# Deriving geodesic equation from energy-momentum conservation

1. May 18, 2015

### dpdt

Hi all,

I am trying to follow the calculation by samalkhaiat in this thread: https://www.physicsforums.com/threa...n-from-the-stress-energy-tensor.547502/page-2 (post number 36). I am having some difficulty getting the equation above equation (11) (it was an unnumbered equation) from the preceding equations:

In particular, the equation states that:
$\frac{dX^c}{dx^0} \int d^3 x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_b (\sqrt{-g} T^{ab} + X^c \Gamma^a_{bd}\int d^3x \sqrt{-g} T^{bd} ) = \int d^3x \sqrt{-g} T^{ac}$ (*)

I am confused to where the $\frac{d X^c}{dx^0}$ comes from. I managed to massage equations so that I obtain equation (*), except that instead of

$\frac{dX^c}{dx^0} \int d^3 x \sqrt{-g} T^{a0}$ (**)

$\int d^3 x \partial_0 (\sqrt{-g} T^{a0} \delta x^c)$ (***)

Help!!! Can someone help me see why the two equations above (equations (**) and (***) ) are equal to each other?
Thank you so much for any help, this calculation is frying my brain!!! I can present my calculation up to this point if it is helpful at all.

Any help will be much appreciated!!!

2. May 20, 2015

### samalkhaiat

You did not copy this correctly.
The equation in question is $$\int d^{3} x \left[ \partial_{0} (\sqrt{- g} x^{c} T^{a 0}) + \partial_{j} (\sqrt{- g} x^{c} T^{a j}) + \sqrt{- g} x^{c} \Gamma^{a}_{b d} T^{b d} \right] = \int d^{3} x \sqrt{- g} T^{a c} . \ \ (R)$$

After that, I explained every step very carefully. Use Eq(4) and Eq(5), which are $$x^{c} = X^{c} + \delta x^{c} , \ \ \ \ (4)$$$$\Gamma^{a}_{b d} ( x ) = \Gamma^{a}_{b d} ( X ) + \delta x^{e} \partial_{e} \Gamma^{a}_{b d} . \ \ \ (5)$$ After the substitutions in (R), use Eq(7) and Eq(8), which are $$\int d^{3} x \ \sqrt{- g} \ \delta x^{c} \ T^{a b} = 0 , \ \ \ \ (7)$$ $$\int d^{3} x \ \partial_{j} ( \sqrt{- g} \ \delta x^{c} \ T^{a j} ) = 0 . \ \ \ (8)$$

Now, you do this and show me your working. Remember to do the $\partial_{0}$ differentiation in the first term of Eq(R). This to get you started: $$\frac{d}{ d x^{0}} \left( \int d^{3} x \ \sqrt{- g} \ \delta x^{c} \ T^{a 0} \right) + \frac{d}{ d x^{0}} \left( X^{c} \int d^{3} x \ \sqrt{- g} \ T^{a 0} \right) .$$ Now in this, the first term is zero because of (7), the second term give you $$\frac{d X^{c}}{d x^{0}} \int d^{3} x \ \sqrt{- g} \ T^{a 0} + X^{c} \int d^{3} x \ \partial_{0} ( \sqrt{- g} \ T^{a 0} ) .$$

Sam

3. May 21, 2015

### dpdt

Dear Samalkhaiat,

Thank you, both for your reply and your original post. I still have two questions. For your ease of reading, I am highlighting my 2 questions in boldfont.

Starting from equation (R) of your post:
$\int d^3x [\partial_0 (\sqrt{-g} T^{a0} x^c) + \partial_j (\sqrt{-g} T^{aj} x^c) + \sqrt{-g}x^c \Gamma^a_{bd} T^{bd}] = \int d^3 x \sqrt{-g} T^{ac}$ (R)

I believe I have the correct derivation for the third term. In particular, once I plug in the expansion of $x^c$ and $\Gamma^a_{bc}$ the third term gives me to first order:
$\int d^3x \sqrt{-g}x^c \Gamma^a_{bd} T^{bd}] = X^c \Gamma^a_{bd} \int d^3x \sqrt{-g} T^{bd}$. (A)
EDIT: Please see the end of my post!

The first two terms, however, are still giving me problems. Proceeding from the first term of equation (R):

$\int d^3x \partial_0 (\sqrt{-g} T^{a0} x^c) = \int d^3x \partial_0 (\sqrt{-g} T^{a0} X^c) + \int d^3x \partial_0 (\sqrt{-g} T^{a0} \delta x^c)$

The second term is zero because of equation (7) in your post. The first term of equation (R) is therefore:

$\int d^3x \partial_0 (\sqrt{-g} T^{a0} X^c) = \frac{d X^c}{d t} \int d^3x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_0 (\sqrt{-g} T^{a0})$. (B)

The second term of equation (R) gives:
$\int d^3x \partial_j (\sqrt{-g} T^{aj} x^c) = \int d^3x \partial_j (\sqrt{-g} T^{aj} X^c) + \int d^3x \partial_j (\sqrt{-g} T^{aj} \delta x^c)$ .

