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The last step of this derivation of Hamilton's Equations is what's making me doubt it. It is as follows:

Assuming the existence of a smooth function $\mathcal{H}(q_i,p_i)$ in $(q_i(t), \,p_i(t))$ phase space, such that it obeys the following (taken as a postulate): \begin{equation} \frac{d\mathcal{H}}{dt}=0 \end{equation} Therefore: \begin{equation} \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}=0 \end{equation} Each additive term in the above equation must therefore either be constant or, more generally, a function of $(q_i,\,p_i)$. So: \begin{equation} \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}=\alpha(q_i,p_i) \end{equation} and: \begin{equation} \dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}=-\alpha(q_i,p_i) \end{equation} (This is the part I wish to ask about) The nontrivial case implies therefore that: \begin{equation} \dot{q_i}=f(q_i,p_i;t)\frac{\partial\mathcal{H}}{\partial p_i} \end{equation} and \begin{equation} \dot{p_i}=-f(q_i,p_i;t)\frac{\partial\mathcal{H}}{\partial q_i} \end{equation} For some function $f(q_i,p_i;t)$.

Is that step justified according to you? If not, why?

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  • $\begingroup$ @AccidentalFourierTransform Those things you divide by might be zero, in fact for a free particle, you are dividing by zero. $\endgroup$ – Timaeus Jan 19 '16 at 20:45
  • $\begingroup$ @Timaeus : For a free particle, $\alpha$ will be zero so it doesn't make sense to go further there does it? $\endgroup$ – Prish Chakraborty Jan 19 '16 at 20:51
  • $\begingroup$ @AccidentalFourierTransform Which means the argument from one to the other is necessary to include both cases. And frankly this wouldn't be of too much concern except for the famous cases where people do divide by zero in situations with different systems, don't fix it, and then falsely claim different systems are actually equivalent. Isn't it Hamilton-Jacobi that fails to have circular motion because it isn't actually the same as Hamiltonian Mechanics? $\endgroup$ – Timaeus Jan 19 '16 at 20:57
  • $\begingroup$ It fails for a constant H. Does it work for velocity dependent forces? And do you really expect to get equations of motion just from knowing that you have a constant of the motion and nothing else? $\endgroup$ – Timaeus Jan 19 '16 at 21:54
  • $\begingroup$ @PrishChakraborty Could you give a reference for this derivation? This looks very atypical (e.g. from the outset, the Hamiltonian doesn't need to be a constant of the motion). $\endgroup$ – Arturo don Juan Jan 19 '16 at 22:15
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Assuming the existence of a smooth function $\mathcal{H}(q_i,p_i)$ in $(q_i(t), \,p_i(t))$ phase space, such that it obeys the following (taken as a postulate): \begin{equation} \frac{d\mathcal{H}}{dt}=0 \end{equation} Therefore: \begin{equation} \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}=0 \end{equation}

That's fine, that's your assumption.

Each additive term in the above equation must therefore either be constant or, more generally, a function of $(q_i,\,p_i)$.

That's wrong. Actually, it could be a function of time as well. The easiest example can be based on the additional fact that there isn't a unique equation of motion if you have just the one postulate above.

You have a vector field in phase space: $$(q_i,p_i)\mapsto (\frac{\partial \mathcal{H}}{\partial q_1}, \dots, \frac{\partial \mathcal{H}}{\partial q_n}, \frac{\partial \mathcal{H}}{\partial p_1}, \dots, \frac{\partial \mathcal{H}}{\partial p_n}).$$

And you have a curve in phase space: $$t\mapsto \left(q_1(t),\dots,q_n(t),p_1(t),\dots, p_n(t)\right).$$

And you are basically saying that the tangent of the curve is orthogonal to the gradient of $\mathcal{H}$ so that you stay at a constant $\mathcal{H}.$ Now if you just imagine $\mathcal{H}$ as height then you can imagine you are walking around in a $2n$ dimensional space, always staying at the same elevation.

In your problem you only have two assumptions, the curve moves along a constant height, and the height function is smooth. In particular you don't have that there is a unique tangent assigned to every point or that the curve doesn't cross itself at an angle. But you can imagine walking on a smooth height in a way that crossed itself. If you can't imagine it on your own, here is a way to make an example. Start with a region of non constant height, and then take two curves that cross each other at an angle, then in a region disjoint from where they cross, smoothly adjust the height function to be level in that region. And then in that region we are free to connect the curves.

Why? because you have only two assumptions, that height is a smooth function of your $2n$ variables and that your tangent to your curve is orthogonal to the gradient so the height stays constant. There is no rule that what you call $q$ and what you call $p$ are related to each other in any way, or that you have unique curves. Normally, if there was a Lagrangian and an action and the Hamiltonian was a Legendre transformation of that Lagrangian as usual. Then you can try to have unique dynamics and a particular relationship between the $q$ and the $p.$

So what does this mean. It means that you need time. You can't get the tangent from the point in phase space, and so while the gradient of $\mathcal{H}$ depends on just the point in phase space, the tangent can change in time and the two parts are still equal and opposite, but they could be different at different times. Therefore your next step

So: \begin{equation} \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}=\alpha(q_i,p_i) \end{equation} and: \begin{equation} \dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}=-\alpha(q_i,p_i) \end{equation}

is wrong too, for the same reason, you'd need time as well.

The nontrivial case implies therefore that: \begin{equation} \dot{q_i}=f(q_i,p_i;t)\frac{\partial\mathcal{H}}{\partial p_i} \end{equation} and \begin{equation} \dot{p_i}=-f(q_i,p_i;t)\frac{\partial\mathcal{H}}{\partial q_i} \end{equation} For some function $f(q_i,p_i;t)$.

Almost every step of your reasoning was incorrect so far. And your conclusion is wrong anyway, so there isn't a correct version of the reasoning. Originally you had a tangent to your curve be orthogonal to your gradient. Then you broke it into two vectors that added up to your tangent and said the inner products of those vectors with your gradient needed to give equal and opposite scalars (and the scalars depend on time since the tangent can depend on time). But each of those inner products involved a sum from $1$ to $n.$ And now in your conclusion you act as if $\dot q_1$ has to be zero if $\partial \mathcal{H} /\partial p_1$ is zero. But your assumptions made no relationship between any of the $2n$ variables $q_1,$ $\dots,$ $q_n,$ $p_1,$ $\dots,$ $p_n.$ So you can have 6 variables and have the height not depend on one of them and call that one $p_1$ and then you can pick one variable that does change in time and call it $q_1.$ Because your assumptions never required that any variable be related to any other variable.

Basically if you make fewer assumptions than normal, you shouldn't give things names that make you think they have properties you didn't assume they had. And a proper proof should appeal only to your assumptions. In this case a smooth dependency on $2n$ variables and a tangent that is orthogonal to the gradient so the height stays constant. That's all you assumed, so almost everything you concluded failed to follow from your assumptions.

And it's really poor form to ask why something doesn't follow. In a proof you are supposed to explain why your results follow from your assumptions. You have results that don't follow from your assumptions, hence your proof has to be wrong. And your proof never even appealed to your assumptions, you just had line after line of untrue things that neither follow from your previous untrue things nor follow from your assumptions.

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  • $\begingroup$ Thanks for the detailed answer. Clears up many holes. If you don't mind, could you explain what the physical significance of this equation is: $dH/dt=\partial H/\partial t$? Does it imply that time appears in H as a variable and not a parameter? $\endgroup$ – Prish Chakraborty Jan 24 '16 at 18:11

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