# General parameterisation of the geodesic equation

1. Mar 14, 2014

### victorvmotti

Hello all,

In Carroll's on page 109 it is pointed out that for derivation of the geodesic equation, 3.44, a "hidden" assumption is that we have used an affine parameter.

Some few lines below we see that "any other parametrization" could be used, called alpha, but in that case the general form of geodesic equation will be 3.58 and not 3.44.

However, when plugging 3.59 into 3.58 and rearranging I can only recover the geodesic equation, 3.44, provided that alpha is itself a linear function of λ, and not "some general parameter alpha."

So unlike what is said below 3.59, we cannot "always" find an affine parameter λ.

Also, related to this confusion is what we call "a null geodesic."

To make it clear for myself I need to see an example of a null path which is NOT a geodesic.

Can someone please introduce or describe such a null path?

2. Mar 14, 2014

### Bill_K

How about a null spiral? A particle travels endlessly in a circular path, with tangential velocity c.

x = R cos Ωt
y = R sin Ωt
z = 0

with RΩ = c

3. Mar 14, 2014

### stevendaryl

Staff Emeritus
I don't have Carroll's book, but unless I made a mistake, here's the general equation:

$g_{\mu \nu} A^\nu + \Gamma_{\mu \nu \lambda} U^\lambda U^\nu = g_{\mu \nu} U^\nu \dfrac{d}{ds} (ln \mathcal{L})$

where $s =$ the parameter,
$g_{\mu \nu} =$ the metric tensor,
$U^\mu = \dfrac{d}{ds} x^\mu(s)$
$A^\mu = \dfrac{d}{ds} U^\mu(s)$
$\mathcal{L} = \sqrt{g_{\mu \nu} U^\mu U^\nu}$
$\Gamma_{\mu \nu \lambda} =$ the connection coefficients, computed from derivatives of $g_{\mu \nu}$

If you choose the parameter $s$ so that $\mathcal{L}$ is constant, then the right-hand side is zero, and you get the usual form of the geodesic equation.

To make $\mathcal{L}$ constant, you just choose $s$ to satisfy:

$ds = \sqrt{g_{\mu \nu} dx^\mu dx^\nu}$

which is proper time (or any other linearly related parameter).

This only works for non-null geodesics. If we're talking about a null geodesic, then $\mathcal{L}$ is identically zero, so I'm not sure what the condition is on the parametrization.

4. Mar 14, 2014

### pervect

Staff Emeritus
I'm not quite seeing an equation # match to the online version of Caroll at http://arxiv.org/pdf/gr-qc/9712019v1.pdf or http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

This online version seems much clearer than I remember it being, perhaps Caroll has revised it?

Anyway, a quote to enable one to find the section in question, which might be helpful.

and

5. Mar 14, 2014

### stevendaryl

Staff Emeritus
My equation for a general parametrization is basically the same as Carroll's equation 3.59, except that the arbitrary function is explicitly written as $\dfrac{d}{ds} (ln \mathcal{L}$. In that form, we can easily see that the right-hand side vanishes if you choose a linear function of proper time. The function $\mathcal{L}$ is just $\dfrac{d \tau}{ds}$ where $\tau$ is proper time. So if $\tau$ is a linear function of $s$, then that's a constant, and so is $ln (\dfrac{d \tau}{ds})$, and so its derivative is zero.

6. Mar 15, 2014

### victorvmotti

Hello Again,

Thanks for the feedback.

What is still unclear for me is where we have made the implicit assumption of an affine parameter to derive the usual form of the geodesic equation.

We start with parallel transport of the tangent vector to the path and demand that its directional covariant derivative vanishes. This gives the usual form of the geodesic equation.

So the question is that where we have derived from the general parametrization of the geodesic equation in the first place?

Also, Carroll says that for any parameter in the general form you can always find an affine parameter to satisfy the usual form of the geodesic equation. But how?

