# What is the relative speed of two near-light speed particles headed towards each other?

I understand that nothing can move faster than light due to time dilation. I want to build upon my understanding of Einstein's theory of Special Relativity, so I came up with this hypothetical problem for myself:

If both particles are moving at 0.9c towards each other (according to an observer on Earth), what speed does each particle have relative to the other?

I'm stumped since I am only familiar with Newtonian classical mechanics to solve this (which would be wrong).

I am aware of Lorentz' factor $\gamma =\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ and the time dilation equation $t_m = \frac{t_o}{\sqrt{1 - \frac{v^2}{c^2}}}$, but I'm not sure how to apply it in this instance. I'm just looking for a gentle push in the right direction.

Edit: My prior research:

1. This article, https://what-if.xkcd.com/1/, talks about the effect of a baseball thrown at 0.9c would have as it traveled towards a batter. This is not what I'm looking for since it doesn't explain the math behind finding the collision speed between two high velocity entities.

2. This question on Yahoo, https://answers.yahoo.com/question/index?qid=20110529201519AAxbxvm, asks a similar question to mine, but no math is involved to demonstrate the calculations behind the reasoning.

3. This question on Physics.SE, Collision between a photon and a massive particle, does not answer my question either. I have not been able to find a similar Physics.SE question which can provide me to clues to my own hypothetical question.

4. I have also Googled the following phrases:

i. "collision faster than the speed of light"

ii. "collision of high speed particles"

iii. "lorentz transformation in a collision"

iv. "collision analysis at light speed"

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In special relativity you have to be very careful about what you mean by "speed" and who is observing it. You've specified the observer, but then what do you mean by "collision speed"? – webb May 21 '14 at 21:54
Hi Klik - this is a very well written question, except that it doesn't suggest you've done any prior research. Could you elaborate on why the explanations you should have found by searching Wikipedia, Google, Hyperphysics, etc. and also elsewhere on this site aren't sufficient? (see the link I gave for more information on what sort of prior research we expect) – David Z May 21 '14 at 21:58
@webb What I mean is at what speed do they collide? Perhaps it is more accurate to say, if both particle are moving at 0.9c towards each other (according to an observer), what speed does each have relative to each other? – Klik May 21 '14 at 22:16
@DavidZ I've included some information to show some of the articles I have looked over for insight into this question. – Klik May 21 '14 at 22:17
OK, well... I suppose that is what I suggested you do, but the point is not really for you to show us that you passed some arbitrary threshold of how much effort you need to put in. The point is for you to be able to find the answer on your own. In other words, you should be doing prior research not because you have something to prove to us, but because you really want to know the answer to your question, and trying to find where it's already been posted is more convenient than asking a whole new question. (cont.) – David Z May 21 '14 at 23:01

Classically we would add the speeds to get $1.8c$, which is obviously not allowed. In relativity you simply use the relativistic velocity addition formula:

$$V = \frac{u+v}{1+uv/c^2}$$

Where $u$ and $v$ are the velocities of the particles as seen from some reference frame, and $V$ is the velocity of one particle in the rest frame of the other, i.e., the relative velocity when we consider one of the particles to be stationary. Plugging in $u = v = 0.9c$, we get $V = \frac{180}{181}c \approx 0.9945c$.

Edit: As pointed out by Alfred Centauri, the above explanation is perhaps too simplistic. A more rigorous version would be the following:

Let's take our particles to be moving in the $x$ direction, with particle 1 moving in the positive direction and particle 2 moving in the negative direction. As seen by particle 1, our velocity is $-0.9c$ (note the sign!). As seen by us (that is, the lab frame), particle 2 is moving with a velocity equal to $-0.9c$. The velocity addition formula tells us how to find the velocity of particle 2 with respect to particle 1. If $V$ is this velocity that we are trying to find, $u$ is our velocity with respect to particle 1 and $v$ is particle 2's velocity with respect to us, then:

$$V = \frac{u+v}{1+uv/c^2} \approx -0.9945c$$

This time we get the correct sign, since, relative to particle 1, particle 2 is moving in the negative $x$ direction.

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Thanks Javier, this is exactly what I was looking for. – Klik May 21 '14 at 22:28
This is close but, actually, incorrect as described. The velocity of the particles in the lab frame are equal and opposite. Thus, $u$ and $v$ in the above are not the "velocities of the particles as seen from some reference frame". If they were, we would get $V = 0$. – Alfred Centauri May 21 '14 at 22:35
@AlfredCentauri: Duly noted, will correct. Thanks! – Javier May 21 '14 at 23:11

To correctly apply the relativistic velocity addition formula, one must properly identify the velocities $u$ and $v$ in the formula.

The setup is as follows:

From the reference frame of the lab, there is a particle, particle 1, with velocity $u$.

From the reference frame of another particle, particle 2, the lab has a velocity $v$.

To find the velocity $w$ of particle 1 in the reference frame of particle 2, the relativistic velocity addition formula is

$$w = \frac{u + v}{1 + \frac{uv}{c^2}}$$

For example, let particle 1 have a velocity of $u = -0.9c$; the particle is travelling to the left in the lab frame.

In the lab frame, let particle 2 be travelling to the right with speed $0.9c$.

It follows that, in the reference frame of particle 2, the lab has a velocity of $v = -0.9c$.

Applying the relativistic velocity addition formula yields

$$w = \frac{-0.9c - 0.9c}{1 + \frac{0.81c^2}{c^2}} = -0.9945c$$

Thus, according to particle 2, particle 1 is moving to the left with speed $0.9945c$.

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