# Energy-Momentum Tensor Algebra

Tags:
1. Feb 15, 2015

### unscientific

1. The problem statement, all variables and given/known data

(a) Show acceleration is perpendicular to velocity
(b)Show the following relations
(c) Show the continuity equation
(d) Show if P = 0 geodesics obey:

2. Relevant equations

3. The attempt at a solution

Part (a)

$$U_{\mu}A^{\mu} = U_{\mu}U^v \left[ \partial_v U^{\mu} + \Gamma_{v\epsilon}^{\mu} U^{\epsilon} \right]$$
$$= \delta_\mu^{v} \left[ \partial_v U^{\mu} + \Gamma_{v\epsilon}^{\mu} U^{\epsilon} \right]$$
$$= \partial_{\mu} U^{\mu} + \Gamma_{\mu \epsilon}^{\mu} U^{\epsilon}$$
$$= \partial_{\mu} U^{\mu} + \Gamma_{\mu mu}^{\mu} U^{\mu}$$
$$= \partial_{\mu} U^{\mu} + \left[ \frac{1}{2} g^{\mu \mu} (\partial_{\mu} g_{\mu \mu} + \partial_{\mu} g_{\mu \mu} - \partial_\mu g_{\mu\mu} ) \right] U^{\mu}$$
$$= -\partial_{\mu} U^{\mu} - g^{\mu\mu} \partial_{\mu} g_{\mu\mu} V^{\mu}$$

How do I show it equals 0?

2. Feb 16, 2015

### Orodruin

Staff Emeritus
Your second step is not correct. $U_\mu U^\nu \neq \delta^\nu_\mu$.

3. Feb 16, 2015

### unscientific

Why is that wrong? I thought it is normalized such that $U_\nu U^\nu = -1$.

4. Feb 16, 2015

### Orodruin

Staff Emeritus
Yes, which does not imply $U_\mu U^\nu = \delta^\nu_\mu$. In fact, if it was true, then $U_\nu U^\nu = \delta^\nu_\nu = 4$.

5. Feb 16, 2015

### unscientific

Then $U_\mu U^\nu = -\frac{\delta_\mu^\nu}{4}$ to make it work? Even then, I don't see how $= -\frac{ \left( \partial_{\mu} U^{\mu} + g^{\mu\mu} \partial_{\mu} g_{\mu\mu} V^{\mu} \right) }{4}$ equals to 0..

6. Feb 16, 2015

### Orodruin

Staff Emeritus
No, you still cannot draw this conclusion and you will have to find another way around. Think of a 4-velocity in SR and express it in the rest frame: $U^\mu = (1,0,0,0)$. Now, obviously, $U_2 U^2 = 0$, which would not be true if $U_\mu U^\nu = - \delta^\nu_\mu/4$. Clearly the only non-zero component of this tensor is the 00 component. This has a straight forward generalisation to GR.

I suggest you instead try to think of how you wokld prove this in SR and try to generalise the result to GR.

7. Feb 16, 2015

### unscientific

I can do it in SR, but that's not the point of this exercise. I need to work with the form I'm given: $U_\mu A^\mu = U_\mu U^v\nabla_v U^\mu$ and show it is equal to 0..

8. Feb 16, 2015

### Orodruin

Staff Emeritus
My point is that you can do it exactly in the same way in GR, with some generalisations required due to no longer having simply $U^\nu \partial_\nu$, but instead having $U^\nu \nabla_\nu$. This is why I suggest that you first recall how you do it in SR.

9. Feb 16, 2015

### unscientific

Ok, let's try it here.

The 4-velocity is given by $U = \gamma(c, \vec v)$. In the rest frame, $U' = c(1,0)$.
The 4-acceleration is given by $\frac{dU}{d\tau}$. In rest frame, $A' = (0,\vec a_0)$.

Using invariance, $U \cdot A = U' \cdot A' = 0$.

10. Feb 16, 2015

### Orodruin

Staff Emeritus
Ok, that is cheating a bit. You have essentially used that $A$ is orthogonal to $U$ when writing it down in the rest frame in order to show that $A$ is orthogonal to $U$.

How about instead trying to show it using the definition of the 4-acceleration, i.e., show that
$$U \cdot A = U \cdot \frac{dU}{d\tau} = 0?$$

11. Feb 16, 2015

### unscientific

$$\frac{d}{d\tau} (U \cdot U) = \frac{d}{d\tau}( -1) = 0$$

12. Feb 16, 2015

### unscientific

I think I got a quick way to show it:

$$\frac{D}{D \tau} (U \cdot U) = 0$$
$$2U \cdot \frac{D}{D\tau} U = 0$$
$$U_\mu U^{\alpha} \nabla_{\alpha} U^{\mu} = 0$$

13. Feb 16, 2015

### unscientific

I tried to do part (c), but I feel like I'm missing something:

$$\nabla_\mu \left[ (\rho + P) U^\mu U^v + g^{\mu v} P \right] = 0$$

Using $\nabla_\mu g^{\mu v} = 0$ and $\nabla_\mu U^\mu = 0$:

$$\nabla_\mu \left[ (\rho + P) U^\mu U^v \right] + g^{\mu v} \nabla_\mu P = 0$$

$$U^\mu U^v\nabla_\mu (\rho + P) + (\rho + P)U^\mu \nabla_\mu U^v + g^{\mu v} \nabla_\mu P = 0$$

Multiplying $U_v$ throughout on the left to cancel out the middle term:

$$U^\mu (U_v U^v)\nabla_\mu (\rho + P) + g^{\mu v}U_v \nabla_\mu P = 0$$

$$-U^\mu \nabla_\mu (\rho + P) + g^{\mu v}U_v \nabla_\mu P = 0$$

Last edited: Feb 16, 2015
14. Feb 17, 2015

### Orodruin

Staff Emeritus
Who says $\nabla_\mu U^\mu =0$? If this was true there would not be much sense in having such a term in the expression you are supposed to derive (and if you assume it you are going to miss this term). Apart from that it looks like you are done already.