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Let's say we have $2$ particles facing each other and each traveling (almost) at speed of light.

example

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

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

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    $\begingroup$ Related: physics.stackexchange.com/q/7446/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jun 21, 2011 at 17:23
  • $\begingroup$ The LHC could measure the difference. It could measure The results of a collision between two particles moving at almost the speed of light and then compare it to the results of two particles each moving at half speed of light and colliding $\endgroup$
    – Bill Alsept
    Commented Nov 15, 2017 at 18:36

6 Answers 6

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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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    $\begingroup$ +1 for pointing out you can't really measure the velocity of another paticle when you're moving at c. $\endgroup$
    – luksen
    Commented Jun 21, 2011 at 19:03
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    $\begingroup$ 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. $\endgroup$
    – Marek
    Commented Aug 20, 2011 at 12:15
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    $\begingroup$ Indeed the most important point to state is that Lorentz transformation is not applicable at $v=c$. $\endgroup$
    – Omar Nagib
    Commented Nov 15, 2015 at 16:27
  • $\begingroup$ How many particles can all get arbitrarily close to light speed relative to all the others simultaneoulsy? $\endgroup$
    – Robert Frost
    Commented Jun 20, 2023 at 16:05
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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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    $\begingroup$ it's just example, let's remove me, in point of view of #1 participle, the second one moves at c+c=2c speed..? $\endgroup$
    – Templar
    Commented Jun 21, 2011 at 17:11
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    $\begingroup$ who said that relative velocity is obtained from adding scalars like that? $\endgroup$
    – lurscher
    Commented Jun 21, 2011 at 17:13
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    $\begingroup$ 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. $\endgroup$
    – Peter Morgan
    Commented Jun 21, 2011 at 17:21
  • $\begingroup$ One can cater for this objection of yours, without defeating the original object of hte question, by simply using some speed $c-\epsilon$ which might be called "arbitrarily close to light speed. $\endgroup$
    – Robert Frost
    Commented Jun 20, 2023 at 16:03
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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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I am assuming you have 2 particles facing each other, and that they are approaching each other ?

enter image description here

First, as mentioned elsewhere on this page, "..a particle moving at the speed of light does not experience time, and thus is unable to make any measurements."

Instead, let's change the particle #1 that you are sitting on to having a specific velocity that is less than c, and particle #2 will remain as v=c.

If your v=0, the combined velocity between you and particle #2 = c, and if you measure its velocity, you will measure a velocity of c.

If you accelerate to a new velocity, and v=100,000 km/s, the combined velocity between you and particle #2 will be 100,000 km/s + 300,000 km/s = 400,000 km/s, but you will still measure the velocity of particle #2 as c.

If you accelerate to a new velocity, and v=260,000 km/s, the combined velocity between you and particle #2 will be 260,000 km/s + 300,000 km/s = 560,000 km/s, but you will still measure the velocity of particle #2 as c.

In other words, you can change your velocity to a vast range of different velocities, yet when it comes to "measuring" the velocity of particle #2, the result is always c. This can be verified if you use the velocity addition equation. The important point is that neither particle has exceeded the speed of light. The combined velocity between 2 moving particles is a different matter altogether.

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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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