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Relativistic Energy and Momentum

We seek a relativistic generalization of momentum (a vector quantity) and energy. We know that in the low speed limit, $v « c$,

{\bf p} = m{\bf u}
\end{displaymath} (15.82)

E = E(0) + \frac{1}{2} m u^2
\end{displaymath} (15.83)

where $E(0)$ is a constant allowed by Newton's laws (since forces depend only on energy differences).

The only possible form for this generalization of these equations consistent with our requirement that the laws of nature remain invariant are:

{\bf p} = {\cal M}(u){\bf u}
\end{displaymath} (15.84)

E = {\cal E}(u) ,
\end{displaymath} (15.85)

that is, the mass and the energy must become functions of the speed only, and leave the vector character of the velocity alone. A boost cannot change the direction of the momentum of a particle, and any (scalar) functional variation in its magnitude can be thrown into the ``mass'' term.

This immediately yields the limiting forms:

{\cal M}(0) = m
\end{displaymath} (15.86)

\frac{\partial {\cal E}}{\partial u^2}(0) = \frac{m}{2}
\end{displaymath} (15.87)

where we have assumed that there is no pathology in the functions at the origin.

There are several possible ways to evaluate the full forms of these functions. Jackson's (based on scattering theory) is tedious and conceals the structure of the result. Furthermore, after telling us that selecting clever initial directions with an eye to simplifying the algebra ``lacks motivation'' he derives a result by selecting particular initial directions. The guy loves algebra, what can I say. Feel free to study his approach. It works.

I, on the other hand, am too lazy to spend most of a period deriving a result that is ``obvious'' in the correct notation. I am therefore going to ``give'' you the result and motivate it, and then verify it trivially be expressing it as a four-vector. This works nearly as well and is not anywhere near as painful.

We begin by considering elastic scattering theory. An elastic collision of two identical particles must conserve momentum and energy in all inertial frames. In the center of mass frame (which we will consider to be $K'$)

{\bf p}_{ia}' + {\bf p}_{ib}' = {\bf p}_{fa}' + {\bf p}_{fb}'
\end{displaymath} (15.88)

E_{ia}' + E_{ib}' = E_{fa}' + E_{fb}'
\end{displaymath} (15.89)

relate the intial and final momenta and energy of the two identical particles.


{\bf u}_{ia}' = {\bf v} = - {\bf u}_{ib}'
\end{displaymath} (15.90)

{\bf u}_{fa} = {\bf v}' = {\bf u}_{fb}'
\end{displaymath} (15.91)

by definition in the center of mass system.

A moments quiet reflection (egad, another pun!) should convince you that in terms of the general transformation:

{\cal M}(v){\bf v} - {\cal M}(v){\bf v} = {\cal M}(v'){\bf v}' - {\cal
M}(v'){\bf v}'
\end{displaymath} (15.92)

{\cal E}(v) + {\cal E}(v) = {\cal E}(v') + {\cal E}(v') .
\end{displaymath} (15.93)

For what it is worth, if the collision is elastic and the particles are identical before and after the collision, $v$ = $v'$ and all the mass terms are the same. We will denote the scattering angle in $K'$ as $\theta'$.

Figure 15.4: $\theta ' = 30^\circ $ and $\beta^2 = \frac{1}{3}$. The dashed lines are the results of a Gallilean transformation from $K'$ to $K$. Note that the scattering is more forward than expected because of the Lorentz contraction of the longitudinal distances seen by the particles.

We thus begin with

{\cal M}(v){\bf v} - {\cal M}(v){\bf v} = {\cal M}(v){\bf v} - {\cal
M}(v){\bf v}
\end{displaymath} (15.94)

{\cal E}(v) + {\cal E}(v) = {\cal E}(v) + {\cal E}(v)
\end{displaymath} (15.95)

where $v$ is the speed of the incoming and outgoing particles. Now, ${\cal
M}(v)$ must be a scalar function of $v$, and in the limit $v \to
0$ must turn into
\lim_{ v \rightarrow 0} {\cal M}(v) = m .
\end{displaymath} (15.96)

The only scalar function of $v$ we have encountered so far with this behavior is $\gamma (v)$, so we should sensibly guess
{\cal M}(v) = \gamma(v) m
\end{displaymath} (15.97)

which has the exactly correct limiting behavior.


{\bf p} = \gamma m {\bf u}
\end{displaymath} (15.98)

is a reasonable guess to be the generalization of momentum we seek. It is easy to verify that this is a consistent choice, and that it indeed results in conservation of momentum in all inertial frames.

To get the energy equation, we use the same approach. Recall that a binomial expansion of $\gamma$ is given by

\lim_{ v \rightarrow 0} \gamma(v) = \left( 1 - \frac{v^2}{c^2}
\right)^{-1/2} = 1 + \frac{1}{2} \frac{v^2}{c^2} + \ldots
\end{displaymath} (15.99)

We need to keep the first non-constant term because we recall that physics is always ``independent'' of absolute energy scale. Then it should be clear that
\lim_{v \rightarrow 0}{\cal E}(v) = \gamma(v) {\cal E}(0) \...
...(0) \frac{v^2}{c^2} \approx {\cal E}(0) +
\frac{1}{2} m v^2
\end{displaymath} (15.100)

as it must in order to yield the low velocity limit of kinetic energy if and only if
{\cal E}(0) = mc^2 .
\end{displaymath} (15.101)

There are several questions to be answered at this point, some experimentally and some theoretically. We need to measure the rest masses and theoretically verify that only this transformation correctly preserves the energy momentum conservation laws in elastic collisions as required. Beyond that, there are still some uncertainties. For example, there could in principal be an additional constant energy added to the energy term that was not scaled by $\gamma$ and the laws of physics would still be expressible, since they are not sensitive to absolute energy scale. We will take advantage of that freedom in several instances to add or subtract an infinite theoretical constant in order to make the rest mass come out to the observed experimental mass m. This is called renormalization.

To obtain the same result a different way, we turn to the notation of 4-vectors. We observe that the common factor of $\gamma$ above in both $E$ and ${\bf p}$ also occurs when one makes velocity into a four vector. This suggests that energy and momentum can similarly be made into four vectors that transform like the coordinates under a boost. If we try the combination

$\displaystyle p_0$ $\textstyle =$ $\displaystyle m c U_0 = \frac{E}{c}$ (15.102)
$\displaystyle {\bf p}$ $\textstyle =$ $\displaystyle m {\bf U}$ (15.103)

we see that it works exactly. It results in an invariant
p_0^2 - {\bf p} \cdot {\bf p} = (m_c)^2 .
\end{displaymath} (15.104)

It is easy to see the value of the invariant when $v = 0$; you should verify explicitly that it holds when $v \ne 0$ as well. Practically speaking, it suffices to show that this length is invariant when one wishes to show that its components transform like the coordinates under the action of a boost (why is that?).

The total energy can thus be expressed in terms of the three momentum as

E = \sqrt{c^2p^2 + m^2 c^4} .
\end{displaymath} (15.105)

Finally, it is sometimes convenient to be able to get the velocity of the particle in terms of its energy and momentum
{\bf u} = \frac{c^2 {\bf p}}{E}
\end{displaymath} (15.106)

which follows directly from the definitions.

This completes our review of ``elementary relativity theory''. We shall now proceed to develop the theory in a new, geometric language which is suitable to our much more sophisticated needs. To do this, we will need to begin by generalizing the notion of a four dimensional vector space with a set of transformations that leave an appropriately defined ``length'' invariant.

next up previous contents
Next: The Lorentz Group Up: Special Relativity Previous: Addition of Velocities   Contents
Robert G. Brown 2007-12-28