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Classic and relativistic 3-velocity

  1. Apr 10, 2012 #1

    Wox

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    I'm trying to understand how the geometry of Minkowski space is related to physical observations, in particular, measurements of the velocity of an object. In the attempt below, I got stuck at the meaning of the relativistic 3-velocity. Can anyone get me back on track?
     
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  3. Apr 10, 2012 #2

    Mentz114

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    The final 4-velocity looks like the Lorentz transform of [itex]d\omega_0/dt[/itex] with parameter [itex]d\bar{x}/dt[/itex] so it must be [itex]\omega[/itex] seen from [itex]\omega_0[/itex]. Probably.
     
  4. Apr 10, 2012 #3

    robphy

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    This may be helpful.
    Regard an inertial observer's 4-velocity [itex]\hat u[/itex] as a unit-vector in Minkowski space.
    Let [itex]\hat t[/itex] be the four-velocity of another inertial frame (call it the "rest" frame).

    The rest-frame can break up [itex]\hat u[/itex] into temporal and spatial components.
    The temporal component is [itex]\gamma \hat t[/itex] (that is, [itex]\cosh\theta \hat t[/itex]), which essentially captures the idea of time-dilation.
    The spatial-component is the part left over (essentially [itex]\sinh\theta \hat s[/itex], where [itex]\hat s[/itex] is the purely-spatial-unit-vector-according-to-[itex]\hat t[/itex], and is sometimes called the celerity).

    If the tip of [itex]\hat u[/itex] is interpreted as one-tick of a standard-clock carried by [itex]\hat u[/itex], then [itex]\gamma \hat t[/itex] refers to the event on [itex]\hat t[/itex]'s worldline that is simultaneous with that distant tick. The spatial-component is then the spatial-separation between those two events.

    So, one could interpret that spatial-part of the 4-velocity as
    u's spatial-displacement-from-t-according-to-t divided by u's elapsed-time-according-to-u, that is [itex](\sinh\theta\mbox{ light-ticks})/(1\mbox{ tick})[/itex].

    Compare this to the ordinary velocity:
    u's spatial-displacement-from-t-according-to-t divided by u's elapsed-time-according-to-t, that is [itex](\sinh\theta\mbox{ light-ticks})/(\cosh\theta \mbox{ tick})[/itex].
     
  5. Apr 11, 2012 #4

    Wox

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    That's really helpful! So "u's spatial-displacement-from-t-according-to-t" is the distance between two events that aren't simultaneous according to u, but are simultaneous in the rest frame? So we're taking distance in the rest frame and divide it by time in the other reference frame to get relativistic 3-velocity [itex]\gamma\frac{d\bar{x}}{dt}[/itex]? This would mean that the relativistic 3-velocity can't be measured by one observer, right (only calculated)?

    The reason for asking is that the relativistic 3-velocity is used for the relativistic momentum [itex]\bar{p}=m\gamma\frac{d\bar{x}}{dt}[/itex], which is then used in the three-vector expression of Newton's second law of motion [itex]\bar{F}=m\frac{d\bar{p}}{dt}[/itex]. To understand what this means, I wanted to understand what the relativistic 3-velocity meant. However dividing distances in one inertial frame by times in another inertial frame doesn't seem to clarify anything...
     
  6. Apr 11, 2012 #5

    robphy

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    I'm not sure what your definition of "measured" is.
    An inertial observer using radar methods, for example, can make all sorts of measurements of spacetime intervals between distant events.... including intervals on other observer worldlines. These intervals can be used in the determination of some quantity of interest.
     
  7. Apr 12, 2012 #6
    Could it be that the problem is caused by the fact that your force equation is wrong?

    F = d(γmv)/dt

    For straight-line acceleration:
    F = m d(γv)/dt = γ3m dv/dt

    And for circular motion:
    FN = γm dv/dt = γm aN

    These are the things that one actually measures: the rest mass m one can measure beforehand, one can measure v as function of t in a linear accelerator, and the relativistic mass γm as well as aN one can measure in motion inside a cyclotron.
     
    Last edited: Apr 12, 2012
  8. Apr 13, 2012 #7

    Wox

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    You're right. I was just referring to the mixing of distance in one frame and time in another frame. I can't get my head around the implications of doing that. What does this mean for the 3-force in [itex]\bar{F}=\frac{d(m\gamma \bar{v}(t))}{dt}[/itex]? For example, how would you measure this force?
     
