Are all null geodesics also affine?

[kkz23691]

Hi

according to the text I am reading

a curve is geodesic if these conditions are met
$\frac{d}{ds}(2g_{mi} \frac{dx^{i}}{ds})-\frac{\partial g_{jk}}{\partial x^{m}}\frac{dx^{j}}{ds}\frac{dx^{k}}{ds} = 0$, where $m=1,...,N$

a curve is a null geodesic if exactly the same conditions are met, just replace $s$ by $\lambda$ and call it an "affine parameter". It doesn't seem right...

On another forum someone had stated that as long as the above differential equations are solved, the curve resulting from them will be a null geodesic and the parameter will be an affine one.

Is this correct?

 

[martinbn]

It depends on the initial conditions. You need a point and a vector. If the vector is null, then the resulting geodesic will be null. If it is time-like so will be the geodesic.

 

[Mentz114]

according to the text I am reading

What is is the text ? I have never seen that formula. Have some additional assumptions been made about the metric ?

Usually one finds geodesics with the methods shown here http://en.wikipedia.org/wiki/Geodesics_in_general_relativity

May be your equation is the usual equation - I don't have the time to work it out right now.

 

[kkz23691]

It is the same expression as shown in the quoted Wikipedia page

$\frac{d^2x^i}{ds^2} + \Gamma^i_{jk} \frac{dx^j}{ds}\frac{dx^k}{ds} = 0$

one only need multiply the expression by $2g_{mi}$ and sum w/respect to $i$

 

[Mentz114]

It is the same expression as shown in the quoted Wikipedia page

Can you show the working ? I'm not sure what you mean because 'i' is not a free index in your expression, it is already summed out.

Finding null geodesics is something I've always found tricky.

 

[Matterwave]

A null geodesic is simply a geodesic with null tangent vectors. There is a result that one can work out (pretty simple one actually) that a geodesic which is initially time-like/null/space-like will remain respectively time-like/null/space-like.

The point the book is trying to make about the null geodesics is that unlike for time-like or space-like geodesics, one can not use arc-length as a parameter along null geodesic curves since that parameter is 0 along the entire curve. So one must use some (arbitrary) affine parameter to parametrize the curve.

 

[kkz23691]

Can you show the working ?

The conditions for null geodesic are

$\frac{d^2x^i}{d\lambda^2} + \Gamma^i_{jk} \frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} = 0$
Multiply by $2g_{mi}$ and sum w/respect to $i$
$2g_{mi}\frac{d^2x^i}{d\lambda^2} + 2g_{mi}\Gamma^i_{jk} \frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} = 0$
But
$\frac{d}{d\lambda}(2g_{mi}\frac{dx^i}{d\lambda})=$
$=2( \frac{dg_{mi}}{d\lambda} \frac{dx^i}{d\lambda}+g_{mi}\frac{d^2x^i}{d\lambda^2})=$
$=2(\frac{\partial g_{mi}}{\partial x^k}\frac{dx^k}{d\lambda}\frac{dx^i}{d\lambda}+g_{mi}\frac{d^2x^i}{d\lambda^2})=$
$=\frac{\partial g_{mi}}{\partial x^k}\frac{dx^k}{d\lambda}\frac{dx^i}{d\lambda}+\frac{\partial g_{mi}}{\partial x^k}\frac{dx^k}{d\lambda}\frac{dx^i}{d\lambda}+2g_{mi}\frac{d^2x^i}{d\lambda^2}=$
In the first term, rename $i$ to $j$
In the second term, rename $i$ to $k$ and $k$ to $j$:
$=(\frac{\partial g_{mj}}{\partial x^k}+\frac{\partial g_{mk}}{\partial x^j})\frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} + 2g_{mi}\frac{d^2x^i}{d\lambda^2}$
Then
$2g_{mi}\frac{d^2x^i}{d\lambda^2}=\frac{d}{d\lambda}(2g_{mi}\frac{dx^i}{d\lambda})-(\frac{\partial g_{mj}}{\partial x^k}+\frac{\partial g_{mk}}{\partial x^j})\frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda}$
Plug this in the starting expression for the null geodesics $\frac{d^2x^i}{d\lambda^2} + \Gamma^i_{jk} \frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} = 0$:

