# How can one motivate the relativistic momentum?

Motivating the classical momentum $\mathbf{p} = m\mathbf{v}$ is quite easy: it is meant to represent the quantity of motion of the particle, and since the mass is one measure of quantity of matter it should be proportional to mass (how much thing is moving) and should be proportional to velocity (how fast and to where it is moving).

Now, in Special Relativity the momentum changes. The new quantity of motion becomes

$$\mathbf{p} = \dfrac{m\mathbf{v}}{\sqrt{1-\dfrac{v^2}{c^2}}}$$

Or, using $\gamma$ the Lorentz factor $\mathbf{p} = \gamma(v) m\mathbf{v}$ where I write $\gamma(v)$ to indicate that the velocity is that of the particle relative to the frame in which the movement is being observed.

The need for this new momentum is because the old one fails to be conserved and because using the old one in Newton's second law leads to a law which is not invariant under Lorentz transformations. So the need for a new momentum is perfectly well motivated.

What I would like to know is how can one motivate that the correct choice for $\mathbf{p}$ is the $\gamma(v)m\mathbf{v}$. There are some arguments using the mass: considering a colision, requiring momentum to be conserved, transform the velocity and then find how mass should transform. Although this work, it doesn't seem natural, and it is derived in one particular example.

On my book there's even something that Einstein wote saying that he didn't think it was a good idea to try transforming the mass from $m$ to $M = \gamma(v)m$, that it was better to simply keep $\gamma$ on the new momentum without trying to combine it with the mass.

So I would like to know: without resorting to arguments based on transformation of the mass, how can one motivate the new form of momentum that works for special relativity?

• I've used arxiv.org/abs/physics/0402024 instead of the limiting glancing collision argument. I don't think it worked terribly well for the students I presented it to, but I like it. Mind you the authors claim to not used the work-energy theorem, but it's differential limit is used. – dmckee Mar 18 '15 at 21:57
• It occurs to me that your question might be simpler than that. Something along the lines of "How do I convince the students that we might need a new momentum rule?". In that case I go with "Change in velocity is the thing that controls change in momentum, but we already have a new velocity composition rules, so we may need a new rule for the evolution of momentum." – dmckee Mar 18 '15 at 22:43
• Actually I think your problem starts with your motivation of classical momentum. What you are presenting as motivations, is actually backwards. It is more what you read into it after you know that momentum is mv. That total momentum is a conserved quantity is known after showing that Newtons laws with translationinvariant potential lead to this conserved quantity. – lalala Apr 18 at 10:28

What I would like to know is how can one motivate that the correct choice for p is the γ(v)mv

In Newtonian mechanics, the momentum of a particle of mass $m$ is given by

$$\mathbf p = m\frac{d {\mathbf r}}{dt} = m \mathbf v$$

where $\mathbf r$ is the position vector and $t$ is a universal parameter. However, in relativistic mechanics, $t$ is a coordinate, not a parameter, and is thus a component of a four-vector, the four-position $\mathbf R = (ct, \mathbf r)$.

The four-velocity is then defined as

$$\mathbf U = \frac{d \mathbf R}{d \tau} = \frac{d \mathbf R}{d t}\frac{dt}{d\tau} = \frac{d \mathbf R}{d t}\gamma = \gamma (c, \mathbf v)$$

where $\tau$ is the proper time parameter. In analogy with Newtonian mechanics, the four-momentum is then

$$\mathbf P = m \mathbf U = \gamma m(c, \mathbf v)$$

and then we see that the relativistic momentum is simply the spatial part of the four-momentum.

Special relativty is about Minkowski spacetime. A line element is given by $$ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2$$ A free particle will move on a straight line, that is, it will minimize the path length $$L = \int ds = \int \sqrt{c^2 \left(\frac{dt}{d\lambda}\right)^2 - \left(\frac{dx}{d\lambda}\right)^2 - \left(\frac{dy}{d\lambda}\right)^2 - \left(\frac{dz}{d\lambda}\right)^2} \ d\lambda$$ where $\lambda$ is an arbitrary parametrisation of the path. We set $$I(\lambda) := \sqrt{c^2 \left(\frac{dt}{d\lambda}\right)^2 - \left(\frac{dx}{d\lambda}\right)^2 - \left(\frac{dy}{d\lambda}\right)^2 - \left(\frac{dz}{d\lambda}\right)^2}$$

The Euler-Lagrange-equations give: $$\frac{d}{d\lambda} \left( \frac{\delta I}{\delta \left( \frac{d(ct)}{d\lambda} \right)} \right) - \frac{\delta I}{\delta (ct)} = 0$$ $$\frac{d}{d\lambda} \left( \frac{\delta I}{\delta \left( \frac{dx}{d\lambda} \right)} \right) - \frac{\delta I}{\delta x} = 0$$ etc.

