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Speed in general relativity

  1. Mar 5, 2010 #1
    Hi all,

    I'm starting to learn a bit of general relativity now, but I'm a bit confused as to the measurement of speed.
    Let's say we use the Schwarzschild metric.
    Now, I can parametrize the wordline of a certain object by giving a
    parametrisation (t(tau),r(tau),theta(tau),phi(tau)), where tau is the proper time of
    that object. Now, I would like to know how to measure the speed of the object;
    that is, the magnitude of its 3-velocity. I can find the 4-velocity, but can I deduce in a meaningful way the speed, i.e. as observed by an observer 'at rest' in the frame
    (of course, 'at rest' is a relative concept - I'm referring to the observer at infinity, where the Schwarzschild metric is more or less flat)?

    From special relativity, I recall that you could write the 4-velocity
    as (gamma,gamma*V) where gamma is the usual gamma factor and V is the three-velocity.
    I'm looking for something analogous - if that's meaningful.


    Thanks,

    APhysicist
     
  2. jcsd
  3. Mar 5, 2010 #2

    George Jones

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    This is meaningful in general relativity if the two observers are coincident.
     
  4. Mar 5, 2010 #3
    In general relativity, all we know about the proper velocities [tex]u^{\alpha}=\frac{dx^{\alpha}}{d\tau}[/tex] is that they satisfy

    [tex]g_{\alpha\beta}u^{\alpha}u^{\beta}=c^2.[/tex]

    For instance, if we have

    [tex]ds^2=c^2dt^2-dx^2-dy^2-dz^2,[/tex]

    then

    [tex]c^2(u^{0})^2-(u^{i})^2=c^2.[/tex]

    AB
     
  5. Mar 5, 2010 #4

    George Jones

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    But if two observers are coincident at an event, then the proper velocity of one observer can be used to express the proper velocity of the other observer in such a way that gamma and 3-velocity naturally appear.
     
  6. Mar 5, 2010 #5
    I talked about a general way of exposing 4-velocity in GR! I don't know how one can make the gamma and 3-velocity appear in GR through an assumption like yours, but I know if we take the particles to be moving on geodesics for which all three spacelike components remain constant by assuming that the observer is co-moving with particles, then the 4-velocity takes a form

    [tex]u^{\alpha}=(c,0,0,0).[/tex]

    We can only have the gamma and 3-velocity appear if we are in a locally inertial frame, measuring the 4-velocity of a particle from there.

    AB
     
  7. Mar 5, 2010 #6

    George Jones

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    So am I!
    You should :wink::biggrin:. This is related to the way I solved the SR elastic collision problem.
    In both my previous posts in this thread, I said the that two observers are coincident. Are you saying that if observer A is coincident with observer B, then A can't measure B's physical speed (relative to A)?

    For two such observers in general relativity,

    [tex]\gamma = \mathbf{g} \left( \mathbf{u}_A , \mathbf{u}_B \right).[/tex]

    Notice that this expression does not explicitly make a choice of coordinate, but physical speed is used in the definition of [itex]\gamma[/itex].
     
  8. Mar 5, 2010 #7

    DrGreg

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    For the benefit of the original questioner, let me spell out more explicitly George's method.

    Let v be the 4-velocity of an object to be measured, and u be the 4-velocity of any local observer "at rest" at the event where the measurement is required.

    In the locally-Minkowski coordinates of a free-falling and momentarily-at-rest local observer at that event, [itex]u^\alpha = (c, 0, 0, 0)[/itex] and [itex]v^\alpha = (\gamma c, \gamma v, 0, 0)[/itex], where v is the magnitude of the 3-velocity that you want to measure (let's assume it's along the x-axis) and [itex]\gamma = (1-v^2/c^2)^{-1/2}[/itex]. Therefore

    [tex]g_{\alpha\beta}u^{\alpha}v^{\beta} = \gamma c^2[/tex]​

    But [itex]g_{\alpha\beta}u^{\alpha}v^{\beta}[/itex] is invariant, i.e. the same in all coordinate systems, so you can also calculate it in Schwarzschild coordinates, and equate the two values.
     
