Double light speed

Let's say we have 2 participles facing each other and traveling at speed of light

Let's say I'm sitting on #1 participle so in my point of view #2 participle's speed is c+c=2c, double light speed? Please say why I am incorrect :)

EDIT: About sitting me is just example, so in point of view of #1 participle, the second one moves at c+c=2c speed?

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Possibly related: physics.stackexchange.com/q/7446/2451 –  Qmechanic Jun 21 '11 at 17:23
Michelson-Morley tried doing the similar thing! –  Pratik Deoghare Jun 21 '11 at 18:24
You could try looking at "Elementary analysis of the special relativistic combination of velocities, Wigner rotation and Thomas precession", arxiv.org/abs/1102.2001, which has just been published in the European Journal of Physics (which is largely dedicated to the teaching of Physics). A number of calculations and some helpful and interesting graphical methods are given quite explicitly. Some of the paper is intended to be "utterly elementary". Perhaps let us know how you get on. –  Peter Morgan Jun 22 '11 at 13:36

One of the results of special relativity is that a particle moving at the speed of light does not experience time, and thus is unable to make any measurements. In particular, it cannot measure the velocity of another particle passing it. So, strictly speaking, your question is undefined. Particle #1 does not have a "point of view," so to speak. (More precisely: it does not have a rest frame because there is no Lorentz transformation that puts particle #1 at rest, so it makes no sense to talk about the speed it would measure in its rest frame.)

But suppose you had a different situation, where each particle was moving at $0.9999c$ instead, so that that issue I mentioned isn't a problem. Another result of special relativity is that the relative velocity between two particles is not just given by the difference between their two velocities. Instead, the formula (in one dimension) is

$$v_\text{rel} = \frac{v_1 - v_2}{1 - \frac{v_1v_2}{c^2}}$$

If you plug in $v_1 = 0.9999c$ and $v_2 = -0.9999c$, you get

$$v_\text{rel} = \frac{1.9998c}{1 + 0.9998} = 0.99999999c$$

which is still less than the speed of light.

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+1 for pointing out you can't really measure the velocity of another paticle when you're moving at c. –  luksen Jun 21 '11 at 19:03
At the level of this question I think it's also useful to add that the above formula for addition of velocities reduces to the Gallilean $v_1 - v_2$ since for $v_1, v_2$ small we can neglect the $v_1 v_2 \over c^2$ term. Thinking in this way, the formula should be a little less mysterious, it's just an extension of the classical formula to high speeds. –  Marek Aug 20 '11 at 12:15

This is what special relativity is all about..

In special relativity you cannot simply state that particle 2 is moving at c+c=2c in a reference frame where particle 1 is at rest.

Speeds add like this (easily found in wikipedia):

$$v_2^{'} = \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$$

i.e. the speed of particle 2 $v_2'$ in a reference frame where particle 1 is at rest is

$$v_2^{'} = \frac{c+c}{1+1} = c$$

you cannot move faster than at the speed of light in the vacuum.

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