Is Acceleration Perpendicular to Velocity in Energy-Momentum Tensor Algebra?

[unscientific]

Homework Statement

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

Homework Equations

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?

 

[Orodruin]

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

 

[unscientific]

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

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

 

[Orodruin]

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

 

[unscientific]

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

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..

 

[Orodruin]

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..

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.

 

[unscientific]

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.

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..

 

[Orodruin]

I can do it in SR, but that's not the point of this exercise.

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.

 

[unscientific]

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.

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

 

[Orodruin]

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
$$
U \cdot A = U \cdot \frac{dU}{d\tau} = 0?
$$

 

[unscientific]

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
$$
U \cdot A = U \cdot \frac{dU}{d\tau} = 0?
$$

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

 

[unscientific]

I think I got a quick way to show it:

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
$$
U \cdot A = U \cdot \frac{dU}{d\tau} = 0?
$$

\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

 

[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

 

[Orodruin]

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.