1
$\begingroup$

Landaus "derivation" of the Lagrangian for a free particle has intrigued me for a while yet I cannot completely convince myself to trust his derivation. My main qualm lies with the $q$ and $t$ independence of the Lagrangian. Intuitively we know that the position and time of a particle in free space do nothing to affect how it will continue to move, but I dont see how this can allow us to jump to the conclusion that the Lagrangian doesn't depend on $q$ or $t$, as its definition seems much too abstract to just make such a leap. I tried to "prove" this from Hamiltons principle but I am not sure if my approach is good:

If we start out with Hamiltons principle, we want to find a function such that whenever we supply a path, q, a particle can take from time $t_i$ to time $t_f$ (times measured with respect to some other fixed point in time) in a certain system, $S=\int_{t_i}^{t_f}L(q,\dot q, t)\mathrm{dt}$ is maximized so that for any other path, $\eta$, that satisfies $\eta(t_i)=\eta(t_f)=0$, we have $\frac{\mathrm{d}}{\mathrm{d} \epsilon}S[q+\epsilon \eta]=0$ when evaluated at $\epsilon=0$. From experiment/common sense we know that, in free space, if we were to observe the same particle that only had its initial position translated, its trajectory would remain the same, but translated by that amount. In other words, for all $a$, $q+a$ is a path that should extremize the action. Ill further assume that $S$ and $L$ are analytic so that we can write $$S[q+a]=\int_{t_i}^{t_f}L(q+a,\dot q, t)\mathrm{dt}=\int_{t_i}^{t_f}L(q,\dot q, t)+aL_q(q,\dot q, t)...\mathrm{dt}.$$ Now if we vary the path by $\epsilon \eta$ again, we should have $$\frac{\mathrm{d}}{\mathrm{d} \epsilon}\int_{t_i}^{t_f}aL_q(q+\epsilon \eta,\dot q+\epsilon \dot \eta, t)...\mathrm{dt}=0.$$ Since $a$ was arbitrary (in whatever region this series converges), we should have $\frac{\mathrm{d}}{\mathrm{d} \epsilon}\int_{t_i}^{t_f} \frac{\partial^k}{\partial q^k} L(q+\epsilon \eta,\dot q+\epsilon \dot \eta, t)=0$ for all integer values of $k$. I want to say that this is a promising step towards showing $L_q=0$ or that $L$ is independent of $q$ but I have no idea how to proceed. If this approach is correct I'm assuming the proof for the independence of $t$ and direction isn't much different.

Id appreciate if anyone could help me finish my argument or point out errors in my understanding/approach--Id like to show this mathematically rather than heuristically so I'd appreciate if arguments along the line of what I was describing could be provided.

Edit: I realize that my last observation implies that the Euler Lagrange equation is satisfied for all partial derivatives of $L$ with respect to $q$. Therefore, every $\frac{\partial^{k+1}}{\partial q^{k+1}}L=\frac{d}{dt}\frac{\partial}{\partial \dot q}\frac{\partial^{k}}{\partial q^{k}}L$, i.e. a total time derivative. Therefore, $L(q+a,\dot q,t)=L(q,\dot q,t)+$ more total time derivatives. However I cannot show they are independent of $\dot q$ and further am unsure how to advance to showing they would sum to 0.

$\endgroup$
4
  • $\begingroup$ Are you referring to a specific textbook when you write "Landaus [sic] 'derivation'"? Like Landau and Lifshitz Mechanics Volume 1? Like section 4? Or something else? $\endgroup$
    – hft
    Jun 14, 2023 at 0:47
  • $\begingroup$ @hft Yes, volume 1. Its one of the first things covered in the book but he covers it in a sentence without any justification $\endgroup$
    – user62783
    Jun 14, 2023 at 1:01
  • $\begingroup$ Well, once you show they are equivalent to total time derivatives, can't you discard them? Since they wouldn't affect the equations of motion. $\endgroup$
    – G. Paily
    Jun 14, 2023 at 3:30
  • $\begingroup$ @G.Paily I dont believe its enough to show they differ by a total time derivative. We should have that L is completely independent from q. When we perform a gallelain boost we see that while the Lagrangian produces the same equations of motion (as it only differs by a total time derivative), there's other changes such as energy content. Clearly a ball flying through empty space will have the same energy content regardless of where it was launched from $\endgroup$
    – user62783
    Jun 14, 2023 at 3:39

1 Answer 1

0
$\begingroup$

It's simply false. A system can be invariant by translation but still have a $q$ dependence in the Lagrangian. What I mean by translation invariance is that the Euler-Lagrange equations are translation invariant: $$ \frac{d}{dt}\frac{\partial L}{\partial \dot q}-\frac{\partial L}{\partial q} = 0 $$ namely, it $q(t)$ is solution to the Euler Lagrange equation, then so is $q_a(t) = q(t)+a$ for any $a$.

