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B ______________ A

       C

I throw light from A to B, reflect it to C. Similarly I throw a ball at 0.1c at same time.

Now I see from C. Measure distance from A to C and time it took to reach me. It's speed was c. Now , if I compare ball's speed to that of light while standing at C, will I find it slower than 0.1c?

If I relate two speeds while standing at C , what will be their comparative speeds?

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  • $\begingroup$ It seems that A, B and C are fixed points? And you are able to measure the distances from the stationary reference frame? In that case, nothing funny goes on: The ball moves with $0.1 \, c$ in your system, so it will take ten times the travel duration compared to the light. You only need special relativity when you start traveling with the ball. The ball will experience a bit less proper time than you because it moves fast. $\endgroup$ – Martin Ueding Oct 3 '16 at 14:56
  • $\begingroup$ @MartinUeding So, relative speed of light is 0.9c here? $\endgroup$ – Anubhav Goel Oct 3 '16 at 16:17
  • $\begingroup$ You will observe the relative speed to be $0.9 \, c$, yes. The ball will observe the speed of light as the speed of light. See the answers for a full explanation. $\endgroup$ – Martin Ueding Oct 3 '16 at 16:58
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In your stationary system the ball and the photon will have a velocity difference of

$$\text{c}-\text{v}=0.9\text{ c}$$

In the moving system of the ball it's relative velocity to the photon is

$$\frac{\text{c}-\text{v}}{1-\frac{\text{c v}}{\text{c}^2}} = \text{c}$$

because the speed of light is the same in every inertial frame of reference (see Einstein's velocity addition formula).

If you want to use the formula in 2 or 3 spatial dimensions you have to split into parallel and transversal components:

$$v_{\text{x}}=\frac{\text{v}_{\text{x}}-\text{v}}{1-\frac{\text{v}_{\text{x}} \text{v}}{\text{c}^2}}$$

for the parallel and

$$v_{\text{y}}=\frac{\text{v}_{\text{y}}\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}{1-\frac{\text{v}_{\text{x}} \text{v}}{\text{c}^2}}$$

for the transversal component:

velocity addition formula, components

where the $\text{v}$s are the velocities in the stationary (relative to A, B, C) system and the $v$s for the velocities in the system that moves with $\text{v}$.

So if the ball is on it's way from A to B while the photon is already on it's way from B to C the relative velocity of the photon in the system of the ball is with Pythagoras

$$v=\sqrt{v_{\text{x}}^2+v_{\text{y}}^2}$$

which is still $\text{c}$ in your scenario where

$$\text{v}=0.1\text{ c}, \ \text{v}_{\text{x}} = \frac{\text{c}}{\sqrt{2}}, \ \text{v}_{\text{y}} = \frac{\text{c}}{\sqrt{2}}$$

when we assume a 45° angle.

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  • $\begingroup$ Thank you for answer, but spatial treatment was not in my mind. $\endgroup$ – Anubhav Goel Oct 3 '16 at 16:23
  • $\begingroup$ Your A,B,C are aligned as a triangle. $\endgroup$ – Yukterez Oct 3 '16 at 16:42
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You will calculate that the ball is travelling at 10% of the speed of light.

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  • $\begingroup$ So relative speed of light and ball is 0.9c? $\endgroup$ – Anubhav Goel Oct 3 '16 at 16:25
  • $\begingroup$ Yes, in your reference frame. In the reference frame of the ball, the relative speed will be $c$. $\endgroup$ – Martin Ueding Oct 3 '16 at 16:58
  • $\begingroup$ @AnubhavGoel From your reference frame the light will be moving at .9c relative to the ball, which is moving at .1c $\endgroup$ – Yogi DMT Oct 3 '16 at 17:21

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