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Why does the potential energy of the system decrease when two charge balls are connected using a connecting wire I am confused because I've seen in textbooks and online solutions to questions that when connected,the potential energy of a system of two charged spheres decreases.But according to law of conservation of energy shouldn't it remain the same?
| The energy is lost as heat through the wire resistance. Basically when you connect two conductors at different potentials through a wire they will exchange charge until they have the same potential. Energy is lost in this process but charge is conserved. This will become more clear once you learn about capacitors and current electricity, more specifically RC circuits.
| {
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Quantum field theory, interpretation of commutation relation Let $\phi$ be the quantum field
$$
\phi(x) = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}} \Big[ b_\mathbf{p}e^{-ip\cdot x} + c_\mathbf{p}^\dagger e^{ip\cdot x} \Big]
$$
with commutation relations
$$
[b_\mathbf{p}, b_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
$$
[c_\mathbf{p}, c_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
all other commutators zero. Let $Q$ be the charge operator
$$
Q = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \Big[c_{\mathbf{p}}^\dagger c_{\mathbf{p}} - b_{\mathbf{p}}^\dagger b_{\mathbf{p}} \Big].
$$
We calculate the commutator $[Q,\phi] = \phi$. The question is what is an interpretation of this commutation relation? We know that $Q$ is the number of antiparticles minus the number of particles.
| One interpretation is as follows: $[Q,\phi(\vec{x})] = \phi(\vec{x})$ means that $Q\phi(\vec{x}) = \phi(\vec{x})(Q + 1)$. Thus, if $\vert q \rangle$ is a charge eigenstate with eigenvalue $q$ (i.e. $Q\vert q \rangle = q\vert q \rangle$) then
$$ Q \phi(\vec{x}) \vert q \rangle = \phi(\vec{x}) (Q + 1) \vert q \rangle = (q + 1) \phi(\vec{x}) \vert q \rangle, $$
which means that $\phi(\vec{x}) \vert q \rangle$ is a charge eigenstate with eigenvalue $q + 1$. Thus, acting with $\phi(\vec{x})$ increases the charge by $1$.
In fact, a common interpretation of the operator $\phi(\vec{x})$ is that it creates a particle at position $\vec{x}$ (being a kind of Fourier transform of the creation operator $c^\dagger_\vec{p}$, which creates a particle with momentum $\vec{p}$). So the commutation relation says that if you add a particle, the total charge will increase by $1$.
| {
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Is it possible to conserve the total kinetic energy of a system, but not its momentum? It is possible to conserve momentum without conserving kinetic energy, as in inelastic
collisions. Is it possible to conserve the total kinetic energy of a system, but not its momentum? How?
To clarify, I am not necessarily talking about an isolated system. Is there any scenario which we could devise in which momentum is not conserved but kinetic energy is?
| Yes, it is possible to conserve the total kinetic energy of the system but not the momentum.
Let me give you an intuitive explanation.
*
*Suppose we have two charges, in which one is fixed and other is free to move and then they are released at some distance. In this case since one charge is fixed so the net external force on the system is not zero, so momentum of the system will not remain conserved, but kinetic energy + potential energy of the system will remain conserved, as there are no dissipative non conservative forces involved in the system.
*Suppose a particle is tied to a point with the help of a string and is performing uniform circular motion on the horizontal surface. In this case the kinetic energy of the system will remain same but not the momentum.
| {
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How to understand transfer of momentum between motorcycle seat and rider? I have an app that samples acceleration at a motorbike seat. I want to be able to compare how comfortable these accelerations feel for two different motorcycles.
Let's say I have the following scenarios:
*
*A 75 kg rider on 300 kg motorcycle experiencing vertical acceleration of 1g from the seat
*A 75 kg rider on 100 kg motorcycle experiencing
vertical acceleration of 1g from the seat
Will the subjective experience of the rider on both bikes be identical?
- OR -
Will the heavier bike feel stronger because it transfers more momentum to the rider?
| The mass of the motorcycle is irrelevant. If the motorcycle (and hence rider) is experiencing the same acceleration, then they will "feel" the same thing. The rider will have the same momentum in either case.
The difference just comes into how much force is needed from the motor for the same acceleration. More mass means more force is needed.
| {
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Calculating the Hamiltonian of complex scalar field I am working on the Peskin's QFT problems, and i'm finding difficulties in calculating the Hamiltonian of complex scalar field. The Hamiltonian read:
$$H = \int d^3x (\pi^*\pi+\nabla\phi^*\nabla\phi+m^2\phi^*\phi)$$
Then I substitute the expression of $\phi$ and $\phi^*$ into it:
$$\phi(x)=\int\frac{d^3p}{(2\pi)^3}\frac 1{\sqrt{2E_p}}(a_pe^{-ipx}+b_p^+e^{ipx})$$
$$\phi^*(x)=\int\frac{d^3p}{(2\pi)^3}\frac 1{\sqrt{2E_p}}(b_pe^{-ipx}+a_p^+e^{ipx})$$
Integrate over $x$ and get the delta function, thereby cancel one of the integral variable q, I get:
$$H=\int\frac{d^3p}{(2\pi)^3}[(-\frac{E_p}2+p^2+\frac{m^2}{2E_p})(b_{-p}a_pe^{-i2E_pt}+a_p^+b_{-p}e^{i2E_pt})+(\frac{E_p}2+p^2+\frac{m^2}{2E_p})(b_{p}b_{p}^++a_p^+a_p)]$$
I don't know how to continue. In some answer book I find the next step change the integral variable back to $x$, and get $\int d^3x\frac{E_p^2+p^2+m^2}{2E_p}(b_{p}b_{p}^++a_p^+a_p)$ but I don't know how and why.
| The other answer can't be correct because it doesn't have units of energy $\omega_k$. Making an expansion using
$$ \phi(x) = \int \frac{d^3 k}{(2 \pi)^3 \sqrt{2 \omega_k}} \ \big(a_{\vec{k}} e^{-ik\cdot x} + b^{\dagger}_{\vec{k}} e^{ik\cdot x} \big )$$
I found (after neglecting the infinite constant)
$$ H = \int \frac{d^3k}{(2\pi)^3} \omega_k (a_{\vec k}^\dagger a_{ \vec k} + b_{ \vec k}^\dagger b_{ \vec k}) $$
| {
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How could the baseline of atmospheric neutrinos be as much as $10^7$ meters while exosphere (outer part of atmosphere) is at most 8e5 meters? Atmospheric neutrinos correspond to neutrinos produced by the interaction of cosmic rays in the Earth atmosphere. The Earth atmosphere is at most $800$ km=$8 \cdot 10^5$ meters.
So how could the "baseline" (=distance of flight) of the atmospheric neutrinos be as much as $10^7$ meters, as stated in the Table of properties of atmospheric neutrinos in the famous (most cited book in particle physics) Particle Data Group review:
http://pdg.lbl.gov/2019/reviews/rpp2019-rev-neutrino-mixing.pdf
And bonus question: how could solar neutrinos have a baseline (it is stated $10^{10}$ meters) higher than 150 millions km= order of $10^8$ meters: that looks impossible?
| Neutrinos produced in Earth's atmosphere have two opportunities to interact with detectors on Earth's surface: once on their way down from the sky, and again when they emerge unscathed on the other side of the planet. So the baseline is more like the diameter of Earth, which is about $12.7×10^6$ meters.
Famously, the IceCube detector in Antarctica looks down to see neutrinos from the north celestial sphere.
Re your bonus question: "million kilometers" is a stupid unit, and everyone makes off-by-thousand exponent-counting errors when using it. An astronomical unit is $1.5×10^{11}$ meters.
| {
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Why is simple harmonic motion considered to be a sinusoidal function? Motion of a pendulum is said to be an harmonic motion. What do we exactly mean by harmonic and why is that the displacement and velocity etc. Represented by a sine function?
| They result from physical systems with a peculiar type of restoring force: the acceleration is proportional to the displacement, but directed to the opposite side.
$x^{(2)}(t) = -k^2x(t)$
One way to examine what types of function have that property is testing as a solution a generic function expanded in a Taylor series. Depending on the boundary conditions it will be a sin, a cos, or combination of both:
$x(t) = x(0) + x^{(1)}(0)t + (1/2!)x^{(2)}(0)t^2 + (1/3!)x^{(3)}(0)t^3 + (1/4!)x^{(4)}(0)t^4 + (1/5!)x^{(5)}(0)t^5 + \dots$
For boundary conditions: $x(0) = 0\;$ and $\;x^{(1)}(0) = k$
$x^{(2)}(0) = -k^2x(0) = 0$
$x^{(3)}(0) = -k^2x^{(1)}(0) = -k^3$
$x^{(4)}(0) = -k^2x^{(2)}(0) = 0$
$x^{(5)}(0) = -k^2x^{(3)}(0) = k^5$
$x(t) = 0 + kt + 0 - (1/3!)k^3t^3 + 0 + (1/5!)k^5t^5 + \dots = \sin(kt)$
for boundary conditions: $x(0) = 1\;$ and $\;x^{(1)}(0) = 0$
$x^{(2)}(0) = -k^2x(0) = -k^2$
$x^{(3)}(0) = -k^2x^{(1)}(0) = 0$
$x^{(4)}(0) = -k^2x^{(2)}(0) = k^4$
$x^{(5)}(0) = -k^2x^{(3)}(0) = 0$
$x(t) = 1 + 0 - (1/2!)k^2t^2 + 0 + (1/4!)k^4t^4 + 0 + \dots = \cos(kt)$
| {
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Why is the normal contact force horizontal on an inclined ladder?
There is only one force acting on the ladder which is its weight and it acts vertically downwards. Then why does the normal contact force from the vertical wall act horizontally on the ladder? There must be a horizontal force acting on the wall to exert a horizontal force on the ladder. What causes the horizontal force on the wall and what is it called?
| Actually if you see the diagram , body is in rotational inertia, and rod is not perfectly vertical or horizontal, thing which work here is frictional force between rod and surfaces, friction always try to maintain the inertia, here the friction which acts is static friction.
| {
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Why is there a "blue hour" after the "golden hour"? There's a great story about why the sky is blue during the day, and turns golden during sunsets:
Rayleigh scattering affects blue light more. During the day, blue light from the Sun is scattered towards us from all directions, causing a blue sky. During the sunset, the length of atmosphere the light from the Sun has to travel through becomes so long that the blue is depleted, giving the sky a golden color.
However, it seems this can't be the full story, because photographers know that after sunset, there is a so-called blue hour where the color of the sky becomes a deep blue again. Why would the color go from blue to golden to blue again?
Wikipedia states very strongly that explanations of this in terms of Rayleigh scattering are wrong, and that the real explanation is the absorption of blue light by ozone. But it doesn't explain why that would create the effect. If the blue light is not reemitted, then this just amounts to having less blue light, so it can't explain why the blue hour is more blue. And if the blue light is reemitted, then the effect of ozone should qualitatively be very similar to the effect of Rayleigh scattering, since it's just another scattering route that favors blue light, leading us back to the original puzzle.
What's the right explanation for the blue hour?
| The reason is the same as in rainbow; different colours (wavelenghts)bends differently (sorry my language).
This Blue hour is the sunlight which arrives below the horizon. The reason why other light doesn't is on their wavelength. The Blue light can bend the most, and thus it can still reaches the Earth surface an hour after sunset, while green, yellow and red are already going above our heads.
This picture form german language Wikipedia on topic "Grüner Blitz" pretty much explains the issue. So there actually even is a green flash between "golden hour" (Red-orange-Yellow) and the blue hour.
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Dimensionless expression for differential equation I am working through Nonlinear Dynamics and Chaos by Steven H Strogatz. In chapter 3.5 (overdampened beads on a rotating hoop), a differential equation is converted into a dimensionless form. I am trying to work out which dimensions the initial equations had, and why the converted form is dimensionless.
Initial equation:
$mr \ddot{\phi} = -b \dot{\phi} -mg \sin\phi + mr \omega^2 \sin\phi \cos \phi $
$m$ is mass, $r$ is radius, $\phi$ is an angle, $b,g$ are arbitrary, positive constants, and $\omega$ is angular velocity.
Using a characteristic time $T$, a dimensionless time $\tau$, with $\tau = \frac{t}{T}$ is introduced.
$\dot{\phi}$ and $\ddot{\phi}$ then become $\frac{1}{T}\frac{d\phi}{d\tau}$ and $\frac{1}{T^2}\frac{d^2\phi}{d\tau^2}$, respectively.
Then the initial equation becomes
$\frac{mr}{T^2}\frac{d^2\phi}{d\tau^2} = -\frac{b}{T}\frac{d\phi}{d\tau} - m g \sin\phi + mr \omega^2 \sin\phi \cos \phi$
This is made dimensionless by dividing through a force $mg$:
$(\frac{r}{gT^2})\frac{d^2\phi}{d\tau^2} = (-\frac{b}{mgT})\frac{d\phi}{d\tau} - \sin\phi + (\frac{r \omega^2}{g}) \sin\phi \cos \phi$
And all the expressions in the brackets are dimensionless.
I understand why the expressions in the brackets are dimensionless, but what about the differentials?
$\phi$ is dimensionless.
but would
$\dot{\phi}$ not have dimension $\frac{1}{s}$, and $\ddot{\phi}$ $\frac{1}{s^2}$?
| You substitute $\dot{\phi} = \frac{1}{T} \frac{d\phi}{d\tau}$. Now let's consider the dimensions
\begin{align}
[\dot{\phi}] &= [\frac{d\phi}{dt}] = \frac{rad}{s}\\
[\dot{\phi}] &= [\frac{1}{T} \frac{d\phi}{d\tau}] = \frac{1}{s} \cdot [\frac{d\phi}{d\tau}] \\
\Rightarrow [\frac{d\phi}{d\tau}] &= rad
\end{align}
and analog for $\ddot{\phi}$. Thus, by substituting $dt \to T d\tau$ we obtain a dimensionless time $\tau$. Just look at the dimension of the last expression
\begin{align}
s = [dt] = [T d\tau] &= s \; [d\tau] \\
&\Rightarrow [d\tau] = 1
\end{align}
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Precise definition of the vertex factor Just a short question about the vertex factor in QFT. When I have an interaction Lagrangian
$$\mathcal{L}_{\mathrm{int}}=-\frac{\lambda}{3!}\phi^3$$
with a real scalar field $\phi$, is the vertex factor given by $-i\lambda$ or $-i\frac{\lambda}{3!}$?
Because as far as I learnt, the vertex factor is $-i\frac{\lambda}{3!}$ and $3!$ is the symmety factor of the diagram. But I saw in many books that they claim that $-i\lambda$ is the vertex factor of this interaction.....
| It seems that OP is not questioning the standard convention to divide each term in the Lagrangian with its symmetry factor, e.g., $${\cal L}~=~-\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{3!}\phi^3.$$
Rather OP is assuming the above standard convention, and asks if the vertex factor$^1$ is $-\frac{i}{\hbar}\lambda$ or $-\frac{i}{\hbar}\frac{\lambda}{3!}$? The answer depends on context:
*
*On one hand, viewing the 3-vertex as an amputated Feynman diagram (say, as the leading contribution to the 1PI 3-point vertex function), the vertex factor is $-\frac{i}{\hbar}\lambda$. An amputated Feynman diagram is typically stripped of the symmetry factor of the external legs because the legs are distinguishable -- they carry different momenta for starters.
*On the other hand, e.g. in the source picture
$$\phi\quad\longrightarrow\quad\frac{\hbar}{i}\frac{\delta}{\delta J} ,$$
where interaction terms
$$ -\frac{i}{\hbar}\frac{\lambda}{3!}\phi^3\quad\longrightarrow\quad-\frac{i}{\hbar}\frac{\lambda}{3!}\left(\frac{\hbar}{i}\frac{\delta}{\delta J}\right)^3$$
are differentiating propagator terms $\frac{i}{2\hbar}J\Delta J$ to build Feynman diagrams, the vertex factor is
$-\frac{i}{\hbar}\frac{\lambda}{3!}$. For the source picture, see eq. (3) in my Phys.SE answer here.
References:
*
*M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; first paragraph on p. 93.
--
$^1$ We have restored the factors of $\hbar$.
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Is Young's modulus a measure of stiffness or elasticity? Young's modulus seems like a modulus of stiffness. It tells us how difficult is it (how much stress is required) to produce longitudinal strain in a solid. It does not tell anything about how an object will react when the deforming force is removed.
How can one on the basis of young's modulus decide that steel is more elastic than a rubber band?
| In physics elasticity is defined as the ability of a material to resist a distorting influence and return to its original size and shape when the distorting influence is removed (source Wikipedia).
Young’s Modulus is the ratio of applied stress to resulting strain in the linear elastic region of behavior. Therefore, they greater Young’s modulus the stiffer a material is, that is, the greater the materials ability to resist a distorting influence (applied stress). For this reason, given the above general definition of elasticity, Young’s modulus is also called the modulus of elasticity.