$\frac{d}{d\lambda}(2g_{mi}\frac{dx^i}{d\lambda})-(\frac{\partial g_{mj}}{\partial x^k}+\frac{\partial g_{mk}}{\partial x^j})\frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda}+ \Gamma^i_{jk} \frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} = 0$
But from the definition of Cristoffel symbol of the 1st kind $[jk,m]=2g_{mi}\Gamma^i_{jk}$
$2g_{mi}\Gamma^i_{jk} = 2[jk,m] = 2\frac{1}{2} (\frac{\partial g_{km}}{\partial x^j}+\frac{\partial g_{mj}}{\partial x^k}-\frac{\partial g_{jk}}{\partial x^m}) =$
$=\frac{\partial g_{mk}}{\partial x^j}+\frac{\partial g_{mj}}{\partial x^k}-\frac{\partial g_{jk}}{\partial x^m}$
because the metric tensor is symmetrical, $g_{km}=g_{mk}$;
Plug this in the main equation and get
$\frac{d}{d\lambda}(2g_{mi}\frac{dx^i}{d\lambda})-(\frac{\partial g_{mj}}{\partial x^k}+\frac{\partial g_{mk}}{\partial x^j})\frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda}+(\frac{\partial g_{mk}}{\partial x^j}+\frac{\partial g_{mj}}{\partial x^k}-\frac{\partial g_{jk}}{\partial x^m})\frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} = 0$
Cancel and arrive at the condition for null geodesic
$\frac{d}{d\lambda}(2g_{mi}\frac{dx^i}{d\lambda})-\frac{\partial g_{jk}}{\partial x^m}\frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} = 0$,
$m=1,...,N$

 

[stevendaryl]

Can you show the working ? I'm not sure what you mean because 'i' is not a free index in your expression, it is already summed out.

Finding null geodesics is something I've always found tricky.

The geodesic equation can be obtained from the "lagrangian" $L = \frac{1}{2} g_{\mu \nu} U^\mu U^\nu$. (where $U^\mu = \frac{dx^\mu}{ds}$). The usual Lagrangian equations of motion are:

$\dfrac{d}{ds}(\dfrac{\partial L}{\partial U^\mu}) - \dfrac{\partial L}{\partial x^\mu} = 0$

which leads to:
$\frac{d}{ds}(g_{\mu \nu} U^\nu) - \frac{1}{2}\frac{\partial g_{\alpha \nu}}{\partial x^\mu} U^\alpha U^\nu = 0$

which is the same as the OP. To see that it's the same as the usual geodesic equation, write

$\frac{d}{ds}(g_{\mu \nu} U^\nu) = \frac{\partial g_{\mu \nu}}{\partial x^\alpha} U^\alpha U^\nu + g_{\mu \nu} \frac{d}{ds} U^\nu$

Rearranging gives:

$g_{\mu \nu} \frac{d}{ds} U^\nu + [\frac{\partial g_{\mu \nu}}{\partial x^\alpha} - \frac{1}{2}\frac{\partial g_{\alpha \nu}}{\partial x^\mu}] U^\alpha U^\nu = 0$

Raising the indices gives:

$\frac{d}{ds} U^\mu + [g^{\mu \tau} (\frac{\partial g_{\tau \nu}}{\partial x^\alpha} - \frac{1}{2}\frac{\partial g_{\alpha \nu}}{\partial x^\tau})] U^\alpha U^\nu = 0$

The expression in the "[ ]" is not literally equal to $\Gamma^\mu_{\alpha \nu}$, but the difference is antisymmetric in $\alpha$ and $\nu$, and so it vanishes when you multiply by $U^\alpha U^\nu$.

What's sort of surprising about this way of getting the geodesic equations is that the starting expression, $L$ is ZERO for light-like paths. So it doesn't make sense to vary it in that case to get the lagrangian equations of motion. But it gives you the right answer.

