Four-vector problem

  1. The four-velocity as defined for example here, is given by
    [tex]
    U=\gamma(c,\bar{u})
    [/tex]
    but I get
    [tex]
    U=\gamma(1,\frac{\bar{u}}{c})
    [/tex]
    Consider the timelike curve [itex]\bar{w}(t)=(ct,\bar{x}(t))[/itex] with velocity [itex]\bar{v}(t)=(c,\bar{x}'(t))\equiv (c,\bar{u}(t))[/itex] and the arc-length (proper time)
    [tex]
    \tau\colon I\subset \mathbb{R}\to \mathbb{R}\colon t\mapsto \int_{t_{0}}^{t}\left\| \bar{v}(k)\right\|dk
    [/tex]
    for which (by First Fundamental Theorem of Calculus (1), the Minkowskian inner product (2) and the definition of the Lorentz factor (3) )
    [tex]
    \Leftrightarrow \frac{d\tau}{dt}=\left\| \bar{v}(t)\right\|=\sqrt{c^{2}-\bar{u}^{2}(t)}\equiv \frac{c}{\gamma}
    [/tex]
    then the velocity of the curve after arc-length (proper time) parameterization, is given by
    [tex]
    \bar{v}(\tau)=\frac{d\bar{w}}{d\tau}=\frac{d\bar{w}}{dt}\frac{dt}{d\tau}=\frac{\bar{v}(t)}{\left\| \bar{v}(t)\right\|}=\frac{(c,\bar{u}(t))}{\frac{c}{\gamma}}=\gamma(1,\frac{\bar{u}(t)}{c})
    [/tex]
    I would think that my [itex]\bar{v}(\tau)[/itex] is the four-velocity but in fact [itex]\bar{v}(\tau)=\frac{U}{c}[/itex] where U the four-velocity as defined in textbooks. What am I missing?
     
  2. jcsd
  3. Matterwave

    Matterwave 3,853
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    If you are using units where c is not 1, then certainly you want the 4-velocity to be normalized to c (or -c depending on signature of the metric), and not 1 which is not in units of velocity (if, again, c is not set to 1).
     
  4. PAllen

    PAllen 5,577
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    I think it is purely a convention. I learned that U is meant to be a unit vector, always, even when c is not taken to be 1. Some people like norm of U = c, some like 1. Norm of one amounts to units of time rather than distance for positions.

    However, since you start with 4-position in units of distance, you should get a U whose norm is c. The flaw is your d tau/ dt computation. It is 1/gamma not c/gamma. Then you get U with norm of c. Specifically, your integral formula is not right. Given your units for v, the integral is c * tau, not tau. This, then, is the initial (and only) error, from which all else follows.
     
    Last edited: Mar 8, 2012
  5. PAllen

    PAllen 5,577
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    I'll add one more thing. The idea of norming U to c has led to Brian Greene's "speed through space-time equals c", which has led to numerous confusions and debates on these forums, as well as facilitating cranks. The convention of Einstein and Bergmann that U has norm 1, irrespective of the value of c sidesteps all of this lunacy.
     
  6. Matterwave

    Matterwave 3,853
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    If you parameterize your curve with the proper time in seconds, and your proper distances are measured in meters, don't you necessarily get the norm condition in U to be c?

    What I mean is, if you use units of distance the same as your units of time, aren't you necessarily setting c=1?
     
  7. PAllen

    PAllen 5,577
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    No. The conventions:

    position: (t, x/c, y./c, z/c)
    line element: d tau^2 = d t^2 - (dx^2 + dy^2 + dz^2)/c^2
    and the consequence that U = d(position) / d tau has norm 1

    in no way have c=1.
     
  8. Matterwave

    Matterwave 3,853
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    Are you talking about adding a factor of c into the metric? o.o
     
  9. PAllen

    PAllen 5,577
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    Don't know what you are asking. The two common conventions for the metric are:

    ds^2 = dx^2 + dy^2 + dz^2 - c^2 t^2

    and

    d tau^2 = dt^2 - dx^2/c^2 - dy^2/c^2 - dz^2/c^2

    I have always used the latter.
     
  10. Ok, so because [itex]\bar{w}(t)=(ct,\bar{x}(t))[/itex] is in space units, [itex]\left\| \bar{v}(t)\right\|[/itex] has units space/time and the integral is in space units. Then the arc-length (proper time) parameterization in time units is given by
    [tex]
    \tau\colon I\subset \mathbb{R}\to \mathbb{R}\colon t\mapsto \frac{1}{c}\int_{t_{0}}^{t}\left\| \bar{v}(k)\right\|dk
    [/tex]
    and the four-velocity in space/time units
    [tex]\bar{v}(\tau)=\gamma(c,\bar{u}(t))[/tex]
    Is this the correct explanation? As I understand, the four-velocity in the other convention (norm=1) is unitless, isn't it? So how is it used then?
     
  11. PAllen

    PAllen 5,577
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    Yes, this is correct.

