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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.

Emilio Pisanty
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user
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    $m:=\sqrt{E^2-p^2}$ is the definition of mass, from which $E^2=p^2+m^2$ trivially (and rigorously) follows. That's pretty much it... – AccidentalFourierTransform May 21 '18 at 18:28
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    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. – Qmechanic May 21 '18 at 19:34
  • Why won't you accept the second derivation, given that it's mathematically identical to the first? – Physiks lover May 24 '18 at 01:55
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    @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. – user May 25 '18 at 13:36
  • @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 $(d/dt) v = dv/dt$ – juanrga Oct 12 '18 at 10:50

3 Answers3

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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.

knzhou
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    @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! – AccidentalFourierTransform May 21 '18 at 18:31
  • 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 $E^2=p^2+m^2$. – my2cts May 21 '18 at 18:35
  • @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. – knzhou May 21 '18 at 18:35
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    @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. – knzhou May 21 '18 at 18:37
  • You are right. I read more into the question than was actually there. – my2cts May 21 '18 at 19:13
  • @AccidentalFourierTransform Here is Einstein's original argument for it. I imagine most arguments are going to involve a body emitting light, because we can figure out how the energy of light behaves by $E = hf$ and the Doppler effect. – knzhou May 21 '18 at 20:22
  • There is a mildly cleaner version of this argument by Terence Tao, though I'm not as much of a fan because it has bulky mathematical notation. – knzhou May 21 '18 at 20:25
  • @my2cts Why $E=p/v$ in classical mechanics? I know $c=1$ here, but why this is classically true? – Ma Joad Oct 27 '19 at 22:22
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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

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

user
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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.

sagittarius_a
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