Due to the highly overall non linear behavior of rubber I believe Young’s modulus for rubber is usually quoted forces small loads. The values for rubber are much lower than steel meaning that rubber is much less able to resist a distorting influence than steel as would obviously be expected.
Given the above definitions of elasticity and Young’s modulus we would conclude that the “elasticity” of rubber is less than steel. It is admittedly counter intuitive because on the one hand we think of rubber bands as being highly elastic in the sense that they’re easy to stretch. But on the other hand if they’re easier to stretch that means they’re less able to resist distortion, which is consistent with the physics definition of elasticity and Young’s modulus.
Hope this helps.
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Internal force disintegrating a solid body? Let $M$ be a block on a frictionless surface. Now let us mentally divide (not physically) the block into 2:1 ration (i.e $1/3$ of the left be called $M_1$ and $2/3$ right be called $M_2$). So $M_1$ applies force $F_1$ on $M_2$ and $M_2$ applies force $F_2$ on $M_1$ and by 3rd law they are equal. Hence acceleration of $M_1$ would be $2a$ and that of $M_2$ would be $a$. Shouldn't this deform the block?
| If the body is isolated, then $a=0$ and the internal forces are $0$. This is a trivial case where there is no longer a contradiction or possibility of the body being torn apart. Therefore, let's consider the scenario of a body that has external forces acting on it.
You are forgetting about external forces being applied to the object. In other words, the net force acting on section $i$ in general is not going to just be $F_i$. You need to specify what external forces are acting on the entire object (and where these forces are being applied). Once you do this, you will find that treating the system as a whole or as many parts should still give you consistent results.
Let's take a simple example of a rigid cube of mass $M$ on a frictionless surface, and we push on the left side of this cube to the right with a force of magnitude $F$. Then by Newton's second law the acceleration of the cube is $a=F/M$.
Now let's hypothetically partition our cube into two pieces like you describe, with $1/3$ of the cube as one part on the left and $2/3$ of the cube as the other part on the right. Let's call the interaction force between the two parts $F_{\text{int}}$. Applying Newton's second law to each partition with acceleration $a=F/M$ we have
$$F-F_{\text{int}}=\frac M3\cdot\frac FM$$
$$F_{\text{int}}=\frac{2M}{3}\cdot\frac FM$$
Both equations show that $F_{\text{int}}=2F/3$. Therefore, we don't have any issues here, and this shows the mistake you were making. Each partition has an acceleration of $a=F/M$, and the partition with mass $2M/3$ mass has twice the net force to the right in order for it to have the same acceleration as the partition with mass $M/3$. No deformation occurs due to how we hypothetically partition the system.
In fact, you can generalize the above example for the case of partitioning the left side to have mass of $xM$ and the right side to have mass $(1-x)M$ for $0\leq x\leq1$. You can use the same above analysis to show that the interaction force has a magnitude of $(1-x)F$. This makes sense: the father you are from where the external force is applied, the weaker the interaction force becomes. It even drops to $0$ once we consider the entire block as a single partition ($x=1$), and it is equal to $F$ right at the point of application ($x=0$).
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Why does the second law of thermodynamics prevent 100% efficiency? So far in my thermodynamics lecture course, my understanding of the laws of thermodynamics is that the first law is about the conservation of energy, the second law says entropy must always increase or stay the same which apparently results in the fact you can never achieve 100% efficiency of heat engines, unless at $T = 0\,\mathrm K$, and the last law says that you can't get to $T= 0\,\mathrm K$.
I have never explicitly seen why the fact that entropy must always increase or stay the same results in the prevention of achieving 100% efficiency. The only proof I have is showing the Carnot cycle is the most efficient and that is only 100% efficient if the cold reservoir is at absolute zero, which it can not be at.
Is there any way to work from the statement: $\Delta S \geq 0$ (for any process in a closed system), to some result which says you can not achieve 100% efficiency?
| If you can convert all of the heat to work, you're reducing entropy by definition ($\Delta S = \frac{Q}{T}$ , If $Q<0$ then $\Delta S < 0$).
If you allow yourself to let some heat flow into somewhere cold (heating something up instead of using all of the heat to work) you raise the entropy in the cold substance enough to let you not defy the second law, and the rest can go to useful work.
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Usually, how much does a phonon travel without scattering? Phonons propagate without problems in a lattice, until they scatter on something, like a defect, an electron, or another phonon. But in a typical solid at room temperature, how much (or how long) is the mean free path of a phonon? I know that depending on the temperature one type of scattering can be more probable than the other, but do mean free paths of these different types of scattering tend to be very different from each other? (With different orders of magnitude?)
| That depends strongly on many factors. The reason for the decrease in thermal conductivity at high temperature is that the mean free path goes down because of phonon-phonon scattering (interaction with the lattice). At some point the phonons do not propagate much further than a lattice period and the lattice melts. So those are the shortest mean free paths.
At lower temperatures, phonon mean free paths can be very long. Some researchers synthesized isotopically pure C-12 diamond with high thermal conductivities. The mean free path can be millimeters.
This is for high-frequency phonons that conduct most thermal energy. Phonons with longer wavelength have longer scattering lengths.
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Path integral for fermions on circle and background gauge fields This question is continuation of Path integral for fermion on circle.
I'm reading Witten article Anomaly Inflow and the $\eta$-Invariant and wanna to understand some technical details.
In section 4.1 authors consider even number $n$ of Majorana fermions:
$$
I = \int dt \frac{i}{2} \sum_{i=1}^{n} \chi_i \frac{d}{dt} \chi_i
$$
To calculate path integral authors want use equations (as I understand, they use (74) and (75) from Path integrals for fermions, susy quantum
mechanics, etc..):
$$
Tr_{\mathcal{H}} e^{iH T} = \int_{NS} D\chi_i \;e^I
$$
$$
STr_{\mathcal{H}} e^{iH T} = \int_{R} D\chi_i \;e^I
$$
At this moment, using $H=0$, we can easily obtain results, by calculation of Hilbert space dimension:
$$
\int_{NS} D\chi_i \;e^I = 2^{\frac{n}{2}}
$$
$$
\int_{R} D\chi_i \;e^I = 0
$$
But authors chose another way:
Though we could add additional terms to the action to get a nonzero Hamiltonian, instead we will turn on a background $SO(n)$ gauge field on the circle.
As I understand, this mean
$$
I \to I_A = \int dt \frac{i}{2} \left(\sum_{i=1}^{n} \chi_i \frac{d}{dt}\chi_i - \chi_i A_{ij} \chi_j \right)
$$
$$
H_A = \frac{i}{2} \chi_i A_{ij} \chi_j
$$
After authors use holonomy U, and say that U is block diagonal matrix. Form of block:
$$
U_k = \begin{pmatrix}
\cos\theta_k& \sin\theta_k \\
-\sin\theta_k & \cos\theta_k
\end{pmatrix}
$$
I don't understand this step. After they immediately obtain results:
$$
Tr_{\mathcal{H}} U = \prod_{k=1}^{n/2} 2 \cos(\theta_k/2)
$$
$$
STr_{\mathcal{H}} U = \pm\prod_{k=1}^{n/2} 2i \sin(\theta_k/2)
$$
Why did they use background gauge field?
How did they calculate this traces?
How this result correspondence straight calculation of dimension of Hilbert space?
| This can be calculated by rotating to the Cartan subalgebra of the gauge symmetry group (which is why the matrices turned out diagonal) and then computing the functional determinants.
See, for example,
*
*https://arxiv.org/abs/1607.04230
which is part of the list of references I gave on your other question on the worldlien formalism of QFT.
| {
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On the derivation of the north-south aberration angle In A.P. French's Special relativity, page $39$, the author said the north-south aberration angle $\alpha$ below, in figure (b), is equal to $v\sin(\theta_{0})/{c}$, where $\theta_{0}$ is the angle when the earth is stationary (no aberration).
How did $v\sin(\theta_{0})/c$ came about?
This value seems rather suspicious, especially because of the $\sin(\theta_{0})$. According to this Wikipedia article, the angle $\theta$ above (named $\phi$ in the article) satisfies $$\tan(\theta)=\frac { \sin(\theta_{0})}{v/c + \cos (\theta_{0})},$$
thus,
$$\tan(\alpha)=\tan(\theta_{0}-\theta)=\frac{\frac{\sin(\theta_{0})}{\cos(\theta_{0})}-\frac { \sin(\theta_{0})}{v/c + \cos (\theta_{0})}}{1+\frac{\sin(\theta_{0})}{\cos(\theta_{0})}\times \frac { \sin(\theta_{0})}{v/c + \cos (\theta_{0})}}.$$
I highly doubt that taking the $\tan^{-1}$ of the above would result in $v\sin(\theta_{0})/c$.
Is there a simpler way to compute the aberration angle to check whether it conforms with $v\sin(\theta_{0})/c$?
|
If you construct the right-angle triangle as shown in red on fig. (b) then it is easy to see that $\sin\alpha = \text{something}/c$. And by breaking $v$ into components you can see (from the doodle on the right) that the something is equal to $v\sin\theta_0$.
Then, as mentioned in my comment, all that's left is to say that $\alpha$ is small (i.e. $v\ll c$) so that $(v\sin\theta_0) /c=\sin\alpha\approx\alpha$.
(Please excuse the shoddy MS Paint editing.)
| {
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Could speed of light be variable and time be absolute? I get my "demonstration" of time dilation from the textbook thought experiment.
A laser is mounted on a cart with a reflective ceiling. At $t=0$ the cart starts moving and the laser is fired. When the laser is reflected back at the starting point the (thought) experiment stops.
Now, two different observers, one sharing the frame of the cart and another standing on the ground perpendicular to the cart will observe two different things. For the first one, the laser bounces back and then down in a straight line. For the second one, the light travels in a triangular pattern which is longer than the path observed by the first guy.
Given that the speed of light is constant, the time has to dilate/contract.
Why is the speed of light held constant here? Could we work out a physics where time is absolute but the maximum speed of light variable?
| No. You're neglecting that time dilation is only one effect explained by relativity. You'd also need to take into account length contraction or, more generally, Lorentz invariance. Your "absolute time" is not going to capture the fact that there are properties of spacetime (not just time) at play here.
In light of the comments (pun intended!), a further point: The idea that the speed of light is constant is a consequence of the Maxwell equations. Relativity follows that logically and historically, it doesn't precede it. If you open the question to spacetime, then you still have the issue that the Maxwell equations are and were well-known. Special relativity is about exploring the consequences of that experimentally verified fact. It's not clear what you mean by "absolute" here, but any interpretation that comes to my mind implies a preferred frame for Maxwell, and we know that does not exist due to experiment.
| {
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Is Griffiths simply wrong here? (Electrostatic Boundary Conditions)
In the above illustration, shouldn't $E_{above}$ and $E_{below}$ be in opposite directions? If not, how did Griffiths end up the following equation? From the above directions, shouldn't the flux add up?
| The illustration and the equation may be confusing. It corresponds to a situation with an external field pointing upward that is larger than that due to $\sigma$. However the two arrows are of the same length, so if this were intended then $\sigma=0$. If there is no such field then $E_{above}$ and $E_{below}$ point in opposite directions and their magnitudes should be added, not subtracted.
| {
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Conservation of Angular Momentum for Rigid Bodies I have question about Conservation of Angular Momentum of Rigid Bodies. I've been doing some examples from Hibbeler's book, and noticed that in this chapter about Conservation of Angular Momentum of Rigid Bodies, there are some examples where we sum all the angular momentums about some fixed point O, but when they are writing equation, they write $I_G$ (inertion in point G, center of mass). Why $I_G$ ? Why they didn't write $I_O$ (inertion in point O, the point about which we are writing this equation of conservation). So this confuses me. I mean, there are also examples where they do it like I expect (Inertion about point O).
Look at this example.
As you can see, we are writing conservation of angular momentum about point A. So:
$\sum{H_{A1}}$ = $\sum{H_{A2}}$.
And as you can see they calculated $I_G$. Why? Why they didn't wrote about point A, and wrote $I_A$ = $I_G + md^2$ (where $d$ is distance of center of mass of body from point A).
So this confuses me.
Whoever helps, thanks in advance!
| Finding the rotational inertia of the disk about point A would involve a complex integration. Instead, they use the parallel axis theorem. The disk is treated like a point mass revolving about point A, but then you have to add its resistance to also rotating about its own center. Either approach would work for the rod. They are treating the bullet like a point mass.
| {
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How did early radar determine range/ distance precisely? Wikipedia talks about precise timing of the returned radar pulse, with an animation of a clock.
But they didn't have atomic clocks and such before or during WWII.
So how did they determine distances and (possibly) velocities back rhen?
| Early radar uses analog signals and displays.
For example, the WWII Chain Home system would send each pulse at the same instant that an electron beam started across a CRT. The beam would be deflected by any received signal, and the left-right offset gave the distance from the return time:
Chain Home display showing several target blips between 15 and 30 miles distant from the station.
Later, the familiar circular display was created. The beam starts at the center and runs outward in a direction corresponding to where the antenna points at that instant, brightening when an echo is received:
Although the radar pulse moves out and back at the speed of light, radar is used over long distances. A target a kilometer away is a six microsecond echo; 100 miles is six hundred microseconds. The tracking signals involved are MHz or less, often much less.
| {
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D'Alembert's principle and the work done by constraint forces in Atwood's machine From what I understand, constraint forces do no work because they are perpendicular to the allowed virtual displacements of the system. However, if you consider an unbalanced Atwood machine, in which both masses are accelerating in opposite directions, you'll find that the tension force of the wire (a constraint force), which pulls the lighter mass up, is parallel to the displacement, which means it does work (right?).
Now, I understand that the same is true for the other side: the tension force on the heavier mass is parallel to the displacement, but in the opposite direction, so that if you add the work done by the tension force on the heavier side to the work done by the tension force on the lighter side you get zero.
So my question is: would it be correct for me to say that individual constraint forces can do work, but it's the sum of the work done by all the constraint forces which is always equal to zero? If this is true, it's a bit different from the notion I had before, which was that individual constraint forces never do work because they are always perpendicular to the displacement.
| Gravity is pulling the weights down, the cable provides a directly opposing restraint, to keep the weights from accelerating downwards at 9.8 meters per second per second. The actual work is done by using the heavier weight's gravitational potential.
| {
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Postulate of constancy of speed of light in vacuum I'm not of course questioning the constancy of the speed of light, just the way the postulate about it is worded. It is often stated that the speed of light is independent of the motion of the source. Einstein himself said "and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite speed $c$ which is independent of the state of motion of the emitting body."
But even in classical physics, isn't the speed of a wave independent of the motion of the source? Once the wave is emitted, its speed just depends on the medium. If I have a loudspeaker moving toward me, the frequency of the sound will be higher but not its speed. If I am moving toward the loudspeaker, both the frequency and the speed of the sound will be higher.
This postulate of relativity is sometimes stated in terms of the state of motion of the observer, but it is frequently stated just in terms of the source, which is what I am questioning.
| The doppler effect in classical physics works just as well if you are moving towards or away from the source or if the source is moving towards or away from you. Relative motion is what matters.
| {
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Why does the rope fly up from the nail against gravity?
Why does the rope fly up from the nail against gravity? As shown in the picture, the blue arrow is the nail fixed on the wall, the yellow arrow is the rope, and the green arrow is the iron block. The rope on the right side of the nail moves downward under the action of the iron block, and the rope on the left side of the nail moves upward. At some point, the rope goes up and away from the nail. Why? How to calculate the height away from the nail?
Video of rope rising into the air
| According to @weeeeliam's answers to the following questions:
What is the formula for calculating the tension of the rope section?
There is a circle of rope that rotates at a uniform angular velocity $ω$. What is the formula for calculating the tension of the rope section? Without gravity, the density of the rope is $ρ$, the radius of the rope circle is $R$, and the section radius of the rope is $r$.
@weeeeliam:
Consider an infinitesimally small section of the of the string $d\theta$. The following diagram illustrates this:
The tension is of the same magnitude throughout the rope, and it acts perpendicular to the vector from the center of the string to the point of action.
From this diagram, you can tell that only the x-components to the left matter, since the y-components of the tensions cancel out. The x-components of the two tension vectors must be equal to the force required for centripetal acceleration.
$$2T\text{ sin}(d\theta/2) = Td\theta = (dm)\omega^2R$$
The small bit of mass can be found as follows:
$$dm = \rho dV = \rho A dx = \rho \pi r^2 (r d\theta) = \rho \pi r^3 d\theta$$
Then, cancelling out the $d\theta$ on both sides of the equation:
$$\boxed{T = \pi \rho r^3 \omega^2 R}$$
Note: This solution assumes $r << R$.
