# Elaborating on stress-energy tensor

1. Jan 5, 2010

### gnieddu

Hi,

I'm trying to work out the logic behind a statement I found in the GR book I'm currently studying. It says that from the conservation equation $$\nabla_aT^{ab}=0$$, one could deduce the following two equations:

$$(\varrho+p)\dot{u}^a = \nabla^ap - u^a\dot{p}$$

$$\dot{\varrho} + (\varrho+p)\nabla_au^a = 0$$

I've tried calculating $$\nabla_aT^{ab}$$ from:

$$T^{ab}=(\varrho+p)u^au^b - g^{ab}p$$

and the best I could make out of it are the following two lines:

$$\nabla_aT^{ab}=(\varrho+p)\nabla_a(u^au^b) + [\nabla_a(\varrho+p)]u^au^b -g^{ab}\nabla_ap =$$

$$=(\varrho+p)[u^b\nabla_au^a+u^a\nabla_au^b] + [\nabla_a(\varrho+p)]u^au^b - g^{ab}\nabla_ap$$

Now I'm stuck here, and, what's worse, I'm not even sure these two lines are correct...

Thanks to whoever could provide some light here!

2. Jan 5, 2010

### George Jones

Staff Emeritus
You're almost there. Rearranging your last line gives

$$\begin{equation*} \begin{split} 0 &= \nabla_aT^{ab}\\ &= \left(\varrho+p\right)\left[u^b\nabla_au^a+u^a\nabla_au^b\right] + \left[\nabla_a \left(\varrho+p\right)\right]u^au^b - g^{ab}\nabla_ap \\ &= \left(\varrho+p\right)u^a\nabla_au^b - \nabla^bp + \left(u^a\nabla_a p\right)\right]u^b + \left(u^a\nabla_a \varrho\right)\right]u^b + \left(\varrho+p\right)u^b\nabla_au^a \end{split} \end{equation*}$$

You also need to use both $1 = u_a u^a$ and its covariant derivative. Think about this, and, if you are still puzzled, ask for more hints.

Last edited: Jan 5, 2010
3. Jan 6, 2010

### George Jones

Staff Emeritus
Okay, I'm with you now. From another thread, I see that you're reading Luvigsen. I didn't have Ludvigsen home with me when I posted last night, but I'm looking at it now, and I see that you're trying to reproduce some stuff from page 70, which is before Ludvigsen introduces covariant derivatives.

Assume that we are working in flat Minkowski space. Look at the bottom of page 48, and set $u = dx/d\tau$. By the chain rule,

$$\frac{d}{d\tau} = \frac{dx^a}{d\tau}\frac{\partial}{\partial x^a} = u^a \frac{\partial}{\partial x^a} = u^a \nabla_a$$

along a streamline. Thus, $u^a \nabla_a$ can be replaced by an overdot.

Now,what do you get when you apply $u^a \nabla_a$ to $1 = u_b u^b$?

4. Jan 6, 2010

### gnieddu

Hi George,

just trying to follow you here. So since, as you say $u^a \nabla_a$ is equivalent to an overdot, applying it to $1 = u_b u^b$ should give:

$$0 = \dot{u}_bu^b + u_b\dot{u}^b$$

but it does not lead me too far. I was also trying to follow your previous hint (i.e. to use the fact that $$u_au^a=1$$). The idea I got was to "multiply" both sides of your re-arranged equation by $$u_a$$ (assuming this is legal -- I'm always a bit scary of doing such things), and get:

$$u_a[(\varrho+p)u^a\nabla_au^b - \nabla^bp + u^au^b\nabla_a\varrho + u^au^b\nabla_ap + (\varrho+p)u^b\nabla_au^a] = 0$$

$$(\varrho+p)u_au^a\nabla_au^b - u_a\nabla^bp + u_au^au^b\nabla_a\varrho + u_au^au^b\nabla_ap + (\varrho+p)u_au^b\nabla_au^a] = 0$$

$$(\varrho+p)\nabla_au^b - u_a\nabla^bp + u^b\nabla_a\varrho + u^b\nabla_ap + (\varrho+p)u_au^b\nabla_au^a] = 0$$

After this, I had the idea that $$u_a\nabla^bp=u^b\nabla_ap=\dot{p}$$ (and doing the same for rho), so that I could simplify the above expression to:

$$(\varrho+p)\nabla_au^b + \dot{\varrho} + (\varrho+p)u_au^b\nabla_au^a] = 0$$

which would lead to one of the two results in Ludvigsen's book, if I just could get rid of the last term before the equal sign...

I have the suspect that, since u is a 4-velocity, $$u_au^b=0$$, but I'm not so sure.

5. Jan 6, 2010

### George Jones

Staff Emeritus
This is correct, but the expression can be simplified.

$$0 = \dot{u}_bu^b + u_b\dot{u}^b = \dot{u}_bu^b + \dot{u}^bu_b = \dot{u}_bu^b + \dot{u}_bu^b = 2 \dot{u}_bu^b = 2 \dot{u}^bu_b$$
This definitely is the right idea, but notice that the index $a$ occurs three times in some terms, which is not legal. Try "multiplying" by $u_b$.
No, $u_au^b$ is a non-zero, second-rank tensor. Keep trying!

