- #1

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- Summary:
- This derivation uses collisions in 1 dimension, plus rapidity

I've been noodling around with derivations of the relativistic energy and momentum, and I almost got it down to just a few lines. But not quite.

I'm going to work in one spatial dimension, for simplicity (even though some derivations require a second spatial dimension)

Let's assume that there is an energy associated with a moving object of mass m given by:

##E=m g(v)##

and a corresponding momentum

##p= m f(v)##

We want to recover the nonrelativistic forms for the case ##\frac{v}{c} \rightarrow 0##, so we should have:

##g(v)=g(0)+\frac{1}{2} v^2+ ## higher-order terms in vc

##f(v)=v+ ## higher-order terms

The key insight is to look at conservation of energy and momentum in different frames. For that reason, it's actually more convenient to switch from velocity v to "rapidity" ##\theta##, related through ##v= c tanh(\theta)## (##tanh## is the hyperbolic tangent). In terms of rapidity, the Lorentz dilation factor ##\gamma = \dfrac{1}{1-\frac{v^2}{c^2}} = cosh(\theta)##.

The reason rapidity is more convenient is the addition law. Under a change of rest frames, velocity changes as follows: ##v′=\dfrac{v+u}{1+\frac{uv}{c^2}}##, where ##u## is the relative velocity between frames. Kind of messy. But if we let ##v= c tanh(\theta)## and ##u=c tanh(\phi)## and let ##v′= c tanh(\theta')##, then the addition law becomes just: ##\theta' = \theta + \phi##.

Since for small values of ##\theta##, ##tanh(\theta)\approx \theta## and so ##v \approx c \theta##, we can rewrite our limiting cases for g and f in terms of ##\theta##:

##g(\theta) \approx g(0)+\frac{1}{2} c^2 \theta^2##

##f(\theta) \approx c \theta##

Let's imagine a collision in which a number of masses collide inelastically to form a large composite mass. Let's choose a frame in which the large mass is at rest after the collision. Then conservation of energy and momentum become:

##\sum_j m_j g(\theta_j) = M g(0)##

##\sum_j m_j f(\theta_j) = 0##

Now, what we do is shift to another frame traveling at relative speed ##u = tanh(\phi)## relative to the first. Let's assume ##\phi## is small, so that we can use a Taylor series about ##\phi## and keep just the first few terms. (Note: we are thus assuming that ##M## is moving nonrelativistically, but not the smaller masses that collided.)

##\sum_j m_j g(\theta_j + \phi) = M g(\phi) \approx M g(0) + \frac{1}{2} M c^2 \phi^2##

##\sum_j m_j f(\theta_j + \phi) = M f(\phi) \approx M c \phi##

At this point, let's also expand the left-hand sides in powers of ##\phi##:

##g(\theta_j + \phi) \approx g(\theta_j) + g'(\theta_j) \phi + \frac{1}{2} g''(\theta_j) \phi^2##

##f(\theta_j + \phi) \approx f(\theta_j) + f'(\theta_j) \phi##

where ##'## means take the derivative of the function with respect to its argument.

Equating equal powers of ##\phi## gives:

We can plug equation 1 into equation 3 to get:

##\sum_j m_j g''(\theta_j) = \frac{c^2}{g(0)} \sum_j m_j g(\theta_j) ##

This has to be true for any collection of masses, so the only solution is if for each particle,

##g''(\theta) = \frac{c^2}{g(0)} g(\theta)##

Equation 5 is just like equation 3, except for a factor of ##c##, so we also conclude:

##f'(\theta) = \frac{c}{g(0)} g(\theta) = \frac{1}{c} g''(\theta)##

This implies that ##f(\theta) = \frac{1}{c} g'(\theta)##. (There is an arbitrary constant involved in integrating, but it must be zero if ##f(0) = 0##, which must be true for momentum).

So the equation for ##g''(\theta)## has an immediate solution:

##g(\theta) = g(0) cosh(\frac{c}{\sqrt{g(0}} \theta)##

Since ##cosh(\theta) = \gamma##, we would be home free if we could show that ##g(0) = c^2##. But information about collisions doesn't seem to give enough information to conclude this. (This post is about collisions in one dimension. There is an alternative derivation in 2 dimensions that allows you to deduce this.)

But there is a final trick up my sleeve, which is the work-energy equation. In one dimension.

