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The Elegant Way

We can write the free particle Lagrangian using only scalar reductions of suitable 4-vectors:

\begin{displaymath}
L_{\rm free} = - \frac{mc}{\gamma} \sqrt{U_\alpha U^\alpha}
\end{displaymath} (17.20)

(which is still $- mc^2 \gamma^{-1}$). The action is thus
\begin{displaymath}
A = -mc \int_{\tau_0}^{\tau_1} \sqrt{U_\alpha U^\alpha} d\tau .
\end{displaymath} (17.21)

The variations on this action must be carried out subject to the constraint
\begin{displaymath}
U_\alpha U^\alpha = c^2
\end{displaymath} (17.22)

which severely limits the allowed solutions. We write this as
$\displaystyle \frac{d(U_\alpha U^\alpha)}{d\tau}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle \frac{dU_\alpha}{d\tau} U^\alpha + U_\alpha \frac{dU^\alpha}{d\tau}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle \frac{dU_\alpha}{d\tau} g^{\alpha\beta}g_{\beta\alpha}U^\alpha + U_\alpha \frac{dU^\alpha}{d\tau}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle U_\beta \frac{dU^\beta}{d\tau} + U_\alpha \frac{dU^\alpha}{d\tau}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle 2 U_\alpha \frac{dU^\alpha}{d\tau}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle U_\alpha \frac{dU^\alpha}{d \tau}$ $\textstyle =$ $\displaystyle 0$ (17.23)

Now,

\begin{displaymath}
\sqrt{U_\alpha U^\alpha} d\tau = \sqrt{\frac{dx_\alpha}{d\t...
...ha}{d\tau}} d\tau = \sqrt{g^{\alpha\beta} dx_\alpha dx_\beta}
\end{displaymath} (17.24)

which is an infinitesimal length in four-space. The latter expression does not explicitly contain $d \tau$. We can thus parameterize the action in terms of a path-parameter $s$ that increases monotonically with $\tau$ but is otherwise arbitrary. Then
\begin{displaymath}
A = -mc \int_{s_0}^{s_1} \sqrt{g^{\alpha\beta} \frac{dx_\alpha}{ds}
\frac{dx_\beta}{ds}} ds .
\end{displaymath} (17.25)

We are clearly making progress. We have taken a perfectly good expression and made in unrecognizable. To make you a little happier, note that this has now got the form of
\begin{displaymath}
A = \int \tilde{L} ds
\end{displaymath} (17.26)

where $\tilde{L}$ is a scalar ``Lagrangian'' written in terms of an independent free parameter. This might be progress after all, since we have quashed the annoying $\gamma^{-1}$.

If we now do the calculus of variations thing and get the Euler-Lagrange equations in four dimensions:

\begin{displaymath}
\frac{d~}{ds} \left( \frac{ d\tilde{L} }{ \partial \left(
...
...\alpha}{ds} \right) } \right) - \partial^\alpha \tilde{L} = 0
\end{displaymath} (17.27)

(for $\alpha = 0,4$). Applying them to the Langrangian in this action, they turn out to be:
$\displaystyle mc \frac{d~}{ds}
\frac{\partial
\left\{g^{\delta\beta} \frac{dx_\...
...elta}{ds} \right\}^{\frac{1}{2}}}
{\partial\left( \frac{dx_\alpha}{ds} \right)}$ $\textstyle =$ $\displaystyle 0$ (17.28)
$\displaystyle \frac{mc}{2}\frac{d~}{ds} \left\{
\frac{ \frac{dx^\alpha}{ds} + \frac{dx^\alpha}{ds} }
{ \sqrt{\frac{dx_\beta}{ds} \frac{dx^\beta}{ds}} } \right\}$ $\textstyle =$ $\displaystyle 0$ (17.29)
$\displaystyle mc \frac{d~}{ds} \left\{ \frac{\frac{dx^\alpha}{ds}}
{\sqrt{\frac{dx_\beta}{ds} \frac{dx^\beta}{ds}} } \right\}$ $\textstyle =$ $\displaystyle 0.$ (17.30)

This still does not have the constraint above imposed on it. We impose the constraint by identifying $ds$ with $d \tau$ in such a way that the constraint is simultaneously satisfied:

$\displaystyle \sqrt{\frac{dx_\alpha}{ds} \frac{dx^\alpha}{ds} } ds$ $\textstyle =$ $\displaystyle c d\tau$  
$\displaystyle \frac{d~}{d\tau}$ $\textstyle =$ $\displaystyle \frac{c}{\sqrt{\frac{dx_\alpha}{ds}\frac{dx^\alpha}{ds}}}
\frac{d~}{ds}$ (17.31)

(which requires both $ds = d\tau$ and $U_\alpha U^\alpha = c^2$). If you like, this constraint picks out of all possible path parameterizations the one that follows the proper time while keeping the four vector velocity scalar product Lorentz invariant. For free particles this is a lot of work, but it is paid back when we include an interaction.