 

[kkz23691]

When starting this thread I was looking for some practical advice how to understand the null geodesic equations. This "affine parameter" in it is hard to understand. Solve the null geodesic, equations, you get the parametric equations of a curve in 3D. What is the difference between this curve parameter and the "affine" one which is supposed to enter the null geodesic equation.

Perhaps solving one very simple problem of this kind (and posting it here) can illustrate the "affine" parameter...

 

[Mentz114]

When starting this thread I was looking for some practical advice how to understand the null geodesic equations. This "affine parameter" in it is hard to understand. Solve the null geodesic, equations, you get the parametric equations of a curve in 3D. What is the difference between this curve parameter and the "affine" one which is supposed to enter the null geodesic equation.

Perhaps solving one very simple problem of this kind (and posting it here) can illustrate the "affine" parameter...

If you solve the equations in $\lambda$, an affine parameter, you will get the same answer you would with any affine parameter. The curve lengths measured by affine parameters are only different by a constant factor which cancels from the solution. It is just that with null geodesics one cannot use $\tau$ ( proper length) because it is not affine, being always zero.

 

[Matterwave]

When starting this thread I was looking for some practical advice how to understand the null geodesic equations. This "affine parameter" in it is hard to understand. Solve the null geodesic, equations, you get the parametric equations of a curve in 3D. What is the difference between this curve parameter and the "affine" one which is supposed to enter the null geodesic equation.

Perhaps solving one very simple problem of this kind (and posting it here) can illustrate the "affine" parameter...

The parameters you use to parametrize time-like and space-like geodesics are affine parameters too. Don't get too caught up on the terminology. The reason the parameter is called affine is that it can be related to any other affine parameter by an affine transformation. Any affine parameter $t$ can be related to any other affine parameter $\lambda$ via a transformation $t=a\lambda+b$ where $a,b\in \mathbb{R}$.

If you try to parametrize the geodesic with a parameter $\mu=t^2$; however, you will end up with a non-affine geodesic. In GR, we only ever deal with affine geodesics, one does not have to worry about geodesics with non-affine parameters. The arc-length of curves is an affine parameter (for non-null curves).

 

[stevendaryl]

When starting this thread I was looking for some practical advice how to understand the null geodesic equations. This "affine parameter" in it is hard to understand. Solve the null geodesic, equations, you get the parametric equations of a curve in 3D. What is the difference between this curve parameter and the "affine" one which is supposed to enter the null geodesic equation.

Perhaps solving one very simple problem of this kind (and posting it here) can illustrate the "affine" parameter...

The geodesic equation in the form you wrote it is only valid for an affine parameter $s$. Or rather, I should say that any solution to that differential equation will be affinely parametrized. The more general kind of parametrization has a more complicated equation:

$\frac{d}{ds} (g_{\mu \nu} U^\nu) - (\frac{\partial}{\partial x^\alpha} g_{\mu \nu}) U^\mu U^\nu = g_{\mu \nu} U^\nu \frac{d}{ds} log(\sqrt{g_{\alpha \beta} U^\alpha U^\beta})$

(or something close to that).

 

[stevendaryl]

The geodesic equation in the form you wrote it is only valid for an affine parameter $s$. Or rather, I should say that any solution to that differential equation will be affinely parametrized. The more general kind of parametrization has a more complicated equation:

$\frac{d}{ds} (g_{\mu \nu} U^\nu) - (\frac{\partial}{\partial x^\alpha} g_{\mu \nu}) U^\mu U^\nu = g_{\mu \nu} U^\nu \frac{d}{ds} log(\sqrt{g_{\alpha \beta} U^\alpha U^\beta})$

(or something close to that).

I'm not 100% sure that that more general equation holds for null geodesics. For slower-than light geodesics, the affine parameter can be characterized by saying that the "velocity" vector $U^\mu = \frac{dx^\mu}{ds}$ has constant magnitude. That is $g_{\mu \nu} U^\mu U^\nu$ is a constant. For null geodesics, that doesn't help, because the "magnitude" of the "velocity" vector is always zero, no matter what the parametrization.

