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Four velocity

  1. Sep 5, 2015 #1
    Why is the magnitude of object four velocity always equal to the square of c?
  2. jcsd
  3. Sep 5, 2015 #2


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    This is by definition. The trick is to define a velocity that is a four-vector. To that end you describe the trajectory of a (massive) particle as a function of its proper time, i.e., the time which is measured in the "comoving" frame of the particle, where the particle is always at rest.

    In an inertial frame you have a time coordinate ##t## and space coordinates ##\vec{x}##. The trajectory in the usual sense is then written as ##\vec{x}(t)##, and the usual three-velocity is given by
    $$\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t},$$
    but this quantity has pretty complicated transformation properties under Lorentz transformations. That's why one uses the trick to use the proper time of the particle as parameter.

    The proper time is defined by its infinitesimal change, when the time ##t## changes infinitesimally by ##\mathrm{d} t##:
    $$\mathrm{d} \tau^2=\frac{1}{c^2} (c^2 -\vec{v}^2) \mathrm{d} t^2.$$
    This makes only sense if ##\vec{v}^2<c^2##. Then ##\tau## is monotonously increasing with ##t## and it is invariant under Lorentztransformations.

    Now it's easy to define a Minkowski four-vector which describes the velocity of the particle in a covariant manner. You describe the particle's trajectory in Minkowski space by ##x^{\mu}(\tau)##, where ##x^{\mu}=(c t,\vec{x})## is the space-time four vector in an arbitrary inertial frame of reference. The four-vector is then given as
    $$u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
    Since ##\mathrm{d} \tau## is a Lorentz scalar and ##\mathrm{d} x^{\mu}## a four-vector (more precisely the components of a four-vector wrt. an inertial reference frame), ##u^{\mu}## is one. From the definition of the proper time you immediately get
    $$u^{\mu} u_{\mu}=c^2.$$
    It's also easy to get the relation between the four-velocity and the usual three-velocity:
    $$u^{\mu} = \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau}=\frac{1}{\sqrt{1-\vec{v}^2/c^2}} \begin{pmatrix} c \\ \vec{v} \end{pmatrix}.$$
    For more details, see my (still unfinished) FAQ article on special relativity:

  4. Sep 5, 2015 #3


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    One good way to think about this is that a four-velocity in relativity is essentially what we mean by a frame of reference. It doesn't make sense to talk about the magnitude of a frame of reference. This also explains why adding two four-velocities doesn't do what you'd expect from Newtonian mechanics; it doesn't make sense to add two frames of reference.

    Another thing to realize is that the magnitude of a vector is frame-invariant, and therefore it wouldn't make sense to imagine that the magnitude of an object's four-velocity would produce some number telling you how fast the object was going. How fast relative to what?
  5. Sep 8, 2015 #4


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    Note, there are different conventions for Minkowski metric, leading to different results for the magnitude of a 4-velocity. The more common modern convention says the Minkowski metric is diag(1,-1,-1,-1) or diag(-1,+1,+1,+1). With these, the magnitude squared is c2. However, I often use an older convention where Minkowski metric is taken as (1,-1/c2,-1/c2,-1/c2). With this convention, a 4-velocity is a unit vector whose magnitude is 1, and this has nothing to do with choosing units where c is 1.

    I would slightly modify bcrowell's comment to say that a 4-velocity defines the timelike basis vector of a frame (rather than defining a frame). With the unit vector convention, it even more clearly makes no sense to expect the magnitude of 4-velocity to mean anything since it is identically 1. The rest of bcrowell's comment applies equally well to considering 4-velocity as a timelike basis vector. A time like basis vector is, if you will, a 'time direction'. You wouldn't expect adding two time directions to make much sense.
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