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Proper velocity

  1. Aug 28, 2007 #1
    1. The problem statement, all variables and given/known data


    [tex] \mathbf{\eta} = 1/ \sqrt{1 - u^2/c^2} \mathbf{u} [/tex] represents the proper velocity in terms of the ordinary velocity. Find vector u in terms of vector eta.

    Then find the relation between proper velocity and rapidity.


    2. Relevant equations



    3. The attempt at a solution

    Scalar u is the magnitude of the vector u, so I cannot just bring the denominator to the other side, right? Do I have to get scalar eta in terms of scalar u?
     
    Last edited: Aug 28, 2007
  2. jcsd
  3. Aug 28, 2007 #2

    George Jones

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    The other way round. Find scalar u in terms of scalar eta.
     
  4. Aug 28, 2007 #3
    Yes. That is what I meant. Those velocity magnitude's so I am not sure how to do that...
     
  5. Aug 28, 2007 #4

    dextercioby

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    Remember that both [itex] \mathbf{u} [/itex] and [itex] \mathbf{\eta} [/itex] are ordinary vectors in R^{3} in which the norm of a vector is given by a scalar product. So compute [itex] \left \langle \mathbf{\eta},\mathbf{\eta}\right\rangle [/itex].
     
  6. Aug 28, 2007 #5
    [itex] \left \langle \mathbf{\eta},\mathbf{\eta}\right\rangle = \eta_x^2 + \eta_y^2 + \eta_z^2[/itex]

    But we have

    [tex] \sqrt{1 - u^2/c^2} \mathbf{\eta} = \mathbf{u} [/tex] so how do you get that scalar u in terms of the scalar eta?

    Let me check my understanding of what u and eta are. The usage of these two terms implicitly uses an inertial reference frame S and an object in motion with respect to S. vector u is the displacement measured in S per unit time measured in S and vector eta is the displacement measured in S per unit time measured by a clock attached to our object. All good?
     
    Last edited: Aug 28, 2007
  7. Aug 29, 2007 #6

    dextercioby

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    [tex]\left\langle \vec{\eta},\vec{\eta}\right\rangle =\left\langle \vec{u},\vec{u}\right\rangle \frac{1}{1-\frac{u^{2}}{c^{2}}}\Rightarrow \eta ^{2}=\frac{u^{2}}{1-\frac{u^{2}}{c^{2}}}\Rightarrow u^{2}=\frac{\eta ^{2}}{1+\frac{\eta ^{2}}{c^{2}}}[/tex]

    [tex]\vec{\eta}=\vec{u}\frac{1}{\sqrt{1-\frac{u^{2}}{c^{2}}}}\Rightarrow \vec{u}=\vec{\eta}\sqrt{1-\frac{u^{2}\left(\eta \right)}{c^{2}}}=\vec{\eta}\sqrt{1-\frac{1}{c^{2}}\frac{\eta ^{2}}{1+\frac{\eta ^{2}}{c^{2}}}}=\vec{\eta}\frac{c}{\sqrt{c^{2}+\eta ^{2}}} [/tex]
     
    Last edited: Aug 29, 2007
  8. Aug 29, 2007 #7
    Thanks. But is that paragraph I wrote correct? I am very confused about proper velocity. It is supposedly proper distance divided by proper time. But what is proper distance? Isn't the distance an object moves in a reference frame attached to it always 0?
     
    Last edited: Aug 29, 2007
  9. Aug 30, 2007 #8
    Let me check my understanding of what u and eta are. The usage of these two terms implicitly uses an inertial reference frame S and an object in motion with respect to S. vector u is the displacement measured in S per unit time measured in S and vector eta is the displacement measured in S per unit time measured by a clock attached to our object. All good?

    I think I got it. Also, is the relation between proper velocity and rapidity the following:

    tanh theta = eta/ sqrt(c^2 + eta^2)

    ?
     
    Last edited: Aug 31, 2007
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