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Having a bit of trouble applying what I know about tensor manipulation, given,

$T^{\mu \nu} = \left( g^{\mu \nu} - \frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n} \right)$,

I need to compute quantities such as

(i) $T^{\mu \nu}T_{\mu \nu}$, (ii) $T^\mu{}_\nu$ and (iii) $ T^\mu{}_\mu $.

Start with (iii)

$ T^\mu{}_\mu = g_{\mu\nu}T^{\mu\nu}$ I don't think this can be correct because both indices appear twice. Just by inspection I would guess that

$T^\mu{}_\mu = \left( g^{\mu}{}_\mu - \frac{p^\mu n_\mu + p_\mu n^\mu}{p \cdot n} \right)$

$g^\mu{}_\mu =2$ here I summed all the diagonal terms

$\frac{p^\mu n_\mu + p_\mu n^\mu}{p^{\mu} n_\mu} $= $\frac{2 p\cdot n}{p \cdot n} = 2$

$T^\mu{}_\mu = 0$? If this is the case then I get the feeling that if the indices are doing things like that, i.e., if you have repeated indices on the tensor, then you will end up with a scalar - and maybe that that scalar will be zero.

(ii) $T^\mu{}_\nu = g_{\nu\sigma}T^{\mu\sigma}=g_{\nu\sigma}\left(g^{\mu\sigma} - \frac{p^\mu n^\sigma + p^\sigma n^\mu}{p\cdot n}\right)=\left(g_{\nu\sigma}g^{\mu\sigma} - \frac{p^\mu n_\nu + p_\nu n^\mu}{p \cdot n}\right)$

I think that $g_{\nu\sigma}g^{\mu\sigma} = \delta^\nu_\mu$ and $\frac{p^\mu n_\nu + p_\nu n^\mu}{p \cdot n}$ I think, is simplified as far as it can go.

(i) I wanted to start with $T_{\mu\nu}T^{\mu\nu}=T^{\mu\nu}g_{\mu\sigma}g_{\nu\rho}T^{\sigma\rho}$ but I can sort of see that this will, firstly, immediately be:

$\left( g^{\mu \nu} - \frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n} \right)\left( g_{\mu \nu} - \frac{p_\mu n_\nu + p_\nu n_\mu}{p \cdot n} \right)$.

And also that there are tricks involved in getting the answer quickly, but if I multiply out the terms I get

$\left( g^{\mu \nu}g_{\mu \nu} - g_{\mu \nu}\frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n} -g^{\mu \nu}\frac{p_\mu n_\nu + p_\nu n_\mu}{p \cdot n} + \frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n}\frac{p_\mu n_\nu + p_\nu n_\mu}{p \cdot n}\right)$

Here's where my understanding breaks down, how do I evaluate

$g_{\mu\nu} p^\mu n^\nu$ in terms of raising and lowering, which index do I choose? It must be a scalar, since it is a double sum. $g_{\mu\nu} p^\mu n^\nu = p \cdot n$.

$\left( g^{\mu \nu}g_{\mu \nu} - g_{\mu \nu}\frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n} -g^{\mu \nu}\frac{p_\mu n_\nu + p_\nu n_\mu}{p \cdot n} + \frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n}\frac{p_\mu n_\nu + p_\nu n_\mu}{p \cdot n}\right)=\delta_\mu^\nu -2-2+ 4 \frac{(p\cdot n)^2}{(p \cdot n)^2} = \delta_\mu^\nu $

Which seems like it could be the right answer, but is it?

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Start with (iii) $ T^\mu{}_\mu = g_{\mu\nu}T^{\mu\nu}$ I don't think this can be correct because both indices appear twice.

What's wrong with $ g_{\mu\nu}T^{\mu\nu}$? Both indices are contracted. Explicitly it means

$$ \sum_{\mu=0}^3\sum_{\nu=0}^3 g_{\mu\nu}T^{\mu\nu}$$

which is a perfectly good scalar.

