Deriving geodesic equation from energy-momentum conservation

[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

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

I have instead

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

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!

 

[samalkhaiat]

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

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

 

[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?

Again, I really appreciate your help. Thank you so much for your answer!

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?

 

[samalkhaiat]

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.

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

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

Again, I really appreciate your help. Thank you so much for your answer!

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 ).

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?

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 let's 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.

 

[dpdt]

Thank you again for your answer!
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!

 

[samalkhaiat]

Thank you again for your answer!
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 $\partial_j X^c = 0$,

Well, that is all what you need. It is like writting x = 3 f(s) + \delta x.

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!

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.

 

[dpdt]

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.

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!