# Derivatives of Lagrangian for relativistic massive point particle

For a relativistic point particle with mass $$m$$ whose worldline is parameterized by $$x(\lambda)$$ the standard Lagrangian is:

$$L(\dot{x}) = -mc\sqrt{g_{ab}\dot{x}^a \dot{x}^b} \tag1$$ where $$g$$ is a Lorentzian metric with signature $$(+,-,-,-)$$, and "$$\dot{}$$" $$=$$"$$\text{d}/\text{d}\lambda$$".

If $$\lambda = \tau = \text{proper time}$$, $$\sqrt{g_{ab}\dot{x}^a \dot{x}^b} = c \implies$$ $$L(\dot{x}) = -mc^2 = \text{constant}$$. This should imply that $$\partial L/\partial \dot{x}^a =\partial mc^2/ \partial \dot{x}^a= 0$$ for example, and so each term in the EL equations should be identically zero leading to equations "$$0=0$$". I know this incorrect because the Lagrangian actually leads us to the famous geodesic equations, so I must be missing some really important point here about Lagrangian formalism. Where's my mistake?

• Possible Duplicate : From Euler-Lagrange equation to non-affine geodesic equation. Commented Sep 5, 2021 at 3:32
• ^ I don't think the above link is a duplicate, since it doesn't address where OP's mistake is; it just gives a derivation of the geodesic equation (which OP has already seen, as mentioned in the second last sentence of the post). Commented Sep 5, 2021 at 4:47

This is what happens when you overload notation and mix up what is the curve in spacetime and what are the coordinates on spacetime, and mixup several functions by giving them the same names.

My first suggestion is that you use the notation $$v^a$$ for the velocity term, so that $$x^a,v^a$$ are independent. More formally, the Lagrangian is a function (on the tangent bundle of a manifold, whose representation in terms of a chart is) given by \begin{align} L(x,v)&=-mc\sqrt{g_{ab}(x)v^av^b} \end{align} So, \begin{align} \frac{\partial L}{\partial v^a}(x,v) &= mc\frac{g_{ab}(x)v^b}{\sqrt{g_{kl}(x)v^kv^l}}\tag{*} \end{align} Note that here $$L$$ is (as mentioned above) simply a function of two variables $$(x,v)\in\Bbb{R}^n\times\Bbb{R}^n$$, so it makes sense to talk about the $$2n$$ partial derivatives $$\frac{\partial L}{\partial x^1},\dots,\frac{\partial L}{\partial x^n},\frac{\partial L}{\partial v^1},\dots,\frac{\partial L}{\partial v^n}$$.

Next, what we require in the Euler-Lagrange equations is a curve $$\gamma:\Bbb{R}\to TM$$, which we lift to $$\gamma':\Bbb{R}\to TM$$. Anyway without getting too much into the differential geometry, once you choose charts (i.e coordinates), you can think of them as standard curves $$\gamma:\Bbb{R}\to\Bbb{R}^n$$ and its velocity $$\dot{\gamma}:\Bbb{R}\to\Bbb{R}^n$$. The 'least' action principle states that a curve $$\gamma$$ satisfies the equations of motion if and only if for every parameter value $$\lambda\in\Bbb{R}$$, one has \begin{align} \frac{d}{ds}\bigg|_{s=\lambda}\left(s\mapsto \frac{\partial L}{\partial v^a}\bigg|_{(\gamma(s), \dot{\gamma}(s))}\right)-\frac{\partial L}{\partial x^a}\bigg|_{\gamma(\lambda), \dot{\gamma}(\lambda)}&= 0 \end{align} In other words, YOU PERFORM THE PARTIAL DERIVATIVES FIRST, AND ONLY AFTERWARDS PLUG IN THE CURVE (and do $$\frac{d}{d\lambda}$$). Excuse the caps, but this is I think a point which is completely glossed over in any treatment of Lagrangian mechanics and is one of the biggest sources of confusion. Also, I'm clearly aware that the above expression is very cumbersome to write, but atleast it is 100% unambiguous and is the explicit/precise way of writing things (and not appreciating this is contributing to your confusion).

