# Deriving Hamilton's Equations from Conservation of Energy

1. Aug 3, 2013

### crastinus

This is an attempt by a novice to do something neat. Please feel free to correct me.

I am trying to see if one can derive the Hamiltonian equations of motion from a simple statement of the conservation of energy for mechanical systems.

My attempt:
1. For a system S with position q and a constant mass m, a change in q from a to b is q(a) + Δq = q(b).
2. If we set the time from a to b equal to 1, then Δq is the time derivative of q, so the velocity v.
3. We can define mv as momentum p.
4. The total energy H of S is equal to a function of momentum, kinetic energy T, plus a function of position, potential energy V: H = T + V. H then is a function of position q and momentum p.
5. We can vary q or p and discover how H varies with each.
a). When we vary q, the rate of change of q is equal to some rate of change of H.
b). When we vary p, the rate of change of p is equal to some rate of change of H.
6. When q varies with time, the rate of change of q is equal to the rate of change of H with respect to p, since p is the only other quantity that can vary. (The rate of change of q with respect to q would always be 1, no matter the value of q.) Thus dq/dt = ∂H/∂p.
7. When p varies with time, the rate of change of p is equal to the rate of change of H with respect to q, since q is the only other quantity that can vary. (Usually, one of these two rates of change is represented as negative with respect to the other. This is because F is defined opposite to the rate of change of potential energy, F = -(dV/dt), but one could presumably write such an equation with a different convention to make both rates of change represented as positive.) Thus, dp/dt = ∂H/∂q.

That completes the derivation.

Does that work? It seems almost too simple.

Or is there a better, more physics-y way to do this?

Thanks,
Tim

Last edited: Aug 3, 2013
2. Aug 8, 2013

### BucketOfFish

Some of the things you said are a bit iffy. For instance, you set time from a to b equal to 1 when it should actually be infinitesimal to get the correct time derivative of q. In your case, Δq would only be the correct time derivative if the velocity was constant within that time span. Also, you say that when we vary p, the rate of change of p is equal to some rate of change of H. This is definitely not true, since ∂H/∂p = (∂H/∂T)(∂T/∂p)+(∂H/∂V)(∂V/∂p) = (∂H/∂T)(∂T/∂p) = ∂T/∂p, which is not a constant but rather depends on the value of p (since T is quadratic in p). There are various other errors.

I think a simpler way of thinking about the Hamiltonian equations is just to use F=dp/dt. Since H=T+V=(1/2)p^2/m+V(q) when not relativistic, we see that F=-(∂U/∂q)=-(∂H/∂q)=dp/dt. We also see that (∂H/∂p)=p/m=v=dq/dt. I think I saw somewhere that Newton's laws can be made equivalent with energy conservation, but I cannot remember why at the moment. I hope this helped.

3. Aug 8, 2013

### crastinus

It did help. Thanks! I figured it was all muddled up.