Wikipedia:Reference desk/Archives/Mathematics/2012 July 27

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July 27

Lagrangian mechanics

This is probably a dumb question, and I'm probably over thinking this, but it's been nagging me for a while and for some reason I can't resolve it.

This question is part physics and part math, but I think the math desk is the most appropriate place for it.

Consider a (classical mechanical) system, describable by the generalized coordinates $(q_{1},q_{2},....,q_{N})={\vec {q}}$ and the Lagrangian $L({\vec {q}},{\dot {\vec {q}}},t)$. Suppose we have a trajectory ${\vec {q}}(t)$ which satisfies the equations of motion, which is to say that ${\frac {\partial L}{\partial {\vec {q}}}}({\vec {q}}(t),{\dot {\vec {q}}}(t),t)={\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\vec {q}}}}}({\vec {q}}(t),{\dot {\vec {q}}}(t),t)\right)$. I'm using a funny but common notation where ${\frac {\partial L}{\partial {\vec {q}}}}=\left({\frac {\partial L}{\partial q_{1}}},{\frac {\partial L}{\partial q_{2}}},...\right)$ and ${\frac {\partial L}{\partial {\dot {\vec {q}}}}}=\left({\frac {\partial L}{\partial {\dot {q}}_{1}}},{\frac {\partial L}{\partial {\dot {q}}_{2}}},...\right)$.

Now we transform the trajectory ${\vec {q}}(t)$ to a new trajectory ${\vec {Q}}(t)$. I'd like to show that if the Lagrangian remains invariant under this transformation ie $L({\vec {q}}(t),{\dot {\vec {q}}}(t),t)=L({\vec {Q}}(t),{\dot {\vec {Q}}}(t),t)$, then ${\vec {Q}}(t)$ also obeys the equations of motion ie ${\frac {\partial L}{\partial {\vec {q}}}}({\vec {Q}}(t),{\dot {\vec {Q}}}(t),t)={\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\vec {q}}}}}({\vec {Q}}(t),{\dot {\vec {Q}}}(t),t)\right)$. 65.92.7.148 (talk) 05:53, 27 July 2012 (UTC)

It depends on the transformation. Check out the articles canonical transformation and symplectomorphism. — Fly by Night (talk) 14:52, 27 July 2012 (UTC)

The symplectomorphism article was above my head, but I think I understand the canonical transformation article, and I don't think it's what I'm looking for. I don't dispute that only some special types of transformations (like canonical transformations) preserve the equations of motion despite changing the Lagrangian and Hamiltonian. But what I'd like to show is that all transformations which do preserve the Lagrangian also preserve the equations of motion. 65.92.7.148 (talk) 18:56, 27 July 2012 (UTC)

So you're trying to prove the statement: "If a transformation preserves the Lagrangian then it preserves the equations of motion." To be honest, my knowledge of this subject is more mathematical, e.g. symplectic geometry. Have you got access to the book V. I. Arnold (1989). Mathematical Methods of Classical Mechanics. Springer.? That text's a gold mine. It starts from the basics and ends up on some very deep stuff, including Hamiltonian mechanics. — Fly by Night (talk) 20:01, 27 July 2012 (UTC)

Yes, I have a copy of that book. Most of it is Arabic to me, but fortunately I was able to find something relevant. On page 88, in the section on Noether's theorem, he writes: "First, let $M=\mathbb {R} ^{n}$ be coordinate space. Let $\phi :\mathbb {R} \rightarrow M,\mathbf {q} =\phi (t)$ be a solution to Lagrange's equation. [If] $h_{*}^{s}$ preserves L, the translation of a solution, $h^{s}\circ \phi :\mathbb {R} \rightarrow M$ also satisfies Lagrange's equations." $h^{s}:M\rightarrow M,s\in \mathbb {R}$ is defined as a "one-parameter group of diffeomorphisms", whatever that means. I'm not sure how $h_{*}^{s}$ is defined. At any rate, he seems to be asserting what I'm trying to prove, but unfortunately he himself does not provide a proof, unless it's buried somewhere else in the text. That said, perhaps with the problem translated into the language of higher mathematics you can solve it, and then translate the solution into something that I, with my feeble knowledge of math, can understand. 65.92.7.148 (talk) 21:43, 27 July 2012 (UTC)

We have a diffeomorphism article. A one parameter group of diffeomorphisms is just the integral curves of a vector field, i.e. the paths followed by a particle under the incluence of smoothly varying force field. Take a look at Integral curve, diffeomorphism and Flow (mathematics). — Fly by Night (talk) 23:23, 27 July 2012 (UTC)

