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Solving this space-time Metric

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    (a) Find ##\dot \phi##.
    (b) Find the geodesic equation in ##r##.
    (c) Find functions g,f,h.
    (d) Comment on the significance of the results.

    2014_B5_Q1.png

    2. Relevant equations

    The metric components are:
    ##g_{00} = -c^2##
    ##g_{11} = \frac{r^2 + \alpha^2 cos^2 \theta}{r^2 +\alpha^2}##
    ##g_{22} = r^2 + \alpha^2 cos^2 \theta ##
    ##g_{33} = (r^2+\alpha^2)sin^2 \theta ##

    3. The attempt at a solution

    Part (a)
    The geodesic equation can be written as
    ##\dot u_\mu = \frac{1}{2} \left( \partial_\mu g_{v\sigma} \right) u^\sigma u^v ##

    The 3rd component reads:
    [tex]\dot u_3 = \frac{1}{2}\left( \partial_3 g_{v\sigma} \right) u^\sigma u^v [/tex]
    Since none of ##g_{ii}## depends on ##\phi##, ##\dot u_3## is simply zero.
    Thus, ##\dot u_3 = 0##. Integrating, ##u_3 = J##. But ##J = u_3 = g_{33}u^3##. Therefore
    [tex]u^3 = \dot \phi = \frac{u_3}{g_{33}} = \frac{J}{(r^2+\alpha^2)sin^2\theta} [/tex]

    Part(b)
    Before we start, let's express LHS in terms of ##u^1##. By the product rule, ##\dot u_1 = \dot g_{11} u^1 + g_{11} \dot u^1##. Also, ##\theta = \frac{\pi}{2}## and ##\dot \theta = 0##.

    Thus the geodesic equation reads:
    [tex]\dot g_{11} u^1 + g_{11} \dot u^1 = \frac{1}{2} \dot r^2 \frac{\partial (g_{11})}{\partial r} + \frac{1}{2} \dot \phi^2 \frac{\partial (g_{33})}{\partial r}[/tex]
    [tex]g_{11} \dot u^1 = \frac{1}{2} \dot \phi^2 \frac{\partial (g_{33})}{\partial r} - \frac{1}{2} \dot r^2 \frac{\partial (g_{11})}{\partial r} [/tex]
    [tex]\left( \frac{r^2}{r^2 + \alpha^2} \right) \ddot{r} = \frac{1}{2} \frac{J^2}{\left( r^2 + \alpha^2 \right)^2} (2r) - \frac{1}{2} \dot r^2 \frac{2\alpha^2 r}{\left( r^2 + \alpha^2 \right)^2} [/tex]
    [tex] r \ddot r = \frac{J^2}{r^2 + \alpha^2} - \dot r^2 \frac{\alpha^2}{r^2 + \alpha^2} [/tex]

    Can't see how this would work..
     
  2. jcsd
  3. Apr 13, 2015 #2

    MarcusAgrippa

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    Find dx, dy, dz from the transformation equations, and substitute into the metric. After expanding the squares and simplifying, you should get what you were asked to find.
     
  4. Apr 15, 2015 #3
    I'm referring to part (b), which is to get the equations in ##r##.
     
  5. Apr 16, 2015 #4
  6. Apr 18, 2015 #5

    MarcusAgrippa

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    Ok. I have a bit of time now. I am not sure what, exactly, you are asking. You have a set of questions, followed by what appears to be an extract from a book, followed by your identification of the components of the metric. As far as (b) is concerned, what exactly are you looking for? Are you trying to find the differential equation for r of the geodesic? Are you looking for the solution of this equation? Are you attempting to solve the problem as stated in the extract, which requires you to make use of another piece of information? Be a bit more explicit about what you are trying to do, and state why you are stuck.
     
  7. Apr 18, 2015 #6
    Thanks very much for taking time out. I'm stuck on part (b) as they want us to show the geodesic equation for ##r## is ##\frac{r^2}{r^2 + \alpha^2} \dot r^2 + \frac{J^2}{r^2+\alpha^2} = B^2##. I tried to do that in part (b), but the form is so different.
     
  8. Apr 18, 2015 #7

    MarcusAgrippa

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    I am not certain what type of geodesic you seek. However, whatever the geodesic type, the quantity
    [itex] -c^2 \dot{t}^2 + \displaystyle \frac{R^2}{r^2+\alpha^2} \ \dot{r}^2 + R^2 \dot{\theta}^2 + (r^2+\alpha^2) \sin^2 \theta\ \dot{\varphi}^2 [/itex]
    is a constant of the motion. Call it C. Its value is zero for light-like geodesics, -1 for time like geodesics, and +1 for space-like geodesics.

