# Finding the mass of a relativistic particle resulting from decay

I'm studying relativistic kinematics for the first time. I’m working with a problem from a book but I’m stuck: A relativistic particle of rest mass $$m_1$$ decays at rest into a particle of rest mass $$m_2$$ and another massless particle.

I'm trying to find the KE of the $$m_2$$ particle but I'm struggling.

I understand the derivation of the total energy of a free particle from the Hamiltonian being $$E=\sqrt{p^2c^2+m^2c^4}$$, and that in a set inertial frame, total momentum and energy is conserved; but I can't seem to manipulate the terms to give the answer in terms of the mass of the particles and $$c$$.

• Do you know more initial and final states such as the momentum of one of the final particles? – Quantumness Apr 22 at 0:06
• @Quantumness nope; it's a question from a textbook, and was presented as above! – Jonathan Keogh Apr 22 at 0:22
• Was mass $m_1$ at rest or was moving . Secondly what was the energy of the massless particle? If you aren't given these things then you can't solve the problem. – Nobody recognizeable Apr 22 at 12:40
• @Nobody recognizeable ah yes sorry I forgot to include that point - the masses are the rest mass, and the decay is happening at rest. I’ll edit. – Jonathan Keogh Apr 22 at 13:05

## the concept

I think your problem may be coming from a misunderstanding about when you can apply conservation of energy and momentum.

In fixed inertial frame energy and momentum are conserved separately, so $$E_i = E_f$$ and $$\vec{p}_i = \vec{p}_f$$. If you select the rest frame of the initial particle as your working frame for the whole calculation, you can use this fact.

When using different reference frames for the before and after pictures, energy and momentum are not conserved separately. Only the magnitude of the total energy-momentum four-vector is: $$\tilde{p}_i\cdot\tilde{p}_i = \tilde{p}_f\cdot\tilde{p}_f$$, where $$\tilde{p}\rightarrow(E/c, \vec{p})$$ is a four-vector.

The four-vector statement is where your energy equation comes from:

$$\tilde{p}\cdot\tilde{p} = -\frac{E^2}{c^2} + p^2.$$

Working in the the rest frame of the system in question, $$\vec{p}=0$$ and $$E = m c^2$$:

$$\tilde{p}\cdot\tilde{p} = -m^2\, c^2.$$

Finally since $$\tilde{p}\cdot\tilde{p}$$ is the same in all reference frames:

$$E^2 = p^2 c^2 + m^2 c^4.$$

## Applying it to your problem

To start your problem you need to pick a reference frame for each state, before/after decay. Lets pick the rest frame of the initial particle, so $$E_i = m_1 c^2$$ and $$\vec{p}_i = 0$$.

If we use the same reference frame for the final state, we can set up conservation separately. After the decay particle 2 is moving so $$E_2 = \gamma m_2 c^2$$, where $$\gamma$$ is the Lorentz factor. The massless particle 3 has energy. We can get it by applying the energy-momentum equation to it, alone:

$${E_3}^2 = {p_3}^2 c^2 + 0$$

So

$$E_i = E_f \implies m_1 c^2 = \gamma m_2 c^2 + |p_3| c.$$

We still need particle 3's momentum, but we can get that from momentum conservation:

$$\vec{p}_i = \vec{p}_f = 0 \implies \vec{p}_2 = -\vec{p}_3.$$

All of the motion happens in one dimension, and $$p_2 = \gamma m_2 v$$

From here we can put it all together and solve for the speed of particle 2 (or $$\gamma$$ if you prefer).

• thanks! From this I can see now that KE$=\frac{(m_1^2 -m_2^2)c^2}{2m_1}$. I just have one more question: why is it that $E_2=\gamma m_2 c^2$? This doesn't seem to come from the energy-momentum equation. I understand why the Lorentz factor is there, but how does it reconcile with the energy-momentum equation, which seems to give a different answer? – Jonathan Keogh Apr 22 at 14:17
• That's the definition of relativistic total energy. Also, if you plug $p_2 = \gamma m_2 v$ into the energy-momentum equation you can derive it. It's a mess of algebra, but you can show $(\gamma^2 v^2 + c^2) = \gamma^2 c^2$ – Paul T. Apr 22 at 15:19