Here is my explanation: suppose the rotation is uniform, then
As shown in the figure, if
$T=πρr^3ω^2R＞ρHπr^2$
$⇒$
$rω^2R > H$
Then the rope will go up.
| {
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How does light speed in a medium affect refraction? Light rays hit a convex lens and converge at a focal point after passing through as shown by the diagram below:
It is obvious by looking here that the light rays travel different distances to the focal point but they arrive at the same time. Photons travelling along the Principal axis travel in a straight line and spend the longest time in the medium. I read that the medium (of glass in this case) slows down light due to weak interaction. An airplane slowing down would lose altitude like the light changing course in the top half of the lens. I just don't understand how the direction of the light rays gets changed in an upward direction for the bottom half of the lens.
To be very blunt lets say 5 individual photons travel along each of the 5 axes shown hitting the lens at the same time ...
A) Do they arrive at the focal point at the same time?
B) How does the slowing down in the medium cause the photons to change direction both downwards and upwards yet allows the photon on the principal axis to continue unaffected?
P.S. I can see the light rays are refracted but can't seem to link this to speed
In advance, thank you for explaining this
| Ray optics is a simplification tool useful for making calculations but not very useful for understanding light behavior. If you want to understand why light bends when it changes medium, you need to consider light as a wave, not as a particle.
The Huygens–Fresnel principle will explain the rest.
| {
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Why does a weather vane arrow point in the direction of the wind? It seems that a weather vane will rotate in order to minimize energy and thus orient itself parallel to the wind.
What I do not understand is why it is implied that the weather vane arrow should point in the direction of the wind.
I do not understand why the arrow pointing in the opposite direction of the wind is also not a minimum energy solution.
| The vane has to be designed so that it has a preference to point in the right direction. In the example that you included, this is implemented by the flag at the back providing a broader cross section than the arrow head and also by the rooster standing slightly to the back half of the arrow.
You are correct that if the vane became perfectly anti-aligned to the wind, it might stay there for a bit. That solution, however, is an unstable equilibrium solution. If the wind shifts even a little, assuming the vane is well-designed, it should snap around to the proper direction for the reasons above.
| {
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Are the muon/tau neutrinos produced in the Sun? If not, then where? I was reading about Solar Neutrinos, and apparently they are all Electron Neutrinos.
However, there are two other types of neutrinos, the Muon and Tau Neutrinos. Does the Sun produce them? If not, are they produced naturally anywhere in the universe?
| The neutrino creating reactions (fusions, decays, and fissions) that take place in the core of the sun are all low enough in energy that they can only produce electron neutrinos.
However, neutrinos mix (because they propogate in mass states which do not correspond to the flavor states in which they are produced and detected), and because the energy of the neutrinos is low (a few MeV or less) the length scale for that mixing is much less than a the solar radius.
The result is that the neutrino flux emerging from the sun is fully mixed and any experimental sampling of the flvors (say by the SNO experiment) will report that only about one third of them are electron flavored.
| {
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What happens to matter when it is converted into energy? According to Einstein’s equation
$$E=mc^2$$
Matter can be converted into Energy. An example of this is a nuclear reaction. What happens to the matter in the process? Do the atoms/subatomic particles just vanish? Any insights into this process are appreciated.
| Not to take away from other more detailed answers, I think there's something that might be more to your point.
Matter is not converted into energy.
Mass is converted into energy.
This can be observed in, for example, nuclear fusion of hidrogen, resulting in a helium nucleus. With two protons (of, let's say, mass=1) you'd expect the resulting nucleus to have mass=2.
But it doesn't. The resulting mass is smaller than 2.
That difference in mass is the energy that was released in the process.
Incidentally, you can get energy out of fusion up until you make an iron nucleus. Anything heavier than that will actually absorb energy (which is then transformed into mass, following the same principle). The consequence is that you can get energy out of fission of those heavier elements (which is how current nuclear power plants work).
| {
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Modified quantum harmonic oscillator: hamiltonians unitarily equivalent and energy spectrum Here is my Hamiltonian:
$H_{\alpha, \beta} (q,p) = \frac{p^2}{2m} + \frac{1}{2}m \omega ^2q^2 + \alpha q + \frac{\beta}{2}(pq^2 + q^2p) + m \frac{\beta^2}{2}q^4$.
How can I prove that $H_{\alpha, \beta}$ and $H_{\alpha, \beta‘}$ are unitarily equivalent and how I can compute the energy spectrum of the theory?
It seems that the energy levels can be computed only perturbatively, because of the presence of term like $q$ or $q^4$, but in the text of the problem there no reference to perturbation theory.
| The hamiltonian can be put in the form $H_{\alpha \beta} = \frac{1}{2m}(p + \beta m q^2)^2 + \frac{m \omega^2}{2}\left(q + \frac{\alpha}{m \omega^2}\right)^2 - \frac{\alpha^2}{m^2 \omega^4}$. Now define $P = p + \beta m q^2$ and $Q = q + \frac{\alpha}{m \omega^2}$. Since $[p,q] = -i \hbar$, it is easy to prove that also $[P,Q] = -i \hbar$, thus $P$ and $Q$ are canonically conjugate as well. In terms of these new variables the hamiltonian is $H_{\alpha \beta} = \frac{1}{2m}P^2 + \frac{m \omega^2}{2}Q^2 - \frac{\alpha^2}{m^2 \omega^4}$, which is a standard harmonic oscillator. This means that two hamiltonians with $\beta \neq \beta'$ are unitarily equivalent in the sense that they display the same spectrum (since they can always be rewritten as a harmonic oscillator). This also means that the spectrum of the theory is $E_n = \hbar \omega (n + 1/2) - \frac{\alpha^2}{m^2 \omega^4}$.
I hope I did not mess with the completion of squares, but in that case I hope the argument still holds.
| {
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Reducing the Aerodynamic loss of a Flywheel in Air I have a hypothetical flywheel spinning at 3000rpm. The steel flywheel is around 800mm in diameter with a "Smooth" finish. I was wondering whether it would be beneficial to implement golf ball dimples on the flywheel surface to reduce the laminar drag. There are many similar questions around golf ball dimples on aeroplane wings but no questions around an object which is statically rotating and therefore has no dominant drag. The surface speed is 125m/s
| Ideally, the flywheel would run in a vacuum.
In air, its surface should be as smooth as possible. With aircraft surfaces you have a stagnating point where air "hits" the aircraft straight on and then a flow path with a clear end at the trailing edge. The length of this flow path is used for calculating the Reynolds number, a similarity number which characterizes the ratio of inertial to viscous forces. Golf ball dimples only make sense at low Reynolds numbers (which is 100,000 or less) when a laminar boundary layer needs to be tripped to a turbulent one in order to delay flow separation caused by the pressure rise on the back side of the golf ball.
For the spinning wheel no such flow path length exists, nor does a steep pressure rise. Surface friction will accelerate the air in the boundary layer of the flywheel on a circular path such that centrifugal forces will create a flow towards the rim of the flywheel. In the end, a circulation shaped like a flattened donut will develop on both sides of the flywheel where air close to the surface is being carried away towards the rim and air outside of this boundary layer is flowing back towards the hub of the flywheel. This whole circulation system will also spin with the flywheel, and it is very unlikely that laminar flow can be maintained in that boundary layer. Also, there is no need for a forced transition because no steep pressure rise has to be mastered. Any surface imperfection will only increase the size of the boundary layer and, consequently, the viscous losses from it.
| {
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Is the capacitance of all types of capacitors increased by a factor of dielectric constant like parallel plate capacitors? I've learnt that the capacitance $C$ of a parallel plate capacitor is given by:
$$C=\frac{A\epsilon_0}{d}$$
where, $A$ is the area of cross section of the plates and $d$ is the separation between the two plates. When the space in between the two plates is filled by a dielectric of dielectric constant $K$ the new capacitance is given by:
$$C'=KC=K\frac{A\epsilon_0}{d}$$
When the space between the plates is filled by a dielectric of dielectric constant $K$, the capacitance is increased by a factor of $K$. Is this applicable for all types of capacitors (spherical, cylindrical, etc.)? If yes, what is the reason behind this fact? For example, the capacitance of a spherical capacitor is given by:
$$C=\frac{4\pi\epsilon_0 r_1 r_2}{r_2-r_1}$$
where, $r_1$ and $r_2$ are the radii of inner and outer metallic shells respectively. If we fill the entire region between the capacitor with a dielectric of dielectric constant $K$ will the resultant capacitance be given by:
$$C'=KC=K\frac{4\pi\epsilon_0 r_1 r_2}{r_2-r_1}\ \ ?$$
|
. . . . the capacitance of a spherical capacitor is given by $C=\dfrac{4\pi\epsilon_0 r_1 r_2}{r_2-r_1} . . . .$
If $r_2 - r_1 = d$ and $d\ll r_1$ and $r_2$ then $4\pi r_1r_2$ is approximately equal to the surface area of a sphere $A$.
The capacitance of a spherical capacitor with these characteristics can now be written as $C=\dfrac{A\epsilon_0}{d}$ which is the capacitance of a parallel plate capacitor which can be thought of as a small part of a spherical capacitor.
Introducing a dielectric between the plates of a capacitor results in the dielectric being polarised ie the atoms being distorted so electric dipoles are set up.
These dipoles produce electric field which are in the opposite direction to the electric field which produced them and so the overall electric field is reduced which in turn results in a increase in the capacitance of a capacitor.
This effect does not depend on the type of capacitor which is being considered and so your factor $K$.
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Effective mass as a consequence of energy band structure The Wikipedia article on effective mass defines it as follows:
In solid state physics, a particle's effective mass (often denoted ${\textstyle m^{*}}$) is the mass that it seems to have when responding to forces, or the mass that it seems to have when interacting with other identical particles in a thermal distribution.
In semiconductor physics, due to the presence of, and, therefore, interactions with, the atoms of a semiconductor, electrons in a semiconductor are unable to move as freely as they can in a vacuum. To account for this decreased mobility, we say that the electron has an effective mass.
I have then seen it said that effective mass is a consequence of energy band structure: (1) It is determined by the curvature of the energy band, (2) it depends on the material, (3) it depends on the band. I also found the following related diagram:
(http://www.ioffe.ru/SVA/NSM/Semicond/Si/bandstr.html)
What I'm struggling with is interpreting/understanding this graph in terms of the description of effective mass as a consequence of energy band structure. I would greatly appreciate it if people would please take the time to explain this graph in terms of the description that I gave of effective mass as a consequence of energy band structure.
| So what is illustrated in this figure is the energy band structure, these represent the dispersion, i.e. the energy $E$ in function of the wave vector $k$. Now, you can find in standard textbooks on solid state physics that the effective mass is defined as $$(m^*)^{-1} =\frac{1}{\hbar^2}\frac{\partial^2 E}{\partial k^2}$$
So the curvature of these (approximate) parabola in the band structure determine the effective mass. Also note that the concave parabola will have a corresponding negative effective mass, these correspond to holes. Furthermore, the curvature of the parabola is inversely proportional to the effective mass, hence, these strongly curved parabola will correspond to a low effective mass, that's why they call it 'light hole', and the other way around.
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How does water vapour replace air molecules? I know that density of moist air is less than density of dry air becuase water molecules replace air molecules, and hence as the average molecular mass of water is less than that of air, the density decreases.
Now my doubt is why do the water molecules replace air molecules, why don't they just get mixed up with air molecules without replacing the already existing ones?
I assume the answer might be due to atmospheric pressure( to maintain it almost constant). But I am not able to find a logic to it.
Please explain. Also suggest if I have to add any extra tags relating to the topic.
| The volume of the Earth’s atmosphere isn’t fixed. It can vary slightly for a number of different reasons, including a slight increase due to water evaporation and So locally, dry air molecules can move out of the way to make room for gaseous water molecules during vaporization and evaporation.
Hope this helps
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Origin of terms in the Nernst-Planck equation We know the Nernst Planck equation is
$$ \frac{\partial c}{\partial t} = - \nabla \cdot J \quad | \quad J = -\left[ D \nabla c - u c + \frac{Dze}{k_\mathrm{B} T}c\left(\nabla \phi+\frac{\partial \mathbf A}{\partial t}\right) \right] $$ $$\iff\frac{\partial c}{\partial t} = \nabla \cdot \left[ D \nabla c - u c + \frac{Dze}{k_\mathrm{B} T}c\left(\nabla \phi+\frac{\partial \mathbf A}{\partial t}\right) \right] $$
We know this is basically a continuity equation where a partial of a conserved density quantity is equal to the divergence of its flux. The first term is derived from Ficks Law, the second term is simply a convection term flux=velocity*density and the final term is due to Einstein's Relation stating
$$ D = \mu \, k_\text{B} T$$
Where
μ is the "mobility", or the ratio of the particle's terminal drift velocity to an applied force, μ = vd/F and we know F=EQ and V=ED so the third term is derived from these equations.
I am having trouble with the modified Nernst Planck equation for porous media where we have
$$ \nabla (\kappa_{eff}\:\ln C)=\mbox{Current Density}$$
where $$ \kappa_{eff}=\frac{2RT\kappa}{F}\left(t_+-1\right)\left(1+\frac{d \ln f}{d \ln C}\right)=D\left(1+\frac{d \ln f}{d \ln C}\right)$$
Where does this term come from? Logarithm of the activity would imply a chemical potential but how is this related to diffusion?
| The Nernst-Planck equation assumes an ideal thermodynamic behavior of ions, i.e., unit activity coefficients. Additionally, it uses self diffusivities, $D$, to describe the response of ions to an applied electric field.
This interpretation is at best valid for dilute electrolytes.
A generalized treatment of ion motion employs Stefan-Maxwell diffusivities, $\mathfrak{D}_{ij}$, and non-ideal energies of the ions, i.e., $f \left(c\right) \neq 1$. This leads to concentrated solution theory (https://onlinelibrary.wiley.com/doi/abs/10.1002/bbpc.19650690712). In this description, the total ionic current density is defined as
\begin{equation}
i = - \kappa \left( \nabla \phi - \frac{2RT}{F} \left(1-t_+^0\right) \left(1 + \frac{\mathrm{d} \ln f}{\mathrm{d} \ln c}\right) \nabla \ln c \right)
\end{equation}.
Note that this equation is only valid for an electrolyte with a monovalent cation, monovalent anion, and a solvent. One can derive equivalent expressions for other electrolytes.
Starting from the constitutive relations of the concentration solution theory, one can derive conservation equations in porous electrodes as described in https://iopscience.iop.org/article/10.1149/1.2221597/meta.
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Why is work defined as $W=Fd$? I am trying to understand what work really means in physics. I seem to be missing the conceptual link. Every resource says that $W=Fd$ but that does not make sense to me.
If, say, an elastic object suspended in space where there is no drag or resisting force of any kind is pushed by a force of a certain magnitude, then it will accelerate. The amount of 'useful' energy spent would completely go into accelerating this body of a particular mass for as long as the force is applied.
First of all, why isn't work $W = mat$ (which is the equation of momentum) for some time $t$.
Why is work $W = mas$ (for a force acting in the same direction of the motion) for some displacement $s$.
Since momentum and energy are both conserved, could it have been that it was a matter of convention how these two quantities were defined?
(I.e why wasn't work defined as $W=mat$)?
|
I am trying to understand what work really means in physics. I seem to
be missing the conceptual link.
The short answer is work in physics is a means of energy transfer between objects that is the result of the action of a net force on an object through a distance. A differential amount of work is $dW=F.ds$. (The other means of energy transfer is heat, which results from a temperature difference between objects). The units of work (and heat) are energy units (N.m; J). Your equation $W=mat$ is for momentum, with units (N.s; kg.m/s$^2$).
An important distinction between momentum and work is that momentum is a property of a system that is conserved, whereas work is not a property of a system. The energy that the work transfers is a property of the systems undergoing the transfer, and is overall conserved.
Hope this helps.
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Confused about what a wave is When a wave of something, let's say light or some electromagnetic wave is given, I am confused because I do not understand if shape of a wave represents projectile of it or some value that possess at certain positions. I researched a lot but I have no idea what a wave really is.
My question is: what is a wave?; what defines a wave?
| The “shape” of the wave is exactly what a wave is. It is a “shape” formed in a medium in which it moves while maintaining the shape (ideally). The shape formed is a property of the medium and of the object that caused the shape to form in the medium.
For example think of the medium as water in a pond and the object as a stone dropped in it. This causes the water around it to ripple and form a shape that moves radially outwards. This is what a wave is. At least one kind of wave. And since it is only dependent on the medium and the source, we can describe the entire thing in terms of a relation called the wave equation.
As for light, the medium turns out to be something called as the Electromagnetic Field and the source is movement of charged particle.
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Constructing W-algebras I am following the algorithm in W-algebras with two and three generators, in order to construct consistent (anti-)commutator relations for a particular W-algebra.
