# Collisions in special relativity

I am wanting to teach some basic trigonometry to school age kids using relativity and would like help to check my reasoning. First, is my problem formulation correct with respect to special relativity, and second am I correct in how I'm solving the problems?

## Problem 1.

I will pose a problem that a $$1kg$$ mass traveling at $$0.5c$$ collides with a stationary $$1kg$$ mass, and the masses merge.

I assume that a mechanism enables the masses to not disintegrate on impact, and that all excess energy is emitted equally in all directions as photons. [Edit: the previous italicised phrase is implausible. As is pointed out by Mark H below, the photon spray would be symmetrical in the rest frame of the combined mass system. In my chosen frame, this would be mostly in the direction of the initially travelling mass. This would decelerate the merged masses to $$0.268c$$, less than my calculated $$0.277c$$. $$0.268c$$ is simply the average velocity of the system, $$\frac{\sum{p}}{\sum{E}}$$ .]

Using the energy/mass/momentum equivalency triangle and Pythagorus' theorem, [to be clear, I will present this without $$c$$ factors, as it is all about the ratios.]

$$E^2 = (mc^2)^2 + (pc)^2$$

and with the velocity being

$$\frac{v}{c} = \frac{pc}{E}$$

assuming conservation of mass and momentum, the final velocity is $$0.277c$$. (30 degree triangle to 16.1 degree triangle by doubling its base - mass)

## Problem 2.

Then assuming that energy is retained inside the system, and the collision adds to the internal energy of the new system, using Pythagorus and using the ratio of the tangents (it is actually really easy when you draw it out) tells us that the new system's mass is $$2.08kg$$ and velocity is 0.27c. [Edit: this appears to be correct, as it is free from assumptions about the direction of emitted photons.]

I've tested it with a student who has just learned the Pythagorean theorem, and it worked fine, but I really want to double check my very rusty Physics.

I understand that this is not the standard way that relativity is taught at university, but I'm not teaching at university :-)

# Problem 1:

Oftentimes, for collision problems with relativity, it helps to look a the problem in a frame where the total momentum is zero. For this problem, we create a new observer traveling at such a speed that Mass 1 and Mass 2 are heading toward each other at the same speed (since they have the same mass). We'll call this speed $v_{cm}$. The velocities of Mass 1 and Mass 2 in this new frame are $v'$ and $-v'$, respectively. So, transforming back into the first frame using the velocity addition formulas, we have:

Mass 1: $$v_1 = \frac{v_{cm} + v'}{1 + \frac{v_{cm}v'}{c^2}}$$ Mass 2: $$0 = \frac{v_{cm} - v'}{1 - \frac{v_{cm}v'}{c^2}}$$ Note that $v_1$ and $v_{cm}$ are measured in the frame where Mass 2 is stationary, whereas $v'$ and $-v'$ are measured in the center-of-momentum frame where Mass 1 and Mass 2 move with the same speed.

From the Mass 2 equation, we conclude that $v_{cm} = v'$. This makes sense because the center of mass has to follow this frame because it has to have zero velocity before and after the collision in this frame. It is only in the frame that any photons emitted will have a uniform spread of motion. For example, if two photons are emitted (the smallest number), they will have equal energy and travel in exact opposite directions.

So, $$v_1 = \frac{2v'}{1 + \frac{v'^2}{c^2}}$$ Solving for $v'$: $$\frac{v'}{c} = \frac{c}{v_1}-\sqrt{\left(\frac{c}{v_1}\right)^2-1}$$ Substituting $v_1 = .5c$ results in $$v' = .268c$$

Notice that this result is independent of the masses of the colliding masses and the final merged mass. Where the mistake lies is assuming that

"all excess energy is emitted equally in all directions as photons."

This can only happen in the center-of-mass/center-of-momentum frame of reference. In the original frame, with Mass 2 stationary, you would observe the photons be emitted in the same direction as the original velocity of Mass 1, taking away momentum from the masses--hence why the real momentum is less than what you calculated. The only exception would be when no photons are emitted and the merged mass gets all the energy and momentum, which I'll analyze in the next section.

# Problem 2:

Mass is not conserved. As an extreme example, if equal masses of matter and antimatter collide, the resulting amount of mass will be zero. Energy and momentum, on the other hand, are conserved.

To save on typing, I'll use units of velocity such that $c=1$:

For mass 1: $$E_1^2 = m_1^2 + p_1^2$$

For mass 2: $$E_2^2 = m_2^2 \implies E_2 = m_2$$

For the combined mass 3, we use the conserved energy and momentum: $$E_3 = E_1 + E_2 = \sqrt{m_1^2 + p_1^2} + m_2$$ and $$p_3 = p_1$$

Therefore: $$m_3 = \sqrt{E_3^2 - p_3^2}$$ $$m_3 = \sqrt{\left(\sqrt{m_1^2 + p_1^2} + m_2 \right)^2 - p_1^2}$$ Simplifying: $$m_3 = \sqrt{m_1^2 + 2m_2\sqrt{m_1^2 + p_1^2} + m_2^2}$$ Substituting $p_1 = \gamma_1 \beta_1 m_1$ and further simplifying results in: $$m_3 = \sqrt{m_1^2 + 2 \gamma _1 m_1 m_2 + m_2^2}$$

You can see that it is only when $\gamma_1 \approx 1$ that $m_3 \approx m_1 + m_2$.

