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

Relativistic centripetal force is a byproduct of time

  1. May 18, 2015 #1
    From special relativity, we know that
    t=t0/(1-v2/c2)1/2
    For the object moves as a circle with radius r, the velocity v = w r
    Therefore, time varying along radius r as
    t=t0/(1-w2 r2/c2)1/2
    In this case
    ds2= dr2- c2 dt2 = dr2- c2 dt02 /(1-w2 r2/c2)
    Since object will move in shortest path in space-time, Dif[ds2,dr] = 0,
    and Dif[r,dr] = 1, the above equation will be
    0 =2 dr - 2 w2 r dt02 /(1-w2 r2/c2)2
    That means the object will move dr due to time varying along radius r as
    dr = w2 r dt02 /(1-w2 r2/c2)2 = w2 r dt2 /(1-w2 r2/c2)
    Or the object will move with accelleration as
    dr/dt2 = w2 r /(1-w2 r2/c2)

    This accelleration is look like acting by centrifugal force.
     
    Last edited: May 18, 2015
  2. jcsd
  3. May 18, 2015 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This is not correct. Objects that are in free fall move on the longest paths in spacetime, not the shortest. But the object you are considering is not in free fall.

    Also, I'm not clear on what question you are asking.
     
  4. May 18, 2015 #3
    I try to proof that this accelerate due to time varying along r. No centrifugal force.
     
    Last edited: May 18, 2015
  5. May 18, 2015 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Proper acceleration can't be explained by "time variation" (to the extent that even makes sense). Nonzero proper acceleration means a force is acting on the object; there's no way to transform away that force by changing coordinates.

    Your math does not look correct in any case. First, as I pointed out before, you are assuming that the object travels on the "shortest path" in spacetime, whereas in fact its path is not extremal at all, since it is not in free fall. Second, you claim to be deriving an equation for ##dr / dt## that says it is nonzero; but you started out with the assumption that the object is moving in a circle with constant ##r##, i.e., ##r## does not change with time. So your analysis looks self-contradictory.
     
  6. May 18, 2015 #5

    Mentz114

    User Avatar
    Gold Member

    I'm not sure what this means. This good Wiki article covers circular motion in Minkowski spacetime.

    http://en.wikipedia.org/wiki/Born_coordinates

    The general geodesic condition for circular motion is ##2g_{tt}r\omega^2=\partial_r g_{tt}## which in Minkowski spacetime has no solution. If ##g_{tt}## is from the Schwarzschild vacuum solution, then there is a solution which looks like a relativistic Keplers law.
     
    Last edited: May 18, 2015
  7. May 18, 2015 #6

    Dale

    Staff: Mentor

    Are you perhaps talking about a circular orbit in the Schwarzschild spacetime? If not then I don't know of a way to get a circular path without a centripetal force.
     
  8. May 18, 2015 #7

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I suspect that if we identify "centripital force" with ##\Gamma^r{}_{tt}##, or possibly ##\Gamma^\hat{r}{}_{\hat{t}\hat{t}}##, using a rotating metric like
    rotmetric.png
    a detailed calculation would wind up show that the "force" (christoffel symbol) is proportional to ##\partial_r \sqrt{|g_{00}|}##, i.e. the partial derivative with respect to r of the "time dilation factor". Another way of saying this would be that I'd expect the force (christoffel symbols) to be proportional to the gradient of some "effective potential", a relativistic version of the effective potiential one uses to find the figure of the rotating Earth (by setting the effective potential equal to a constant). But I haven't worked out the details.
     
  9. May 18, 2015 #8

    Mentz114

    User Avatar
    Gold Member

    I did the detailed calculation some ago and you are correct as the equation I cited above shows. (My ##g_{tt}## is a positive number which gets its sign from the signature)
     
  10. May 18, 2015 #9
    May I explain more by this example... A ball is in a slot on a disk. This slot is in the radial direction of the disk. When we spin the disk, the ball will be accelerated by centrifugal force. I think actually there is no centrifugal force, but this acceleration is due to time varying along r.
     
  11. May 18, 2015 #10
    Sorry that I have a little bit English confusion. Now I've changed from "centripetal force" to "centrifugal force".
     
  12. May 18, 2015 #11

    Mentz114

    User Avatar
    Gold Member

    I think you are saying that the centrifugal force can be explained by a potential.

    This is what GR already shows. In holonomic coordinates the acceleration is a kinematic effect ( depending on velocity) but in the local frame calculation it becomes a velocity-dependent potential by the transformation of a Christoffel symbol to a Ricci rotation coefficient.

    I've calculated the potential and I can show the working if you want to see it.
     
    Last edited: May 18, 2015
  13. May 18, 2015 #12

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Ok, this is a different scenario from what it seemed like you were describing in the OP. Note that there are two "forces" that the ball is subjected to: one in the radial direction (the one you are talking about), and the other in the tangential direction (because the slot is pushing sideways on the ball).

    It's due to picking a non-inertial frame; "time varying" only applies in that frame. In an inertial frame in which the center of the disk is at rest, there is no "time varying", and also no centrifugal force; the only force acting on the ball is the tangential force of the slot pushing sideways on it.
     
  14. May 18, 2015 #13
    Please show your work. Thank you.
     
  15. May 19, 2015 #14
    I think it's quite interesting how the force in the radial direction, centrifugal force, came from. Thereforce, I try to proof that actually no centrifugal force, but the acceleration is due to time varying along r.
     
