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

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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Related: physics.stackexchange.com/q/7446/2451 and links therein. –  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

6 Answers 6

up vote 45 down vote accepted

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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Easy. you cannot sit on a particle moving at the speed of light. If you could then you would be massless, and unable to sum properly

In any case, there is no reference-frame moving with either photon, so no operational way to measure relative velocities between them. relative velocities have meaning only from a inertial frame. There are no inertial frames moving with the photon, otherwise this frame would measure that photon to be at rest

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it's just example, let's remove me, in point of view of #1 participle, the second one moves at c+c=2c speed..? –  Templar Jun 21 '11 at 17:11
who said that relative velocity is obtained from adding scalars like that? –  lurscher Jun 21 '11 at 17:13
Galileo, Descartes, Newton, etc.? Perhaps not Aristotle. Though of course they're off the pace these days. There are so many varying accounts out there that address this question that it seems unlikely we can come up with something new here, now. There's a Wikipedia's page that's specially for this: en.wikipedia.org/wiki/Velocity-addition_formula, though it may be too mathematical. –  Peter Morgan Jun 21 '11 at 17:21

Perhaps your question is whether the speed of approach of the two particles is 2c, is this so? Yes, it is 2c and this does not violate the principles of relativity, because such speed is not the speed of a particle, but it is just a derived value. On the other hand, the speed of a photon is c regardless of the inertial frame, and is calculated by the relative speed formula, given in the previous answers.

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You are both correct and wrong.

If - sitting on one photon - you would measure the velocity of the approaching photon, the figure received would be exactly c.

However, if two photons separated by the distance of 1 light-year are sent toward each other, they will meet after exactly six months and exactly in the middle of this distance i.e. 1/2 lightyear. Go figure what the relative speed of these photons was :)

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Because these particles would be photons- photons have a momentum of h/wavelength, so what would happen is that due to momentum conservation, one would have 2 times the momentum, and have 1/2 the wavelength (possibly absorbing the other). It would have twice the frequency, so twice the energy. Therefore, the photons speed remains the same.

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