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Rotations in special relativity

  1. Feb 11, 2006 #1


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    Hi, I have something which bothers me some time and I hope some-one can relieve me from this burden.

    If one takes the space-time interval in Cartesian Coordinates, one gets


    Ofcourse we could write this in polarcoordinates etcetera. Now, if we want to describe a circular motion, some people claim that this can be done by the special theory of relativity. But, I've learned at a course GR that, if we have a force, the geodesic equation of motion should get a force-term. Carroll for instance states that if we have a charged particle with a Lorentzforce, he writes the tensorial force-term in the geodesic equation ( page 70 of his lecture notes on GR ); instead of 0 the geodesical term equals the Lorentzforce. How should we describe this case in special relativity? I see that the connectionterms become zero in flat space time and stuff, but I don't know how to proceed. Also because I'm not very familiar in describing accelerations in special relativity.
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  3. Feb 11, 2006 #2


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    Circles are, of course relative. But in a given inertial reference frame, circular motion is simply a world-line of the form:

    x = a cos bt
    y = a sin bt
    z = 0

    (after applying the appropriate Euclidean motion to the spatial coordinates)

    But I suspect that you mean to ask about a "frame of reference" for an observer engaged in circular motion?

    Minowski space is just a particular Lorentzian manifold. So all the techniques of differential geometry are perfectly applicable to special relativity. (In other words, the math you learned for GR can also be used for SR)

    Another thing that might be interesting to do is to set up a reference-frame-like thing for the accelerating observer. At each point, you take the time axis to be tangent to his world-line, and the corresponding spatial "slice" to be orthogonal to his world-line.

    Of course, this wouldn't be a real coordinate-chart, because generally you will have many coordinates all referring to the same event in space-time. (Which can be intuited as saying that clocks are capable of running backwards according to these accelerated "reference frames")

    There is no "good" reason to use such things -- as you know, reference frames really have no business extending beyond an infinitessimal neighborhood of the observer. But it has the nice feature that the details can be worked out using elementary calculus, without having to invoke the machinery of differential geometry.
  4. Feb 11, 2006 #3


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    In SR, we must describe the geometry of space-time in a Minkowskian reference frame.

    Thus if we have a particle moving in a circle, we typically pick the center of the circle as the origin of our inertial frame and then we just write

    x = cos(wt), y=cos(wt)

    as Hurkyl mentioned.

    Anything more exotic, involving any sort of non-Minkowskian coordinates, involves GR, not SR. SR describes the motion of object in Minkowskian reference frames, just like Newtonian mechanics describes the motion of objects in inertial frames.

    The acceleration of an object in SR is described by it's acceleration 4-vector.

    To compute the acceleration 4-vector, we start by specifying the curve that the object is following parametrized as a function of proper time, i.e. we have 4 functions

    [tex][t,(\tau),x(\tau),y(\tau),z(\tau)] [/tex]

    For your example of a circle, I think the appropriate parametrization would be

    t(tau) = t/sqrt(1-(w*r)^2/c^2)
    x(tau) = r*cos(w*tau)
    y(tau) = r*sin(w*tau)

    but I haven't double-checked this

    Given this parametrization of the curve, one then computes the acceleration 4-vector as follows:


    Like the velocity 4-vector, the accelration 4-vector is a geometric object. It's talked about at some length in MTW's section on the accelerated observer. (MTW = "Gravitation", Misner Thorne Wheeler).

    Because it is a geometric object and because we are doing SR, the acceleration 4-vector can be boosted just like any other 4-vector.

    Some other useful properties of the acceleration 4 vector are:

    In an inertial frame co-moving with the accelerating object, it is equal to
    (0,ax,ay,az), where ax,ay,az is the tradiational "3-vector" form of the acceleration

    The acceleration 4-vector is always perpendicular to the velocity 4-vector (This is proved in detail MTW, it happens because the velocity 4-vector always has a constant length)

    The Geodesic equation that you were talking about earlier actually gives the GR equivalent of SR's acceleration 4-vector.

    When we use 4-vectors, we can say that F^i = ma^i, the force 4 vector is the mass (invariant mass!) times the acceleration 4-vector.

    In GR, we say that

    F^i = m (\frac{d^2 x^i}{d\tau^2} + \Gamma^i{}_{jk} \frac{dx^j}{d\tau}\frac{dx^k}{d\tau})

    m is again the invariant mass.

    You might think of this as replacing the ordinary derivative in the 4-acceleration formulas of SR with the covariant derivative. But I'm not aware of any special name usually given to the vector Fi/m offhand. It's what laypeople usually mean when they talk about "gravity", though, after one converts the 4-vector into a 3-vector. "gravity" can thus result either from coordinate acceleration, or from Christoffel symbols - and it's a quantity associated with a specific path through space-time.
    [end add]

    The force on an object following a geodesic is zero.

