optics - Spherical aberration in concave mirrors

I saw this posted on the forum with an answer: Cause of Spherical Aberrations

However, the answer helps explain aberration in lenses. Why is there spherical aberration in concave mirrors?

Thanks, -Prasad

special relativity - Rotation in the x-t plane

I am currently studying special relativity using tensors. My lecture notes (which happen to be publicly accessible, see top of page 99) say that the standard configuration can be viewed as a rotation in the x-t plane. Can anyone explain this a bit? Is there a good way to visualize it?

The Standard configuration;

I would prefer an intuitive (not rigorous) explanation over a maths one, but if you have a maths one then for completeness it would be good to post it.

All other notes from these lectures are available here.

Answer

You probably already understand that merely rotating a coordinate system does not change the physical system you're modeling. Given one set of axes, any rotation of those axes just represents some freedom of choice to do math how you see fit.

So with that in mind, if you were pointing northeast (and chose a pair of coordinate axes to point northeast and northwest) and had a compass (pointing north), you'd have the freedom to rotate your coordinate axes to point north instead, simplifying the description of the compass direction.

Similarly, an object traveling with nonzero velocity traces out a diagonal line on the $tx$-plane, but there is a single, well-defined group of operations that could align the $t$-axis with such a diagonal line. This is analogous to a rotation.

I say analogous it's not exactly a rotation the way we typically think of it. We think of rotations as tracing out circles. That's because rotations preserve distance, and a circle is a uniform distance from its center.

But in spacetime, on the $tx$-plane, we have to think of "distance" as being the spacetime interval, and a curve of constant interval isn't a circle--it's a hyperbola. And in this case, we're talking about a hyperbola with diagonal asymptotes. Those asymptotes are the lines that light follow, unreachable by any "rotation" on the $tx$-plane.

This operation is every bit a "rotation" as our usual 3d rotations are; perpendicular vectors stay perpendicular under it, and so on. It's just the equivalent of a rotation for a different kind of underlying space(time) than regular old 3d space.

And so, it is convenient that we don't typically call this a "rotation" in so many words, but we give this operation its own name: the Lorentz transformation.

Using the Lorentz transformation to put some align a four-velocity with the $t$-direction is no different from aligning one's $xyz$-directions with east, north, and up. It's an arbitrary choice, one that you don't have to abide by if it's inconvenient, but just as there always exists a rotation to align one's $xyz$-directions that way, there always exists a Lorentz transformation to align the four-velocity of a massive object with the $t$-direction.

electromagnetism - How do I find the polarity of a U-shaped electromagnet?

How do I find the polarity of a U-shaped electromagnet ?

Current flowing clockwise --> South Pole

Current flowing anticlockwise --> North Pole

However the direction of flow of current changes when seen from top as compared to bottom. From the examples, I find that one should view the direction from the bottom. So why is this so ?

In this image:

I believe the polarity at P is North and that at Q is South. Am I right ?

Is there any other method to determine the polarity of a U-shaped electromagnet ?

Answer

The current direction $I$ is from the positive terminal of the voltage source to the negative terminal.

Look end on along the axis of the electromagnet.

Clockwise current $I$ direction $\Rightarrow$ south pole
Anticlockwise current $I$ direction $\Rightarrow$ north pole

electrostatics - Why do charges on the sphere not fly?

Consider a uniformly charged conducting sphere. The $E$ just outside the sphere is $\sigma/\epsilon_0$ and inside the sphere is zero. What about at a point $ON$ the sphere? Here we consider the smallest unit of charge is $dq$ not $e$ (A classical electrostatics case).

What is the electric field experienced by the charge element $dQ$ (the red dot on the sphere)

I have included five representative charge elements on the sphere and their respective fields on the red dot. By principle of superposition of electric fields one can assume that at the red dot there is a non-zero radial field. So, shouldn't the charge element just fly off?

$E_n$$_e$$_t$ at red dot = E due to charge at yellow dot + E due to charge at blue dot and so on

electromagnetism - Poynting's theorem - energy conservation

Poynting's theorem:

$$\int_V\left(\vec{E}\cdot\vec{J}\right)\,\mathrm dV = -\dfrac{\partial}{\partial t}\int_V\dfrac{1}{2}\left(\epsilon_0 E^2 + \dfrac{1}{\mu_0}B^2\right)\,\mathrm dV - \dfrac{1}{\mu_0}\oint_S \left(\vec{E}\times\vec{B}\right)\cdot\,\mathrm {d\vec S}$$

My question is with regards to how this proves conservation of energy?

The first integral gives the power gained by charges in a volume. The second integral gives the power lost in the electric and magnetic fields. It can be shown independently that these interpretations for these mathematical terms makes sense. It's the third integral that bothers me.

I've looked at proofs online... and they all make the step of simply "interpreting" the last term as the power entering the surface of the volume. So they make the assumption that energy is conserved and hence interpret the last term accordingly.

But this is not a proof of conservation of energy. A proof would require independent reasons to consider the last term as the power flux entering the volume.

Contrast this with conservation of charge within electromagnetism. Conservation of charge is a consequence of Maxwell's equations, it does not have to be assumed independently.

