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Rocket accelerating to relativistic speeds

  1. Feb 4, 2014 #1
    1. The problem statement, all variables and given/known data
    A rocket is accelerated for 10 years at a constant rate of 10 m/s². Then starts braking at -10 m/s². Then goes all the way back to its starting point by again first accelerating 10 years at 10m/s² and then braking at the opposite rate for 10 years. How much time has it been for an observer on the resting frame situated on the departure point?

    2. Relevant equations

    The Lorentz boost transformation in the x direction: http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction

    3. The attempt at a solution

    I think a possible solution would be to substitute dt and dt' in the Lorentz transformation for the time coordinate and integrate dt', but I feel like something is wrong here:
    [itex]dt' = \gamma(t-\frac{vdx}{c^2}) = \frac{dt-\frac{v}{c^2}\frac{dx}{dt}dt}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{dt-\beta^2 dt}{\sqrt{1-\beta^2}} = \sqrt{1-\beta^2} dt

    Can you tell me if I am using the right approach?
  2. jcsd
  3. Feb 4, 2014 #2


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    The most simple realization of what's the action of a constant force in the non-relativistic limit is the motion of a charged particle in a homogeneous electric fields. Restriction to the motion along one dimension as indicated in the question, you have
    [tex]\frac{\mathrm{d} p}{\mathrm{d} t}=m \frac{\mathrm{d}}{\mathrm{d} t} \frac{v}{\sqrt{1-v^2/c^2}}=q E.[/tex]
    The equation is written in terms of the resting observer's frame ("lab frame"), i.e., [itex]t[/itex] is the lab time. Note that here we use the constant (speed-independent!) invariant mass [itex]m[/itex]. You can integrate this equation very easily to get [itex]v(t)[/itex]. Then you just calculate the proper time of the rocket, i.e., the time a travelor inside the rocket measures on his wristwatch by using
    [tex]\tau=\int_0^{t_{\text{end}}} \mathrm{d} t' \sqrt{1-v^2(t')/c^2}.[/tex]
    Note that you have to solve the equations of motion piecewise for [itex]a=q E/m=\pm 10 \mathrm{m}/\mathrm{s}^2[/itex]!
  4. Feb 4, 2014 #3


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    I'm not sure if the increase of momentum with time is constant for observers on earth, if the ship sees the same acceleration all the time.

    I would solve this problem with the concept of rapidity, that is nicely linear in ship time.
  5. Feb 4, 2014 #4
    Thanks for your answer.

    I'm actually given a hint, but it does not help me at all. It tells me that "A masspoint whose "instantanean" inertial system is accelerated at a constant rate a, fulfills the equation[itex]\frac{d(\gamma(v)v)}{dt} = a[/itex]." With your answer I can finally understand where that comes from. more or less. However can you explain what has this to do with a charge on electrical field? You totally lost me there.

    Thank you too mfb, I'll take a look at that page.
  6. Feb 4, 2014 #5


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    It's not so clear what's meant when you say a relativistic particle is moving with constant acceleration. The question is, what's meant by acceleration. Thus, it's better to clearly define it in terms of a physical setup. The easiest equations of motion for relativistic point particles come out as those describing the motion in external electric and magnetic fields. They look like Newton's equation of motion under influence of the usual Lorentz force, but you have to put the relativistic momentum on the left-hand side. The result is, what I've written down for the special case that you only have an electric field. If you set [itex]E=\text{const}[/itex] you precisely get, what you have been given as a hint, because
    Then for a constant electric field you have
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} [\gamma(v) v]=\frac{q E}{m}=a=\text{const}.[/tex]
    Now, it's easy to integrate this wrt., then solve for [itex]v[/itex] and integrate it again to get the position as a function of time! Just try it!
  7. Feb 4, 2014 #6


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    How can something accelerate at 10 m/s² for 10 years in any reference frame?

    Would it then be travelling at $$3.15 * 10^9 m/s$$ (10 times the speed of light) in that reference frame?
  8. Feb 4, 2014 #7

    D H

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    Presumably that 10 m/s² is the rocket's proper acceleration.
  9. Feb 4, 2014 #8
    The 10 m/s^2 is the rocket's proper acceleration, as DH points.

    From the first integral i get [itex]v(t)^2 = \frac{(at)^2}{1+ (\frac{at}{c})^2}[/itex], and substituing i get [itex]\tau = c \int dt' \frac{1}{1+(at')^2} = \frac{c}{g}tan ^{-1}(\frac{gt'}{c})[/itex]. Is that simple? The results are a bit weird: 1,4 years for each 10 year period, when it should be greater thatn 10, right? What am I missing?
  10. Feb 4, 2014 #9

    D H

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    That's correct.

    That's incorrect. Your lost a square root, and you also have ##1+(at')^2##, which is dimensionally incorrect (but that's probably just a typo).

    You'll get a bit more than twice that 1.4 years if you fix that calculation.

    However, what is this value you calculated? The answer is that this is time measured on a clock on the accelerating spacecraft assuming that the ten years represents time as measured by the stationary observer. Your second integral is wrong if that ten years represents time as measured on the spacecraft clock.
  11. Feb 5, 2014 #10
    Ok, now I got i right, the result is [itex]\tau = \frac{c}{a}sinh(\frac{a}{c}t')[/itex]. I've inverted the formula and found an expression for t', but the calculator raises an error... why is the result so extremely high?
  12. Feb 5, 2014 #11

    D H

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    That's the right expression for converting time measured on the accelerating spaceship to time measured by a stationary observer. Why is the result so high? It's only 1.001 months with your t'=1 month, but it's 6.28 months with t'=6 months (a factor of 1.05), 1.195 years with t'=1 year. The spaceship gets up to relativistic speeds after a few months, and from then on it only keeps getting ever close to c. With t'=10 years, the spaceship's velocity relative to the stationary observer is 0.99999999854 c.
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