Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constant Acceleration

  1. Jun 10, 2007 #1
    I have this "thought" experiment. Imagine a large rigid object, a thin disc (dime) with the ("heads") surface area of the earth, accelerating in the "heads" direction for many years such that individuals on the surface (of the "head" side) experience the normal sensation of "gravity". This object is the only object in the Universe.

    How long can the disc undergo this constant "felt" acceleration? (I believe the answer is "forever".) Does the force required to maintain the constant "felt" acceleration ever need to increase? (I believe the answer is "no".) Does the acceleration of the leading surface need to differ from the trailing surface in order for the disc to remain rigid at relativistic velocities? (I believe the answer is "yes".)

    Thanks so much!
  2. jcsd
  3. Jun 10, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, yes, and yes, assuming you are measuring the acceleration with a local accelerometer mounted on the leading and tailing edges.
  4. Jun 10, 2007 #3

    Chris Hillman

    User Avatar
    Science Advisor

    Ditto pervect (who probably meant "in principle forever, no, yes", i.e. agreeing with your guesses), except that I'd add:

    1. The acceleration must produced by some force via the expenditure of energy, which presumably limits how long you can keep up constant acceleration,

    2. For a thin disk accelerated parallel to its axis of symmetry, the variation in acceleration in the direction of motion would be very small; the effect is more pronounced if you consider a thin rod accelerated parallel to its axis of symmetry, which you can imagine being pulled with constant tension by a string,

    3. This kind of discussion can quickly become surprisingly technical, since there are a lot of subtle issues lurking in the weeds. For example, in the rod, I am assuming the force is being measured at the pulled end of the string by an observer comoving with the pulled end.
    Last edited: Jun 10, 2007
  5. Jun 10, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    I'd better clarify the force question. In a frame comoving with the accelerating dime, the force would be constant.
  6. Jun 10, 2007 #5
    Thanks. But here's my problem: Since the dime is accelerating, that frame is not valid for inertially sound measurements. So... in reality this constantly accelerating dime needs exponentially increasing force (and therefore energy), right?
  7. Jun 10, 2007 #6

    Chris Hillman

    User Avatar
    Science Advisor

  8. Jun 10, 2007 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    One of the best ways to describe the situation is to use four-vectors. On the minus side it's a bit technical, on the plus side ambiguities of language don't affect the answers.

    Using the 4-vector approach, geometric units, and setting the constant acceleration to unity, the results from for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html would appear as below.

    Note that I'm only writing out two of the components of the 4-vector, the other two components are identically zero.

    [itex]\tau[/itex] = proper time

    The position vector can be written as a function of proper time as follows:

    [itex]x^i(\tau)[/itex] :

    [tex]t = x^0 = sinh \, \tau \hspace{.5 in} d = x^1 = cosh \, \tau[/tex]

    I've chosen this as the simplest way of writing down the equations of motion, note however that [itex]x^1(0)[/itex] is not equal to zero, but is rather equal to 1.

    The 4-velocity vector can be writen as a function of proper time as follows:

    [tex]v^0 = \frac{d x^0}{d \tau} = cosh \, \tau \hspace{.5 in} v^1 = \frac{d x^1}{d\tau} = sinh \, \tau[/tex]

    The energy-momentum 4-vector is just
    [tex]p^0 = m v^0 = m \, cosh \, \tau \hspace{.5 in} p^1 = m v^1 = m \, sinh \, \tau [/tex]

    Here m is the invariant mass of the dime. The simplest case to analyze is where the invariant mass m is constant. This requires the dime to be propelled by an external source, for instance the 'dime' might really be a laser powered light sail.

    Note that [itex]p^0[/itex] can be interpreted as the energy of the dime (it's written here as a function of proper time), and [itex]p^1[/itex] can be interpreted as the momentum of the dime.

    Note that the 4-velocity of the dime is the rate of change of the position of the dime with regard to proper time tau, rather than the coordinate time t.

