Here is what blogging my mind.

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

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 indestructable. 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.

$$ 2\left({\sqrt {\frac {g_{tt}}{g_{tt}+g_{ss}v^{2}}}}-1\right)\ $$

So, If we say that the system is in the flat space (gtt = -c^2 and gss = 1) 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 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? Where did I get it wrong?

Edited: Correct my bad grammar.

Edit 2:

Eh? 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?

Edit 3:

Corrected my false calculation, but the strange result is still the same.


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 $m_0^2c^2=\frac{E^2}{c^2}-p^2$.

  • $\begingroup$ Alright, I understand what you mean. It's like a ball in a box is free falling. The kinetic energy is different when we consider reference frames as a box or earth. However, what I don't get is, when the thing crash and release the energy (sound, vibration or heat), like that particle A and B collided. Isn't it supposed to release an equal amount of energy, no matter where I observed from? See: physics.stackexchange.com/questions/352048/… $\endgroup$ – Branya The Great Aug 17 '17 at 12:03
  • $\begingroup$ @BranyaTheGreat If you do the calculation in 2 different frames, you'll find their answers are related by a suitable transform between them (whether you use special or general relativity or Newtonian mechanics). $\endgroup$ – J.G. Aug 17 '17 at 15:55
  • $\begingroup$ @BranyaTheGreat Yeah. You create an uranium nucleus by smashing two lighter particles together, it's an uranium nucleus in all frames!! :) The uranium nucleus has the energy of an uranium nucleus in all frames. $\endgroup$ – stuffu Aug 17 '17 at 19:56

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.

  • $\begingroup$ Still, you can say it release half, quarter or tenth of the total kinetic energy. It still not answers the question. $\endgroup$ – Branya The Great Aug 17 '17 at 12:05
  • $\begingroup$ I pointed out one huge error in calculation 2. If you correct the error, the answer of calculation 2 becomes more like the answer of calculation 1. Try calculating the two calculations using low speeds and Newton's mechanics first. $\endgroup$ – stuffu Aug 17 '17 at 17:49

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