Therefore if we evaluate the differentials and multiply by $I$: $$c \frac{d^2t}{d\lambda^2} = 0$$ $$- \frac{d^2x}{d\lambda^2} = 0$$ $$- \frac{d^2y}{d\lambda^2} = 0$$ $$- \frac{d^2z}{d\lambda^2} = 0$$

Now we parametrisate by proper time $d\lambda = d\tau = \frac{1}{c} ds$, introduce $x_\mu = (ct,-x,-y,-z)^T$ and multiply by $m$. This leaves us $$m \frac{d^2x_\mu}{d\tau^2} = 0$$ the covariant equation of motion of a free particle if we combine all 4 equations. Using $$d\tau = \frac{1}{c} ds = \frac{1}{c} \sqrt{c^2 dt^2 - dx^2 - dy^2 - dz^2} \\ = \frac{1}{c} dt \sqrt{c^2 -\left(\frac{dx}{dt}\right)^2 - \left(\frac{dy}{dt}\right)^2 - \left(\frac{dz}{dt}\right)^2} = dt \frac{1}{\gamma(v)}$$ to express by system time $t$, this is equal to: $$\frac{d}{dt} \left( m \cdot \gamma(v) \cdot \frac{dx_\mu}{dt} \right) \hat{=} \frac{d}{dt} \left( \matrix{\gamma(v) \cdot m c \\ - m \cdot \gamma(v) \cdot \frac{dx}{dt} \\ - m \cdot \gamma(v) \cdot \frac{dy}{dt} \\ - m \cdot \gamma(v) \cdot \frac{dz}{dt}} \right) \hat{=} \frac{d}{dt} \left( \matrix{ \gamma(v) \cdot m c \\ - m \cdot \gamma(v) \cdot \vec{v} } \right) = \left( \matrix{0 \\ \vec{0}} \right)$$

The new dynamical quantities are $\vec{p} = m \cdot \gamma(v) \cdot \vec{v}$, which we may call momentum, and $\frac{E}{c} = \gamma(v) \cdot m c$ where $E$ is energy.

One can now try to add forces on the right side of the equation of motion.

In short: If we start by the assumption that a free particle moves on a straight line in Minkowski space, we are led to new dynamical quantities $\vec{p}$ and $E$ that can be used to describe properties of motion in a similar way as they did in newtonian mechanics.

If one tries to describe nature on basis of tensors, the quantity $\gamma(v) \cdot m$ is not a "good" quantity, as it does not transform like a tensor (e.g scalar). However the quantities $m$ and $(\frac{E}{c}, \vec{p})^T$ are tensors (scalar and contravariant tensor of first rank). So these are the "better" quantities according to the criterion.

• This is neat. Alas, the student for whom I have to motivate this stuff have not seen Lagrangian or Hamiltonian mechanics yet. – dmckee Mar 19 '15 at 2:57
• @dmckee: Maybe you could motivate it by the fact that, according to fermat's principle, light traveles along the line of minmal optical path length. This leads to effects like refraction that are calculated on the same basis. Perhaps this makes it a bit more familiar to your students. – image Mar 19 '15 at 10:54
• Well, that's how I motivate Lagrangian mechanics: first the mirror rule and then refraction from the minimum time principle (using a lifeguard on a beach as a model for refraction), but I don't see launching into a long discussion to motivate the motivation. – dmckee Mar 19 '15 at 14:52

The Hamiltonian $H$ generates time translations and the momentum $\mathbf p$ generates space translations. In a relativistic theory time and space can mix, so we should consider the 4-vector $$p^\mu = (H, \mathbf p).$$ Now whatever the $\mathbf p$ is in terms of mass, velocity and so on, certainly it is $\mathbf 0$ in a rest frame. Then so that there can be momentum in any frame at all, the time component can't vanish. Thus in a rest frame $$p^\mu = (m, \mathbf 0)$$ for some quantity $m$. Boost to a general frame to conclude that $$p^\mu = (\gamma m, \gamma m \mathbf v).$$ Comparison with the non-relativistic limit shows that $m$ is the mass.