# Deriving geodesic equation using variational principle

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1. Feb 10, 2016

### dwellexity

I am trying to derive the geodesic equation using variational principle.
My Lagrangian is $$L = \sqrt{g_{jk}(x(t)) \frac{dx^j}{dt} \frac{dx^k}{dt}}$$
Using the Euler-Lagrange equation, I have got this.
$$\frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = \frac{\frac{dx^u}{dt}}{g_{pq}\frac{dx^p}{dt}\frac{dx^q}{dt}} \frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$

How do I prove the right hand side to be zero to get the geodesic eqaution?

I know that the derivation might be simpler using a different Lagrangian but I want to do it using this one.

2. Feb 10, 2016

### Orodruin

Staff Emeritus
In general, you will not get the geodesic equations from the variational principle. The reason for this is that there is an inherent ambiguity in the parametrisation of a curve, which does not affect its length. If you just look at the EL equations based on curve length, you would end up with something like
$$\nabla_{\dot \gamma} \dot\gamma = \alpha(t) \dot \gamma,$$
where $\dot\gamma$ is the tangent vector of the curve (or in other words, the change in the tangent vector is proportional to the tangent vector itself). In order to ensure that you obtain the geodesic equations, you need to require that the parametrisation is affine, which corresponds to a reparametrisation such that $\alpha(t) = 0$. This is mathematically equivalent to letting $\mathcal L = g(\dot\gamma,\dot\gamma)$ rather than $\mathcal L = \sqrt{g(\dot\gamma,\dot\gamma)}$.

3. Feb 10, 2016

### dwellexity

I didn't understand it. You are differentiating $\dot \gamma$ wrt $\dot \gamma$? Also $g(\dot\gamma,\dot \gamma)$?

Having found the equation I have given above, how do I proceed to do this reparametrisation?

4. Feb 10, 2016

### Orodruin

Staff Emeritus
Yes, the definition of a geodesic is that its tangent vector is parallel transported along the curve. The parallel transport equation for a general vector $X$ is given by $\nabla_{\dot\gamma} X = 0$, the only difference here is that the geodesic tangent vector replaces $X$. The $g(\dot\gamma,\dot\gamma)$ is the value of the metric tensor (which is a linear map from the tensor product of the tangent space with itself to real numbers) evaluated when both its arguments is the tangent vector $\dot\gamma$.

5. Feb 10, 2016

### dwellexity

I am just a beginner in these things. Can you guide me on how to proceed from the equation I have to get the geodesic equation?

6. Feb 10, 2016

### Orodruin

Staff Emeritus
It is a bit unclear to me exactly how you arrived at your formula. Can you give more details? In particular, how did you get the Christoffel symbols in there? They should result from the computation, not be there from the beginning.

7. Feb 10, 2016

### dwellexity

I put the above mentioned Lagrangian $$L = \sqrt{g_{jk}(x(t)) \frac{dx^j}{dt} \frac{dx^k}{dt}}$$
in the EL equation $$\frac{d}{dt}\frac{\partial L}{\partial \dot x^l} = \frac{\partial L}{\partial x^l }$$
and solved to get this.

Please see the attached images. I hope it is legible.

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8. Feb 10, 2016

### Orodruin

Staff Emeritus
The point is that you have not really solved for $\ddot x^\mu$. There is a term in the RHS which contains time derivatives of $\dot x^\mu$, which then becomes $\ddot x^\mu$. If you solve for $\ddot x^\mu$ correctly, you should get something on the RHS which is proportional to $\dot x^\mu$ as well as the Christoffel symbols.

9. Feb 10, 2016

### stevendaryl

Staff Emeritus
I don't quite understand that comment. His right-hand side IS proportional to $\dot{x^\mu}$

10. Feb 10, 2016

### stevendaryl

Staff Emeritus

Your original equation, in the first post can be written as:

$\frac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{dt} \frac{dx^\lambda}{dt} = \frac{dx^\mu}{dt} \frac{d\ log(L)}{dt}$

where $L = \sqrt{g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}}$. (Remember: $\frac{d log(X)}{dt} = \frac{dX}{dt}/X$)

If you make the change of parameters from $t$ to $s$, you have to use the chain rule:

$\frac{d x^\mu}{dt} = \frac{ds}{dt} \frac{d x^\mu}{ds}$
$\frac{d^2 x^\mu}{dt^2} = (\frac{ds}{dt})^2 \frac{d^2 x^\mu}{ds^2} + \frac{d^2s}{dt^2} \frac{d x^\mu}{ds}$

If you make those sorts of changes, then in terms of $s$, your equation becomes (after rearranging):

$\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\lambda}{ds} = \frac{dx^\mu}{ds} [ \frac{d\ log(L)}{ds} - \dfrac{\frac{d^2 s}{dt^2}}{(\frac{ds}{dt})^2}]$

So to get the geodesic equation, you have to make the right-hand side equal to zero, which means that $s$ has to solve the equation:

$\frac{d\ log(L)}{ds} - \dfrac{\frac{d^2 s}{dt^2}}{(\frac{ds}{dt})^2} = 0$

That can be rewritten (going back to derivatives with respect to $t$) as:

$\frac{d\ log(L)}{dt} - \dfrac{\frac{d^2 s}{dt^2}}{(\frac{ds}{dt})} = 0$

which looks complicated, but implies the simple equation:

$\frac{ds}{dt} = K L$ where $K$ is some constant.