6. Jan 6, 2010

### gnieddu

I think I got it. If I "multiply" by $u_b$ as you suggest, I'm still able to get rid of the two $\dot{p}$ terms and the equation reduces to:

$$(\varrho+p)u^au_b\nabla_au^b+u^a\nabla_a\varrho+(\varrho+p)\nabla_au^a=0$$

Now the first term is zero, since:
$$(\varrho+p)u^au_b\nabla_au^b=(\varrho+p)u_bu^a\nabla_au^b=u_b\dot{u}^b=0$$

as per your other suggestion. The rest of the equation reduces to:

$$\dot{\varrho}+(\varrho+p)\nabla_au^a=0$$

which is one of the two results. And now I can use it and substitute in:

$$(\varrho+p)u^a\nabla_au^b - \nabla^bp + u^au^b\nabla_a\varrho + u^au^b\nabla_ap + (\varrho+p)u^b\nabla_au^a =$$
$$(\varrho+p)u^a\nabla_au^b-\nabla^bp+u^b[u^a\nabla_a\varrho+(\varrho+p)\nabla_au^a]+u^bu^a\nabla_ap=0$$

and get:
$$(\varrho+p)u^a\nabla_au^b-\nabla^bp+u^bu^a\nabla_ap=0$$
$$(\varrho+p)\dot{u}^b=\nabla^bp-u^b\dot{p}$$

which is the other result I was looking for.

I really hope I'm correct now. Thanks!

7. Jan 7, 2010

### George Jones

Staff Emeritus
Yup.

8. Jan 7, 2010

### Altabeh

Sorry to chime in unexpectedly, but I have a question about the above calculation. If we write $$u_b\dot{u}^b = \dot{(u_bu^b)} - \dot{u}_bu^b= - \dot{u}_bu^b$$, so isn't it false to put $$u_b\dot{u}^b=\dot{u}_bu^b$$?

Thanks
AB

9. Jan 7, 2010

### George Jones

Staff Emeritus
I'm going to assume that the derivative operator is metric-compatible, and I don't like using the overdots, so I'm going to switch to nabla notation.

$$\left( u^a \nabla_a u_b \right) u^b = \left( u^a \nabla_a g_{cb} u^c \right) u^b = \left( u^a \nabla_a u^c \right) g_{cb} u^b = \left( u^a \nabla_a u^c \right) u_c = u_b \left( u^a \nabla_a u^b \right)$$

$$u_b\dot{u}^b = \dot{(u_bu^b)} - \dot{u}_bu^b= - \dot{u}_bu^b.$$

Why not?

10. Jan 7, 2010

### Altabeh

I think that is not a contradiction iff $$u_b\dot{u}^b=0$$. But how can we get this?

AB

11. Jan 7, 2010

### gnieddu

Hi George,

while thinking about the issue introduced by Altabeth, there's another question that your reply above suggests. In the chain of equalities, you deal with $g_{cb}$ as if it were a constant, and so you can take it out of the scope of the nabla operator. This is what I normally do as well, but I've always this doubt buzzing in my head.

After all, $g_{cb}$ is a tensor (albeit a constant one), isn't it? Can you please comment on why we're allowed to treat it as a constant, and, perhaps more interestingly, whether this is allowed on every space geometry?

Thanks

Gianni

12. Jan 7, 2010

### Altabeh

This is what I was thinking about the whole time until I hit an early post of George's in which he said he is gonna keep using $$g_{{\mu \nu }}=\eta_{{\mu \nu }}[/itex] after then. Now I feel like the dot-setup is really useful since at least it does not make any such giddinesses. The only thing is that you need to undestand that [tex]\dot{(u_bu^b)}$$ vanishes. But I'm still puzzled and am not able to grasp George's 'no contradiction' stuff!!

13. Jan 8, 2010

### George Jones

Staff Emeritus
Assuming that $\nabla_a g_{bc} =0$ (more on this below), what I wrote in post #9 is correct. What Altabeh wrote in post #8 also is correct, and, taken together, the two posts give two equations of the form

$$\begin{equation*} \begin{split} A &= B \\ A &= -B. \end{split} \end{equation*}$$

These equations have unique solution $A = B = 0$.

What about $\nabla_a g_{bc} = 0$?

Assume special relativity. With respect to inertial coordinates, the components of the metric tensor $\mathbf}{g}$ are constant, and the relation in question clearly is true. With respect to curvilinear coordinates the components of the metric tensor are, in general, not constant, even in special relativity.

For example, with respect to spherical coordinates, (a single component; no sum) $g_{\theta \theta} = r^2$, and it would seem that

$$\nabla_r g_{\theta \theta} = \frac{\partial \left( r^2 \right)}{\partial r} = 2r.[/itex] This, however, is incorrect. Even in special relativity, with respect to curvilinear coordinate systems, $\nabla_a$ is not just a partial derivative. Ludvigsen's equation (9.5), [tex]\nabla_a g_{bc} = \partial_a g_{bc} - g_{ec} \Gamma^e{}_{ab} - g_{be} \Gamma^e{}_{ac},$$

is true in all coordinate systems, including curvilinear coordinate systems in special relativity. With respect to inertial coordinate systems, all the components $\Gamma^e{}_{ab}$ are zero, but with respect to curvilinear coordinates, the components $\Gamma^e{}_{ab}$ are non-zero in such a way that $\nabla_a g_{bc} = 0$.

This is true in special relativity and in standard general relativity.

14. Jan 8, 2010

### Altabeh

Actually $\nabla_a g_{bc}$ is because $g_{bc}$ is covariantly constant. I first thought that George's nabla is supposed to be a partial differentiatial operator in any specetime but later saw his early post that we are in a Minkowski spacetime and that it is literally a covariant differential operator I like to show it by a ';'. :tongue2:

Anyways, Goerge, thanks for your clarification.

Also George I really think you are so knowledgeable in GR so would you please give your idea about my question https://www.physicsforums.com/showthread.php?t=367845"? I'll appreciate any help.

Last edited by a moderator: Apr 24, 2017
15. Jan 8, 2010

### gnieddu

Hi George,

thanks very much for the clear explanation! That clarified my doubts.

Best regards

Gianni