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Lorrentz transform question about velocity

  1. Dec 30, 2011 #1


    The book states to stop particle B and calculate the velocity of A in B's frame.
    i.e. transform velocity to a frame S' that means with velocity v(x) = UB = 0.7c


    Transform UA to UA'

    UA' = (UA - V) / [1 - (v/c^2)(UA)]

    They then fill in values and obtain an answer of 0.23c. But I'm wondering where has this equation come from? I can see that it looks a bit like Lorrentz transformation gamma.

    But if someone could explain exactly what's going on to obtain this equation? Has gamma been applied? Thank you
     
    Last edited: Dec 30, 2011
  2. jcsd
  3. Dec 30, 2011 #2
    I think you mite have written it down wrong. Start with the lorentz transformation
    and note u = dx/dt and u' = dx'/dt'


    x' = gamma (x-vt)
    t' = gamma (t-vx/c^2)

    gamma = (1-(v/c)^2)^-1/2
    v = constant velocity between two frames.

    Now using chain rule: u' = dx'/dt' = (dx'/dt)*(dt/dt')

    You should be able to derive u' now.

    I think the equation should be u' = (u-v)/(1-uv/c^2)

    The gamma's should cancel.
     
  4. Dec 30, 2011 #3
    Yes, you're correct. I've edited the equation.
     
  5. Dec 30, 2011 #4
    Can you explain a bit about how you obtained u' = (u-v)/(1-uv/c^2)

    I'm not sure I follow. Thank you.
     
  6. Dec 30, 2011 #5
    You know [itex]x' = \gamma (x-vt)[/itex] and [itex]t' = \gamma (t - \frac{v}{c^2}x)[/itex]

    and you need to find out u' in terms of u

    so its just simple maths ...

    for the above 2 eqn's find dx' (by finding dx'/dt and taking dt to right side)and dt' (by finding dt'/dt and taking dt to right) and just divide them ...
     
  7. Dec 30, 2011 #6
    So you differentiate x' with respect to t to obtain dx'/dt.
    Then differentiate t' with respect to t to obtain dt'/dt.
    Then you can just use chain rule as i mentioned above.
    Remember to substitute u = dx/dt.

    If you don't understand how to get it still, do say and i'll write it out if you want?
    Also do you understand why this equation is used to solve the problem?
     
  8. Dec 30, 2011 #7
    That would be great. Thanks very much!

    To be honest, no I don't.
     
  9. Dec 30, 2011 #8
    So x' is the distance to an object in frame S'.
    x is the distance to an object in frame S.
    Frame S' moves at constant velocity v wrt frame S

    Lorentz transformations are:
    x' = gamma (x-vt)
    t' = gamma (t-vx/c^2)


    Now u = dx/dt, u' = dx'/dt'
    Now we differentiate x' wrt to t.
    So dx'/dt = gamma (dx/dt - v) = gamma (u - v)

    Notice gamma is just a constant as v is constant, so it just stays there, can think of it just like a number.

    Differentiate t' wrt t.
    dt'/dt = gamma(1 - (vdx/dt)/c^2) = gamma (1- (vu)/c^2)

    So by chain rule dx'/dt' = (dx'/dt)/(dt'/dt)

    So we divide these equations.

    dx'/dt' = [gamma(u-v)] / [gamma(1-uv/c^2)] = (u-v)/(1-uv/c^2)

    so therefore
    u' = (u-v)/(1-uv/c^2)

    So u' would be the speed of the object viewed in frame S'. u would be the speed of the object view in frame S.




    Now when considering these sorts of the problems, i like to consider Ship B to just be an inertial frame moving relative to another inertial frame (the Earth).

    Let the Earth (stationary frame) to be frame S.

    Now let Ship B be frame S', which is moving relative to S (Earth) at a constant velocity, v = 0.7c

    Ship A is this object, that is moving at a velocity u = 0.8c in frame S (earth frame), you want to find it's velocity, u' in the frame S' (ship B frame)

    So you just substitute these values in the equations to get your answer.
     
    Last edited: Dec 30, 2011
  10. Dec 30, 2011 #9
    Thank you The1337gamer for that detailed explanation. I'll need to go over it a few times, but I think I'm beginning to understand what you're saying.

    I hadn't thought about this, about considering another point. In this case the earth.

    That's quite an interesting perspective.

    Which equations should these values be substituted into to get the answer?
     
  11. Dec 30, 2011 #10
    You should tell us that ...

    You have speed of object in S frame ie u
    you have speed of S' wrt S ie v

    And you need speed speed of object wrt S' ie u'

    How will you solve it ... ??
     
  12. Dec 30, 2011 #11
    Well, my initial thought is that u' = u/γ
     
  13. Dec 30, 2011 #12
    I meant equation not equations, sorry. The equation i was referring to is the same one you that i was helping you understand/derive from Lorentz transformations.

    You can't use u' = u/gamma.

    I think the relationship for length contraction is L' = L/gamma, you cannot use this to calculate relative velocities.

    Also i accidently wrote v = 0 instead of v = 0.7c in my last post, have corrected that now.
     
  14. Dec 30, 2011 #13
    Well if we go by you, you already found the relation ... So This could be your final not first thought :rofl:

    well to be honest we have already explained you the whole derivation ... try writing post #8 in your notebook properly ... you will understand it ...
     
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