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nonequilibrium
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I've been staring at this for hours now.
Can someone spot an error in the following derivation of the relativistic doppler shift?
Imagine a flashlight (not on) and an eye passing each other in space. Call the frame where the flashlight is fixed S for Source, and that of the eye O for observer. Call v the relative velocity between both. In S the flashlight sends out a flash after a time [tex]\tau[/tex] (after passing). If t is the time in S for that flash to reach the eye, then [tex]ct = v \tau + vt[/tex], because the lhs is the total distance covered by the flash to get to the eye and the first term on the rhs is the distance the eye is at at the moment of the flash; the second term is there since the eye is also moving away while the flash is going toward it.
In O the flashlight sends out a flash after a time [tex]\tau' = \gamma \tau[/tex] after the passing. If t' is the time in O for the flash to reach the eye, then [tex]ct' = v\tau' (= \gamma v \tau)[/tex] because as soon as the flash has "left" the flashlight, all the light has to do is travel the distance crossed by the flashlight during [tex]\tau'[/tex], which is exactly the rhs.
Rewriting the first equation as [tex](c-v)t = v \tau [/tex] and dividing this equation by the latter, we get [tex]\frac{c-v}{c} \frac{t}{t'} = \frac{v}{\gamma v}[/tex] or [tex]t = \frac{\sqrt{1-(v/c)^2}}{1-v/c} t'[/tex]. We can easily rewrite the latter equation as [tex]t = \sqrt{ \frac{c+v}{c-v} } t'[/tex].
If we regard t and t' as the period of the EM-wave in resp. S and O, we get the wrong formula: the minus and plus in front of the v has switched places...
A BIG THANK YOU TO ANYONE WHO CAN FIND IT!
Can someone spot an error in the following derivation of the relativistic doppler shift?
Imagine a flashlight (not on) and an eye passing each other in space. Call the frame where the flashlight is fixed S for Source, and that of the eye O for observer. Call v the relative velocity between both. In S the flashlight sends out a flash after a time [tex]\tau[/tex] (after passing). If t is the time in S for that flash to reach the eye, then [tex]ct = v \tau + vt[/tex], because the lhs is the total distance covered by the flash to get to the eye and the first term on the rhs is the distance the eye is at at the moment of the flash; the second term is there since the eye is also moving away while the flash is going toward it.
In O the flashlight sends out a flash after a time [tex]\tau' = \gamma \tau[/tex] after the passing. If t' is the time in O for the flash to reach the eye, then [tex]ct' = v\tau' (= \gamma v \tau)[/tex] because as soon as the flash has "left" the flashlight, all the light has to do is travel the distance crossed by the flashlight during [tex]\tau'[/tex], which is exactly the rhs.
Rewriting the first equation as [tex](c-v)t = v \tau [/tex] and dividing this equation by the latter, we get [tex]\frac{c-v}{c} \frac{t}{t'} = \frac{v}{\gamma v}[/tex] or [tex]t = \frac{\sqrt{1-(v/c)^2}}{1-v/c} t'[/tex]. We can easily rewrite the latter equation as [tex]t = \sqrt{ \frac{c+v}{c-v} } t'[/tex].
If we regard t and t' as the period of the EM-wave in resp. S and O, we get the wrong formula: the minus and plus in front of the v has switched places...
A BIG THANK YOU TO ANYONE WHO CAN FIND IT!
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