I wish to derive the relativistic energy-momentum relation $E^2 = p^2c^2 + m^2 c^4$ following rigorous mathematical steps and without resorting to relativistic mass.

In one spatial dimension, given $p := m \gamma(u) u$ with $\gamma(u) := (1 - \frac{|u|^2}{c^2})^{-1/2}$, the energy would be given by

$$E = \int{ \frac{d}{dt}p \space dx}$$

I'm having a hard time with this this integration.

How is the relation $E^2 = p^2c^2 + m^2 c^4$ rigorously derived starting from relativistic momentum, without resorting to relativistic mass?

To give an idea of the rigour I expect in an answer, in example, an answer I'd accept for the derivation of $ E = \frac{1}{2} m v^2$ in classical mechanics would have been as follows:

We seek to integrate the differential form $F \space dx$. Parametrising $x$ by $t$, we obtain $dx = \frac{d}{dt} x \space dt$.

The integral of interest is $\int F \space dx = m \int \frac{d^2}{dt^2}x \space dx = m \int (\frac{d^2}{dt^2}x) (\frac{d}{dt} x) dt$ after changing variables.

We recognize the integrand as $\frac{d}{dt} \left( \frac{1}{2} \left(\frac{d}{dt}x \right)^2 \right) $, and so the result $E = \frac{1}{2} m v^2$ follows from the fundamental theorem of calculus.

Again, as an example, a derivation of $E = \frac{1}{2} m v^2$ I would definitely not accept would be as follows:

$ \int F \space dx = m \int a \space dx = m \int \frac{dv}{dt} \space dx = m \int dv \frac{dx}{dt} $ = $ m \int v \space dv = \frac{1}{2}m v^2$.

Please carry out rigorous mathematical manipulations only.

  • 2
    $\begingroup$ $m:=\sqrt{E^2-p^2}$ is the definition of mass, from which $E^2=p^2+m^2$ trivially (and rigorously) follows. That's pretty much it... $\endgroup$ – AccidentalFourierTransform May 21 '18 at 18:28
  • 1
    $\begingroup$ Hi user. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic May 21 '18 at 19:34
  • $\begingroup$ Why won't you accept the second derivation, given that it's mathematically identical to the first? $\endgroup$ – Physiks lover May 24 '18 at 1:55
  • $\begingroup$ @Physikslover it's absolutely not identical. $\frac{d}{dt}$ is notation for the differential operator which is something that maps functions to other functions. On the other hand $F \space dx$ is a differential form, where $dx$ is a function that maps tangent vectors (pairs of vectors) to scalars. Moving the $\frac{1}{dt}$ over from $v$ to $dx$ is mathematical nonsense. $\endgroup$ – user May 25 '18 at 13:36
  • $\begingroup$ @user No one is moving part of the operator $d/dt$ over $v$, what is being moved is the differential in the denominator in the fraction $dX/dt$ because the action of the operator is identical to the fraction of differentials $(d/dt) v = dv/dt$ $\endgroup$ – juanrga Oct 12 '18 at 10:50

Since $P = Fv$ we have $$\frac{dE}{dt} = \frac{dp}{dt} v$$ by Newton's second law. Integrating both sides with respect to $t$ gives $$\int \frac{dE}{dt} \, dt = \int v \frac{dp}{dt} \, dt = \int v \, dp$$ by the chain rule, aka ordinary $u$-substitution. We have $$p = \gamma m v = \frac{m v}{\sqrt{1-v^2}} \quad \Rightarrow \quad dp = \frac{m \, dv}{(1-v^2)^{3/2}}$$ where I set $c = 1$ for convenience and used the quotient rule. Integrating with initial and final velocities zero and $v_0$ gives $$E(v_0) - E(0) = \int_0^{v_0} \frac{mv}{(1-v^2)^{3/2}} \, dv = \frac{m}{\sqrt{1 - v_0^2}} - m.$$ At this point we cannot proceed further since we don't know the constant of integration. One can show by physical arguments that $E(0) = m$. Thus $$E(v) = \frac{m}{\sqrt{1-v^2}}$$ as desired. This isn't a hard derivation, but you're right: a lot of textbooks botch it.

