# Does a Lagrangian being independent to a spacial coordinate imply spatial translation symmetry?

I’m trying to delve into lagrangian mechanics, mainly because i want to have a deeper understand of things lik Noether’s theorem.

What i don’t fully understand is what the lagrangian of a system actually is. I’ve heard it’s related to equations of motion for the system, and that’s it’s something like the kinetic energy minus the potential, but it’s full meaning i only have a limited grasp of.

As i understand it’s an important term in many areas of physics, so i’d like to have a better grasp of it, if anybody here can help.

So there was a relatively simple question i saw that i want to understand better, this is it:

Consider the Lagrangian $$L(q_1,q_2,\dot{q_1}, \dot{q_2})$$ of a conservative mechanical system described by generalised coordinates $$(q_1,q_2)$$. What condition must be satisfied for the momentum $p_1$ associated to $q_1$ to be conserved?

Now i understand that if you substitue in euler-lagrange equation, you get $$\frac{dL}{dq_1}=0$$

And apparently that implies the lagrangian L is independant of $q_1$. After thinking about it for a bit though, does this also imply spatial translational symmetry? I think it would since conservation of momentum is required for this question, but does this equation you end up with actually imply spatial translation symmetry because the lagrangian, which in some respects represents the path of motion, is independent of $q_1$?

• The homogeneity of space leads to conservation of linear momentum of the system. and it will mean dL/dq1+dl/dq2 =0 Aug 12, 2018 at 5:45
• @drvrm I know, i actually made a post about that second thing you thing you mentioned, but my question was if the lagrangian being independant of q1 had a connection to spatial translation. thank you though Aug 12, 2018 at 6:00

I know this question was asked a long time ago, but I think it is very interesting so I will try to give my best answer.

Yes, the Lagrangian is constructed as $$\mathcal L = T-V$$. The goal of using a Lagrangian is to model systems with constraints by eliminating the forces of constraint. The Lagrangian will satisfy the Euler-Lagrange equations over the path of motion. This rule can be derived from D'Alambert's principle, which says that the virtual work of the forces of constraint vanish in systems with holonomic constraints. Here is a derivation. The Lagrangian can also be derived from Hamilton's principle of least action, which establishes a quantity of action that is minimized by the path of motion. I think this derivation is simpler, so I will exhibit it here:

Hamilton's principle says that the action is minimized over the path of motion. Suppose that $$q_i$$ are a set of generalized coordinates for the system. The action is expressed in terms of the Lagrangian with the following: $$I = \int_{t_1}^{t_2}L(\vec q_i, \dot{\vec{q_i}}, t) \ dt$$ The endpoints of the motion are known. In between, we consider arbitrary infinitesimal variations in the path of motion. We express this by letting $$\delta q_i = \epsilon \eta_i(t)$$, where $$\eta_i$$ is an arbitrary continuous function fixed at the endpoints and $$\epsilon$$ is infinitesimal. We wish to find a relation for $$q_i(t)$$ in terms of time and its derivatives such that $$\delta I = 0$$ over the path. Taking the variation of the integral, the chain rule tells us the following: \begin{align*} \delta I &= \int_{t_1}^{t_2}\delta L(\vec q_i, \dot{\vec{q_i}}, t) \ dt\\ &=\int_{t_1}^{t_2}\sum_{i}\frac{\partial L}{\partial q_i}\delta q_i+\frac{\partial L}{\partial \dot q_i}\delta \dot q_i \ dt\\ &= \epsilon \int_{t_1}^{t_2}\sum_i \underbrace{\frac{\partial L}{\partial q_i}\eta_i(t)}_{1}+\underbrace{\frac{\partial L}{\partial \dot q_i}\dot \eta_i(t)}_{2} \ dt \end{align*} Then the second can be evaluated using integration by parts \begin{align*} \int_{t_1}^{t_2} \sum_{i}\frac{\partial L}{\partial \dot q_i}\dot \eta_i \ dt &= \sum_{i} \frac{\partial L}{\partial \dot q_i}\eta_i(t)\Big|_{t_1}^{t_2}-\int_{t_1}^{t_2}\sum_{i}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right)\eta_i(t) \ dt \end{align*} Since we have assumed $$\eta_i(t_1)=\eta_i(t_2)=0$$, the constant term vanishes, and we are left with the following: \begin{align*} &= -\int_{t_1}^{t_2}\sum_i \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right)\eta_i(t) \ dt \end{align*} Plugging this term back into the original expression, \begin{align*} \epsilon \int_{t_1}^{t_2}\sum_i \frac{\partial L}{\partial q_i}\eta_i(t)-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right)\eta_i(t) \ dt &= \epsilon\int_{t_1}^{t_2}\sum_i \left[\frac{\partial L}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right)\right]\eta_i(t) \ dt \end{align*} Taking all $$\eta_i$$ to be independent for any $$t$$, we can apply what is called the "Fundamental Lemma of Calculus" to set each term equal to zero. This gives Lagrange's equations: $$\frac{\partial L}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right)=0$$

Now, in order to answer your question about conservation of momentum, the notion of conjugate momentum needs to be established. Given a generalized coordinate $$q_i$$, the momentum conjugate to that coordinate is defined as $$p_i \equiv \frac{\partial L}{\partial \dot q_i}$$ With this new definition, we can consider Lagrange's equations in the new form $$\frac{\partial L}{\partial q_i} = \frac{d p_i}{d t}$$ If you took the Lagrangian of a free particle for instance, you would find that the momentum conjugate to $$x$$ is $$m\dot x$$, just the standard linear momentum!

Things get interesting once we assume a symmetry. If there is a symmetry over translations of a coordinate $$q_i$$, for instance, then no matter what value of $$q_i$$ we start with, the physics are exactly the same. This implies that the Lagrangian is not explicitly dependent on $$q_i$$. We call $$q_i$$ a cyclic or ignorable coordinate, expressed via

$$\frac{\partial L}{\partial q_i} = 0$$

(note the partial derivative instead of the total derivative). We can see from Lagrange's equations immediately that this implies $$\frac{d p_i}{d t}=0$$, and so $$p_i$$ is a constant of the motion. This establishes the law that momentum conjugate to a cyclic coordinate is conserved.

The full Noether's Theorem is a little more complicated, involving generating functions of continuous symmetries, but this is a good example that establishes the intuition for it. I hope this answer is satisfactory, and not too much to wade through. I learned about this from Classical Mechancis by Herbert Goldstein, although I think Classical Mechanics by John Taylor is more accessible.

Cheers!

I should logically imply that. For momentum associated with q1 to be conserved the Lagrange should be independent of q1. Also translational symmetry implies momentum conservation. So it is like a=>b and b=>c so a=>c..

• Ah, i thought so, thank you. I’m a bit new to this, so i was just kind of trying to figure it out on my own. Aug 12, 2018 at 11:17
• Your reasoning gives "translational symmetry $\implies$ coordinate-independence", but the question talks about the other direction. May 21, 2021 at 11:12