I am considering $W(2,4,4)$ where both dimension four operators are fermionic. I have two questions related to the method introduced in the paper, namely:
*
*They use Jacobi identities (e.g. $\{\Psi_m,\{\Psi_n,\Psi_\ell\}\} + \mathrm{permutations} = 0$) to fix some of the arbitrary constants, but this isn't sufficient to constrain all of them. How do they calculate the rest? (There is some step in the example involving computing determinants but it isn't clear how this relates to the notation of the previous section.)
*For fermionic operators, when computing $\{\Psi_m,\{\Psi_n,\Psi_\ell\}\}$, due to the terms that appear in the inner anti-commutator, you encounter stuff like $\{\Psi_m, L_{n+\ell}\}$ but normally, we use a commutator in this case, as one element is even and one is odd. So when computing the Jacobi identity with fermionic operators, should one actually use $[,\}$ as opposed to $\{,\}$?
| As it turns out, even though the authors of the paper I linked claim there are benefits to working with the commutators rather than the OPEs, I found using the OPEs much simpler, and didn't need the machinery provided in the algorithm in the paper.
In my case, the two fermionic generators carry an $\mathfrak{sl}(2)$ charge which we require the OPE to respect, and so I simply wrote down all possible terms that may arise order by order, i.e.
$$\Psi^+(z) \Psi^{-}(z) = \frac{a_1}{z^8}\mathbb{I} + \frac{a_2}{z^6}T + \frac{a_3}{z^5}\partial T + \frac{a_4}{z^4}\partial^2 T + \frac{a_5}{z^4}T^2 + \dots$$
using combinations of derivatives and $T$ with arbitrary constants $a_i$. Imposing the Jacobi results in some trivial constraints already satisfied (e.g. taking $TTT$) but others (e.g. $\Psi^+ \Psi^{-}T$) led to constraints on the $a_i$ up to null states.
A useful paper for carrying this out is An Algorithmic Approach to Operator Product Expansions, W-Algebras and W-Strings.
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Irreversible, Compression and expansion work for a piston with ideal gas How can we calculate the work done by the piston or on the piston by including the internal pressure of the piston and not the external one. Because if the external pressure is doing work, the internal pressure might be too in the opposite direction, just like friction does. And I am specifically talking about "Irreversible change"
| It is important to be aware that, in an irreversible expansion or compression of an ideal gas, the gas does not satisfy the ideal gas law. The ideal gas law applies only to a gas in thermodynamic equilibrium, or, in the case of a reversible process, a gas passing through a continuous sequence of thermodynamic equilibrium states. In an irreversible process, the force $F_g$ exerted by the gas on the piston also includes a contribution from viscous damping stresses that always act in the direction opposite to the piston movement (and depend on the rate at which the piston is moving).
If we do a force balance on the piston (assumed frictionless) during an irreversible expansion or compression, we obtain (using Newton's 2nd law):
$$F_g-P_{ext}A=m\frac{dv}{dt}$$where A is the area of the piston, m is its mass and $P_{ext}$ is the external force per unit area. Note from this equation that, if the piston is massless, then the force exerted by the gas on the piston must exactly match the external force applied to the piston throughout the process (even when there are viscous damping stresses that contribute to the gas force). If we multiply the above force balance equation by the piston velocity, we obtain: $$F_gv=F_g\frac{dx}{dt}=P_{ext}\frac{dV}{dt}+mv\frac{dv}{dt}$$where dV/dt is the rate of change of gas volume: $\frac{dV}{dt}=Av$. If we integrate this equation with respect to time, we obtain the work done by the gas on its surroundings (the piston): $$W_{g}=\int{F_gdx}=\int{P_{ext}dV}+\frac{1}{2}mv^2$$
Eventually, the piston will come to rest (as a result of viscous damping stresses within the gas) and the system will attain thermodynamic equilibrium. At this point, irrespective of the mass of the piston, we will have simply $$W_g=\int{P_{ext}dV}$$Irrespective of whether the process is reversible or irreversible, this same equation will always apply. It is just that, in a reversible process, the external pressure must also always match the pressure P=nRT/V calculated from the ideal gas law.
| {
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Why are lasers more useful for cutting and similar tasks than non-coherent light? I'm working a lot with lasers the last few years, and trying to understand the basics. So some of what I say next may be wrong or partially correct - if I find that out, that will be valuable.
As I understand it, a laser is a light source (these days it might well be a diode) where the frequency and phase of the light is coherent. If you use suitable optics, you can then make a parallel beam which can then be focused onto a spot. If the spot is small enough and the power high enough, that beam can then be used for applications such as cutting, basically because it heats the surface it falls on to very high temperatures.
My question is this : couldn't you do the same thing with (for instance) a high power LED? You can certainly have a fairly narrow band of frequencies from a non-laser, if that makes optics design easier. So why, then, do we favour lasers for this kind of application? Why do we prefer the light to also be phase-coherent?
| The acronym LASER is short for "light amplification by stimulated emission of radiation". The stimulation part is important, because this way we are able to obtain high energy densities. In contrast to a light bulb or a LED, the laser beam is also rather directional. This enables us to focus the beam onto "small" spot sizes. Thereby the energy density increases further.
As Jon Custer already pointed out, the coherence is not really that important. A $CO_2$ laser is not particular coherent. However, it efficiently uses the available power and transfers it into the laser modes -- all high power lasers must by multi-mode, because the population inversion factor can't be larger than $1/2$.
| {
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Using the uncertainty principle to estimate energies in ground states Suppose, for example, that we want to find the minimum energy of a particle undergoing simple harmonic motion. In classical mechanics, the energy is:
$$E = \frac{p^2_x}{2m} + \frac{1}{2} m \omega_0^2x^2$$
where $m$ is the mass of the particle and $x$ the position. I was taught that one can use Heisenberg's uncertainty principle to find an estimate (which actually turns out to be an exact answer) for the ground state energy of this system. Since the potential energy well is symmetric, one can deduce that the expectation value of $x$ and $p_x$ will be zero. The uncertainty in the momentum will be given by:
$$\Delta p_x \ge \frac{\hbar}{2\Delta x}.$$
Now then I was told that the energy of the system must obey the following:
$$E \ge \frac{1}{2m} \left(\frac{\hbar}{2\Delta x}\right)^2 + \frac{1}{2}m\omega_0^2(\Delta x)^2.$$
I can see that this inequality is obtained plugging in the value $p_x = \frac{\hbar}{2 \Delta x}$ and $x=\Delta x$, but I really don't get how this is "mathematically" possible. The first equation contains on both sides values of energy, momentum and position but the new inequality contains on the LHS a value for the energy and on the RHS uncertainties, how is this possible? Surely the equation must be the following:
$$\Delta E \ge \frac{1}{2m} \left(\frac{\hbar}{2\Delta x}\right)^2 + \frac{1}{2}m\omega_0^2(\Delta x)^2$$
| I totally agree: This "derivation" is mathematically improper.
The energy of the system under consideration is given by
$E = \frac{p^2}{2m} + \frac{m\omega^2 x^2}{2}$. However, as you know this is not true, because the energy levels would not posses any uncertainty. Thus, we should interpret this relationship by assuming that $p$ and $x$ are random variables. For example, we could assume that both are independent and normally distributed with standard deviation $\sigma_p$ and $\sigma_x$. However, as you know this is also not true, because the two uncertainties are correlated.
So, in order to obtain a proper mathematical derivation, we could use a Taylor expansion. By expressing the energy as a function of two variables, $E = E(p, x) = f(q_1, q_2)$, the uncertainty (= standard deviation, $\sigma_E$) becomes (here we use the variance instead of the standard deviation)
\begin{align}
\sigma^2_E &\approx
\sum_{i=1}^{k} \sum_{j=1}^k
\frac{\partial f}{\partial q_i}
\frac{\partial f}{\partial q_j}
\mathrm{cov}[q_i, q_j]
\\
%%%
&=\sum_{i=1}^k
\left(%
\frac{\partial f}{\partial q_i}
\right)^2
\sigma^2_{q_i}
+ 2 \sum_{i=1}^{k-1} \sum_{j=i+1}^k
\frac{\partial f}{\partial q_i}
\frac{\partial f}{\partial q_j}
\mathrm{cov}[q_i, q_j] %+ \mathcal{O}(\sigma^4)
\end{align}
where "cov" is the covariance of the two random variables. However, this get's messy.
So, if we like to approximate the uncertainty in energy the simple phenomenological model
$$
E + \Delta E \approx \frac{(p + \Delta p)^2}{2m} + \frac{m\omega^2 (x + \Delta x)^2}{2}
$$
seams to be reasonable.
Now, if we evaluate this relationship for the point $(p, x) = (0,0)$ in phase-space, we obtain $E=0$ and
$$
\Delta E \approx \frac{(\Delta p)^2}{2m} + \frac{m\omega^2 (\Delta x)^2}{2}
$$
The argument used in your notes reads "the uncertainty in position and momentum provide typical values for position and momentum". Thus, your professor did not try to estimate the uncertainty $\Delta E$, but they estimated the averaged energy $\left\langle E \right\rangle$. In order to do so, they used "typical values". In this case the LHS reads $\left\langle E \right\rangle$.
There are several ways to calculate this average energy. Your professor choose to plug in "typical values". Thus, he/she just replaced $p \to \Delta p$ and $x \to \Delta x$. This is valid, because if the expectation value of a random variable is zero, $\left\langle Q \right\rangle = 0$, then it's variance is equal to it's expectation value squared, $ (\Delta Q)^2 = \sigma_{Q}^2 = \left\langle Q^2 \right\rangle - \left\langle Q \right\rangle^2 = \left\langle Q^2 \right\rangle$. Hence,
\begin{align}
\left\langle E \right\rangle &= \left\langle \frac{p^2}{2m} \right\rangle + \left\langle \frac{m\omega^2x^2}{2} \right\rangle
\\
&= \frac{\left\langle p^2 \right\rangle}{2m} + \frac{m\omega^2 \left\langle x^2 \right\rangle}{2}
\\
&= \frac{\sigma_p^2}{2m} + \frac{m\omega^2 \; \sigma_x^2}{2}
\end{align}
| {
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Is the equation $g=GM_Em/R_E^2$ in Tipler incorrect? I was reading my textbook (Tipler et al.), and I am unsure of one of the expressions they used. On page 374, it says (near Figure 11-10) that $g = GM_Em/{R_E}^2$. Is this even dimensionally correct? I got a units of $m/s^2$ on the left hand side and Newtons on the left hand side. I don't think that they are in agreement. Do they mean to say $g = GM_E/{R_E}^2$?
| Yes it is incorrect. Most probably it is a typo.
It can either be
$$F_g =G \frac {M_E m}{R_E^2}$$
Here they might have misprinted $F_g$ as $g$.
Or
$$g = G \frac {M_E}{R_E^2}$$
Here they probably mistyped the $m$ over there.
| {
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Gauge ghosts & unphysical states in gauge theory I have a general question about a statement from Wikipedia about ghost states as occuring in gauge theory:
"In the terminology of quantum field theory, a ghost, ghost field, or gauge ghost is an unphysical state in a gauge theory."
I learned gauge theory up to now with mainly mathematical beckground. My main reference is this https://arxiv.org/abs/1607.03089 paper by A. Marsh.
Question: What is concretely an "unphysical" state or field from viewpoint of gauge theory?
| TL;DR: Ghost states do not belong to the physical Hilbert space.
This begs OP's next question: What exactly is the physical Hilbert space in a gauge theory?
If we transcribe the gauge symmetry into a BRST symmetry, we can give a precise technical definition: The physical Hilbert space is given by the BRST-cohomology of zero ghost number.
References:
*
*M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994.
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Entropy change in the free expansion of a gas Consider the adiabatic free expansion of a gas since there is no external Pressure hence Work done on the system is 0 and since the walls are insulated (hence adiabatic) the heat absorbed is 0. However since this is a irreversible process then entropy change > 0 hence dQ > 0 . However there is no heat absorption. What am I missing ?
| The equality $dS=dQ/T$ is only valid for reversible processes, so it does not apply in a free expansion.
| {
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Understanding Electrostatic Work $$W=-\int_\infty^\textbf{r}\textbf{F}\cdot\textbf{dl}
=-Q\int_\infty^\textbf{r}\textbf{E}\cdot\textbf{dl}
= Q(V(\textbf{r})-V(\infty))
=QV(\textbf{r})$$
I'm trying to understand how this definition of work turns into a positive
$$W=QV(\textbf{r}).$$
The integration of $-Q\int_\infty^r \textbf {E} \cdot \textbf{dl}$ would just flip the bounds of integration and then the equation would become $W=Q(V(\infty)-V(\textbf{r}))$ which would simplify to $W=-QV(\textbf{r})$? Isn't this correct. Why does Griffiths have it as a positive? What did I miss?
| Keep in mind that electric field is negative gradient of potential.
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Two-Dimensional Lorentz Velocity Transformation Problem
Two spaceships A and B are approaching along perpendicular directions, as seen from earth.
If A is observed by a stationary Earth observer to have velocity $_$ = -0.90c and B to have
velocity $_$ = +0.90c, determine the speed of ship A as measured by the pilot of ship B.
To solve the problem, I split it into two parts: first part for the $x$ component the velocity of A, and second part for the $y$ component. Let $v$ and $v'$ be the velocities of ship A in frame S and S' respectively. Similarly, let $w$ and $w'$ be the velocities for ship B. We set frame S to be attached to Earth while S' is attached to spaceship B. From Lorentz' velocity transformation formula, $$v_x'=\dfrac{v_x-w_x}{1-\dfrac{v_xw_x}{c^2}}=-w_x=-0.9c$$ since $v_x=0$. Similarly for $v_y$, $$v_y'=\dfrac{v_y-w_y}{1-\dfrac{v_yw_y}{c^2}}=v_y=-0.9c$$ since $w_y=0$. Then I used Pythagoras' Theorem to obtain the total velocity of ship A, and...well here's the result: $v=\sqrt{v_x^2+v_y^2}\approx 1.273c$
At this point, I think my problem clear. As we know, nothing can exceed the speed of light, so it is not possible for ship A to be travelling at a speed higher than $c$. What did I do wrong in my steps? Every example I found online was one-dimensional. I looked for answers on this website, and I found mixed answers more confusing than the problem itself. I'm suspecting that my mistake lies in the assumption that Pythagora's Theorem holds for relativistic velocities. Help, please?
| As mentioned by @G.Smith, your second equation is not correct. According to the velocity addition formula, we have:
$$u'_x=\frac{u_x-v}{1-\frac{vu_x}{c^2}}=\frac{0-v}{1-\frac{0}{c^2}}=-v=-0.9c$$
Remember that the velocity of $B$ shall be considered as $v$ in the above link. $(u_x=v=0.9c)$ Therefore, we can write:
$$u'_y=\frac{u_y\sqrt{1-\frac{v^2}{c^2}}}{1-\frac{vu_x}{c^2}}=u_y\sqrt{1-\frac{v^2}{c^2}}=-0.9c×\sqrt{1-0.9^2}=-0.392c$$
And, finally:
$$u'=\sqrt{u^{\prime 2}_x+u^{\prime 2}_y}=0.982c$$
| {
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If a ball is thrown to a person standing on a frictionless surface, is the impulse of the thrower equal to that of the catcher? If a person throws a ball, exerting a given impulse does the person that catches the ball receive the same impulse assuming that the catcher moves. Is the impulse that the catcher receives less than the impulse that the thrower receives because the ball continues to move with the catcher or does the catcher receive all of the impulse, to begin with and then return momentum to the ball as they pull it along in their hand?
| Let both the men have a mass of M and the ball have a mass of m. Now, let's say the first man throws the ball at an angle $\theta$ and velocity v towards the second man.
Now, just after he has thrown the ball we will see this:
The velocity of the first man will be $\frac{mv \cos{\theta}}{M}$ in the opposite direction of the ball's motion by conservation of momentum in the horizontal direction, as no forces act in that direction.
Now, just before the ball reaches the other man this will be the situation:
Since the man and the ball move together after he catches the ball, we can treat it like an inelastic collision, giving us a velocity of the combined ball-man system as $\frac{mv\cos{\theta}}{M+m}$,
Here, we can clearly see that the final velocity of the first man$\left(\frac{mv \cos{\theta}}{M}\right)$ is more than that of the second man-ball system$\left(\frac{mv\cos{\theta}}{M+m}\right)$ and their masses are equal. So, we can say that the impulse given to the second man is lesser than that given to the first man.
Hope this helps!
