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There is a concave mirror, which has somehow maintained the speed $\frac{c}{2}$ toward west ( for instance) , and there is a light wave or you can say a ray of light is moving in N-E direction. So we can say the component of velocity vector of light parallel to principal axis of mirror is the vector involving in the calculation of relative speed between the photon and mirror (frame of reference is mirror) and will be

$v_1$ (parallel) = $\frac{(\frac{c}{2} + \frac{c}{2^1/2})}{(1 + \frac{c^2}{2(2^{1/2})(c^2)})} = c \,\ (0.89180581)$
There will be no change in the velocity vector perpendicular to principal axis, since there is no motion of mirror in that direction, $v_2$( perpendicular) = $\frac{c}{2^{1/2}} = c \,\ (0.70710678)$, then the total velocity of light w.r.t mirror is $(v_1^2 + v_2^2) = c\,\ (1.13812021) > c$;
How??

NOTE: i have used concave mirror because i was also curious about the angle on which the photon will be reflected. But this is more important.

Qmechanic
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Mohammad Anser
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1 Answers1

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In a given frame, there is no problem measuring a relative velocity > $c$. If a mirror moves left at $0.5c$ (relative to the lab frame) and a photon moves right to strike it, we will measure that the two objects have a relative velocity of $c + 0.5c = 1.5c$.

But this is just a calculation. It doesn't say that observers in the frame where the mirror is at rest see the same photon move at that speed. Indeed, any such observer sees that photon move at exactly $c$ as well.

So you are allowed to measure relative speeds in excess of $c$. But never a single object moving relative to your frame in excess $c$.

BowlOfRed
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