Is potential energy real or fictious?

[WannabeNewton]

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.

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 $\mathcal{H} = 2T - \mathcal{L} = 2T - (T - U)) = T + U$.

EDIT: BTW I forgot to mention that this only works for Lagrangians that can be written in the form $\mathcal{L} = T - U$

 

[Feinman]

I have related question about thermodynamic potentials. We learn (and teach) the first law as a conservation law and then the second law describes the requirement to maximize the entropy. We introduce the other thermodynamic potentials and describe how energy needs to be minimized, for example, that deltaG is a indicator of equilibrium and spontaneity and that either the entropy goes up or the enthalpy goes down (or both) for deltaG to be negative. So, how do we get the energy minimization principle into the picture. Many books (I think many) don't directly address eh question and Callen I found suspicious

 

[physwizard]

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 $\mathcal{H} = 2T - \mathcal{L} = 2T - (T - U)) = T + U$.

EDIT: BTW I forgot to mention that this only works for Lagrangians that can be written in the form $\mathcal{L} = T - U$

Wannabe! you are a genius! I didn't realize this could be proved.
I guess the question here is whether the concept of energy and its conservation can be attributed some physical significance. Its true that many of the physical systems are driven by second order differential equations. Integrating these once will give rise to constants of the motion - functions of the state of the system whose values do not change with time. Energy is one such constant of the motion. Whether we should just view this as a calculational convenience or as something having physical significance? I guess most of us (even people without a physics background) have a good intuitive feel of the concept of energy and its conservation, we see it happening often enough even in our daily lives. So its not really an 'abstract' constant of motion which is useful as a calculational tool but difficult to perceive, it is often of substantial help in understanding phenomenon.
But I guess I still haven't been able to answer the original question. To some extent, it may be a philosophical point.

 

[WannabeNewton]

I agree with you that while you can have arbitrary constants of motion that don't have any immediate "physical significance", energy can usually be given some physical interpretation, at the least using correspondence laws. As an example, if we consider a particle falling freely in Schwarzschild space-time, we know its equations of motion satisfy $u^{a}\nabla_{a}u^{b} = 0$ and we also know that Scwarzchild space-time, on account of it being stationary, has a time-like killing vector field $\xi^{a}$ which by definition satisfies $\nabla_{(a} \xi_{b)} = 0$. Combining these together, we see that the quantity $E = -\xi_{a}u^{a}$ is a constant of motion along the worldline of this free falling particle because $u^{a}\nabla_{a}(-\xi_{b}u^{b}) = -\xi^{b}u^{a}\nabla_{a}u^{b} - u^{a}u^{b}\nabla_{a}\xi_{b} = - u^{(a}u^{b)}\nabla_{(a}\xi_{b)} = 0$.

Now this quantity $E$ ostensibly seems abstract and not of any immediate physical significance but if we compute it in the coordinate basis, using the Schwarzschild metric, we see that $E = -u^{\mu}\xi_{\mu} = (1 - \frac{2M}{r})\frac{dt}{d\tau}$. While this may not immediately strike someone as "energy" in its current mathematical form we see that in the far-field limit i.e. $r>>M$ this is nothing more than the special relativistic expression for the energy per unit mass of a particle as measured by a static observer. So even here, even though this energy per unit mass is of course a powerful computational tool, it is also something that can be given physical significance via correspondence laws. It is as you said very important in understanding geodesic motion in Schwarzschild space-time because it, along with the conserved angular momentum, allow for the study of trajectories through potential energy diagrams in much the same way as Newtonian mechanics and as you very well know these potential energy diagrams tell a lot about the physical nature of trajectories.

 

[Feinman]

My understanding is that force is real enough. We can feel it an measure it with m and a. If the force is the gradient of some f(x), that defines a potential energy, no? Where does the energy go as x gets smaller? Either into kinetic energy or some physical disturbance (rock hits the ground). Still trying to morph that into thermodynamic potentisls.