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Velocity addition formula question

  1. Oct 28, 2013 #1


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    In the Galilean formula u=u'+v, the velocities are bold so we know they're vectors, so for instance if u'=(a,b), and v=(c,d), we say that u=(a+c,b+d). But, in SR u=(u'+v)/(1+vu'/c2), none of the velocities are shown in bold, at least not not the way I saw it written. So, if we used the same two dimensional hypothetical vectors as up top, in SR terms, how would you find the added velocity u? Would it be (a+c,b+d)/(1+(ac+bd)/c2), or would you have to break it up into the individual x,y,z components and solve it for each one dimension component.
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  3. Oct 28, 2013 #2


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  4. Oct 28, 2013 #3
    In special relativity, velocities are 4 dimensional rather than 3 dimensional. If you are talking about the velocity addition formula for 4 velocities, then you recover the Galilean formula. But, in practice, you gotta be careful how you do it, by referring the velocities to the same 4D coordinate system.

  5. Oct 29, 2013 #4


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    That formula is only for [itex]{\bf u'}[/itex] parallel to [itex]{\bf v}[/itex].
    There is another formula for [itex]{\bf u'}[/itex] perpendicular to [itex]{\bf v}[/itex].
  6. Oct 29, 2013 #5


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    For the general case, in which v1 and v2 are not co-linear, velocity "addition" is not commutative. In other words, the relative velocity vector VAB of particle A as viewed from particle B is different from the relative velocity VBA of B as viewed from A. (Not just a minus sign!) This is because the Lorentz transformations associated with v1 and v2 do not commute.

    It may also be thought of as an aberration effect. Although the magnitude of the relative velocity will be the same, if |v1| ≠ |v2| the direction of V will be different depending on who is viewing whom.

    It's simpler to express the result not in terms of the v's but rather their γ factors. For the co-linear case it's easy to start with V = (v1 + v2)/(1 + v1v2) and derive Γ = γ1 γ2 (1 + v1v2). For the non-co-linear case this generalizes quite simply to Γ = γ1 γ2 (1 + v1·v2)
    Last edited: Oct 29, 2013
  7. Oct 31, 2013 #6
    Generally when the velocities are not parallel you need the individual components as they transform in different ways. Let us say there is an object A moving with velocity v relative to some reference frame S. Another object B moves with velocity u' as measured in the rest frame of object A. What is the velocity (u) of object B as measured in frame S?

    For convenience, we can align the x axis with the velocity vector v and resolve velocity vector u' into components (u'x, u'y. u'z). (Note that when all measurements are made from a single inertial reference frame that we can resolve or add velocity components in the usual Euclidean way. For a real object the magnitude of the combined components should never be greater than c, as mentioned in the parallel thread.) We can now express u' in the form of a four vector as [a0, a1, a2, a3] = [γ(u'), γ(u')ux', γ(u')uy', γ(u')uz'] where γ(u') represents 1/√(1-(u')2) = 1/√(1-(ux2+uy2+uz2)). Now we can perform a Lorentz boost of -v using Matrix multiplication and obtain a transformed 4 velocity [b0, b1, b2, b3] = [γ(u), γ(u)ux, γ(u)uy, γ(u)uz]. The 3 velocity of B relative to S is then u = (ux,uy,uz) = (b1/b0, b2/b0, b3/b0). The boost is -v so that after the transformation the velocity of object A relative to S is positive. The magnitude of the transformed 3 velocity u is √(ux2+uy2+uz2).

    N.B. The four velocities are usually expressed in a more compact form as γ(u')[1, u'x, u'y. u'z] and γ(u)[1, ux, uy. uz)] where c=1.
  8. Oct 31, 2013 #7


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    It's easier by far to use matrices rather than writing out all those individual components. A Lorentz transformation along the x-axis

    x' = γ(x - vt)
    t' = γ(t - vx)

    can be written as a 4x4 matrix

    [tex]\left(\begin{array}{cccc}\gamma&-\gamma v&0&0\\-\gamma v&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)[/tex]

    The three spatial directions can then be combined into a 3-vector,

    [tex]\left(\begin{array}{cc}\gamma&-\gamma \mathbf{v}\\-\gamma \mathbf{v}&\mathbf{I} + (\gamma - 1)\mathbf{\hat{v}} \mathbf{\hat{v}}\end{array}\right)[/tex]

    where ## \mathbf{\hat{v}} ## is the unit vector in the v direction.

    For the problem at hand, just write ##\mathbf{v_1}## and ##\gamma_1## in place of ##\mathbf{v}## and ##\gamma##, and apply the transformation to the other velocity 4-vector

    [tex]\left(\begin{array}{c}{\gamma_2}\\\gamma_2 \mathbf{v_2}\end{array}\right)[/tex]

    For example the time component of the result is, as stated earlier,

    [tex]\Gamma = \gamma_1 \gamma_2 (1 - \mathbf{v_1 \cdot v_2})[/tex]

    EDIT: For completeness, here's the space components, which are not as pretty: :frown:

    ## - \gamma_1 \gamma_2 \mathbf{v_1} + \gamma_2 \mathbf{v_2} + \gamma_2 (\gamma_1 - 1) \mathbf{\hat{v_1}} (\mathbf{\hat{v_1} \cdot \mathbf{v_2}})##

    Note that the space components are not symmetric (or even antisymmetric!) under the interchange ## \mathbf{v_1} \leftrightarrow \mathbf{v_2}##, which illustrates the important point that the "addition" of velocities is not commutative.
    Last edited: Oct 31, 2013
  9. Nov 1, 2013 #8


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    The simple answer for u' perpendicular to v is
    [tex]{\bf u_\perp}=\frac{\bf u'_\perp}{\gamma(1+{\bf u'\cdot v}}[/tex].
    Last edited: Nov 1, 2013
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