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

Question

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"

Answer 1

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.

Answer 2

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 $0.9945c$.