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World line of a particle

  1. Oct 11, 2017 #1
    1. The problem statement, all variables and given/known data
    A particle is moving along the x-axis. It is uniformly accelerated, in the sense that the acceleration measured in its instantaneous rest frame is always g, a constant. Find x and t as a function of the proper time ##\tau## assuming that the particle passes through ##x_0## at time t = 0 with zero velocity.

    2. Relevant equations


    3. The attempt at a solution
    I am not sure I understand the meaning of "in the sense that the acceleration measured in its instantaneous rest frame is always g, a constant". The way I thought of doing it was ##dt = \gamma d \tau = \frac{d\tau}{\sqrt{1-(gt)^2}}## which implies ##\sqrt{1-(gt)^2}dt=d\tau##, then I integrate and get ##t(\tau)## and a similar reasoning for ##x(\tau)##. However I am not sure this is correct, as this implies that g is constant in the frame where we calculate x and t and I am not sure this is equivalent to what the problem states. Can someone tell me how to approach this correctly? Thank you!
     
  2. jcsd
  3. Oct 11, 2017 #2

    Orodruin

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    This would imply that the velocity was given by ##gt## which is not the case. What it means is that, in the instantaneous rest frame ##S'##,
    $$
    \frac{d^2 x'}{dt'^2} = g.
    $$
    From there it is up to you to relate the quantities in the instantaneous rest frame to the original frame.
     
  4. Oct 11, 2017 #3
    The way I was thinking to do it was to consider the 4 acceleration, which in MCRF would be ##(0,g,0,0)##, while in an inertial frame, call it S (where we want to compute x and t), it would be, based on Lorentz transformations, ##g \gamma (v,1,0,0)##, where v is the instantaneous velocity. Thus, in S we would have for the 1 component ##g \gamma = \frac{dv}{dt}=\frac{dv}{\gamma d\tau}##, which implies ##g d\tau = \frac{dv}{\gamma^2}## and by integration I can get ##v(\tau)## and similar reasoning for x and t. Is this correct?
     
  5. Oct 11, 2017 #4

    Orodruin

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    Well, if you are familiar with 4-vectors it all becomes much simpler. Try to answer the following questions:
    • What is the relationship between 4-acceleration and 4-velocity?
    • What is always true about the 4-velocity?
    • Based on the answer to the previous question, how can you generally parametrise the 4-velocity?
    • What happens when you insert that parametrisation into the relationship between the 4-acceleration and 4-velocity?
    • What happens when you then demand that ##A^2 = -g^2##?
    The answer to those questions should give you the 4-velocity as a function of the proper time, which you can integrate to get the space-time coordinates as a function of the proper time. No reference to the instantaneous rest frame will ever be necessary.
     
  6. Oct 11, 2017 #5
    I tried something related to this, but got stuck on the way. So we have ##a^\alpha = \frac{d u^\alpha}{d\tau}##. In the MCRF, ##u^\alpha = (1,0,0,0)## and in all frames ##u^\alpha u_\alpha = -1## and in another frame, ##u^\alpha = \gamma(1,v_x, v_y, v_z)##. But I am not sure how to connect all these.
     
  7. Oct 11, 2017 #6

    Orodruin

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    So first off, so you are using a different sign convention than I am for the inner product, so the last bullet should read ##A^2 = g^2##.

    Correct. This ticks off the first bullet.

    So the ##u^2 = -1## ticks off the second bullet point. Now, your particle is only moving along the x-axis so you can neglect the y and z components completely. It is true that you can parametrise the 4-velocity as ##u = \gamma(1,v)##, where ##v## is the parameter, but there is a better way for this problem. Consider the components of ##u##. In terms of those components, which form does ##u^2 = -1## take?
     
  8. Oct 11, 2017 #7
    But ##u^2 = - 1## means that ##\gamma^2 v^2 - \gamma^2 = - 1## which is obvious. Parametrizing in terms of proper time ##\tau##, I guess would be better? We have ##u^\alpha = \frac{d x^\alpha}{d \tau}##, but not sure exactly how to proceed.
     
  9. Oct 11, 2017 #8

    Orodruin

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    No, I am saying to write down the constraint ##u^2 = -1## in terms of the components ##u^0## and ##u^1##. Once you have done that we can discuss how to parametrise it better.

    Edit: It is true that ##u^2 = -1## can be parametrised with ##v## in the way you mention and differentiating it would give you a differential equation for ##v## in terms of ##\tau##. However, there is a much more elegant parametrisation that will essentially hand you the answer on a silver platter.
     
  10. Oct 11, 2017 #9
    Do you mean ##u_1^2-u_0^2 = -1##?
     
  11. Oct 11, 2017 #10

    Orodruin

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    Yes. So how can you parametrise that equation in a straightforward manner? Does it remind you of something?

    Edit: If it doesn't remind you of something. Would it remind you of something if you changed the minus signs for plusses?
     
    Last edited: Oct 11, 2017
  12. Oct 11, 2017 #11
    Oh, so I can write ##u^\alpha=(cosh(\tau),sinh(\tau))## and so ##a^\alpha = (sinh(\tau),cosh(\tau))##
     
  13. Oct 11, 2017 #12

    Orodruin

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    Almost. The parameter is not necessarily the proper time ##\tau##, but generally a function of the proper time (since ##u## is - also it would not match dimensionally to put a dimensional argument of the hyperbolic functions). I suggest you call the parameter ##\theta## for nomenclature reasons.
     
  14. Oct 11, 2017 #13
    But still how can I use the initial condition? In the MCRF ##u^alpha## is still ##(1,0,0,0)##
     
  15. Oct 11, 2017 #14

    Orodruin

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    Patience, we will get there. The initial conditions you have on the position and the velocity will turn into integration constants for the coordinates and an initial condition for the parameter ##\theta## .
     
  16. Oct 11, 2017 #15

    Orodruin

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    Also note that we will make no reference whatsoever to the instantaneous rest frame. Everything here is being done in the original inertial frame.
     
  17. Oct 11, 2017 #16
    So zero velocity initially means that ##\theta(0)=0##. Then we would have ##x - x_0 = \int_0^{\theta_1} sinh(\theta) = cosh(\theta_1)-1##
     
  18. Oct 11, 2017 #17

    Orodruin

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    Yes.

    Not so fast. The 4-velocity is the derivative of ##x^\mu## with respect to proper time, not with respect to the parameter we used. In order to integrate ##dx^\mu/d\tau = u##, you must first find ##u## as a function of proper time ##\tau##, which means finding an expression for ##\theta## in terms of ##\tau##. I suggest going back to the bullet list of post #4. You have ticked off the third bullet by letting ##u^\mu = (\cosh(\theta),\sinh(\theta))##. Bullets four and five remain.
     
  19. Oct 11, 2017 #18
    Ok, so for number 4 we have ##a^\alpha = \frac{v^\alpha}{d\tau}=(sinh(\theta)\frac{d\theta}{d\tau},cosh(\theta)\frac{d\theta}{d\tau})##
     
  20. Oct 11, 2017 #19

    Orodruin

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    Indeed. And number 5?
     
  21. Oct 11, 2017 #20
    ##\frac{d\tehta}{d\tau}=g##, so we get ##\theta = g\tau## and then I can integrate ##u^\alpha## to get x
     
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