 

[kkz23691]

It depends on the initial conditions. You need a point and a vector. If the vector is null, then the resulting geodesic will be null.

Could you provide a short example to illustrate the initial point and the initial vector?

 

[pervect]

Could you provide a short example to illustrate the initial point and the initial vector?

Let's use a 2d flat space example, where a geodesic is a straight line.

You can specify a straight line by its slope, which is constant everywhere along the line, and one point on the line. Neither alone is sufficient to specify the line, you need both. If you are considering an unparameterized line, the slope in unparameterized form would be dy/dx. But to compare with the geodesic equations, it's better to think of x and y being functions of some paramter, which we will call s. We can define a general parameterized curve x(s) and y(s), and if we have dx/ds = constant and dy/ds = constant we have a constant slope (because dy/dx = (dy/ds) / (dx/ds), and curve with a constant slope is a straight line.

Specifying only the slope of a line does not specify the line, you need to specify the slope and one point on the line to specify the line.

If you write out the metric for a plane, the Christoffel symbols $\Gamma^{i}{}_{jk}$ are all zero, and as a consequence when we write the geodesic equation for a geodesic on a plane, it becomes, d^2 x / ds^2 = 0 and d^2 y / ds^2 = 0, This directly imples the same condition we described above, since d^x / ds^2 = 0 implies that dx/ds = constant, the same for y.

There's some fine print here: you might argue (correctly) that only the ratio (dy/ds) / (dx/ds) has to be constant to have a straight line. The fine print that I'll give brief lip servie to is that the parameterization where s is constrained such that dy/ds = constant and dx/ds = constant is a special parameterizationof the line called an affine parameteriztion. It' makes the math a lot simpler, and it can be shown that you do not lose any generality by requiring an affine parameterization, though it would be distracting to try to prove this (as well as too much work), so I won't attempt it.

The abstract point of view is that the derivative operated d/dx and d/dy form a vector space. It might be helpful to review the abstract notion of a vector space from your linear algebra textbook, to see why the derivative operators form a vector space. After you are comfortable with that, you should try to wrap your head around the notion that this abstract vector space actually represents the concept of the slope and/or direction of a line.

 

[kkz23691]

I get the idea of this example; it's just the previous poster (post #2) seemed to indicate that choosing an initial point and a null initial vector will guarantee a null geodesic coming out of the solution of $\frac{d}{d\lambda}(2g_{mi}\frac{dx^i}{d\lambda})-\frac{\partial g_{jk}}{\partial x^m}\frac{dx^j}{d\lambda}\frac{dx^k}{d\lambda} = 0$.

So I was wondering what this means in the general case. Say, we have a metric $ds^2=Adr^2+Bd\theta^2+Cdz^2+Ddt^2$. I have seen textbooks give initial conditions $r=R$, $\theta=z=t=0$ and $\dot{z}=\dot{r}=0$ and claiming that using these initial conditions, plugging in the above metric in the above null geodesic condition will output a null geodesic.

Now, according to post#2, somehow the initial point and the initial vector (where is it?) only can guarantee that the geodesic is indeed null. Therefore, somehow in the above textbook example there is an initial null vector given. I see only the time derivatives of $r$ and $z$, but what about $\dot{\theta}$...?

I am trying to understand this in order to understand the "initial null vector". Does this make sense?

 

[PeterDonis]

choosing an initial point and a null initial vector will guarantee a null geodesic coming out of the solution

This is just a special case of the general theorem that a point and a vector at that point uniquely determine a geodesic.

I have seen textbooks give initial conditions $r=R$, $\theta=z=t=0$ and $\dot{z}=\dot{r}=0$ and claiming that using these initial conditions, plugging in the above metric in the above null geodesic condition will output a null geodesic.

Can you give a reference? These conditions look incomplete to me (as they apparently do to you). But without seeing the context of the actual reference, it's hard to know what the authors of the textbook intended.