In your case, you're getting the mysterious "$$0=0$$" (assuming $$g_{ab}$$ doesn't actually depend on $$x$$) because you plugged in the curve first and then (erroneously) tried to do partial derivatives (because your notation makes it "seem ok" when it is infact not). More explicitly, what you said is we can choose a parametrization of the curve such that for each $$\lambda$$, $$g_{ab}(\gamma(\lambda))\dot{\gamma}^a(\lambda)\dot{\gamma}^b(\lambda)=\text{const}$$. Ok, that's certainly true, but this means the function $$\ell:\Bbb{R}\to\Bbb{R}$$ defined as \begin{align} \ell(\lambda)&=L(\gamma(\lambda),\dot{\gamma(\lambda)})\\ &=-mc\sqrt{g_{ab}(\gamma(\lambda))\dot{\gamma}^a(\lambda)\dot{\gamma}^b(\lambda)}\\ &=-mc^2 \end{align} is constant. HOWEVER, $$\ell$$ and $$L$$ are NOT at all the same function, but you're thinking that they are. The correct statement is that $$\ell'=0$$, NOT that $$\frac{\partial L}{\partial v^a}=0$$, and by extension, the mapping $$\lambda\mapsto \frac{\partial L}{\partial v^a}|_{\gamma(\lambda),\dot{\gamma}(\lambda)}$$ is not the zero mapping. It is this abuse of notation which is tripping you up completely (you wrote $$\frac{\partial L}{\partial \dot{x}^a}=-\frac{\partial mc^2}{\partial \dot{x}}=0$$, when if anything you should have written $$\frac{\partial \ell}{\partial \dot{x}^a}=0$$. But of course, $$\ell$$ is a function of one variable so using partial derivatives is just nonsense. Hence, by writing things clearly, you can immediately spot where you're making errors). Just for completeness, the correct formula is (according to $$(*)$$ above) \begin{align} \frac{\partial L}{\partial v^a}\bigg|_{(\gamma(\lambda),\dot{\gamma}(\lambda))}&= mc\frac{g_{ab}(\gamma(\lambda))\,\,\dot{\gamma}^b(\lambda)}{\sqrt{g_{kl}(\gamma(\lambda))\dot{\gamma}^k(\lambda)\dot{\gamma}^l(\lambda)}}\\ &=mc\frac{g_{ab}(\gamma(\lambda))\,\,\dot{\gamma}^b(\lambda)}{c}\\ &=m\cdot g_{ab}(\gamma(\lambda))\,\, \dot{\gamma}^b(\lambda) \end{align} Therefore, you can clearly see now that there is no reason for this to be identically zero for all $$\lambda$$.

This is a very basic notation and definition issue, which I beat to death in the following MSE answer About the "ambiguity" of explicit and implicit function of a variable (yes this is a basic issue, but I really think you could benefit from being more careful with the basics).

Below are some answers I've written which I think could help you clarify some really basic issues (particularly since I explain where exactly the sloppiness of the physics notation is occurring, and where the 'pitfalls'. Once you know these 'pitfalls', you can of course go back to writing sloppily/succinctly)

• Differentiation in the Geodesic Problem. Here OP's question is about how one can think of $$x, \dot{x}$$ as independent variables for partial differentiation (so not directly related to your question here), but I explain what the notation actually stands for, and emphasize the importance of keeping track of what is the function and where it is being evaluated.

• Doubt about ordinary and partial derivative. Here, OP is confused about a simple application of the chain rule (of course not your confusion here), but the mistake the OP makes is the same as the one you make, namely to confuse two functions $$\Lambda$$ and $$L$$, where the former is obtained by a certain composition with the latter.

• +1, good answer. OP – for bonus points, try deriving the geodesic equation from the variational problem given in your question. This answer should be a good resource on how to start, it's only missing the derivation itself. Commented Sep 5, 2021 at 6:56

There is already a good answer from peek-a-boo. Here we will give an equivalent answer from the perspective of the action rather than the Lagrangian.

1. Firstly, recall that the action $$S[x]~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda ~L~=~-E_0 \Delta \tau \tag{1}$$ of a relativistic massive particle is minus the rest energy $$E_0=m_0c^2$$ times the change $$\Delta \tau=\tau_f-\tau_i$$ in proper time, cf. e.g. this related Phys.SE post. So the action principle (1) maximizes proper time! The solution path is a geodesic in that sense.

2. Secondly, recall that the stationary action principle (1) by construction keeps the endpoints $$\lambda_i$$ and $$\lambda_f$$ fixed during the variation.

TL;DR: The problem arises because OP tries to fix the worldline parameter $$\lambda=\tau$$ as proper time before the variation.

1. On one hand, as a consequence, OP then fixes the endpoints $$\lambda_i=\tau_i$$ and $$\lambda_f=\tau_f$$.

2. On the other hand, according to (1) all paths (virtual or not) will now have the same action value given by the endpoints! I.e., the action principle has been rendered useless. It can not be used to determine the actual geodesic.

The same point is also made in my Phys.SE answers here & here.

To actually derive the geodesic equation, see e.g. this Phys.SE post.