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Derivation of momentum & energy formulas

  1. Oct 29, 2009 #1


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    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.

    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]​


    (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 momentum p and energy E are given by

    (F1).....[tex]p = m c F(\phi)[/tex]
    (F2).....[tex]E = m c^2 G(\phi)[/tex]​

    for some smooth functions F and G to 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 constant K. 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 mass m as 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 function H to 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 that H is 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, both F and G are solutions for f of 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)

    Attached Files:

  2. jcsd
  3. Oct 29, 2009 #2


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    (Continued from previous post)

    Relation between momentum and energy

    From (A6) and (A7) we have

    [tex]\frac{dp}{dt} = F = \frac{dE}{dx} = \frac{dE/dt}{dx/dt}[/tex]
    [tex]\frac{dp}{d\phi} \frac{d\phi}{dt} = \frac{(dE/d\phi) (d\phi/dt) }{c \tanh \phi} [/tex]​


    [tex] \frac{dp}{d\phi} c \tanh \phi = \frac{dE}{d\phi} [/tex]​

    whenever [itex]d\phi/dt \neq 0[/itex]. On grounds of continuity it must always be true.

    Substituting (F1) and (F2) we have

    (F16).....[tex] F'(\phi) \tanh \phi = G'(\phi) [/tex]​

    Solving the equations

    This section is just maths, no physics involved.

    Apply [itex]\partial^2/\partial\phi^2[/itex] to (F15):

    (M1).....[tex]f''(\psi + \phi) + f''(\psi - \phi) = 2 H''(\phi) f(\psi)[/tex]​

    Apply [itex]\partial^2/\partial\psi^2[/itex] to (F15):

    (M2).....[tex]f''(\psi + \phi) + f''(\psi - \phi) = 2 H(\phi) f''(\psi)[/tex]​

    Equate the two and divide by [itex]H(\phi)[/itex] which is never zero.

    (M3)....[tex] f''(\psi) = \frac{H''(\phi)}{H(\phi)} f(\psi) [/tex]​

    As the left hand side is independent of [itex]\phi[/itex] then

    [tex] \frac{H''(\phi)}{H(\phi)}[/tex]​

    must be some constant k i.e.

    (M4)....[tex] H''(\phi) = k H(\phi)}[/tex]
    (M5)....[tex] f''(\psi) = k f(\psi)}[/tex]​

    As the solution for f = G must satisfy G''(0) = 1, we can reject k = 0.

    If k < 0 the solution for H is

    (M6)....[tex] H(\phi) = H(0) \cos(\phi\sqrt{|k|}) + \frac{H'(0)}{\sqrt{|k|}} \sin(\phi\sqrt{|k|}) [/tex]​

    which is incompatible with the requirement that [itex] H(\phi) > 0 [/itex].

    Therefore [itex] k = \alpha^2 > 0 [/itex], for some non-zero constant [itex]\alpha[/itex] yet to be determined, giving the solutions

    (M7)....[tex] H(\phi) = H(0) \cosh(\alpha\phi) + \frac{H'(0)}{\alpha}\sinh(\alpha\phi) = \cosh(\alpha\phi) [/tex]

    (M8)....[tex] f(\psi) = f(0) \cosh(\alpha\psi) + \frac{f'(0)}{\alpha} \sinh(\alpha\psi) [/tex]​

    When solving for momentum f = F, we have F(0)= 0 and F'(0) = 1, whence

    (M9)....[tex] F(\psi) = \frac{1}{\alpha} \sinh (\alpha \psi) [/tex]​

    When solving for momentum f = G, we have G'(0)= 0 and G''(0) = 1, so [itex]G(0) = 1/\alpha^2 [/itex], whence

    (M10)....[tex] G(\psi) = \frac{1}{\alpha^2} \cosh (\alpha \psi) [/tex]​

    All that remains is to determine the value of [itex]\alpha[/itex], and for this we use (F16) which gives

    (M11)....[tex] \cosh (\alpha \phi) \tanh \phi = \frac{1}{\alpha} \sinh (\alpha \phi) [/tex]​


    (M12)....[tex] \tanh \phi = \frac{1}{\alpha} \tanh (\alpha \phi) [/tex]​

    and so [itex]\alpha = 1[/itex].


    [tex] p = m c \sinh \phi [/tex]

    [tex] E = m c^2 \cosh \phi [/tex]​

    which are the standard equations for momentum and energy expressed as functions of rapidity. Converted into velocity they take the more familiar form

    [tex] p = \gamma m v [/tex]

    [tex] E = \gamma m c^2 [/tex]​


    My main question is, are assumptions (A1) to (A8) above sensible?
  4. Oct 29, 2009 #3


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    The most elegant and general way of thinking about conserved quantities is based on Hamilton's principle and symmetries. I don't know if you know this approach but I'll outline it anyway.

    Let ξ be a vector which represents an infinitesimal translation of spacetime in an arbitrary direction. Let γ be the actual path of a system in spacetime from a to b. Symmetry under spacetime translations implies that a neighboring path γ' (γ translated along ξ) from a + εξ to b + εξ must have the same action. Now create a new path by joining a to a + εξ and b + εξ to b, which together with γ' gives a path ζ from a to b, but which is not the actual path. This is an alternative path which is infinitesimally close to the actual path from a to b, so it must have the same action, which follows from the fact that S is stationary near γ. Clearly, this implies that dS(a)⋅ξ = dS(b)⋅ξ, i.e. the rate of change of S along the segments from a to a + εξ and b to b + εξ must be equal. This must hold for all a and b, and arbitrary tranlations ξ, so it follows that dS must be constant along the path, i.e. it is conserved. dS is the four-momentum, with components ∂S/∂xµ; the time component is -E, and the space components constitute the ordinary 3-momentum. This argument is easily seen to apply to multiple particles, where dS is replaced by ∑dSi, i.e. the quantity is additive.

    For a free particle, S is -m times the proper time, so we see that -E = -m ∂τ/∂t = -m/√(1 - v2), and similarly for the other components.

    I have done this derivation here geometrically, without assuming Noether's theorem, but if the theorem is assumed the conserved quantity corresponding to the symmetry can be written down immediately.
    Last edited: Oct 30, 2009
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