$g^\mu{}_\mu =2$ here I summed all the diagonal terms

What dimension spacetime are you working in? The general result is

$$ g^\mu_\mu \equiv g_{\mu\nu} g^{\mu\nu} = \delta^\mu_\mu = D, $$

which equals 4 in 4 dimensions.

if you have repeated indices on the tensor, then you will end up with a scalar - and maybe that that scalar will be zero.

Contracting indices on a tensor does not necessarily give you a scalar and if it does that scalar is not necessarily zero. An important example is given by the contractions of the Riemann tensor:

$$ \begin{array}{lcl} R_{\mu\nu} &\equiv& R^\alpha_{\ \mu\alpha\nu} \\ R &\equiv& R^\mu_\mu = g^{\mu\nu} R_{\mu\nu} = g^{\mu\nu} R^\alpha_{\ \mu\alpha\nu} \end{array}$$

neither of which necessarily vanish. (They are physically important because they enter into Einstein's field equation of general relativity: $ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 8\pi G T_{\mu\nu}$).

I think that $g_{\nu\sigma}g^{\mu\sigma} = \delta^\nu_\mu$ and $\frac{p^\mu n_\nu + p_\nu n^\mu}{p \cdot n}$ I think, is simplified as far as it can go.

I think you mean $\delta^\mu_\nu$, which is numerically equal to $\delta^\nu_\mu$, but still different. :) Otherwise you're correct.

Here's where my understanding breaks down, how do I evaluate

$g_{\mu\nu} p^\mu n^\nu$ in terms of raising and lowering, which index do I choose? It must be a scalar, since it is a double sum. $g_{\mu\nu} p^\mu n^\nu = p \cdot n$.

Either index - you get the same answer both ways. Your answer is right: $p\cdot n$.

$\left( g^{\mu \nu}g_{\mu \nu} - g_{\mu \nu}\frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n} -g^{\mu \nu}\frac{p_\mu n_\nu + p_\nu n_\mu}{p \cdot n} + \frac{p^\mu n^\nu + p^\nu n^\mu}{p \cdot n}\frac{p_\mu n_\nu + p_\nu n_\mu}{p \cdot n}\right)=\delta_\mu^\nu -2-2+ 4 \frac{(p\cdot n)^2}{(p \cdot n)^2} = \delta_\mu^\nu $

There is an error in your first term: both indices are contracted, so it should be $\delta_\mu^\mu = D$. There is also an error in your last term. Expanding out:

$$ (p^\mu n^\nu + p^\nu n^\mu)(p_\mu n_\nu + p_\nu n_\mu) = p^\mu n^\nu p_\mu n_\nu +p^\mu n^\nu p_\nu n_\mu+p^\mu n^\nu p_\mu n_\nu +p^\nu n^\mu p_\nu n_\mu. $$

If you look carefully you'll see that you get $(p\cdot n)^2$ terms but also $ p^2 n^2 $ terms as well.

Which seems like it could be the right answer, but is it?

No, because the expression should be a scalar and you have a tensor with uncontracted indices.

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does the position of the index matter? Is there a distinction between $g_\mu^\nu$, $g^\nu{}_\mu$ and $g^\mu{}_\nu$ for any tensor? I can see where I have overlooked the multiplication of the very last term you mentioned now, that makes sense. But with $g^\mu{}_\mu$ I just thought we would have $-1+1+1+1$, but actually we have $g_{\mu\nu}g^{\mu\nu}$ is this $ \left( \begin{array}{c} -1 \\ 1 \\ 1 \\ 1 \end{array} \right)\left( \begin{array}{cccc} -1 & 1 & 1 & 1 \end{array} \right) $? –  shilov Mar 14 '13 at 10:40
    
In general there is a difference between say $A_{\ \ \mu}^\nu$ and $A_\mu^{\ \ \nu}$ (defined by which index you raise), but if the tensor is symmetric $A_{\mu\nu}=A_{\nu\mu}$ you don't need to worry about position. To raise one of the indices on $g_{\mu\nu}$ requires the inverse metric. Is it clear if it write $g^\mu_\nu = g^{\mu\rho} g_{\rho\nu}$ and then set $\mu=\nu$ and sum? –  Michael Brown Mar 14 '13 at 11:39
    
Thank you for the help –  shilov Mar 25 '13 at 15:50
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