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Trajectories in spacetime

  1. Feb 2, 2008 #1
    1. The problem statement, all variables and given/known data

    The equations of motion are:
    [tex] \frac{d^2 x^\mu}{d\tau ^2} = 0 [/tex]
    Write down the tra jectory that’s the most general solution to them. What are the conditions
    on your solution if the particle is:
    (a) Massive?
    (b) Massless?

    3. The attempt at a solution

    I am not too sure what is meant by write down, but judging from equation I can assume that the path is straight worldlinefor a massive particle in spacetime, since there is no four-force acting on it. What my main problem is the massless particle, as it is obvious that the acceleration is 0 from the fact that massless particles should always travel at c. Any idea what is meant by that. I am really sorry for asking, but the prof is the kind of person who states that everything is supposed to be from kindergarten.. string theorists don't you love them.


  2. jcsd
  3. Feb 2, 2008 #2


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    The most general solution to the differential equation:

    [tex] \frac{d^2 y}{dx^2} = 0 [/tex]


    [tex] y(x) = ax + b [/tex]

    So the solution to the equation you have is just something of the above form for every coordiante (ie, every value [itex]\mu[/itex] can take). Yes, this is a straight line through spacetime.

    This will give 2D arbitrary parameters (where D is the dimension of spacetime, which is probably 4, but you never know when a string theorist is teaching), specifying two vectors: an initial position and a velocity. The velocity must be timelike (ie, v^2 = - v_0^2 + v_1^2 + ... + v_D^2 < 0) for massive particles and lightlike ( v^2 = 0) for massless ones. This doesn't follow from the equation, but from the definition of mass:

    [tex] m^2 = - p^2 \propto - v^2 [/tex]

    where the proportionaliy constant is positive.
    Last edited: Feb 2, 2008
  4. Feb 2, 2008 #3
    That is what i have figured. It would be something along the lines of

    [tex] x^\mu = a^\mu \tau + b^\mu [/tex]

    Where a^\mu and b^\mu are the parameters, which we can not find cause of lack of information.

    And D = 0 in our class
  5. Feb 2, 2008 #4


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    D=0? That's funny dimension for even a string theorist to be working in. In zero dimensions, no one can hear you scream. :)
  6. Feb 2, 2008 #5
    ups sry i really should wear my glasses, did not see what, well i thought it was the 0, 1, 2, 3 thing.

    Even when i read his replies to my emails no one can hear me scream....
  7. Feb 3, 2008 #6
    I was just working on the problem of the massless particle again, if it is moving at c, shouldn't it be a null like dependence rather then a timelike?
  8. Feb 3, 2008 #7


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    Sure. StatusX just said 'lightlike' instead of null. It's the same thing.
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