. The non-zero mass of the pseudo goldstone boson would be very small if there is only a tiny bit tilting.
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Observed speed of a receding light source Let’s say there’s a planet 4 light years away from Earth and we send a rocket ship towards that planet at 99.9% light speed. We stay behind on Earth and watch the rocket ship travel towards the other planet.
Eventually we should be able to see our rocket ship reach it’s destination. How much time will have elapsed for us until we see that occur?
My intuition would say about 4 years. But I also know that when we observe such a far-away planet, we are ”seeing it as it was 4 years ago”. Well 4 years ago the rocket was still on Earth, so how can I be seeing it landing on the planet now?
Something has to give, but what? Will it appear as if the trip took 8 years to complete?
| It takes the spaceship 4 years to get there, and then it takes light from the spaceship getting there 4 years to get back, i.e. 8 years for an observer on Earth to see it land on the planet.
| {
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What exactly causes particles to drift toward the bottom of a tube in a centrifuge? As I understand, centrifuges can effectively create gravitational fields that are a thousand times stronger than the Earth's. And it is this pseudo-gravitational force that causes particles to sink toward the bottom of a test tube, but this gravitational force within the tube is nothing but the centrifugal force, which is a fictitious force that appears only when we are working within the rotating reference frame.
So by which mechanism exactly do the particles sink toward the bottom of the tube? And how can we have a fictitious force, yet with real effects (the sinking of the particles)?
| Rotating reference frame
In the rotating reference frame the centrifugal force explains the sinking of dense particles, just as gravity does on the surface of earth. Unfortunately, the terminology “fictitious” makes it seem as though it cannot do anything. For that reason I prefer the term “inertial force” instead of “fictitious force”.
In a non-inertial frame fictitious forces (or inertial forces) are necessary to explain the motion of objects and the stresses, strains, and other physical effects. Therefore, it is perfectly valid to attribute the settling to the centrifugal force.
Inertial reference frame
In the inertial reference frame it is important to note that the particles do not accelerate towards the bottom of the tube. Their acceleration is at all times towards the center. It is not the particles which sink outwards but rather the vial and the less dense materials are accelerated inwards faster.
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Why do we use cross products in physics? We can define cross products mathematically like if we take two vectors, we can find another vector with certain properties but why do we use it in physics, if we consider a hypothetical physical quantity like force which is equal to cross product of certain vectors?
For example, the force exerted on a charge in motion in an uniform magnetic field.
Why is it so? Why does that force have to be a cross product of two vectors?
Is it possible to come up with them when what we do is just observe the nature?
| It's really much simpler than the other answers so far have made it out to be. We use the cross and dot products (and all the other math) because they allow us to create fairly simple mathematical models (that is, the laws of physics) that accurately represent what the universe actually does.
| {
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Light beam vs sound beam Why is it that it's very common to have beams of light but not beams of sound? Laser beams are widely available, and I am aware that it is also possible to direct sound, however, we rarely see examples of it.
Is it more difficult to direct due to longer wavelength or is it more dispersive in air or something?
| Like very well the rest of the contributors commented "wave-beams" (apologies for the slight abuse of the term) are not uncommon at all. Medical imaging is just one field where they are used. Sonars is another possible application (both transmission, and reception).
In general, in acoustics (whether it is ultrasound, underwater, or "conventional acoustics") the basic idea is to somehow use an array of transducers and design their interaction in order to create a beam pattern (most commonly know as beamforming - Wikipedia link). This technology finds applications in both mechanical (acoustical) and electromagnetic waves (telecommunications).
One more example of its use in a "more conventional" application is a long-throw loudspeaker (like the SB-3F™: Sound Field Synthesis Loudspeaker and the SB-2: Parabolic Wide-Range Sound Beam, both by Meyer Sound) similar to what Hadrien has cited.
| {
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What is meant by "information" with regard to general relativity and/or black holes? In The Universe in a Nutshell chapter 4, Hawking explains the warping of spacetime according to general relativity and introduces the basics of black holes.
It surprised me to read about "information" that fell into the black hole, and then Hawking spends several pages on whether that information is lost. Information? Really? Information, like what team won the Super Bowl, or how tall Danny DeVito is?
I think of information as concepts that have meaning for humans in a cultural context, at a more abstract level than particles, waves, matter, and energy. But clearly Hawking (and others, e.g. Susskind) have a different thing in mind when talking about cosomological information. So what is "information" in this sense? Is it just the physical properties of the particles and waves that fall into the black hole? Even these are human, cultural concepts, aren't they? After all information per se doesn't fall into a black hole, a particle does, and I as an observer might have information on the properties of that particle before it fell, but I don't get in what sense "information" could have fallen into the black hole.
| Consider a vibrating violin string. To predict its motion, it suffices if I know two functions. One function, $y(x)$, specifies its shape, and the other, $v(x)$, gives its initial velocity at each point. If I know the functions $y$ and $v$ at some time $t$, then I can find them at any other time $t'$ using Newton's laws. In this sense, there is information that is never lost.
As a rough analogy for what Hawking is describing, the black hole keeps us from seeing part of the violin string, and the part we can't see gets bigger and bigger.
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Physical interpretation for wave function in infinite square well If you look at the wave function of a particle in infinite square problem for some specific energy level, say for n =1, then the probability of particle to be found in middle of the well is higher than at any other point. Similarly for higher energy levels, there are points called nodes where the particle can't be found.
