Is potential energy real or fictious?

[sophiecentaur]

Science has no answer to this question, so it's more a question of philosophy.

The real "why" questions are, as you say, not Science and can only be 'answered' by those pesky Philosophers. I find it very annoying that Science education assumes that there are answerable 'why' questions when what they really mean is "how does this work?" or, even more basic, "what happens when ...?"
I am often amazed that so few people in this world (with the exception of some PF members haha) seem to accept this basic fact of life.
I can feel this thread getting moved away from its place in General Physics.

 

[Berney123]

When we try to explain why, almost everything will resort to philosophy and as more and more questions are answered the more and more difficult it will be for these answers to be explained. I am also amazed that some people can just accept things without asking why. Thanks so much for the thoughtful answers to this very difficult questions.

 

[sophiecentaur]

When we try to explain why, almost everything will resort to philosophy and as more and more questions are answered the more and more difficult it will be for these answers to be explained. I am also amazed that some people can just accept things without asking why. Thanks so much for the thoughtful answers to this very difficult questions.

You are not making the distinction between "explaining why" and 'describing roughly how it works'.
The 'how it works' approach actually has an answer because it uses a model that you already understand to describe a new phenomenon. The 'why' question can be answered at so many levels that it is almost meaningless.

See the famous 'Why" question that Richard Feynman dealt with in his quirky fashion.

Everyone asks 'why' about things but they all have different frames of reference. If the answer they get is enough to satisfy them (or you) then they and you will get a nice feeling about it. The reason that many people stop asking 'why' is that the answers they keep getting have got too hard for them to cope with or they are actually just too plain busy to find the time to pursue it further.

 

[JayJohn85]

Consider someone on the top floor of a 400 floor building, and someone on the ground floor.

If they both have the same human body and are going to push 100kg desks, then they both have the same potential energy to do work on these desks, regardless of their gravitational potential energy with respect to each other. The way you asked your original question about their capacity to do work while being at different heights, and your implication that you wonder if potential energy is not real lead me to think that your contention is that the increase in gravitational potential energy gives the person a higher absolute potential energy with respect to the objects that they might do work on, as in your example objects on the same floor as the person. If that were true, which it isn't, then it would make sense for you to question if potential energy is not real since this is not what we experience.

Just because one person has a higher gravitational potential energy does not mean that that potential energy is relevant when you compare two people's ability to do work. When you talk about their capacity to do work, such as pushing a desk, you are using 2 different potential energy references for the two cases, and so it does not make sense to say a person on a higher floor has more potential energy when any work they do on that floor will be at the same gravitational potential energy, and thus cancel out when compared to someone on a different floor in the building.

This is an awesome post. Basically your saying the energy is different potential chemical energy and potential gravitational energy. That the gravitational energy of the guy on the high floor gets canceled out by the floor? Same as the guy on the bottom. The top guy would only have more potential if he was to come crashing down of the top floor onto like a spring or something.

But if they had the same mass and potential for work they would have the same chemical potential energy. Did I understand this correctly?

 

[physwizard]

All forms of energy are introduced just to make conservation of energy possible. The is no other purpose to the whole concept of energy, than to have a conserved quantity.

this sounds interesting. but wouldn't the very fact that it is possible to maintain such a conserved quantity be of physical significance?
the lagrangian and hamiltonian formalisms give a good physical significance to energy. in fact i would say that these formalisms give a very concrete definition of, and significance to the concept of energy. if the lagrangian is time translationally invariant, then energy is conserved. before these formalisms were known, it had been the fashion to define new forms of energy whenever it was observed that a particular form of energy was disappearing.
if energy in the universe is conserved, then it has a very strong physical significance. it means that the laws of physics do not change with time!
as per relativity, energy contributes to the mass, though i don't know how much this has been tested in the laboratory for the different forms of energy.
not sure of how energy conservation would shape up in the light of quantum mechanical uncertainty though.

 

[WannabeNewton]

the lagrangian and hamiltonian formalisms give a good physical significance to energy. in fact i would say that these formalisms give a very concrete definition of, and significance to the concept of energy. if the lagrangian is time translationally invariant, then energy is conserved.

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.

if energy in the universe is conserved, then it has a very strong physical significance. it means that the laws of physics do not change with time!

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]

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.

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) .

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).

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]

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 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) .

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!

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?

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]

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!

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.
"real world" would mean something which can be built and tested quite easily in the laboratory, in other words, not some kind of theoretical thought experiment.

 

[Johnahh]

"real world" would mean something which can be built and tested quite easily in the laboratory, in other words, not some kind of theoretical thought experiment.

Isn't GR based on the thought experiment of a light clock? I'm pretty sure the effects of GR are "real".

 

[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.