# Quick one-line on lowering indices

Tags:
1. Mar 1, 2015

### unscientific

At low speeds and assuming pressure $P=0$,
$$T^{\alpha \beta} = \rho U^\alpha U^\beta$$
$$g_{\alpha \mu} g_{\gamma \beta} T^{\alpha \beta} = \rho g_{\alpha \mu} g_{\gamma \beta} U^\alpha U^\beta$$
$$T_{\gamma \mu} = \rho U_\mu U^\beta g_{\gamma \beta}$$

Setting $\gamma = \mu = 0$:

$$T_{00} = \rho U_0 U^\beta g_{0 \beta}$$

Since $g_{0 \beta} \backsimeq \eta_{0 \beta}$ and the only non-zero term is $\eta_{00} = -1$, combined with $U_\alpha U^\alpha = -c^2$:

$$T_{00} = \rho U_0 U^0 g_{00} = \rho c^2$$

I'm still learning tensor calculus, would that be considered a proper derivation?
Also, is $g_{ij} \backsimeq \eta_{ij}$ the reason why $T_{ij} \backsimeq 0$?

2. Mar 1, 2015

### Staff: Mentor

It's correct as far as it goes, but you left out two things.

First: why do you assume $P = 0$? The place you got this from only said a time-independent source at low speeds; it didn't say zero pressure. What eliminates the pressure terms?

Second: how do you know that $U_0 U^0 = U_{\alpha} U^{\alpha}$ for this case (which is what lets you substitute $- c^2$ for $U_0 U^0$)? In general, $U_{\alpha} U^{\alpha} = U_0 U^0 + U_1 U^1 + U_2 U^2 + U_3 U^3$. What eliminates the other three terms?

No. The correct reason is connected to the questions I asked above.

3. Mar 2, 2015

### unscientific

Kinetic energy $\propto k_BT$ is much less than rest mass energy

$$P \backsimeq n k_B T << nMc^2 \backsimeq \rho c^2$$
So $P \backsimeq 0$.

$U_\alpha U^\alpha = -c^2$ because of invariance when you evalue $U = (c,0)$ in the rest frame.

4. Mar 2, 2015

### ChrisVer

I also have a question, why do you take $g_{\mu \nu} \approx \eta_{\mu \nu}$?

5. Mar 2, 2015

### unscientific

In the weak field limit of gravity, $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ where $|h_{\mu \nu} << 1$.

6. Mar 2, 2015

### ChrisVer

aha, so you take the weak field limit...because I didn't see that in written in the text you attached. I would just say that in the low speed limit:
$U^a = \begin{pmatrix} c \\ u_x \\ u_y \\ u_z \end{pmatrix} \approx c \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$
and this gives immediately the components of $T^{ab}$

7. Mar 2, 2015

### Staff: Mentor

Not in all cases. I was asking what makes it true in this case.

I wasn't asking why $U_\alpha U^\alpha = -c^2$. I was asking why $U_0 U^0 = - c^2$, i.e., why $U_0 U^0 = U_\alpha U^\alpha$. I think ChrisVer gave the answer to that, though.

Small nitpick: in the low speed limit, and in the rest frame of the fluid...

8. Mar 2, 2015

### ChrisVer

wouldn't you agree that in the low speed limit this approximation holds (I just used $u_i/c \approx 0$)? The thing is that the text says "taking the low speed limit" (together with time independence of the source)

9. Mar 2, 2015

### Staff: Mentor

Sure, but "low speed" requires a choice of frame; "speed" is relative. I was only pointing out that the implicit choice of frame being made is the rest frame of the fluid. I said it was a small nitpick.