popjinx Posted September 5, 2005 Share Posted September 5, 2005 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 Link to comment Share on other sites More sharing options...
Tom Mattson Posted September 12, 2005 Share Posted September 12, 2005 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. Link to comment Share on other sites More sharing options...
DQW Posted September 13, 2005 Share Posted September 13, 2005 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). Link to comment Share on other sites More sharing options...
Tom Mattson Posted September 13, 2005 Share Posted September 13, 2005 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. Link to comment Share on other sites More sharing options...
DQW Posted September 13, 2005 Share Posted September 13, 2005 Ah, I see. Link to comment Share on other sites More sharing options...
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