Recent content by antibrane

Not quite, that is a lot more work. You don't have to actually find the generator to determine what type it should be. Say, for the sake of argument, the generator F is a function of p_x,y,X,Y (it's not) so that F=F(p_x,y,X,Y). This means that you would be able to write the other coordinates...

If you were integrating just a line of charge from 0 to a then you are adding up infinitesimal lengths dr. As you said earlier in your description, a disc of charge is like many rings, with infinitesimal width dr, stacked up from 0 to a. Thea area of such a ring would just be the perimeter of...

Yea, it is zero by symmetry. The integrand is odd (it obeys f(-x)=-f(x)) so the "negative area" accumulated from -a to 0 cancels out the "positive area" accumulated from 0 to a.

Looking at it, I get: \log \left[\sqrt{i}\right] = \log\left\{\exp\left[\frac{i}{2}\left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = \frac{i}{2}\left(\frac{\pi}{2} + 2\pi n\right)\log e = i\left(\frac{\pi}{4} + \pi n\right) where n is an integer. It looks like you might have used a period of...

Looking at Equation (24), \left\langle n^{(0)} \right.\left|n^{(1)}\right\rangle = -\frac{1}{2} \sum_{k=1}^{1-1}\left\langle n^{(1-k)}\right.\left| n^{(k)}\right\rangle = 0 so the \left\langle n^{(0)} \right.\left|n^{(1)}\right\rangle term in what you wrote is zero.

Yes, you are allowed to invert both sides of an equation. In general, if the same variables are held constant, you can do this. I'll show you what I mean; take the total derivatives of your functions, and assume that they are all functions of each other for generality: \begin{align} dx &=...

When I look at this in Mathematica I get a derivative of the delta function, in other words: \mathcal{L}^{-1}\left\{s\right\} = \frac{d}{dt}\delta(t)

It sounds like your confusion has to do with computing the cross product d\mathbf{\vec{l}}\times \mathbf{\vec{r}}. I would suggest breaking the wire up into the three separate sections, for example, the integral for the longest part of the half-square would have d\mathbf{\vec{l}}=...

It is giving you a solution, y=\frac{1}{5x}\qquad z=\frac{4}{5x} In fact that is an infinite number of solutions, since x is just a free-parameter. You get this because not all of your equations are independent. In other words, one of the equations can be written as a linear combination...

The intensity of an EM wave is written in terms of the amplitude: I=cn \frac{\epsilon_0}{2}\left|E_0\right|^2 so these aren't really independent things.

When doing a change of coordinates with multiple variables, the way to find the general volume element dv_A dv_B in terms of the new variables is the Jacobian determinant; you can look at what I am referring to here. What you are really looking for is, dv_{r}dv_{c}=...

You could use Cramer's rule.

That's an interesting question, but in order for the cosmic microwave background to be damaging, considering that gamma-rays are at about 10^18 Hz and higher, using the Doppler shift: f' = f \sqrt{\frac{1+\beta}{1-\beta}} you can find that you would need (since the CMB is about 160...

Energy and power are indeed, as you thought, frame dependent quantities; something can be moving relative to me at one velocity in one frame and I say it has a different kinetic energy than it does when I am in another frame where it has a different relative velocity. In your problem, depending...

No that is not consistent. In base SI units, charge is give in amp-seconds and the other side of your equation would be kilograms per second.