What is the physical interpretation for this? Why are some points more probable than others?
| I am going to focus on the last sentence of your question, which sums it up: "why are some points more probable than others?"
The answer is that the waves have to satisfy Schrodinger's equation, and that equation includes that higher kinetic energy goes with higher $d^2 \psi/dx^2$. Meanwhile the boundaries of the box exert their influence, which is that the wavefunction has to go to zero there. So the overall shape of the wavefunction is a combination of these two properties. The maths here is essentially the same as that which describes standing waves on a classical string such as the string of a violin or guitar. In the case of the violin, each part of the string is the same as other parts, but when there is a vibration all at a single frequency, then there is a fixed wavelength, and there is only one way to fit these waves into the region between the two ends. Similarly, for the quantum particle/wave in a box, for a given energy one has focussed on a solution where all parts of the wavefunction oscillate at the same frequency, and this implies a fixed wavelength, and the ends determine how the waves fit into the box. It is the combination of these features which produces the nodes and hence the fact that if one were to detect the location of the entity (particle) then one is more likely to find it in some places than others.
You should note that the presence of nodes at a fixed places is a special property of the states of well-defined energy. But the entity doesn't have to have a completely precisely defined energy. If it is in a superposition of energy eigenstates then its wavefunction will not necessarily have these nodes, except at the walls.
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Is the weight of a projectile launched from earth and while still flying in the atmospheric sky transferred to the ground? Raised by this question.
Q1. Is the weight of a projectile like a bullet or a ballistic missile (atmospheric flight trajectory) transferred to the ground?
Of course i know that if the projectile was out of sky there is no doubt that it can be seen as a separate object like satellites or the moon but if a mass goes in a ballistic path inside the atmosphere (specially near ground) does it's wight transferred to the ground?
Q2. what about jet propelled rockets like space launch vehicle while they are near ground?
| Any object in the atmosphere, that is not in freefall, is ultimately transferring it's weight to the ground. A normal bullet fired from a gun is essentially in freefall once it leaves the gun. After leaving the gun, it has no lift, it starts falling. It may have upward momentum which it will immediately start losing. It's upward acceleration is transferred to the ground by the gun's recoil, until it leaves the barrel. While in a freefall trajectory it does not transfer weight to the ground. A powered missile, or rocket, is a different matter. While under power, or using lift surfaces, it will be pushing gasses downwards to create lift, these downward moving gasses will eventually cause increased pressure on the ground. Even lighter than air balloons displace air which will increase pressure on the ground. The ground pressure differences involved will eventually equal the amount of lift created.
| {
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Fermi energy definition Ok, so I'm having a hard time understanding the definition of Fermi Energy. Several sites basically repeat each other, saying that it is the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero 1, and others say that it is the highest energy that the electrons assumes at 0K 2. Are these 2 concepts the same, and I'm just not getting it? Is the top level of an electron at 0K equal to the difference in energy between highest and lowest occupied states? Any clarification would be greatly appreciated.
| From the Schrödinger equation, energy for a bound electron is quantized, such that only certain energy levels are allowed. Because electrons are fermions, they obey the Pauli exclusion principle, which states that no two electrons can have all their quantum numbers (such as energy level, orbital, spin) equal.
This means that on each energy level, there are only a certain amount of electrons that can occupy that energy level (depending on how many orbitals that level has). If another electron would be added, that electron would have to occupy a different energy state (usually a higher one). The Fermi energy is then the energy of the highest occupied state, when the system is in the ground state.
When the system is in the ground state implies that all levels under the highest occupied state are also occupied.
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Justification for Loop de Loop minimum speed I was trying to figure out the minimum speed an object would have to travel on a loop not to lose contact with the loop. Setting the centripetal force equal to gravity $m\frac{v^2}{r} = mg$ gives $v = \sqrt{gr}$ that explanation is valid and makes sense to me but I was wondering why a conservation of energy approach wasn't. Entering the loop with speed $v$ and setting Kinetic energy equal to gravitational potential $0.5mv^2 = mgR$ gives $v = \sqrt{2gr}$ which obviously is not the same. Why is this explanation not correct?
| First case :- In the first case(where you used centripetal force), the velocity $v=\sqrt{gr}$ is, in fact, the velocity of the object at the top of the loop. To find the velocity of the object at the bottom of the loop, you will need to use energy conservation.
$$\frac{1}{2}m((\sqrt{gr})^2-v_{bottom}^2)=-2mgr \Rightarrow v_{bottom}=\sqrt{5gr}$$
Second case :- In the second case, you assumed the velocity at the topmost point of the loop to be $0$. So, by energy conservation,
$$\frac{1}{2}m(0-v_{bottom}^2)=-2mgr \Rightarrow v_{bottom}=\sqrt{4gr}$$
Well, you can see that in the second case, the bottom velocity comes out to be lesser than that in the first case. It is because, the assumption that the velocity at the topmost point being zero is wrong. Imagine it this way: If, somehow, the velocity at the topmost point becomes zero(well, that would never be the case as the object, when thrown with bottom velocity $=\sqrt{4gr}$, will surely leave the contact from the loop before reaching the topmost point), then how do you expect the object to complete the loop. The object would stop and gravity would take over and pull the object back to the ground without letting it complete the loop. So you need to throw the object a little faster($v=\sqrt{5gr}$) such that it always stays with the loop and does not fall down midway between the loop.
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