Now, in your problem, $m_1 = m_2 = m$, so the combined mass is $$m_3 = 2m\sqrt{\frac{1+\gamma_1}{2}}$$ Since, $\beta_1 = v/c = .5 \implies \gamma_1 = \frac{1}{\sqrt{1-\beta_1^2}} = \sqrt{4/3}$ and $m = 1\,kg$, $$m_3 = 2(1\,kg)\sqrt{\frac{1+\sqrt{4/3}}{2}} = 2.076\,kg$$ This value matches yours because $\gamma$ is small, so the results are nearly classical. The final velocity will differ, as show below.

You can figure out the velocity using momentum conservation $$p_3 = p_1$$ and $$p = \gamma\beta m.$$ So, $$\gamma_3 \beta_3 m_3 = \gamma_1 \beta_1 m_1$$ Skipping the arithmetic ... $$\beta_3 = \frac{1}{\sqrt{(m_3/p_1)^2 + 1}}$$ Using $m_3 = 2.076\,kg$ and $p_1 = \gamma_1 \beta_1 m_1 = (\sqrt{4/3} (.5) (1\,kg) = 1/\sqrt{3}\,kg$ (with $c=1$, velocity is unitless): $$\beta_3 = .268$$ or $$v_3 = .268c$$ This is the same answer as in Problem 1.

On another note, as a former physics and math teacher, the notion of using special relativity to teach "basic trigonometry" is bizarre to me. Trigonometry is already hard enough for a beginner--I know I had my troubles in school. I think bringing in special relativity, which doesn't use triangles in any obvious way, will only obscure what you're trying to teach. I don't doubt that your student could follow along, but I wonder what he or she actually takes from the lesson beyond a series of seemingly nonsensical steps involving triangles whose sides aren't measured in inches or meters.

If you want a powerful examples of trigonometry in physics:

• Please specify, are you answering problem 1 or 2. If it is problem 2, you have arrived at the exact same solution that my 11 year old did in testing out this problem, using only Pythagorus' theorum (m=2.08, v=0.27). I did not give him a series of steps, he worked it out from first principles. Problem 1 demonstrates that energy must be lost if mass is conserved. The point of problem 2 is that an alternative solution is for energy to be conserved and mass increases. The children I am teaching general respond well to bizarre teaching techniques :-) – Richard Dec 12 '14 at 8:37
• BTW, thanks for your other ideas that also demonstrate Pythagorus' theorem, and thanks for the time spent checking my model answer using more formal techniques. – Richard Dec 12 '14 at 8:49
• @Richard I missed the part about why the masses would be held constant in problem 1. I'll edit my answer. – Mark H Dec 12 '14 at 8:49
• While I'm editing, you'll probably enjoy this article by a Harvard physics professor: profmattstrassler.com/articles-and-posts/… – Mark H Dec 12 '14 at 9:17
• you're amazing! I had found this article a few weeks ago, looked for it earlier today to have a proper read but found that I had lost the link. – Richard Dec 12 '14 at 9:32

"First, is my problem formulation correct with respect to special relativity, and second am I correct in how I'm solving the problems?"

Your equations are perfectly ok. However, your scenario 1 seems suspect. Why would the sum of the rest masses be conserved in a relativistic inelastic collision? Scenario 2 seems fine to me.

I like your quest for unconventional approaches in teaching math and physics. Teaching special relativity to students who have just mastered the Pythagorean theorem seems perfectly possible (and a fun proposition) to me. Thinking about this I would, however, use the "space-time Pythagorean relation" (rather than the "energy-momentum Pythagorean relation") for this purpose. Idea would be to to use the relation that specifies proper duration $\tau$ in terms of the distance $x$ traveled and duration $t$ as observed by an external observer: $$\tau^2 + (x/c)^2 = t^2$$

This equation relates the time $\tau$ lapsed on the wristwatch of a traveler on a moving train to the time $t$ lapsed on the railway station clock and the railway track distance $x$ covered. Together with the relationship for the velocity $v$ measured by an observer at the railway station: $$x/t=v$$

this allows one to derive the equations for time dilation etc.

Next comes the definition for the velocity $w$ measured as the railway track distance covered per unit time on the traveler's wristwatch:

$$x/\tau=w$$

This velocity is often referred to as the "proper velocity", and arguably is a more physical generalization of Newtonian velocity to relativistic speeds. Firstly, proper velocities times mass always equal momentum, and proper velocities can therefore grow without limit. And secondly, proper velocities equal the velocities that would be measured by accelerometers.

Using the above equations, the students can work out observer times and traveler times for trips to Aplha Centauri and beyond.

• I like the use of the Pythagorean theorem with time dilation. I sometimes tell people that you are always moving at the speed of light. When you are at rest, you are moving towards the future. In order to move through space, you have to divert some speed in towards the space direction. Using a right triangle with velocity through space and velocity through time as the legs and the speed of light as the hypotenuse, you can work out time dilation *($v_t/c$).. – Mark H Dec 12 '14 at 20:14
• @Johannes - Your suggestion is excellent. I think it is the E-p-m triangle multiplied by (proper time)/(mass*c2). It offers many more possibilities for interesting calculations without facing the difficulties of Problem 1. – Richard Dec 13 '14 at 2:31
• @MarkH - your triangle (c - v_x - v_t) is the E-p-m triangle scaled by c/E. As Pythagorus works in n dimensions, it can be also be viewed as a 4D pyramid: c^2 = v_x^2 + v_y^2 + v_z^2 + v_t^2. This stuff is just too beautiful - and all it takes is Year 10 maths. – Richard Dec 13 '14 at 3:09