Recent content by kyp4

Hi all, I posted this awhile back in the homework sections of the forums and received only one reply, which suggested that I post it here instead, though I understand that it belongs in the homework section. The fundamental problem is not this particular exercise but about integration of...

Thanks, I will try my luck over there.

Yes the radicals can make it confusing. Is it clear that \frac{\sqrt{\frac{\frac{5\pi}{2} + 2}{A}}}{\sqrt{\frac{A}{\frac{5\pi}{2} + 2}}} = \sqrt{\frac{\frac{5\pi}{2} + 2}{A}}\sqrt{\frac{\frac{5\pi}{2} + 2}{A}} = \frac{\frac{5\pi}{2} + 2}{A} and \frac{\sqrt{\frac{A}{\frac{5\pi}{2}...

You simply substitute: \frac{h}{r} = \frac{1}{\sqrt{\frac{A}{\frac{5\pi}{2} + 2}}}\left[\frac{A}{2}{\sqrt{\frac{\frac{5\pi}{2} + 2}{A}} - \frac{\pi}{4}\sqrt{\frac{A}{\frac{5\pi}{2} + 2}\right] = \frac{A}{2}\frac{\left(\frac{5\pi}{2}+2\right)}{A}-\frac{\pi}{4} = \pi + 1

Regarding the arbitrary variable c, it appears in C(r) as a coefficient of all the terms, that is C(r) = \frac{5\pi cr}{2} + 2cr + \frac{Ac}{r} and is canceled out when solving for r after setting the derivative to zero, arriving at the same result you got. Try it. You've got the...

Homework Statement The problem I am facing is 2.9 in Sean Carroll's book on general relativity (Geometry and Spacetime) I should note that I am not studying this formally and so a full solution would not be unwelcome, though I understand that forum policy understandably prohibits it. The...

I apologize for revealing the complete solution. I am new to the forums and did not realize this was against forum policy. Makes sense though. While setting c=1 works out, it is more proper to consider it as some arbitrary variable c and it will disappear if it turns out not to be relevant as...

Moderator's note: giving complete solutions to homework problems is against our forum guidelines. (Solution deleted.)

Yes, I think I see now. The components, in matrix form, of the resulting two-form would then be (d\theta \wedge d\phi)_{\mu\nu} = \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0 \end{array}\right] in the dx^\mu \otimes dx^\nu basis in...

Yes, I see that now. Thanks dx! Anyone have any insight as to the meaning of d\theta \wedge d\phi? I guess my confusion is what are the "components" of these basis dual vectors to which to apply the definition of the wedge product?

I am studying general relativity from Sean Carroll's text and I have a simple question about the wedge product. According to the text, the wedge product of two one-forms (dual vectors) is (A \wedge B)_{\mu\nu} = 2A_{[\mu}B_{\nu]} = A_\mu B_\nu - A_\nu B_\mu I understand the why the first two...

Yes I find the matrix notation simpler but I think when dealing with higher ranked tensors later, the matrix analogy becomes difficult if not impossible since, for example, you might be dealing with a 4D matrix (in the case of a (0, 4) tensor) and multiplying such things with other tensors and...

Yes, I understand perfectly now, I was summing over the row when I should have been summing over the column. Correcting this, the primed basis vectors line up perfectly with the primed coordinate system in the same way the unprimed basis vectors line up with the unprimed coordinate system and...

Yes, I see, though I'm not sure how it helps resolve the original problem.

I don't think the basis vectors technically have components in themselves (as they are abstract and invariant geometric entities) without reference to some coordinate system. A vector can be expressed through it's components with respect to some coordinate system and/or with respect to some...