I don't think that QM is essential for talking about PE, but PE is essential for talking about QM. So if you think that you are more comfortable with QM than with PE then I think you are kidding yourself.Nano-Passion said:
sophiecentaur said:
Science has little to do with fondness - it has more to do with what we learn from nature, if we like it or not. So, you may not like such ideas as energy, the relativity principle and the equivalence principle which indeed emerged from classical mechanics, but such things turned out to be very useful. Note that such concepts as energy only "explain" in a limited sense: they started out as book keeping concepts but turned out to be at least related to invisible but nevertheless physical entities. And I'm afraid that currently that's about as far as we can look "inside" - so that we have little choice but to stick to such superficial descriptions.Nano-Passion said:
PE is a basic concept in QM. If you are uncomfortable with PE then you simply cannot be comfortable with QM. It has nothing to do with intuition, just logic. It is an illogical position to say that you are comfortable with QM but not with PE when PE is part of QM.Nano-Passion said:
Velocity also depends on an arbitrary reference frame that is useful for calculation.Nano-Passion said:
Define "more real".Nano-Passion said:
A.T. said:
Nano-Passion said:
Khashishi said:
Berney123 said:
DragonPetter said:
A.T. said:
This is not entirely correct. If we have that ##\partial_{t}L = 0## then we know ##H = \text{const.}## where ##H## is the Hamiltonian but in the lab frame the Hamiltonian is not necessarily the total energy of the system. The standard example is that of a bead sliding on a hoop that rotates with constant angular velocity about the z-axis. In this case the Lagrangian is time translation invariant and thus implies the Hamiltonian is constant but it is not equal to the total energy in the lab frame (the total energy isn't even conserved for this system!). If you transform to the frame co-rotating with the hoop and associate a "potential energy" with the centrifugal force then this gives a physical interpretation of the Hamiltonian for this system. It should be stressed however that in the lab frame the total energy is not necessarily the Hamiltonian.physwizard said:
Energy conservation is very complicated in General Relativity. Non-stationary space-times do not have in general a globally conserved energy current. The RW metric (for the RW cosmological model) is an example of a non-stationary metric (no time-like killing vector field).physwizard said:
quite true. but here there are unknown forces of constraint involved. it would be really interesting if you could provide a real world classical mechanics example where all the forces involved are accounted for in the potential energy function (Edit: and the relation between generalized and cartesian coordinates is not does not have an explicit dependence on time.) and still the conservation of the energy function (which works out to be numerically equal to the hamiltonian in the hamiltonian formalism(just to avoid using the term 'hamiltonian' when dealing with the lagrangian formalism)) works out to be a different idea from the conservation of energy (T + V) .WannabeNewton said:
interesting. does this mean that the global energy conservation principle should be abandoned in GR? is there any experimental/observational evidence of this? (for eg. a star just disappearing suddenly without its energy showing up in any other form.)how would an observer perceive this loss/gain of energy?WannabeNewton said:
Hi phys! Thanks for responding. Let me think about this for a bit because I can't immediately grasp what you are asking for with regards to "real world" examples, thanks!physwizard said:
It has its uses when it can be applied (e.g. in the case of stationary space-times) so no it isn't abandoned but you should note that local energy conservation always applies i.e. ##\nabla^{a}T_{ab} = 0## always holds, so that prevents such things. The failure of the existence of global energy conservation for general space-times is rooted in the issue of isometries (or lack thereof) of general curved space-times and is quite an interesting textbook matter. I don't think a forum post could do it much justice unfortunately.physwizard said:
okay, let me just add that i would like an example where the relation between the generalized coordinates and cartesian coordinates does not have an explicit time dependence. i will edit my earlier forum post to include this.WannabeNewton said:
Ah but in this case the Hamiltonian would have to be the total energy. This can be seen as follows: let ##r_{\alpha}## be the cartesian coordinates and ##q_{\alpha}## the generalized coordinates so that ##r_{\alpha} = r_{\alpha}(q_{1},...,q_{n})## i.e. there is no explicit time dependence. ##\dot{r_{\alpha}} = \frac{\partial r_{\alpha}}{\partial q_{\beta}}\dot{q_{\beta}}## (implied summation over the ##\beta##'s) hence ##T = \frac{1}{2}\sum _{\alpha}m_{\alpha}\frac{\partial r_{\alpha}}{\partial q_{\beta}}\frac{\partial r_{\alpha}}{\partial q_{\gamma}}\dot{q_{\beta}}\dot{q_{\gamma}}##. Defining ## T^{\beta \gamma} = \sum _{\alpha}m_{\alpha}\frac{\partial r_{\alpha}}{\partial q_{\beta}}\frac{\partial r_{\alpha}}{\partial q_{\gamma}}## for convenience, we have that ##p^{\beta} = \frac{\partial \mathcal{L}}{\partial \dot{q}_{\beta}} = \frac{1}{2}(T^{\beta\gamma}\dot{q}_{\gamma} + T^{\gamma\beta}\dot{q_{\gamma}}) = T^{\beta\gamma}\dot{q}_{\gamma}## (here I have used symmetry of ##T^{\beta\gamma}##). Therefore we see that ##p^{\beta}\dot{q}_{\beta} = 2T## thus ##\mathcal{H} = 2T - \mathcal{L} = 2T - (T - U)) = T + U##.physwizard said:
Wannabe!WannabeNewton said:
you are a genius! I didn't realize this could be proved.
