0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Note that this site has MathJax enabled, so you can write your equations out in a human-readable format. $\endgroup$ – Kyle Kanos Oct 14 '15 at 16:31
  • 1
    $\begingroup$ I find it hard to follow your reasoning and hard to read your equations but you might want to read Travelling faster than the speed of light (addition of velocities in special relativity) $\endgroup$ – RedGrittyBrick Oct 14 '15 at 16:33
  • $\begingroup$ Where are some of those $2^{1/2}$ coming from? $\endgroup$ – Kyle Kanos Oct 14 '15 at 17:04
  • $\begingroup$ cos45 =sin45 = ${ \frac{1}{2^{1/2}}}$ the component on north and east axis $\endgroup$ – Mohammad Anser Oct 14 '15 at 17:20
  • $\begingroup$ The speed of light relative to the mirror frame is still c. Are you asking about the perceived speed of convergence viewed by an observer who sees the mirror moving at c/2? $\endgroup$ – Bill N Oct 14 '15 at 20:41
2
$\begingroup$

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 $c$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ for higher speeds the relative velocity is not $v1$ +$v2$ it is $v3$ (relative) = ${ \frac{ v1 + v2}{1 + \frac{v1.v2}{c^1/2}}}$ using this formula the relative velocity between $c$ and $c/2$ is $c$ $\endgroup$ – Mohammad Anser Oct 15 '15 at 2:32
  • $\begingroup$ It is not "higher speeds" that need the formula, but when adding velocities between frames (for instance to determine what the speed will be measured by some other observer). As long as we stay in a single frame, simple vector addition is correct. $\endgroup$ – BowlOfRed Oct 15 '15 at 2:49
  • $\begingroup$ in above formula i have given please $c^1/2$ by $c^2$ $\endgroup$ – Mohammad Anser Oct 15 '15 at 3:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.