11. Feb 10, 2016

### lavinia

The formula

$\frac{d^2 x^u}{dt^2} + \Gamma^u_{mk} \frac{dx^m}{dt} \frac{dx^k}{dt} = 0$ is the equation for a geodesic.

So setting the variation equal to zero says that a geodesic is a critical point of the variation of your Lagrangian. BTW: The square root in the Lagangian is unnecessary. Try your equations without it.

To see that this is the geodesic equation, write the the definition of a geodesic,

$\nabla_{\dot \gamma} \dot\gamma = 0$ in your local coordinate system $x^{j}$

One thing that is not determined in this formula is the speed of your geodesic. It may not be pararameterized by arc length. But it is of constant speed. This follows because the connection is a Levi-Civita connection, in particular the connection is compatible with the metric. So

$\dot\gamma<\dot\gamma,\dot\gamma> = <\nabla_{\dot \gamma} \dot\gamma,\dot\gamma> + <\dot\gamma,\nabla_{\dot \gamma} \dot\gamma> = 0$ which says that the length of the tangent vector,$\dot\gamma$ ,is constant.

Last edited: Feb 10, 2016
12. Feb 10, 2016

### Orodruin

Staff Emeritus
You are right of course. For some reason it did not look familiar to me, probably because I would generally write it as $\dot x^\mu d\ln\mathcal L/dt$ instead.

13. Feb 10, 2016

### Orodruin

Staff Emeritus
The square root allows him to express the minmal distance between two points. As I already mentioned in #2, this does not fix the parametrisation of the curve and this is solved by using the squared integrand. Technically these are two different problems, but of course several curves are equivalent wrt curve length (forming an equivalence class consisting of the different possible parametrisations of the curve). The option of using the squared version of the integrand just fixes the parametrisation to be the curve with an affine parameter rather than an arbitrary one. The affine parametrisation also happens to be just the geoedesic condition that the tangent vector is parallel transported along the curve.

14. Feb 10, 2016

### pervect

Staff Emeritus
Other people have answered this, but I'm going to try and make a shorter answer.

Look at $$\frac{d}{dt}\left[g_{rs}\frac{dx^r}{dt}\frac{dx^s}{dt}\right]$$

The term inside the brackets is just the square of the length of the tangent vector $u$ where $u^i = \frac{dx^i}{dt}$. We can write $||u||^2 = g_{rs} u^r u^s$

If the curve is parameterized such that this length (and hence the square of the length) of it's tangent vector is constant, then you have your proof, because d/dt of a constant is zero. As other posters have implied, you need a particular parameterization of the curve to make this happen. The correct parameterization that makes the length of the tangent vector constant is usually called an affine parameterization. For a time-like curve, the affine parameterization is one where the curve ((the $x^i$)) are parameterized by proper time $\tau$ so that $u^i$ is the four-velocity.

The geodesic equation will only give you a geodesic if you have an affine parameterization, if you have a general parameterization of the curve, you need to re-parameterize the curve affinely first.

I'm assuming that the rest of the math here is correct, I haven't double checked it. I vaguely recall seeing ||u|| appear in the denominator (in MTW), not the numerator, but I haven't gone to the length of tracking this down.

15. Feb 10, 2016

### Orodruin

Staff Emeritus
Just to add to the previous post: The affine parameter for timelike geodesics (and geodesics in Riemann spaces) can be taken as the proper time (curve length). However, for null geodesics the curve parameter is also affine, but not equal to the proper time (the proper time is zero since it is a null curve).

16. Feb 10, 2016

### lavinia

Let's see if I understand your point.

For the Riemannian case at least, if one uses the Lagrangian without the square root then it follows directly that the tangent to the critical curve is parallel along the curve. So the solution is already an affinely parameterized geodesic.

If one uses the square root Lagrangian on the other hand, the solution equation merely says that the geodesic curvature of the critical curve is zero. This means that if it were to be reparameterized by arc length, then its tangent would be parallel along the curve. or what is equivalent if it were reparameterized to have constant speed.

In each case one derives the equation for the geodesic from the condition that the geodesic curvature is zero. In the second case, the equation holds for the curve after reparameterization.

17. Feb 10, 2016

### Orodruin

Staff Emeritus
That seems to summarise it, yes.

18. Feb 10, 2016

### stevendaryl

Staff Emeritus
I don't understand that paragraph. If you are using "parallel" in the sense that two vectors are parallel if one is a positive multiple of the other, then reparametrizing the path doesn't change whether its tangent is parallel or not. Does it?

19. Feb 10, 2016

### Orodruin

Staff Emeritus
Parallel has a different meaning in differential geometry. A tensor field $T$ is parallel along a curve $\gamma$ if $\nabla_{\dot\gamma} T = 0$.

The defining property of a geodesic is that its tangent vector is parallel along the geodesic, i.e., $\nabla_{\dot\gamma} \dot\gamma = 0$.

20. Feb 10, 2016

### PAllen

The standard definition of parallel transport (as given e.g. by Orodruin, above), forces an affine parameter (not just in the geodesic case), because it forces preservation of the vector norm as well as direction, and this is only possible with affine parameters. You need to use a more general definition of parallel transport to allow for arbitrary parameter for the curve. Specifically, you need to allow the RHS of the covariant derivative be a scalar function of the parameter times the vector. This allows the magnitude to vary, with direction being locally preserved. Starting from curve affinely paremetrized with a vector parallel transported on it, if you change parameter in a non-linear way, the simple equation for parallel transport will no longer be satisfied.