  9. Mar 5, 2010 #8
    I think your explanation is clear and catchy (though I see something wrong), I want to do the calculations my way that sounds more comfortable at least for me. I take the EP for granted as an accurate principle locally and assume the free-falling observer is comoving with the observer, so [tex]u^{\alpha}=\frac{dx^{\alpha}} {d\tau}=(1,0,0,0).[/tex] The observer is measuring the velocity of a particle moving along x axis. Now we know that

    [tex]g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}=c^2,[/tex]

    On the other hand the particle is experiencing the local Minkowski frame, thus

    [tex]ds^2=c^2dt^2-dx^2-dy^2-dz^2=c^2dt^2(1-v^2/c^2),[/tex]

    where [tex]v[/tex] is the particle's 3-velocity (according to the observer) and by definition, [tex]v^0=dt/dt=1.[/tex] So for the particle we can write

    [tex]\gamma^2=\frac{dt^2}{d\tau^2},[/tex]

    with [tex]\gamma[/tex] being the Lorentz factor. Introducing this result and the velocity [tex]u^{\nu}[/tex] into the first equation yields

    [tex]g_{\mu\nu}v^{\mu}u^{\nu}=\gamma^{-1}c^2.[/tex]

    This is my method and yet is different than yours! Now let's set an example for this. I would like to take into account the linearized isotropic Schwarzschild metric,

    [tex]ds^2=c^2(1-2m/r)dt^2-(1+2m/r)(dx^2+dy^2+dz^2).[/tex]

    For this metric, our formula gives

    [tex]c^2(1-2m/r)v^{0}u^{0}=\gamma^{-1}c^2[/tex] or
    [tex](1-2m/r)v^{0}=\gamma^{-1}[/tex]

    and finally

    [tex]\frac{v^2}{c^2}=\frac{2m}{r}.[/tex]

    (I don't this gives the right answer. Could someone check it?)

    Remember that this is the same as the coordinate velocity because we are assuming the observer to be comoving with particle. I don't know if I've not made a mistake here and of course it would be great to tell me if there's any! I am definitely not familiar with George's method so my understanding of how actually he calculates things like this is so limited.

    AB
     
    Last edited: Mar 6, 2010
  10. Mar 7, 2010 #9

    DrGreg

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    I don't understand your logic for saying this. You just showed in the line above that [itex]dt/d\tau = \gamma[/itex] for the observed particle (v0), and earlier that [itex]dt/d\tau = 1[/itex] for the observer (u0) from which it follows that

    [tex]g_{\mu\nu}v^{\mu}u^{\nu}=\gamma c^2[/tex]​

    Now, in Schwarzschild coordinates [itex](t, r, \theta, \phi)[/itex], [itex]g_{00} = c^2(1 - 2GM/rc^2)[/itex] and the other components of the metric don't matter in this case. We have

    [tex]u^{\alpha} = \left(\frac{1}{\sqrt{1 - 2GM/rc^2}}, 0, 0, 0\right)[/tex]​

    (The last three components must be zero and the first is chosen to ensure that [itex]g_{\mu\nu}u^{\mu}u^{\nu} = c^2[/itex].) And

    [tex]v^{\alpha} = \left(\frac{dt}{d\tau}, \frac{dr}{d\tau}, \frac{d\theta}{d\tau}, \frac{d\phi}{d\tau}\right)[/tex]​

    And so

    [tex]\frac{c^2}{\sqrt{1-v^2/c^2}}= g_{\mu\nu}v^{\mu}u^{\nu} = c^2\frac{dt}{d\tau}\sqrt{1 - 2GM/rc^2}[/tex]​

    if I haven't made a silly mistake in the algebra.

    That tells you how to calculate v from [itex]dt/d\tau[/itex] and r for any particle.

    I don't really follow what you were trying to do in the last part of your post. You seem to have made some assumption about the motion of the particle, whereas the original questioner did not.
     
