What I meant is this : suppose I have a function f(x,y). Can you write down the total derivative of f w.r.t x i.e df/dx in terms of the partial derivatives \partial_{x}f and \partial_{y}f ? If you can do that for f and x, go ahead and do it for rho and t.
If you aren't sure of which equation...
For part a, how'd you write the total time derivative of rho in terms of its partial derivatives?
For part b, what do you think the relevant equations are?
Turin, can you please tell me how you interpret ' bounded region of each spacelike hypersurface x^0 = const.
Does it mean that x^0 = const. are closed hypersurfaces ? Or is it something else?
I don't like this solution either. For one, it does not really use the formula mentioned. The only way out is to say that the surface isn't closed. But if that's not what Schutz meant, what did he mean by that sentence?
If it vanishes, that means it's a constant and independent of x^0
In your first post you did not use the fact that we are dealing here with a closed hypersurface, which is what Schutz seems to be implying.
Maria, I suspect this question would be better answered were you to post it in the Solid State forum. After all, this is not a homework question. Anyways, here's what I can tell you:
We do know that in solids ions are periodically arranged, and we can expect that their electrostatic potentials...
I checked the question on Schutz and it says that the stress energy tensor is non zero 'only in some bounded region of each spacelike hypersurface x^0 = const. '. I don't understand what that's supposed to mean. Does it mean we're working with closed surfaces and Gauss' law applies. What do you...
Walia, your derivation seems correct to me. I can't think of a case of your derivation being invalid except for the metric being non-symmetric. But I don't know if the metric can be non-symmetric at all.
Oh. Your previous expression was actually incorrect. The last 2 terms in the rhs should have contained single derivatives of both rho and phi, and not double derivativs (compare with the expression prior to that).
Substitute that in your original expression and see. Incidentally you won't get back the original term ( the first expression you wrote) but the modified term for z tending to infinity. This is because you've calculated the field using that condition.
No. If you have a complex equation of the form
f+ ig = p +is where f,g,p,s are real, it follows that f=p and g=s. Here you have to equate the coefficients of the number i on both sides.
While evaluating the surface integral keep in mind that there is no current outside the cylinder, so all contributions to the integral come from inside the cylinder.
Suppose that the ring has a very small radius dr . Then b-a = dr. Can you manipulate the above equation (factorising it, for example) and get something in terms of dr.
I would suggest forgetting fancy conventions and working it out in some co-ordinate system with all x s and y s, or r s and thetas, whichever you like.
Think of the electrostatic case. An isolated charged body creates a field. Now if the field acts on this charged body itself, the isolated charged body begins to move. This violates conservation of momentum. Does this make things clearer?
No, it can't produce a force on itself for the same reason you can't pull yourself up on a bucket - the source can't move itself w/o some external force appearing.
Oops, that had been bad advice. My apologies, I should have paid more attention.
Here's some better advice : you want to take the curl of this \frac{\mathbf{J(r ^')}}{|\mathbf{r - r^'}|}
This is a vector multiplied by a scalar. So you got to use the product rule for curl in such...
Yes, but that field won't act on the source itself.
We know B is constant from the cylindrical symmetry of the situation. No matter where you stand on the circle, the situation looks just the same. You can see the same amount of current flowing through the wire, and at the same distance. So it...
I didn't mean the first term as in first term of the binomial expansion. Rather the first surviving term in the whole expression. If you had expanded further there would be other surviving terms, but we are saying that they for large z they are far too small compared to the z term we already...
Firstly, your topic name is a bit of a misnomer, isn't it? This doesn't have anything to do with either SR or 4-vectors, but perhaps you didn't know that.
Anyways here's a hint : we can write x \frac{dx}{dt} = \frac{1}{2} \frac {d}{dt} x^2
See if you can use it here.
It's not very clear to me what you are asking. The proton has a certain velocity relative to the lab which is given. You can work out the x and y components of that velocity. The other frame's velocity relative to the lab is given. This has only an x component. You know how to calculate relative...
There's some problem with this logic of putting v = 0 . If there's a current, it means charged particles are moving and have some velocity. But is there a magnetic field acting on this current?
I remembered that the metric was derived with some assumption about the field being slowly varying, in which case time derivative of phi is small compared to space derivatives. Check equation 8.46 of Schutz for instance.
On the third term on the rhs you got an extra rho which wasn't there in your previous steps. Also you can multiply the i s in the first term of the rhs to get -1.
Yes, it is. Only because a/z is such a small term you could ignore the higher order terms. The idea is to get the first term with 1/z and leave the rest, because for large z they would be too small compared to this term.
If you have done (ii) you can also work out (iii). The field...