 

[kkz23691]

L. Godinho, J. Natario "An Introduction to Riemannian Geometry With Applications to Mechanics and Relativity" contain similar statements... I can try to look up others too.

But clearly the goal in these was to derive a parametric equation (in 3d) of the null geodesic curve, given the metric and initial conditions. Would you think that the derivatives are incomplete, or something else?

 

[PeterDonis]

clearly the goal in these was to derive a parametric equation (in 3d) of the null geodesic curve, given the metric and initial conditions.

First, it would be in 4d, not 3d, correct? The metric you gave is for a 4d spacetime.

If the goal was as you say, then the initial conditions look incomplete; a unique initial vector has not been specified (unless there are other conditions given elsewhere). Also, the metric is not fully specified, because it's not clear what $A$, $B$, $C$, and $D$ are functions of, or what properties those functions are supposed to have (for example, do they have particular limits as, say, $r$ goes to infinity?).

 

[martinbn]

To me it seems that the initial conditions are complete in the following sense. The conditions for the vector are $\dot{r}=\dot{z}=0$, which makes it look as if two more things need to be specified, what $\dot{\theta}$ and $\dot{t}$ are. But reparametrizing the curve would change the vector at the given point by a scalar multiple, so only one more pies of information is needed. But the vector is null, which is one more condition.

 

[PeterDonis]

the vector is null, which is one more condition.

I agree that if it is specified that the initial vector is null, you can find values of $\dot{\theta}$ and $\dot{t}$ that will ensure that. But there will be two such values, not one (corresponding to the two possibilities for the relative signs of $\dot{\theta}$ and $\dot{t}$). And that assumes that the metric coefficients are known, which, as I noted in a previous post, I don't see from the information I have. I don't have the textbook in question so I can't check it directly.

 

[martinbn]

I assumed that the coefficients are given. But you make a good point about the signs. I have a pdf of the book but couldn't find this specific examples.

 

[PeterDonis]

I assumed that the coefficients are given.

Yes, but without knowing what they are, or what their functional dependencies are, you can't compute the Christoffel symbols, so you can't evaluate the geodesic equation. The form of the metric means that some of the Christoffel symbols must vanish, but it gives no help at all regarding the non-vanishing ones.

 

[martinbn]

Yes, but the question was in principle. If the point and the vector at the point are given, there is a geodesic through that point with the given vector as a tangent vector, f
or any metric.

 

[PeterDonis]

If the point and the vector at the point are given, there is a geodesic through that point with the given vector as a tangent vector, for any metric

Yes, agreed. I'm just wondering about the specific example in the textbook, because if that's all they wanted to say, they could have just said it and pointed out that it's a general theorem. It seems like there was more to that specific example, but it's hard to be sure.

 

[kkz23691]

I'm just wondering about the specific example in the textbook...

Sorry to be a bit scattered, I have a number of relativity texts and have been looking at them, just don't have the time today to go through them to find a close example. I thought the problem and the question made sense in general, as martinbn said:

Yes, but the question was in principle. If the point and the vector at the point are given, there is a geodesic through that point with the given vector as a tangent vector, for any metric.

This is what I thought martinbn meant in his initial post#2. My issue is with the null vector. I understand the meaning of "initial point"; what is the meaning of initial "null vector"? I would assume this is an initial "zero tangent". Is this a correct assumption? The expression for the zero tangent is
$g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}=0$ (the vector has zero length)
which is confusing, because in it I see derivatives of the coordinates w/respect to the parameter $\lambda$, while the initial conditions given in that example were time derivatives $\dot{r}=\dot{z}=0$.
I guess then we can still write
$g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}=g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}(\frac{dt}{d\lambda})^2=0$

But the vector is null, which is one more condition.

martinbn, could you explain the condition for the "null vector"...

 

[martinbn]

I think the derivatives, denoted by the dots, are derivatives with respect to the affine paramter of the curve. That parameter need not have any interpretation as time. As PeterDonis said, perhaps there is more in the example, may be something is understood from the context it was given in.