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The Brute Force Way

Recall that the Lagrangian path to the dynamics of a particle (which is most easily made covariant, since it uses $(q(t),\dot{q}(t),t)$ as its variables) is based on the Action

\begin{displaymath}
A = \int_{t_0}^{t_1} L(q(t),\dot{q}(t),t)dt .
\end{displaymath} (17.3)

By requiring that $A$ be an extremum as a functional of the system trajectory, we obtain the Euler-Lagrange equations
\begin{displaymath}
\frac{d~}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i}
\right) = 0 .
\end{displaymath} (17.4)

These are equivalent to Newton's law for suitable definitions of $L$ and the force.

The simplest way to make this relativistic is to express it in terms of the proper time and then require that the action $A$ be extremal in all frames. Then,

\begin{displaymath}
A = \int_{\tau_0}^{\tau_1} \gamma L d\tau
\end{displaymath} (17.5)

is the action, since $dt = \gamma d\tau$. We now must remark
  1. If the extremal condition is to be invariant with respect to LT's, then $A$ must be invariant (and hence a 4-scalar);
  2. Therefore (since $d \tau$ is invariant) $\gamma L$ must be invariant;
  3. Finally, since $\gamma$ is just a number, $L$ must be a 4-scalar.
This final conclusion greatly constrains the possible forms of $L$.

Note well that it is not clear that this argument (from Jackson) is valid. $\gamma$, while a number, is not invariant under boosts - indeed, it is parametrically related to the boost parameter! It is also perfectly clear that the first statement is false - while it is true that if $A$ is a four-scalar with respect to boosts that its extremums must be preserved, it is equally true that this condition is not necessarily unique - all that is necessary is that a boost monotonically scale the action in such a way that the extremal property is preserved!

A weaker (but still sufficient) argument might then be:

If $L$ is a 4-scalar, and $\gamma$ is a monotonic function independent of the 4-coordinates, 4-velocities, and $\tau$, then the property of a given trajectory resulting in an extremum of the action is preserved.
In my opinion this is clearer and still adequate for our purposes. $L$ being a 4-scalar (0th rank tensor in 4-space w.r.t. the Lorentz transformation) is sufficient to produce an invariant extremum of the action $A$, even if the numerical values of $A$ vary under a boost. To prove that it is also necessary very likely involves an exercise in the calculus of variations that distributes the derivatives over $\gamma L$ - similar exercises are already in the homework for this chapter in other contexts.

Either way, we will now assert that the Lagrangian of a free particle must be a 4-scalar (and hence must be formed out of the full contraction of tensors of higher rank), and will remain alert in the work below for any sort of inconsistency that might be related to $\gamma$.

Obviously, we want it to reproduce the classical non-relativistic theory in the appropriate limit, that is, a free particle should have constant energy and momentum, or, equivalently, 4-velocity. The simplest (not ``only'' as Jackson states) Lorentz invariant function of the 4-velocity is it's quadratic form:

\begin{displaymath}
U^\alpha U_\alpha = c^2
\end{displaymath} (17.6)

Of course, any polynomial functional of this quadratic is also a possible scalar, but they are not the thing to try first. Thus a reasonable guess for the Lagrangian is
\begin{displaymath}
L = (\mbox{\rm constant})c^2\gamma^{-1} = - m c^2 \sqrt{1 -
\frac{u^2}{c^2}} .
\end{displaymath} (17.7)

If we now crunch through the Euler-Lagrange equation we find that this choice for the constant leads to

\begin{displaymath}
\frac{d~}{dt}(\gamma m {\bf u}) = 0
\end{displaymath} (17.8)

which is indeed Newton's law for a free particle, but with the relativistic form of the three-momentum.

If one chooses a frame where the particle is initially at rest, a trajectory where it remains at rest will yield the least action (you should check this). This is because $\gamma^{-1}$ is maximal when $\beta = 0$ (and the Lagrangian has a minus sign).

Now, suppose that the charged particle is in a electromagnet potential. If it were moving slowly in a scalar potential $\Phi$ only, its potential energy would be $V = q \Phi$. The non-relativistic Lagrangian in this case should be just $T - V$ (where $T$ is the free particle Lagrangian). The interaction part of the relativistic Lagrangian must therefore reduce to $-q\Phi$ in this non-relativistic limit.

We must find a Lorentz invariant (scalar) form for $\gamma L_{\rm int}$ that reduces to $-q\Phi$ for non-relativistic velocities. Since $\Phi$ is the time component of a four vector potential, we can guess that the correct generalization must involve the four vector potential $A^\alpha$. Since it must be a scalar, it must involve the scalar product of $A^\alpha$ with some four vector(s). The only ones avaliable are $x^\alpha$ and $U^\alpha$.

