Death by Physical Chemistry...PLEASE HELP!

[popjinx]

So, I was assigned this homework problem about Hamiltonians, and I have no idea where I should begin to even attempt to solve it...

 

Problem: A simple one-dimensional problem in physics is dropping an object of mass m from a height Xo under the influence of gravity, for which the Force=-mg, and the well-known trajectory is X=Xo-(1/2)gt^2. Find the Hamiltonian for this problem by explicitly evaluating the potential energy V, taking V=0 at X=0 (ground level), and show that dH/dt=0, or that H=E (total energy), a constant.

 

Potential Energy V= mass*gravity*height or in this case V= mgXo

[Tom Mattson]

Sorry this has gone on so long without a response, but here goes.

 

Potential Energy V= mass*gravity*height or in this case V= mgXo

 

No, that means that V is a constant, which isn't true. You are supposed to make V(x) such that V(0)=0. That means that V(x)=mgx.

 

The canonical way to start these problems is to begin by writing down the Lagrangian:

 

[imath]L(x,\dot{x},t)=T-V=(1/2)m\dot{x}^2-mgx[/imath].

 

Note that our Lagrangian does not actually have an explicit time dependence.

 

Now the Hamiltonian is a function of the momenta and the coordinates, not the velocities and the coordinates. So we find the momentum as follows:

 

[imath]p=\frac{\partial L}{\partial \dot{x}}[/imath]

[imath]p=m\dot{x}[/imath]

[imath]\dot{x}=p/m[/imath].

 

I solved for [imath]\dot{x}[/imath] in the last step because when we write down the Hamiltonian we will use that relation to eliminate [imath]\dot{x}[/imath].

 

How we get the Hamiltonian.

 

[imath]H=p\dot{x}-L[/imath]

[imath]H=p\dot{x}-(1/2)m\dot{x}^2+mgx[/imath].

 

Make the substitution [imath]\dot{x}=p/m[/imath] and simplify.

[DQW]

Alternatively, noticing that the Langrangian is not explicitly time-dependent, you can directly write H=T+V.

 

PS : This is a purely classical problem - nothing quantum about it (in fact, I don't see any physical chem in it either).

[Tom Mattson]

P-chem covers quantum mechanics. Since your average chemistry student does not take an intermediate course in mechanics, embedding in p-chem a crash course in classical mechanics is pretty standard.

[DQW]

Ah, I see.