P.S.: Sorry for the bad drawings
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Left and right Weyl representations are inequivalent representations We introduce the two-component spinors in the following representations:
$$
\psi_\alpha \rightarrow\psi'_\alpha=\mathcal{M}_\alpha^\beta\psi_\beta$$
$$
\bar\psi_\dot{\alpha}\rightarrow\bar\psi_\dot{\alpha}'=\mathcal{M^*}_\dot{\alpha}^\dot{\beta}\bar{\psi}_\dot{\beta}
$$
Where the matrix $\mathcal{M}\in SL(2,\mathbb{C})$
These two representations are not equivalent because there is no matrix for which $\mathcal{M^*}=\mathcal{CMC^{-1}}$. I don't quite see how that is true. Since the entries in $\mathcal{M^*}$ are just the complex conjugates of $\mathcal{M}$, isn't it trivial to find a matrix $\mathcal{C}$ that satisfies that condition?
In other words, how can we prove that there is no such matrix $\mathcal{C}$?
| This is the claim that left-handed Weyl spinors and right-handed Weyl spinors are not the same.
*
*An abstract way to see that there cannot be such a matrix is to first show that the two representations correspond to $(1/2,0)$ and $(0,1/2)$ in the usual parlance of expressing representations of the complexification $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}\cong \mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$ by representations of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, classified by pairs of half-integers $(j_1,j_2)$. Since the $j_i$ are values of Casimirs but an equivalence of representations must preserve Casimirs, the two representations cannot be equivalent.
*(This is cribbed from a deleted answer by Chiral Anomaly here) A concrete way to see this as follows:
$$ M^\ast = CMC^{-1}$$
implies that $\mathrm{tr}(M^\ast) = \mathrm{tr}(M)^\ast = \mathrm{tr}(M)$ by cyclicity of the trace, and so $\mathrm{tr}(M)\in\mathbb{R}$. But $M=\mathrm{diag}(z,z^{-1})\in\mathrm{SL}(2,\mathbb{C})$ for arbitrary $z\in\mathbb{C}$, and $z + z^{-1}$ is clearly not real for such arbitrary $z$.
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Given fluids expand non-linearly, how were physicists able to make a linear temperature scale? Materials expand with increase in temperature. As far as I know this property for fluids was put in use to make initial thermometers. We know that expansion of fluid is given by the following formula (at a given pressure) (Wikipedia)
$$\Delta V = \alpha_V (T) V \Delta T$$
Now from this equation we get that the expansion of fluid is not linear. Therefore for a unit change in temperature (i.e., $\Delta T = 1\ \text {unit}$) at different volumes and temperature the expansion of fluid is different. This means the thermometer so made is not a linear one (in which equal spacing represent equal change in temperature).
So
*
*How do we know now that our temperature scale now is a linear one?
If possible do tell how this problem was overcome by physicists?
Note that my problem is not that we cannot divide the scale on my own equal sections but rather if the fluid isn't expanding linearly then how would we be able to get the correct measurement.
| Physicists were using gas-based thermometers, where linear range is looking like -150 to 2000 Celsius. They were quite lucky that gases are not very easy to liquify in XVII century.
| {
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What does the arbitrary constant in cosine equation of displacement in S.H.M say? The phase and phase constant in a displacement time equation show from where the particle has started.
In my school textbook, first the displacement equation was given as :-
$$x= A\sin(\omega t+\phi)$$
where $\phi$ is the phase constant.
But then it said if the particle is at extreme position then we add $\pi/2$ because obviously displacement is maximum at $\pi/2$
So now the equation at extreme should be :-
$$x=A\sin\left(\omega t+\frac{\pi}{2}\right)$$
$$x=A\cos(\omega t)$$
But in my textbook the equation is :-
$$x=A\cos(\omega t + \phi')$$
It says that $\phi '$ is another arbitrary constant. But technically $\phi$ is $\sin ^{-1} (x/A)$, here $x$ will be $A$ and we get $\pi/2$ so no constant remains. But what is this $\phi '$ constant and on which thing it depends?
| At $t=0$ the displacement $x$ is not necessarily $0$:
$$
x(0)=A\sin(\phi) \tag{1}
$$
In addition, the velocity at $t=0$
$$
\dot{x}(0)=\omega A \cos(\phi) \tag{2}
$$
(1) and (2) are two equations for your 2 unknowns $A$ and $\phi$. Thus, as you alluded to
$$
\frac{x(0)}{\dot x(0)}= \omega\tan(\phi)
$$
from which you can determine $\phi$, and plug it back into either (1) or (2) to obtain $A$ if you need the amplitude.
| {
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On the "spectrum" of an operator in quantum mechanics Very simple question, I'm new to this. I'm reading Griffiths book on QM and have a question about the "spectrum" of an observable operator. Does the spectrum of an operator require specification of a particular system? Or is the spectrum of an operator just every possible eigenvalue that can be obtained by every possible eigenfunction of an operator?
| Each operator has it's own spectrum. For the hamiltonian the spectrum is the set of allowed energy levels. This depends on masses and potentials.
| {
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Commutator of $B$, $C$ vanishes if $A$, $B$, $C$, $AB$, $AC$ are Hermitian
Suppose 3 operators $A$, $B$, $C$ are Hermitian operators. Assume $A$ has a non-degenerate spectrum, and $AB$ and $AC$ are also Hermitian. Show that $$[B,C] = 0$$
From the conditions $A$, $B$, $C$, $AB$ and $AC$ are Hermitian operators, one can derive that $$[A,B]=[A,C]=0$$
How can one proceed to show that $[B,C]=0$?
| You are nearly there. If $A$ commutes with $B$ it means that they can be diagonalized simultaneously. Now use the fact the the eigenvalues of $A$ are non-degenerate. This means that also $B$ is diagonal in the same basis. Repeat with $C$ and you are done.
| {
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$V$-$I$ characteristic of a solar cell please explain the VI characteristics of a solar cell. The characteristics is given in my book without any explanation. How can the Voltage decrease on increasing current shouldn't it be opposite.
Solar Cell I-V characteristics
(Image from Electrical 4 U - Characteristics of a Solar Cell and Parameters of a Solar Cell)
| A simple (but sufficient) model is to consider the sunlight as a stream of photons, each with their own energy depending on the colors in the sunlight. Each photon can give its energy to a single electron.
If you try to get out a small current from the solar cell, there are more than enough photons, but you can't change the energy of the photons. That limits the voltage.
If you try to get a high current out, you need a high voltage, but that gives a problem. Red photons only carry a limited amount of energy. Purple photons carry only a little bit more. So as you try to achieve a higher voltage, you quickly lose usable colors of photons. That's the sharp drop in current.
| {
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How do the unit vectors in spherical coordinates combine to result in a generic vector? I might be missing the obvious, but I can't figure out how the unit vectors in spherical coordinates combine to result in a generic vector. In cartesian coordinates, we would have for example $ \mathbf{r} = x \mathbf{\hat{i}} + y \mathbf{\hat{j}} + z \mathbf{\hat{k}}$. But in spherical coordinates, the position vector is actually a multiple of the unit vector $ \mathbf{\hat{e}_{r}} $, since $ \mathbf{r} = r \mathbf{\hat{e}_{r}} $ and not a linear combination of $\mathbf{\hat{e}_{\theta}} $, $ \mathbf{\hat{e}_{\phi}}$ and $ \mathbf{\hat{e}_{r}}$ (attached picture). Do we actually combine all 3 unit vectors in spherical coordinates to obtain a certain vector, or are $\mathbf{\hat{e}_{\theta}} $ and $ \mathbf{\hat{e}_{\phi}}$ just an indication of direction?
| I think the problem comes from confusing the radial unit vector for spherical coordinates and a trajectory vector. They are different, to build the second one you can use the $\hat{r}$ only if the path is radial, but it can have another components if the trayectory is not.
| {
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How does the current remain the same in a circuit? I understand when we say current, we mean charge (protons/electrons) passing past a point per second. And the charges have energy due to the e.m.f. of the power supply.
Now tell me, if a lamp has resistance and you hook it in the circuit, how will the current stay the same? The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?
[Edit]
All answers explained a bit of everything, so it was hard to choose one. If YOU are looking for an answer, please check the others too, in case the accepted one doesn't answer your question.
| Ok , so I think that confusion lies in the concept of current. When we apply potential difference through a circuit having resistance, a single electron does not move from one end to the other. What actually happens is that it replaces electron next to it which in turn replaces the electron next to it and this happens till the last electron reaches the other terminal. The resistance from beginning itself shows its effect on the movement of charge and not when the charge passes through it. What u say is that a charge when passing through a resistance should lose energy and hence current shoould slow down is partially correct. Yes, resistance does slow down the movement of electron but that effect is observed in overall circuit and not just in the path where resistance is connected.
| {
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Feynman Propagator for fermionic field $S_F(x-x)=0$ When doing some calculations in my QED course, my tutor used that for
$$S_F(x-y)= - \int \frac{d^4p}{(2\pi)^4}\frac{i( \gamma_{\mu}p^{\mu}+m)}{p^2-m^2+i\epsilon}e^{-ip(x-y)}$$
the Feynman Propagator of a fermionic field,
$$S_F(x-x) =0.$$
He shortly mentioned some symmetry argument (odd function over a symmetric interval?) but I didn't really get what he meant. I would be very grateful for some explanation here!
| This is not true. See section 3 of this article where the full analytic expression is derived. For finite mass, we can write the $x^2\rightarrow 0^+$ fermionic propagator (the $x^2\rightarrow 0^-$ limit gives the same answer, but with Bessel instead of Hankel functions when $x^2\neq 0$):
$$
\begin{align}
S_F(x)&=(i\gamma^{\mu}\partial_{\mu}+m)G_F(x)\\
&=-\frac{1}{4\pi^2}\left(\frac{-2i\gamma^{\mu}x_\mu}{(x^2)^2}+\frac{m}{x^2}\right)
\end{align}$$
Notice that the above expression diverges (i.e. it does not approach zero) in both of the following related limits:
*
*$x^{\mu}\rightarrow 0 \,\,\,\implies x^2\rightarrow 0$
*$x^2\rightarrow 0$ (but not necessarily $x^{\mu}\rightarrow 0$)
| {
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To what extent is standard quantum mechanics actually non-relativistic? I often hear that while QFT is a relativistic theory, standard quantum mechanics is not. But fields aren't inherently relativistic, you can easily construct non-relativistic QFTs, a relativistic QFT is one with a Lorentz invariant Lagrangian.
In the same way, it doesn't seem to me that there is anything inherently non-relativistic about the standard description of quantum mechanics, and that the actual non-relativistic part is the Schrödinger equation.
As a matter of fact if we substitute the Schrödinger equation axiom with dynamics described purely by quantum channels, with the only requirement that an evolved quantum state is still a quantum state and that the theory is linear, we do recover some statements that resemble locality in the Lorentz sense, like the no communication theorem.
Are fields necessary to obtain a relativistic theory? To what extent is standard QM non-relativistic? If QM is inherently non-relativistic even without the Galilean dynamics given by the Schrödinger equation, why should one care about non-locality in Bell inequalities and "spooky action at a distance"? Those would be expected in a non-relativistic theory, just like the gravitation of the Sun on the Earth is spooky action at a distance in Newtonian gravity.
| The Klein Gordon equation , the Dirac equation were developed partially to give quantum mechanical equations that were also Lorenz invariant, to replace the Schrodinger one. Maxwell's equation when used for the photons as a quantized equation is of course Lorentz invariant.
QFT depends on the plane wave solutions of the above equations in its construction, that is why it is Lorentz invariant by construction.
| {
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Normalization of the action in Special Relativity The action for a massive point particle in Special Relativity is given as
$$A =-mc^2\int d\tau,$$
Where $\tau$ represents the proper time, and $m$ represents the (rest) mass. From what I could understand, the Action must not change with respect to the reference frame, and hence it can be written as $$something\int d\tau$$ but why must the something need to be proportional to mass?
| For a single particle, it does not matter what prefactor you use, the equations of motion and everything else stays the same. The factors only start to matter when you couple different systems to each other. For example, consider a charged particle in an electromagnetic field described by a vector potential $A_\mu$. The right action describing its movement is
$$ S = S_{\text{EM}}+\int d\tau \left[ -m - q A_\mu \frac{dX^\mu}{d\tau} \right] , $$ where $S_\text{EM}$ is the action for the electromagnetic field.
This only works out to give the correct dynamics if the proportionality factor $m$ here corresponds to the mass of the particle.
Similarly, when one tries to extract meaningful quantities like the energy-momentum tensor from the action, only one choice of the factor will give the correct result.
I personally like to think of it as the particle having an always present static contribution to its potential energy, coming from its rest-mass energy (hence the negative sign), but don't take that too seriously.
| {
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Definition of reducible representation A reducible representation of a group $g \rightarrow D(g)$ is one which leaves a subspace $U$ invariant, i.e. $D(g)|u\rangle \in U, \space \forall |u\rangle \in U$.A completely reducible representation is one that can be broken down into a direct sum of irreducible representations.
In Howard Georgi's book "Lie Algebras in Particle Physics", he defines irreducible representations in terms of projection operators (page 5 Equation 1.11) in terms of projection operators P that project onto the invariant subspace:
$$ PD(g)P = D(g)P$$ where, presumably
$$ P = \sum_{\alpha} |\alpha \rangle \langle\alpha|$$.
Furthermore, Georgi defines completely reducible representations to be those in which both $P$ and $1-P$ project on to an invariant(under the action of $D(g)$) subspace.
I'm struggling to see how Georgi's definitions are equivalent to the first.
| *
*Leaving a subspace invariant means that $D(g)u\in U$ for all $g\in G, u\in U$. Since $P_U v \in U$ for the projector $P_U$ onto $U$ and any $v\in V$, you have that $D(g)P_U v \in U$ for all $v\in V$. So applying $P_U$ again to $D(g)P_U v$ does nothing, since the latter is already in $U$, there is nothing to project away.
*$1 - P$ is the projector onto a subspace $U^\ast$ disjoint from $U$, and $V = U \oplus U^\ast$ (this is a standard fact about projectors).
| {
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The Calculation of work in different frames Say there is a man running on a road. Friction is applying a force hence he is accelerating, but the friction is static hence does no work. We know that the change in kinetic energy is due to internal work done by his muscles.
Now say that the same man is running on a wooden plank kept on a frictionless surface. As the man moves ahead the plank goes behind. At the point of contact, the friction applies force in the forward direction accelerating the man ahead, but since the plank itself is moving behind, the point of contact is going behind. Hence, work done from ground frame is negative. But if I change my frame to the plank then work is zero again. Thus work is frame dependent.
Is my conclusion correct? And I always think if the calculation for work is frame dependent than when we theorize physics we might be missing out on different types of work which may lead to concepts like dark energy. How to physicists think about this.
|
Thus work is frame dependent.
Is my conclusion correct?
Yes. Work is frame dependent. This can be easily seen by noting that distance is frame dependent and since $W=F\cdot d$ then work is also frame variant.
Note that the plank frame that you describe is non inertial. So energy is not trivial and is actually not even conserved in that frame. However, despite that complication your conclusion was correct.
if the calculation for work is frame dependent than when we theorize physics we might be missing out on different types of work which may lead to concepts like dark energy
No, there is no link between the two. Dark energy is a specific term in the current best model of the universe.
| {
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If magnetic field lines don't exist, what are these iron filings doing around a magnet? Obviously the iron filings can be seen aligning themselves along the virtual magnetic field lines produced by the permanent magnet, the virtual magnetic field line is made of electromagnetic field due to the alignment of electrons in the magnet but why the patterns, why lines? Do these lines have thickness? Are they due to interference pattern?
|
That does look like there are real lines, doesn't it?
It's because each iron filing becomes its own magnet which affects the others.
Notice how they're all crowded together close to the ends of the magnet, and then there's a region where they're thin, and then they get closer together farther away?
I think that's because close up, the magnetic field of the bar magnet is very strong and is the most important thing. But a little distance from there, the iron filings are each strong magnets and they repel each other to the sides while they strongly attract each other at their ends. If you could cut the bar magnet in half the long way without making enough heat to destroy its magnetic fields, the halves would spring apart because they repel each other. The metal magnet holds itself together too strongly to let that happen. But if you cut it in half the short way, the two halves would cling to each other.
Farther away the iron filing magnets are weaker and so they don't repel each other as much and the lines get closer.
| {
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Boyle's Law and hot air balloon A bit dumb question because it is really difficult to imagine it. :-
Pressure is force per area. Talking about gases, the pressure is said to be the force molecules exert on walls of let's say a balloon.
Usually in examples of Boyle's law, our teachers mention hot air balloon, that the size of balloon increases as pressure decreases.
But at as height increases pressure decreases because there are less molecules above us.
