Recent content by redstone

I guess nobody seems to know. So for posterity... My own research into it, it appears to not be necessary for the metric itself to work, but when solving for the stress energy tensor, it looks like it makes terms, at least the T00 term, simpler, since there are d/dt terms that act on the x...

Reading over Alcubierre's paper on his "warp" drive (http://arxiv.org/abs/gr-qc/0009013), the metric in equation 3 has a velocity term, v, that doesn't seem to be needed anywhere. Even in the one spot where it seems potentially valuable, equation 12, he just call it =1 and essentially ignores...

I'm trying to understand what kind of relation the metric can have with a general tensor B. d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}=d{{s}^{2}} \frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=1 \frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=\frac{1}{D}g_{a}^{a}...

given the vector in the first equation below, does that necessarily imply the third equation, as shown? {{u}_{a}}{{e}^{a}}={{x}_{a}}{{e}^{a}} {{u}_{a}}{{e}^{l}}g_{l}^{a}={{x}_{a}}{{e}^{l}}g_{l}^{a} {{u}_{a}}{{e}^{l}}={{x}_{a}}{{e}^{l}}

Ah, yes, of course. Thank you.

If you know that {{x}^{a}}{{g}_{ab}}={{x}_{b}} is it proper to say that you also know {{g}_{ab}}=\frac{{{x}_{b}}}{{{x}^{a}}}

Makes sense. thanks for all the help

I've read that the metric tensor is defined as {{g}^{ab}}={{e}^{a}}\cdot {{e}^{b}} so does that imply that? {{g}^{ab}}{{g}_{cd}}={{e}^{a}}{{e}^{b}}{{e}_{c}}{{e}_{d}}={{e}^{a}}{{e}_{c}}{{e}^{b}}{{e}_{d}}=g_{c}^{a}g_{d}^{b}

Looking for a check on my tensor math to make sure I've done this correctly... Where D equals the dimension of the metric - Step 0: {{A}^{ab}}=\frac{1}{D}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}} Step 1: {{g}_{ab}}{{A}^{ab}}={{g}_{ab}}\frac{1}{D}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}} Step 2...

I left out the steps going between step 1 and step 2, but that's what I thought I did. Here are the steps I had. Step 1: A=g_{i}^{a}g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}} Step 1a: \frac{V}{V}A=g_{i}^{a}g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}} Step 1b...

Maybe you can help me with the following. I I'm pretty sure the final answer I get is wrong, but every step I took looks reasonable to me. Do you know which step I did something illegal on? Start: A={{g}_{ij}}{{x}^{i}}{{x}^{j}}\] Step 1: A=g_{i}^{a}g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}...

Ah, yes, I think I see. That does give me a missing V factor. So would you consider this correct then: {u^i} = {g^{kj}} A _{kj}^i {u^i} = g_a^j{g^{ka}} A _{kj}^i \frac{g_j^a{u^i}}{V} = {g^{ka}} A _{kj}^i where, V=g_j^j=\delta_j^j Which when run in reverse...

Homework Statement {u^i} = {g^{kj}} A _{kj}^i just trying to modify it, not sure of my tensor algebra. Is this right? {u^i} = {g^{kj}} A _{kj}^i {u^i} = g_a^j{g^{ka}} A _{kj}^i g_j^a{u^i} = {g^{ka}} A _{kj}^i Just not sure if there should have been a metric contraction, with the resulting D...

This confuses me a bit. The only difference between this step and step 2, is that here the RHS includes g^e_a and g^a_e which I thought were both just identity matrices? so doesn't that mean it should be the same as step 2?

OK, so I guess there's something wrong with the following too then? 1) g_{ab}A^{ab}=g_{cd}A^{cd} 2) g_{ab}g^a_eA^{eb}=g_{cd}A^{cd} 3) g_{ab}g^a_eA^{eb}=g^e_ag^a_eg_{cd}A^{cd} 4) g_{ab}A^{eb}=g^e_ag_{cd}A^{cd} 5) g^f_bg_{af}A^{eb}=g^e_ag_{cd}A^{cd} 6)...