Energy released from two particles colliding at relativistic velocities

Question

Here is what confusing me.

For simplicity, let's assume that there are only three particles in the universe, particle A, B, and C.

Particle A and B going on the head on collision at near speed of light let's say 0.9C or something while using particle C which stands at almost the middle as the frame of reference.

According to equation $$ E_{\text{k}}=mc^{2}\left({\sqrt {\frac {g_{tt}}{g_{tt}+g_{ss}v^{2}}}}-1\right)\ = mc^2\left(\frac{1}{\sqrt{1-v^2/c^2}} - 1 \right) = mc^2(\gamma -1) $$

The total energy is for two particles, A and B is equal to 2Ek. (On C as the observer).

So let's says this three particle are indestructible. And they immediately stop when A and B particles have collided very close (infinitesimally close to or just barely touch) the C particle). And for simplicity again let's say the mass of each particle is about 1/c^2. They should release energy about.

$$ 2 \ mc^2\left(\gamma-1\right)\ $$

So, If we plug v as 0.9C into equation, I got about 2.59.

However, if we look from the perspective of A or B particle, another particle wouldn't be heading in at 1.8C according to the equation: $$ V = \frac{u+v}{1+uv/c^2} $$

Which I got new V equals to 0.9945C. However, when I plug new V into the equation above (without 2, because one particle is used as the reference frame.)

I got about 8.55. Why are they not equal?

I'm pretty sure that I'm wrong somewhere, but if I'm not wrong with the equation, and if it can be concluded that energy is related to the reference frame, what does it mean?

Why wouldn't the particle C be able to observe the other part of energy (8.55-2.59)?

Is it because of the other part of the energy released is faster than the speed of causation/light with particle C as the frame of reference?

Answer 1

I think your confusion is you expect conserved (unchanging over time) quantities to be "invariant" (the same in all reference frames), which isn't true. Energy is one component of the conserved four-momentum, but Lorentz transformations only preserve its norm, not any one component. Explicitly $m_0^2c^2=\frac{E^2}{c^2}-p^2$.

Answer 2

All of particle's kinetic energy is released when the particle collides inelastically into a massive object.

A small part of particle's kinetic energy is released when the particle collides inelastically into a low-mass object. Like when a car collides into a bird.

If a bird were to calculate that its collision with a 1000 kg car moving 10 m/s releases 50000J energy, then the bird would calculate incorrectly.

Answer 3

The total energy of a particle or system is given by:

$E_t = \sqrt{(pc)^2 + (m c^2)^2}$

where $pc = \gamma m v c$ is the momentum energy and $m c^2$ is the rest mass energy. This quantity is conserved in a given reference frame. If photons are involved, we use $p = \hbar/\lambda$ to calculate the energy of the photons where $\lambda$ is the wavelength and $\hbar$ is Planck's reduced constant. For example if the particle was to decay into some other particles and photons, then this equation for $E_t$ correctly predicts that the total energy would be the same before and after the decay.

If you were to smash two particles together in an inelastic collision, the total energy before and after the collision, remains the same when you take the energy of the released photons into account (as long as we remain in the same reference frame).

However, if you were to switch reference frames, this total energy is different, because it is not an invariant quantity. This is easy to see if you consider a stationary baseball with $E_t = Mc^2$. When you switch reference frames to the point of view of an observer moving relative to the ball its total energy is rest mass energy plus a component due to its momentum in the new reference frame.

The invariant quantity when switching reference frames is actually the rest energy of the particles ($mc^2$):

$mc^2 = \sqrt{E_t^2 - (pc)^2}$

The kinetic energy is equal to the total energy minus the rest energy, so this quantity is not invariant.

$E_k = E_t - mc^2$.

To show this is correct, we can substitute the expanded expression for $E_t$ and solve:

$E_k = \sqrt{(pc)^2 + (m c^2)^2} - mc^2$

$E_k + mc^2 = \sqrt{(pc)^2 + (m c^2)^2}$

$(E_k +mc^2)^2 = (pc)^2 + (m c^2)^2$

$E_k^2 + 2mc^2 E_k + (mc^2)^2 = (pc)^2 + (m c^2)^2$

$E_k^2 + 2mc^2 E_k = (pc)^2 $

$E_k^2 + 2mc^2 E_k = (ymvc)^2 $

$E_k^2 + 2mc^2 E_k - (ymvc)^2 =0$

Solving the quadratic gives us the expected result:

$E_k = mc^2(\gamma -1)$

In summary, rest energy $mc^2$ is invariant with a change of reference frames, but the total energy, kinetic energy and momentum energy are not.

P.S. To be more complete the total energy should include a term for potential energy but I did not want to complicate things too much here.