How to prove that kinetic energy equals $K=\frac{p^2}{(\gamma+1)m}$?

Question

In this video at timestamp 1:40:34, the lecturer states that the general form for kinetic energy expressed with momentum is

$$K = \frac{p^2}{ (\gamma + 1)m}$$

with $\gamma$ as the Lorentz factor.

which reduces in the non-relativistic case to the form we're all familiar with $K=(1/2)mv^2$ (since $p=\gamma m v$).

I've tried to prove this many times now, but for some reason I can't get to this result. Also I can't find this definition of kinetic energy anywhere else expressed like this. I can only find a way to express kinetic energy with momentum using the energy-momentum relation, which doesn't (I think) easily reduce to the form I stated here.

Here's something I tried:

\begin{align} &K^2=(\gamma - 1)^2(mc^2)^2 \\ &=(\gamma - 1)^2(\gamma^2 m^2 c^4 - (pc)^2)\\ &= (\gamma - 1)^2c^2(\gamma^2 m^2 c^2 - p^2)\\ &\iff \\ &K = \frac{(\gamma - 1)^2 c^2(\gamma^2 m^2 c^2 - p^2)}{(\gamma - 1)c^2} \\ &= \frac{(\gamma - 1)(\gamma^2m^2c^2 - p^2)}{m} \end{align}

which has some common terms, so I tried to expand and multiply (top & bottom) with $\gamma + 1$ but there are terms that do not vanish.

Can someone help me get the same result?

Answer 1

Use rapidity $\theta$, so that $(v/c)=\tanh\theta$ and $\gamma=\displaystyle\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$.

These identities [derived from hyperbolic trigonometry] might help. $$\gamma^2\left(1-(v/c)^2\right)=1 \quad \quad \mbox{from the definition of $\gamma$}$$ which can be solved for $(v/c)^2$ and written as $$\mbox{as a difference of squares.... details omitted} $$

Start with the usual expression [in terms of $\gamma$] $$K=(\gamma-1)mc^2.$$ Can you take it from here?

(There is actually a more interesting way to write it. I may update this post later.)

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Here are some references for that expression you asked about.

D.E. Fahnline, "Parallels between relativistic and classical dynamics for introductory courses," Am.J.Phys. 43, 492--495 (1975). https://doi.org/10.1119/1.9775

P.J.Riggs, "A Comparison of Kinetic Energy and Momentum in Special Relativity and Classical Mechanics," The.Phys.Teach. 54, 80--82 (2016). https://doi.org/10.1119/1.4940169

T.A.Moore, Six Ideas that Shaped Physics, Unit C, 2e. (McGraw Hill, New York, 2003), Eq.(C8.16) on p.143.

R.Chabay and B.Sherwood, Matter and Interactions, 4e (Wiley, Hoboken NJ, 2015), p.220.

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update:

Here's the underlying trigonometric identity (implied by the OP's problem): $$(\cosh\theta-1)=\frac{\sinh^2\theta}{\cosh\theta+1},$$ which follows from $$1\quad=\quad\cosh^2\theta-\sinh^2\theta \quad=\quad \cosh^2\theta(1-\tanh^2\theta).$$

Answer 2

One way that I found is to use the following two facts (along with $K = (\gamma-1)mc^2$): $$E = \sqrt{p^2c^2 + m^2c^4}$$ $$E = \gamma mc^2$$

We thus have $$\gamma mc^2 = \sqrt{p^2c^2 + m^2c^4} \implies \gamma^2m^2c^4 = p^2c^2 + m^2 c^4$$

Arranging this to get $p$ in terms of everything else, $$(\gamma^2-1)m^2c^2 = p^2$$

Now, let's divide by $m$ on each side and note that $\gamma^2-1 = (\gamma+1)(\gamma-1)$ by simple expansion.

We then have $$(\gamma+1)(\gamma-1)mc^2 = \frac{p^2}{m}$$

Using the fact that we have $K = (\gamma-1) mc^2$ it then follows that $$K = \frac{p^2}{(\gamma + 1)m}$$