About null paths that are not geodesic is it possible to illustrate that an endlessly circular path does not satisfy the geodesic equation?

But then what is the geodesic equation for a null path?

Should we use the form based on the four-momentum that is derived and used for timelike paths?

7. Mar 15, 2014

### pervect

Staff Emeritus
I thought the section I quoted from the online version of caroll was pretty clear. A geodesic can be regarded as a curve that parallel transports itself (and this appears to be the preferred definition for mathemeticians). If you only require that a curve parallel transport itself in the same direction, allowing parallel transport to change the length of a vector, there is no requirement on how you parameterize the curve doing the transporting. When you require that parallel transport not change the length of a transported vector, there is an implied requirement as to how the transporting curve must be parameterized.

This requirement of unchanging length of the tangent vector can be used for timelike, spacelike, or null geodesics. Parameterizing by proper time is a special case that applies only to timelike geodesics.

To go a bit beyond Caroll and provide a physical intuitive reaso for the affine parameteriztion of null geodesics (rather than a formal mathematical one)

If you consider an actual plane electromagnetic wave following a null geodesic, the EM wave will have regularly spaced nulls where the electric and magnetic fields both vanish. I.e. for a plane EM wave the electric field is E sin (omega t) and the magnetic field B sin( omega t). At certain times E and B are both zero, creating a null. These nulls must be equally spaced if the geodesic is parameterized affinely.

8. Mar 15, 2014

### WannabeNewton

The geodesic equation under any parametrization reads $\xi^{\nu}\nabla_{\nu}\xi^{\mu} = \alpha \xi^{\mu}$ for some smooth scalar field $\alpha$ defined on the geodesic. So what the geodesic equation under any parametrization really says is the tangent vector's direction is parallel transported along the geodesic but the length isn't. This is easy to see as $\xi^{\nu}\nabla_{\nu}\frac{\xi^{\mu}}{(-\xi_{\gamma}\xi^{\gamma})^{1/2}} = \frac{\alpha \xi^{\mu}}{(-\xi_{\gamma}\xi^{\gamma})^{1/2}} + \frac{\xi_{\gamma}\xi^{\nu}\nabla_{\nu}\xi^{\gamma}}{(-\xi_{\gamma}\xi^{\gamma})^{3/2}} = 0$.

However it is also easy to show that given the above, one can always reparametrize the geodesic so as to satisfy $\xi^{\nu}\nabla_{\nu}\xi^{\mu} = 0$. This is called an affine parametrization.

See here: https://www.physicsforums.com/showpost.php?p=4381515&postcount=4

9. Mar 15, 2014

### stevendaryl

Staff Emeritus
There are two ways to go about deriving the geodesic equation. They are equivalent for the connection used in General Relativity.

One approach is to maximize proper time:

$\tau = \int \mathcal{L} ds$

where $\mathcal{L} = \dfrac{d\tau}{ds} = \sqrt{g_{\mu \nu} U^\mu U^\nu}$
and where $U^\mu = \dfrac{dx^\mu}{ds}$

Using the calculus of variations, we maximize proper time for a path $x^\mu(s)$ satifying:

$\dfrac{d}{ds} (\dfrac{\partial \mathcal{L}}{\partial U^\mu})= \dfrac{\partial}{\partial x^\mu} \mathcal{L}$

This doesn't assume anything about the parameter $s$ being affine, but the resulting equation is only the usual geodesic equation in the case where the parameter is affine.

The other approach to deriving the geodesic equation is to start with the vector equation

$\dfrac{D}{D s} U = \alpha U$

where $U$ is the velocity 4-vector, and where $\alpha$ is an arbitrary function of $s$ and where $\dfrac{D}{D\tau}$ is the operator defined by the following procedure: (I can never remember what the name of it is--it's not a covariant derivative, but it's related to it)

$(\dfrac{D}{D s} A)^\mu = \dfrac{d A^\mu}{ds} + \Gamma^\mu_{\nu \lambda} A^\nu U^\lambda$

The two approaches give the same equation, except in the first approach, the function $\alpha$ is not arbitrary, but is equal to $\dfrac{d}{ds}(ln \mathcal{L})$.