    You have already shown that
    1. [itex] \dot{\varphi} = \displaystyle \frac{J}{(r^2 + \alpha^2) \sin^2 \theta} [/itex]
    2. The theta-equation is satisfied by [itex] \dot{\theta}=0, \ \theta = \pi/2 [/itex]
    Inserting these values in the above constant for the motion gives
    [itex] C = -c^2 \dot{t}^2 + \displaystyle \frac{R^2}{r^2+\alpha^2} \ \dot{r}^2 + 0 + (r^2+\alpha^2) \ \left( \displaystyle \frac{J}{(r^2 + \alpha^2) \cdot 1} \right)^2 [/itex]

    The metric does not contain t, so [itex] \dot{t} [/itex] is a constant of the motion.

    Unless I have misunderstood the question, this appears to give the answer that you want.

    A bit of advice: the metric ds^2 can be used to construct a Lagrangian function L by replacing all d(variable) objects on the RHS of the metric by that same variable dotted. In particular, your metric gives
    [itex] L = -c^2 \dot{t}^2 + \displaystyle \frac{R^2}{r^2+\alpha^2} \ \dot{r}^2 + R^2 \dot{\theta}^2 + (r^2+\alpha^2) \sin^2 \theta\ \dot{\varphi}^2 [/itex]
    Putting this Lagrangian into the Euler-Lagrange equations delivers the equations for geodesics. Any (undotted) variable that does not feature explicitly in this Lagrangian yields a constant of the motion, since [itex] \partial L/\partial (variable) =0 [/itex]. Also, since the proper time does not appear in L - never! - L is itself a constant of the motion. Constants of the motion permit you to find first integrals of the geodesic equations without doing any integration.
     
    Last edited: Apr 18, 2015
  9. Apr 18, 2015 #8
    The metric is given by ## -c^2 d\tau^2 = -c^2 dt^2 + \left( \frac{R^2}{r^2 + \alpha^2} \right) dr^2 + R^2 d\theta^2 + (r^2 + \alpha^2)sin^2 \theta d\phi^2##. Dividing both sides by ##dt^2##, we have
    [tex] -c^2 \left( \frac{d\tau}{dt} \right)^2 = -c^2 + \displaystyle \frac{R^2}{r^2+\alpha^2} \ \dot{r}^2 + 0 + (r^2+\alpha^2) \ \left( \displaystyle \frac{J}{(r^2 + \alpha^2) \cdot 1} \right)^2 [/tex]

    So I suppose the constant ##B^2 = c^2 \left(1 - \frac{d\tau}{dt} \right)^2 ##?

    Subsituting ##r = \sqrt{D^2 + (vt)^2}## into the geodesic for ##r##, I have
    [tex]J^2 + (v^2t)^2 = B^2\left[ D^2 + (vt)^2 + \alpha^2 \right] [/tex]

    How do I solve for ##B## and ##J##?
     
  10. Apr 18, 2015 #9

    MarcusAgrippa

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    Presumably, the dot-notation you were using means d/ds, not d/dt. Check it in your notes or text book.
     
  11. Apr 18, 2015 #10
    The geodesic equation is
    [tex]\dot u_\mu = \frac{1}{2} \left( \partial_\mu g_{v\sigma} \right) u^\sigma u^v [/tex]

    where the dot represents a derivative to any affine parameter.

    Taken from Hobson:
    image.png


    In this question, I let it be ##t## to find ##\dot \phi## in the first part. But isn't the metric always
    [tex]c^2 d\tau^2 = -c^2 dt^2 + \left( \frac{R^2}{r^2 + \alpha^2} \right) dr^2 + R^2 d\theta^2 + (r^2 + \alpha^2)sin^2 \theta d\phi^2 [/tex]
    where ##\tau## is the proper time?
     
  12. Apr 18, 2015 #11

    MarcusAgrippa

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    t is not an affine parameter in general.
     
    Last edited: Apr 18, 2015
  13. Apr 18, 2015 #12
    Then strangely the answer to my first part is correct somehow..How did that work?
     
  14. Apr 18, 2015 #13

    MarcusAgrippa

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    Let [itex]\lambda[/itex] be an affine parameter. Another affine parameter is obtained by taking [itex] \lambda' = a \lambda +b [/itex]. s is an affine parameter. In your case, it may have worked because [itex] \dot{t} [/itex] is a constant of the motion. But it is not so in general.

    The metric is not always the one you have quoted. This metric is a particular example.

    All you need do is decide what kind of geodesic your are looking for, then substitute into the equation I gave you and identify terms.

    I have given you enough help. Try thinking.
     
