# Homework Help: [General Relativity] Prove that a tensor is a co-tensor

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1. Oct 6, 2014

### mef51

Hello! I'd appreciate any help or pokes in the right direction.

1. The problem statement, all variables and given/known data
Show that a co-tensor of rank 2, $T_{\mu\nu}$, is obtained from the tensor of rank 2 $T^{\mu\nu}$ by using a metric to lower the indices:
$$T_{\mu\nu} = g_{\mu\alpha}g_{\nu\beta}T^{\alpha\beta}$$

2. Relevant equations
I know that a vector is a covector if its components transform from one frame to another as:
$$B_\alpha '= \frac {\partial x^\beta}{\partial x'^\alpha } B_\beta$$

3. The attempt at a solution
Analogous to a co-vector, I figure that a tensor of rank 2 is a co-tensor if
$$T_{\alpha\beta} '= \frac {\partial x^\mu}{\partial x'^\alpha } \frac {\partial x^\nu}{\partial x'^\beta } T_{\mu\nu}$$

So I'll start on the right-side and try to simplify it to the left side...
$$RS = \frac {\partial x^\mu}{\partial x'^\alpha } \frac {\partial x^\nu}{\partial x'^\beta } T_{\mu\nu} = \frac {\partial x^\mu}{\partial x'^\alpha } \frac {\partial x^\nu}{\partial x'^\beta } g_{\mu\delta} g_{\nu\epsilon} T^{\delta\epsilon}$$
Now I'll substitute in the Jacobians...
$$= ({\Lambda^{-1}})^\mu_\alpha ({\Lambda^{-1}})^\nu_\beta g_{\mu\delta} g_{\nu\epsilon} T^{\delta\epsilon}$$

And I'm not sure where to go from here. I suspect I can contract the jacobians with the metric tensors but I'm not sure how to handle that.

Thanks,
mef

2. Oct 6, 2014

### Matterwave

So I think what the problem is asking you is to show that $T_{\mu\nu}\equiv g_{\mu\alpha}g_{\nu\beta}T^{\alpha\beta}$ defined in this way is in fact a co-vector. Is that your interpretation of the problem as well?

In that case, by changing a coordinate system, we should have still $T_{\mu'\nu'} = g_{\mu'\alpha'}g_{\nu'\beta'}T^{\alpha'\beta'}$. What does the right hand side of this equation look like in terms of un-primed indices?

3. Oct 6, 2014

### mef51

Ok so then we can express the right-side in terms of the unprimed terms using
$$g_{\mu'\alpha'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\alpha}^{\epsilon}g_{\delta\epsilon}$$
$$g_{\nu'\beta'}=(\Lambda^{-1})_{\nu}^{\eta}(\Lambda^{-1})_{\beta}^{\kappa}g_{\eta\kappa}$$
And we'll get
$$T_{\mu'\nu'}=g_{\mu'\alpha'}g_{\nu'\beta'}T^{\alpha'\beta'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\alpha}^{\epsilon}(\Lambda^{-1})_{\nu}^{\eta}(\Lambda^{-1})_{\beta}^{\kappa}g_{\delta\epsilon}g_{\eta\kappa}T^{\alpha'\beta'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}g_{\delta\epsilon}g_{\eta\kappa}(\Lambda^{-1})_{\alpha}^{\epsilon}(\Lambda^{-1})_{\beta}^{\kappa}T^{\alpha'\beta'}$$ $$=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}g_{\delta\epsilon}g_{\eta\kappa}T^{\epsilon\kappa}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}T_{\delta\eta}$$

So we see that $T_{\mu'\nu'}$ transforms as $T_{\mu'\nu'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}T_{\delta\eta}$ which means that it's a co tensor!

4. Oct 6, 2014

### Matterwave

Some of your indices should still be primed. Each $(\Lambda^{-1})^\mu_{~~\nu'}$ should have a prime on the bottom index (or else they wouldn't sum over with the T indices in the 3rd and 4th parts of the expression, and they wouldn't match the primped indices on the left hand side of the equation),. Also, usually one would be a bit more general and just leave it in terms of $\frac{\partial x^\mu}{\partial x^{\nu'}}$ so that we can consider more than just Lorentz transformations. But other than that, it looks fine to me.

5. Oct 9, 2014

### mef51

You're right.
My prof has a convention where you only write the primes on the object instead of the indices (so $T_{\mu\nu}'$ instead of $T_{\mu'\nu'}$ and he also defined the transformation $\Lambda$ so that $\Lambda$ has a prime on the top and $\Lambda^{-1}$ has a prime on the bottom:
$\Lambda_{\nu}^{\mu}=\frac{\partial x'^{\mu}}{\partial x^{\nu}}$ and $(\Lambda^{-1})_{\nu}^{\mu}=\frac{\partial x^{\mu}}{\partial x'^{\nu}}$

I prefer priming the indices instead because then you don't have to remember that $\Lambda^{-1}$ has primes on the bottom and vice versa

Last edited: Oct 9, 2014