Special theory of relativity > Conservation of mass and momentum > Problem 14.5.19

Statement

$14.5.19.$ [Insert the problem statement]

Solution

### Condition
$14.5.19$

$ a. $ At what speed did an excited nucleus of mass $M$ move if, after emitting a $\gamma$-quantum of mass $m$, it came to rest? How much do the mass and energy of the excited and unexcited nuclei differ?
$b. $ In what range of velocities of the excited nucleus from part $a$ is the following event possible: the $\gamma$-quantum emitted by the excited nucleus is absorbed by an unexcited stationary nucleus.

### Solution
$\textbf{a}. $ As in some other problems of this chapter, I will assume that the author uses mass as a measure of total energy, i.e., the masslessness of the photon and the constancy of the rest mass are not cancelled. And the question about the difference in mass and energy in part $a$ can be reduced to a difference only in masses.

So, if the nucleus stops after emitting the quantum, its energy becomes equal to the rest energy:
$$
E = Mc^2 = mc^2 + M_r c^2 \tag{1}
$$
We can immediately answer the second question:
$$
\Delta m = M - M_r = m \tag{2}
$$

For the momentum of the emitted photon we have
$$
p = \frac{E}{c} = mc
$$
After the photon emission, the nucleus stops, so from conservation of momentum it follows that it initially had the same (including direction) momentum:
$$
p = \gamma M_r v = M v = m c
$$
Then obviously
$$
v = \frac{m}{M}c \tag{3}
$$
$Note:$
When trying to solve the problem in another way, one might make a mistake (I initially did so) by writing
$$
M = \gamma M_r = \gamma (M - m)
$$
and then finding the speed. The answer would be different (you can check). But why is this incorrect? After all, the formula seems to agree with our definition of mass as total energy... Here I want to draw attention to the fact that although the nucleus emits a $\underline{massless}$ particle, its own rest mass decreases. That is, not only kinetic energy is converted into photon energy. Let us try to show this:

Let:  
- \\(M_{r1}\\) – the initial rest mass of the nucleus,  
- \\(M_r\\) – the final rest mass of the nucleus,  

We have already written the conservation laws of energy and momentum:

$$
M = M_r + m, \quad M v = m c
$$

From the first equation \\(m = M - M_r\\). Substituting into the second:

$$
M v = (M - M_r) c \quad\Rightarrow\quad \frac{v}{c} = 1 - \frac{M_r}{M}. \tag{4}
$$

On the other hand, from the definition \\(M = \gamma M_{r1}\\):

$$
\frac{v}{c} = \sqrt{1 - \left(\frac{M_{r1}}{M}\right)^2}. \tag{5}
$$

Equating and squaring:

$$
\left(1 - \frac{M_r}{M}\right)^2 = 1 - \left(\frac{M_{r1}}{M}\right)^2.
$$

Expanding the brackets and simplifying:
$$
2M_r - \frac{M_r^2}{M} =  \frac{M_{r1}^2}{M},
$$

$$
2 M_r M - M_r^2 = M_{r1}^2 
$$

$$
M_{r1}^2 - M_r^2 = M_r(2M - M_r) - M_r^2 = 2M_r(M - M_r).
$$

Since \\(M - M_r = m > 0\\) and \\(M_r > 0\\), we obtain:
$$
M_{r1}^2 - M_r^2 > 0 \quad\Rightarrow\quad M_{r1} > M_r.
$$
Thus, $\textbf{the rest mass of the nucleus decreases upon emitting a photon}$.

$\textbf{b}. $ 
Photon absorption occurs only if the rest energy of the nucleus changes by exactly a certain amount (and due to "recoil", the kinetic energy also changes).

A photon that can be absorbed by a stationary unexcited nucleus must have energy $mc^2$ in the laboratory frame (LS) (reversibility of the process from $(a)$).

Suppose the excited nucleus moves with speed $u$, and the energy of the photon in its rest frame is $E_0$. Then in the LS
$$

E_\gamma = E_0\\, \gamma_u(1+\beta\cos\theta), \qquad

\beta=\frac{u}{c},\quad \gamma_u=\frac{1}{\sqrt{1-\beta^2}}.
$$

For a fixed $u$, the energy varies from $E_{\min}=E_0\gamma_u(1-\beta)$ to $E_{\max}=E_0\gamma_u(1+\beta)$. For there to exist an angle $\theta$ such that $E = mc^2$, it is sufficient that $mc^2$ lies in this interval.

From part $(a)$ at $u=v_1=\dfrac{m}{M}c$ and $\theta=0$ the photon has energy exactly $mc^2$, i.e., $E_{\max}(v_1)=mc^2$.

The function $E_{\max}(u)$ increases monotonically, $E_{\min}(u)$ decreases monotonically, and as $\theta$ changes, the energy changes continuously, therefore:

- if $u<v_1 $, then $E_{\max}<mc^2$ and absorption is impossible;

- if $u\ge v_1$, then $E_{\max}\ge mc^2$ and there exists an angle $\theta$ such that $E_\gamma=mc^2$.

Thus, the event is possible for all velocities of the excited nucleus not less than $v_1$:
$$
u \geqslant \frac{m}{M}c.
$$

#### Answer
$$
a. \ v=\frac{m}{M}c, \qquad \Delta m=m;
$$
$$
b. \ u \geqslant \frac{m}{M}c
$$

Answer

[Insert a concise answer or boxed result]