# Energy Momentum Tensor - EM - Derivative

1. Aug 8, 2013

### ProfDawgstein

I have a short question about the derivative of a given EM-Tensor.

$\rho$ = mass density
$U^\mu$ = 4 velocity

$T^{\mu\nu} = \rho U^\mu U^\nu$

Now I do $\partial_\mu$

Should I get

a) $\partial_\mu T^{\mu\nu} = (\partial_\mu \rho) (U^\mu U^\nu) + \rho (\partial_\mu U^\mu) U^\nu + \rho U^\mu (\partial_\mu U^\nu)$

b) $\partial_\mu T^{\mu\nu} = (\partial_\mu \rho) (U^\mu U^\nu) + \rho (\partial_\mu U^\mu) U^\nu$

This should be the product rule, right?
So I should get 3 terms?

$(\partial_\mu \rho) (U^\mu U^\nu) = 0$

because

$(\rho \partial_\mu U^\mu) = 0$ (mass conservation)

Solution is

$\partial_\mu T^{\mu\nu} = \rho (U^\mu \partial_\mu) U^\nu$

where $\partial_\mu (\rho U^\mu) = 0$

What am I missing?

2. Aug 8, 2013

### Staff: Mentor

a) looks correct, all 3 terms should be there.

No, this is not correct. Local conservation of energy is

$$\partial_{\mu} T^{\mu \nu} = 0$$

(Note, btw, that that only holds in flat spacetime; in curved spacetime local energy conservation is $\nabla_{\mu} T^{\mu \nu} = 0$.)

3. Aug 8, 2013

### ProfDawgstein

I should have added that this is:
-special relativity
-comoving frame

I just want to know how to get from

$\partial_\mu T^{\mu\nu} = (\partial_\mu \rho) (U^\mu U^\nu) + \rho (\partial_\mu U^\mu) U^\nu + \rho U^\mu (\partial_\mu U^\nu)$

to

$\partial_\mu T^{\mu\nu} = \rho (U^\mu \partial_\mu) U^\nu$

Cheng used

(mass conservation) $\partial_\mu (\rho U^\mu) = 0$

(comoving) $U^\mu = \gamma (c, 0, 0, 0)$

$U^\mu \partial_\mu = \gamma \partial_t = \partial_\tau$

$\rho \partial_\tau U^\nu$ is the 4 force density

This is for a system of EM field and charges.

So this really is $T^{\mu\nu}_{charge}$

4. Aug 8, 2013

### Staff: Mentor

Ok, good.

You mention "Cheng"; can you give a more specific reference? Several things that you quote in what follows don't look right to me, but I can't tell for sure without seeing the context.

5. Aug 8, 2013

### ProfDawgstein

https://www.amazon.com/Relativity-G...74631208&sr=1-1&keywords=cosmology+relativity

page 399 - 12.15 - visible on book preview

EDIT:

I think $j^\mu = \rho U^\mu$ can be considered to be the 'mass current'.

$\partial_\mu j^\mu = 0$ leads to mass conservation.

Which is the same as $\partial_\mu (\rho U^\mu) = 0$.

Last edited by a moderator: May 6, 2017
6. Aug 8, 2013

### Staff: Mentor

Where are you getting that from? Cheng uses $j^{\mu}$ to represent the charge-current density 4-vector, which is not the same as mass current. So $\partial_{\mu} j^{\mu} = 0$ represents charge conservation, not mass conservation.

After looking at the page you referenced, I'm still not sure I understand Cheng's notation. I need to look at this some more.

Last edited: Aug 8, 2013
7. Aug 8, 2013

### ProfDawgstein

I just used $j^\mu$ to illustrate that mass density equation.

First he does mass, and then he switches to charge/current (charge per mass).

8. Aug 8, 2013

### Staff: Mentor

Yes, I see that, but I can't see the actual page where the problem is posed in the book preview (the page you referenced is where the solution is given), so I'm not sure how much headway I'm going to make in understanding his notation.

9. Aug 8, 2013

### WannabeNewton

Consider $T_{ab} = (\rho + p)u_{a}u_{b} + g_{ab}p$. I will prove something just a bit more general that holds for fluids with non-vanishing pressure. Now $\nabla^{a}T_{ab} = 0 \Rightarrow (\rho + p)u_{a}\nabla^{a}u_{b} + (\rho + p)u_{b}\nabla^{a}u_{a} + u_{a}u_{b}\nabla^{a}(\rho + p) + \nabla_{b}p$
so contracting with $u^{b}$ and using the fact that $u^{b}\nabla^{a}u_{b} = 0$ we find that $(\rho + p)\nabla^{a}u_{a} + u_{a}\nabla^{a}\rho = 0$. In your case we have a pressureless fluid so we get $\nabla^{a}(\rho u_{a}) = 0$. Hence $\nabla^{a}T_{ab} = \rho u_{a}\nabla^{a}u_{b}$. Note that this works for any space-time $(M,g_{ab})$ and metric compatible derivative operator $\nabla_{a}$.

