E=MC^2 Mass and Energy, synonymous?

[WannabeNewton]

Do you have a reference as to whether the Lagrangian for a particle interacting with any classical field propagating on background flat Minkowski space-time always lends to a Hamiltonian that is the total energy? It is certainly true in the specific case of the electromagnetic field (Goldstein Edition 3 page 342). Regardless, I do not see how this justifies using this narrow definition of energy as an all encompassing definition of said quantity.

 

[Dale]

I'm not a forum regular nor do I ever intend on being one.

That's too bad.

This was already explained above and rexplained when DaleSpam was unable to understanding it the second and third time.

Reasons why that attempt is flawed
1) Not all systems can be described by a Lagrangian

I understood that, and explicitly mentioned that as a reason why the definition of energy was theory-specific and even formulation-specific. In turn, this was, I think, what Feynman was describing.

2) The quantity you speak of is known as, among many other names, Jacobi's integral and given the letter h. h does not always equal the energy. It may even happen that h is constant but not the energy

The Jacobi integral is much more limited in scope (three-body gravity) than the time-symmetry of the Lagrangian. Your arguments against the Jacobi integral are a straw man fallacy since I never listed that as even a potential definition.

3) In those instances where h is the energy of the system then its only mechanical energy, which is a well defined quantity. Energy, on the other hand, comes if many other forms.

The Lagrangian for an isolated system of EM is also time invariant, leading to a conserved EM energy also, not just a conserved mechanical energy. So while your criticism may be valid for your strawman Jacobi, it is not valid for the actual Noether definition of energy.

I suggest that you take a look in The Feynman Lectures on Physics, Vol I, Feynman, Leighton, and Sands, Addison Wesley, (1963)(1989). pages 4- to 4-2

Where he never makes your specific claim that "energy defies definition".

DaleSpam - Please understand that I can't be insulted into responding to your attempts at an argument nor can I be coerced into posting just because you make a claim about how right you are.

Please . I never insulted you. Your argument is disproven and for some reason you choose to mention how satisfied your responses made you feel. That is irrelevant and I merely pointed it out.

You repeatedly dismiss the "work" definition of energy, but the only actual argument you provided against it was that momentum could qualify, which I rebutted and apparently you agreed with the rebuttal since you didn't even attempt to refute it and didn't bring it up again.

Your only remaining argument is your claim that the definitions define "forms of energy" rather than "energy itself". That doesn't seem correct at first glance since none of the definitions of energy claim to be definitions of "forms of energy" nor do they seem limited to any specific set of known forms of energy, but you haven't been able to clarify your meaning well enough to tell if this final argument has any merit.

In any case, I have done far more than merely claim how right I am. I have disproven your primary position by counterexample and rebutted the bulk of your arguments on logical grounds. I do understand your unwillingness to proceed, but you have not been mistreated, nor insulted, nor ignored, only refuted.

 

[Dale]

The simplest example is that of a hoop rotating about the z-axis with constant angular velocity and a bead sliding without friction on the hoop. The Lagrangian will be time translation invariant and will thus lead to a conserved Hamiltonian but the total mechanical energy is not conserved; in this system the Hamiltonian energy is not even equal to the total mechanical energy.

I am interested in this example. So what is the conserved Noether charge, if not energy?

Without working it out it does seem that mechanical energy should be conserved since the hoop rotates at constant angular velocity and the bead experiences no friction so it should also rotate with constant angular velocity. At first glance there appears to be no mechanical potential energy and no change in kinetic energy, so what is not conserved?

The point is that there are many different types of energies so you suggesting that the energy be unequivocally defined as the quantity that comes out of time invariant Lagrangians is nonsensical I'm afraid. Furthermore, energy makes sense for systems where a Lagrangian can't even be defined.

Agreed.

 

[WannabeNewton]

Mornin' DaleSpam! Consider the setup again: we have a hoop of radius $R$ and mass $M$ rotating about the $z$-axis with prescribed constant angular velocity $\Omega$ and we have a bead of mass $m$ sliding without friction along the hoop. Fix the origin of the coordinate system to the center of the hoop and let $\varphi$ be the angle that the position vector to the bead (the vector from the origin to the bead) makes with the rotation axis of the hoop. The potential energy of the bead is then $U = -mgR\cos\varphi$ and the kinetic energy is $T = \frac{1}{2}mR^{2}\dot{\varphi}^{2} + \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2}$. The first term in $T$ is just the kinetic energy of the bead coming from its velocity tangential to the hoop. Since the hoop is itself rotating with angular velocity $\Omega\hat{z}$, the bead acquires an additional velocity tangential to the rotation of the hoop with magnitude $v = \left \| \mathbf{r} \times \mathbf{\Omega}\right \| = (R\sin\varphi) \Omega$ which gives rise to the second term in $T$.

