# General parameterisation of the geodesic equation

• victorvmotti
In summary: Anyway, I'm not sure exactly which part of this conversation you want me to summarize, so I'll just provide a general summary of the main points.In summary, the conversation discusses the derivation of the geodesic equation and how it is dependent on the choice of parameterization. It is pointed out that the use of an affine parameter in the equation assumes that the tangent vector is parallel transported. Any other parametrization will result in a different form of the geodesic equation. The discussion also mentions the concept of a null geodesic and how it differs from a general geodesic. Overall, the conversation highlights the importance of choosing the proper parameterization when dealing with geodesics in curved space.
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?

victorvmotti said:
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?
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

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.

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.

Caroll said:
Having boldly derived these expressions, we should say some more careful words about the parameterization of a geodesic path. When we presented the geodesic equation as the requirement that the tangent vector be parallel transported, (3.47), we parameterized our path with some parameter ##\lambda##, whereas when we found the formula (3.56) for the extremal of the spacetime interval we wound up with a very specific parameterization, the proper time. Of course from the form of (3.56) it is clear that a transformation

Equation 3.58 (3.58) [affine transformation]

for some constants a and b, leaves the equation invariant. Any parameter related to the proper time in this way is called an affine parameter, and is just as good as the proper time for parameterizing a geodesic. What was hidden in our derivation of (3.47) was that the demand that the tangent vector be parallel transported actually constrains the parameterization of the curve, specifically to one related to the proper time by (3.58). In other words, if you start at some point and with some initial direction, and then construct a curve by beginning to walk in that direction and keeping your tangent vector parallel transported, you will not only define a path in the manifold but also (up to linear transformations) define the parameter along the path.

Of course, there is nothing to stop you from using any other parameterization you like, but then (3.47) will not be satisfied. More generally you will satisfy an equation of the form

Equation 3.59 (3.59)

for some parameter ##\alpha## and some function f (## \alpha##). Conversely, if (3.59) is satisfied along a curve you can always find an affine parameter ## \lambda##(## \alpha##) for which the geodesic equation (3.47) will be satisfied.

With parallel transport understood, the next logical step is to discuss geodesics. A geodesic is the curved-space generalization of the notion of a "straight line" in Euclidean space. We all know what a straight line is: it's the path of shortest distance between two points. But there is an equally good definition -- a straight line is a path which parallel transports its own tangent vector. On a manifold with an arbitrary (not necessarily Christoffel) connection, these two concepts do not quite coincide, and we should discuss them separately.

and

Thus, on a manifold with metr ic, extremals of the length func-
tional are curves which parallel transport their tangent vector with respect to the Christoffel
connection associated with that metric. It doesn’t matter if there is any other connection
defined on the same manifold. Of course, in GR the Christoffel connection is the only one
which is used, so the two notions are the same.

pervect said:
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

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.

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?

victorvmotti said:
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.

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.

victorvmotti said:
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.

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

victorvmotti said:
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.

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.

victorvmotti said:
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?

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

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.

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

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

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

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.

stevendaryl said:
it's not a covariant derivative, but it's related to it)

Seems that Carroll calls it directional covariant derivative.

stevendaryl said:
So if you choose $Q$ to satisfy

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

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.

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stevendaryl said:
$\dfrac{D}{D s} U = \alpha U$

where $U$ is the velocity 4-vector

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.

stevendaryl said:
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.

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?

stevendaryl said:
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:

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 $\dfrac{D}{Ds} A = {lim}(\delta s \rightarrow 0) \dfrac{\delta A}{\delta s}$

victorvmotti said:
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?

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
$(\dfrac{dt}{ds})^2 - (\dfrac{dz}{ds})^2 - (\dfrac{d\rho}{ds})^2 - \rho^2 (\dfrac{d\phi}{ds})^2 = 0$

stevendaryl said:
The connection terms don't vanish for a null path.

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)}$

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

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

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.

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stevendaryl said:
If you do that, you'll see that $g_{\nu \lambda}$ has no effect, and you can just divide through by it.

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?

victorvmotti said:
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?

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 $s$.

stevendaryl said:
The first approach (extremizing proper time) doesn't work for null geodesics, since every null path has the same proper time, zero.

But why Carroll in section 3.5 on pages 113-116 when establishing the null geodesic equation, first uses this approach, that is extremizing proper time, to compute the Christoffel symbols.

Then instead of the general form of the geodesic equation that has $\alpha$ on the right side, uses the same usual geodesic equation obtained from extremizing proper time that vanishes on the right side, and just asks us to remember exchanging $\tau$ with $\lambda$ in that equation. After a few lines it is demonstrated that energy of the photon is inversely proportional to the scale factor in an expanding universe.

Based on you wrote it seems that we should not use the usual geodesic equation for null paths.

So I wonder if we were to use the general geodesic equation for null paths can we show the same relation between energy and scale factor?

victorvmotti said:
But why Carroll in section 3.5 on pages 113-116 when establishing the null geodesic equation, first uses this approach, that is extremizing proper time, to compute the Christoffel symbols.

Then instead of the general form of the geodesic equation that has $\alpha$ on the right side, uses the same usual geodesic equation obtained from extremizing proper time that vanishes on the right side, and just asks us to remember exchanging $\tau$ with $\lambda$ in that equation. After a few lines it is demonstrated that energy of the photon is inversely proportional to the scale factor in an expanding universe.

Well, although the derivation of the geodesic equation by maximizing proper time doesn't work for null geodesics, the resulting equation is in fact valid for null geodesics, as well. I don't immediately know why that should be the case, unless there is some kind of continuity argument.