  16. May 19, 2015 #15

    Mentz114

    User Avatar
    Gold Member

    OK I've attached a cut-down section of my notes. This is not an article although it has that format because I don't have a suitable Latex document template (!).
    I've done my best to get everything right but I cannot be certain there aren't mistakes. Also I never finished section 4 because it looks like circular logic (pun).
     

    Attached Files:

  17. May 19, 2015 #16

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    As I said before, it comes from using a non-inertial frame.

    All this "proof" shows is the consequences of using a non-inertial frame. If you use an inertial frame, the "time varying along r" disappears. But you can't change physics by changing frames.
     
  18. May 19, 2015 #17

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I think the following is vague enough to be true. I don't really have the inclination to work out all the details at the moment to say anything more specific, so take this whole post with a grain of salt, it's an idea that needs more development and testing. The idea is that there is some effective potential (velocity dependent in the case of a stationary metric, a scalar potential in the more familiar static metric case). The exact form of the effective potential depends on the coordinate choice, for instance if you have a non-rotating flat space metric, the effective potential is zero, but if you have a rotating flat metric, which is just a different choice of coordinates, the effective potential is not zero. The claim is basically that the gradient of the effective potential can be related to the "force" and the value of the potential itself can be related to time dilation. In the rotating case, the claim is that the value of the effective potential when the velocity-dependent part is zero gives the "centrifugal force". The velocity-dependent part would be related to the coriolis force, but I don't think anyone is exploring this aspect.

    Anyway, since both "force" and time dilation can be derived in the static/stationary case from the effective potential, the claim is that there is a close relationship between the two. Interpreting this relation as "time dilation causes gravity" appears attractive, but is certainly not equivalent to Einstein's field equations. One difference that comes to mind is that Einstein's field equations predict non-Euclidean spatial geometries, this would not be a prediction of the "time dilation causes gravity" idea. However, this doesn't come about in the rotating case, it's more of a concern in the Schwarzschild case.

    Experiment favors GR, of course, if one regard the two as competing theories by exploring their predictions. The non-zero value of the PPN parameter gamma pretty much rules out an Euclidean spatial geometry for the solar system metric, which means that spatial geometry of the solar system isn't Euclidean.
     
  19. May 19, 2015 #18

    Mentz114

    User Avatar
    Gold Member

    Having decided to look at circular motion in flat spacetime one is constrained to look at world lines of bodies moving in circles, like the Born and Langevin congruences. So I don't understand your point about anything I claim being dependent on choice of coordinates. For these congruences the acceleration is proper and all observers will agree about that.( I don't know if I am amongst those claiming)
    When writing the Hamiltonian of particle moving on a given worldline, the 'time-dilation' ##dt/d\tau## is treated as a potential. There is unusual about this.

    What I claim is that the physical effects ( eg proper acceleration) which are put down to the velocity in the holonomic basis, must be moved into the potential sector in the local basis or be lost. And this is what happens, which is no surprise.

    I think the OP noticed that ##\partial_r \dot{t}## for some worldline gives the correct acceleration and then used faulty logic to 'prove' this.

    But by an amazing coincidence, this is actually what happens. In the local frame basis the kinematic effect has become a potential equal to ##\dot{t}##. And since the OP asked to see the working, I gave them both barrels.

    I hope it had a salutory effect. But I suspect it has been a waste of time.

    [edit]
    I cut out some not relevant stuff and changed a few words in my doc.
     

    Attached Files:

    Last edited: May 19, 2015
  20. May 20, 2015 #19

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I was thinking about this issue some more, and I came to the conclusion that we can relate the link between time dilation and force to the existence of a conserved energy. Specifically, the Komar energy in any static or stationary space-time - the rotating space-time being an example of the stationary case.

    So, the rate of change of energy with distance is related to force, via the principle that energy = force*distance. But this "force" can and usually will be different from the proper acceleration that a local observer with constant coordinates measures. There is some discussion in Wald on this (see the disucssion on the force at infinity) in his discussion of energy in his book "General Relativity". This creates the link between energy and force.

    At the risk of creating more confusion, I'll describe how we could measure the "force". If we had a masless string (a string that was pure tension in its rest frame), we would measure the "force" on a rotating object by the tension in the massless string at the center of rotation. The continuity equations and time dilation make the "force" we measure this way different than the "force" (tension) at the end of the string (the end attached to the rotating object). The two tensions will be different, only the later value would be equal to the reading of an accelerometer on the object multiplied by its mass. Hence the ambiguity in what we mean by "force".

    The conservation of energy implies that there will be a constant and reciprocal red/blue shift between any pairs of points with constant coordinates via the usual arguments based on the principle of equivalence and "falling photons". These red/blue shifts are interpreted as time dilation, so we have the link from energy to time dilation.

    Thus we can link both time-dilation and "force" to energy, explaining the observed relationship, which is not a coincedence, but necessary - as long as we meet the required conditions of a static or stationary space-time.
     
    Last edited: May 20, 2015
  21. May 20, 2015 #20

    Mentz114

    User Avatar
    Gold Member

    In as much as I understand the above I agree.

    But I think one ought to concentrate on the local physics. If I spin a disk in my lab then I can analyse my results without giving any thought to boundaty conditions at infinity. If we analyse always in terms of proper time and restrict our conclusions to the spacetime region in which our ( Fermi-Walker) coordinates have acceptably small deviations from the ideal. In that case the Newtonian relationships between energy, momentum, work and force must be taken to hold. Energy conservation can only be local in any case without action at a distance.
     
    Last edited: May 20, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relativistic centripetal force is a byproduct of time
  1. Relativistic force (Replies: 1)

  2. Relativistic forces (Replies: 6)

Loading...