  11. Mar 7, 2010 #10
    I just took [tex]v^{\alpha}[/tex] to be the coordinate velocity just because the comoving observer would be found at rest relative to the particle, so

    [tex]g_{\mu\nu}\frac{dx^{\mu}}{d\tau}u^{\nu}=g_{\mu\nu}\gamma v^{\mu}u^{\nu}= c^2.[/tex]

    I think your equation sounds flawless! But it is clear that, for example, in the case of an orbiting particle, we must be given the value of [tex]\frac{d\phi}{d\tau}[/tex]. But if the motion occurs in the equatorial plane [tex]\theta={\pi}/ 2[/tex] and the particle is hovering at [tex]{ (r_0,\theta_0,\phi_0)},[/tex] then

    [tex]v^2/c^2 = 2GM/c^2{r_0}=2m/r_0,[/tex]

    which is the same as the result given in my earlier post:

    But as I said, I don't think this is true as it gives such a very high value for the speed [tex]v[/tex] which is about 11,250 m/s for a particle hovering at the surface of earth unless my algebra would have went wrong somewhere! Does this sound logical?

    AB
     
    Last edited: Mar 7, 2010
  12. Mar 7, 2010 #11

    DrGreg

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    I understand your confusion now. I am using [itex]u^{\alpha}[/itex] and [itex]v^{\alpha}[/itex] to refer to 4-velocity which is defined to be [itex]dx^{\alpha}/d\tau[/itex], not [itex]dx^{\alpha}/dx^0[/itex]. (The expression [itex]dx^{\alpha}/dx^0[/itex] does not transform correctly to be a tensor.)

    Note that as

    [tex]g_{\alpha \beta} \, \frac{dx^{\alpha}}{d\tau} \, \frac{dx^{\beta}}{d\tau} = c^2[/tex]​

    then the value of [itex]d\phi/dt[/itex] has an effect on the value of [itex]dt/d\tau[/itex], and therefore affects the value of v in the formula I gave before.
     
  13. Mar 7, 2010 #12

    George Jones

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    According to Newton, what is the escape speed for a particle that starts on the Earth's surface?
     
  14. Mar 7, 2010 #13
    A genius point, man! :tongue:

    is equivalent to what you, DrGreg, said about my confusion! Anyways, thank you guys!

    AB
     
  15. Mar 7, 2010 #14

    George Jones

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    Far from it! I've worked with this expression many times in a variety of ways; for example, in terms of frames in

    https://www.physicsforums.com/showthread.php?p=848684#post848684.

    In this, [itex]\mathbf{e}_0[/itex] is the 4-velocity of the hovering observer, and [itex]\mathbf{e}'_0[/itex] is the the 4-velocity of an observer who falls freely from rest at infinity.
     
  16. Mar 7, 2010 #15

    George Jones

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  17. Mar 7, 2010 #16
    My guess is that the 3 velocity of a particle in Schwarzschild coordinates according to an observer at infinity is:

    [tex] \sqrt{ g^2 u_x^2 + g u_y^2 +g u_z^2 } [/tex]

    where g = [itex] (1-2m/r) [/itex], r is the radial location of a stationary local observer that measures the local velocity of the particle to be

    [tex] \sqrt{ u_x^2 + u_y^2 + u_z^2 } [/tex]

    and u_x is the radial component and u_y and u_z are the horizontal components of the particle's velocity.

    I hope that is some help (and correct).
     
    Last edited: Mar 7, 2010
  18. Mar 7, 2010 #17

    DrGreg

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    kev, [STRIKE]it's not clear to me how you are defining ux, uy, uz. You need to spell it out a bit more explicitly. In what coordinate system and relative to whom?[/STRIKE] EDIT: I didn't read your post carefully enough, I'll now go away and think about it.

    (The original questioner asked for something in terms of a parameterisation [itex](t(\tau), r(\tau), \theta(\tau), \phi(\tau))[/itex].)
     
    Last edited: Mar 7, 2010
  19. Mar 7, 2010 #18
    My original equation for the 3 velocity (V) of a particle in Schwarzschild coordinates according to an observer at infinity:

    [tex]V = \sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } [/tex]

    where g = [itex] (1-2m/r) [/itex]

    could be re-expressed as:

    [tex] \sqrt{ g^2 (dr/dt)^2 + g (r d\theta/dt)^2 + g (r sin(\theta) d\phi/dt)^2 } [/tex]

    if that makes things clearer, but to me it is a bit messy and detracts from the trivial point I was trying to make that the vertical velocity component is treated differently to the horizontal velocity components.