These are actually two different things: an extremal path, versus a path whose velocity vector is unrotated by parallel transport, but they turn out to be equal in GR.

10. Mar 15, 2014

### stevendaryl

Staff Emeritus

$\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha \dfrac{dx^\mu}{ds}$

Now, change parameters to $\tau$. Let $Q = \dfrac{d \tau}{d s}$
The equation becomes (after canceling factors of $Q$ from both sides):

$\dfrac{1}{Q} \dfrac{d Q}{d\tau} \dfrac{dx^\mu}{d \tau} + \dfrac{d^2}{d\tau^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{d\tau} \dfrac{dx^\lambda}{d\tau} = \dfrac{\alpha}{Q} \dfrac{dx^\mu}{d\tau}$

So if you choose $Q$ to satisfy

$\dfrac{d Q}{d \tau} = \alpha$

or, in terms of $s$

$\dfrac{1}{Q} \dfrac{d Q}{d s} = \dfrac{d (ln Q)}{ds} = \alpha$

then you get the usual form of the geodesic equation.

Sure, you just plug it into the geodesic equation, and show that it doesn't work.

The second approach to deriving a geodesic equation works exactly the same whether or not the path is null. So the same equation applies. The first approach (extremizing proper time) doesn't work for null geodesics, since every null path has the same proper time, zero.

11. Mar 15, 2014

### victorvmotti

Seems that Carroll calls it directional covariant derivative.

12. Mar 15, 2014

### victorvmotti

Doesn't this imply that alpha should be zero given how we defined Q above?

You seem to be so close to what Carroll has written in the book, that is relating the general parameter to an affine parameter, using your equation above I can recover the equation he suggests to derive a usual form of the geodesic from the general form, but a difference here is that he says alpha, used in your general form, should be a function of s.

Last edited: Mar 15, 2014
13. Mar 15, 2014

### victorvmotti

This was particularly helpful to make sense of a parallel transport. Because Carroll, when introducing the parallel transport, demands the norm of the vector, or in general "the components" of an arbitrary rank tensor, should not change which means that its "directional covariant derivative" should vanish.

When you say that fundamental to parallel transport notion is that "velocity vector is unrotated" the general parametrization of the geodesic equation makes a lot more clear sense.

14. Mar 16, 2014

### victorvmotti

If we are in locally inertial coordinates then connection coefficients vanishes and therefore both null and timelike geodesic paths should be described by coordinates that are exponential functions of the general parameter s. Am I correct?

If we are in general coordinates then on null paths the term involving connection coefficients in the general form of the geodesic equation vanishes because the inner product of vectors on a null path is zero. So again the null geodesic path is described by coordinates that are exponential functions of the general parameter s. Am I correct?

15. Mar 16, 2014

### stevendaryl

Staff Emeritus
Oh, I didn't actually describe the procedure, I just gave the equation. The procedure is this, in a little more detail:

1. Start with a parametrized path $\mathcal{P}(s)$.
2. Let $A(s)$ be an arbitrary 4-vector that is defined along the path $\mathcal{P}(s)$
3. Let $\tilde{A}(s + \delta s)$ be the result of parallel-transporting $V(s + \delta s)$ back along the path $\mathcal{P}$ from the point $\mathcal{P}(s + \delta s)$ to the point $\mathcal{P}(s)$
4. Let $\delta A = \tilde{A}(s+\delta s) - A$
5. Then $\dfrac{D}{Ds} A = {lim}(\delta s \rightarrow 0) \dfrac{\delta A}{\delta s}$

16. Mar 16, 2014

### stevendaryl

Staff Emeritus
The connection terms don't vanish for a null path. Let's look at a specific simple example: cylindrical coordinates: $(t, z, \rho, \phi)$

