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Let's say there are two objects going nearly the speed of light, in opposite directions.

Object 1 and 2 moving in opposite directions

Obviously from the vantage point of #1, #2 would be moving at about the speed of light, c. But assuming #1 thought it was stationary, what would be the perceived energy of #2 from the perspective of #1? In my mind it seems like this would make sense since the remaining energy wouldn't go into making it move faster

$E^{2} = (mc^2)^2 + (pc)^2$

$E^{2} = (mc^2)^2 + (mc*c)^2$

$E = \sqrt{2}mc^2$

But I could also see the answer just being this:

$E = 2mc^2$

And I could also see myself being completely wrong on both fronts.

Thanks for your time. I apologize if this has already been asked before, but all I could find were questions like this, which, while the question sounds similar it is fundamentally a different question.

Qmechanic
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thansen0
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2 Answers2

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Rapidity:

$$ \omega = \cosh^{-1}{\gamma} = \cosh^{-1}{\frac E m}$$

is additive, so:

$$ \omega = \omega_1 + \omega_2 $$

solves the problem.

Note that thinking about what newtonian physics says is not helpful, nor is being imprecise with statements like "moving at about the speed of light". Generally people say "ultra-relativistic" for that concept.

Regarding energy, for a given rapidity and mass:

$$ E = T+m = m\cosh{\omega}$$

That is the total energy (including rest mass) of a moving mass. While energy depends on reference frame, it is still just energy, and not "perceived energy".

JEB
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If they are massive objects travelling at $c$ then the Lorenz factor will diverge and they won't have a well defined energy.

To find the energy at smaller but still relativistic speeds use $E=\gamma mc^2$ twice.

bemjanim
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