We know that the velocity of an object as measured from two different frames of reference transform accodring to the Lorentz transformation. This should have consequences for how we measure momentum, \(\vec{p}=m\vec{v}\), and energy, \(\vec{E} = \frac{1}{2}m\vec{v}^2\) from two different frames of reference. Is there a corresponding Lorentz transformation for momentum and energy, and if so, what is it?
Let's try to construct a four vector for momentum, in other words we describe momentum in spacetime, where we know that \(s = (t,x)\) (for now we stick to the x-direction only). If we write this for velocity, we have \(v_\mu = \gamma(1,\vec{v})\). By multiplying this with m, we should get an expression for momentum:
\(P_\mu = m\gamma(1,\vec{v})=\gamma(m,\vec{p})\)
where \(\vec{p}\) is the Newtonian momentum.
The little ? sign is to show that we are using Einstein notation, where greek letters show that we are using 4-vectors.
What about the time component? This is the relativistic energy, \(E_{rel} = m\gamma\)!
This means that the 4-vector becomes a momenergy (momentum-energy) 4-vector, where the time component is the energy and the space component is the momentum.
A neutron moves along the positive x-axis in the laboratory frame with a velocity close to the speed of light. It disintegrates spontaneously and a proton and an electron is seen to continue in the same direction as the neutron. We will attempt to use momenergy in order to calculate the speed of the proton and the electron in the lab-frame.
- Lab-frame (t,x): The neutron begins moving at the origin with a velocity \(v_n=-v_{rel}\). We have \(P_\mu(n)=P_\mu(p)+P_\mu(e)\), where n represents the neutron, p represents the proton and e represents the electron.
- Rest frame (t',x'): The neutron is always at the origin, moving with a velocity \(v_n'=0.852=v_{rel}\) relative to the ground. We have \(P'_\mu(n)=P'_\mu(p)+P'_\mu(e)\).
Using the expression we found for momenergy, we write expression for the momenergy 4-vector of the
Proton: \(P'_\mu(p)=\gamma'_p(m_p,\vec{p_p})=\frac{1}{\sqrt{1-(v_p')^2}}(m_p,m_p\vec{v}'_p)\)
Electron: \(P'_\mu(e)=\gamma'_e(m_e,\vec{p_e})=\frac{1}{\sqrt{1-(v_e')^2}}(m_e,m_e\vec{v}'_e)\)
both in the frame of the neutron.
Now, as the neutron is in the rest frame we have that the velocity and thus the momentum is equal to zero, meaning that \(\gamma = 1/(1-v^2)=1/(1-0^2)=1\), which means that the expression for the momenergy of the neutron can be written as:
Neutron: \(P'_\mu(n)=\gamma'_n(m_n,\vec{p_n})=(m_n,0)\)
We can now use conservation of momenergy, \(P_\mu'(n) = P_\mu'(p) + P_\mu'(e)\), in order to find expressions for the velocties of the electron \(v_e'\) and the proton \(v_p'\) in the neutron frame. If we set this up componentwise, we get two equations:
\(E_n = m_n = \gamma_p'm_p+\gamma_e'm_e\) and \(p_n = 0 = \gamma_p'm_pv_p'+\gamma_e'm_ev_p'\).
Note here how mass isn't conserved!
Once we've done this we see that
\(\gamma_p' = \frac{m_n^2+m_p^2-m_e^2}{2m_pm_n}\) -> \(v_p'=\sqrt{1-1/\gamma'_p}\)
\(\gamma_e'= \frac{m_n - \gamma_p'm_p}{m_e}\) -> \(v_e'=\sqrt{1-1/\gamma'_e}\)
Keep in mind that the masses must be in natural values if we want to set \(c=1\). We can now find what the lab measures in their frame of reference by using the transformation properties for 4-vectors: \(P'_\mu = c_{\mu\nu}P_{\nu}\)
This is really a matrix, which when multiplied gives us the values as seen from the other system. In our case, we have a neutron not accelerating and moving with respect to the ground, with a velocity \(v=v_{rel}=0.848000c\). When we use the transformation matrix, we get relative velocity in the opposite direction, \(v_{rel}=-0.848000c\).
We use the expression for energy and momentum of the electron and proton in the lab frame, \(E=\gamma_{rel}E'\) and \(\vec{p}=\gamma_{rel}\vec{p}'\), in order to find the velocity of the electron and proton in the planet frame.
We mentioned earlier to note that the mass isn't conserved. Did you? Let's assume that the mass was conserved. This means that we could have tried to find the velocities of the proton and electron by setting \(m_n=m_e+m_e\). What happens if we try?
\(\gamma_p' = \frac{m_n^2+m_p^2-m_e^2}{2m_pm_n}=\frac{(m_p+m_e)^2+m_p^2-m_n^2}{2m_pm_e}=\frac{2m_pm_n}{2m_pm_n}=1\)
This implies that the velocity of the proton is zero, meaning that, as the velocity of the neutron in the rest system is zero, the electron also has no velocity relative to the neutron (recall that \(p_n = 0 = \gamma_p'm_pv_p'+\gamma_e'm_ev_p'\) has to be true)! This isn't possible. Therefore the mass cannot be conserved! What has happened here, is that parts of the mass has been converted into energy.
We can test this by using the relativistic formula for addition of velocities to obtain the speed of the two particles in the laboratory frame:
\(v'=\frac{v-v_{rel}}{1-vv_{rel}}\) and \(v = \frac{v'+v_{rel}}{1+vv_{rel}}\)
These should be the same as the result we got using energy and momentum.