Collision in special relativity

[exponent137]

Let us assume that we have inelastic collision of masses $m_1=1$ and $m_2=k$
This means $m_2=k m_1$.
($k>>1$)
$v_1=v$, $v_2=0$, Velocity after collision is $v'$.
Units are such that $c=1$. Let us assume that $v_1$ is close to one.
At inelastic collision we respect conservation of energy and momentum. I assume that non-kinetic energy go to thermal energy.

If the moving body is much less massive than rest body, my calculation gives:
$\Delta W=\gamma (1-v/v') +k$
$\Delta W\approx \gamma-\gamma'\approx\gamma-1$
Let us assume that speed after collision is small.

If the moving body is much more massive than rest body,
$m_1=k$ and $m_2=1$
$\Delta W=k\gamma (1-v/v') +1$ and it follows:
$\Delta W\approx 1$

This means that in the rest system of the small body (before collision) we see much smaller thermal energy than in the rest system of the larger body (before collision.)

I do not understand, how to explain this.

 

[PeterDonis]

Velocity after collision is $v'$.

I assume you mean the velocity, in the original frame, of the single object that exists after the collision?

I assume that non-kinetic energy go to thermal energy.

It goes into the rest mass of the single object that exists after the collision; "thermal energy" is part of rest mass. You can't get things right if you don't include that. I don't see it anywhere in your equations; I would suggest reposting your analysis with all masses explicitly included. Trying to separate out kinetic energy from rest energy is usually not worthwhile in relativity, and often leads to errors.

 

[exponent137]

I assume you mean the velocity, in the original frame, of the single object that exists after the collision?

Yes, in the original frame of the single object that exists after the collision.

It goes into the rest mass of the single object that exists after the collision; "thermal energy" is part of rest mass. You can't get things right if you don't include that. I don't see it anywhere in your equations; I would suggest reposting your analysis with all masses explicitly included. Trying to separate out kinetic energy from rest energy is usually not worthwhile in relativity, and often leads to errors.

I did not separate kinetic energy from rest energy. But, yes, I think that thermal energy is part of rest mass, but I do not know how to include this?

Let us assume that the moving mass is the smaller one , ($m_1$), and $m_2/m_1=k$. We are in the frame, where $m_2$ is at rest.
My derivation was:
conservation of momentum:
$m_1\gamma_1 v_1=(m_1+m_2)\gamma'v'$
$m_1\gamma_1 v_1=m_1(1+k)\gamma'v'$
conservation of energy:
$m_1\gamma_1+m_2=(m_1+m_2)\gamma'+\Delta W$
The result is
$\Delta W/m_1 =\gamma_1 (1-v_1/v')+k$
I do not know how to better include $\Delta W$ as the part of the rest mass?
$v'$ is much smaller than $v_1$.

The other option is:
Let us assume that moving mass is the larger one ($m_2$) and $m_2/m_1=k$. We are in the frame, where $m_1$ is at rest.
My derivation was:
conservation of momentum:
$m_2\gamma_1 v_1=(m_1+m_2)\gamma'v'$
$m_1 k \gamma_1 v_1=m_1(1+k)\gamma'v'$
conservation of energy:
$m_2\gamma_1+m_1=(m_1+m_2)\gamma'+\Delta W$
The result is
$\Delta W/m_1 =k \gamma_1 (1-v_1/v')+1$
$v'$ is almost the same as $v_1$, and both a little smaller than 1.

And so on. But in the rest sistem of $m_1$ the result is not the same as in the rest system of the $m_2$.
Why?

 

[PAllen]

You are postulating incorrectly that which should be derived.

Using m1, m2 for the initial masses, you must treat the result mass M as something to derive from conservation. In all cases, I refer to invariant masses. Then, you have, for your set up:

m1 γ1 v1 = M γ' v' for momentum, and

m1 γ1 + m2 = M γ' for energy.

You have to solve for M. When doing approximations for high mass ratio, you need to be careful keep enough terms for significance in the final result, because M will differ only by a tiny amount from m1 + m2. If you do it right, all works out. For example, for γ = 2, m1 = .01, m2 = 1 and then with interchange of m1 and m2, you get M = 1.01985 either way. This tells you that .00985 of the mass of the resultant object is thermal.

 

[PAllen]

Closing the circle on the OP, I provide a few more hints on how to exploit high mass ration for approximation in a correct way.

A key technique to know is that if you have a problem involving energy and momentum, it is almost always better to use those as your variables rather than gamma and velocity. A few key relations are (continuing with c=1):

E² = P² + m², or E² - P² = m²

and E/m = γ.

Then the description of this problem becomes simply:

E₁ + m₂ = E'
P₁ = P'

Solving for M is trivial: M² = (E₁ + m₂)² - P₁²

From here, one limiting cases is E₁, P₁ << m₂. If you drop squares of 'smalls', use common approximations for square roots and quotients, you get:

M ≈ m₂ + m₁ γ₁

while if you assume m₂ is very small compared to the other quantities, you end up with:

M ≈ m₁ + m₂ γ₁

equivalent as expected. Whichever frame you use, the result is saying (in the high mass ratio limit) all of the KE of the smaller body relative to the larger body is added as thermal energy to the combined object, contributing to its mass.

 

[exponent137]

I will calculate the case with $m_2$ is small, (small mass is in rest at the start) that you will see that I understand:
$$
M^2=E_1^2+2E_1m_2+m_2^2-(E_1v_1)^2=E_1^2/\gamma_1^2+2E_1m_2+m_2^2
$$
$$
=m_1^2+2m_1\gamma_1 m_2+m_2^2
$$
If we drop $m_2^2$ and we make approximate square root we obtain your last equation.
Thanks for help.

 

[exponent137]

I will add still the case, where the $m_2>>m_1$ and $m_1$ is moving and $m_2$ is at rest.
$$
M^2=E_1^2+2E_1m_2+m_2^2-(E_1v_1)^2=E_1^2/\gamma_1^2+2E_1m_2+m_2^2
$$
$$
=m_1^2+2m_1\gamma_1 m_2+m_2^2
$$
The result is the same as above, because in both cases the same mass is moving, $m_1$, and the same mass is at rest, $m_2$, and the velocity is the same. It is not important that their values are swaped. The result is symmetric according to $m_1$ and $m_2$, so this symmetry is the key, that the termal energy is the same in our both cases.

If we drop $m_1^2$ and we make approximate square root we obtain Pallen's eq. before his last equation.