Our Lagrangian is then $L = \frac{1}{2}mR^{2}\dot{\varphi}^{2} + \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} + mgR\cos\varphi$. Notice that $\frac{\partial L}{\partial t} = 0$ therefore the Hamiltonian $H = \sum p_{i}\dot{q_{i}} - L = \text{const.}$ and is given by $H = \frac{1}{2}mR^{2}\dot{\varphi}^{2}- \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} - mgR\cos\varphi$ whereas the total mechanical energy is given by $E = \frac{1}{2}mR^{2}\dot{\varphi}^{2}+ \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} - mgR\cos\varphi$ so the total energy is not equal to the Hamiltonian in this case. The reason $E$ is not conserved is because whatever is keeping the hoop rotating at a prescribed constant speed must be doing work on the system.

If you however let the hoop rotate freely, so that its azimuthal position is no longer prescribed but rather a generalized coordinate, then you will find that $E$ for this new system is conserved and that it will be equal to the Hamiltonian $H$.

 

[Dale]

I will have to look into the math in detail tomorrow as I have a full day of work and an assignment due. However, I had a few notational quenstions.

Mornin' DaleSpam! Consider the setup again: we have a hoop of radius $R$ and mass $M$ rotating about the $z$-axis

Is the z axis parallel to gravity or perpendicular?

with prescribed constant angular velocity $\Omega$ and we have a bead of mass $m$ sliding without friction along the hoop. Fix the origin of the coordinate system to the center of the hoop and let $\varphi$ be the angle that the position vector to the bead (the vector from the origin to the bead) makes with the rotation axis of the hoop.

How can $\varphi$ ever be anything other than 90º if the bead is constrained to be along the hoop? I would think that you would want an angular variable specifying the angle around the hoop.

The potential energy of the bead is then $U = -mgR\cos\varphi$

OK, from your initial description I didn't realize that you were doing this in the presence of gravity. This makes more sense. This potential makes sense for a horizontal z and for $\varphi$ the angle from the horizontal perpendicular to the axis of rotation.

and the kinetic energy is $T = \frac{1}{2}mR^{2}\dot{\varphi}^{2} + \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2}$. The first term in $T$ is just the kinetic energy of the bead coming from its velocity tangential to the hoop. Since the hoop is itself rotating with angular velocity $\Omega\hat{z}$, the bead acquires an additional velocity tangential to the rotation of the hoop with magnitude $v = \left \| \mathbf{r} \times \mathbf{\Omega}\right \| = (R\sin\varphi) \Omega$ which gives rise to the second term in $T$.

I don't see how the rotation of the hoop gives any kinetic energy to the bead. I would think that the rotation of the hoop only gives a constant KE to the hoop itself.

 

[WannabeNewton]

Here is the picture of the system DaleSpam: http://s21.postimg.org/vz4nttwqu/IMG_0568.jpg

The extra kinetic energy term arises because the bead is also swirling around with the hoop, since it is constrained to stay on the hoop as the hoop rotates. So on top of the kinetic energy the bead has from sliding around on the hoop, which would be there even if the hoop wasn't rotating, it now also has an extra kinetic energy term due to the hoop actually swirling around. You are correct that the hoop itself also has a kinetic energy term but I abused the Lagrangian a little by excluding this since it will drop out of the equations of motion anyways. You can include the kinetic energy of the hoop itself if you want, the discrepancy between the Hamiltonian vs Total Energy won't change.

 

[VantagePoint72]

This is a neat example, WannabeNewton, but I don't think it supports your larger point that the Noether definition of energy is limited. I think I agree with the point, I just don't think this supports it.

As you pointed out, the reason for the failure of the Hamiltonian to agree with the mechanical energy in this problem is that the condition that the hoop rotates with constant angular frequency effectively "smuggles in" an outside force. Hence, all it really demonstrates is that if you're not very careful, it's possible to construct a system in which work is being done by/on the environment without that being immediately obvious. That the Hamiltonian-as-energy definition fails in such cases in not surprising.

If you construct a Lagrangian for the total isolated system—the bead, the hoop, and whatever is interacting with the hoop to make it rotate with constant angular velocity—then, as far as I can tell, the Hamiltonian should once again agree with the total energy. So, I take this example more as a cautionary tale about what system constraints can subtly imply for outside forces then a genuine refutation of the Hamiltonian definition of mechanical energy.