    Thus the various components of the geodesic equation serve as GR's equivalent of the SR acceleration 4-vector.

    Note that IF an object is following a worldline where all the spatial components of x^i are constant, then

    [itex]F^i = m \Gamma^i{}_{00} (dt/dtau)^2[/itex]

    0 representing the time component. (All the other first and second derivatives are zero, except for dt/dtau)
    Last edited: Feb 11, 2006
  5. Feb 12, 2006 #4


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    Ok, thing are getting clear now, thank you both for the replies! So, if I have a circular motion, all I have to do is choose smart an inertial frame, parametrise the curve by [itex] \tau [/itex] to get something like

    [tex] x^0 = \frac{\tau}{\sqrt{1-(\frac{\omega*r}{c})^2}} [/tex]

    [tex] x^1 = r\sin(\omega*\tau) [/tex]

    [tex] x^2 = r\cos(\omega*\tau) [/tex]

    [tex] x^3 = 0 [/tex]

    Then there is a potential responsible for the circular motion, and this one I can put equal to

    [tex] \frac{d^2x^i}{d\tau^2} = F^i/m [/tex]

    Does this make sense?
  6. Feb 12, 2006 #5


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    It's close, but there's something important missing. Because we are in minkowski space

    dt^2 - dx^2 - dy^2 -dz^2 = dtau^2

    (using geometric units where c=1)

    Therfore in our parameterized curve, we must have

    (dt/dtau)^2 - (dx/dtau)^2 - (dy/dtau)^2 - (dz/dtau)^2 = 1

    I didn't think to mention that the last go-around.

    This isn't true in your set of equations, so I think your expression for t(tau) is wrong, as was my earlier expression.

    It might be less confusing to write (using geoemtric units again)

    t = a * tau
    x = r*cos(wt) = r*cos(w a tau)
    y = r*sin(wt) = r*sin(w a tau)

    then a^2(1-w^2*r^2) = 1

    part of the reason my previous expression was wrong (and misled you as well) was that w was not playing it's traditional role as an angular frequency in spite of its name.

    Also I'm not quite sure why you call a "force" a potential, hopefully that's not important.
  7. Feb 13, 2006 #6


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    Ofcourse, I've been sloppy there, and also with the Force-part; what I meant was that the force responsible for the circular motion is the force which appears in the geodesic equation on the right. But what do you mean with "w does not play the traditional role as an angular momentum"?

    Also, I have another question which concerns general relativity and is something totally different ( I hope you excuse me for this ). A time ago I asked here on the forum if it can be justified to put in the field equations of GR the energy momentum tensor for the electromagnetic field. So

    [tex] G_{\mu\nu} = 8\pi T_{\mu\nu} [/tex]

    which, for an electromagnetic field, becomes

    [tex] G^{\mu\nu} = 2 ( -F^{\mu\lambda}F^\nu_{{}\lambda} + \frac{1}{4}\delta^{\mu\nu}F_{\lambda\sigma}F^{\lambda\sigma} ) [/tex]

    Someone ( can't remember who it was ) said that it was OK to put it this way. But then I get some ethical problems...Can we say that electromagnetism is some sort of gravity? I mean, gravity doesn't account as a force in GR, it's just a consequence of distribution of matter and energy, and an electromagnetic field is just that; an energetic field which curves space-time. Ofcourse I know that the two forces are different, but still...Can someone comment on that? And could it be observed then that photons get deflected in electromagnetic fields? I hope I've been clear about this :)
    Last edited: Feb 13, 2006
  8. Feb 13, 2006 #7


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    It's traditional to write x = r cos(wt) which gives v = r*w.

    When x = r cos (w tau), v is NOT equal to r*w.

    That's all I meant, nothing very deep.

    Electromagnetic fields are a source term for gravity. That doesn't mean electromagnetism IS gravity, it just means that it's a source term for gravity, i.e. electromagnetism has a stress-energy tensor, and the stress energy tensor (I haven't double checked the tensor expression you wrote) is just what you plug into the right hand side of Einstein's equation.

    The effect is very small, but electromagnetic fields do contribute to gravity. For instance, the Schwarzschild metric for an uncharged black hole is diferent from the Reisser-Nordstrom metric for a charged black hole. This means, for example, that an _uncharged_ particle (or a light beam, for that matter) will follow different geodesic paths for a black hole with charge compared to an otherwise identical black hole without charge. In the first case the uncharged particle / light beam follows the geodesics of the associated R-N metric, in the second case it follows the geodisics of the associated Schwarzschild metric.
  9. Feb 16, 2006 #8


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    Ok, it's clear now, thank you very much :smile:
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