Proof that $v=\frac{dE}{dp}$ at relativistic speeds?

Question

This is a formula I have seen stated in two different places in a relativistic context, but I do not know the proof and seem unable to derive it on my own. $E$ is the energy of a free particle, $v$ is its speed and $p$ its momentum.

Answer 1

The fundamental energy-momentum relation in special relativity is $E^2=p^2+m^2$ (let’s ignore factors of $c$, since they will factor out in the end — that is, let’s work in natural units).

Taking the derivative with respect to $p$ on both sides gives

$$2E\frac{\partial E}{\partial p}=2p,$$

since $m$ is a constant. Thus,

$$\frac{\partial E}{\partial p}=\frac{p}{E}.$$

Now, noting that $E=m\gamma$ and $p=mv\gamma$, where $\gamma=1/\sqrt{1-v^2}$, the result is proven.

In the last step, we could have instead used the fact that the four-momentum is given by

$$p^{\mu}=m\frac{\mathrm{d}x^{\mu}}{\mathrm{d}\tau},$$

So that $E=m\,\mathrm{d}t/\mathrm{d}\tau$ and $p=m|\mathrm{d}\vec{x}/\mathrm{d}\tau|$ (this only works in the massive case of course, but proving your identity is trivial in the massless case, since $E=p$). Thus,

$$\frac{\partial E}{\partial p}=\frac{p}{E}=\frac{|\mathrm{d}\vec{x}/\mathrm{d}\tau|}{\mathrm{d}t/\mathrm{d}\tau}=\left|\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}\right|,$$

which is the definition of the speed. Note that in the second step we used the chain rule.

I hope this helps!

Answer 2

We have $E=\gamma mc^2$ (relativistic total energy) and $p=\gamma mv$, where $\gamma=1/\sqrt{1-v^2/c^2}$. Therefore, we can write:

$$E=\frac{pc^2}{v} \space,$$

differentiating both sides WRT $p$ implies:

$$\frac{dE}{dp}=c^2\left(\frac{v-p\frac{dv}{dp}}{v^2}\right)=\frac{c^2}{v^2}\left(v-\frac{p}{\frac{dp}{dv}}\right) \space. \tag{*}$$

Moreover, we have:

$$\frac{dp}{dv}=\frac{d(\gamma mv)}{dv}=m\frac{d}{dv}\left(\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\right)=\frac{m}{\left(1-\frac{v^2}{c^2}\right)^\frac{3}{2}}={\gamma^3}m \space. \tag{**}$$

Substituting both $p=\gamma mv$ and its derivative [Eq. $(**)$] into Eq. $(*)$ infers:

$$\frac{dE}{dp}=\frac{c^2}{v^2}\left(v-\frac{\gamma mv}{\gamma^3 m}\right)=\frac{c^2}{v}\left(1-\frac{1}{\gamma^2}\right)=\frac{c^2}{v}\left(1-1+\frac{v^2}{c^2}\right)=v \space.$$

Answer 3

I use the signature convention $(+,-)$.

Any spacetime-displacement vector can be written as $$\tilde T=a\hat t+ \tilde s,$$ where $\hat t$ is unit and future-timelike and $\tilde s\cdot\hat t=0$ (that is, $\tilde s$ is purely spacelike to the observer with 4-velocity $\hat t$). Assume $a \gt 0$. Since $a$ is the "temporal displacement" and $\tilde s$ is the "spatial displacement", then we can define the "spatial velocity according to the $\hat t$-observer" as $$\tilde v\equiv \displaystyle\frac{\tilde s}{a}.$$ So, $\tilde T=a(\ \hat t+\tilde v\ )$. For non-spacelike vectors ($\tilde T\cdot\tilde T \geq 0$), we have $1=\left\|\hat t\right\|^2=\hat t\cdot \hat t\ \ \geq\ -\tilde v \cdot \tilde v=\left\|\vphantom{\hat t}\tilde v\right\|^2=v^2$.

Similarly, any [future-pointing non-spacelike] energy-momentum vector can similarly be decomposed into $$\tilde P=E\hat t+ \tilde p,$$ where $E$ is the relativistic-energy and $\tilde p$ is the relativistic momentum. When $\tilde T$ is chosen to be parallel to $\tilde P$ (so the particle with this constant momentum $\tilde P$ will traverse this displacement $\tilde T$), we have $$\tilde v =\displaystyle\frac{\tilde p}{E},$$ as suggested by @Ben Crowell .

Now, as @Bob Knighton does, \begin{align} m^2 \equiv \tilde P\cdot \tilde P &=E^2(\ \hat t\cdot \hat t\ )+(\ \tilde p\cdot \tilde p\ )\\ &=E^2 - p^2 \end{align} and, by differentiation, \begin{align} 0 &= 2 E \frac{dE}{dp} - 2p, \end{align} so that \begin{align} \frac{dE}{dp} = \frac{p}{E} =v, \quad \mbox{where the last equality uses the relation above}. \end{align} This should be in line with @Ben Crowell , who hints that a good proof should be able to cover both the timelike ($m^2>0$) and lightlike ($m^2=0$) cases.