Most of the texts I've seen (admittedly not a large number) which introduce the concept of relativistic momentum usually just pull a formula out of nowhere and then prove the formula must be correct. There is nothing technically wrong with this approach, but it's aesthetically unsatisfying, as it doesn't show how you would get the formula if you didn't already know the answer. The few occasions where I've seen a derivation that didn't assume an answer (e.g. in Einstein's own papers) contain some rather ugly and tedious mathematics.(adsbygoogle = window.adsbygoogle || []).push({});

I thought I'd have a go at a more elegant method, and after some weeks of effort in my spare time, here it is. I wouldn't be at all surprised if someone else hasn't published something similar to this before, so I claim no originality.

I do assume a fair amount of background knowledge. You will need to know about rapidity and how to add it. You need familiarity with hyperbolic functions and their derivatives, partial differentiation and the solution of second-order differential equations.

I am confident of the logic that leads from my assumptions to my conclusions. The reason I am posting this is to ask whether the assumptions that I make are reasonable in this context. Could I simplify my argument still further with even better assumptions?

I will restrict consideration to a single dimension of space, i.e. motion along a single straight line in the absence of gravity.

We already know the Newtonian definitions of mass, momentum, force and energy. We want to derive relativistic versions which are compatible with the Newtonian versions at low speeds, but which are also consistent with the postulates of relativity.

General Assumptions

(A1)The mass of a particle is invariant (i.e. independent of relative motion of particle and observer).

(A2)Total energy is conserved in a closed system.

(A3)Total momentum is conserved in a closed system.

Otherwise neither of these concepts would be useful.

(A4)The energy of any particle is proportional to its mass.

(A5)The momentum of any particle is proportional to its mass.

This follows from conservation. Consider the limit as any two particles collide and coalesce at a relative speed approaching zero. In this Newtonian limit, mass is additive, and so too are momentum and energy.

We will also assume that the Newtonian relationships between force, momentum and energy

(A6)....[tex]F = \frac{dp}{dt}[/tex]

and

(A7)....[tex]E = \int F dx[/tex]

are still valid (relative to any inertial frame) in relativity.

The specific scenario

Consider two particles of equal mass which collide and coalesce to form a single particle, with no loss of energy to the surroundings.

(A8)Measured in the rest frame of the final particle, the two initial particles have equal but opposite velocities.

On grounds of symmetry.

Note also that as mass is invariant, the mass of the final particle depends only on the two initial particles and their relative velocity and is independent of the motion of any observer.

Mathematical formulation

I will measure motion using rapidity [itex]\phi[/itex] instead of velocity [itex]v = c \tanh \phi[/itex]. This will simplify the maths because the composition law for rapidity (in one spatial dimension) is just ordinary addition.

Thus, from (A4) and (A5), we postulate that momentumpand energyEare given by

(F1).....[tex]p = m c F(\phi)[/tex]

(F2).....[tex]E = m c^2 G(\phi)[/tex]

for some smooth functionsFandGto be determined.

In the Newtonian limit as [itex]\phi \rightarrow 0[/itex] we know these functions are approximately [itex]F(\phi) = \phi + O(\phi^2)[/itex] and [itex]G(\phi) = K + \tfrac{1}{2}\phi^2 + O(\phi^3)[/itex], for some unknown constantK. To be more precise:

(F3).....[tex]F(0) = 0[/tex]

(F4).....[tex]F'(0) = 1[/tex]

(F5).....[tex]G'(0) = 0[/tex]

(F6).....[tex]G''(0) = 1[/tex]

So, considering our two colliding particles of massmas described above, moving with rapidities [itex]\phi[/itex] and [itex]-\phi[/itex] relative to the final rest frame, let the mass of the final particle be

(F7).....[tex]M = 2 m H(\phi)[/tex]

for some smooth functionHto be determined. In the Newtonian limit, mass is conserved so

(F8).....[tex]H(0) = 1[/tex]

Note also that

(F9).....[tex]H(\phi) > 0[/tex]

We have, relative to the final rest frame, conservation of momentum and energy

(F10).....[tex]m c F(\phi) + m c F(-\phi) = 2 m c H(\phi) F(0)[/tex]

(F11).....[tex]m c^2 G(\phi) + m c^2 G(-\phi) = 2 m c^2 H(\phi) G(0)[/tex]

Note that either of the above equations implies thatHis an even function, and so, in particular,

(F12).....[tex]H'(0) = 0[/tex]

More generally, relative to a frame in which the final particle has rapidity [itex]\psi[/itex], we have

(F13).....[tex]m c F(\psi + \phi) + m c F(\psi - \phi) = 2 m c H(\phi) F(\psi)[/tex]

(F14).....[tex]m c^2 G(\psi + \phi) + m c^2 G(\psi - \phi) = 2 m c^2 H(\phi) G(\psi)[/tex]

making use of the additivity of rapidities.

To summarise, bothFandGare solutions forfof the equation

(F15).....[tex]f(\psi + \phi) + f(\psi - \phi) = 2 H(\phi) f(\psi)[/tex]

each satisfying the appropriate boundary conditions as above.

(Due to length, the derivation will continue in the next post)

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# Derivation of momentum & energy formulas

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