I appreciate your help, but unfortunately those articles don't help me much. Really, this question should be solvable by simple calculus. I think learning about diffeomorphisms is a bit overkill. 65.92.7.148 (talk) 04:15, 28 July 2012 (UTC)

I had a quick look in Landau and Lifshitz's Mechanics (3rd Ed) - s45 discusses canonical transformations but as a prelude it mentions point transformations, they talk about the choice of generalised coordinates being unrestricted and how we can change from $q_{1},q_{2},$ to $Q_{1},Q_{2},$ by $Q=Q(q,t)$. There's no detail but I web-searched for point transformations and got this PDF [1] titled "Coordinate Invariance of Lagrange's Equations" which looks helpful, maybe give that a look? Although I don't think the Langrangian itself is invariant but the Euler-Lagrange equations should be...83.100.173.200 (talk) 15:01, 28 July 2012 (UTC)

I had a look through the PDF, and I don't believe it solves the problem (though I confess I may be mistaken). From what I understand, the author expresses the same system in terms of a different set of generalized coordinates (ie makes a passive transformation), and shows that the trajectory of the system as expressed in terms of these new coordinates still obeys Lagrange's equation. What I'm trying to do is make an active transformation of the coordinates ie change the system, and show that Lagrange's equations are still applicable if the Lagrangian is the same. That said, I appreciate the help. 65.92.7.148 (talk) 03:48, 29 July 2012 (UTC)

Sorry it wasn't helpful. Diffeomorphisms are a very common topic and I'm suprised you're doing Hamiltonian mechanics but haven't been taught about diffeomorphisms. They're just transformations that, along with their inverses, are differentiable, i.e. the smudge things in a smooth way. The beauty of the general setting, i.e. using vector fields and differential forms, means that everything is coordinate free: you don't even need to use coordinates! But I understand that it might be bewildering trying to learn it on your own. — Fly by Night (talk) 17:19, 28 July 2012 (UTC)

Don't be sorry. I suspect you mistook me for a math major, but alas I'm just a lowly physics undergrad. And when we learn analytical mechanics, we don't use any fancy-shmancy mathematics, just you're everyday multivariable cal. So that's why I was a bit put-off by this diffeomorphism business. 65.92.7.148 (talk) 03:48, 29 July 2012 (UTC)

What do you mean by "transform the trajectory ${\vec {q}}(t)$ to ${\vec {Q}}(t)$"? Taking your proposition at face value, it isn't true. Let $L={\tfrac {1}{2}}m{\dot {x}}^{2}+{\tfrac {1}{2}}m{\dot {y}}^{2}$. Then $x(t)=vt,y(t)=0$ (for some constant v) satisfies the equations of motion, but $X(t)=r\sin(v/r)t,Y(t)=r\cos(v/r)t$ doesn't, even though $L({\dot {x}}(t),{\dot {y}}(t))\equiv L({\dot {X}}(t),{\dot {Y}}(t))$. -- BenRG (talk) 17:42, 28 July 2012 (UTC)

You're right, I didn't state the question well. Here's a second try: suppose we make the transformation ${\vec {q}}\rightarrow {\vec {Q}}({\vec {q}},t),\ {\dot {\vec {q}}}\rightarrow {\dot {\vec {Q}}}({\vec {q}},{\dot {\vec {q}}},t)$ which preserves the Lagrangian. Then, if ${\vec {q}}(t)$ is a solution to Lagrange's equations, then so is ${\vec {Q}}({\vec {q}}(t),t)$. In this case, your counter-example no longer works, because the transformation $x\rightarrow X=r\sin {\frac {x}{r}},\ y\rightarrow Y=r\cos {\frac {x}{r}},\ {\dot {x}}\rightarrow {\dot {X}}={\dot {x}}\cos {\frac {x}{r}},\ {\dot {y}}\rightarrow {\dot {Y}}=-{\dot {y}}\sin {\frac {x}{r}}$ doesn't preserve the Lagrangian. 65.92.7.148 (talk) 03:36, 29 July 2012 (UTC)