By equation (8) in your post, the second term is zero. The second term of equation (R) is therefore:

$\int d^3x \partial_j (\sqrt{-g} T^{aj} X^c) = X^c \int d^3x \partial_j (\sqrt{-g} T^{aj}) + \int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai})$. (C)

Adding equations (A), (B), and (C) gives me that equation (R) becomes:

$\frac{d X^c}{d t} \int d^3x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_0 (\sqrt{-g} T^{a0}) + X^c \int d^3x \partial_j (\sqrt{-g} T^{aj}) + \int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai}) + X^c \Gamma^a_{bd} \int d^3x \sqrt{-g} T^{bd} = \int d^3 x \sqrt{-g} T^{ac}$.

Combining the second and third terms, I obtain that equation (R) becomes:

$\frac{d X^c}{d t} \int d^3x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_b (\sqrt{-g} T^{ab}) + \int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai}) + X^c \Gamma^a_{bd} \int d^3x \sqrt{-g} T^{bd} = \int d^3 x \sqrt{-g} T^{ac}$. (D)

Which is equal to the equation above equation (11) in your original post if the following is true:

$\int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai}) = 0$ .

So, my first question: is there a reason why said term is zero?

EDIT:
Once I looked at my derivation of the third term of (R), I realized that I did not do it properly. The third term of (R) is
$\int d^3x \sqrt{-g} x^c \Gamma^a_{bd} T^{bd} = \int d^3x \sqrt{-g} T^{bd} (\Gamma^a_{bd} + \delta x^d \partial_d \Gamma^a_{bd})(X^c + \delta x^c)$.

To first order in $\delta x$ this is:

$\int d^3 x \sqrt{-g} T^{bd} (X^c \Gamma^a_{bd} + \delta x^c \Gamma^a_{bd} + X^c \delta x^d \partial_d \Gamma^a_{bd} )$.

The second term is zero because we can write it as:

$\int d^3x \sqrt{-g} T^{bd} (\delta x^c \Gamma^a_{bd}) = \Gamma^a_{bd} \int d^3 x \sqrt{-g} T^{bd} \delta x^c$,
which is zero because of equation (7) of your post.

So the third term of (R) is equal to:

$\int d^3 x \sqrt{-g} T^{bd} (X^c \Gamma^a_{bd} + X^c \delta x^d \partial_d \Gamma^a_{bd} ) = X^c \Gamma^a_{bd} \int d^3 x \sqrt{-g} T^{bd} + \int d^3 x \sqrt{-g} T^{bd} (X^c \delta x^d \partial_d \Gamma^a_{bd} )$.

So my second question: is there a reason for the last term,

$\int d^3 x \sqrt{-g} T^{bd} (X^c \delta x^d \partial_d \Gamma^a_{bd} )$,

to be equal to zero?

Last edited: May 21, 2015
4. May 21, 2015

### samalkhaiat

For the same reason that allows us to pull $X^{c}$ out of the volume integral. $X^{c}$ is a function only of whatever we choose to parameterize the geodesic path. So, you could think of it as $X^{c} ( s )$.
Notice that you used the same index contraction over [d] twice!! You need to be careful not to use the same index. Any way back to your question. Again for the same reason that lets you pull $\Gamma$ out of the integral. In the expansion of $\Gamma ( x )$, the term $\partial \Gamma$ is evaluated at $x^{c} = X^{c}$, so you can pull it out together with $X$ of that integral.

Last edited: May 21, 2015
5. May 21, 2015

### dpdt

Would you mind elaborating on the reason of why
$\partial_j X^c = 0$ ?

I understand that seeing $X^c$ as a function of the geodesic parameter $s$ allows one to set $\partial_j X^c = 0$, but isn't also true that
$\partial_j X^c = \partial_j (x^c - \delta x^c) = \delta^c_j$,
where $\delta^c_j$ is the delta function?

Thanks!

6. May 22, 2015

### samalkhaiat

Well, that is all what you need. It is like writting $x = 3 f(s) + \delta x$.
Why is that? First of all, you should not apply naive algebra on a deninition. Second, $\partial_{j} x^{a} = \delta^{a}_{j}$, and $\delta x^{a}$ is a function of $x$, so $\partial_{j} ( \delta x^{a}) \neq 0$.
Look, as I said in the original post, one should think of $X^{a}$ as a spatially fixed point $P$ in the $x$-coordinate system. Then, one tries to see what kind of curve does the point $P$ trace when the energy-momentum is covariantly conserved.

Last edited: May 22, 2015
7. May 22, 2015

### dpdt

This statement makes sense to me; I always got myself confused when dealing with partial vs. total derivatives. I think I understood the derivation now. Thank you for your help, and for your original post!