  15. Apr 18, 2015 #14
    Ok, after reading through your comments I realized I have been doing this the wrong way completely. Here's my new attempt:

    Part(a)
    Letting the lagrangian be ##L = -c^2 \dot t^2 + \frac{R^2}{r^2 + \alpha^2} \dot r^2 + R^2 \dot \theta^2 + (r^2 + \alpha^2)sin^2 \theta \dot \phi^2##, where ##\frac{d}{dt} \left( \frac{\partial L}{\partial \dot q_i} \right) = \frac{\partial L}{\partial q_i}## where ##q_i## is a coordinate in the metric. Can I choose the 'dot' to be with respect to ##t##? Choosing ##q_i = \phi##, we have
    [tex] \frac{d}{dt} \left[ 2(r^2+\alpha^2) sin^2 \theta \dot \phi \right] = 0 [/tex]
    [tex] \dot \phi = \frac{J}{(r^2+\alpha^2)sin^2 \theta}[/tex]


    Part(b)
    For ##\theta = \frac{\pi}{2}## and ##\dot \theta = 0##, the lagrangian becomes ##L = -c^2 \dot t^2 + \frac{R^2}{r^2 + \alpha^2} \dot r^2 + \frac{J^2}{(r^2+\alpha^2)}##. Thus shifting the constants to the left gives
    [tex]L + c^2 = B^2 = \frac{R^2}{r^2 + \alpha^2} \dot r^2 + \frac{J^2}{(r^2+\alpha^2)} [/tex]
    This is the first equation.

    For the second equation, I use ##\frac{d}{dt} \left( \frac{\partial L}{\partial \dot r} \right) = \frac{\partial L}{\partial r}##, giving:
    [tex]r \ddot r + 2 \dot r^2 - \frac{2 r^2 \dot r^2}{r^2 + \alpha^2} = \frac{ \alpha^2 \dot r^2}{r^2 + \alpha^2} - \frac{J^2}{r^2 + \alpha^2} [/tex]

    I can replace the term ##\frac{J^2}{r^2 + \alpha^2}## and solve for ##B^2##.

    Part(c)
    Do I simply find ##dx^2##, ##dy^2## and ##dz^2## and compare it to the original metric to find ##f,g,h##? Does the condition ##\dot \theta = 0## and ##\theta = \frac{\pi}{2}## still hold?
    For example, we have
    [tex]dx = \frac{r}{\sqrt{r^2 + \alpha^2}} cos \phi (dr) - \sqrt{r^2+\alpha^2} sin \phi (d\phi)[/tex]
    [tex]dy = \frac{r}{\sqrt{r^2 + \alpha^2}} sin \phi (dr) + \sqrt{r^2+\alpha^2} cos \phi (d\phi)[/tex]
    [tex]dz = cos \theta (dr) - r sin\theta (d\theta) = 0[/tex]

    Substituting them instantly gives back the metric. Does this mean that ##f = g = 1## and ##h## can be anything?
     
    Last edited: Apr 18, 2015
  16. Apr 19, 2015 #15
  17. Apr 20, 2015 #16
    bumpp on part (c)
     
  18. Apr 21, 2015 #17

    MarcusAgrippa

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    Part (a): not quite, though in this particular calculation, it makes no difference in the end. The total derivative in the Euler-Lagrange equation is with respect to an affine parameter: proper time for time-like geodesics, proper distance if geodesic is space-like, and an arbitrary affine parameter if the geodesic is light like. Similarly, the dot on the dotted coordinate must be differentiation with respect to an affine parameter. In the case of part (a), [itex] \dot{\varphi} [/itex] is with respect to an affine parameter for the geodesic. Same for [itex] d/d\tau [/itex], but when you integrate this particular equation, the answer is the same whatever you take the differentiation variable to be. In general, however, replacing [itex] \tau [/itex] with t is not valid unless you use the change rule for differentiation to get the correct derivative.

    (b) In principle correct, but the comment about the derivative and the dot is the same as above. You must use the correct variable, and it is the affine parameter, which t is not, or you will not get the correct answer. In your solution, the over-dot saves you, since the formal calculation looks the same whatever the over-dot represents.

    (c) I think that they want you to substitute the squared d(variable) into the original metric.
     
  19. Apr 21, 2015 #18
    Good point. Yeah this question is a special case, as ##\dot t = constant##, so ##dt = A d\lambda## and the constant ##A## cancels out throughout on both sides of the equation when input into the metric, so I can let ##\lambda## be ##t## in this case.

    So I take it that my method in part (b) is correct, for reasons mentioned above.

    I did that, using
    [tex] dx = \frac{r}{\sqrt{r^2 + \alpha^2}} cos \phi (dr) - \sqrt{r^2+\alpha^2} sin \phi (d\phi) [/tex]
    [tex]dy = \frac{r}{\sqrt{r^2 + \alpha^2}} sin \phi (dr) + \sqrt{r^2+\alpha^2} cos \phi (d\phi) [/tex]
    [tex] dz = cos \theta (dr) - r sin\theta (d\theta) = 0 [/tex]

    My only concern is that I use ##d\theta = d\phi = 0## and ##\theta = \frac{\pi}{2}## to simplify ##dx,dy,dz## before squaring and putting them in the metric. That gives ##f=g=1## and ##h## can be anything.
     
  20. Apr 22, 2015 #19

    MarcusAgrippa

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    No. Substitute into the original metric.
     
  21. Apr 22, 2015 #20
    Do you mean I can't simplify?
     
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