I think you need to be clear about what you mean by "preserves the Lagrangian". Your transformation seems to be a change of trajectory, but not a change of generalised co-ordinates. The Lagrangian is a function of the generalised co-ordinates (and, sometimes of time too). So the form of the Lagrangian is preserved since you are not changing your system of co-ordinates. In BenRG's example, he has two trajectories, both described in Cartesian (x,y) co-ordinates, and along both trajectories the value of the Lagrangian is constant, and has the same value. One trajectory is $(vt,0)\,$ i.e. particle moves along the x axis at constant speed v. The other is $\left(r\sin(vt/r),r\cos(vt/r)\right)\,$ i.e. the particle travels around a circle of radius r at constant speed v. The Lagrangian in this example is just the kinetic energy of the particle (there are no external forces or constraints on the particles motion), which has a constant value of $mv^{2}/2$ along both trajectories. However, the first trajectory satisfies the equations of motion (which are $m{\ddot {x}}=0,\ m{\ddot {y}}=0\,$), and the second one doesn't. In other words, a particle moving in the absence of external forces or constraints will have constant kinetic energy, but the converse is not true. Gandalf61 (talk) 10:29, 29 July 2012 (UTC)

By "preserves the Lagrangian" I mean this: if you define a transformation that changes the values of the generalized coordinates (and hence the generalized velocities as well, usually), the new value of the Lagrangian will always be numerically equal to the original value of the Lagrangian. By always I mean for any values of the original generalized coordinates and velocities.

Example: suppose you have a particle moving in a 2D plane, with a potential proportional to the distance from the origin. So, in Cartesian coordinates, $L={\frac {1}{2}}m({\dot {x}}^{2}+{\dot {y}}^{2})-U\left({\sqrt {x^{2}+y^{2}}}\right)$. If I make the transformation $x\rightarrow X=x+a,\ y\rightarrow y$ ie move the particle in the x direction by an amount a, then the Lagrangian will in general by different, because the value of the potential energy will change. In the particular case that $x=-{\frac {a}{2}}$, the Lagrangian will be the same under this transformation, but that's just a special case.

On the other hand, if I make the transformation $x\rightarrow X=x\cos \theta +y\sin \theta ,\ y\rightarrow Y=y\cos \theta -x\sin \theta$, then the Lagrangian will be the same no matter what x, y, dx/dt, and dy/dt are: ${\frac {1}{2}}m({\dot {X}}^{2}+{\dot {Y}}^{2})-U\left({\sqrt {X^{2}+Y^{2}}}\right)={\frac {1}{2}}m({\dot {x}}^{2}+{\dot {y}}^{2})-U\left({\sqrt {x^{2}+y^{2}}}\right)$ for all x, y, dx/dt, dy/dt.

In BenRG's example, $L(X,Y,{\dot {X}},{\dot {Y}})={\frac {1}{2}}m({\dot {X}}^{2}+{\dot {Y}}^{2})=m{\dot {x}}^{2}$, which is only equal to the original Lagrangian $L(x,y,{\dot {x}},{\dot {y}})$ if dy/dt = 0. 65.92.7.148 (talk) 16:00, 29 July 2012 (UTC)

You need to think carefully about the difference between changing the trajectory of a particle and changing the generalised co-ordinates that are used to describe that trajectory. In BenRG's example, he described two different trajectories:

$(x_{1}(t),y_{1}(t)=(vt,0)$

and

$(x_{2}(t),y_{2}(t)=\left(r\sin(vt/r),r\cos(vt/r)\right)$

because you appeared originally to be asking about changes to trajectories that preserve the value of the Lagrangian along the trajectory. The form of the Lagrangian is the same for both trajectories:

$L={\tfrac {1}{2}}m({\dot {x}}^{2}+{\dot {y}}^{2})$

because the form of the Lagrangian depends only on the system of generalised co-ordinates, not on the trajectory. BenRG does not change the system of generalised co-ordinates, he only changes the trajectory through phase space. So along the first trajectory we have

$L_{1}(t)={\tfrac {1}{2}}m({\dot {x_{1}}}^{2}(t)+{\dot {y_{1}}}^{2}(t))={\tfrac {1}{2}}mv^{2}$

and along the second trajectory we have

$L_{2}(t)={\tfrac {1}{2}}m({\dot {x_{2}}}^{2}(t)+{\dot {y_{2}}}^{2}(t))={\tfrac {1}{2}}m\left(v^{2}\sin ^{2}(vt/r)+v^{2}\cos ^{2}(vt/r)\right)={\tfrac {1}{2}}mv^{2}$

and so the value of the Lagrangian is the same (and is constant) along both trajectories. But the first trajectory satisfies the equations of motion, whereas the second one doesn't.

