- #36
jcsd
Science Advisor
Gold Member
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a vbiew from one rest frame:
A view from another:
It should be clear that in the second rest frame in order for them to be detected simulatenously the two phtons cannot of been emitted simulateously.
lets do some maths for this
Let [itex]x^{\mu}_1[/itex] and [itex]x^{\mu}_2[/itex] (where [itex]x^0 = t[/itex])be the postion 4 -vector of the emitters as it emits the light where the orgin is half way between the emitters in the the rest frame of the emitters, where c is the speed of light and l the diatnce between the the two emitters:
[tex]x^{\nu}_1 = \left(\begin{array}{c}0\\{\frac{l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x^{\nu}_2 = \left(\begin{array}{c}0\\{\frac{-l}{2}}\\0\\0\end{array}\right)[/tex]
Let [itex]x'^{\mu}_1[/itex] and [itex]x'^{\mu}_2[/itex] be the postion 4 vectors of the emitters as they emit light in rest frame of someone who is moving with velocity u relative to the emitters:
the following realtionships apply:
[tex]x'^{\mu}_1 = {\Lambda^{\mu}}_{\nu}x^{\nu}_1[/tex]
[tex]x'^{\mu}_2 = {\Lambda^{\mu}}_{\nu}x^{\nu}_2[/tex]
therefore:
[tex]x'^{\mu}_1 = \left(\begin{array}{c}{\frac{-\gamma l\beta}{2c}}\\{\frac{\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x'^{\mu}_2 = \left(\begin{array}{c}{\frac{\gamma l\beta}{2c}}\\{\frac{-\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
So in the first frame the photons are emitted both at t = 0
but in the second frame the're emitted at [itex]\frac{-\gamma l\beta}{2c}[/itex] and [itex]\frac{\gamma l\beta}{2c}[/itex] respectively which are only equal if u = 0.
They're not the same clock as they're not local.
Code:
|> o <|
Code:
|-------->o<--------|
A view from another:
Code:
|> 0 |
Code:
----|-------->0<--------|
It should be clear that in the second rest frame in order for them to be detected simulatenously the two phtons cannot of been emitted simulateously.
lets do some maths for this
Let [itex]x^{\mu}_1[/itex] and [itex]x^{\mu}_2[/itex] (where [itex]x^0 = t[/itex])be the postion 4 -vector of the emitters as it emits the light where the orgin is half way between the emitters in the the rest frame of the emitters, where c is the speed of light and l the diatnce between the the two emitters:
[tex]x^{\nu}_1 = \left(\begin{array}{c}0\\{\frac{l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x^{\nu}_2 = \left(\begin{array}{c}0\\{\frac{-l}{2}}\\0\\0\end{array}\right)[/tex]
Let [itex]x'^{\mu}_1[/itex] and [itex]x'^{\mu}_2[/itex] be the postion 4 vectors of the emitters as they emit light in rest frame of someone who is moving with velocity u relative to the emitters:
the following realtionships apply:
[tex]x'^{\mu}_1 = {\Lambda^{\mu}}_{\nu}x^{\nu}_1[/tex]
[tex]x'^{\mu}_2 = {\Lambda^{\mu}}_{\nu}x^{\nu}_2[/tex]
therefore:
[tex]x'^{\mu}_1 = \left(\begin{array}{c}{\frac{-\gamma l\beta}{2c}}\\{\frac{\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
[tex]x'^{\mu}_2 = \left(\begin{array}{c}{\frac{\gamma l\beta}{2c}}\\{\frac{-\gamma l}{2}}\\0\\0\end{array}\right)[/tex]
So in the first frame the photons are emitted both at t = 0
but in the second frame the're emitted at [itex]\frac{-\gamma l\beta}{2c}[/itex] and [itex]\frac{\gamma l\beta}{2c}[/itex] respectively which are only equal if u = 0.
They're not the same clock as they're not local.
Last edited: