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Einstein's postulates of Special Relativity are:

  1. The laws of Physics look the same in all inertial reference frames;

  2. The speed of light is always observed with the same value $c$ in every reference frame regardless the motion of the source;

Now, from these two postulates we can deduce that no massive particle travels faster than light.

So the postulates imply that no massive particle travels faster than light.

On the other hand, sometimes I've seem people saying it the other way around, as if saying that particles with mass having always less speed than light should be the starting point.

In the case, the converse is true? Requiring that no massive particle travels faster than light implies the two postulates?

In that sense, the postulates of relativity are essentially the same thing as requiring that no particle with mass travels faster than light?

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  • $\begingroup$ If you can get to the usual velocity addition formula I think you can get the rest from there. But I'm not sure that your conditions uniquely implies $v' = (u + v)/(1 + uv/c^2)$. $\endgroup$ – dmckee Sep 26 '16 at 19:34
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No; the postulate that no massive particle travels faster than light does not imply Einstein's two original postulates.

Let's pick units in which the speed of light is equal to 1. Consider a Universe in which all massive particles satisfy the energy-momentum relationship $$ (1 - \delta)^2 E^2 - p^2 = m^2 $$ with $0< \delta < 1$, while the photon satisfies the usual momentum-energy relationship $E = p$. The speed of any particle is $v = p/E$, so in particular, the speed of a photon is equal to 1. However, the speed of a massive particle in this Universe is $$ v = \frac{p}{E} = \frac{\sqrt{(1 - \delta)^2 E^2 - m^2}}{E} = (1 - \delta) \sqrt{1 - \frac{m^2}{(1 - \delta)^2 E^2}}, $$ which will always satisfy $v < 1 - \delta$ (approaching this bound asymptotically as $E/m \to \infty$.)

Effectively, what we have done here is to postulate a Universe with two different spacetime metrics, one for massive particles and one for massless. The four-momentum of a photon satisfies the relationship $\eta_{\mu \nu} p^{\mu} p^{\nu} = 0$, with $$ \eta_{\mu \nu} = \begin{Bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{Bmatrix}; $$ meanwhile, the four-momentum of a massive particle satisfies the relationship $\tilde{\eta}_{\mu \nu} p^{\mu} p^{\nu}= -m^2$, where $$ \tilde{\eta}_{\mu \nu} = \begin{Bmatrix} -(1-\delta)^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{Bmatrix}. $$ While the normal Lorentz transformations would leave $\eta_{\mu \nu}$ invariant (and therefore the speed of light would be the same in all reference frames), they would not leave $\tilde{\eta}_{\mu \nu}$ invariant. Thus, the relationship between a massive particle's energy and its momentum would be different in different reference frames, and postulate #1 would be violated. One could also try to define a set of deformed Lorentz transformations that would leave $\tilde{\eta}_{\mu \nu}$ invariant; but this would result in $\eta_{\mu \nu}$ changing between reference frames, violating postulate #2.

This construction relies pretty heavily on the fact that the speed of massive particles is not only always less than $c$, but is also bounded away from $c$: the limit of a particle's velocity as its energy $E \to \infty$ is some number strictly less than $c$. I believe that if you want to get Einstein's postulates out of the behavior of massive particles, it is necessary that the limiting speed of matter be equal to the speed of light; I'm unsure whether this is also a sufficient condition as well (though I suspect not.)

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"Now, from these two postulates we can deduce that no massive particle travels faster than light."

I haven't seen a valid deduction so far but let us assume there is one. Still the answer to your question is "no". Generally, you cannot deduce the antecedent from the consequent. In logic, such an invalid deduction is called "the fallacy of affirming the consequent".

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