 

[Dale]

Here is the picture of the system DaleSpam: http://s21.postimg.org/vz4nttwqu/IMG_0568.jpg

D'oh! Thanks for the picture. I was totally misunderstanding your intended scenario. I thought that the hoop was rotating about its own axis. You mean the hoop rotating perpendicular to its own axis, so the normal to the plane of the hoop is rotating in a horizontal plane.

OK, tomorrow I will look at the details again.

 

[PeterDonis]

In what situations does this fail?

It fails for any system that is not time translation invariant. For example, the universe as a whole is not; it's expanding. So there's no time translation symmetry, hence no invariant scalar derived from it.

 

[WannabeNewton]

If you construct a Lagrangian for the total isolated system—the bead, the hoop, and whatever is interacting with the hoop to make it rotate with constant angular velocity—then, as far as I can tell, the Hamiltonian should once again agree with the total energy. So, I take this example more as a cautionary tale about what system constraints can subtly imply for outside forces then a genuine refutation of the Hamiltonian definition of mechanical energy.

But how would you take into account the potentially non-conservative external forces doing work on the system in the Lagrangian if you were to include them in the system? There are ways of taking into account non-conservative forces in the Euler-Lagrange equations themselves using virtual work but how would you incorporate them into the Lagrangian itself?

Personally, I would say the concept of energy is much better codified in its various forms within the field of thermodynamics.

Interestingly, and unrelated to my previous comment, one can define notions of momentum and energy at spatial infinity for an entire asymptotically flat space-time using the Hamiltonian formulation of general relativity. This is called the ADM energy-momentum.

EDIT: I found the newly revised version of the original paper by Arnowitt et al. yey! Here it is: http://arxiv.org/pdf/gr-qc/0405109v1.pdf

 

[VantagePoint72]

But how would you take into account the potentially non-conservative external forces doing work on the system in the Lagrangian if you were to include them in the system? There are ways of taking into account non-conservative forces in the Euler-Lagrange equations themselves using virtual work but how would you incorporate them into the Lagrangian itself?

What makes you so sure the forces would be non-conservative? It's just a cyclic process of the hoop doing work on the (whatever) and the (whatever) doing work on the hoop. Admittedly, I can't, off hand, imagine what (whatever) needs to look like in order to drive the hoop at a constant angular frequency, but it looks to me like it would be a non-dissipative process (in the idealization, naturally).

Besides, non-conservative forces are generally just a short-cut. A full (and I mean full) Lagrangian of the bead+hoop+environment would not have them: friction is nothing more than the transfer of kinetic energy of a macroscopic object to a bunch of microscopic particles. However, rather than writing down a Lagrangian for every single particle in the system, we just take a short cut and say that friction causes some of the kinetic energy to be lost to thermal energy. But thermal energy is just kinetic energy by a different name! Of course, to write down this full Lagrangian, you have to go well beyond classical mechanics, which is why I think it's fair to criticize the Hamiltonian-as-energy definition—at least, it's fair to do so in classical physics.

So, to sum up, I don't think it's immediately clear that a Lagrangian for the full, isolated system would require dissipative forces. Even if it did, that's just because we're only using an approximate description of the system in which the composition of matter is neglected. So, as I said, I think interpreting your example as proof that the Hamiltonian is not a good definition of the total energy is a mistake.

 

[WannabeNewton]

I should have been clear in that I wasn't referring to the hoop scenario necessarily but rather a general scenario.

 

[VantagePoint72]

I see. While, from the very beginning I think I was clear that I criticizing your particular example, even going as far as saying that nonetheless "I think I agree with the point". With respect to the question of whether the Hamiltonian is suitable as the definition of energy, I am arguing that the bead and hoop example do not prove anything either way because they don't fulfill the conditions of the proposed definition (one of which is that the system be isolated). Do you agree?

 

[WannabeNewton]

I see. While, from the very beginning I think I was clear that I criticizing your particular example, even going as far as saying that nonetheless "I think I agree with the point". With respect to the question of whether the Hamiltonian is suitable as the definition of energy, I am arguing that the bead and hoop example do not prove anything either way because they don't fulfill the conditions of the proposed definition (one of which is that the system be isolated). Do you agree?

Oh sure I wasn't using it as an example refuting the universality of the Hamiltonian within classical systems. I was just giving an example of a situation where you cannot naively assume the Hamiltonian and the total mechanical energy of the assumed system are the same which is what I interpreted HomogenousCow as asking an example of. I think my above post regarding the ADM energy-momentum if anything provides somewhat more of a support that the Hamiltonian notion of energy even carries over to entire space-times (albeit restricted to asymptotically flat ones).