If we multiply the Euler-Lagrange equation (in terms of $s$) from the left by:

\begin{displaymath}
\frac{c}{\sqrt{\frac{dx_\alpha}{ds}\frac{dx^\alpha}{ds}}}
\end{displaymath} (17.32)

and use the constraint to convert to $\tau$, the result (for the equation of motion) is:
\begin{displaymath}
m\frac{c}{\sqrt{\frac{dx_\beta}{ds}\frac{dx^\beta}{ds}} }
\...
...ds}
\frac{dx^\beta}{ds}} } \frac{d~}{ds}x^\alpha \right\} = 0
\end{displaymath} (17.33)

or
\begin{displaymath}
m \frac{d^2x^\alpha}{d\tau^2} = 0
\end{displaymath} (17.34)

which certainly looks it has the right form.

We can include an interaction. Just as before, $\gamma L_{\rm int}$ must be a Lorentz scalar. When we make a parameterized version of the Lagrangian, the part under the integral must be a 4-scalar. The covariant form of the result is (hopefully obviously)

\begin{displaymath}
A = - \int_{s_0}^{s_1} \left\{ mc \sqrt{g^{\delta\beta}
\f...
...{ds}} + \frac{q}{c}
\frac{dx_\beta}{ds}A^\beta \right\} ds .
\end{displaymath} (17.35)

The ``four Lagrangian'' in this equation is
\begin{displaymath}
\tilde{L} = - \left\{ mc \sqrt{g^{\delta\beta} \frac{d_\del...
...eta}{ds}} + \frac{q}{c} \frac{dx_\beta}{ds}A^\beta \right\}.
\end{displaymath} (17.36)

As before we construct the Euler-Lagrange equation.

$\displaystyle mc \frac{d~}{ds} \left\{ \frac{\frac{dx^\alpha}{ds}}
{\sqrt{\fra...
...c{q}{c}A^\alpha \right\} -
\frac{q}{c}\frac{x_\beta}{ds}\partial^\alpha A^\beta$ $\textstyle =$ $\displaystyle 0$ (17.37)

Again we multiply through from the left by

\begin{displaymath}
\frac{c}{\sqrt{\frac{dx_\alpha}{ds}\frac{dx^\alpha}{ds}}}
\end{displaymath} (17.38)

and convert to $\tau$ to get:
\begin{displaymath}
m \frac{d^2x^\alpha}{d\tau^2} + \frac{q}{c} \frac{dA^\alpha...
...\frac{q}{c}\frac{dx_\beta}{d\tau} \partial^\alpha A^\beta = 0
\end{displaymath} (17.39)

The derivative $dA^\alpha/d\tau$ is a bit jarring. However, if we express this total derivative in terms of partials we observe that:

\begin{displaymath}
\frac{dA^\alpha}{d\tau} = \frac{dx_\beta}{d\tau} \frac{\par...
...ta}
A^\alpha = \frac{dx_\beta}{d\tau} \partial^\beta A^\alpha
\end{displaymath} (17.40)

Substituting, the equation of motion becomes:
\begin{displaymath}\frac{d (mU^\alpha)}{d\tau} = m \frac{d^2x^\alpha}{d\tau^2} =...
...frac{dx_\beta}{d \tau} = \frac{q}{c}F^{\alpha
\beta} U_\beta.
\end{displaymath} (17.41)

which is, as expected, the Lorentz force law in covariant form! How lovely!

To make a Hamiltonian in this notation, we must first make the canonical momentum:

\begin{displaymath}
P^\alpha = - \frac{\partial \tilde{L}}{ \partial\left( \frac{x_\alpha}{ds}
\right)} = m U^\alpha + \frac{q}{c} A^\alpha
\end{displaymath} (17.42)

which is a covariant version of the complete set of interaction equations from the previous section (it does both energy and 3-momentum).

There are several ways to make a Hamiltonian (recall that in general there is what amounts to gauge freedom, minimally the ability to add an arbitrary constant which naturally does not affect the resulting differential equations). One is17.1:

\begin{displaymath}
\tilde{H} = U_\alpha P^\alpha + \tilde{L}
\end{displaymath} (17.43)

