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Relativity - Constant Acceleration

  1. Jul 16, 2006 #1

    eep

    User Avatar

    Hi, I'm working out of Hartle's book Gravity and had a question about Ch. 5 problem 7. The problem states that a particle has constant acceleration along the x-axis, that is, the acceleration when measured in it's instantaneous rest frame is always the same constant. They want me to get the parametric equations for x and t as functions of proper time [itex]\tau[/itex], assuming at t=0 that x=0 and v=0.

    I assumed that since the acceleration is constant, the three-force is also constant. I can then use

    [tex]
    \frac{dp}{dt} = F = ma
    [/tex]

    for some constant a. Since F is constant I can integrate once to get

    [tex]
    {\gamma}m\frac{dx}{dt} = mat
    [/tex]

    Or, solving for [itex]\frac{dx}{dt}[/itex]

    [tex]
    \frac{dx}{dt} = \frac{at}{\sqrt{1+a^2t^2}}
    [/tex]

    Using the relationship [itex]{d\tau}^2 = dt^2 - dx^2[/itex] I can arrive at

    [tex]
    d\tau = dt\sqrt{(\frac{1}{1+a^2t^2})}
    [/tex]

    Integrating and solving for t and letting [itex]\tau[/itex] = 0 when t=0 I get

    [tex]
    t(\tau) = \frac{{\sin}h(a\tau)}{a}
    [/tex]

    I can then solve for x and get

    [tex]
    x(\tau) = \frac{{\cos}h(a\tau)}{a} - \frac{1}{a}
    [/tex]

    due to the initial conditions. Did I do this right?

    EDIT: I really just need to know if my first couple steps are right, as everything follows from that.
     
    Last edited: Jul 16, 2006
  2. jcsd
  3. Jul 22, 2006 #2
    Are you assuming d^2x/dt^2 = a, or d^2x/dtau^2 = a ? It doesn't look to me as if your solution satisfies either criterion. I have to go, but once I've thought about this problem a bit more I'll try to give me own ideas for a solution.
     
  4. Jul 23, 2006 #3

    eep

    User Avatar

    Neither.

    [tex]
    \frac{dp}{dt} = ma
    [/tex]

    but

    [tex]
    p = {\gamma}m\frac{dx}{dt}
    [/tex]

    so

    [tex]
    a = \frac{d({\gamma}\frac{dx}{dt})}{dt}
    [/tex]
     
  5. Jul 23, 2006 #4
    Well then I disagree with your interpretation of the problem.
    [tex] \frac{d({\gamma}\frac{dx}{dt})}{dt} = \frac{d(\frac{dx}{d\tau})}{dt} [/tex]
    which is the rate of change an external observer would measure in the velocity
    of the particle as measured in it's rest frame. Doesn't seem like that's what they're asking. As you stated it would give the result if a constant force were applied to a particle; that's not the same as constant acceleration in it's rest frame.

    My interpretation would be that[itex] \frac{d^2x}{d\tau} = a [/itex]
    Working this out a bit:

    [tex]
    \frac{d^2\vec(x)}{d\tau^2} = \frac{d\vec(u)}{d\tau} =
    \gamma \frac{d^2x}{dt^2} ( \gamma^3 \frac{dx}{dt}, 1 - \gamma^3 \frac{dx}{dt})
    [/tex]
    [tex]
    a = \gamma \frac{d^2x}{dt^2} (1 - \gamma^3 \frac{dx}{dt})
    [/tex]

    I haven't worked it any further than that. Hartle covers 4-acceleration in that chapter, right? I know it was in that book; I wish I had it handy though.
     
    Last edited: Jul 23, 2006
  6. Jul 23, 2006 #5
    How can F = ma work for an non-inertial frame? :confused:
     
  7. Jul 23, 2006 #6

    eep

    User Avatar

    If

    [tex]\frac{d^2x}{d{\tau}^2} = a[/tex]

    then...

    [tex]\frac{dx}{d\tau} = a\tau[/tex]

    and

    [tex]x(\tau) = \frac{1}{2}a\tau^2[/tex]

    due to initial conditions, no? 4-vectors are covered in the chapter...
     
  8. Jul 23, 2006 #7
    Yeah, that seems all correct to me. And from the normalization condition you can find [itex] \frac{dt}{d\tau} [/itex], and integrate. In fact, it's the same as what you had above.

    I know it seems wrong, because [itex] \frac{dx}{d\tau} [/itex] should go 1 and not infinity, but this isn't a physical situation. The more physical case would've been what you had above, with a constant three force, and in that case the limits check out.
     
  9. Jul 23, 2006 #8

    eep

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    I think that's what was confusing me, I had arrived at this answer before but thought I did something wrong since the limit approaches infinity! Let me make sure I'm understanding the notation right,

    [itex]\frac{dx}{d\tau}[/itex] is the velocity of the particle as measured in it's "rest frame", correct? That confuses me as I always think of the rest frame as having no velocity, as the coordinate system is attached to the particle. [itex]\frac{dx}{dt}[/itex] would be the velocity measured by some other inertial frame, and depends on the relationship between [itex]t[/itex] and [itex]\tau[/itex]. I'm self-teaching myself out of this book so I'm never quite sure if I'm understanding things correctly.
     
  10. Jul 23, 2006 #9
    I misspoke before. Define

    [tex] x' = x - vt [/tex]
    [itex] \frac{dx'}{d\tau} [/itex] is the velocity of the particle as measured in its rest frame, and will always be 0 by definition.
    [itex] \frac{dx}{d\tau} [/itex] is the velocity at which the particle would see the origin moving away, which I think is the proper way to work this problem.

    I'm starting to re-think my solution. It doesn't bother me that
    [itex] \frac{dx}{d\tau} [/itex] goes to infinity as tau does, because tau will be bound by t due to time dilation effects. I worked it out, and

    [tex] \frac{dt}{d\tau} = \sqrt{ 1 + (a\tau)^2} \rightarrow
    t(\tau) = (\tau / 2) \sqrt{1 + (a\tau)^2} + (1/2a)log( (a\tau) + \sqrt{1 + (a\tau)^2})
    [/tex]
    [tex]
    \lim_{\tau\rightarrow\infty} t(\tau) = \sqrt{(2t/a)}
    [/tex]
    This gives

    [tex]
    \lim_{\tau \rightarrow \infty} x(t) = t \rightarrow \frac{dx}{dt} = 1
    [/tex]
    as it should be.

    I'd still feel a bit iffy handing this in, but I can rationalize it in my mind.
     
    Last edited: Jul 23, 2006
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