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General relativity, geodesic question

  1. Jun 25, 2017 #1
    1. The problem statement, all variables and given/known data

    question attached

    geodesicgrii.png

    2. Relevant equations


    3. The attempt at a solution

    Attempt :

    Check if ##V^{\alpha}\nabla_{\alpha}V^u=0##

    Since Minkowski space, connection tensors/christoffel symbols are zero so this reduces to:

    ##V^{\alpha}\partial_{\alpha}V^u=0##

    where ##\partial_{\alpha}=\frac{\partial}{\partial x^{\alpha}}##

    Where ##V^{\alpha}=\frac{\partial x^{\alpha}}{\Lambda}## is the tangent vector

    ##\frac{\partial x^{\alpha}}{\Lambda} \partial_{\alpha}V^u = \frac{\partial x^{0}}{\Lambda} \partial_{0}V^u + \frac{\partial x^{1}}{\Lambda} \partial_{1}V^u +\frac{\partial x^{2}}{\Lambda} \partial_{2}V^u +\frac{\partial x^{3}}{\Lambda} \partial_{3}V^u
    = 4 \frac{\partial V^u}{\Lambda} ## (using the chain rule)

    which is 4 separate equations for ## u=0,1,2,3,4 ##, obviously these are not zero.

    Have I done something wrong and should I have expected a summation ?

    Or would I argue that these can not simultaneously all be zero for the same ##\Lambda## since cosh and sinh are independent?

    Many thanks.
     
  2. jcsd
  3. Jun 26, 2017 #2
    Apologies typo edit:

    Where ##V^{\alpha}=\frac{\partial x^{\alpha}}{ \partial \Lambda}## is the tangent vector

    ##\frac{\partial x^{\alpha}}{\partial \Lambda} \partial_{\alpha}V^u = \frac{\partial x^{0}}{\partial\Lambda} \partial_{0}V^u + \frac{\partial x^{1}}{\partial\Lambda} \partial_{1}V^u +\frac{\partial x^{2}}{\partial\Lambda} \partial_{2}V^u +\frac{\partial x^{3}}{\partial\Lambda} \partial_{3}V^u
    = 4 \frac{\partial V^u}{\partial\Lambda} ## (using the chain rule)
     
  4. Jun 28, 2017 #3

    strangerep

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    Which part of the question are you trying to answer in your attempt in post #1? (If you're trying to answer part (c), that depends on the answer to part (a) which you haven't shown.)

    Also, which textbook are you working from?

    [And btw, I notice in several of your HW posts you make no attempt to write down any "relevant equations". That's not helping.]
     
  5. Jun 29, 2017 #4
    I have not listed relevant equations since they are either included in the question or attempt.

    I am attempting part c)

    For part a) you simply differentiate w.r.t ##\Lambda## and then compute ##g_{uv}V^vV^u## and check it's sign. ## < 0 ## here being time-like from the chosen metric convention.

    Therefore I would identify a failure to make a connection between these two parts of the question as a mistake in the attempt, rather than purposely not showing to attempt question a, because I thought this wasn't as issue to me. Pretty much then, my approach to the question is wrong, which is all needed to be said.
     
  6. Jun 29, 2017 #5

    strangerep

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    Are you sure? What are your components ##V^\mu## explicitly?

    For part (c), can't you simply evaluate ##dV^\mu/d\lambda## (since the connection is 0 in Minkowski spacetime) ?

    And you didn't tell me which textbook the question is from. Or is it from private lecture notes?
     
  7. Jul 1, 2017 #6
    My second post was exactly this? To evaluate ##dV^\mu/d\lambda=0##. My question was this is 4 seperate equations? So to check whether it's a geodesic or not you check whether all four equations are consistent in the ##\lambda## which solves them?

    For part a)
    ##V^u=( sinh \Lambda, \frac{1}{\sqrt 3} coshh \Lambda, \frac{1}{\sqrt{2}}cosh \Lambda, \frac{1}{\sqrt{6}}cosh \Lambda) ##

    So the check for part c) returns the components of the original vector as four seperate equations. My question was concerning solving them anyway. THIS: So to check whether it's a geodesic or not you check whether all four equations are consistent in the ##\lambda## which solves them?
     
  8. Jul 1, 2017 #7

    strangerep

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    I think you're overcomplicating this. To be a geodesic you'd need ##dV^\mu/d\lambda=0## for all ##\lambda##.
     
  9. Jul 2, 2017 #8
    but ##\mu## runs over ##\mu=0,1,2,3## so need to solve, there are four equations to check, what am i over-complicated, don't we need to solve the following system of equations? To which the answer will be inconsistency due to the non-linear dependence of sinh and cosh?

    ## \frac{dV^{0}}{d\Lambda}=0##
    ## \frac{dV^{1}}{d\Lambda}=0##
    ## \frac{dV^{2}}{d\Lambda}=0##
    ## \frac{dV^{3}}{d\Lambda}=0##
     
  10. Jul 3, 2017 #9

    strangerep

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    I get the feeling you're not reading my reply hints carefully enough. I'll try one more time:
    Note the "for all ##\lambda##" part.
     
  11. Jul 4, 2017 #10
    oh okay.
    but there are four equations?
    so this would have only happened if ## V^{u} ## components were constants indepedent of ##\Lambda##?
     
  12. Jul 4, 2017 #11

    strangerep

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    So? It's a single vector (which has 4 components).

    If by "this", you mean "the curve would be a geodesic only if..." then yes -- because you're working with Minkowski spacetime in this question.
     
  13. Jul 5, 2017 #12
    As in this is the expected result since we are in flat space-time?
    I don't think I'm picking up what you are saying properly, could you explain a tad more?
    greatly appreciated, thanks
     
  14. Jul 5, 2017 #13

    strangerep

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    In Minkowski spacetime, with metric ##(-1,+1,+1,+1)##, the Christoffel connection coefficients are all 0. So the equation for a geodesic curve (in terms of an arbitrary parameter ##\lambda##) is just $$0 ~=~ \frac{d^2 x^\mu(\lambda)}{d\lambda^2} ~\equiv~ \frac{dV^\mu(\lambda)}{d\lambda} ~.$$ So if this equation is not satisfied (for any component), then the curve is not a geodesic.
     
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