    Last edited: Apr 13, 2012
  9. Apr 13, 2012 #8

    Wox

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    Yes, that's a typo, it should be [itex]\bar{F}=\frac{d(m\gamma \bar{v}(t))}{dt}[/itex], but since [itex]\gamma \bar{v}(t)[/itex] (the relativistic 3-velocity) mixes distances and times in different frame, I'm not sure what it means.
     
    Last edited: Apr 13, 2012
  10. Apr 14, 2012 #9
    There are really two parts to applying the relativistic version of Newton's second law of motion. There is the 4 force acting on the particle, and there is the rate of change of (mass times 4 velocity) with respect to proper time. You have focused exclusively on the second part. The first part determines the actual 4 force from other considerations.

    For example, consider a charged particle in an electromagnetic field. The first part of the second law equation (i.e., the right hand side) is the 4 force acting on the particle. This is proportional to the Faraday tensor contracted with the 4 velocity vector. The second part of the equation (the left hand side) is equal to the rate of change of 4 momentum with respect to proper time (i.e., proper time of the particle). If you reckon both these sides of this equation with respect to an arbitrary inertial frame of reference, then there will be a factor of γ present on each side of the equation. These γ factors can be cancelled, leaving the form of the 3 force given by your relationship on the left hand side of the resulting equation.

    Chet
     
  11. Apr 15, 2012 #10
    3-velocity contains no mix of distances and times in different frames... In my reply, everything referred to a single reference system: x, t, v as well as γ all refer to S.

    PS: 3-velocity is simply the normal velocity that one measures: v=dx/dt
     
    Last edited: Apr 15, 2012
  12. Apr 16, 2012 #11

    Wox

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    Yes, but relativistic 3-velocity ([itex]\frac{d\bar{x}}{d\tau}=\gamma\bar{v}(t)[/itex]) does contain a mixture, because [itex]d\bar{x}[/itex] is the displacement with respect to the reference frame of an observer, while [itex]d\tau[/itex] the elapsed time along the world line of the particle and since [itex]\tau=t_{o}[/itex] where [itex]t_{o}[/itex] the time coordinate in the particle's rest frame, [itex]\frac{d\bar{x}}{d\tau}[/itex] mixes distance in the frame of the observer and time in the frame of the particle. This is what I understood from robphy's post.
     
  13. Apr 16, 2012 #12
    Sorry, common relativistic coordinate velocity (which I presume is what people usually mean with "3-velocity") is simply v=dx/dt as determined with a single reference system, similar to the one defined by Einstein in his 1905 paper:
    http://www.fourmilab.ch/etexts/einstein/specrel/www/ section 2

    There isn't any difference between "classic" and "relativistic" (3-)velocity.
    - see the first sentence of http://en.wikipedia.org/wiki/4-velocity

    However, from the Wikipedia article, it seems that you are instead discussing "proper velocity" or celerity.

    PS. Then the application suggestions here may be interesting:
    http://en.wikipedia.org/wiki/Proper_velocity#Applications
     
    Last edited: Apr 16, 2012
  14. Apr 16, 2012 #13

    Wox

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    Aha, thanks! I didn't know there was already a name for that.
     
  15. Apr 17, 2012 #14

    Wox

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    Thanks again for putting me on the right track guys! With respect to the spatial component of the 4-velocity (called the proper velocity as pointed out above), I also found this article to be very useful: www.umsl.edu/~fraundor/aps96n07.pdf
     
  16. Apr 20, 2012 #15

    Wox

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    Apart from the proper velocity issue, I had the relativistic law of motion issue:

    So if I understand correctly, the 4-force on a charged particle in an electromagnetic field [itex]\bar{F}=q\Phi\cdot(\gamma c,\gamma \bar{v})[/itex] and the 4-force according to relativistic law of motion [itex]\bar{F}=\frac{dp}{d\tau}=\gamma\frac{dp}{dt}= \gamma(c\frac{dm\gamma}{dt},\frac{dm\gamma\bar{v}}{dt})[/itex] so that

