Relativistic Particle Decay

[PsychonautQQ]

Homework Statement

A K_0 particle has a mass of 497.7 MeV/c^2. It decays into a -∏ and +∏, each having a mass of 139.6 MeV/c^2. Following the decay of the K_0, one of the pions is at rest in the laboratory. Determine the kinetic energy of the other pion after the decay and of the K_0 prior to the decay

Homework Equations

E = γmc^2 = mc^2 + K

The Attempt at a Solution

This is the exact wording of the problem from my textbook. I feel like in a chapter meant to be an introduction to special relativity they would specify whether we are talking about rest mass's or relativistic masses. The semi-general consensus amongst my peers is to do the problem as if all mass's given are rest masses.

Kinetic energy of K_0 particle before decay:
(γ-1)mc^2 = Kinetic Energy
and the total energy is γmc^2

Since the total energy before the decay is γmc^2, and energy has to be conserved in all reference frames, in the reference frame of the lab where one of the ∏ particles is at rest
I think I can put
Rest mass energy + kinetic energy of K_0 particle = rest mass energy of both ∏ particles + the kinetic energy of one of them.

So now I have two equations and three unknowns, being γ and the two kinetic energies...

This may not be the most clearest thing I've ever typed but special relativity isn't exactly clear in my head either X_x.

 

[mfb]

Mass is always "rest mass". Forget "relativistic mass". Scientists got rid of that concept some decades ago.

Momentum conservation gives you the missing equation. I would not use three unknowns, it is sufficient to consider γ (or v, or p, does not matter) for the moving pion and the kaon.

 

[PsychonautQQ]

i'm still a bit confused, is the point your trying to make that I only need two equations to solve for the two kinetic energies and gamma because of how gamma and kinetic energy are naturally connected or something?

 

[vanhees71]

Yep, the only thing you need (on this level of discussion) are energy-momentum conservation and the on-shell conditions for the particles, i.e.,
$$E_1=E_1'+E_2', \quad \vec{p}_1=\vec{p}_1'+\vec{p}_2', E_1^2=m_{\mathrm{K}}^2 c^4 + c^2 \vec{p}_1^2, \quad E_j'{}^2=m_{\pi}^2 c^4 + c^2 \vec{p}_j'^2.$$
Here $(E_1/c,\vec{p}_1$ denotes the four-momentum of the Kaon and $E_j'/c,\vec{p}_j'$ for $j \in \{1,2 \}$ are the four momenta of the pions, and indeed relativistic mass is only a confusing idea from the very early times of relativity. This was abandoned in 1908 by Minkowski in analyzing the mathematical structure of special relativity. Unfortunately the old concepts still survived some (even otherwise very good) textbooks.

 

[Nickie]

Dear student,

Thank you for sharing your thought process and attempting to solve this problem. I would like to provide some clarification and guidance to help you better understand and solve this problem.

Firstly, let's define some terms and concepts to ensure we are on the same page. In special relativity, mass is not constant and can change with velocity. However, it is still a fundamental property of matter and is often referred to as "rest mass" or "invariant mass." This is the mass measured in the rest frame of an object, meaning the frame where the object is at rest. On the other hand, "relativistic mass" is a concept that is no longer used in modern physics and can lead to confusion. It refers to the mass of an object as measured in a frame of reference where the object is moving. For this problem, we will use the concept of rest mass.

Now, let's look at the problem at hand. We are given the rest mass of a K_0 particle and the rest mass of the two pions it decays into. We are also told that one of the pions is at rest in the laboratory frame. This means that in the laboratory frame, the total energy before and after the decay must be the same, as energy is conserved in all frames of reference. This allows us to set up the following equation:

Total energy before decay = Total energy after decay
γmc^2 = m_πc^2 + K_π + m_πc^2

Where γ is the Lorentz factor, m is the rest mass, and K is the kinetic energy. We can solve for the kinetic energy of the pion (K_π) by rearranging the equation:

K_π = γmc^2 - 2m_πc^2

To find the kinetic energy of the K_0 particle before the decay, we can use the fact that the total energy before decay is equal to γmc^2. However, we need to find the value of γ. To do this, we can use the conservation of momentum, which states that the total momentum before and after the decay must also be the same. Since one of the pions is at rest, the initial momentum before the decay is zero. After the decay, the two pions will have equal and opposite momenta, meaning the total momentum is also zero. This allows us to set