The metric components are:
$g_{tt} = 1$
$g_{zz} = g_{\rho \rho} = -1$
$g_{\phi \phi} = - \rho^2$

The non-zero connection coefficients are (if I computed them correctly)
$\Gamma^\rho_{\phi \phi} = \rho$
$\Gamma^\phi_{\rho \phi} = \Gamma^\phi_{\phi \rho} = \dfrac{1}{\rho}$

So the geodesic equation is (in components):

$\dfrac{d^2 \phi}{ds^2} + \dfrac{2}{\rho}\dfrac{d \phi}{ds} \dfrac{d \rho}{ds} = \alpha \dfrac{d \phi}{ds}$

$\dfrac{d^2 \rho}{ds^2} + \rho (\dfrac{d \phi}{ds})^2 = \alpha \dfrac{d \rho}{ds}$

$\dfrac{d^2 z}{ds^2} = \alpha \dfrac{d z}{ds}$

$\dfrac{d^2 t}{ds^2} = \alpha \dfrac{d t}{ds}$

Being a null geodesic means that
$(\dfrac{dt}{ds})^2 - (\dfrac{dz}{ds})^2 - (\dfrac{d\rho}{ds})^2 - \rho^2 (\dfrac{d\phi}{ds})^2 = 0$

17. Mar 16, 2014

### victorvmotti

Actually I meant this, not sure if I am correct:

$g_{\nu \lambda}\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda}g_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha g_{\nu \lambda}\dfrac{dx^\mu}{ds}$

Null path, then

$g_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = 0$

$g_{\nu \lambda}\dfrac{d^2}{ds^2} x^\mu = \alpha g_{\nu \lambda}\dfrac{dx^\mu}{ds}$

$\dfrac{d^2}{ds^2} x^\mu = \alpha\dfrac{dx^\mu}{ds}$

$x^\mu = \alpha\exp{(s\alpha)}$

18. Mar 16, 2014

### stevendaryl

Staff Emeritus
That is not the correct equation. The geodesic equation is:

$\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha \dfrac{d x^\mu}{ds}$

Typically in equations involving indices, there should never be repetitions of indices in a term except in the case of one raised index and the corresponding lowered index. In your equation, the indices $\nu$ and $\lambda$ on $\Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds}$ are dummy indices. They should be replaced by $\nu'$ and $\lambda'$ to avoid being confused with the indices on $g_{\nu \lambda}$. If you do that, you'll see that $g_{\nu \lambda}$ has no effect, and you can just divide through by it.

Last edited: Mar 16, 2014
19. Mar 16, 2014

### victorvmotti

Thanks, you know I multiplied both sides of the general form the geodesic equation by the metric to cancel the connection term. I learned here that it is not allowed if we do not change the indices of the metric tensor.

But still the general form of the geodesic equation, I mean when we are not using an affine parameter, makes me confused.

Suppose when we are in Riemann normal coordinate, or locally inertial coordinates,

For both timelike and null geodesic paths we have

$\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha \dfrac{dx^\mu}{ds}$

Assuming a Riemann normal coordinate then $\Gamma^\mu_{\nu \lambda}$ vanishes and then

$\dfrac{d^2}{ds^2} x^\mu = \alpha \dfrac{dx^\mu}{ds}$

$x^\mu = \alpha\exp{(s\alpha)}$

Right?

20. Mar 16, 2014

### stevendaryl

Staff Emeritus
Sort of. I think you mean:

$x^\mu = a^\mu \exp{(s\alpha)} + b^\mu$

where $a^\mu, b^\mu$ are constant vectors. That's the general solution to the equation

$\dfrac{d^2}{ds^2} x^\mu = \alpha \dfrac{dx^\mu}{ds}$

But locally inertial coordinates are only good locally. So that solution is only valid for small values of $s$.