    The magnitude of U really has no real meaning in either convention. All information about measured velocity in any basis (frame) is contained in the direction of U as a tangent vector. The norm 1 convention makes this explicit: it is literally a unit tangent vector to a world line.

    There are pros and cons to either convention.
     
  12. Thanks, you've been a great help!

    And what about the magnitude of space-component [itex]\gamma\bar{u}(t)[/itex] of the four-velocity? I'm not quite sure how all this relates to some physical reality. Can I interpret the space-component of the four-velocity as the classical velocity (at least when not choosing the norm=1 convention)?
     
  13. PAllen

    PAllen 5,577
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    Let's say you have U as a tangent vector, considered a coordinate independent quantity. You want to know the spatial velocity measured in some frame defined by a 4 orthonormal unit vectors, one timelike the others spacelike. You take U dot <x unit vector>/ U dot <t unit vector>, same for y and z. Clearly, the norm of U drops out.

    In the coordinates you initially used to express U, the spatial velocity is just the u you started with (which you would get by executing the procedure above). The quantity gamma*u would be rather meaningless: the rate of change of distance in a given frame by a particle's proper time. This could exceed c by a large factor.
     
  14. PAllen

    PAllen 5,577
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    Error here. In this convention you simply have position: (t,x,y,z)

    Then U, with norm 1, becomes: gamma * (1,u)

    This is not necessarily taking c=1, because the covariant metric diagonal is still (1, -1/c^2, -1/c^2,-1/c^2). Contravariant diagonal obviously (1,-c^2,-c^2,-c^2).
     
    Last edited: Mar 9, 2012
  15. Chestermiller

    Staff: Mentor

     
  16. PAllen

    PAllen 5,577
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    Probably no one interested anymore, but I have clarified a few things in my own mind.

    Given the desire to express a 4 - tangent vector to a world line in terms of u = (dx/dt, dy/dt, dz/dt) with conventional meanings, two separate conventions affect the form it takes:

    - how you norm it
    - is your metric canonic (all +1,-1) or not (you have c^2 or 1/c^2 in your metric).

    The signature of the metric is irrelevant for this situation.

    With canonic metric, you have a factor of c in your tangent vector components, so you have two natural choices:

    gamma * (c, u) // norm c; dimensions distance/time; from d (ct,x,y,z) / d tau

    gamma * (1, u/c) // norm 1; dimensionless; from d (t, x/c, y/c, z/c) / d tau

    With non-canonic metric, you have only one natural form, with norm 1:

    gamma * (1, u) // mixed dimensions, as is characteristic of non-canonic metric
    // from d (t,x,y,z) / d tau

    With non-canonic metric, a form with norm c is simply unnatural.

    It happens that I learned SR with non-canonic metric, 4-velocity being a unit vector, and the concept of 'speed through spacetime' not remotely meaningful. It appears that almost all modern books use canonic metric.

    [EDIT: For emphasis, note something I derived in an earlier post: the norm of c or 1 plays no role at all in computing any observable. You could even normalize to 42 and it would make no difference. Only the direction in 4-space of the tangent vector plays any role in computing observables.]
     
    Last edited: Mar 10, 2012
  17. Matterwave

    Matterwave 3,853
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    Yes, this what I meant when I said "are you talking about adding a factor of c to the metric?" I always learned to us diag(-1,1,1,1) for my metric so any factors of c's I need are in the 4-vectors themselves.
     
  18. Not sure what you mean: there is no dot product in Minkowskian space-time... Do you mean that [Ux/Ut,Uy/Ut,Uz/Ut] is the spatial 3-velocity? Why?
     
  19. PAllen

    PAllen 5,577
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    Sure there is a dot product. It is defined by the metric. For example, if the metric is diag(1,-1/c^2, -1/c^2,-1/c^2), then the dot product of X and Y is:

    x0*y0 - x1*y1/c^2 - x2*y2/c^2 - x3*y3/c^2
     
  20. Depends on your definition of the dot product, but I see what you mean. But I don't see why [Ux/Ut,Uy/Ut,Uz/Ut] would correspond to a spatial velocity.
     
  21. PAllen

    PAllen 5,577
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    In a metric space there is one definition of the dot product. The Euclidean one looks the way it does solely because the Euclidean metric is diag(1,1,1).

    If U is some tangent vector, and x,y,z,t are unit vectors for some frame, then the dot product of U with such unit vectors expresses U in that frame basis. Then Ux/Ut gives the x speed (well, actually, x-speed/c , but that is just as good). Look at the tangent vector itself expressed in your starting coordinates (c-normed, canonic metric; works the same in any other convention):

    U = gamma(c,u)

    Ux = gamma * ux is not the x speed; but note Ux/Ut = ux/c. This feature will be true in any other basis. In particular, in an orthonormal basis with U itself taken as the time unit vector, you get spatial speed of zero - the particle has no spatial speed in its own basis.
     
    Last edited: Mar 13, 2012
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