# Can relativistic kinetic energy be derived from Newtonian kinetic energy?

Relativistic kinetic energy is usually derived by assuming a scalar quantity is conserved in an elastic collision thought experiment, and deriving the expression for this quantity. To me, it looks bodged because it assumes this conserved quantitiy exists in the first place, whereas I'd like a derivation based upon using KE $= \frac12 mv^2$ in one frame, and then summing it in another frame say to get the total kinetic energy. Can this or a similar prodecure be done to get the relativistic kinetic energy?

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Newtonian and Einsteinian physics obey fundamentally different symmetry conditions, so it is not clear to me why you would expect to be able to do this. As for the existence of a conserved quantity looking funny to you, we can appeal to Noether's theorem. Both theories have a time translation symmetry, so both should have a conserved quantity related to it. Just not he same quantity. – dmckee Apr 13 '12 at 20:05

Assuming energy conservation isn't "bodged" because at the most fundamental level, energy is defined as the quantity that is conserved as the result of the time-translational symmetry. All specific formulae for energy, such as $mv^2/2$ in nonrelativistic mechanics, are just solutions to the problem "find a conserved quantity linked to that symmetry".

Still, you can try to achieve what you have defined. First, you must realize that $K=mv^2/2$ only holds if $v\ll c$: it's just not a valid formula in relativity for large velocities. It seems that you believe that $E=mv^2/2$ is correct in some frames even in relativity but it's not. Your formula is just an approximation, via Taylor expansions, $$\frac{mc^2}{\sqrt{1-v^2/c^2}} = mc^2 + \frac{mv^2}{2} + \frac{3mv^4}{8c^2} + \dots$$ If you want to use $E=mv^2$ for small $v$ and deduce what is $E$ for an arbitrary $v$ comparable to the speed of light $c$, you must use infinitely many interpolating inertial systems.

In this SE question

How to deduce the theorem of addition of velocities?

Ron Maimon explained how velocities add. So if you want to switch to an inertial system moving by velocity $v$, you may calculate a rapidity from $$\tanh a = \frac vc$$ These rapidities behave as angles so if you're boosting by some incremental speeds many times, the rapidities just add up (much like angles for rotations). The total energy is then $mc^2\cdot \cosh a$ which is equal to the usual relativistic formula but the derivation of this fact will have to use some conservation of energy argument similar to one you know. The relativistic formula is the only one that reduces to $mv^2/2$ for infinitesimal $v$ and that conserves the energy while the object is boosted.

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apologies for not making it clear that I'm looking for a derivation of relativistic kinetic energy. I'm thinking of relativistic kinetic energy of a particle as consisting of an infinite number of infinitesimal Newtonian kinetic energies, and this total must be conserved, but I don't know how to get at it for a derivation. – Physiks lover Apr 13 '12 at 17:38
@Physikslover: There is a problem with this idea, because the KE is not frame invariant, so the first infinitesimal increment is Newtonian, but the increments near relativistic speed are not given by Newtonian energy increments, only in the rest frame, and then you need to know how to boost an energy increment. If you know how to boost an energy increment, you know how to boost energy, and you get the relativitic energy. Lubos is right (and links to an answer of mine, hey, thanks) +1. – Ron Maimon Apr 13 '12 at 19:02
Thanks for the agreement, Ron. I was trying to fill in the gaps so that it would make more sense and resemble what Physiks lover wants - but it's so redundant. The resulting derivation is just a differential version of normal derivations; the beef has to be relativistic and it may be repulsive for the OP, anyway. – Luboš Motl Apr 14 '12 at 5:01