My overall point, examples aside, was that one cannot take the Hamiltonian as the end all be all, unequivocal "definition" of energy.

 

[HomogenousCow]

Is the hamiltonian always equal to total energy when we have a closed system where the particles only interact with each other through the field?

 

[WannabeNewton]

There are explicit conditions that the Lagrangian must satisfy in order for the Hamiltonian to equal to the total energy for a given system. See Goldstein edition 3 page 339.

 

[HomogenousCow]

What makes you so sure the forces would be non-conservative? It's just a cyclic process of the hoop doing work on the (whatever) and the (whatever) doing work on the hoop. Admittedly, I can't, off hand, imagine what (whatever) needs to look like in order to drive the hoop at a constant angular frequency, but it looks to me like it would be a non-dissipative process (in the idealization, naturally).

Besides, non-conservative forces are generally just a short-cut. A full (and I mean full) Lagrangian of the bead+hoop+environment would not have them: friction is nothing more than the transfer of kinetic energy of a macroscopic object to a bunch of microscopic particles. However, rather than writing down a Lagrangian for every single particle in the system, we just take a short cut and say that friction causes some of the kinetic energy to be lost to thermal energy. But thermal energy is just kinetic energy by a different name! Of course, to write down this full Lagrangian, you have to go well beyond classical mechanics, which is why I think it's fair to criticize the Hamiltonian-as-energy definition—at least, it's fair to do so in classical physics.

So, to sum up, I don't think it's immediately clear that a Lagrangian for the full, isolated system would require dissipative forces. Even if it did, that's just because we're only using an approximate description of the system in which the composition of matter is neglected. So, as I said, I think interpreting your example as proof that the Hamiltonian is not a good definition of the total energy is a mistake.

This is what I was thinking of, in a full relativistic theory all the interactions should be mediated by the field.

 

[Dale]

Our Lagrangian is then $L = \frac{1}{2}mR^{2}\dot{\varphi}^{2} + \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} + mgR\cos\varphi$. Notice that $\frac{\partial L}{\partial t} = 0$ therefore the Hamiltonian $H = \sum p_{i}\dot{q_{i}} - L = \text{const.}$ and is given by $H = \frac{1}{2}mR^{2}\dot{\varphi}^{2}- \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} - mgR\cos\varphi$ whereas the total mechanical energy is given by $E = \frac{1}{2}mR^{2}\dot{\varphi}^{2}+ \frac{1}{2}m(R\sin\varphi)^{2}\Omega^{2} - mgR\cos\varphi$ so the total energy is not equal to the Hamiltonian in this case. The reason $E$ is not conserved is because whatever is keeping the hoop rotating at a prescribed constant speed must be doing work on the system.

So you have basically gone over all of this with other posters, but I thought I would put in my thoughts anyway.

In this case, it seems to me that H is the correct expression for the total energy of the system. You are absolutely correct that it is not equal to the mechanical energy of the bead. The difference between the mechanical energy of the bead and the total energy is some other unspecified energy that must be present in order for the system to have a time-invariant Lagrangian. (It cannot be a non-conservative force or the Lagrangian could not be time invariant since eventually the energy lost to a non-conservative force will be used up.)

This definition of energy is still not general since it only applies to systems described by a Lagrangian and whose Lagrangian is time-invariant. But I think that where it does apply it correctly defines energy.

One thing that I like about your example is that it is a counter-example to Popper's assertion that the posted definitions define forms of energy. In this Lagrangian the form of the additional energy is completely undefined. We don't know what form it is, but we nonetheless know that it is present. He could still take a philosophical stance that the definitions define the amount of energy but not what energy itself "actually really" is (which is what I believe Feynman was doing). In any case, the "forms" of energy argument is disproven.

 

[WannabeNewton]

Interestingly enough, you can give meaning to the term in $H$ that differs from the term in $E$ if you identify that term in $H$ as the "potential energy" of the centrifugal force on the bead in the frame co-rotating with the hoop. This allows you to interpret $H$ in the "conventional" way.

 

[Dale]

Interestingly enough, you can give meaning to the term in $H$ that differs from the term in $E$ if you identify that term in $H$ as the "potential energy" of the centrifugal force on the bead in the frame co-rotating with the hoop. This allows you to interpret $H$ in the "conventional" way.

That is interesting. I was aware that centrifugal forces etc. do come out naturally when you have an angular coordinate, I guess that it should not be surprising that they come out when you have an angular constraint that is not a coordinate.