Hamilton's Equations of Motion

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The discussion centers on the relationship between generalized forces and Hamilton's equations of motion. It asserts that when the generalized force F is not zero, the equation dH/dq = -pdot remains valid, as generalized forces are inherently included in the Lagrangian formulation. The Lagrangian, defined as L = T - V, incorporates the effects of forces through its derivatives, eliminating the need to explicitly include F in Hamilton's equations. The derivation shows that dL/dq leads to the momentum p, confirming that pdot = -dH/dq is consistently true. Overall, the analysis emphasizes that the force is fundamentally embedded within the Lagrangian framework.
jhat21
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Here's a quick one:

If the generalized force F is not zero,
does the equation
dH/dq = -pdot

become
dH/dq = F- pdot
?
 
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What have generalized forces have to do with the hamiltonian formulation of dynamics ??
 
dextercioby said:
What have generalized forces have to do with the hamiltonian formulation of dynamics ??
Everything. Hamiltonian dynamics concerns itself with positions and generalized momenta. A generalized force is the derivative of a generalized momentum.
 
Hamilton's equations are always:

\dot{p} = -\frac{\partial H}{\partial q} = f
\dot{q} & = & \frac{\partial H}{\partial p} = v
 
Last edited:
Oh because dL/dq always equals d/dt[ dL/dqdot ]!
no matter what the force is, the force is inclusive in dL/dq
since
L = T - V
and
dL/dq = dT/dq - dV/dq ,
where the force, F = -dV/dq
taking the partial derivative of L w/respect to position q takes care of our generalized force F, so we don't need to write it explicitly in the Hamiltonian equations. it's already covered by the Lagrangian in dL/dq, which we can express as the time derivative of the momentum, pdot.

the idea is that the force is in the Lagrangian!

Then when u derive the Hamiltonian equations from Lagrange's
u have dH/dq = - dL/dq
and using the above observation about dL/dq, we have
dL/dq = d/dt[ dL/dqdot ]

going to inside the brackets we have
dL/dqdot = d/dqdot [ 1/2 m qdot^2] = m qdot = p
which is always - always true.
and that means its time derivative is
d/dt[ dL/dqdot = p ] = pdot

so the conclusion is the same
the equation is always

pdot = - dH/dq

oh my god
dT/dq = 0
it's so obvious
then
dL/dq = F
lol
 
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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