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Circular Motion in SR/weak gravitational field limit

  1. Apr 6, 2010 #1
    This scenerio originated from
    where is was intended to argue that in the weak field limit where Newtonian gravity is a good approximation gravitational mass is gamma*m0=E/c2, as opposed to the rest mass argued for by several in that thread.

    The System

    Consider an object moving in a circle with constant angular velocity w, radius R and rest mass m0 from the frame of the center of rotation. If wR << c the classical results can be used.

    dw/dt = w2
    V = wR
    a = w2R

    P = m0wR
    F = dP/dt = m0dw/dtR = m0w2R

    The Observer

    The physics of this system should be the same for all observers. Let us consider this system from an observer moving at V parallel to the axis of rotation. The motion is now a helical spiral along a cylinder with constant acceleration towards the axis.

    gamma = 1/sqrt(1-(V/c)2) = constant
    dt/dt' = 1/ gamma

    R' = R //Distance perpendicular to the motion independent of motion.
    w' = w / gamma // Angular rotation acts like hands on a clock. (corrected by starthaus)
    m0 = m0 // Rest mass is invariant.

    dw'/dt' = d w/gamma /dt' = dw/dt dt/dt' / gamma = w2/gamma2 = dw/dt/gamma2 = w'2
    a' = Rdw'/dt' = Rdw/dt/gamma2 = a/gamma2 = Rw'2

    Because the motion is perpendicular to the original motion the magnitude of the momentum is:

    |P'| = gamma * |P|

    The axial momentum is constant for a set V and the radial momentum is invariant.

    P'r = Pr = m0wR = gamma * m0w'R
    F' = dP'/dt' = gamma * m0dw'/dt'R = 1 / gamma m0dw/dtR = F / gamma

    The Force

    So far we have said nothing about the nature of the force other than how it must transform in SR. In the limit of a weak gravitational field the correspondence limit states that GR can be approximated by Newtonian gravity.

    Let Mg by the barycenter mass needed to cause the motion above. Since wR << c we can ignore the time delay of the light. The subscript g indicates gravitational mass, as the point of this exercise is to determine if gravitational mass is the rest mass or gamma * rest mass.

    |F| = GMgmg/r2

    For the observer the time delay for an arbitrary V might be significant. By examining a triangle with hypotenuse C and leg along the cylinder of V we can determine the effects of this delay.

    One effect is that that the observed distance between m0 and where M was when the force originated should be stretched by gamma. Another is that the force should have components in the axial and radial directions.

    r' = gamma R
    |F'| = GMg'mg'/r2 = GMg'mg'/R2 / gamma2
    Fr' = |F'| / gamma = GMg'mg'/R2 / gamma3

    but from above we find that
    Fr' = F / gamma
    GMg'mg'/R2 / gamma3 = GMgmg/R2 / gamma
    Mg'mg' / gamma2 = Mgmg
    Mg'mg' = Mgmg * gamma2

    This suggests mg = gamma * m0, not the rest mass (but consistent with rest energy + kinetic energy / c2)

    SR + weak field -> Frame Dragging?
    F' above has a component in the radial direction that is not balanced by another force, requiring a pseudo force to restore the original dynamics.

    Fa' = |F'| V/c / gamma = GMg'mg'/R2 V/c / gamma3

    This force is in the direction of motion, similar to frame dragging.

    The Force II

    What if this force is from electromagnetism, where the Q is invariant? Parallel to above we find that:

    Fr' = |F'| / gamma = kQq/R2 / gamma3 = kQq/R2 / gamma / gamma2 = kQq/R2 (1-(V/c)2) / gamma

    This force can be restored by adding a pseudo force
    PFr' = kQq/R2 (V/c)2 / gamma = k/c2QVqV/R2 / gamma
    which clearly is magnetism, as expected from the original derivation of SR from Maxwell's laws.

    Both the electric force and magnetic force have a frame dragging component as above.

    Originally I had thought that the difference in the optical eclipse and tidal maximum was evidence for the different formulations above. Now I am not so sure they are in fact different.

    Is there a mathematical difference between these ontologically different statements:
    1) an inverse square law of some property of an object causes that property to scale by gamma in SR, with no pseudo forces.
    2) an inverse square law of some property of an object in SR keeps the property invariant but produces a magnetism-like pseudo force.