Again, we must eliminate:
\begin{displaymath}
U^\alpha = \frac{1}{m}\left(P^\alpha - \frac{q}{c}A^\alpha\right)
\end{displaymath} (17.44)

in favor of $P^\alpha, A^\alpha$. Thus:
\begin{displaymath}
\tilde{L} = -mc\sqrt{\frac{1}{m^2}\left(P_\alpha -
\frac{q}...
...rac{q}{mc}\left(P_\alpha - \frac{q}{c}A_\alpha\right)A^\alpha
\end{displaymath} (17.45)

and
$\displaystyle \tilde{H}$ $\textstyle =$ $\displaystyle \frac{1}{m}\left(P_\alpha -
\frac{q}{c}A_\alpha\right)P^\alpha -m...
...\alpha -
\frac{q}{c}A_\alpha\right)\left(P^\alpha - \frac{q}{c}A^\alpha\right)}$  
    $\displaystyle \quad\quad\quad - \frac{q}{mc}\left(P_\alpha - \frac{q}{c}A_\alpha\right)A^\alpha$ (17.46)
  $\textstyle =$ $\displaystyle \frac{1}{m}\left(P_\alpha - \frac{q}{c}A_\alpha \right)
\left(P^\...
...\alpha -
\frac{q}{c}A_\alpha\right)\left(P^\alpha - \frac{q}{c}A^\alpha\right)}$ (17.47)

This Hamiltonian in four dimensions is no longer an energy since it is obviously a 4-scalar and energy transforms like the time-component of a four vector. However, it works. Hamilton's equations (in four dimensions) lead again directly to the relativistic Lorentz force law:

$\displaystyle \frac{dx^\alpha}{d\tau} = \frac{\partial \tilde{H}}{\partial P_\alpha}$ $\textstyle =$ $\displaystyle \frac{1}{m}\left(P^\alpha - \frac{q}{c}A^\alpha\right)$ (17.48)
$\displaystyle \frac{dP^\alpha}{d\tau} = - \frac{\partial \tilde{H}}{\partial x_\alpha} = -
\partial^\alpha\tilde{H}$ $\textstyle =$ $\displaystyle \frac{q}{mc}\left(P_\alpha - \frac{q}{c}A_\alpha\right) \partial^\alpha
A^\beta$ (17.49)

There is a bit of algebra involved in deriving this result. For example, one has to recognize that:

\begin{displaymath}
p^\alpha = mU^\alpha = P^\alpha - \frac{q}{c}A^\alpha
\end{displaymath} (17.50)

and $p_\alpha p^\alpha = m^2c^2$ and apply this to eliminate unwanted terms judiciously, that is after differentiation. If you apply it too early (for example at the beginning) you observe the puzzling result that:
$\displaystyle \tilde{H}$ $\textstyle =$ $\displaystyle \frac{1}{m}p_\alpha p^\alpha - c \sqrt{p_\alpha
p^\alpha}$  
  $\textstyle =$ $\displaystyle \frac{1}{m} m^2c^2 - c \sqrt{m^2c^2}$  
  $\textstyle =$ $\displaystyle m c^2 - mc^2$  
  $\textstyle =$ $\displaystyle 0$ (17.51)

which leads one to the very Zen conclusion that the cause of all things is Nothing (in four dimensions, yet)!

We are left with a rather mystified feeling that the algebraic hand is quicker than the eye. Somehow an equation whose four-scalar value is zero has a functional form, a structure, that leads to non-zero, covariant equations of motion. Also (as already remarked) this Hamiltonian is not unique. Obviously one can add an arbitrary four-scalar constant to the equation and get no contribution from the derivatives (just as one can in nonrelativistic classical physics). There are other gauge freedoms - ultimately there several other ways of writing the Hamiltonian associated with the given Lagrangian; all of them yield a constant value that is not the energy when evaluated and yield the correct equations of motion when processed.

Finally there exist what are called singular Lagrangians - Lagrangians for which the generalized coordinates do not always map into generalized conjugate variables! Dirac was (unsurprisingly) the first to formally identify this in the context of constrained systems (systems described by a Lagrangian and constraints with Lagrange multipliers for which the Hesse determinant vanishes); Bergmann (at Syracuse) also made major contributions to the formal development of the concept. However the roots of the problem date much further back to e.g. Noether's theorem. I have a couple of papers on this that I've collected from the web, although the idea is also discussed in various monographs and textbooks on mechanics.

It is worth pointing out that there was at one point considerable work being done here at Duke on the idea - N. Mukunda, Max Lohe, (both friends of mine) worked on the idea with Larry Biedenharn (my advisor); Biedenharn also published work with Louck on the subject, and of course Mukunda and Sudarshan's book on classical mechanics remains a ``classic''. Since Dirac's time the notion that the ``right'' unified field theory will have certain interesting properties related to this has been batted around.

This points out an ongoing problem in relativistic quantum theories. These theories are generally based on a Hamiltonian, but manifestly covariant Hamiltonians for a given system cannot in general be uniquely derived from first principles as the mapping between velocities and momenta is not always one-to-one. Thus even when a covariant Lagrangian density can be constructed, the associated Hamiltonian is not obvious or necessarily unique. This is just one (although it is one of the most fundamental) obstacles to overcome when developing a relativistic quantum field theory.


next up previous contents
Next: Motion of a Point Up: Covariant Field Theory Previous: The Brute Force Way   Contents
Robert G. Brown 2013-01-04