    The 4-force on the dime is
    [tex]f^0 = \frac{d p^0}{d \tau} = m \, sinh \, \tau \hspace{.5 in} f^1 = \frac{d p^1}{d \tau} = m \, cosh \, \tau[/tex]

    The 4-acceleration of the dime is
    [tex]a^0 = \frac{d v^0}{d \tau} = sinh \, \tau \hspace{.5 in} a^1 = \frac{d v^1}{d \tau} = cosh \, \tau[/tex]

    Some comments here are in order:

    0) There's no problem in transforming the 4-vectors into an instantaneous comoving frame - we can transform both the 4-acceleration and the 4-force in this manner, because 4-vector transformations depend only on the relative velocity, i.e. all 4-vectors transform according to the Lorentz transform, which is indepenent of acceleration and depends only on velocity.

    1) using the 4-vector approach, we can say that that the 4-force is directly proportional to the 4-acceleration, and the constant of proportionality is the invariant mass m of the dime.

    2) In the frame of the dime, the 4-acceleration is constant. If you do the appropriate Lorentz boost, the 4-acceleration in an instantaneously comoving frame is just (0,1).

    3) In the frame of the dime, the 4-force is also constant, and equal to m*a, and since a=1, the 4-force is equal to m.

    5) We could write the 4-velocity and energy -momentum 4-vector in the instantaneous frame of the dime just as we did for the 4-acceleration and the 4-force, but they aren't very interesting.

    6) The idea of constant proper acceleration is the same as the idea that the 4-acceleration in the instantaneously comoving frame is just (0,a). I've made a=1 for simplicity in writing the above expressions down.
    Last edited: Jun 10, 2007
  9. Jun 11, 2007 #8
    Another Great Job.

    Aha! The key is that to the observer in an inertial frame of reference the force would remain the same, the mass would increase, and the acceleration would decrease. However to the observer of the dime, all three would be constant.

    You really helped. That math was great and simple. Very enlightening. You don't know how many pages I've read (both on the web and in my "old" textbooks) and you manage to explain in so well. Good job.

    Now... about the rigid body problem... I probably need to tell you where I'm coming from first.

    Please don't think I'm lame here, but I'm on a mission. My personal hero is James Randi, randi.org. He has attempted to debunk the Flat Earth Society, theflatearthsociety.org. I have spent weeks and thousands of dollars to make sure that young minds get the proper scientific background to make a solid conclusion. I'm setting up a "counter" website and posting on their forums. They hold that the earth is like a dime with the surface on the "heads" side, accelerated through space at 1g.

    I'm convinced now, thanks to you and others here, that arguing limitation on acceleration, velocity, and force are unwarranted. The force is applied solely to the "tails" side (by dark energy, they claim). I suspect that I may be able to debunk this model by examining the effect of a constant acceleration to the trailing end of a rigid body. I believe that we have a large acceleration (1g) and such a thickness (of miles) that the relativistic effects would create too great of stress in the dime.

    Forgive me if you think I'm wasting your time. I truly wish to help others by fighting those who don't apply scientific methods in their decision-making and then advocate their beliefs in an attempt to influence others.

    (BTW, anyone who has the time and interest to join me on any such endeavor is most welcome. I really could use the help.)

    Thank you so much.
  10. Jun 11, 2007 #9


    User Avatar
    Staff Emeritus
    Science Advisor

    Threads on this "flat earth socieity" have appeared before, see for instance https://www.physicsforums.com/showthread.php?p=1067430

    These threads tend to get locked for terminal silliness, however, as the thread above was. The consensus appears to be that the people arguing about this are just arguing for the sake of argument. So while we don't necessarily mind discussing the physics of an accelerated observer in SR, we'd rather avoid getting sucked into this "flat earth" debate.

    The relativistic mass would increase, but note that the invariant mass m I use in the formulas I use above would not increase.

    Note also that four-forces are [tex]\frac{dp}{d \tau}[/tex], the rate of change of the energy-momentum with respect to proper time.

    "Standard" forces are defined as [tex]\frac{dp}{dt} [/tex], the rate of change of energy momentum with respect to coordinate time.

    Four-forces are not compatible with the rather old-fashioned notion of relativistic mass, but are compatible with invariant mass.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Constant Acceleration