Relativistic Mass in this fission decay

[mtuck]

Homework Statement

A nucleus of mass 𝑀 which is at rest undergoes nuclear fission and splits into two smaller nuclei of masses 𝑚1 and 𝑚2. Mass 𝑚1 flies off to the left at speed 𝑣1 and mass 𝑚2 flies off to the right at speed 𝑣2. The speeds of both masses are non-relativistic, 𝑣1⁄𝑐 ≪ 1 and 𝑣2⁄𝑐 ≪ 1. (a) Is there a non-zero mass difference ∆(𝑚𝑎𝑠𝑠) = 𝑀 − (𝑚1 + 𝑚2) ? (b) If yes for (a), then find an expression for the mass difference ∆(𝑚𝑎𝑠𝑠). (c) What is the physical interpretation of the result in (a)? That is, why is there a mass difference?

Relevant Equations

m = gamma(m_0)

I'm not sure where to start

 

[etotheipi]

As is almost always the case, I don't think it's helpful to think about the so-called and rarely useful "relativistic mass" $m_r \equiv \gamma m$, since it just introduces an additional layer of confusion. Forget that concept immediately!

Instead, you just need to consider conservation of energy. Use that a free particle of mass $m$ at speed $v$ has a (total) energy of$$E = \gamma_v mc^2 = mc^2 \left(1-\frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$and equate the total energy of the system before and after the disintegration. Since the speeds are small compared to $c$ you can try binomial expanding the RHS to first order in $v^2/c^2$ and see if you recover anything familiar from classical mechanics.

 

[italicus]

Another way to solve the problem is the use of 4-vectors formalism, if you know it. Given a particle of mass M, having velocity $\vec v$ in an inertial frame ( the “laboratory”) , its 4-impulse is defined as :

$$P = (\gamma M c , \gamma M \vec v ) = (\frac{E}{c} , \vec p) $$

Note that mass is invariant , doesn’t change with velocity. This is an old argument of discussion, if mass is to be considered increasing with velocity or not . The modern point of view is that mass is invariant with velocity. But this is a good subject for another discussion, now doesn’t matter.

In your exercise, the given mass M is at rest in the reference frame of the laboratory, so its 4-impulse is:

$P=( Mc,0)$

because its velocity is zero and $\gamma =1$ . So its total energy is only rest energy , that is : $E = Mc^2$, and the time component of P is just $E/c = Mc$ . The space component is zero : $\vec p = 0$ .

Now M splits into $m_{1}$, which goes left with speed $v_{1}$ and relevant factor $\gamma_{1}$; and $m_{2}$, which goes right with speed $v_{2}$ and relevant factor $\gamma_{2}$.

This implies that they have 4-impulse :

$P_{1} = (\gamma_{1} m_{1} c , \gamma_{1} m_{1} \vec v_{1} )$

$P_{2} = (\gamma_{2} m_{2} c , \gamma_{2} m_{2} \vec v_{2} )$

Now , it must be : $P = P_{1} + P_{2}$

from which , first of all, we see that the vector sum of spatial terms , that is momenta, is zero ; and it is almost obvious because the system is isolated, as in non relativistic mechanics.

But more important is that :

$$ Mc = \gamma_{1} m_{1} c + \gamma_{2} m_{2} c = (\gamma_{1} m_{1} + \gamma_{2} m_{2} ) c $$

Now we can do two considerations .

First : multiplying for c both members , obtain that the rest energy $Mc^2$ of the original particle equals the sum of total energies of the two masses generated by the splitting. And this is logical : where does the energy of each particle come from ? Obvious , from the rest energy of the original particle. But each particle must have rest energy $m_{i}c^2$ ( i = 1,2) and kinetic energy too.

If splitting were relativistic, that is the two generated particles had a relativistic speed, the solution would be rather difficult. But we have small velocities, so the energy of each particle must be the sum of rest energy $mc^2$ and traditional kinetic energy $1/2 *m *v^2$ .

second: consider that, for small speed, we can assume:

$\gamma \approx (1+ \frac {v^2}{2c^2}$ , as already suggested in reply #2. Do this for each of the two masses; in the end, you will have that:

$$M = m_{1} + m_{2} + \frac {1}{c^2} * (K_{1} + K_{2} )$$
where (...) are the classical kinetic energies.

And now you can answer your questions.

 

[Steve4Physics]

Relevant Equations:: m = gamma(m_0)

That equation is not relevant! I think the questions are looking for very short/simple answers, using your understanding of the basic physics of fission.

Where does the kinetic energy of the 2 smaller nuclei come from?
Can you write an expression for the kinetic energy of each particle? (Remember velocities are non-relativistic.)
Now can you answer the questions?