But now you seem to have switched to asking about changing the system of generalsied co-ordinates - your last post does not mention trajectories at all. You have simply observed that given a potential that is radially symmetric about a point P, a translation of co-ordinates changes the form of the Lagrangian, but a rotation of co-ordinates about the point P does not change the form of the Lagrangian. And, of course, if the form of the Lagrangian is unchanged then the equations of motion are not only satisfied by the same trajectories through phase space (which we would expect on physical grounds), but they also take the same form in the new generalised co-ordinates system i.e. you can just substitute X for x and Y for y. Which also tells us that if we find a trajectory (x(t), y(t)) that satisfies the equations of motion, then any rotation of that trajectory through an angle θ about P will also satisfy the equations of motion - just as we would expect from the symmetry of the system. Gandalf61 (talk) 17:51, 29 July 2012 (UTC)

Yes, you're right, after reading BenRG's counterexample I realized that I had made a mistake and didn't ask the question I was trying to ask. So, here's what I'm really asking: suppose the configuration of a system is described by the vector ${\vec {q}}$ in configuration space. Next, we define some transformation ${\vec {q}}\rightarrow {\vec {Q}}({\vec {q}},t)$ which changes the configuration of the system from q to a new configuration Q eg by rotating the entire system, or by shifting the entire system, etc. Now we assume that the Lagrangian of the system in the new configuration has the same value as the system in the old configuration. My question is, if q(t) obeys Lagrange's equations, will Q(t)=Q(q(t),t) too? This is a little different than the way you phrased my question, because you're talking about a change of coordinate system (eg rotating the x-y axes) rather than changing the system (eg rotating the system).

That said, despite what you wrote I still think the claim that Q(t)=Q(q(t),t) satisfies the equations of motion is a non-trivial one. Now if I were talking about a change in the coordinate system, then yes it would be trivial because Lagrange's equations hold for any coordinate system: that's why they're so powerful! But it doesn't seem so obvious if we consider an active transformation of the coordinates.

Let's look at the previous example. Lagranges' equations are $m{\ddot {x}}=-{\frac {\partial U}{\partial x}}$ and $m{\ddot {y}}=-{\frac {\partial U}{\partial y}}$. The question is, if x(t) and y(t) satisfy the EOMs, do X(t) = x(t)cosθ + y(t)sinθ and Y(t) = y(t)cosθ - x(t)sinθ satisfy them as well? I can verify that this is the case by plugging them in, but it isn't obvious to me that this should always be true. 65.92.7.148 (talk) 22:59, 29 July 2012 (UTC)

If you rotate or translate the whole system configuration, including external forces or constraints, then in general the form of the Lagrangian and the equations of motion will change, but a new rotated or translated trajectory will produce the same values of the Lagrangian along its path as the original trajectory. If the original trajectory satisfied the original equations of motion then the new trajectory will satisfy the new equations of motion. This is clear because a rotation or translation of the whole system is equivalent to a rotation or translation of the co-ordinate system of the same magnitude but in the opposite direction.

If you just rotate or translate the positions and velocity vectors of the particle(s), but do not rotate or translate forces and constraints, then the form of the Lagrangian and equations of motion do not change (because they do not depend on the particle configuration), but a new rotated or translated trajectory will, in the general case, produce different values of the Lagrangian along its path from the original trajectory. The new trajectory will not necessarily satisfy the equations of motion, even if the original trajectory did.

In your example, the potential (representing external forces) is radially symmetric about some point P, so U has the same form whether we rotate it or not. So a rotation of the system configuration about P leaves the form of the Lagrangian unchanged (because we can regard U as not being rotated), and a rotated trajectory also produces the same values of the Lagrangian along its path as the original trajectory (because now we can switch our view point and regard U as having rotated along with the trajectory).

On the other hand, a translation of particle positions and velocity vectors without translating U leaves the form of the Lagrangian unchanged, but the values of the Lagrangian along a translated trajectory are different from the values along the original trajectory, and in general the translated trajectory will not satisfy the equations of motion.

To sum up:

1. If two trajectories x₁(t) and x₂(t) produce the same values of the Lagrangian at the same time t along the whole of their paths, must one trajectory satisfy the equations of motion if the other does ? Answer: No - BenRG has shown a counterexample.
2. If two trajectories x₁(t) and x₂(t) produce the same values of the Lagrangian at the same time t along the whole of their paths because of a symmetry in the external forces or constraints, must one trajectory satisfy the equations of motion if the other does ? Answer: Yes, because we can exploit the symmetry of the external forces (or, if you like, the symmetry of the Lagrangian) to regard these forces as having moved along with the trajectory, but nevertheless still leaving the form of the Lagrangian unchanged. Gandalf61 (talk) 07:41, 30 July 2012 (UTC)