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Geodesic vs null geodesic

  1. Mar 9, 2015 #1
    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?
     
  2. jcsd
  3. Mar 9, 2015 #2

    martinbn

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    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.
     
  4. Mar 9, 2015 #3

    Mentz114

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    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.
     
  5. Mar 9, 2015 #4
    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##
     
  6. Mar 9, 2015 #5

    Mentz114

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    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.
     
  7. Mar 9, 2015 #6

    Matterwave

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    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.
     
  8. Mar 10, 2015 #7
    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##
     
  9. Mar 10, 2015 #8

    stevendaryl

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

    [itex]\dfrac{d}{ds}(\dfrac{\partial L}{\partial U^\mu}) - \dfrac{\partial L}{\partial x^\mu} = 0[/itex]

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

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

    [itex]\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[/itex]

    Rearranging gives:

    [itex]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[/itex]

    Raising the indices gives:

    [itex]\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[/itex]

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

    What's sort of surprising about this way of getting the geodesic equations is that the starting expression, [itex]L[/itex] 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.
     
  10. Mar 10, 2015 #9
    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...
     
  11. Mar 10, 2015 #10

    Mentz114

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    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.
     
  12. Mar 10, 2015 #11

    Matterwave

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    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).
     
  13. Mar 10, 2015 #12

    stevendaryl

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    The geodesic equation in the form you wrote it is only valid for an affine parameter [itex]s[/itex]. 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:

    [itex]\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})[/itex]

    (or something close to that).
     
  14. Mar 10, 2015 #13

    stevendaryl

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    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 [itex]U^\mu = \frac{dx^\mu}{ds}[/itex] has constant magnitude. That is [itex]g_{\mu \nu} U^\mu U^\nu[/itex] 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.
     
  15. Mar 22, 2015 #14
    Could you provide a short example to illustrate the initial point and the initial vector?
     
  16. Mar 22, 2015 #15

    pervect

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    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.
     
  17. Mar 22, 2015 #16
    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?
     
  18. Mar 22, 2015 #17

    PeterDonis

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    This is just a special case of the general theorem that a point and a vector at that point uniquely determine a 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.
     
  19. Mar 22, 2015 #18
    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?
     
  20. Mar 22, 2015 #19

    PeterDonis

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    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?).
     
  21. Mar 22, 2015 #20

    martinbn

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    To me it seems that the initial conditions are complete in the following sense. The conditions for the vector are [itex]\dot{r}=\dot{z}=0[/itex], which makes it look as if two more things need to be specified, what [itex]\dot{\theta}[/itex] and [itex]\dot{t}[/itex] 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.
     
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