Tensor Algebra: Checking {u^i} = {g^{kj}} A _{kj}^i

[redstone]

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 factor in there somewhere.

 

[dextercioby]

The 3rd equation is wrong in the sense that it doesn't imply the 2nd. In the 2nd equation both a and j are summed over, so to get there you need to sum/contract the 3rd one with g_{a}^{j}. When doing that, the RHS becomes the RHS from the 2nd equation, while in the first appears the term \delta_{j}^{j} which is V, the dimension of the tangent/cotangent space.

 

[redstone]

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...
\frac{g_j^a{u^i}}{V} = {g^{ka}} A _{kj}^i
\frac{g_a^j g_j^a{u^i}}{V} = g_a^j {g^{ka}} A _{kj}^i
\frac{\delta_j^j {u^i}}{V} = g_a^j {g^{ka}} A _{kj}^i
\frac{\delta_j^j {u^i}}{V} = {g^{kj}} A _{kj}^i
\frac{V {u^i}}{V} = {g^{kj}} A _{kj}^i
{u^i} = {g^{kj}} A _{kj}^i

So I guess the rule is that whenever you move that mixed metric to the other side, you swap indices, and put in that V factor.

Unless I did something wrong here that needs to be corrected, thanks for the help!

 

[dextercioby]

Yes, it's correct. Just pay attention with the indices, so you don't end up with more than 2 appearances of the same index on the same side of an equality.

 

[redstone]

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}}

Step 2: \frac{1}{V}\frac{1}{V}g_{a}^{i}g_{b}^{j}A={{g}_{ab}}{{x}^{i}}{{x}^{j}}

Step 3: \frac{1}{V}\frac{1}{V}{{g}^{ab}}g_{a}^{i}g_{b}^{j}A={{g}^{ab}}{{g}_{ab}}{{x}^{i}}{{x}^{j}}

Step 4: \frac{1}{V}\frac{1}{V}{{g}^{ab}}g_{a}^{i}g_{b}^{j}A=V{{x}^{i}}{{x}^{j}}

Step 5: \frac{1}{V}\frac{1}{V}\frac{1}{V}{{g}^{ij}}A={{x}^{i}}{{x}^{j}}

I believe the correct answer is:
\frac{1}{V}{{g}^{ij}}A={{x}^{i}}{{x}^{j}}

but I'd like to know where I went wrong above.

 

[fzero]

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}}

Step 2: \frac{1}{V}\frac{1}{V}g_{a}^{i}g_{b}^{j}A={{g}_{ab}}{{x}^{i}}{{x}^{j}}

Step 2 is incorrect. In the expression for step 1, you're summing over all indices, so there's no operation that brings factors of the metric over to the LHS. If you can't reduce a step to multiplying left and right-hand sides of an equation by the same factor, chances are that you did something wrong.

 

[redstone]

Step 2 is incorrect. In the expression for step 1, you're summing over all indices, so there's no operation that brings factors of the metric over to the LHS. If you can't reduce a step to multiplying left and right-hand sides of an equation by the same factor, chances are that you did something wrong.

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: \frac{1}{V}g_{a}^{a}A=g_{i}^{a}g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}
Step 1c: \frac{1}{V}g_{i}^{a}g_{a}^{i}A=g_{i}^{a}g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}
Step 1d: \frac{1}{V}g_{a}^{i}A=g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}

Step 1e: \frac{1}{V}\frac{1}{V}g_{b}^{b}g_{a}^{i}A=g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}
Step 1f: \frac{1}{V}\frac{1}{V}g_{b}^{j}g_{j}^{b}g_{a}^{i}A=g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}
Step 2: \frac{1}{V}\frac{1}{V}g_{a}^{i}g_{b}^{j}A={{g}_{ab}}{{x}^{i}}{{x}^{j}}

Is one of these steps illegal?

 

[fzero]

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: \frac{1}{V}g_{a}^{a}A=g_{i}^{a}g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}

You already have an index a on the RHS that's being summed over, so you shouldn't use the same index in the sum on the LHS. That's the source of your confusion. You should write:

\frac{1}{V}g_{c}^{c}A=g_{i}^{a}g_{j}^{b}{{g}_{ab}}{{x}^{i}}{{x}^{j}}

and then it's clear that you can't cancel g's out on both sides.

You make the same mistake in some of the other steps. Just never use the same dummy index twice and it'll cut down on mistakes.