The correct $\gamma L_{\rm int}$ depends only on the $U^\alpha$. If it depended on the coordinates as well, then the physics would not be translationally invariant and the results of our calculation might well depend on where we chose the origin. This does not seem reasonable. Once again, this does not uniquely determine it, it only determines the simplest (linear) form to within a sign and a constant:

\begin{displaymath}
\gamma L_{\rm int} = -\frac{q}{c} U_\alpha A^\alpha
\end{displaymath} (17.9)

or
\begin{displaymath}
L_{\rm int} = - q \Phi + \frac{q}{c} \vec{u} \cdot \vec{A}.
\end{displaymath} (17.10)

There could be additional terms involving polynomials of this quantity, the product $A^\alpha A_\alpha$ (which is indeed present in some theories) and other scalar reductions of field and charge/field tensor quantities. Linearity, in either the vector potential or the velocity, is an axiom and not logically necessary.

The complete relativistic Lagrangian for a charged particle is thus

\begin{displaymath}
L = -mc^2 \sqrt{1 - \frac{u^2}{c^2}} + \frac{q}{c} \vec{u} \cdot \vec{A} -
q \Phi.
\end{displaymath} (17.11)

It should take you about one hour to show that this yields the correct relativistic Lorentz force law. The free particle part is obvious, the electric field is obvious. You will have to work a bit, using
\begin{displaymath}
\frac{d~}{dt} = \left( \frac{\partial ~}{\partial t} + \vec{u} \cdot \mbox{\boldmath$\nabla$}\right)
\end{displaymath} (17.12)

to squeeze $- \vec{u} \times (\mbox{\boldmath$\nabla$}\times \vec{A})$ out of the remainder. I suggest that you simply work out the terms by expanding them as far as they go and reassembling the pieces, but some of you may know enough vector algebra to do it better ways. This will be on the next assignment, so feel free to start.

The canonical momentum $\vec{P}$ conjugate to the position coordinates $x$ is obtained (as usual) from

\begin{displaymath}
P_i = \frac{\partial L}{\partial u_i} = \gamma m u_i + \frac{q}{c} A_i.
\end{displaymath} (17.13)

This result,
\begin{displaymath}
\vec{P} = \vec{p} + \frac{q}{c} \vec{A}
\end{displaymath} (17.14)

(where $\vec{p}$ is the relativistic kinetic momentum of the particle) is extremely important to remember, as it is a necessary ingredient in the construction of either a quantum theory or an elegant classical theory. Placing the particle in a field alters its canonical ``momentum''.

We make the Hamiltonian out of the Lagrangian and the canonical momentum via

\begin{displaymath}
H = \vec{P} \cdot \vec{u} - L
\end{displaymath} (17.15)

The basic result here has too many variables. We must eliminate $\vec{u}$ in favor of $\vec{A}$ and $\vec{P}$. Note that
\begin{displaymath}
\vec{u} = \frac{c\vec{P} - q \vec{A}}{\sqrt{\left( \vec{P} - \frac{q}{c}
\vec{A} \right)^2 + m^2 c^2} }
\end{displaymath} (17.16)

(something that is a wee chore to prove, of course, but it is straightforward algebra). With even more tedious algebra, you can show that the Hamiltonian is:
\begin{displaymath}
H = \sqrt{(c \vec{P} - q \vec{A})^2 + m^2c^4} + q \Phi = W.
\end{displaymath} (17.17)

From this result, Hamilton's equations of motion should reproduce the Lorentz force law. See that it does (although the relationship between the EL equations and Hamilton's equations makes the result obvious). Note that if we interpret the Hamiltonian (as usual) as the total energy $W$ of the particle, this result is related to the free particle energy by $\vec{p} \to (\vec{P} -
\frac{q}{c} \vec{A})$ and the addition of the scalar potential energy $q
\Phi$. This is actually just a single change in the four-vector momentum:

\begin{displaymath}
(W - q\Phi)^2 - (c \vec{P} - q \vec{A})^2 = m^2c^4 = p^\alpha p_\alpha
\end{displaymath} (17.18)

(which has the usual form if
\begin{displaymath}
p^\alpha = \left( \frac{E}{c}, \vec{p} \right) = \left( \frac{1}{c}(W -
e\Phi), \vec{P} - \frac{q}{c} \vec{A} \right)
\end{displaymath} (17.19)

). This also makes the invariance properties of the Hamiltonian manifest.

It is really annoying to obtain the invariance properties of things after the fact. It is also annoying (although perhaps useful) to have the three vector coordinates $\vec{x},\vec{u}$ hanging around at this point. So let us rederive these results using only four-vectors and suitable scalar reductions.


next up previous contents
Next: The Elegant Way Up: Covariant Field Theory Previous: Covariant Field Theory   Contents
Robert G. Brown 2013-01-04