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Relativistic Velocity

  1. Mar 15, 2013 #1
    I saw the following scenario on an episode of Cosmos: An interstellar space ship accelerates at 1g for the 1st half of a journey to a distant star and then decelerates at 1g for the 2nd half. So with constant acceleration, the velocity would go beyond c at some point, which is not possible. So how would one find the relativistic versions of

    v(t)=at

    and

    d(t)=0.5at2

    (assuming x0=v0= 0)

    Thanks!
     
  2. jcsd
  3. Mar 15, 2013 #2

    Bill_K

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    The world line of a uniformly accelerating particle is given by

    x = (c2/a) cosh(aτ/c)
    t = (c/a) sinh(aτ/c)

    where τ is the particle's proper time. The velocity is v = x/t = c tanh(aτ/c), which is always less than c.

    For small τ we can expand the functions and recover the Newtonian formulas:
    x ≈ (c2/a)(1 + ½(aτ/c)2 + ...) = c2/a + ½aτ2
    t ≈ (c/a)(aτ/c + ... ) = τ
    v ≈ c(aτ/c - ...) = aτ

    (Note that these nice formulas displace the origin, and use x0 = c2/a)
     
  4. Mar 16, 2013 #3

    Bill_K

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    Sorry, should be v = dx/dt = (dx/dτ)/(dt/dτ) = (c sinh(aτ/c))/(cosh(aτ/c)) = c tanh(aτ/c)
     
  5. Mar 16, 2013 #4

    jtbell

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    See also http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Mar 16, 2013 #5
    If an agent in the initial rest frame expends energy to accelerate the traveler beyond C, he finds increasing (classically unexpected) inertial resistance, with diminishing returns as the traveler appears to get "heavier" as it accelerates.
     
  7. Mar 16, 2013 #6

    Bill_K

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    1977ub, I believe you misread the question. We're not talking about the effect of an applied external force, presumably constant in the initial rest frame. This is a spaceship accelerating under its own thrust, constant in its own instantaneous rest frame.
     
    Last edited: Mar 16, 2013
  8. Mar 16, 2013 #7

    Meir Achuz

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    This is a standard problem. The acceleration g is in the instantaneous rest system of the spaceship, not in the original fixed system.
    For an acceleration g in the instantaneous rest system of the spaceship, the velocity is (with c=1)
    [tex]v=\frac{gt}{\sqrt{1+g^2t^2}}=\frac{dx}{dt}[/tex].
    The distance traveled is
    [tex]x=\left(\sqrt{1+g^2t^2}-1\right)/g[/tex].
     
    Last edited: Mar 16, 2013
  9. Mar 16, 2013 #8
    Actually I didn't finish my thought with that post because I couldn't figure out / remember certain things about the issue in the ship's perspective. It's pretty straightforward to decide the velocity of the ship from the perspective of the original frame, and issues which stand in the way of success of getting to c for an external agent.

    For someone on the ship, how do they determine what is standing in the way of getting to c? How do they determine their rate wrt the original frame at any given moment?

    If they trust / use the original frame's mileposts, which are encountered to be coming closer and closer together, will the decide that they must be > c wrt the original frame?
     
    Last edited: Mar 16, 2013
  10. Mar 16, 2013 #9

    Dale

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    If they divide distance in the launch frame by their own proper time then they will indeed get a number larger than c. This is not a violation of SR because the specified quantity is not a velocity.
     
  11. Mar 16, 2013 #10
    Ok Thanks that's what I thought. [everyone on the ship breaks out the champagne]
     
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