Collision in special relativity

• I
• exponent137
In summary, the velocity after collision for two masses, one much larger than the other, can be calculated using the equations of conservation of momentum and energy. The resulting mass of the combined object will be equal to the sum of the initial masses, plus a very small amount of thermal energy. This applies to both cases where the smaller mass is initially moving and the larger mass is at rest, and vice versa.
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.

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exponent137 said:
Velocity after collision is ##v'##.

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

exponent137 said:
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
PeterDonis said:
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.

PeterDonis said:
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?

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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.

exponent137
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, Nugatory and PeterDonis
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.

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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.

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1. What is the principle of relativity in special relativity?

The principle of relativity in special relativity states that the laws of physics are the same for all observers in uniform motion, regardless of their relative velocities. This means that there is no preferred frame of reference in the universe.

2. How does time dilation affect collisions in special relativity?

Time dilation, a consequence of special relativity, causes time to appear to pass at different rates for observers in different frames of reference. This can affect collisions in two ways: objects moving at high speeds will experience time slower than stationary observers, and the duration of a collision will appear longer to an outside observer than to the objects involved in the collision.

3. Can two objects collide at the speed of light in special relativity?

No, it is impossible for two objects to collide at the speed of light in special relativity. As an object approaches the speed of light, its mass increases and the amount of energy required to accelerate it further also increases. This means that it would require an infinite amount of energy to accelerate an object to the speed of light, making it impossible to reach that speed.

4. How does the concept of mass-energy equivalence apply to collisions in special relativity?

The famous equation E=mc^2, derived from special relativity, shows that mass and energy are equivalent and can be converted into each other. This means that in a collision, some of the mass of the objects involved can be converted into energy, and vice versa. This is important in understanding the effects of high-speed collisions, such as those in particle accelerators.

5. What is the difference between elastic and inelastic collisions in special relativity?

In an elastic collision, the total kinetic energy of the objects involved is conserved, meaning that no energy is lost during the collision. In an inelastic collision, some of the kinetic energy is lost due to other forms of energy, such as heat or sound. In special relativity, the conservation of energy and momentum still apply, but the equations used to calculate the resulting velocities and energies are modified to account for time dilation and mass-energy equivalence.

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