Translated to the action, this means that $S(q_a)$ and $S(q)$ differ by a boundary term for any path $q$ (not only restricted to on-shell paths, i.e. satisfying Euler-Lagrange).

Physically, this means that the "force" is translation invariant, but this does not mean that the potential is. Take for example: $$ L = \frac{\dot q^2}{2}+E\cdot q $$ i.e. a massive particle in a uniform electric field with potential $V=-E\cdot q$. Another classic example would be: $$ L = \frac{\dot q^2}{2}+\dot q\cdot \frac{q\times B}{2} $$ which is a particle in a uniform magnetic field of vector potential $A = \frac{q\times B}{2}$.

In order to get $\partial_qL=0$ from symmetry, you'll need to add isotropy. At best, you'll only be able to prove that it is linear in $q$.

Similarly, the time independence of the equations of motion does not guarantee that $L$ is explicitly time independent. The former mean that if $q(t)$ is a solution, then so is $q_a(t)=q(t+a)$. For the action, it similarly means that $S(q_a)$ and $S(q)$ differ by boundary term.

Again, a linear dependence in time can be killed when evaluating the EL equations (only it must involve velocity). Take for example: $$ L = \frac{\dot q^2}{2}-t\dot q \cdot E $$ which are obtained from the previous Lagrangian by adding a total time derivative: $\frac{d}{dt}tq\cdot E$.

In general since $L$ is defined up to a time derivative, your attempt was trivially false. However, the previous example show that even if you reason up to an additive time, $L$ can still have an explicit $q$ (reps. $t$) dependence despite the EOM's invariance by spatial (resp. temporal) translation. Indeed, if the previous equations are of the form: $$ L = L_0(\dot q)+\frac{d\Lambda}{dt} $$ then the EL equations would give: $$ \frac{d}{dt}L_0'(\dot q) = 0 $$ i.e. (assuming $L_0'$ to be invertible): $$ \dot q = 0 $$ so would only give uniform linear motion. Since in the previous examples the motion isn’t uniform linear (in fact it is neither for the first one), the Lagrangians are not equivalent.

Hope this helps.

$\endgroup$
7
  • $\begingroup$ I guess I should’ve specified that I wanted to show that L was independent of q/t up to a total time derivative. You mention that the dependence on time can only be thrown out when we evaluate the EL equations. While it might be trivial I’m unsure how we actually show this. Clearly if L is independent of t the EOM will be, but the converse seems less trivial $\endgroup$
    – user62783
    Jun 14, 2023 at 15:20
  • $\begingroup$ As I've written, even up to an additive a total time derivative, $L$ is not necessarily space/time independent as the previous examples show. The last example is tangible case of $L$ having explicit time dependence while the EOM's are autonomous. When you are unsure about how to do this, are you having issues deriving the EL equations? $\endgroup$
    – LPZ
    Jun 14, 2023 at 15:39
  • $\begingroup$ not sure if I’m missing something but your example was with a nonzero potential. In your example it still looks like we can say L is independant of time (or space depending on which we use) up to a total time derivative. $\endgroup$
    – user62783
    Jun 14, 2023 at 16:50
  • $\begingroup$ We could write the lagrangian of free space as T+f’ but an equivalent lagrangian would just be T which is independant of time/space. Landau claims isotropy+homogeneity of space and time is enough to arrive at this. $\endgroup$
    – user62783
    Jun 14, 2023 at 16:52
  • $\begingroup$ I added some details on why my examples are not a constant Lagrangian with a time derivative. In your question, you only asked whether homogeneity forces the Lagrangian to be spatially invariant (up to a total derivative), you did not ask about isotropy. $\endgroup$
    – LPZ
    Jun 14, 2023 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.