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I Uniform acceleration and hyperbola

  1. Jul 6, 2016 #1
    hi, There is a situation that if t^2-x^2=constant in minkowski space we have uniform acceleration(hyperbola). But, when I look at the derivation of this circumstance, the only thing I have found is: if there is a gravitational field uniformly, then we have uniform acceleration. But I can not make any connection with hyperbola. How does hyperbola prove a uniform acceleration ????? Thanks in advance...
     
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  3. Jul 6, 2016 #2
    Define what you mean by a gravitational field.

    The "gravitational field" you describe has zero spacetime curvature.
     
  4. Jul 6, 2016 #3

    PeterDonis

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    Where have you looked? Have you looked at how the proper acceleration of an object traveling on a particular worldline (the hyperbola in this case) is computed?
     
  5. Jul 6, 2016 #4

    PeterDonis

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    Also, as you have written this formula, the constant will have to be negative for the curve to be timelike (which it must be to be a valid worldline for an observer). You might want to write it as ##x^2 - t^2 = \text{constant}## so the constant can be positive; that will turn out to make more sense physically.
     
  6. Jul 6, 2016 #5
    Let me a little bit clear, I just want to know under which circumstances ##x^2 - t^2 = \text{constant}## means a uniform acceleration ???
     
  7. Jul 6, 2016 #6

    pervect

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    Here is a hint. ##dx^2 / dt^2## is not constant, what is constant is ##dx^2 / d\tau^2## where ##\tau## is proper time. http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html may be somewhat helpful, though it just gives the formulas, and not the derivation.

    So a more precise statement would be that the hyperbola is a curve of constant proper acceleration.

    To work out the problem for yourself, you need the following definition of proper time ##\tau## from special relativity.

    $$c^2 \,(\Delta \tau)^2 = c^2 (\Delta t)^2 - (\Delta x)^2$$

    [add] While this would work, I can be clearer. Let

    $$\beta = \frac{v}{c} \quad \gamma = \frac{1}{\sqrt{1-\beta^2}}$$

    Then ##\Delta t = \gamma \Delta \tau##, proper time is different from coordinate time because of "time dilation".


    The rest is a lot of algebra. Probably the easiest thing to do is to verify the pre-existing results from "The Relativistic Rocket" link. If you're not familiar with proper time, you may have to research that a bit first, though I tried to give you the equation that you really needed.

    A detour into "rapidity" could be helpful, see for instance https://en.wikipedia.org/wiki/Rapidity

    Rapidities, denoted by w in the Wiki article, have the property that rapidities add, unlike velocities. So a curve of constant proper acceleration is a curve where, rather than v= at, we have ##w = a \tau##.

    [add]
    Note that from the wiki
    $$ \beta = \tanh w \quad \gamma = \cosh w$$

    When v <<c, ##w \approx \beta## because ##\tanh w \approx w - w^3/3 + ...## via a Taylor series expansion. Similiarly when v<<c, ##d\tau = dt##.

    To calculate proper acceleration, we can calculate dv/dt in a frame in which the object is at rest. But this is the same as ##dw/d\tau## in a frame where the object is at rest.

    Because velocities do not add ##\left( v(t+\Delta t) \,(minus)\, v(t) \right) / \Delta t ## requires a relativistic velocity subtratction operation when the object is not at rest, because velocities do not add in special relativity. But rapidities do add, so ##\left( w(\tau + \Delta \tau) - w(\tau) \right) / \Delta \tau## uses a simple arithmatical subtraction. Furthermore ##\tau## is observer independent, so the calculation is much easier.
     
    Last edited: Jul 6, 2016
  8. Jul 6, 2016 #7

    PeterDonis

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    Actually it's ##w = a \tau##, where ##\tau## is the proper time of the observer following the curve.
     
  9. Jul 6, 2016 #8

    pervect

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    I fixed the typo - and I added a bit more explanation as well. Then I fixed the typos I added in the expansion. Hopefully it's all good now.
     
    Last edited: Jul 6, 2016
  10. Jul 6, 2016 #9
  11. Jul 6, 2016 #10

    stevendaryl

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    This is not a derivation of the equation of the hyperbola, but it's an argument that the hyperbola does represent constant (proper) acceleration.

    First, if a rocket travels so that [itex]x^2 - t^2 = X^2[/itex], then [itex]x = \sqrt{X^2 + t^2}[/itex]. When [itex]t \ll X[/itex], we can expand the square root in a Taylor series to get:

    [itex]x \approx X + \frac{1}{2} \frac{t^2}{X}[/itex]
    [itex]\frac{dx}{dt} \approx \frac{1}{X} t[/itex]
    [itex]\frac{d^2 x}{dt^2} \approx \frac{1}{X}[/itex]

    So the hyperbola [itex]x^2 - t^2 = X^2[/itex] gives rise to motion which, for small [itex]t[/itex], when the rocket is moving nonrelativistically, means an acceleration of [itex]\frac{1}{X}[/itex].

    But what about when [itex]t[/itex] gets large? No problem. After a certain amount of time, the rocket is traveling at some speed [itex]v[/itex]. The proper acceleration is the acceleration as measured in the frame in which the rocket is momentarily at rest. So we need to compute things from the frame that is traveling at speed [itex]v[/itex] with respect to the original frame. Let [itex]x', t'[/itex] be the coordinates for this frame. Now, what does the rocket's position as a function of time look like in this frame? Amazingly, it's exactly the same as in the original frame!

    [itex]x' = \gamma(x - vt)[/itex]
    [itex]t' = \gamma(t - vx)[/itex]
    [itex](x')^2 - (t')^2 = [/itex] (after some algebra, using that [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]) [itex]x^2 - t^2[/itex].

    So in this new frame, the rocket is also traveling according to the equation [itex]x' = \sqrt{X^2 + (t')^2}[/itex]. So we can conclude that when [itex]t' \ll X[/itex], the rocket is moving nonrelativistically, with an accleration of [itex]\frac{1}{X}[/itex].

    So in every frame, when the rocket comes momentarily at rest in that frame, it will have an acceleration of [itex]\frac{1}{X}[/itex]. So [itex]\frac{1}{X}[/itex] is always the proper acceleration (the acceleration as measured in its momentary rest frame).

    The other interesting thing about hyperbolic motion is that when the rocket comes to rest in any rest frame, the time in that rest frame is always [itex]t=0[/itex]. (Assuming that all rest frames have the same origin for their coordinate systems.)
     
    Last edited: Jul 7, 2016
  12. Jul 7, 2016 #11
    ok guys thanks for all your nice answers.....
     
  13. Jul 8, 2016 #12

    vanhees71

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    Physically you get it from the motion of a charged particle in a homogeneous static electric field. See p.30 of

    http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf
     
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