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Does everything that has to do with velocity

  1. Aug 13, 2008 #1
    Given that you have a constant force applied to any mass, what is the rate at which the mass' acceleration decreases as its velocity approaches c?

    Does everything that has to do with velocity in some way, have a relativistic transformation i.e. t'=f(t)?
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  3. Aug 13, 2008 #2


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    I would define a constant force as something that causes constant proper acceleration. The world line of an object that has constant proper acceleration is a hyperbola (-t2+x2=constant). I have to go to bed, so I'm not going to work out the details now, but maybe this thread will be helpful. (See the first question I asked and DrGreg's answer).

    I'm not sure how to interpret that question. There is a coordinate system called an inertial frame associated with each velocity, and everything that happens in spacetime can be described using the coordinates of any inertial frame. A "relativistic transformation" (a Lorentz transformation) is a coordinate change from one inertial frame to another, i.e. if you know how something looks in one inertial frame, you would use a Lorentz transformation to find out how it looks in another.
  4. Aug 13, 2008 #3
    Since force obviously decreases with distance could you explain what you mean when you say you are applying a constant force to a mass that is rapidly moving away from you?
  5. Aug 13, 2008 #4
    Constant force applied to a mass causes the mass to have a constant acceleration.
  6. Aug 13, 2008 #5


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    Yah, I think you're misunderstanding the question. Pretend he's asking about a rocket with constant thrust.
  7. Aug 13, 2008 #6


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    Here is a relativity calculator you can play with and draw your own conclusions.
  8. Aug 13, 2008 #7
    In that case the rocket acceleration will never descrease and the speed relative to something else will never be c.
  9. Aug 13, 2008 #8
    To further my first question, if we have a relativistic transformation of t, t', as well as of x, x', why can't we take [tex]\frac{x'}{t'^2}[/tex], and have that be a relativistic transformation of a, which then simplifies to


    Does the above equation correctly represent the boundary of a constant acceleration as velocity approaches c?
  10. Aug 13, 2008 #9
    The acceleration has to decrease as the velocity approaches c.
  11. Aug 13, 2008 #10
    No that is not true at all.
  12. Aug 13, 2008 #11
    by a factor of 1/sqrt[1 +(at/c)^2]. So delta v = at / sqrt[1 + (at/c)^2] where a is proper acceleration (ie force applied/proper mass), t is time in the rest frame in which the mass was initially at rest, and v is relative velocity between that rest frame and the accelerated mass.

  13. Aug 14, 2008 #12


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    Acceleration is the force you feel when pushed against your chair in an accelerating rocket. This force is always present when the engines are firing. That's one of the main points of special relativity.
  14. Aug 14, 2008 #13


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    The OP is asking why - if the first hour of his trip gets him from .98c to .99c (delta v=+.01c), but the second hour of his trip gets him from .99c to only .999c (delta v=+.001c) - why that doesn't constitute a decease in acceleration.
  15. Aug 14, 2008 #14
    It might not constitute a decrease in acceleration in an objects inertial frame, but I think it obviously constitutes a decrease in observed acceleration of that object, if not, then I'm definitely missing something.
  16. Aug 14, 2008 #15


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    This is the difference between proper acceleration (loosely speaking, what the object itself "feels") and coordinate acceleration in some inertial reference frame.
  17. Aug 14, 2008 #16
    Heh, please explain
  18. Aug 14, 2008 #17


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    The proper acceleration (what the object "feels", what it measures with an accelerometer) is the coordinate acceleration ([itex]d^2x/dt^2[/itex]) in a co-moving inertial frame. So "constant proper acceleration" doesn't mean that [itex]d^2x/dt^2[/itex] is constant in one particular inertial frame. It means that if we calculate [itex]d^2x/dt^2[/itex] in the inertial frame that's moving with the same velocity as the object we get the same result at every point on the object's world line. (Note that this means that we're using a different inertial frame at each point).

    When the proper acceleration is constant, the coordinate acceleration (in one specific inertial frame) is decreasing.
  19. Aug 14, 2008 #18
    That's what I thought coming into this topic, the acceleration of an object as viewed by an observer decreases as the object's velocity approaches c, but my real question was, at what rate does it decrease? Is this the correct way to show the change from proper acceleration to coordinate acceleration?

    [tex] a_c=a_p\sqrt{1-(\frac{a_pt}{c})^2}[/tex]

    Or is it more complicated than this? I got that equation using lorentz transformations, solving for x'/(t'^2).
  20. Aug 14, 2008 #19


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    Acceleration, like almost everything else in relativity, is a relative concept. Although everyone agrees what zero acceleration is, they disagree on the magnitude of a non-zero acceleration.

    Thus an acceleration that is constant according the object feeling the acceleration (constant "proper acceleration") would, according to an inertial observer, be a decreasing acceleration.

    It turns out, surprisingly, that, for linear motion of a constant mass, the "force" is invariant, i.e. all inertial observers agree on its value. (The appropriate relativistic equation for force is [itex]\textbf{f} = d\textbf{p}/dt[/itex], not [itex]\textbf{f} = m \, d^2\textbf{x}/dt^2[/itex].)

    Thus a constant force applied to a constant mass in the same direction as its velocity causes constant proper acceleration but decreasing "coordinate acceleration" relative to an inertial observer.

    The two accelerations are linked by

    [tex]a_{proper} = \gamma^3 \, a_{coord}[/tex]

    This is proved in this thread, posts #13,#14,#15 equation (12) (with a correction in post #28).

    For an example of a force acting on a non-constant mass, consider friction. Friction may cause a body to heat up, and that gain in energy is accounted for as an increase in mass (i.e. rest mass). Another, more obvious, example is a rocket ejecting burnt fuel and therefore losing mass.


    Proof that, for linear motion of a constant mass, force is invariant:

    In the notation of the posts I referred to above

    [tex] p = m \, dx/d\tau = mc \, \sinh\phi[/tex]
    [tex]\frac {dp}{d\tau} = mc \, \frac{d\phi}{d\tau} \, \cosh\phi[/tex]
    [tex]f = \frac {dp}{dt} = \frac {dp/d\tau}{dt/d\tau} = mc \, \frac{d\phi}{d\tau} = m \, a_{proper}[/tex]​
  21. Aug 14, 2008 #20
    I read your proof and understood it, but my question was more directed at the at the equation [tex] a_c=f(a_p) [/tex], not [tex] a_p=f(a_c) [/tex].

    Before I go further let me ask you a simple question, in your equation, [tex]a_{p} = \gamma^3 \, a_{c}[/tex], isn't the velocity equal to [tex]\frac{x}{\tau}[/tex]? So in that equation, you will need to know both proper time and coordinate time in order to execute the function, which doesn't really show [tex]\frac{da_c}{da_p}[/tex].

    Back to what I was saying before, here are the relativistic transformations for x and [tex]\tau[/tex]:


    My question is why can't we use this to calculate coordinate acceleration, in terms of proper velocity/acceleration i.e. solve for x'/t'^2?

    [tex]a_c = a' = \frac{x'}{t'^2} = \frac{x\gamma}{\tau^2\gamma^2} = \frac{x}{\tau^2\gamma} = a_p\sqrt{1-(\frac{a_p\tau}{c})^2}[/tex]

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