How will the pressure of molecules outside the balloon effect the pressure of molecule inside the balloon.
| Like you said, pressure arises from the force exerted by the molecules on stuff. In case of the balloon whose pressure inside is higher than the pressure outside, the number of collisions per second of the molecules outside the balloon is lower than that of the ones inside. Thus the balloon feels a net force that is outward. So the balloon expands until the net force becomes zero. Thus happens when the rate of collision outside equals that of inside. This is nothing but the condition where the pressure is equalised.
| {
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Explicit calculation of spin connection through Cartan's first structure equation Given the metric
$$
ds^2 = F(r)^2dr^2 + r^2d\theta^2 + r^2 \sin^2(\theta)\, d\phi^2,
$$
I'm trying to find the corresponding spin connections $\omega^a_{\ b}$ using the first structure equation:
$$
de + \omega e = 0.
$$
I found the vielbeins $e$ and their exterior derivatives $de$:
$$
de^1 = 0, \quad de^2 = drd\theta, \quad de^3 = sin(\theta)drd\phi + r\cos(\theta)d\theta d\phi,
$$
but I am stuck on actually working out the $\omega$.
I read through Zee's 'GR in a nutshell', and he does the same calculation but just says: "In general, write $\omega^a_{\,b} = \omega^a_{\,bc}e^c = \omega^a_{\,b\mu}dx^\mu$. Plug this into the first structure equation and match terms. How do I actually go about calculating $\omega^1_{\, 2}$, $\omega^1_{\, 3}$, and $\omega^2_{\, 3}$ at this point?
| There is also an explicit procedure that is often better if the vielbein is simple.
We have $$ \mathrm d e^a=-\frac{1}{2}C^a_{bc}e^b\wedge e^c, $$ where the $C^a_{bc}$ are the vielbein commutators. We can invert the first structure equation explicitly as $$ 0=\mathrm de^a+\omega^a_{\ b}\wedge e^b \\ =-\frac{1}{2}C^a_{bc}e^b\wedge e^c+\omega^a_{c\ b}e^c\wedge e^b \\ \frac{1}{2}C^a_{bc}e^b\wedge e^c=\frac{1}{2}\left( \omega^a_{b\ c}-\omega^a_{c\ b} \right)e^b\wedge e^c, $$ so $$ C^a_{bc}=\omega^a_{b\ c}-\omega^a_{c\ b}. $$
Lowering the index, we get $$ C_{a,bc}=\omega_{b,ac}-\omega_{c,ab} \\ C_{b,ca}=\omega_{c,ba}-\omega_{a,bc} \\ -C_{c,ab}=-\omega_{a,cb}+\omega_{b,ca}, $$ now sum these up: $$ C_{a,bc}+C_{b,ca}-C_{c,ab}=2\omega_{c,ba} \\ \omega_{c,ab}=\frac{1}{2}\left(C_{c,ab}-C_{a,bc}-C_{b,ca}\right) \\ \omega_{ab}=\frac{1}{2}\left(C_{c,ab}-C_{a,bc}-C_{b,ca}\right)e^c. $$
If the veilbein is simple, then the $\mathrm de^a=-\frac{1}{2}C^a_{bc}e^b\wedge e^c$ will only involve a few terms at most, and the spin connection is very easy to calculate from this.
| {
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What would the surface temperature of the Earth be without the Sun? If the Sun would not heat the earth, and never had heated it in the past,
its surface would only be heated from the inside.
What would be the current temperature of the surface?
| 57 degrees Fahrenheit
And without sunlight, the Earth would get very, very cold. Earth's surface temperature now averages about 57 degrees Fahrenheit, but by the end of the first week without the sun, the average surface temperature would be below the freezing point.
| {
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If sound passes through material, vibration is produced. So are electromagnetic waves produced too? Sound means vibration of molecules and vibration produces electromagnetic waves. So, this means that sound produces electromagnetic waves directly.
Is this possible?
| Motions of neutral atoms do not radiate
@Cyclone's answer provides very useful insight, but is ultimately irrelevant. As long as the medium that is vibrating is neutral, no energy is radiated. The EM waves emitted by the protons in the air interfere destructively with those emitted by the electrons and cancel out.
Put another way, the emitted radiation is proportional to the integral of the net current density. If positive and negative charges are moving the same, the net current density is 0.
In order to radiate, you would have to accelerate the atoms fast enough that they move faster than the electron cloud can catch up, causing a separation of the center of positive and negative charge (i.e. forming a dipole).
| {
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Entanglement between what? According to the standard definition of "Entropy of Entanglement"
https://en.wikipedia.org/wiki/Entropy_of_entanglement
one starts from the density matrix of a pure state
$$
\rho=|\psi\rangle\langle\psi|,
$$
then divides the system into two parts, $A$ and $B$, traces away the degrees of freedom of one of the two subsystems, say subsystem B, and thus obtains the reduced density matrix of the remaining subsystem
$$
\rho_A=\mathrm{Tr}_B(\rho).
$$
Eventually, the entanglement between subsystem $A$ and subsystem $B$ is given by the Von Neumann entropy of $\rho_a$:
$$
S(\rho_a)=-\mathrm{Tr}[\rho_A\,\log\rho_A].
$$
My question is: does the choice of the bipartition play an essential role? I think that the final result strongly depends on how one chooses subsystem $A$ (and, of course, the complementary subsystem $B$). To my knowledge, in fact, books do not emphasize how important is the choice of the partition. Is there a region?
| Indeed, the definition is pointless without a choice of partition of the underlying space. It is true that sometimes this is not made explicit. For example, people often talk about "entangled particles", but they should be really talking about specific properties of the particles being entangled, not the particles themselves. Similarly, you can have entanglement between different degrees of freedom of the same particle (though you might not have "nonlocality" in such cases).
| {
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How to visualize newtonian objects in term of harmonic oscillators? From the group $U(1)$ in QED, via group representation we get the harmonic oscillators. However, I still have a hard time to imagine newtonian objects (computer, headphone, etc) as a combination of oscillators. I have take a look at classical limit but the visualization is still vague.
Let's say I have a glass breaking. How to visualize it in oscillations?
| Suppose the object comprises $n$ degrees of freedom, and we work in a Cartesian coordinate system for these degrees of freedom, so each coordinate $q_i$ is $0$ at a stable equilibrium. The kinetic energy is of the form $\sum_i\frac12m_i\dot{q}_i^2$; the potential energy is of the form $\sum_i\frac12k_iq_i^2$, plus higher-order terms if an exact description of the system is anharmonic. (Cross terms, e.g. a term proportional to $q_1q_2$, can be deleted by rotating the coordinate axes.) Close enough to the equilibrium, such terms can be neglected. Such a harmonic approximation is of $n$ uncoupled harmonic oscillators.
| {
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Torque and force Does the force due to torque on the edge of a wheel depend on the mass of the the wheel or is it always $\tau / R$?
| Yes, it does depend on wheel mass, but indirectly. Here is how =>
By definition torque is:
$$ \tau = r\,F_\perp $$
And Newton second law expressed for rotation is :
$$ \tau=I\,\alpha $$
where $I$ is moment of inertia and $\alpha$ is angular acceleration. Putting these equations together and solving for a force gives
$$ F_\perp = \frac{I\,\alpha}{r} $$
In your case, wheel can be seen as a solid disk, which has moment of inertia:
$$ I_\textrm{disk} = \frac{m\,r^2}{2} $$
Substituting moment of inertia back into force equation gives final solution :
$$ F_\perp = \frac{m\,r\,\alpha}{2} $$
So you need wheel mass $m$, radius $r$ and angular acceleration $\alpha$ to calculate resulting force. However, if you have already calculated/extracted wheel's torque, then you can calculate same thing by just $\tau\,/\,r$. Still you need to remember that torque according to second Newton law depends on rotating object moment of inertia. And moment of inertia depends on an object mass AND on object shape. You can look for common shapes moment of inertia here.
In a general case if body is continuous, then moment of inertia can be found integrating infinitesimal masses in a body:
$$ I = \int r^2 \,dm $$
EDIT :
If you need to include ground resistance effect on wheel, then just decouple net force :
$$ F_\perp = F_{\perp\text{engine}} - F_{\perp\text{resistance}}
\\= F_{\perp\text{engine}} - \mu_{s}\,m\,g
$$
| {
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Why does the same proportion of a radioactive substance decay per time period? (half life) Just wondering, if decay is random, why does the activity half every half life, as in, why does it have to reduce by the same proportion in the same time period?
| An example that might help:
Start with a big pile of coins. Flip them. Remove the heads. About half remain.
Take the remainder and flip them. Remove the heads. About half remain.
Take the remainder and flip them. Remove the heads. About half remain.
The analogy: An atom has a 50% chance of decaying in some particular interval $T_{1/2}$. After each of those intervals, half are left.
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How fast does a particle need to go to have a significant probability to quantum tunnel through a star? According to this answer tunneling probability depends, among other things I don't know, on the length of the barrier.
Due to length contraction when going at relativistic speeds, it seems it should be theoretically possible to contract lengths arbitrarily by going at an arbitrarily high fraction of the speed of light. I suppose that means the probability of tunneling can get arbitrarily high if we just go fast enough.
At what point does the probability of a particle (say, a proton) tunneling through, say a planet or a star, significant. Let's say >10%?
Is this a case where quantum mechanics and special relativity still play nicely together, or not?
| For quantum tunnelling density and velocity is arbitrary. Tunnelling might be the wrong word as it involves a spherical "wave" starting at the particle extending to infinity, the Schrodinger wave. It is the probability of the particle existing at any point on the wave. The wave-function "collapses" when someone observes the particle, and the particle ends up at one place on the wave, almost always the point of highest likelihood, however it can also pop-up anywhere on the wave however most of the time on or near the point of highest likelihood. To cross something as massive as a star the probability would be probably (depending on the wave) approaching zero however not quite zero, MUCH lower than 10%. I only gave a brief explanation, I would recommend looking at other sources that can explain this topic much better than I can.
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Why is Newton's First Law not considered a vacuous statement? How did Newton come to postulate the law of inertia when every inertial body that he was considering was (by his own other hypotheses) being acted on by the force of universal gravitation.
In other words, he simultaneously posits the law of Inertia and then theorizes that the contrapositive of his statement is not testable ("If no forces act upon a body at rest, then it will stay at rest", but there is never a situation in which no forces are acting upon a body).
Did he get to this by considering the case of how the inertial velocity of something like a ball being swung on a string in a circle takes off in the direction of the tangent to the circle when the centripetal force is removed, or what were the observations that led to postulate this law? That's the only immediately testable example that I have been able to think of so far, as the same experiment with planets orbiting the sun is not/was not feasible to test in his day.
I'm guessing this law may have some historical context related to the philosophical question of "what makes things move" in Aristotle, and Newton used his law to resolve paradoxes that I am not familiar with.
Or is it something entirely different than what I'm thinking, in that what he actually meant was "it takes less force to keep a body in motion than it does to start a body in motion"?
| Physics from the time of Newton to now is the discipline where mathematical differential equations are used , whose solutions fit the experimental data points and are predictive of new data. In order to do this , a subset of the possible mathematical solutions of the differential equations is picked by use of postulates/laws/principles , the axioms that relate data to solutions of the mathematical equations.
That is the purpose of Newtons laws of motion, they are the axioms of the kinematic theory distilled from observations . Nothing is redundant. Note that the concept of Force is defined in the second law. He took words used in everyday language and postulated a mathematical meaning for them, and set up the axioms. His theory has been validated innumerable times, though it had to be modified for the very high masses and velocities ( general and special relativity developed for these regions) and at the quantum dimensions.
| {
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Double slit for electrons (two beams or one)? I understand the double slit for waves but for electrons do we have a beam for each slit so each beam is responsible le to shoot electrons through its own slit.
Or do we have just one beam? Which slit do we place the beam to? If it’s in the middle, wouldn’t all the electrons hit the wall between the two slits?
Sorry Ian very confused as how to carry out the experiment for electrons
| In the Copenhagen interpretation, the principle is that there is a wave and that wave difracts through the holes like any other wave. The electron as a particle then appears at the screen with a probability that depends on the square magnitude of the wave. In this context, it is not suggested that the electron definitely went through one or other slit. Rather - if you adjust the experiment to determine which one it went through then you adjust the wave and so change the pattern on the screen.
The interference of "particles" in quantum mechanics is, mathematically, the interference of some kind of waves.
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Clarification on a statement Bernevig's textbook on Topological insulators On page 11 of the aforementioned book Bernevig claims after doing some calculation that the integral of Berry curvature over a sphere containing a monopole is $2\pi$. Now my question is the following: the berry curvature is defined as the curl of the berry connection and hence its integral over a closed surface should vanish by the divergence theorem. How is this possible.
| It hinges on the fact that the berry connection is not globally defined on the sphere. For Stokes' theorem(or Divergence theorem as you stated) to work, you need a vector field ${A}$ defined on the whole manifold.
But if you choose a coordinate system on the sphere (e.g. spherical coordinate) and express the berry connection in that coordinate, you will find a point where the connection becomes singular. Even if you choose a different gauge to try removing the singularity, there is always one. See the second page of this lecture note for an example.
In sum, if the sphere in question encloses a monopole, the connection(hence curvature also) cannot be globally defined on the sphere, and this is the obstruction against application of Stokes' theorem.
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Travelling to the future while seemingly standing still? Some time ago I thought about this possibility, knowing about the time dilation from general relativity. What would theoretically happen if I travelled at near the speed of light in a circle that has such a small radius so that my brain and the people around me wouldn't notice I am moving at all? If possible, that could allow someone to travel to the future by seemingly not moving at all. I hope the question is clear and please let me know if it isn't, so that I can later fix it and make it more comprehensive.
| That's funny, I had the exact same thought while I was first taught special relativity. This can definitely work, though you would require enormous amounts of energy. An example would be a satellite in orbit around Earth. Given high enough speed, the people inside the satellite could undergo significant time dilation(everyone who is moving undergoes time-dilation).
Also from @knzhou comment. You would require a large enough radius so as to avoid tense forces on people that are in orbit(that's why the satellite would be more appropriate).
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How is a 25-year-old can of soda now empty without having been opened or poked? I just discovered in my parents' basement a Sprite can from 1995* and also a Coca-Cola can probably from the same year.
Both cans are unopened and have no visible damage or holes. The Coca-Cola can feels "normal", but the Sprite can is empty!
You can hear in my video that there is no liquid sloshing around in it.
I can't find any way that the soda could have escaped.
We also don't see any mess near where the cans were, but I don't know for certain that the cans stayed in the same place for 25 years, so maybe there could have been a mess of liquid leaked out somewhere else if the cans had been stored somewhere else earlier. But nothing feels sticky or looks like there has been a leak of any kind.
What are possible explanations for how a carbonated drink could disappear from a sealed aluminum can?
*I had kept it as a "collector's item" from when the Houston Rockets won their second championship.
| The more likely explanation of any of the described scenarios is probably in the realm of corrosion chemistry.
In time, the protective Al coating is penetrated in the presence of dissolved salts.
The highly reactive exposed Aluminum will chemically react (even with the water content, not too mention any acids presence including carbonic or phosphoric, found in select brands):
2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2 (g)
This will further create pressure.
The form of the attack most likely results in so-called pit corrosion. To quote a source:
Corrosion of aluminum in the passive range is localized, usually manifested by random formation of pits. The pitting-potential principle establishes the conditions under which metals in the passive state are subject to corrosion by pitting. Pitting corrosion is very similar to crevice corrosion.
As a pinhole leak (or leaks) from a pressurized can empty the contents of a soda can, it may further leave the false impression that it was always empty.
| {
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Age of Universe from Friedmann Equation - How to actually solve the integral? The Friedmann equation for a flat universe can be written as
$$
H(t)=\frac{\dot{a}}{a}=H_0\sqrt{\Omega_{m,0}\cdot a^{-3}+\Omega_{\Lambda,0}}=H(a)
$$
To calculate the age of the universe, many books jump directly to the result. But there should be some sort of integral in between. I assume one can do the following:
$$
t=\int_0^{t_0}\!\mathrm{d}t=\int_0^{1}\!\mathrm{d}a\frac{\mathrm{d}t}{\mathrm{d}a}=\int_0^{1}\!\frac{\mathrm{d}a}{\dot{a}}
$$
with $\dot{a}$ from above expressin for $H$.
But how is this integral solved? Mathematica did something for hours but did not came up with a result. Most books and wikipedia pages skip directly to the result
$$
t_0=\frac{1}{3H_0\sqrt{\Omega_\Lambda}}\log{\frac{1+\sqrt{\Omega_\Lambda}}{1-\sqrt{\Omega_\Lambda}}}
$$
which leads to the well known result of ~13 billion years (depending on the DM density).