EDIT: As Peter remarked, $\nabla^{a}T_{ab} = 0$ and $\nabla^{a}j_{a} = 0$ are different physical conditions; the latter is a simple consequence of Maxwell's equations $\nabla^{a}F_{ab} = j_{b}$ and $\nabla_{[a}F_{bc]} = 0$. As an aside, if $T_{ab}$ is the energy momentum tensor for the electromagnetic field in particular then what one can show is that $\nabla^{a}T_{ab} = j^{a}F_{ab}$ (proof on request).

Last edited: Aug 8, 2013
10. Aug 8, 2013

### ProfDawgstein

without pressure (p = 0)

$\nabla^{a} T_{ab} = \rho u_a (\nabla^{a} u_b) + \rho u_b (\nabla^{a} u_a) + u_a u_b (\nabla^{a} \rho)$

1st term: $\rho u_a (\nabla^{a} u_b)$ where $\nabla^{a}$ applies only to $u_b$ ?

Later you do $\nabla^{a}(\rho u_{a}) = 0$, where you pull in $\rho$.

When can you do that, and when is it not allowed?

If $\rho$ is just a scalar, you can pull it in, because of the linearity of $\nabla^{a}$.

Why do you contract with $u^{b}$ ?

How do we know that $u^{b}\nabla^{a}u_{b} = 0$?

Slightly confused here, or just too much of it...

Last edited: Aug 8, 2013
11. Aug 8, 2013

### WannabeNewton

Ok so $\nabla^{a}T_{ab} = \rho u_{a}\nabla^{a}u_{b} + \rho u_{b}\nabla^{a}u_{a}+ u_{a}u_{b}\nabla^{a}\rho = 0$. Yes $\nabla^{a}$ is only being applied to $u_{b}$ in the first term. $u^{b}u_{b} = -1$ by definition which immediately implies that $u^{b}\nabla^{a}u_{b} = 0$; contracting both sides of $\nabla^{a}T_{ab} = 0$ with $u^b$ will then simplify the expression down to $\rho\nabla^{a}u_{a}+ u_{a}\nabla^{a}\rho = \nabla^{a}(\rho u_{a}) = 0$. All I did in that last step was use the Leibniz (product) rule for $\nabla_{a}$.

12. Aug 8, 2013

### ProfDawgstein

now contract with $u^{b}$

$\nabla^{a}T_{ab} = \rho u_{a} (\nabla^{a} u^{b} u_{b}) + \rho u^{b} u_{b}\nabla^{a}u_{a}+ u_{a}u_{b} u^{b} \nabla^{a}\rho = 0$

$\nabla^{a}T_{ab} = \rho u_{a} (\nabla^{a} [-1]) + \rho u^{b} u_{b}\nabla^{a}u_{a}+ u_{a}u_{b} u^{b} \nabla^{a}\rho = 0$

$\nabla^{a}T_{ab} = 0 + \rho [-1]\nabla^{a}u_{a}+ u_{a} [-1] \nabla^{a}\rho = 0$

$\nabla^{a}T_{ab} = -2 \rho \nabla^{a}u_{a} = 0$

$\nabla^{a}T_{ab} = \nabla^{a} \rho u_{a} = 0$ (scalar / linearity)

did I insert the $u^{b}$ correctly?

what if $\rho = \rho(x)$ ?

13. Aug 8, 2013

### WannabeNewton

No the first term vanishes because it contains $u^{b}\nabla^{a}u_{b}$ which is zero because $\nabla^{a}(u^b u_b) = 2u^b\nabla^a u_b = \nabla^a(-1) = 0$. You can't pull the $u^b$ into the derivative like that. Also, $\rho$ is not a constant. What you have in the end is $u^a\nabla_a \rho + \rho\nabla^a u_a = 0$. This becomes $\nabla^a (\rho u_a) = 0$ because of the usual product rule from calculus.

14. Aug 8, 2013

### ProfDawgstein

So we have

$\rho = \rho(x^\mu)$ mass density varies from point to point

$u_a = u_a(x^\mu)$ velocity of particle different from point to point

$X_a = X_a(x^\mu) = \rho u_a = \rho(x^\mu) u_a(x^\mu)$ can be considered 'mass current/flow', which also depends on position

