6.2.1Exercises, examples & solutions

Prior to the disintegration, particle $M$ has 4-momentum:

After the disintegration, we have two particles with 4-momentum:

and

From conservation of momentum we get:

Take the ratio of the last two equations:

and from this we find

Thus, we see that the mass after the disintegration is $\frac{1}{4}M + \frac{4\sqrt{7}}{16}M \approx 0.911 \lt M$.

Before the absorption of the photon the 4-momentum is:

After emitting the photon, the particle has mass $M$ and velocity $u$. The emitted photon has as frequency $\tilde{f}$ and 4-momentum $\left ( \frac{h\tilde{f}}{c}, -\frac{h\tilde{f}}{c} \right ) = (\frac{1}{4}mc, -\frac{1}{4}mc)$. The total momentum after the process is:

Since 4-momentum is conserved, we find:

We rearrange the two above equations:

If we divide the second equation by the first, we have:

The mass of the particle is:

Initially, the 4-Momentum is

with

After the decay, we have

with $\hat{f}$ a unit vector pointing in the $\pm x$-direction. We know $\frac{u}{c} = \frac{4}{5} \rightarrow \gamma(u) = \frac{5}{3}$. Conservation of 4-momentum now leads to::

We still need to find out which direction the photon travels: parallel to $m$ or in the opposite direction. According to the above conservation of 4-momentum both seem possible. We require that in the above $f \gt 0$.

First we inspect the negative sign of ±:

If we solve for $f$, we get $f \lt 0$, which conflicts our requirement. That leaves us with the $+$sign:

Solving for $m$ gives: $m = \frac{3}{5}M$. If we plug this back in, we find for the photon $hf = \frac{8}{5}Mc^2$.

The total 4-momentum before the collision is

After the collision, we have only one particle with 4-momentum

In this process, 4-momentum is conserved.

From $P^1$ and $P^2$ we get

hence, $u_x = u_y$. The new particle moves over the line $x=y$.

If we combine $P^0$ with $P^1$, we find:

This gives $\frac{u_x}{c} = \frac{1}{4}$. Thus, the new particle moves with velocity $\vec{u} = \frac{1}{4}c \hat{x} + \frac{1}{4}c \hat{y}$. We find its mass by calculating $\gamma_f = \frac{1}{\sqrt{1-2\frac{1}{16}}} = 2\sqrt{\frac{2}{7}}$ and using this in the $P^0$ equation:

a. In classical mechanics, we use -for this type of collision- conservation of momentum and of kinetic energy. This gives us:

This set has as solution (not surprising): $u = -\frac{4}{5}c, U = \frac{3}{5}c$.

b. Now we use 4-momentum conservation:

Note: the spatial part of momentum is thus non-zero, in contrast to the classical case.

After the collision we have:

with

Next, we use conservation of 4-momentum: $P_i^\mu = P_f^\mu$. This is, however, hard to do analytical! Thus we use either a graphical or numerical method. If you do this, you will find: