# Lagrange's equation is form invariant under ANY coordinate transformation. Hamilton's equations are not under ANY phase space transformation. Why?

When we make an arbitrary invertible, differentiable coordinate transformation $$s_i=s_i(q_1,q_2,...q_n,t),\forall i,$$ the Lagrange's equation in terms of old coordinates $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right)-\frac{\partial L}{\partial q_i}=0,\forall i$$ changes to $$\frac{d}{dt}\left(\frac{\partial \hat{L}}{\partial\dot{s}_i}\right)-\frac{\partial \hat{L}}{\partial s_i}=0, \forall i$$ where $$\hat{L}(s,\dot{s},t)$$ is obtained from $$L(q,\dot{q},t)$$ by $$\hat{L}(s,\dot{s},t)=L(q(s,t),\dot{q}(s,\dot{s},t),t).$$

When we go to the Hamiltonian framework, things do not remain so simple. Under an arbitrary transformation in phase space, $$~~~Q_i\to Q_i(q_1,q_2...,p_1,p_2,..,t),\forall i,\\ P_i\to P_i(q_1,q_2...,p_1,p_2,..,t), \forall i$$ the Hamilton's equations do not remain form invariant. This only happens for a restricted class of transformations, called the canonical transformations. Also the new Hamiltonian is not obtained from the old Hamiltonin by $$\hat{H}(Q,P,t)= H(q_i(Q,P,t),p_i(Q,P,t),t)$$ even when the transformation is canonical, unless that canonical transformation is also a symmetry.

Why is the reason for this?

## 2 Answers

You should really think about the variables we use as being like coordinates on some manifold, the configuration space (roughly the same as the phase space, I won't be careful about the distinction). In this language, changing variables is equivalent to changing coordinates on this manifold. The action is some scalar function on this space, and we can take coordinate derivatives, $$\frac{\delta S}{\delta q_a}$$ with respect to whatever coordinates $$q_a$$ we are using on the space. As in multivariable calculus, we can form the directional derivatives $$D_v S = v_a\frac{\delta S}{\delta q_a}$$ for any vector with components $$v_a$$ we like. If you want something more formal and geometric, the directional derivative is a Lie derivative on the configuration manifold.

Now, remember that when we vary the action, we demand that the variation be stationary. That is, we demand that all directional derivatives vanish, meaning $$D_v S=0$$ for all vectors $$v$$. This statement about the vanishing of directional derivatives, you'll note, it entirely agnostic to the coordinates we use, but nonetheless implies that if we use coordinates $$q_a$$, that $$\frac{\delta S}{\delta q_a}=0$$ for each $$a$$. But any coordinate system would result in the same condition that all the coordinate derivatives vanish. This should also be in some sense familiar from multivariable calculus.

So this is casting the invariance of the Euler-Lagrange equations in a geometric language. Aside from being nice, this is also going to be the right language to understand what's going on in the Hamiltonian picture.

The phase space is normally coordinatized by pairs of coordinates $$(p^a,q_a)$$, but this is really not necessary. At the end of the day, the phase space is again a manifold and the $$(p,q)$$ are simply a special coordinate system on that manifold (Darboux's theorem implies that we can always, at least locally, find such a coordinate system). The thing that defines these special coordinates, really, is that the symplectic form takes a very special form.

In case you are less than familiar with symplectic forms, let me do the following to motivate the idea. Instead of using the coordinates $$(p^a,q_a)$$, instead use a collective coordinate $$\xi^a = \langle p^a, q_a\rangle$$, so all I've really done is put the $$p$$'s and $$q$$'s into one big vector. Just to be clear, if $$q_a$$ and $$p^a$$ were $$n$$-dimensional vectors, then $$\xi^a$$ is a $$2n$$-dimensional vector formed by concatinating the components (well, any way of putting the components together will do...this will just change the precise form of the $$\Omega$$ I introduce in a moment by permuting it's rows and columns appropriately). In terms of this, the Hamilton's equations may now be written $$\frac{d\xi^a}{d t}=\Omega^{ab}\frac{\partial H}{\partial \xi^b}$$ where $$\Omega$$ is some matrix. Usually, this looks something like $$\Omega=\left(\begin{array}{cc} 0 & -1\\ 1 & 0\end{array}\right).$$ This $$\Omega$$ is typically known as the inverse of the symplectic form. Though sometimes you'll just hear it called the symplectic form (an abuse of language, but not uncommon) or, more accurately, a Poisson bivector. These names are not so important to what I want to say, but I figure I may as well mention the correct terminology for anyone who wants to try searching online themselves.

Now, the symplectic form does, in fact, transform under coordinate changes just like you might expect any tensorial object over a manifold to do. If we take on faith that the symplectic form should transform as a tensor under coordinate changes, then we already know how the right-hand side of the rewritten Hamilton's equation transforms if we were to transform to some other coordinate system. Let's investigate the left-hand side.