    If these statements are equivalent, by convention the second statement prevails.
  2. jcsd
  3. Apr 6, 2010 #2
    If w=constant, dw/dt=0. Wrong off the bat.
  4. Apr 6, 2010 #3

    Jonathan Scott

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    Gold Member

    I don't fully understand what the original post is about, but it sounds as if it's trying to work with a semi-Newtonian model of weak gravity. If you want to do this for orbits, SR corrections and GR corrections are or similar importance, so at the very least you need to take curved space into account. The motion of a test body is unaffected by its mass and energy, so the question of whether you want to base forces on masses or total energy depends on the convention you assume for your force law.

    The simplest way of looking at gravity in a relativistic but Newtonian-friendly way in a central static field is to consider the rate of change of momentum relative to isotropic coordinates. One must be careful to keep consistent factors of c, by which here I mean the coordinate speed of light, as in these coordinates the effective value of c varies with potential, but at least in isotropic coordinates it is the same value in all directions, varying approximately as (1-Gm/rc2)2 times the standard value.

    In terms of the Newtonian local field g = -Gm/r2 and other conventional quantities (expressed in the isotropic coordinate system), the equation of motion is then as follows, for all directions of motion (radial, tangential or in between) and all speeds (including relativistic), provided that the gravitational field is weak enough (as in the solar system) that we can approximate both the time dilation and ruler shrinking factors as (1-Gm/rc2):

    (I've rewritten this all without using LaTeX because for some odd reason the preview is showing rubbish when I use LaTeX).

    dp/dt = E/c2 g (1 + v2/c2)

    For a static field, the total energy E of the test particle is constant (as in Newtonian theory, where it is the sum of the potential and kinetic energy), and p = Ev/c2, noting that c in this context varies slightly with potential. This means that we can drop the energy from the equation, as follows:

    d(v/c2)/dt = 1/c2 g (1 + v2/c2)

    In both of the above equations, the coordinate rate of change of momentum only depends on the speed, and not on the direction of motion.

    This shows that a slow-moving particle accelerates in a Newtonian way but that the momentum of a particle moving at or around the speed of light varies twice as fast, because of the curvature of space (if moving tangentially) or because of the varying coordinate speed of light (if moving radially). However, the mass or energy of the particle does not affect the motion.
  5. Apr 6, 2010 #4
  6. Apr 6, 2010 #5
    It looks like you are interpreting my intent as trying to calculate orbits for fast moving objects.

    What I am attempting to do is observe a slow moving orbit from a fast moving observer to see how that observer would measure the forces in the nearly classical original context.

    I don't see how a nearly flat reference of slow moving objects in one frame becomes more curved in another frame under a Lorentz boost. Isn't the space-time curvature invariant?
  7. Apr 7, 2010 #6

    Jonathan Scott

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    Gold Member

    OK, thanks, that clarifies it.

    In that case, everything transforms (of course) in the usual way for SR - length, acceleration, force, energy - but note that as usual there is a big difference between effects parallel to the direction of motion (including Lorentz contraction) and perpendicular to it. (Curvature can mean lots of different things; most of them are affected by Lorentz transformations, but in the case of motion of a system perpendicular to the plane of an orbit the effect of spatial curvature cancels out).

    As you note, the force law transformation has to be the same as for electromagnetism overall because for example you could use electromagnetic forces to cancel out the gravitational force and that cancellation would have to remain valid from the transformed viewpoint.

    As there is no "standard" Newtonian gravitational force law which copes with SR transformations, the question of whether you would base such a force on the rest mass or on the total energy mainly depends on your conventions, for example whether you use the Lorentz force law (including magnetism) as your semi-Newtonian model of gravity, or just the normal GmM/r^2 force, roughly equivalent to the electrostatic part only.

    This is similar to the choice between whether momentum is Ev/c^2 or mv/sqrt(1-v^2/c^2). Both expressions are correct, but one can be more useful than the other in a given context.

    This isn't very useful in understanding gravity, because the interesting terms in gravitational interactions (especially those involving spatial curvature) effectively arise from relative motion, and SR doesn't have anything to say about that.
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