- #1
Mangoes
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I'm being asked to derive the velocity transformation between vy and vy' and my result isn't exactly matching my goal but I don't know what I'm doing wrong. It's an introductory modern physics course and we're covering special relativity.
Assume a reference frame S' moving in some constant velocity vx with respect to a stationary frame S. Since time isn't synchronized (dt ≠ dt') then the velocities vy and vy' aren't the same even though y and y' aren't being affected by length contraction (y = y'). The aforementioned statement isn't necessary for the following but I'm just wondering if the above is a correct statement.
Anyways, this is my logic:
Since y = y', dy = dy'. I want to know dy/dt'. From the chain rule:
[tex] \frac{dy}{dt'} = \frac{dy}{dt}\frac{dt}{dt'} [/tex]
Letting [tex] λ = \frac{1}{(1 - (v/c)^2)^{1/2}} [/tex], the Lorrentz transformation for time is:
[tex] t' = λ(t - \frac{vx}{c^2}) [/tex]
Since there is no fundamental difference between two inertial reference frames, the inverse transformation must be of the same form:
[tex] t = λ(t' + \frac{vx'}{c^2}) [/tex]
In differential form,
[tex] dt = λ(dt' + \frac{vdx'}{c^2}) [/tex]
Now differentiating the above wrt t',
[tex] \frac{dt}{dt'} = λ(\frac{dt'}{dt'} + v\frac{dx'}{dt'}/c^2) [/tex]
[tex] \frac{dt}{dt'} = λ(1 + vv_x'/c^2) [/tex]
Going back to my original goal:
[tex] \frac{dy}{dt'} = \frac{dy}{dt}\frac{dt}{dt'} = v_y\frac{dt}{dt'} [/tex]
[tex] \frac{dy}{dt'} = v_y(λ(1 + vv_x'/c^2)) [/tex]
...which is wrong. Where am I going wrong with this? Would appreciate some clarity.
Assume a reference frame S' moving in some constant velocity vx with respect to a stationary frame S. Since time isn't synchronized (dt ≠ dt') then the velocities vy and vy' aren't the same even though y and y' aren't being affected by length contraction (y = y'). The aforementioned statement isn't necessary for the following but I'm just wondering if the above is a correct statement.
Anyways, this is my logic:
Since y = y', dy = dy'. I want to know dy/dt'. From the chain rule:
[tex] \frac{dy}{dt'} = \frac{dy}{dt}\frac{dt}{dt'} [/tex]
Letting [tex] λ = \frac{1}{(1 - (v/c)^2)^{1/2}} [/tex], the Lorrentz transformation for time is:
[tex] t' = λ(t - \frac{vx}{c^2}) [/tex]
Since there is no fundamental difference between two inertial reference frames, the inverse transformation must be of the same form:
[tex] t = λ(t' + \frac{vx'}{c^2}) [/tex]
In differential form,
[tex] dt = λ(dt' + \frac{vdx'}{c^2}) [/tex]
Now differentiating the above wrt t',
[tex] \frac{dt}{dt'} = λ(\frac{dt'}{dt'} + v\frac{dx'}{dt'}/c^2) [/tex]
[tex] \frac{dt}{dt'} = λ(1 + vv_x'/c^2) [/tex]
Going back to my original goal:
[tex] \frac{dy}{dt'} = \frac{dy}{dt}\frac{dt}{dt'} = v_y\frac{dt}{dt'} [/tex]
[tex] \frac{dy}{dt'} = v_y(λ(1 + vv_x'/c^2)) [/tex]
...which is wrong. Where am I going wrong with this? Would appreciate some clarity.
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