    [tex]
    \begin{array}{cc}
    &q\Phi\cdot(c,\bar{v})=(c\frac{dm\gamma}{dt},\frac{dm\gamma\bar{v}}{dt})\\
    \Leftrightarrow & q\left[ \begin{array}{cccc}
    0&-E_{x}&-E_{y}&-E_{z}\\
    E_{x}/c&0&B_{z}&B_{y}\\
    E_{x}/c&-B_{z}&0&B_{x}\\
    E_{z}/c&B_{y}&-B_{x}&0
    \end{array} \right]\cdot\left[ \begin{array}{c}
    c\\
    v_{x}\\
    v_{y}\\
    v_{z}
    \end{array} \right]=\left[ \begin{array}{c}
    c\frac{dm\gamma}{dt}\\
    \frac{dm\gamma v_{x}}{dt}\\
    \frac{dm\gamma v_{y}}{dt}\\
    \frac{dm\gamma v_{z}}{dt}
    \end{array} \right]\\
    \Leftrightarrow & q(\bar{E}+\bar{v}\times\bar{B})=\frac{dm\gamma \bar{v}}{dt}\quad\quad\text{and}\quad\quad \bar{E}\cdot\bar{v}=-c\frac{dm\gamma}{dt}\\
    \Leftrightarrow & \bar{f}=\frac{dm\gamma \bar{v}}{dt}\quad\quad\text{and}\quad\quad \bar{E}\cdot\bar{v}=-c\frac{dE_{tot}}{dt}
    \end{array}
    [/tex]
    where [itex]\bar{f}[/itex] the classical Lorentz 3-force. Is this correct? (I'm not sure about the signs in the electromagnetic tensor). And what does the second equation (originating from the time component) represent?
     
  17. Apr 20, 2012 #16
    Yes. This is right on target!

    The time component says that the rate of change of total energy of the particle (in this case kinetic energy) is equal to the rate at which work is done by the electromagnetic field on the particle. Incidentally, there shouldn't be a c in the very last equation, and there should be a q on the left hand side. Also, recheck the sign.

    Chet
     
  18. Apr 20, 2012 #17

    Mentz114

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    Wox, I don't know if it is only a typo but one of the By terms in the field tensor above should be negative.
     
  19. Apr 23, 2012 #18

    Wox

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    After some corrections:

    [tex]
    \begin{array}{cc}
    & \bar{F}=\frac{d\bar{p}}{d\tau}=\gamma\frac{d\bar{p}}{dt}\\

    \Leftrightarrow &q\Phi\cdot(c,\bar{v})=(\frac{1}{c}\frac{dE_{tot}}{dt},\frac{dm\gamma\bar{v}}{dt})\\

    \Leftrightarrow & q\left[ \begin{array}{cccc}
    0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\
    E_{x}/c&0&B_{z}&-B_{y}\\
    E_{x}/c&-B_{z}&0&B_{x}\\
    E_{z}/c&B_{y}&-B_{x}&0
    \end{array} \right]\cdot

    \left[ \begin{array}{c}
    c\\
    v_{x}\\
    v_{y}\\
    v_{z}
    \end{array} \right]=\left[ \begin{array}{c}
    \frac{1}{c}\frac{dE_{tot}}{dt}\\
    \frac{dm\gamma v_{x}}{dt}\\
    \frac{dm\gamma v_{y}}{dt}\\
    \frac{dm\gamma v_{z}}{dt}
    \end{array} \right]\\

    \Leftrightarrow & q(\bar{E}+\bar{v}\times\bar{B})=\frac{dm\gamma \bar{v}}{dt}\quad\quad\text{and}\quad\quad q\bar{E}\cdot\bar{v}=-\frac{dE_{tot}}{dt}
    \end{array}
    [/tex]

    There is still a [itex]-[/itex] sign in the last equation. I must be making a mistake, but I don't see where. The four-momentum is [itex]\bar{p}=(\frac{E_{tot}}{c},m\gamma\bar{v})[/itex], right?
     
    Last edited: Apr 23, 2012
  20. Apr 23, 2012 #19

    Mentz114

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    In tensor notation
    [tex]
    \ddot{x}^\mu = F^{\mu\nu}J_\nu
    [/tex]
    One of the column vectors is actually a row vector so flip the sign of the first element of one or the other. This is raising or lowering an index in Minkowski spacetime.
     
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