# [Special Relativity] Energy-momentum invariant question

• Inquisitus
The fractional change in the rest mass is then \frac{m' - m}{m} = \sqrt{\frac{2(\gamma+1)}{\gamma^2+1}} - 1 For v=0.8 c, this is about 0.41.In summary, the problem involves an inelastic collision between two identical particles with one particle initially moving at a velocity of 0.8 times the speed of light. The question asks for the fractional change in the system's total energy after the particles coalesce. The student's first attempt at a solution was marked as incorrect by their lecturer, who suggested using 2m instead of m in the denominator. However, the student's reasoning for using m
Inquisitus
I was just wondering why what I've done in a spec rel question is wrong.

## Homework Statement

A particle of mass m is traveling at 0.8c with respect to the lab frame towards an identical particle that is stationary with respect to the lab frame. If the particles undergo an inelastic collision and coalesce, what is the fractional change in the system's kinetic energy as measured from the lab frame?

http://img85.imageshack.us/img85/6922/croppercapture1io7.png

## Homework Equations

$$p=\gamma m v$$
$$E^2 = (mc^2)^2 + (pc)^2$$

## The Attempt at a Solution

http://img183.imageshack.us/img183/6559/croppercapture3vr8.png

My lecturer marked the first line here as being wrong; he said that for the momentum term in the denominator I should have put 2m rather than just m. However, my thinking here was that since the momentum of a system is conserved within any given frame, the momentum prior to the collision should be the same as following the collision, which is why I used $$\gamma m v c$$ rather than $$\gamma m_1 v_1 c$$. This makes things vastly simpler as we know the velocity of the moving particle prior to the collision, but we don't know the velocity of two particles after the collision.

Has my lecturer just misunderstood what I was doing, or am I doing something wrong?

Last edited by a moderator:
Your logic is right, but you've computed the ratio of the total energies, not the kinetic energies. Just subtract out the total rest mass in the numerator and denominator and it should work.

Whoops, I actually meant total energies rather than kinetic energies, sorry

I'll see my lecturer tomorrow and see what he says. Thanks for the help!

Whoops, I actually meant total energies rather than kinetic energies, sorry
The total energy of the system cannot change. And the system's mass is definitely not 2m. The problem seems well stated, but neither your approach nor what your tutor told you seem correct.

Ich said:
The total energy of the system cannot change. And the system's mass is definitely not 2m. The problem seems well stated, but neither your approach nor what your tutor told you seem correct.

Well the question asked what the fractional change in the systems total energy is upon the particle's collision; it was just a type in my post. The question, everything I've learnt, my solution to the problem, and my lecturer all say that this is possible

Sorry, I was assuming that since this is an inelastic collision, we shouldn't expect energy to be conserved (ie, some of it would go into heat, sound, etc). But assuming these are the only two particles around, and there is no radiation, the only way for energy to be conserved is if the rest mass of the particles increases (as would happen, for example, if some of the energy is converted into heat). In this case we have, both before and after the collision:

$$p^{tot}_\mu= ( \gamma m c^2, \gamma m v, 0, 0) + (mc^2,0,0,0)$$

$$= ( (\gamma + 1 ) m c^2, \gamma m v, 0, 0)$$

The velocity of a single particle with four momentum $p_\mu$ can be read off as $v_i = c^2 p_i/p_0$, giving the resulting particle a velocity of:

$$v' = c^2 \frac{\gamma m v}{ (\gamma + 1 ) m c^2} = \frac{\gamma}{\gamma+1} v$$

and a rest mass of

$$m' = \frac{1}{c^2}\sqrt{p_0^2 - c^2p_1^2} = \sqrt{ (\gamma+1)^2 m^2 c^4 - \gamma^2 m^2 v^2 c^2 } = m \sqrt{ (\gamma+1)^2 - \gamma^2 \frac{v^2}{c^2}}$$

$$= m \sqrt{ (\gamma+1)^2 - (\gamma^2 -1)} = m \sqrt{2(\gamma+1)}$$

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## What is the energy-momentum invariant in special relativity?

The energy-momentum invariant in special relativity is a mathematical quantity that remains constant in all frames of reference. It is a combination of the energy and momentum of a particle and is given by the equation E^2 - p^2c^2 = m^2c^4, where E is the energy, p is the momentum, c is the speed of light, and m is the rest mass of the particle.

## Why is the energy-momentum invariant important?

The energy-momentum invariant is important because it is a fundamental property of particles in special relativity. It allows us to accurately describe and predict the behavior of particles at high speeds and in different frames of reference. It also plays a crucial role in the understanding of mass-energy equivalence and the conservation of energy and momentum.

## How is the energy-momentum invariant related to the speed of light?

The energy-momentum invariant is related to the speed of light through the famous equation E=mc^2. This equation shows that the energy of a particle is directly proportional to its mass and the square of the speed of light. This relationship is a fundamental principle in special relativity and helps explain the behavior of particles at high speeds.

## Does the energy-momentum invariant have any practical applications?

Yes, the energy-momentum invariant has many practical applications, particularly in particle physics and high-energy experiments. It allows scientists to accurately calculate and measure the energy and momentum of particles, which is crucial in understanding their behavior. It also helps in the development of technologies like particle accelerators and medical imaging devices.

## Is the energy-momentum invariant always conserved?

In special relativity, the energy-momentum invariant is always conserved. This means that its value will remain constant regardless of the frame of reference or the speed of the particles involved. This principle is a fundamental part of special relativity and is essential in understanding the behavior of particles at high speeds.

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