| cite | improve this answer | |
  • 1
    $\begingroup$ @knzhou I'm not sure how you would show by physical arguments that $E(0)=m$ without assuming that $E^2=p^2+m^2$ (at least implicitly). In reality, the equation OP wants to prove is nothing but the definition of mass, so I don't buy any argument that requires integration. Any such arguments is circular. But please do try and convince me I'm wrong! $\endgroup$ – AccidentalFourierTransform May 21 '18 at 18:31
  • $\begingroup$ This is circular. By assuming $p=m\gamma v$ you have assumed and since $E=p/v$ in classical mechanics and special relativity, even for light, you have assumed $E=m\gamma$ so $E^2 (1-v^2)=m^2$ so $E^2=p^2+m^2$. $\endgroup$ – my2cts May 21 '18 at 18:35
  • $\begingroup$ @AccidentalFourierTransform There are a number of good physical arguments for this (i.e. as rigorous as the light clock argument for time dilation), I’ll dig up some links when I get to a computer. $\endgroup$ – knzhou May 21 '18 at 18:35
  • 2
    $\begingroup$ @my2cts I started from the starting point the OP wanted. Of course the logic could also go the other way. Or one could also establish the result of $p$ first using physical arguments. $\endgroup$ – knzhou May 21 '18 at 18:37
  • $\begingroup$ You are right. I read more into the question than was actually there. $\endgroup$ – my2cts May 21 '18 at 19:13

For completeness, here's an arguably cleaner and simpler formulation of @knzhou 's answer:

We obtain

$$E = \int_{0}^{x_0} (\frac{d}{dt} p) \space dx = \int_{0}^{t_0} (\frac{d}{dt} p) \space v \space dt = \int_{0}^{p_0} v \space dp = \int_{0}^{v_0} v \space (\frac{d}{dv} p) \space dv$$

by applying a sequence of reparametrizations $dx = v \space dt$, $dp = (\frac{d}{dt} p) \space dt$ and $dp = (\frac{d}{dv} p) \space dv$ to the integral for $E$. Since $ \frac{d}{dv} p = m \space (1 - \frac{v^2}{c^2})^{-3/2}$, it follows that

$$ E = \int_{0}^{v} \dfrac{m v}{(1-\frac{v^2}{c^2})^{3/2}} dv = \frac{mc^2}{(1 - \frac{v^2}{c^2})^{1/2}} - mc^2.$$

Defining the total energy $\Sigma = E + mc^2$, since $\Sigma = \gamma m c^2$ and $p = \gamma m v$, it is easy to see by direct computation that $\Sigma^2 - c^2 p^2 = m^2 c^4$, hence

$$\Sigma^2 = m^2 c^4 + c^2 p^2 \space .$$

| cite | improve this answer | |

I want to elaborate a little bit inspired by the comment made by AccidentalFourierTransform and how it relates to the answer by knzouh. You start from the four-velocity

$$ u = \gamma~( c, \mathbf{v}), $$

on which you base your definition of the four-momentum as

$$ p = m_0\gamma~( c, \mathbf{v}) = ( m_0\gamma c,m_0\gamma \mathbf{v}) = ( m_0\gamma c,m_0\gamma \mathbf{v}) . $$

Something you know about the four-velocity is, that is always squares to $u^\mu u_\mu=\gamma^2 (c^2-v^2) = c^2$ from which you immediately conclude that $p^\mu p_\mu = m_0 c^2$, thus representing a Lorentz scalar. On the other hand, if you recognize $ m_0\gamma c^2$ as the energy $E$ and the term $m_0\gamma \mathbf{v}$ as the momentm $\mathbf{p}$ this can as well be written as $p^\mu p_\mu = E^2 /c^2 - \mathbf{p}^2$ which thus leads to desired formula

$$ E^2 /c^2 - \mathbf{p}^2 = m_0c^2 . $$

In order for this argument to work we need to justify why we interpret $p^0$ as the energy. One possible way to proceed is the relativistic action $S$ of a free particle and define $E=- \partial_t S$ and $\vec{p} = \nabla_\mathbf{x} S$ as outlined for example here.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.