##\Delta E = F \Delta x = \dfrac{\Delta p}{\Delta t} \Delta x = \Delta p \dfrac{\Delta x}{\Delta t} = \dfrac{\Delta p}{\Delta \theta} \Delta \theta \dfrac{\Delta x}{\Delta t} ##

Or dividing through by ##\Delta \theta## and taking the limit as ##\Delta t \rightarrow 0## and ##\Delta \theta \rightarrow 0##

##\dfrac{dE}{d\theta} = \dfrac{dp}{d\theta} v## where I've used ##\frac{dx}{dt} = v##.

Since in terms of ##\theta##, ##v = c \tanh(\theta)##, this becomes:

##\dfrac{dE}{d\theta} = \dfrac{dp}{d\theta} c tanh(\theta)##.

In terms of our functions ##g## and ##f##, this implies

##m \dfrac{dg}{d\theta} = m \dfrac{df}{d\theta} c tanh(\theta)##.

Since ##f(\theta) = \frac{1}{c} g'(\theta)##, we can rewrite this in terms of ##f##

##c f(\theta) = \dfrac{df}{d\theta} c tanh(\theta)##

Which implies that

##\dfrac{\frac{df}{d\theta}}{f} = \frac{1}{tanh(\theta)} = \dfrac{cosh(\theta)}{sinh(\theta)}##

which has the solution:

##f(\theta) = K sinh(\theta)##

for some constant ##K##. The nonrelativistic limit is ##f(\theta) \approx c \theta## tells us that ##K = c##.

So there we have it:

##f(\theta) = c sinh(\theta)##

Since ##f(\theta) = \frac{1}{c} g'(\theta)##, this implies:

##g'(\theta) = c^2 sinh(\theta)##, so ##g(\theta) = c^2 cosh(\theta)## (the constant must be zero from the fact that ##g''(\theta) \propto g(\theta)##)

So we conclude the energy-momentum functions for relativity:

##E = m g(\theta) = m c^2 cosh(\theta) = m c^2 \gamma##

##p = m f(\theta) = mc sinh(\theta) = m c tanh(\theta) cosh(\theta) = m v \gamma##

I'm going to work in one spatial dimension, for simplicity (even though some derivations require a second spatial dimension)

Let's assume that there is an energy associated with a moving object of mass m given by:

##E=m g(v)##

and a corresponding momentum

##p= m f(v)##

We want to recover the nonrelativistic forms for the case ##\frac{v}{c} \rightarrow 0##, so we should have:

##g(v)=g(0)+\frac{1}{2} v^2+ ## higher-order terms in vc

##f(v)=v+ ## higher-order terms

The key insight is to look at conservation of energy and momentum in different frames. For that reason, it's actually more convenient to switch from velocity v to "rapidity" ##\theta##, related through ##v= c tanh(\theta)## (##tanh## is the hyperbolic tangent). In terms of rapidity, the Lorentz dilation factor ##\gamma = \dfrac{1}{1-\frac{v^2}{c^2}} = cosh(\theta)##.

The reason rapidity is more convenient is the addition law. Under a change of rest frames, velocity changes as follows: ##v′=\dfrac{v+u}{1+\frac{uv}{c^2}}##, where ##u## is the relative velocity between frames. Kind of messy. But if we let ##v= c tanh(\theta)## and ##u=c tanh(\phi)## and let ##v′= c tanh(\theta')##, then the addition law becomes just: ##\theta' = \theta + \phi##.

Since for small values of ##\theta##, ##tanh(\theta)\approx \theta## and so ##v \approx c \theta##, we can rewrite our limiting cases for g and f in terms of ##\theta##:

##g(\theta) \approx g(0)+\frac{1}{2} c^2 \theta^2##

##f(\theta) \approx c \theta##

Let's imagine a collision in which a number of masses collide inelastically to form a large composite mass. Let's choose a frame in which the large mass is at rest after the collision. Then conservation of energy and momentum become:

##\sum_j m_j g(\theta_j) = M g(0)##

##\sum_j m_j f(\theta_j) = 0##

Now, what we do is shift to another frame traveling at relative speed ##u = tanh(\phi)## relative to the first. Let's assume ##\phi## is small, so that we can use a Taylor series about ##\phi## and keep just the first few terms. (Note: we are thus assuming that ##M## is moving nonrelativistically, but not the smaller masses that collided.)