$\nabla^{a} T_{ab} = \rho u_a (\nabla^{a} u_b) + \rho u_b (\nabla^{a} u_a) + u_a u_b (\nabla^{a} \rho)$

then we contract with $u^{b}$

$\nabla^{a} T_{ab} = \rho u_a u^{b} (\nabla^{a} u_b) + \rho u_b u^{b} (\nabla^{a} u_a) + u_a u_b u^{b} (\nabla^{a} \rho)$

because $u_b u^{b} = -1$ we get

$\nabla^{a} T_{ab} = \rho u_a u^{b} (\nabla^{a} u_b) + \rho [-1] (\nabla^{a} u_a) + u_a [-1] (\nabla^{a} \rho)$

then

$\nabla^{a} T_{ab} = \rho u_a u^{b} (\nabla^{a} u_b) - \rho (\nabla^{a} u_a) - u_a (\nabla^{a} \rho)$

seeing that the last two terms are $- \nabla^{a} (\rho u_a)$

$\nabla^{a} T_{ab} = \rho u_a u^{b} (\nabla^{a} u_b) - \nabla^{a} (\rho u_a)$

the last term now is equal to $X_a$, our 'mass current/flow' which is conserved, so $\nabla^{a} (\rho u_a) = 0$

now we have

$\nabla^{a} T_{ab} = \rho u_a u^{b} (\nabla^{a} u_b)$

now $\nabla^{a}(u^b u_b) = 2u^b\nabla^a u_b = \nabla^a(-1) = 0$

which means that $\nabla^{a} T_{ab} = \rho u_a [u^{b} (\nabla^{a} u_b)] = 0$

so

$\nabla^{a} T_{ab} = 0$

BUT Cheng gets

$\partial_\mu T^{\mu\nu} = \rho (U^\mu \partial_\mu) U^\nu$

which should be the same as

$\nabla_\mu T^{\mu\nu} = \rho (U^\mu \nabla_\mu) U^\nu$

am I blind? :P

15. Aug 8, 2013

### Staff: Mentor

Shouldn't there be a $\nabla_b \rho$ term as well?

16. Aug 8, 2013

### WannabeNewton

If it holds for any arbitrary metric compatible derivative operator $\nabla_{a}$ then it also holds for $\partial_{a}$.

17. Aug 8, 2013

### WannabeNewton

I can't see there being a $\nabla_{b}\rho$ term myself. The $\nabla_{b}p$ term comes from $\nabla^{a}(g_{ab} p) = \nabla_{b}p$.

18. Aug 8, 2013

### Staff: Mentor

Ah, sorry, I forgot about that term in the original $T_{ab}$. Move along, nothing to see here.

19. Aug 8, 2013

### ProfDawgstein

so my calculation (post #14) should be 100% correct?

Do not forget that

$T_{ab} = \rho u_a u_b$

where

$T_{ab} = T_{ab}^{charge}$ (see post #1) EDIT: #3

there is also

$T_{ab}^{field}$

in the end

$\partial_\mu T^{\mu\nu} = \partial_\mu (T^{\mu\nu}_{field} + T^{\mu\nu}_{charge}) = 0$

I have no idea why you get $\nabla^{a} T_{ab} = 0$ and Cheng

$\partial_\mu T^{\mu\nu}_{field} = - \frac{1}{c} F^{\nu\lambda} j_\lambda$

$\partial_\mu T^{\mu\nu}_{charge} = \frac{1}{c} F^{\nu\lambda} j_\lambda$

I think I just lost the feeling for UP/DOWN

Cheng: $\partial_\mu T^{\mu\nu}_{charge} = \rho (U^\mu \partial_\mu) U^\nu$

written another way

$\nabla^{a} T_{ab}^{charge} = \rho (\nabla^{a} U_a) U_b$

which means that

$\rho U_a (\nabla^{a} U_b) + U_a U_b (\nabla^{a} \rho) = 0$ (they cancel)

if we contract that one with $U^{b}$ we get

$\rho U_a U^{b} (\nabla^{a} U_b) + U_a U_b U^{b} (\nabla^{a} \rho)$

and we use $U_b U^{b} = -1$

$\rho U_a U^{b} (\nabla^{a} U_b) + U_a [-1] (\nabla^{a} \rho)$

so

$\rho U_a U^{b} (\nabla^{a} U_b) - U_a (\nabla^{a} \rho)$

according to WBN the 1st term equals zero, then we have

$U_a (\nabla^{a} \rho) = 0$

which is getting even more confusing...

maybe this will make things clear...

Thanks everybody until now

Last edited: Aug 8, 2013
20. Aug 8, 2013

### WannabeNewton

Can you give the exact page in Cheng where the problem statement is given in detail? Your original post didn't include anything about the energy-momentum tensor of an electromagnetic field, it only talked about the energy-momentum tensor of dust so that's what I thought you were exclusively talking about. If the only energy-momentum is coming from dust i.e. $T_{ab} = \rho u_{a}u_{b}$ then using $\nabla^a T_{ab} = 0$ one can show that $\nabla^a (\rho u_a) = 0$ as I have previously done in this thread. From this we see that $\nabla^a T_{ab} = u_b\nabla^a(\rho u_a) + \rho u_a \nabla^a u_b = \rho u_a \nabla^a u_b = 0$ i.e. $u^a \nabla_a u^b = 0$ which is just the statement that dust particles travel on geodesics, given the above conditions.

Last edited: Aug 8, 2013