Again: But how is the integral solved?
| I found that Mathematica could deal with the integral in a minute or so, giving the result:
where I have used the notation $\Omega_{m,0} = m$ and $\Omega_{\Lambda, 0} = n$. Being a conditional expression, it depends upon these parameters being in certain bounds, but let's go ahead and assume that they do. Since the universe is flat we can simplify the result by using the condition $n + m = 1$, and after some steps of algebra we arrive at: $$ H_0 t_0 = \frac{2}{3 \sqrt{n}} \log \frac{1 + \sqrt{n}}{\sqrt{1 - n}} . $$ This is almost the same as the answer you present, except for the denominator of the logarithm, and a factor of 2. Is it possible there is a typo there?
| {
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Schrödinger equation on operated states I understand that we can apply the Schrödinger equation to any wavefunction. Now, my question is, can we apply it to states that are being operated upon? Because, when we apply an operator on a state, we get another state which might/might not be a linear multiple of the original one, but still should satisfy the Schrödinger equation, right?
Basically, can I apply Schrödinger equation on states like $\hat{A}\psi $
Meaning, can I write:
$$i\hbar \frac{\partial (\hat{A}\psi) }{\partial t} = \hat{H}(\hat{A}\psi)$$
| If you start with $\psi$ that solves the Schrödinger equation:
$$
i\hbar \frac{\partial \psi}{\partial t} = \hat{H}\psi
$$
And acts $\hat{A}$ on $\psi$:
$$
i\hbar \frac{\partial (\hat{A}\psi) }{\partial t} = \hat{H}(\hat{A}\psi)
$$
Then because we work in the Schrödinger picture and if we assume that $\hat{A}$ doesn't explicitly depends on time:
$$
i\hbar \hat{A}\frac{\partial \psi }{\partial t} = \hat{A}\hat{H}\psi + [\hat{H}, \hat{A}] \psi
$$
So $\hat{A}\psi$ will solve the Schrödinger equation if $[\hat{H}, \hat{A}]=0$, i.e. if $\hat{A}$ is a symmetry generator.
| {
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Cyclic invariance of trace of fermions Consider the Green's function of fermion operators with imaginary time,
$$\mathcal{G}(\nu, \nu', \tau) = - \langle T_\tau c_{\nu}(\tau) c_{\nu'}^\dagger(0)\rangle\tag{1}$$
To show it satisfies the periodicity,
$$\mathcal{G}(\nu, \nu', \tau) = - \mathcal{G}(\nu, \nu', \tau+ \beta) \tag{2}$$
one needs to used the following identity, (see eq (71) and (72) on page 16 of http://folk.ntnu.no/johnof/green-2013.pdf)
$${\rm Tr}(ABC \ldots XY Z) = {\rm Tr}(ZAB \ldots XY )\tag{71}$$
$$Tr(e^{-\beta H}c_{\nu'}^\dagger e^{H \tau} c_\nu e^{-H \tau} ) = Tr( e^{H \tau} c_\nu e^{-H \tau} e^{-\beta H}c_{\nu'}^\dagger )\tag{3}$$
which is important for the appearance of the minus sign in equation (2). However, since $c_\nu$ and $c_\nu^\dagger$ are fermions, I doubt equation (3) is not correct. My question is whether (3) is indeed correct or not? If (3) is not correct, how can (2) hold?
| I can see why you might be confused. If the trace were only over the spinor indices on a fermi field, $\psi_\alpha$ say, then there would be an additional minus sign. Here, however, the trace is over the entire many-body Hilbert space and the $c_\nu$'s are just like any other operator and so have a cyclic trace. This is what Ismasou's terse comment is saying.
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Ionizing radiation in thermal radiation According to the black-body radiation equation, the spectrum extends to infinitely high frequency (although its intensity gets small quickly towards high frequency).
(1) How do you roughly estimate the ionizing radiation power in common high power thermal sources like a 2KW heater without fan, then make sure it is safe considering it is used for years instead of 0.1 second/year in the case of DR X-ray scan? (First the portion of power getting radiated instead of convected needs to be estimated, second this seems not to be black-body and will it have a similar radiation curve that extends to infinite frequency?)
(2) In the photoelectric effect, there are things like cut-off frequency; so, why doesn't thermal radiation - which is quantum mechanical in microscopic level - have a cut-off frequency, i.e. it doesn't radiate X-ray and gamma-ray at all?
|
How to roughly estimate the ionizing radiation power in common high
power thermal source like 2KW heater without fan then make sure it is
safe considering it is used for years instead of 0.1 second/year in
case of DR Xray scan?
A common 2 kW resistance heater does not emit ionizing radiation. It's radiant heat is primarily infra red. The range for infrared radiation is about 430 x 10$^{12}$Hz to 300 x 10$^9$ Hz. It is considered non-ionizing radiation.
Ionizing radiation begins around 3 x 10$^{15}$ Hz. X-radiation is in the range of 3 x 10$^{16}$ to 2 x 10$^{19}$ Hz.
Bottom line: Infrared heaters do not pose the health risks associated with ionizing radiation like X-radiation.
Hope this helps.
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Shape complexity of a substance and its effect on heat retention While cooking, I have noticed that a ball of boiled spinach retains its heat for a long time. I understand that spinach is 91% water. It is my hunch that the same volume of water would cool much faster. If 9% of spinach is attributable to other compounds, my first guess might be that water would cool only 9% faster than spinach, but I think spinach would cool at a much slower rate than that. If true, would the complex shape of rolled spinach be the primary factor in its heat retention ability when compared with water?
| You may need to test your hunch to see if you are right. Water has a very high specific heat, so it is a very good reservoir of heat. A timer and a thermometer might say that it cools off just as slowly.
However, two things might make a "ball of water" cool off more quickly:
*
*Evaporation. If the whole surface of the ball is exposed to the air (and especially if the water is well above room temperature), then evaporation can cool the surface quickly. The spinach likely inhibits a lot of this by being between the air and most of the water.
*Convection. If the water was just in a pan or a ball, then currents could move warmer interior water to the surface where it could cool. With the spinach in the way, these currents aren't possible. Any hot water in the center has to cool by conduction. This is much slower than if convection were present.
The shape is probably important, but the complexity of the shape should not be. Just that it inhibits those two processes.
| {
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Stability of plum pudding model My textbook says that the plum pudding model(Thomson's model) should be electrostatically unstable. Why is that so?
| Your textbook is wrong. Earnshaw's theorem does not apply to the plum pudding model: the pudding provides a non-zero divergence.
Actually Thomson was aware of the possible problem of stability, and this was one of the reasons for his choice of the model. In his 1904 paper, he was able to show that, for an increasing number of negative point-like "plums" embedded in a spherical uniform positive charge, regular shapes in the form of rings, or rings with a center, are stable equilibrium configurations. He considered not only rings but also three-dimensional shells of particles at regular positions, arriving to propose a connection between such static shell structures and some properties of the periodic table, anticipating, in a sense, the correct explanation of periodicity of atomic properties.
It is also worth of notice that the analysis of scattering of alpha particles on gold thin foils, performed by Rutherford in 1909, arrived as a shock, since it falsified Thomson's model with its stability, and forced to assume the planetary model, evidently unstable, due to radiation. Without recognizing the stability of Thomson's model, such a key step in the history of atomic model would not be understandable.
A final note is that Thomson's model had a recent resurrection in connection with the theoretical analysis of quantum dots. You may find some reference to this application in the wikipedia page on the model.
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Does Another Universe Where Time Runs Backwards Revive the Idea of a Cyclical Universe? https://www.pbs.org/wgbh/nova/article/big-bang-may-created-mirror-universe-time-runs-backwards/
This article talks about a new theory where the big bang happened in one direction for us and the other direction where time is going backwards (mathematically). Would that revive the idea of a cyclical universe again? I had this little vision in my head when I read this. A universe chain with a big bang on essentially on either side of the start/end of time and in the next big bang our universe is the one going backward in time and the new one is going forward.
| An interesting idea is that at the Big Bang positive matter went forwards, and negative matter went backwards, but time is circular, so we'll meet again.
Since most of space is pretty empty, a collision between the two universes would not wipe everything out instantly. Galaxies would, for the most part, collide with nothing much. But bits of our universe would go forward into their universe, and bits of their universe would go backwards into our universe.
I believe that makes the theory testable, because it should be possible to test whether bits of an antimatter universe are creeping into our universe.
There appears to be some evidence that this is happening, because recent research has shown that the universe is expanding at an increasing speed, when it should be expanding at a reducing speed, because of gravity.
That could be explained by saying that antimatter coming from the future is the fuel to provide the extra energy to keep the universe expanding faster.
That's probably not the true explanation, but it's worth investigating.
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Clarification of the concept "less resistance means less heating" in a wire So my textbook says that the reason cables that are suppose to carry high currents, are thicker that those that are meant to carry lesser current, is that "less resistance (of the wire) means less heating..."? Is this even true?
Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating!
| Well as far as i can i see, you cannot change the current passing through the wire. This is because they are "suppose to carry high current"(as per the question) . So the only possible way to reduce heat losses is to decrease resistance(heat loss is proprtional to resistance of wire).
That being said transmission cables for long distance generally operate at high voltages(not high current) to minimize heat loss
| {
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Do black holes move through space? I know it was already asked here:
Does a black hole move through space? What happens to other things around it?
And it might be a very stupid question, but here it is:
From a relativistic perspective, do black holes move through space, or is it the space around them that is curved in such a way that for us they seem to move?
I know there is no absolute frame of reference in relativity, but let's say the standpoint of one blackhole, I would think time is frozen, so without time how can things move?
| Yes they can. Everything moves in space. If you're looking for proof they can move, the Standard Model says the galaxies (which have black holes) are moving away from us. That’s a lot of black holes moving.
| {
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Gravity, matter vs antimatter I have a simple question regarding matter-antimatter gravity interaction.
Consider the following though experiment:
If we imagine a mass $m$ and an antimass $m^-$, revolving around a large mass $M$
the potential energy of mass $m$ should be:
$$ U_1=-\frac{GmM}{R} $$
and the potential energy of mass $m^-$ should be:
$$ U_2=-\frac{GmM}{R} $$
or:
$$ U_2=\frac{GmM}{R} $$
depending on the sign of the gravity interaction between matter and antimatter.
If the two particles annihilate to energy, then the gravitational field of $M$ will interact with the emitted photons and will change their frequency.
But, as the interaction between gravity and the photons has nothing to do with the question of the gravity between matter and antimatter, can't we simply use the interaction between gravity and photons, and the energy conservation to establish the nature of the gravity interaction between matter and antimatter?
| The short answer is that we don't know. While we have made some small amounts of anti-matter, it has not lasted long enough to measure the very weak force of gravity.
However, people have speculated. Here is a long article on Wikipedia about that, with arguments going both ways.
The introduction to that article states
While the consensus among physicists is that gravity will attract both matter
and antimatter at the same rate that matter attracts matter, there is a strong
desire to confirm this experimentally
The thought experiment you describe is one of the arguments in favor of this consensus. (look for Phillip Morrison in the article) To my layman's mind it is a convincing argument, but we never really know until we have measured it.
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What if the net force provided for a circular motion is larger than the required centripetal force? Will an object be pulled towards the centre linearly if the net force provided for a circular motion is larger than the required centripetal force? And why?
For example, if the object in a circular motion which is connected by a string is pulled towards the centre by hand.
| When pulled harder towards the centre, it cannot start moving linearly towards the centre.
It already has a tangential speed (otherwise there would be no circular motion). In order to start moving linearly towards the centre, the tangential speed must be zero. If you pull directly inwards, then there is no force acting tangentially, so nothing to slow down the tangential speed.
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What does this notation for spin mean? $\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$ In my quantum mechanics courses I have come across this notation many times:
$$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$$
but I feel like I've never fully understood what this notation actually means. I know that it represents the fact that you can combine two spin 1/2 as either a spin 1 (triplet) or a spin 0 (singlet). This way they are eigenvectors of the total spin operator $(\vec S_1+\vec S_2)^2.$ I also know what the tensor product (Kronecker product) and direct sum do numerically, but what does this notation actually represent?
Does the 1/2 refer to the states? Or to the subspaces? Subspaces of what exactly (I've also heard subspaces many times but likewise do not fully understand it). Is the equal sign exact or is it up to some transformation?
And finally is there some (iterative) way to write a product of many of these spin 1/2's as a direct sum?
$$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\dots=\left(\mathbf{1}\oplus\mathbf 0\right)\otimes\mathbf{\frac 1 2}\dots=\dots$$
| The $\otimes$ sign denotes the tensor product. Given two matrices (let’s say $2\times 2$ although they can be $n\times n$ and $m\times m$) $A$ and $B$, then
$A\otimes B$ is the $4\times 4$ matrix
\begin{align}
A\otimes B =\left(
\begin{array}{cc}
A_{11}B&A_{12}B\\
A_{21}B&A_{22}B
\end{array}\right)= \left(\begin{array}{cccc}
A_{11}B_{11}&A_{11}B_{12}&A_{12}B_{11}&A_{12}B_{12}\\
A_{11}B_{21}&A_{11}B_{22}&B_{12}B_{21}&A_{12}B_{22}\\
A_{21}B_{11}&A_{21}B_{12}&A_{22}B_{11}&A_{22}B_{12}\\
A_{21}B_{21}&A_{21}B_{22}&A_{22}B_{21}&A_{22}B_{22}
\end{array}\right) \, .
\end{align}
A basis for this space is spanned by the vectors
\begin{align}
a_{1}b_{1}&\to \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array}
\right)\, ,\quad
a_1b_2 \to \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)\, ,\quad
a_2b_1\to \left(\begin{array}{c} 0 \\ 0 \\ 1 \\0\end{array}\right)\, ,\quad
a_2b_2\to \left(\begin{array}{c} 0\\0\\0\\1\end{array}\right)
\end{align}
In terms of $a_1\to \vert +\rangle_1$, $a_2\to \vert -\rangle_1$ etc we have
\begin{align}
a_1b_1\to \vert{+}\rangle_1\vert {+}\rangle _2\, ,\quad a_1b_2\to \vert{+}\rangle_1\vert{-}\rangle _2 \, ,\quad
a_2 b_1\to \vert{-}\rangle_1\vert {+}\rangle _2 \, ,\quad a_2b_2\to
\vert{-}\rangle_1\vert{-}\rangle_2\, .
\end{align}
In the case of two spin-$1/2$ systems, $\frac{1}{2}\otimes \frac{1}{2}$ implies your are taking $\sigma_x\otimes \sigma_x$, $\sigma_y\otimes \sigma_y$, $\sigma_z\otimes \sigma_z$, since these are operators acting on individual spin-$1/2$ systems. The resulting matrices can be simultaneously block diagonalized by using the basis states $a_1b_1$, $\frac{1}{\sqrt{2}}(a_1b_2\pm a_2b_1)$ and $a_2b_2$. There is a $3\times 3$ block consisting of $a_1b_1, \frac{1}{\sqrt{2}}(a_1b_2+a_2b_1)$ and $a_2b_2$ and a $1\times 1$ block with basis vector
$\frac{1}{\sqrt{2}}(a_1b_2-a_2b_1)$.
The $3\times 3$ block never mixes with the $1\times 1$ block when considering the operators $S_x=s_x^{1}+s_x^{2}$ etc. The basis vectors of the $3\times 3$ block transform as states with $S=1$, in the sense that matrix elements of $S_x$, $S_y$ and $S_z$ are precisely those of states with $S=1$; the basis vector of the $1\times 1$ block transforms like a state of $S=0$. Hence one commonly writes
\begin{align}
\frac{1}{2}\otimes \frac{1}{2} = 1\oplus 0
\end{align}
with the $\oplus$ symbol signifying that the total Hilbert space is spanned by those vectors spanning the $S=1$ block plus the vector spanning the $S=0$ part; note that those vectors are product states of the type $a_1b_1$ etc.
| {
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"url": "https://physics.stackexchange.com/questions/534887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Are the components of the curvature tensor defined by the Robertson-Walker metric constant? In the Robertson-Walker solution the curvature is uniform. However the space is "expanding" and so the matter density is decreasing over time (or not because it's modelled as a perfect fluid?). Does the expansion affect the components of the curvature tensor if this is the case?
| I prefer to put the metric in the form
$$ ds^2 = dt^2 - a^2(t)\left[d\chi^2 + f^2(\chi) d\Omega^2 \right]$$
or alternatively
$$ ds^2 = a^2(\tau)\left[d\tau^2 - d\chi^2 - f^2(\chi) d\Omega^2 \right],$$
where
$$ f(\chi) = \sin\chi,\,\, \chi,\,\, \sinh\chi $$
for positive, zero and negative curvature, respectively. f is the radial coordinate commonly denoted by r in Robertson-Walker coordinates. This makes it explicit that coefficients of the metric change in time, and likewise the space curvature changes in time because the metric describes uniform space curvature - not uniform spacetime curvature.
You can see that coefficients of the curvature tensor change in time directly from Einstein's equation. As you say, the energy/matter density is decreasing, so Einstein's equation is a statement that curvature is decreasing.
diagrams for the above coordinates are shown at Description of singularities
| {
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Where does energy go in joining capacitors of different capacitance charged by different potential, hypothetically assuming no resistance in circuits? I don't understand why there is any change in initial and final energy since we have already assumed a perfectly conductive circuit. I mean, theoretically at least, there should be no change in energy.
