Prove EL Geodesic and Covariant Geodesic Defs are Same via Riemmanian Geometry

[binbagsss]

... via plugging in the Fundamental theorem of Riemmanian Geometry :

$\Gamma^u_{ab}=\frac{1}{2}g^{uc}(\partial_ag_{bc}+\partial_bg_{ca}-\partial_cg_{ab})$

Expanding out the covariant definition gives the geodesic equation as:

(1) $\ddot{x^u}+\Gamma^u_{ab} x^a x^b =0$

(2) Lagrangian is given by $\sqrt{g_{ab}\dot{x^a}\dot{x^b}}$ but assuming affinely parameterised: ( $\frac{dL}{ds}=0$ ) then can treat the Lagrangian as $g_{ab}\dot{x^a}\dot{x^b}$ and E-L equations give geodesic.

Expanding out E-L equations I have: $2g_{uv}\ddot{x^v} +2\partial_a(g_{uv})\dot{x^a}\dot{x^v}-\partial_ug_{ab}\dot{x^b}\dot{x^a}$

I believe i need to multiply christoffel symbol by $g_{uc}$ and expression (1) by $g_{uc}$ doing so and plugging into (1) gives me :

$g_{uc} \ddot{x^u}+2\dot{x^a}\dot{x^b}(\partial_a g_{bc}+\partial_bg_{ca}-\partial_cg_{ab})$

Now I see I can idenitify the middle term in brackets with the last term in (2) ,but, I don't know what to do now, I think I could rename indicies in either term 1 or 3 in the brackets to idenitty with term 2, but i can't rename in both terms as to me it looks like i'd need to do, to sort out the factor of two. even so the first term not in brackets looks factor of 2 out.

thanks

 

[Orodruin]

First of all: Please be more careful with your LaTeX. Some things do not render properly and other things are very difficult to read.

Second, make sure you get all factors of 2 correctly.

Once you have done that, I suggest that you multiply the $\dot x^a \dot x^b$ into the parenthesis so that you can rename summation indices freely in each term. You will also likely need to use that the metric tensor is symmetric.

 

[binbagsss]

Second, make sure you get all factors of 2 correctly.

factors of $2$ in equation (2) are correct, no?

and then the other factor has come from $1/2 g^{uc} g_{uc}$ in the christoffel expression is $2$ because $g_{uc}g^{uc}=4$?

 

[Orodruin]

factors of $2$ in equation (2) are correct, no?

and then the other factor has come from $1/2 g^{uc} g_{uc}$ in the christoffel expression is $2$ because $g_{uc}g^{uc}=4$?

You cannot contract something with $g_{uc}$ that already has $c$ as a summation index in it. Therein lies your error.

 

[binbagsss]

You cannot contract something with $g_{uc}$ that already has $c$ as a summation index in it. Therein lies your error.

ah ofc thanks alot

so Instead let me do:

$1/2g^{uc}_{up} (\partial_a g_{bc} + \partial_b g_{ca}-\partial_cg_{ab}) =1/2 \delta^{c}_{p} (\partial_a g_{bc} + \partial_b g_{ca}-\partial_cg_{ab}) =1/2(\partial_a g_{bp} + \partial_b g_{pa}-\partial_pg_{ab})$

and plugging this into (1) now gives me:

$2\dot{x^u}+\dot{x^a}\dot{c^b}(\partial_ag_{bp}+\partial_bg_{pa}-\partial_pg_{ab})$
$= 2\ddot{x^u}+2\partial_ag_{bp}\dot{x^a}\dot{x^b}-\partial_pg_{ab}\dot{x^a}\dot{x^b}$

where the last name follows by renaming a dummy indice as using symmetry of the metric,so my first term is wrong mising multiplicaiton by the metric.

actually is my application of $\delta ^c _ p$ wrong? is this telling me when replacing $c$ i need to raise it to a $p$? (if so, i guess what this really means is if there is an upper index c to replace it with a p, but it will come equivalent to what i just said via raising or lowering indices on the delta with the metric).

 

[stevendaryl]

Expanding out E-L equations I have: $2g_{uv}\ddot{x^v} +2\partial_a(g_{uv})\dot{x^a}\dot{x^v}-\partial_ug_{ab}\dot{x^b}\dot{x^a}$

If you multiply through by $\frac{1}{2} g^{wu}$ at this point, you get:

$g^{wu} g_{uv}\ddot{x^v} +\frac{1}{2} g^{wu}[ 2 \partial_a(g_{uv})-\partial_ug_{av} ]\dot{x^v}\dot{x^a}$ (replacing $b$ by $v$ in the last term)

and since $g^{wu} g_{uv} = \delta^w_v$, this becomes:

$\ddot{x^w} +\frac{1}{2} g^{wu}[ 2 \partial_a(g_{uv})-\partial_ug_{av} ]\dot{x^v}\dot{x^a}$

So you get a modified version of the connection coefficients: Instead of

$\Gamma^w_{av} = \frac{1}{2} g^{wu}[ \partial_a(g_{uv}) + \partial_v(g_{au}) -\partial_ug_{av} ]$

you have $\tilde{\Gamma}^w_{av} = \frac{1}{2} g^{wu}[ 2 \partial_a(g_{uv})-\partial_ug_{av} ]$

So it doesn't work, exactly. However, look at the difference between the two:

$\Delta^w_{av} = \tilde{\Gamma}^w_{av} - \Gamma^w_{av} = \frac{1}{2} g^{wu}[\partial_a(g_{uv})-\partial_v (g_{au})]$

But since $\Delta^w_{av}$ is anti-symmetric in $a$ and $v$, it follows that:

$\Delta^w_{av} \dot{x^a} \dot{x^v} = 0$

So you don't get exactly the right connection coefficients, but it doesn't make any difference to the equation of motion.

 

[Orodruin]

$\ddot{x^w} +\frac{1}{2} g^{wu}[ 2 \partial_a(g_{uv})-\partial_ug_{av} ]\dot{x^v}\dot{x^a}$

So you get a modified version of the connection coefficients: Instead of

$\Gamma^w_{av} = \frac{1}{2} g^{wu}[ \partial_a(g_{uv}) + \partial_v(g_{au}) -\partial_ug_{av} ]$

you have $\tilde{\Gamma}^w_{av} = \frac{1}{2} g^{wu}[ 2 \partial_a(g_{uv})-\partial_ug_{av} ]$

Precisely because $\dot x^a \dot x^b$ is symmetric, I would not say that you get a modified version of the connection coefficients. The geodesic equations by themselves can never give you the connection coefficients because of this, but only the symmetric part. To really fix the connection coefficients from a geodesic equation you need an additional assumption, such as the connection being torsion free.

However, as I understand the question presented by the OP, this was never a problem since he was not asked to derive the formula for the connection coefficients in terms of the metric, just to verify that the Levi-Civita connection gives the same geodesic equation as that found by minimising the distance. (Other connections generally do not.)