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Studying special relativity and using the idea of four vectors

  1. Mar 23, 2005 #1

    JFo

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    We are currently studying special relativity and using the idea of four vectors.

    The position 4-vector has been defined in class as

    [tex] \vec{X} = (ct, \vec{r}) \ \ \ \ \ \ \ where \ \vec{r} = (x, y, z) \ \ [/tex](ie: the usual 3 space position vector)

    and the velocity 4-vector is then

    [tex]\vec{V} = \frac{d\vec{X}}{d\tau} \ \ \ \ \ \ \ \ \ where \ \tau \ \ is \ the \ proper \ time [/tex]

    substituting [tex]\ \ \ \tau = \frac{t}{\gamma} \ \ \ [/tex] we got

    [tex] \vec{V} = \gamma (c, \vec{v}) [/tex]

    The part I don't understand is why we differentiate with respect to the proper time, rather than the time as seen by an observer in the frame with which we want the velocity with respect to?

    To illustrate my concerns, consider the simplified scenario of an object travelling at constant speed u in the +x direction of a frame S.
    I would think that if we wanted to find its velocity according to an observer in frame S then we would need to calculate the distance it travels, divided by the time that it took with all quantities being measured by an "observer" in S.
    Yet it seems from the definition of a 4-velocity I'm supposed to divide by the time it takes according to the frame stationary to the object (the proper time). This to me doesn't seem logical since were trying to find the velocity wrt S.

    I have tried to come up with an argument to explain why we use the proper time, rather than the time measured by S.
    Perhaps in the limit that [tex] dt [/tex] goes to zero, [tex] d\vec{X} [/tex] goes to zero as well, so the time it takes to move a distance [tex] d\vec{X} [/tex] approaches the proper time.
    Is this the correct explanation, or am I way off?

    Your help is greatly appreciated.
     
    Last edited: Mar 23, 2005
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  3. Mar 24, 2005 #2

    dextercioby

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    In classical mechanics,time is absolute & the velocity is defined as the derivative wrt this absolute quantity.In SR,the proper time (or the Lorentz interval) is absolute and that's why we define the 4velocity as a derivative wrt the proper time...

    Daniel.
     
  4. Mar 24, 2005 #3

    pervect

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    The 4-velocity is not the velocity. If you wanted to calculate the velocity, you'd do exactly what you described. But velocities do not transform as 4-vectors.

    4-velocities do transform as 4-vectors., because 'tau' is the same for all observers

    Take a look at the definition again

    (dt/dtau, dx/dtau, dy/dtau, dz/dtau) transforms exactly like

    (t,x,y,z)

    because tau is the same for all observers. This means you can transform 4-velocities with the Lorentz transform just like any other 4-vector.

    If the rationale of the 4-velocity vector doesn't seem obvious, be patient. Aside from their transformation properties, you'll find out (for instance), that the energy and momentum of an object are given by

    [tex]
    \vec{E} = m \vec{u}
    [/tex]

    where u is the 4-velocity. So it's definitely a useful concept.
     
  5. Mar 24, 2005 #4

    robphy

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    Note that the 4-velocity [tex] \widetilde{V} = \gamma (c, \vec{v}) [/tex] is (1/c) times "a [future] timelike unit 4-vector tangent to the worldline of the particle":

    [tex]
    \begin{align*}
    \widetilde{V}\bullet \widetilde{V}
    &= \gamma^2 (c^2- \vec{v}\cdot\vec{v}) \\
    &= c^2\gamma^2 (1- v^2/c^2)\\
    &= c^2\
    \end{align*}
    [/tex]
    So, [tex]\widetilde{V}/c[/tex] is a unit 4-vector. Without that [tex]\gamma[/tex], the norm of [tex] \widetilde{V}[/tex] would not be a scalar.
     
  6. Mar 24, 2005 #5

    JFo

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    You guys are great!

    What I get from dextercioby's and pervect's posts is that the reason the 4-position is differentiated wrt the proper time, is because the 4-velocity will then be the same for all observers, where if instead, each observer used their respective times, they would get different answers. did I get the right message?

    robphy: I don't quite understand your post right away, but this is because I am still not very familiar with worldlines, timelike intervals, and the geometry aspect of 4 vectors. I'm sure it will become clearer to me once I study these things.

    Question to all: did 4-vectors seem a little strange when you first encountered them?
     
  7. Mar 24, 2005 #6

    dextercioby

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    Not if u're familiar with linear algebra & know a bit o diff.geom...

    I was...

    Daniel.
     
  8. Mar 24, 2005 #7

    JFo

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    I've taken a course in linear algebra but haven't started studying diff geom yet, Part of the problem is that there is no book to reference from the class, and the lectures are not very clear. The class started with 30 people, and there are only 12 remaining and we haven't even gotten our first exam back yet. By far the most disorganized class I've ever been in.
     
    Last edited: Mar 24, 2005
  9. Mar 25, 2005 #8

    dextercioby

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    Okay,then a 4vector is simply an element of a 4D vector space...Changing a basis is done by changing inertial observers and that is achieved through a Lorentz Transformation.That's all there is to 4 vectors.

    Daniel.
     
  10. Nov 23, 2011 #9
    Re: 4-velocity

    The 4-velocity is not the velocity relative to a given frame of reference S. It is the velocity relative to the space-time continuum itself (which can, in a sense, be regarded as stationary). Objects at rest within the S frame of reference, are not at rest with respect to the space-time continuum. They are moving in the t direction of the S reference frame with a velocity equal to c. Every object and every frame of reference has a velocity equal to c relative to the space-time continuum. However, the direction of the 4-velocity vector (in 4D space) is unique to each object and each frame of reference, and is the same only for frames of reference and objects at rest relative to one another.
     
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