    The OP did express an interest in the 3 velocity in Schwarschild coordinates which no one seems to have responded to so far and how the 3 velocity relates to the 4 velocity.

    Locally, the four velocity W is the same as in SR, i.e.:

    [tex]W = c = \gamma_v(c, u_x, u_y, u_z) = \gamma_v\sqrt{ c^2 - u_x^2 - u_y^2 - u_z^2 } [/tex]

    where [itex]\gamma_v = 1/(1-v^2/c^2)^{1/2}[/itex] and v is the local 3 velocity [itex] (u_x, u_y, u_z) = (u_x^2 + u_y^2 +u_z^2)^{1/2}[/itex].

    The transformed four velocity W' in Schwarzschild coordinates to an observer at infinity is:

    [tex]W' = c = \gamma_v'(c, u_x', u_y', u_z') = \gamma_v'\sqrt{ c^2 - u_x '^2 - u_y' ^2 - u_z '^2 } = \gamma_v'\sqrt{ c^2 - g^2 u_x^2 - g u_y^2 - g u_z^2 } [/tex]

    where [itex]\gamma '_v = 1/(1-V^2/c^2)^{1/2} [/itex] and V is the transformed 3 velocity defined at the top of this post.

    Not so sure about the last transformation (but it does satisfy the requirement that W = W' =c) and sorry if the notation or formalism is not correct. I am just feeling my way, so feel free to correct as you see fit.

    P.S. It turns out that the above equations are only true for the 3 and 4 velocity of a photon which affords a simplification because d(tau) can be set to zero. For the motion of a particle with mass and arbitrary velocity it gets a lot more complicated. I will have to come back to that when I have more time.

    P.P.S. Actually, right now I am not sure any of the above is correct. Maybe someone can help sort out my mess :)
     
    Last edited: Mar 8, 2010
  20. Mar 8, 2010 #19
    I have a question here: How do you get the following relations?

    [tex]\begin{align*}\mathbf{e}_{0}^{\prime} & =\left( 1-\frac{2M}{r}\right) ^{-1}\frac{\partial}{\partial t}-\left( \frac{2M}{r}\right) ^{\frac{1}{2}}\frac{\partial}{\partial r}\\\mathbf{e}_{1}^{\prime} & =-\left( \frac{2M}{r}\right) ^{\frac{1}{2}}\left( 1-\frac{2M}{r}\right) ^{-1}\frac{\partial}{\partial t}+\frac{\partial}{\partial r}\\\mathbf{e}_{2}^{\prime} & =\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\\\mathbf{e}_{3}^{\prime} & =\frac{1}{r}\frac{\partial}{\partial\phi}.\end{align*}[/tex]

    Deep down, I did mean that though the escape velocity is something so referred to in the everyday physics, but it didn't really come to my mind at the time I dealt with that problem straightly while it came to yours! We all work in the same branch of science; one is genius and the others are less so!

    AB
     
  21. Mar 8, 2010 #20
    Yeah, and we give a relation of the form

    [tex]g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2[/tex] (1)

    where in the [tex]\gamma_v[/tex], the appearing v is the local 3-velocity

    [tex]v^{i}=(dx^1/d\tau,dx^2/d\tau,dx^3/d\tau).[/tex]

    Take the sides of the last equation in (1) to the power 2 and write [tex](v^0)^2=V^2-v^2[/tex] where [tex]V^{\alpha}[/tex] denotes the 4-velocity of the particle. Hence

    [tex]g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},[/tex]

    and with [tex]u^2=(u^0)^2[/tex] we obtain finally

    [tex]V^2={\frac {-{v}^{4}{g_{{00}}}^{2}{u}^{2}+{c}^{6}+{v}^{2}{g_{{00}}}^{2}{u}^{
    2}{c}^{2}}{ \left( {c}^{2}-{v}^{2} \right) {g_{{00}}}^{2}{u}^{2}}}.[/tex]

    If your method doesn't lead to this equation, then it doesn't work!

    AB
     
    Last edited: Mar 9, 2010
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