But if Carroll did not include the correction on the right-hand side for non-affine parameters, then that means it has to be an affine parameter.

stevendaryl said:
Well, although the derivation of the geodesic equation by maximizing proper time doesn't work for null geodesics, the resulting equation is in fact valid for null geodesics, as well. I don't immediately know why that should be the case, unless there is some kind of continuity argument.

But if Carroll did not include the correction on the right-hand side for non-affine parameters, then that means it has to be an affine parameter.

Let me illustrate why there has to be a nonzero right-hand side for the simplest possible case: flat 2D spacetime, inertial coordinates. In this case, the trajectory is given by a pair of functions:

$x(s)$
$t(s)$

If $s$ is an affine parameter, then since the connection coefficients vanish, the geodesic equations reduce to the trivial pair of equations:

$\dfrac{d^2 x}{ds^2} = 0$
$\dfrac{d^2 t}{ds^2} = 0$

Which has the solution:

$x = A + B s$
$t = C + D s$

A null geodesic is one where $\dfrac{dx}{ds} = \pm c \dfrac{dt}{ds}$, which means that $|\dfrac{B}{D}| = c$

Now, let's rewrite the solution in terms of a new parameter, $q$ which is nonlinearly related to $s$: $s = q^3$. In terms of $q$, the solution looks like this:

$x = A + B q^3$
$t = C + D q^3$

This obviously does not satisfy

$\dfrac{d^2 x}{dq^2} = 0$

$\dfrac{d x}{d q} = 3 B q^2$
$\dfrac{d^2 x}{dq^2} = 6 B q$
$\dfrac{d t}{d q} = 3 D q^2$
$\dfrac{d^2 t}{dq^2} = 6 D q$

So
$\dfrac{d^2 x}{dq^2} = \dfrac{2}{q} \dfrac{dx}{dq}$
$\dfrac{d^2 t}{dq^2} = \dfrac{2}{q} \dfrac{dt}{dq}$

So this has the form:
$\dfrac{d^2 x^\mu}{dq^2} =\alpha \dfrac{dx^\mu}{dq}$

with $\alpha = \dfrac{2}{q}$

Great example, thanks a lot.

stevendaryl said:
Well, although the derivation of the geodesic equation by maximizing proper time doesn't work for null geodesics, the resulting equation is in fact valid for null geodesics, as well. I don't immediately know why that should be the case, unless there is some kind of continuity argument.

Given what you said above it must be straightforward.

Suppose the general form that is valid for the null geodesic.

$\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \sigma} \dfrac{dx^\nu}{ds} \dfrac{dx^\sigma}{ds} = \alpha \dfrac{dx^\mu}{ds}$Now let $Q = \dfrac{d \lambda}{d s}$

$\lambda$ is NOT an affine parameter. So if you choose $Q$ to satisfy

$\dfrac{d Q}{d \lambda} = \alpha$The general form based on this parameter change turns into

$\dfrac{d^2}{d\lambda^2} x^\mu + \Gamma^\mu_{\nu \sigma} \dfrac{dx^\nu}{d\lambda} \dfrac{dx^\sigma}{d\lambda} = 0$

This has the same effect of exchanging $\tau$ with $\lambda$ in the usual form of the geodesic equation obtained from the maximizing proper time approach which is done by Carroll in section 3.5.

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victorvmotti said:
Given what you said above it must be straightforward.

Suppose the general form that is valid for the null geodesic.

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

Now let $Q = \dfrac{d \lambda}{d s}$

$\lambda$ is NOT an affine parameter.

So if you choose $Q$ to satisfy

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

The general form based on this parameter change turns into

$\dfrac{d^2}{d\lambda^2} x^\mu + \Gamma^\mu_{\nu \sigma} \dfrac{dx^\nu}{d\lambda} \dfrac{dx^\sigma}{d\lambda} = 0$

This has the same effect of exchanging $\tau$ with $\lambda$ in the usual form of the geodesic equation obtained from the maximizing proper time approach which is done by Carroll in section 3.5.

Yes, that's right. What I meant was that it's surprising that extremizing the proper time would result in an equation that describes null geodesics, because null geodesics don't extremize the proper time. (Well, they give proper time zero, but so does any other null path, geodesic or not).

So in conclusion one can say that the usual form that is zero on the right side could be used for both timelike and null paths.

However, the parameter needs special care. If the geodesic is timelike we can treat it as an affine parameter also but if the geodesic is null we should treat it as a general parameter which is not an affine parameter.

## 1. What is the geodesic equation?

The geodesic equation is a mathematical expression that describes the path of shortest distance between two points on a curved surface, such as a sphere or a curved space-time in general relativity.

## 2. Why is it important to parameterize the geodesic equation?

Parameterizing the geodesic equation allows us to determine the exact path of an object in space-time, providing valuable information in fields such as astronomy and physics. It also allows for the calculation of important quantities, such as the curvature of space-time.

## 3. What are the main parameters used in the geodesic equation?

The main parameters used in the geodesic equation are the initial position and velocity of the object, as well as the metric of the curved space-time in which the object is moving. Other factors, such as the mass of the object and external forces, may also be included in certain cases.

## 4. How is the geodesic equation derived?

The geodesic equation is derived from the principle of least action in classical mechanics, which states that the path taken by a physical system is the one that minimizes the action. In the case of the geodesic equation, the action is proportional to the length of the path taken by the object.

## 5. Can the geodesic equation be applied to any curved space-time?

Yes, the geodesic equation is a general expression that can be applied to any curved space-time, whether it is described by the theory of general relativity or other mathematical models. It is a fundamental tool in the study of curved space-time and its effects on the motion of objects.

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