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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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haxonek
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    Do you know the formula for relativistic addition of velocity? You simply need to calculate the velocity of particle 2 in particle 1’s frame and calculate the relativistic energy from there – DavidH Jan 01 '20 at 04:51
  • Do you mean w = u + v / (1 + uv/c^2)? I do, however I'm wondering specifically in regards to perceived energy. If you're viewing one item that should have a velocity of 2c in newtonian physics, would it appear that the energy used to get to that point would be sqrt(2)mc^2, or would the constant be 2, or even something else? I apologize if this questions doesn't make sense, I was just reading Susskind's book on it and this question kept bugging me – haxonek Jan 01 '20 at 04:55
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    You can’t have massive objects moving at $c$, the energy of such an object would be infinite. If they are moving at speeds less than $c$ it’s simply an application of the relativistic addition of velocities – DavidH Jan 01 '20 at 05:02
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    Where does that $mc$ come from? The SR formula for momentum (of a body with non-zero mass) is $p=mv\gamma$ – PM 2Ring Jan 01 '20 at 20:23
  • Are you wanting the kinetic energy of #2 or the total energy of #2 or the total energy of the system in the rest frame of #1? Also, the energy will change depending on the reference frame you use. – Bill N Jan 02 '20 at 00:36

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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  • I guess what I'm trying to get at is because of the relationship between E^2 = (mc^2)^2 + (pc)^2, where (pc)^2 => (mcc)^2, it implies a situation where if you had two objects moving at near the speed of light, you could do so with less than double the amount of energy required to move one object near the speed of light, since the energy added after nearing the speed of light is treated as adding energy to energy. This said I'm likely just fundamentally misunderstanding the equation. Thank you for your time though, that second eq is interesting – haxonek Jan 01 '20 at 05:21
  • The first line should be arccosh or inverse-cosh (not just cosh), as used in the final line. – robphy Jan 01 '20 at 16:55
  • @robphy ty. lesson learned: Never post during New Years Eve party. – JEB Jan 01 '20 at 19:54
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    @haxonek So for adding boosts, the rapidity is viewed as the hyperbolic rotation angle, so it is what you can add linearly, and it is the simplest. Regarding piggy backing 2 boosts to cheat the energy equation (a good idea at first), since the energy required to do the second boost contributes to $m_1$'s mass, you can't win. The question isn't that clear but I think that's what your are looking for. I can edit my answer to add that if it is what you are looking for. – JEB Jan 01 '20 at 20:23
  • @haxonek - You're using $p=mv\approx mc$, but it should be $$p=\gamma mv=mv\cosh\omega=mc\sinh\omega\approx\infty$$ (because $v\approx c$ is equivalent to $\omega\approx\infty$). Thus $$E^2=p^2c^2+m^2c^4=m^2c^4(\sinh^2\omega+1)=m^2c^4\cosh^2\omega$$ $$E=mc^2\cosh\omega=\gamma mc^2\approx\infty,$$ not $\sqrt2mc^2$ nor $2mc^2$. Of course, $\omega$ and $E$ are the rapidity and energy of object 2 in the reference frame of object 1. And the energy of object 1 in its own frame is $mc^2\cosh0=mc^2$, so the total energy is $mc^2(1+\cosh\omega)$. – mr_e_man Jan 02 '20 at 02:42
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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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  • $\gamma$ and E are non-linear with $v$ and depend on the reference frame, so you can't simply double everything. – Bill N Jan 02 '20 at 13:38
  • If the two bodies are non-interacting then energy is an additive property of the system, no? – bemjanim Jan 02 '20 at 15:00
  • Yes, but in the reference frame of #1, the $\gamma$s are different for the two objects, and not the same as in the original reference frame. – Bill N Jan 02 '20 at 15:31
  • Oh I thought it was from the lab frame my bad – bemjanim Jan 02 '20 at 16:42