Suppose we perform some transformation $$\xi^a=\xi^a(\zeta)$$ to a new coordinate system $$\zeta$$. Then by chain rule, $$\frac{d \xi^a}{d t}=\frac{\partial \xi^a}{\partial \zeta ^\mu}\frac{d\zeta^\mu}{d t},$$ so we see, perhaps unsurprisingly the time derivative also transforms as a tensor under a coordinate change (I used $$\mu$$ for the indices of the new coordinates just to keep things visually distinct).

So in the end, we find that Hamilton's equations transform as $$\frac{d \xi^a}{d t}=\Omega^{ab}\frac{\partial H}{\partial \xi^b}\implies \frac{\partial \xi^a}{\partial \zeta ^\mu}\frac{d\zeta^\mu}{d t} = \Omega^{ab}\frac{\partial \zeta^\nu}{\partial \xi^b}\frac{\partial H}{\partial \zeta^\nu},$$ which if we move the Jacobian on the left to the right, we find $$\frac{d\zeta^\mu}{dt}=\left(\frac{\partial\zeta^\mu}{\partial \xi^a}\Omega^{ab}\frac{\partial\zeta^\nu}{\partial\xi^b}\right)\frac{\partial H}{\partial \zeta^\nu}$$ and hence we see that if we define a new symplectic form $$\Omega^\prime$$ by $$\Omega^{\prime\, \mu\nu}=\frac{\partial\zeta^\mu}{\partial \xi^a}\Omega^{ab}\frac{\partial\zeta^\nu}{\partial\xi^b},$$ (which is equivalent to my assertion that $$\Omega$$ should be tensorial under coordinate changes) Hamilton's equation actually are invariant in the sense that we still have equations of the form $$\frac{d \zeta^\mu}{dt}=\Omega^{\prime\, \mu\nu}\frac{\partial H}{\partial\zeta^\nu}.$$ The only difference is that the components of $$\Omega$$ and $$\Omega^\prime$$ might not be the same. But really this shouldn't be such a shock after all this setup since we wouldn't expect the components of a tensor to remain the same after a coordinate change.

Consider as an example the Minkowski metric. We know what this looks like in Cartesian coordinates. If we changed to polar coordinates, for example, of course the component entries in the metric change, but it's still, in a very real sense, the same metric. We just have a new representation thereof.

So where to canonical transformations fit into all this? They are simply the very special coordinate transformations which actually leave the components of the symplectic form invariant. Formally, these are coordinate transformations generated by vector fields over phase space whose Lie derivative of the symplectic form vanishes. This is very similar in many respects to a vector field being a Killing vector of some metric.

Finally, I should point out that the way I have framed the entire discussion above, it may seem strange why we should consider canonical transformations at all. After all, we can use any transformation at the cost of a nice form for the symplectic form. Perhaps non-canonical transforms can put the equations in a nice form.

In principle this is of course true. However canonical transformations play a very essential role which is intimately tied to Noether's theorem and symmetry. Essentially one can guarantee that every symmetry of the action corresponds to precisely a canonical transformation. Furthermore, only vector fields which correspond to canonical transformations are guaranteed to have a charge associated to them (like the Hamiltonian is the charge associated to time evolution (which is itself a canonical transformation)).

• It looks like I could learn something here, but I had to stop at $\xi^aa=<p^a,q_a>$ because I can make no sense of it. $\xi$ appears to be a coordinate variable, and $a$ and index. Then why is there an $a$ sitting next to $\xi^a$? And what do your angle brackets denote? – garyp Mar 16 at 11:44
• @garyp That second $a$ is a typo. All I'm doing here is taking the two $n$-dimensional vectors of coordinates $p^a$ and $q_a$ and mashing them together into one big $2n$-dimensional vector $\xi^a$. – Richard Myers Mar 16 at 19:49
• @garyp I have edited my answer to try and make this a little clearer, but since all I'm doing is rewriting the standard Hamiltonian equations of motion, you should be able to use those to ground when I'm saying here before the coordinate transformation. – Richard Myers Mar 16 at 19:54
• Aha. Got it. I was trying to make the angle brackets into a product. If I had read on the context would have cleared it up. I think. – garyp Mar 17 at 18:31
1. A more geometric approach to the Hamiltonian formulation is to consider the $$(2n+1)$$-dimensional contact manifold $${\cal M}$$ with coordinates $$(q^i,p_j,t)$$. The Hamiltonian action functional is $$S_H[\gamma]~=~\int_I \gamma^{\ast} \Theta, \qquad \Theta~=~p_j \mathrm{d}q^j -H \mathrm{d}t, \tag{1}$$ where $$\gamma:I\to {\cal M}$$ is a curve.

From this it becomes clear that the Hamiltonian $$H$$ is not a scalar object but rather the $$t$$-component of a 1-form/co-vector, and therefore transforms non-trivially under coordinate transformations. This answers OP's main question. See also e.g. my related Phys.SE answer here.

2. Furthermore, the Hamiltonian formulation can be generalized to non-canonical coordinates, cf. e.g. my Phys.SE answer here.