##\sum_j m_j g(\theta_j + \phi) = M g(\phi) \approx M g(0) + \frac{1}{2} M c^2 \phi^2##

##\sum_j m_j f(\theta_j + \phi) = M f(\phi) \approx M c \phi##

At this point, let's also expand the left-hand sides in powers of ##\phi##:

##g(\theta_j + \phi) \approx g(\theta_j) + g'(\theta_j) \phi + \frac{1}{2} g''(\theta_j) \phi^2##

##f(\theta_j + \phi) \approx f(\theta_j) + f'(\theta_j) \phi##

where ##'## means take the derivative of the function with respect to its argument.

Equating equal powers of ##\phi## gives:

- ##\sum_j m_j g(\theta_j) = M g(0)##
- ##\sum_j m_j g'(\theta_j) = 0##
- ##\sum_j m_j g''(\theta_j) = M c^2##
- ##\sum_j m_j f(\theta_j) = 0##
- ##\sum_j m_j f'(\theta_j) = Mc##

We can plug equation 1 into equation 3 to get:

##\sum_j m_j g''(\theta_j) = \frac{c^2}{g(0)} \sum_j m_j g(\theta_j) ##

This has to be true for any collection of masses, so the only solution is if for each particle,

##g''(\theta) = \frac{c^2}{g(0)} g(\theta)##

Equation 5 is just like equation 3, except for a factor of ##c##, so we also conclude:

##f'(\theta) = \frac{c}{g(0)} g(\theta) = \frac{1}{c} g''(\theta)##

This implies that ##f(\theta) = \frac{1}{c} g'(\theta)##. (There is an arbitrary constant involved in integrating, but it must be zero if ##f(0) = 0##, which must be true for momentum).

So the equation for ##g''(\theta)## has an immediate solution:

##g(\theta) = g(0) cosh(\frac{c}{\sqrt{g(0}} \theta)##

Since ##cosh(\theta) = \gamma##, we would be home free if we could show that ##g(0) = c^2##. But information about collisions doesn't seem to give enough information to conclude this. (This post is about collisions in one dimension. There is an alternative derivation in 2 dimensions that allows you to deduce this.)

But there is a final trick up my sleeve, which is the work-energy equation. In one dimension.

##\Delta E = F \Delta x = \dfrac{\Delta p}{\Delta t} \Delta x = \Delta p \dfrac{\Delta x}{\Delta t} = \dfrac{\Delta p}{\Delta \theta} \Delta \theta \dfrac{\Delta x}{\Delta t} ##

Or dividing through by ##\Delta \theta## and taking the limit as ##\Delta t \rightarrow 0## and ##\Delta \theta \rightarrow 0##

##\dfrac{dE}{d\theta} = \dfrac{dp}{d\theta} v## where I've used ##\frac{dx}{dt} = v##.

Since in terms of ##\theta##, ##v = c \tanh(\theta)##, this becomes:

##\dfrac{dE}{d\theta} = \dfrac{dp}{d\theta} c tanh(\theta)##.

In terms of our functions ##g## and ##f##, this implies

##m \dfrac{dg}{d\theta} = m \dfrac{df}{d\theta} c tanh(\theta)##.

Since ##f(\theta) = \frac{1}{c} g'(\theta)##, we can rewrite this in terms of ##f##

##c f(\theta) = \dfrac{df}{d\theta} c tanh(\theta)##

Which implies that

##\dfrac{\frac{df}{d\theta}}{f} = \frac{1}{tanh(\theta)} = \dfrac{cosh(\theta)}{sinh(\theta)}##

which has the solution:

##f(\theta) = K sinh(\theta)##

for some constant ##K##. The nonrelativistic limit is ##f(\theta) \approx c \theta## tells us that ##K = c##.

So there we have it:

##f(\theta) = c sinh(\theta)##

Since ##f(\theta) = \frac{1}{c} g'(\theta)##, this implies:

##g'(\theta) = c^2 sinh(\theta)##, so ##g(\theta) = c^2 cosh(\theta)## (the constant must be zero from the fact that ##g''(\theta) \propto g(\theta)##)

So we conclude the energy-momentum functions for relativity:

##E = m g(\theta) = m c^2 cosh(\theta) = m c^2 \gamma##

##p = m f(\theta) = mc sinh(\theta) = m c tanh(\theta) cosh(\theta) = m v \gamma##

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