Now, considering there is a change in energy at all, is it because electrons accelerate in moving from one capacitor to another, so energy gets dissipated in the form of electromagnetic radiation?
Image source: NCERT Physics Textbook for Class XII Part I, page 82
| When charges move due to an electric field (i.e. down the potential gradient) the electric potential energy of the system decreases. When you put in the second capacitor there is a brief moment where an electric field is present that causes excess charges to move to the other plates. Hence you have less energy stored.
Or you can view it as doubling the area of the plates, thus decreasing the field between each plate by half, which decreases the energy by half (two systems each with a quarter of the original energy). Although this is somewhat already covered by the solution you supply, but not explicitly. But this gives a better picture, as it doesn't really matter how we actually make the system, as potential energy only depends on position of the charges, not how they got there.
Any way you look at it, "I mean, theoretically at least, there should be no change in energy" is not a true statement. We are talking about electric potential energy, which can change even if there are no dissipative mechanisms in the system. You don't need to bring in any sort of radiation. You can imagine moving each charge one-by-one very slowly if you want.
| {
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"source": "stackexchange",
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The quantization of the electromagnetic field in Peskin and Schroeder (Eq.9.52) I'm working on the quantization of the electromagnetic field in Peskin (page 294).
However, I'm confused about the Eq.(9.52). Peskin says Eq.(9.51) and Eq.(9.52) are equivalent, but why? Is Eq.(9.52) derived from Eq.(9.51)?
For a real scalar field, I know
$$
D_{F}(x-y)=\left\langle 0\left|T \phi\left(x_{1}\right) \phi\left(x_{2}\right)\right| 0\right\rangle=\left. Z[J]^{-1}\left(-i \frac{\delta}{\delta J\left(x_{1}\right)}\right)\left(-i \frac{\delta}{\delta J\left(x_{2}\right)}\right) Z[J]\right|_{J=0}
$$
If we want to derive the Eq.(9.52), do we need to use it?
| The idea is that the quadratic part of the action should be invertible and the inverse gives the propagator of the field. Eq. 9.52 is trying to solve for the inverse $\tilde{D}^{\nu\rho}_F$, but you cannot because $-k^2g_{\mu\nu} + k_\mu k_\nu$ is singular. That's the exact same as saying that the quadratic action vanishes for too many field configurations for the path integral to converge.
Basically, Peskin is using two different approaches to convince you that you can't naïvely invert the matrix giving the quadratic part of the action to get the propagator—you need some tricks.
| {
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Geometry of Young's experiment for optical path length I am currently studying the textbook Modern Optical Engineering, fourth edition, by Warren Smith. When presenting the concept of optical path length, the author says the following:
With reference to Fig. 1.13, it can be seen that, to a first approximation, the path difference between $AP$ and $BP$, which we shall represent by $\Delta$, is given by
$$\Delta = \dfrac{AB \cdot OP}{D}$$
I'm having difficulty understanding how the mathematics $\Delta = \dfrac{AB \cdot OP}{D}$ corresponds to the figure. I suspect that there is some use of trigonometry and/or geometry that I am not seeing. I would greatly appreciate it if someone would please take the time to explain this to me.
| Let us consider the following diagram:
When the distance $L$ between the slit plane and the screen is large compared to the distance $d$ between the slits, we can assume $S_1P$ and $S_2P$ are parallel to each other. And the $\delta$ in the image represents the optical path difference.
Now consider the light orange coloured triangle. Here $\sin \theta=\delta/d$. As $L>>d$, we can assume $\sin\theta\approx\theta$ and hence $\theta\approx\delta/d$.
Now consider the light blue coloured triangle ($POQ$). Here $\tan\theta=y/L$. Using small angle approximation we can tell $\theta\approx y/L$.
So equating $\theta\approx\delta/d$ and $\theta\approx y/L$, we get $\delta/d\approx y/L$.
If you haven't figured it till now, I have just derived the expression in your question but used different symbols in accordance with the diagram in my answer.
Image taken from the question - In Young's double slit experiment, why are the two theta values equivalent?
| {
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How to explain the Maxwell Boltzmann distribution graph (physically)? While studying the kinetic theory of gases, I came to the section: "The Distribution of Molecular Speed" of a book. The book first (without any explanation) proposes the Maxwell's speed distribution law as
$$P(v) = 4\pi \left ( \frac {M}{2\pi RT} \right)^{3/2} v^2 e ^{-Mv^2/2RT} $$ and then shows the distributional graphs for various gasses at different temperatures as shown in the figure below.
One thing that bothered me was that it didn't provide any physical explanation for the shape of the curve. Therefore, I would like to know how and why, via molecular collisions, we achieve the same (exact) curve at a given temperature even if we start with a random distribution of molecular speed (but given that $ \bar K = \frac {3}{2}k_bT$)?
I would prefer an intuitive explanation rather than a mathematical one.
Note that you needn't to comment on this portion of the question if you wish.
I searched for this over the internet and on Physics.SE but couldn't find any (do tell if you know of any). I also discussed this with my friend came with an ad-hoc explanation but that too raises many questions. If you are interested, the explanation is given below but as I mentioned, it is not a good one and has many loopholes.
We considered a distribution in which the temperature is $T$ and every molecule possesses the same speed (though in random direction) and hence the kinetic energy of each molecule is $\frac {3}{2}k_bT$. When we continue further we see that the molecules collide with one another and (most of the time) the collisions are between glancing and head-on collision. It turns out that some molecules collide with other molecules in such a way that in due time they gain kinetic energy, some other molecules collide in such a way that there is a net loss of their kinetic energy and hence this distribution is achieved.
But as I said earlier there are many questions to be asked and one them is given below.
*
*Why are some type of collisions favoured more over the others?
| Spherical coordinates give R^2 sin
The difference between $$ dN/N \propto \exp(-v_x^2/s^2) dv_x $$ and $$dN/N \propto v^2 \exp(-v^2/c^2)dv$$ lies in the detail about correctly moving from $dv_x$ to $dv$. Consider the infinitesimal volume in cartesian and spherical coordinates $$dv_x dv_y dv_z \propto v^2 \sin\theta\, d\phi\, d\theta\, dv $$ This is how $v^2$ appears before exponent (plus to confirming $v_x^2+v_y^2+v_z^2=v^2$ inside it).
| {
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Can a system reach equilibrium using only reversible processes? Consider an isolated system. We know that it will reach equilibrium after enough time. According to what I have read so far, the system should undergo irreversible processes to reach equilibrium, but is there any counterexample to this statement? i.e a system and a nonequilibrium initial condition such that processes used to reach equilibrium are all reversible?
| Assume an isolated system. Start from an equilibrium state $0$ and remove some internal constraint so that the system spontaneously moves to another equilibrium state $1$. The difference in entropy $\Delta S = S_1 - S_0$ between the entropies of the final $S_1$ and initial $S_0$ is never negative (2nd law), and it is positive for an irreversible process the amount being characteristic and a measure of irreversibility.
| {
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Why is the speed of light in vacuum a universal constant? While getting familiar with relativity, the second postulate has me stuck. "The speed of light is constant for all observers". why can't light slow down for an observer travelling the same direction as the light?
| The short answer is, because if we assume the speed of light is constant then our equations predict experimental outcomes with greater accuracy.
The electromagnetic answer is, because if you plug the measured permittivity and permeability of free space into Maxwell's equations, that is the speed you get.
The Relativistic answer is, the speed of light is assumed to be a law of physics, and therefore a constant for all observers.
| {
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Can the real-time Green's function be written in the form of path integral on the real axis? In every textbook, the path integral of the Green's function is written in imaginary-time. I wonder whether we could write real-time green function in the path integral form.
| You mean the (equilibrium) statistical mechanics textbooks, which deal mainly with the statistical sum, and therefore need only the imaginary path integral.
The real-time path integral is usually derived in quantum mechanics textbooks, where it is sometimes also shown to be the direct solution of the Schrödinger equation with a source term, i.e. a true Green's function in the mathematical sense. Feymann's own QM book is probably a good place to look.
Finally, non-equilibrium statistical mechanics routinely uses Green's functions in real time. For the path integral formulation of the Keldysh formalism you may look here.
| {
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If work is a scalar measurement, why do we sometimes represent it as the product of force (a vector) and distance (scalar)? Consider an object being pushed 3/4 of the distance around a circular track. The work done on the object would be the distance of 3/4 the track’s circumference times the force applied to the object (given that it was pushed at a constant force). Since we are multiplying a vector by a scalar, why is work a scalar measurement? Or would the work done on the object actually just be force times displacement? Thanks.
| The work is the dot product of the force and the displacement, and displacement is a vector; we have to take into account what direction it is pointing. If an object is traveling in a circle, then it has to have a centripetal force, so it doesn't have a constant force. It could have a constant magnitude force, though. The centripetal force is perpendicular to the displacement, so the work done by the centripetal force is zero.
If you look at a ball thrown at an angle of 45 degrees, the ball's velocity starts out with an upwards component, while gravity is pointing down. Since the angle between them is more than 90 degrees, he dot product is negative; gravity is decreasing the kinetic energy of the ball. At the peak of its trajectory, the velocity is perpendicular to gravity, and so, at the very moment, gravity is not doing any work (you can verify that by writing an equation for its kinetic energy in terms of time, then taking the derivative with respect to time). Once the balls starts coming back down, the angle between its velocity and gravity is less than 90 degrees, so gravity is doing work on it, and it is speeding up.
| {
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Why doesn't the orange light filter appear bluish if it transmitts orange and reflects or absorbs all other frequencies? I was going through the absorption and reflection phenomenon of light and this question struck my mind as I was walking through the light filters. Why does orange filter not appear bluish if it transmitts orange light and reflects all other lights except orange?
| There are actually two major types of filters:
*
*Absorption filter
A piece of orange glass (like in a photo camera filter), for example, is an absorption filter. It absorbs higher frequency light, and doesn't absorb much the red-orange frequencies. The result is that due to the usual, Snell's reflection these are reflected, but due to transparency of the glass they are also transmitted.
Here's an example of a photographic absorption filter:
(image source)
*Reflection/scattering filter
Some kinds of opalescent glass, on the other hand, look orange when you look at a white light through a sample, but when viewed from the side, it's blue. This is exactly what you expect. Same result will be with an interference filter, which, instead of scattering the unwanted wavelengths, specularly reflects them.
Here's an example of the opalescent glass:
(image source)
| {
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What does "up to a total derivative" really mean and how should I know when to use it? I am a mathematician who is taking a quantum field theory course without much prior pyhsics. We have had the term "up to a total derivative" a few times, yet every time I asked what it meant I didn't really grasp it.
As an example, for our last tutorial we were given the Lagrangian
$$ \mathcal{L} = i\psi^*\partial_0\psi - \frac{1}{2m}\nabla\psi^*\cdot\nabla\psi,\tag{1}
$$
but then immediately in the tutorial it was given that this is equivalent (up to a total derivative) to
$$ \mathcal{L} = \frac{i}{2}(\psi^*\partial_0\psi - (\partial_0\psi^*)\psi) - \frac{1}{2m}\nabla\psi^*\cdot\nabla\psi.\tag{2}
$$
The things I really don't understand are:
*
*how exactly are these things the same? (/what does "up to total derivative" mean)
*how do I know when I should try to convert something to another thing through a total derivative?
| Since the Lagrangian density (which is confusingly also referred as a Lagrangian) is defined as a function which is integrated on, we may always think the $\mathcal{L}$ in inside a 4D integral, since the action is defined via
$$
S = \exp \left( \int d^4x \mathcal{L} \right).
$$
Now, we may decompose the term to two identical parts and integrating one of them by parts,
\begin{align*}
\int dt i\psi^*\partial_0\psi &= \int dt\frac{1}{2}i\psi^*\partial_0\psi + \int dt\frac{1}{2}i\psi^*\partial_0\psi\\
&=\int dt\frac{1}{2}i\psi^*\partial_0\psi + \left. \frac{1}{2}i\psi^*\psi \right|_{\pm \infty} - \int dt\frac{1}{2}i\partial_0(\psi^*)\psi \\
&= \int dt\frac{i}{2}(\psi^*\partial_0\psi - (\partial_0\psi^*)\psi)
\end{align*}
where the substitution term vanishes, since the field values are considered to vanish at infinity: $\psi(\pm \infty) \rightarrow 0$. This is the total derivative term.
| {
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Conservation of energy of 2 identical Rolling Disks with and without friction My physics book claims that if two identical disks moving at the same velocity travel up nearly identical hills, with the second hill not having friction, then the disk rolling up the first hill will travel to a greater height. Given that the disks started with the same kinetic energy while rolling at the base of the hill, shouldn't they reach the same height (i.e. same potential energy) as a result?
|
Given that the disks started with the same kinetic energy while rolling at the base of the hill, shouldn't they reach the same height (i.e. same potential energy) as a result?
First, let's assume that no slipping occurs on the hill with friction and that both disks are rolling without slipping before they get to their respective hills.
You are forgetting to consider that the disk on the frictionless incline will still be spinning when it gets to it's maximum height. Therefore, it will still have some of its initial kinetic energy that can't be converted into potential energy. This means it can't go as high as the disk on the hill with friction.
Thinking just about the forces, the static friction force points up the incline as the disk rolls up it, so this force counters the acceleration down the incline due to gravity which allows disk to go farther before it stops, as well as providing a torque that causes the disk to eventually stop rotating at the top of it's trajectory.
| {
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"timestamp": "2023-03-29T00:00:00",
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Confusion regarding properties of Poisson Brackets I have just started learning about Poisson Brackets, and came across the following property
$$\{q_i,q_j\}=0$$
And
$$\{p_i,p_j\}=0.$$
Where $p$ and $q$ are respectively the momentum and position coordinates i.e. phase space coordinates.
Now Poisson Brackets are defined as
$$\{F,G\}=\frac{\partial F}{\partial q_i}\frac{\partial G}{\partial p_i}-\frac{\partial G}{\partial q_i}\frac{\partial F}{\partial p_i}$$
$i$ and $j$ here stand for the $i$'th and $j$'th spatial coordinates.
$$\{q_i,q_j\}=0$$
$$\Rightarrow \{q_i,q_j\}=\frac{\partial q_i}{\partial q_i}\frac{\partial q_j}{\partial p_i}-\frac{\partial q_j}{\partial q_i}\frac{\partial q_i}{\partial p_i} =0$$
But I am having a hard time proving it.
I know that the second term $(\frac{\partial q_j}{\partial q_i})$ is zero because the i'th and j'th spatial coordinates are orthogonal and hence, there is no change in $q_i$ on changing $q_j$. However I don't know how to prove the first term to be zero, and that is where I need help.
To summarise, my question is prove that
$$\frac{\partial q_i}{\partial q_i}\frac{\partial q_j}{\partial p_i}=0$$
Any help will be deeply appreciated.
| I think there is an misunderstanding on your side. $$\frac{\partial q_j}{\partial q_i} = \delta_{ij},$$ where $\delta_{ij}$ is the so-called Kronecker delta.
I hope this helps so far.
| {
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Why can vector components not be resolved by Laws of Vector Addition? A vector at any angle can be thought of as resultant of two vector components (namely sin and cos).
But a vector can also be thought of resultant or sum of two vectors following Triangle Law of Addition or Parallelogram Law of Addition, as a vector in reality could be the sum of two vectors which are NOT 90°.The only difference here will be that it is not necessary that components will be at right angle.
In other words why do we take components as perpendicular to each other and not any other angle (using Triangle Law and Parallelogram Law).
| There are two main cases of separating into components: standard coordinates system, and separating into normal and parallel.
In a standard coordinate system, there is a set of coordinates, and points are given in terms of those coordinates. Not all coordinate systems are vector spaces; for instance, the longitude latitude system isn't, since the Earths surface is curved, and in general relativity, the coordinates are only locally vectors, since space-time is curved. But for flat space, we can treat the coordinate system as consisting of an origin, each point being represented by a vector pointing from the vector to that point, and the coordinates being given by the projections of that vector onto basis vectors. Generally, orthogonal basis vectors are chosen because that makes things simpler.
The other case where a vector is separated into components is where there is some reference vector, and other vectors are separated into normal and parallel to that. The most common reference vector is the gravitational vector: the parallel direction is treated as a distinguished axis of up/down, and other directions are sideways. But there are other examples of reference vectors: for instance, if you're pushing a cart up a slope, you might take the slope as a reference vector. Then if you have another vector, such as the force on the cart, you can resolve it into a component parallel to the slope, and another perpendicular to the slope. And of course if one vector is parallel to the reference vector, and another is perpendicular, then they will be perpendicular to each other.
| {
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