Let $(x,y,z,t)$ be a Lorentz frame equipped with the Minkowski metric. Assume 2 particles interact, without external forces applied to them. The total 4-momentum $p_1+p_2$ is therefore conserved. If we differentiate with respect to the frame's time $t$,

$$ p_1 + p_2 = \text{constant}\; \Leftrightarrow \; \frac{d(p_1+p_2)}{dt} = 0 \;\Leftrightarrow \; \frac{m_1}{\gamma_1}a_1 + \frac{m_2}{\gamma_2}a_2 = 0 $$ where $m_1$ and $a_1$ are the mass and 4-acceleration of the first particle, and $$\gamma_1=\frac{dt}{d\tau_1}=\left(1-\frac{v_1^2}{c^2}\right)^{-1/2}.$$

This is strange because $ p_1 + p_2 = \text{constant}$ looks covariant (even coordinate-free), whereas $\frac{m_1}{\gamma_1}a_1 + \frac{m_2}{\gamma_2}a_2 = 0$ doesn't. If the last equality holds in all galilean frames, it means that the ratio $\frac{\gamma_1}{\gamma_2}$ is the same in all Lorentz frames. And I think that's false.

Did I make a mistake somewhere?

EDIT: after more calculations, I don't think $ p_1(t) + p_2(t) = \text{constant}$ is covariant either, because it uses the frame's time $t$. When the 4-momentums $p_1$ and $p_2$ change, the previous equation identifies pairs of events as simultaneous in the frame. Those events won't be simultaneous in another Lorentz frame.

For example, take 2 particles with same mass $m$, going in a circle and diametrically opposed : $\vec{x}_1(t) = (R\cos(\omega t), R\sin(\omega t), 0)$ and $\vec{x}_2(t) = (-R\cos(\omega t), -R\sin(\omega t), 0)$. It might be two moons orbiting around the same planet. In this Lorentz frame, the total 4-momentum is conserved: $$ p_1(t) + p_2(t) \;=\; (\vec{0}, \;2\gamma_1m c^2) $$ However, in another Lorentz frame moving along the $x$-axis, the Lorentz transformation gives $t' =0 \Leftrightarrow ct = \beta x$. This last equation won't give 2 events diametrically opposed in the first frame, so the total 4-momentum won't be preserved with respected to the other frame's time $t'$.

Maybe the problem of $ p_1 + p_2 = \text{constant}$ is that it's an action at a distance. Does it only hold for shocks of particles?

  • $\begingroup$ What if any is the interaction? $\endgroup$
    – my2cts
    Jul 17, 2021 at 1:19

1 Answer 1


You are correct that the equation $p_1 + p_2$ is covariant, and the equation $\frac{m_1}{\gamma_1}a_1 + \frac{m_2}{\gamma_2}a_2 = 0$ is not - because you've picked out a specific Lorentz frame in which to define the $\gamma$'s and the three-accelerations. Despite not being covariant, the second equation is indeed true in all Lorentz frames.

(Well, okay, strictly speaking this depends on your exact definition of a "covariant" equation. Some people define that to mean that both sides of the equation can be expressed as manifestly Lorentz-covariant tensors, while others adopt the weaker definition that the equation's truth value is the same in all Lorentz frames.)

I think your confusion is in thinking that there are "galilean frames" in special relativity, in which there are quantities invariant under Galilean transformations. In special relativity, nothing nontrivial is invariant under Galilean transformations. Therefore, the numerical values of the three-accelerations $a_1$ and $a_2$ will be different in different Lorentz frames, and it's not true that the ratio $\gamma_1/\gamma_2$ will be invariant.

  • $\begingroup$ I'm only considering Lorentz transformations, even between what I call galilean frames. $a_1$ is the 4-acceleration, not 3-acceleration. Can you prove this equation is preserved by Lorentz transformations ? $\endgroup$
    – V. Semeria
    May 24, 2017 at 22:24

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