Basically I want to find out... is energy conservation a consequence of Maxwell's equations or something assumed independently from it?

Answer

Let the electromagnetic field has $u$ as its energy density (amount of energy per unit volume in the field) and let $\bf S$ represents the energy flux- the amount of energy per unit time flowing across a unit area perpendicular to the flow).

Now electromagnetic field can interact with matter and do work on them; so this energy interaction must be considered when discussing energy conservation.

The field energy inside a volume $V$ is $\displaystyle \int_V u\,\mathrm dV\;.$ Amount of energy flowing out of the volume $V$ is given by $\displaystyle \int_\Sigma \mathbf S\cdot n\,\mathrm da \;.$

Now, the work done per unit time by the field on the matter inside the volume $V$ is given by $\displaystyle \int _V Nq(\mathbf E + \mathbf v\times \mathbf B)\cdot \mathbf v\,\mathrm dV$ where $N$ is the number of particles per unit volume; this can be written as \begin{align}\int _V Nq(\mathbf E + v\times \mathbf B)\,\mathrm dV& = \int_V Nq\mathbf E\cdot v\,\mathrm dV\\ &= \int_V \mathbf E\cdot \underbrace{(Nq\mathbf v)}_\textrm{current-density}\,\mathrm dV\\ &= \int_V E\cdot \mathbf J\,\mathrm dV \end{align}

So, the continuity equation is written thus: $$\underbrace{-\frac{\partial}{\partial t}\int_V u\,\mathrm dV}_\textrm{rate of change of energy inside volume $V$}=\underbrace{\int_\Sigma \mathbf S\cdot \mathbf n\,\mathrm da}_\textrm{amount of $\mathbf{field\, energy}$ flowing out of volume $V$ per unit time} + \underbrace{\int_V \mathbf E\cdot \mathbf J\,\mathrm dV}_\textrm{work done per unit time by the field on the matter inside volume $V$} \;. $$ This definitely implies the conservation of energy and moreover, the equation which OP wrote in his query is the offshoot of this same continuity equation.

We can write the differential form of the continuity equation above as:

$$-\frac{\partial u}{\partial t}= \textrm{div}\,\mathbf S + \mathbf E\cdot \mathbf J$$ (since, all the derivation is done in Cartesian coordinates, then $\textrm{div}\equiv \mathbf \nabla$).

Or, we can write $$\mathbf E\cdot \mathbf J= -\frac{\partial u}{\partial t}-\mathbf E\cdot \mathbf J\tag 1$$

In order to find out what $u$ and $\bf S$ actually are, it is assumed that they solely depend on the fields $\bf E$ and $\bf B\;.$

Now, using $$\mathbf J = \epsilon_oc^2\,\mathbf \nabla\times\mathbf B- \epsilon_0\frac{\partial\mathbf E}{\partial t}\,,$$ we can write $\mathbf E\cdot \mathbf J$ as $$\mathbf E\cdot \mathbf J= \epsilon_oc^2\,\mathbf E\cdot (\mathbf \nabla\times\mathbf B)- \epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\;.$$

Now, \begin{align}\mathbf E\cdot (\mathbf \nabla\times\mathbf B)&= (\mathbf \nabla\times\mathbf B)\cdot \mathbf E\\ &= \mathbf \nabla\cdot (\mathbf B\times\mathbf E)+ \mathbf B\cdot (\mathbf \nabla \times \mathbf E)\;.\end{align}

Therefore, \begin{align}\mathbf E\cdot \mathbf J&=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\\ &=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E)+ \epsilon_oc^2 \mathbf B\cdot \left(-\frac{\partial\mathbf B}{\partial t}\right)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E) - \frac{\partial}{\partial t}\left(\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B+ \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E\right) \tag 2\end{align}

Comparing $(1)$ and $(2)\,,$ we get $$u = \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E+\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B\\ \mathbf S=\epsilon_oc^2\,\mathbf E\times\mathbf B \;.$$

This vector $\bf S\,,$ that is the energy flux vector we considered at the beginning of the making up of the continuity equation, is called the Poynting Vector.

The whole derivation is based on the continuity equation, which is the mathematical expression of conservation of energy.

OP can conclude the equation he wrote in the question by converting the differential form of equation $(2)$ into integral form; using the definition of $u$ and $\bf S$ derived above and lastly using Gauss's theorem.

$\bullet$ Lectures on Physics by Feynman, Leighton, Sands.

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you're saying that energy conservation is needed as an independent equation, in order to deduce that the Poynting vector gives energy flux. I don't think this is true. I think the energy continuity equation can be shown as a consequence of Maxwell's equations.

No! Never, ever I've made such statement. I've answered to your query I know the books say it represents energy flux, but how do you prove it represents energy flux?; that's it. What I've done is to clearly present the continuity equation before OP and have tried to clearly interpret each term and how they together imply conservation of energy. The continuity equation can indeed be derived from Maxwell's equation and this is what I've done to define $u$ and $\bf S\;.$ I'm really pretty upset on the allegation OP made.

But this is not a proof of conservation of energy.

