# Lowering tensor indices

1. Jan 27, 2013

Hi
have i got this corresct :
$g_{\mu\nu}g_{\mu\nu}T^{\mu\nu} = T g_{\mu\nu} = T_{\mu\nu}$
and is:
$g_{\mu\nu}g_{\mu\nu}u^{\mu} = g_{\mu\nu} u_{\nu} = u_{\mu}$

2. Jan 27, 2013

### Mentz114

No, it should be

gapgqbTpq = Tab

gapvp = va

3. Jan 27, 2013

how would i go about going from $T^{ab}$ to $T_{ab}$ i.e with the same indices that is mainly why i am confused it needs to be the same indices not different ones.

4. Jan 27, 2013

### WannabeNewton

$g_{\gamma \mu }g_{\delta \nu }T^{\mu \nu } = T_{\gamma \delta }$ and then you are free to relabel the indices back to mu and nu. What you wrote is incorrect because you have 2 mu's on the bottom and a mu on the top which isn't how einstein summation works.

5. Jan 27, 2013

ok i think i understand so to lower:
$T^{\mu\nu} = u^{\nu}u^{\mu}$

would one multiply by $g_{\gamma \mu}g_{\delta \nu}$ and relable T as instructed, but how would the u's become:
$u_{\mu}u_{\nu}$

I get the relabelling thing but wouldn't they have three lower indices i.e:
$u_{\mu \gamma \delta} u_{\nu \gamma \delta}$

6. Jan 27, 2013

### WannabeNewton

$g_{\gamma \mu }g_{\delta \nu }T^{\mu \nu } = g_{\gamma \mu }g_{\delta \nu }u^{\mu }u^{\nu }$ so $T_{\gamma \delta } = u_{\gamma }u_{\delta }$ and since these are free indices I can relabel them accordingly that is $\gamma \rightarrow \mu , \delta \rightarrow \nu$ provided I do so on both sides; this gets the desired result.

7. Jan 27, 2013

thanks you have been really helpful one last question, is $T_{ab} g^{ab}= T^{a}_{b}$ if not how can we get there becuase the metric always has two indices?

8. Jan 27, 2013

### WannabeNewton

$T_{ab} g^{ab} = T_{a}^{a}$ because the metric tensor will first raise the lower index b on $T_{ab}$ up to an a and now you just have $T_{a}^{a} = T$ where T is just the trace. Note that here it is arbitrary whether you choose to raise a or b because in the end you end up summing over the same index on the bottom and on the top and you can relabel that to w\e you want; that is $T_{a}^{a} = T_{b}^{b}$ because you end up summing over all components regardless.

9. Jan 27, 2013

Yep anytime and to answer your final question you could just use $T_{ab}g^{bc} = T_{a}^{c}$