Are Hamiltonian and Energy Always Equivalent in Classical Mechanics?

  • Thread starter Thread starter deadringer
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the relationship between Hamiltonian (H) and energy (E) in classical mechanics, specifically under certain conditions. It is established that H equals E when the potential energy (V) is independent of velocity. The kinetic energy (T) is defined as T = 1/2 m (q dot)², where q dot represents the velocity. The example of a bead sliding on a spinning hoop illustrates the complexities of defining T and V, particularly when considering additional kinetic energy contributions from angular velocity (w). The conclusion emphasizes that energy conservation is not guaranteed due to the time-dependent nature of the system.

PREREQUISITES
  • Understanding of Hamiltonian mechanics and its definitions
  • Familiarity with Lagrangian mechanics and the principle of least action
  • Knowledge of kinetic and potential energy concepts in classical mechanics
  • Basic grasp of angular motion and its effects on energy
NEXT STEPS
  • Study the derivation of the Hamiltonian from the Lagrangian formulation
  • Explore the concept of effective potential energy in rotating systems
  • Learn about the conservation of energy in non-conservative systems
  • Investigate the implications of time-dependent potentials on energy conservation
USEFUL FOR

Students and professionals in physics, particularly those focused on classical mechanics, Hamiltonian dynamics, and Lagrangian systems. This discussion is beneficial for anyone seeking to deepen their understanding of energy relationships in dynamic systems.

deadringer
Messages
32
Reaction score
0
We are asked to differentiate between H and E. I think that they are equal in some cirsumstances but am not sure what these are.
 
Physics news on Phys.org
Start from the definition. How is H(q,p,t) defined?
 
I start from the definition of H, and then plug in that p is the partial derivative of L wrt q dot.
The next stage is a bit iffy. I assume that the Kinetic energy can be assumed to be 1/2 m * ((q dot) squared), where q is the position vector. I argue that any other velocity dependent terms in the total energy should be be attributed to a time dependent potential. Then we get that the partial derivative of T wrt (q dot) is m*(q dot). Therefore H = T + V plus the scalar product of q dot with the partial derivative of V wrt (q dot).

This would indicate that H = E only if V is velocity independent. The only problem is I'm not sure if my assumption of the form of T is correct.

I have another related question.

"A bead of mass m slide without friction under gravity on a massless circular hoop of radius a which is set spinning with angular speed w about a vertical diameter. Show that it H = 1/2 m (a* (theta dot))^2 + Veff and write out Veff."

The bead clearly has two perpendicular velocities; in the plane of the ring (due to theta dot) and perp. to this plane (due to w), as well as the gpe.
We set L = T - V, but I'm unsure whether to consider the kinetic energy due to w as a part of T or V.
According to the (possibly eroneous) definition of T =1/2 m * ((q dot) squared), the w kinetic term should be considered an addition to V, and therefore should be subtracted in the Lagrangian, but obviously if it is considered an addition to T is should be added in L.

We are also asked to "explain why the energy of the bead is not constant". I'm confused about this because even the augmented potential is not velocity dependent, therefore we should get H = E, and the partial derivative of L wrt t is zero, therefore the total derivative of H should be zero, therefore E should be conserved.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K