It's indeed a proof of conservation of energy.

Contrast this with conservation of charge within electromagnetism. Conservation of charge is a consequence of Maxwell's equations, it does not have to be assumed independently.

Okay, let's fix this thing. Conservation of charge implies this continuity equation(this can be derivable from Maxwell's equations):

$$-\frac{\partial \rho_V}{\partial t}= \mathbf \nabla \mathbf J\;.$$

This suggests conservation of energy would look like:

$$-\frac{\partial u}{\partial t}= \mathbf \nabla \mathbf S\;.$$

But that is incomplete as it is the total energy and not just field energy which is conserved; the interaction of field with matter has to be taken into consideration.

is energy conservation a consequence of Maxwell's equations or something assumed independently from it?

Yes. The continuity equation meant for conservation of energy is derived from Maxwell's equation; that's how the definition of $u$ and $\bf S$ have been computed. OP must be confused to see in my answer first the continuity equation as if I'm saying continuity equation should be considered independently in order to consider conservation of energy. No! That was intentionally done in order to conceive that the Poynting Theorem indeed implies conservation of energy.

Is equivalent resistance always lower if we add a resistor to a passive electronic circuit?

How to prove that equivalent resistance of any passive network is always lower if we add a resistor between arbitrary two nodes?

Note that this is not necessarily a parallel circuit, 2 nodes that we connect with a resistor are not the same 2 nodes between we want to measure equivalent resistance but completely arbitrary 2 nodes in passive network.

So imagine a circuit with 4 access points: A, B, C and D. We want to measure equivalent resistance between A and B. How can you prove that Rab will be less if we add a resistor between C and D?

I tried to search the web but didn't have any success.

$pi$, $sigma$ - atomic transitions with respect to the magnetic field axis

I am confused about the atomic transition with different polarized lights. I post the pictures as follows.

There are four cases. In case 1, the right-handed circular polarized light propagates along the $\mathbf{B}$ axis. So it induces the $\delta m=+1$ transition(i.e.,$\sigma$+ transition).

In case 2, the left-handed circular polarized light propagates along the $\mathbf{B}$ axis. So it induces the $\delta m=-1$ transition(i.e.,$\sigma$- transition).

In case 3, the light travels normal to the $\mathbf{B}$ axis and it's linear polarized. So it induces the $\delta m=0$ transition(i.e.,$\pi$ transition).

Question a: In case 4, what kind of transition will happen?

Question b: The linear polarized light can be decomposed into $\sigma$+ and $\sigma$- polarized light. If the linear-polarized light travels along the $\mathbf{B}$ axis, there will be $\sigma$+ and $\sigma$- transitions meanwhile. Is it right?

Answer

a: In case 4, what kind of transition will happen?

In this case, the radiation can excite both $\sigma$ and $\pi$ transitions, although you wouldn't really call them that. More specifically, it can excite transitions which keep $L_z$ constant, and it can also excite transitions which change it by $\Delta L_z=\pm \hbar$.

To see why, simply decompose the circular polarization into a superposition of linear polarizations: one along $z$, and one along (say) $x$. The polarization along $z$ can excite the $\Delta L_z=0$ transitions. The linear polarization along $x$, on the other hand, is itself a superposition of circular polarizations with a $\vec k$ vector along $z$, and each of these excites one $|\Delta L_z|=1$ transition. (Why am I allowed to switch the wave propagation direction? This is all done inside the dipole approximation, which takes the polarization to be uniform, so the atom doesn't actually 'talk' to the propagation direction.)

To do this slightly more rigorously, the transition operator for case 4 is $\hat z+i\hat x$, and it can be decomposed as $$ \hat z+i\hat x=\hat z+i\left(\frac{\hat x+i\hat y}{2}+\frac{\hat x-i\hat y}{2}\right). $$ The overall matrix element for the transition is therefore $$ \left\langle f\middle |\hat z+i\hat x\middle |i\right\rangle = \left\langle f\middle |\hat z\middle |i\right\rangle +i\left\langle f\middle |\frac{\hat x+i\hat y}{2}\middle | i\right\rangle +i\left\langle f\middle |\frac{\hat x-i\hat y}{2}\middle| i\right\rangle. $$ Here $\pi$ transitions are those for which $\left\langle f\middle |\hat z\middle |i\right\rangle$ is nonzero, and $\sigma_\pm$ are those for which $\left\langle f\middle |\frac{\hat x\pm i\hat y}{2}\middle | i\right\rangle$ is nonzero. (Don't trust me on the sign conventions, though.)

This then answers your second question pretty directly,

b: The linear polarized light can be decomposed into $\sigma$+ and $\sigma$- polarized light. If the linear-polarized light travels along the $\mathbf{B}$ axis, there will be $\sigma$+ and $\sigma$- transitions meanwhile. Is it right?

From the above simply take $$ \left\langle f\middle |\hat x\middle |i\right\rangle = \left\langle f\middle |\frac{\hat x+i\hat y}{2}\middle | i\right\rangle +\left\langle f\middle |\frac{\hat x-i\hat y}{2}\middle| i\right\rangle. $$ and you get your claim.