OK, I didn't phrase that right. What I meant was that the real line has a discrete topology. Now I take the group of translations on it. My question was whether this group would be any different from the group of translations on a (real line with usual topology) and I think that they would be...
Hi.
I wanted to know in what way the group of translations on a real line with discrete topology (let's call it Td) will be different from the group of translations on a real line with the usual topology (lets call it Tu)? Is Td a Lie Group? Will it have the same generator as Tu?
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...
So you agree that you need to use another law.
The argument can be stated this way: Let us consider a system of particles. According to N1L none of the momentums of each particle remains constant unless forces act on it. Here the forces are all forces of interaction. So the total rate of...
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.
When you take torque, you take it about a point. First you have to find some point which you like (which can be any point, but you may want to choose something that makes the calculation simple.) Then you look how far the lines of forces are from this point - you drop perpendiculars to the lines...
The book uses g = 9.8 whereas you used 10.
You did well. There is also an equivalent way of doing it which may seem simpler : write down the force equation and the torque equation. For the system to be in equilibrium, both force and torque must be zero. Plug in the values and you're done.
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...