Undeniably, the calculus of variations is an extremely useful and brilliant mathematical tool. It “automatically” selects the optimal function for us from a multitude of candidates, bypassing the need for specific detailed analysis. The principle of least action in physics has provided a vast arena for the calculus of variations, which in turn has driven the development of the method itself. However, a prominent characteristic of the calculus of variations is that, in most cases, the calculations are quite complex. If one were to use “brute force,” it might even be difficult to write down the resulting system of differential equations. Therefore, useful tricks are highly welcomed. This article intends to introduce such a small trick to simplify certain variational problems.
How did I come across this trick? In fact, several months ago, while reading Gravitation and Spacetime, I reached the section on variational principles and simply could not understand it. I couldn’t figure it out. Why could something that seemed clearly wrong to me yield the correct result? My mathematical intuition told me it was definitely the author’s mistake, yet I couldn’t pinpoint exactly where the author went wrong, so I set the problem aside. Recently, I finally arrived at a satisfactory answer, and I suspect that the trick to be discussed in this article has been “misused” by physicists.
The Trick
First, let us look at how we typically handle variational problems. Taking a single-variable function as an example, to find the extremal curve of: S=\int L(x,\dot{x},t)dt we usually substitute it directly into the Euler-Lagrange equation: \frac{\partial L}{\partial x}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=0 This always works, but it is usually not simple. Let’s look at a special case: the integrands of many variational problems are in the form of a square root. For instance, the action in special relativity is S=-\int mc\sqrt{c^2 dt^2-dx^2-dy^2-dz^2}, and the arc length in geodesic problems is l=\int\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}, etc. For this class of problems, if we directly treat one variable as the independent variable and the others as functions to be substituted into the Euler-Lagrange equation, we obtain very complex results. The variables become so “entangled” that we might not even be able to list the differential equations.
However, if we look closely, we find that the source of complexity is the square root, while the expression inside the square root is usually not very complex. It would be great if there were a way to square it. This is indeed achievable, but to find it, we must not start from the Euler-Lagrange equation; we must start from the most fundamental operations of the calculus of variations.
\begin{aligned}\delta S &=\delta \int Ldt \\&=\int \delta(Ldt)\\&=\int \delta\sqrt{(Ldt)^2}\\&=\int \frac{\delta (L^2 dt^2)}{2\sqrt{(Ldt)^2}}\end{aligned}
Let ds=\sqrt{(Ldt)^2}=Ldt, then it becomes: \int \frac{\delta (L^2 dt^2)}{2ds}=\int \frac{\delta (L^2 dt^2)}{2ds^2}ds
At this point, if we during the variation process (note, only during the variation process) treat ds as a pure parameter independent of x and y, then we can rewrite it as: \int \frac{1}{2}\delta \left[L^2 \left(\frac{dt}{ds}\right)^2\right]ds
This means that if we use ds=Ldt as a parameter, the variation of S=\int Ldt yields results (differential equations) equivalent to the variation of S=\int \frac{1}{2} [L^2 (\frac{dt}{ds})^2]ds. Consequently, we only need to substitute L'=\frac{1}{2} L^2 (\frac{dt}{ds})^2 into the Euler-Lagrange equation. Generally speaking, this is simpler than substituting L directly.
For example, when considering geodesics, we want to vary l=\int\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}. Let ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}, then: l=\int g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}ds
Denoting L'=\frac{1}{2} g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}, substituting this into the Euler-Lagrange equation yields: \frac{d}{ds}\left(g_{\mu\nu}\frac{dx^{\nu}}{ds}\right)=\frac{\partial g_{\alpha\beta}}{\partial x^{\nu}}\left(\frac{dx^{\alpha}}{ds}\right)\left(\frac{dx^{\beta}}{ds}\right)
Misuse
However, this trick can potentially be misused. The action for a free particle in special relativity is S=-\int mc\sqrt{c^2 dt^2-dx^2-dy^2-dz^2}. According to the conclusion above, it is equivalent to S=-\int \frac{1}{2}mc[c^2 (\frac{dt}{ds})^2-(\frac{dx}{ds})^2-(\frac{dy}{ds})^2-(\frac{dz}{ds})^2]ds, where ds=\sqrt{c^2 dt^2-dx^2-dy^2-dz^2}. This is correct when no interaction is considered. But if a potential energy term is added to the action, it is easy to fall into the illusion that the free particle term can be replaced by S=-\int \frac{1}{2}mc[c^2 (\frac{dt}{ds})^2-(\frac{dx}{ds})^2-(\frac{dy}{ds})^2-(\frac{dz}{ds})^2]ds in the same way. This will lead to incorrect results. This is because the trick we discussed involves squaring the entire Lagrangian, not just a part of it.
For example, in the case we previously considered: S= -mc^2 \int \sqrt{1-\frac{v^2}{c^2}}dt-\alpha \phi \sqrt{1-\frac{v^2}{c^2}}dt
One cannot simply transform the action into: S=-\int \frac{1}{2}mc\left[c^2 \left(\frac{dt}{ds}\right)^2-\left(\frac{dx}{ds}\right)^2-\left(\frac{dy}{ds}\right)^2-\left(\frac{dz}{ds}\right)^2\right]ds-\frac{\alpha}{c} \phi ds
This will yield incorrect results. In fact, there is no simple shortcut here; the best method is to perform the variation directly rather than relying on the Euler-Lagrange equations.
“Just one more small step, as if in the same direction, and truth becomes error.” I believe this is one manifestation of that saying.
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