Rest energy in classical mechanics

[DrStupid]

[Moderator's note: Post spun off from another thread.]

There's no rest energy in Newtonian physics anyway.

That is correct but it doesn't mean Eo=0. The rest energy is unlimited in classical mechanics. Therefore it is impossible to find a relation between total energy and momentum.

 

[PeterDonis]

The rest energy is unlimited in classical mechanics.

Huh? There is no such thing as "rest energy" in classical mechanics.

 

[haushofer]

Huh? There is no such thing as "rest energy" in classical mechanics.

It reminds me of the procedure of chapter 2 in one of my papers,

https://arxiv.org/abs/1206.5176

It amounts to taking a nonrelativistic limit of a point particle by sending the speed of light c to infinity. In that case "the rest energy E=mc2" becomes infinite (and can be removed by coupling the point particle to a gauge field (before the limit is taken, the divergent term in the action can be regarded as a total derivative). I wouldn't recast such a procedure physically as DrStupid's statement, because infinite amounts of energy are a bit problematic. :P

There are quite some subtleties in this approach, but that's where the statement perhaps comes from.

 

[DrStupid]

Huh? There is no such thing as "rest energy" in classical mechanics.

Can you please be a bit more specific? Do you mean that a system at rest has no energy in classical mechanics?

because infinite amounts of energy are a bit problematic.

That's why I wrote unlimited instead of infinite. However, in the classical limit $c \to \infty$ the relativistic rest energy goes indeed infinite

 

[PeterDonis]

Do you mean that a system at rest has no energy in classical mechanics?

No. It might contain heat energy (although strictly speaking, that is just kinetic energy of the individual atoms or molecules, which are not at rest). But the term "rest energy" is not used to describe that in classical mechanics; it has no meaning at all in classical mechanics since it's not used at all in that context. So the only standard meaning the term "rest energy" has is its meaning in relativity: the energy equivalent of the rest mass. And there is no such thing in classical mechanics.

If you want to use your own idiosyncratic definition of "rest energy", I can't stop you, but you need to explicitly define how you're using the term if you're not using it with its standard meaning.

 

[PeterDonis]

in the classical limit $c \to \infty$ the relativistic rest energy goes indeed infinite

"The classical limit of relativity" is not the same as "classical mechanics". The statement of yours quoted just above is true, but it's not equivalent to saying "the rest energy is unlimited in classical mechanics".

 

[DrStupid]

But the term "rest energy" is not used to describe that in classical mechanics;

That means your objection was not about physics but just about semantics. In order to complete it: Which term is used instead?

"The classical limit of relativity" is not the same as "classical mechanics".

What is the difference between special relativity and classical mechanics if c goes infinite?

 

[Andrew Mason]

I thought Special Relativity was part of "classical mechanics". The distinction is usually drawn between quantum mechanics ("modern") and everything that came before ("classical").

After all, SR follows from Maxwell's laws. Newton did not address the laws of electromagnetism so it is not as if Maxwell conflicted with Newton. So it stands to reason that there is no analogue in Newtonian physics for "rest energy" or the energy momentum relation $E^2 - p^2/c^2 = (m_0c^2)^2$.

$E^2-p^2/2m = 0$ has no analogy at all because it is a tautology. It does not describe a relationship between different quantities because by definition kinetic energy IS $\sqrt{p^2/2m}$. It is no more informative than: "(Isaac Newton) - (Newton, Isaac) = 0".

AM

 

[PeterDonis]

That means your objection was not about physics but just about semantics.

My objection was that I didn't know what you were talking about. Now you have made it clear that you are talking about the limit of special relativity as $c \to \infty$.

In order to complete it: Which term is used instead?

There is no term used in classical mechanics the way "rest energy" is used in relativity, because there is no corresponding concept in classical mechanics.

If you are using a classical mechanics model in which objects can have internal energy (such as heat energy) that is present when the object is at rest, you would use whatever term describes the kind of internal energy in your model (e.g., "heat energy").

What is the difference between special relativity and classical mechanics if c goes infinite?

As noted already in this thread, in SR, there is infinite rest energy present, which you have to remove from your model by hand. In classical mechanics, there is no such thing.

 

[PeterDonis]

I thought Special Relativity was part of "classical mechanics".

The term "classical mechanics" can have different meanings. @DrStupid appears to be using it to mean "Newtonian mechanics". That is how I am using it in my responses.

The distinction is usually drawn between quantum mechanics ("modern") and everything that came before ("classical").

That's one common usage, but another one is drawing the distinction between relativity and pre-relativity, the latter being "classical mechanics".

 

[Sorcerer]

That means your objection was not about physics but just about semantics. In order to complete it: Which term is used instead?
What is the difference between special relativity and classical mechanics if c goes infinite?

The term "classical mechanics" can have different meanings. @DrStupid appears to be using it to mean "Newtonian mechanics". That is how I am using it in my responses.
That's one common usage, but another one is drawing the distinction between relativity and pre-relativity, the latter being "classical mechanics".

As an undergrad I've often heard "classical physics" used to describe physics before quantum mechanics and "classical mechanics" used to describe Newtonian mechanics. I know this is way off topic, but that's what I've heard the most and it doesn't seem as ambiguous.

 

[DrStupid]

As noted already in this thread, in SR, there is infinite rest energy present, which you have to remove from your model by hand. In classical mechanics, there is no such thing.

I'm not aware of any law that limits the energy of a system at rest in classical mechanics.

 

[PeterDonis]

I'm not aware of any law that limits the energy of a system at rest in classical mechanics.

If you mean that, in classical mechanics, you could in principle raise the temperature of a system at rest as high as you want, storing unbounded amounts of heat energy in it, yes, that's true. And it has absolutely nothing to do with the relativistic rest energy going to infinity in the limit as $c \to \infty$. The latter happens for a system that stores no heat energy at all; its only rest energy (relativistically speaking) is its rest mass. That doesn't happen in classical mechanics; in classical mechanics, a system storing zero heat energy (or other internal energy) has zero energy when it's at rest.

 

[DrStupid]

If you mean that, in classical mechanics, you could in principle raise the temperature of a system at rest as high as you want, storing unbounded amounts of heat energy in it, yes, that's true.

That doesn't only apply to thermal energy but to any kind of energy. Another example would be the specific energy of rocket fuels or explosives. In contrast to relativity they have no theoretical limit in classical mechanics.

in classical mechanics, a system storing zero heat energy (or other internal energy) has zero energy when it's at rest.

If it stores zero internal energy. A system that stores infinite internal energy has infinite energy when it's at rest. That's exactly what infinite rest energy implies. In relativity this is not possible for a system with finite mass. But with $c \to \infty$ or in classical mechanics it is. Therefore I don't see a difference but rather a similarity in this regard.

 

[PeterDonis]

Another example would be the specific energy of rocket fuels or explosives.

Yes, specific energy in these cases would be another form of internal energy in classical mechanics.

In contrast to relativity they have no theoretical limit in classical mechanics.

What do you think the theoretical limit on such forms of energy is in relativity?

If it stores zero internal energy.

Yes, which is exactly what I said: "a system storing zero heat energy (or other internal energy)". Did you read what you quoted?

A system that stores infinite internal energy has infinite energy when it's at rest.

If you want to postulate such a system in classical mechanics, yes. Of course, we have no evidence of the existence of any such system.

That's exactly what infinite rest energy implies.

No, it doesn't, because in classical mechanics you just postulate "internal energy" as a separate thing from the mass of the system or its motion. There is no "infinite rest energy" in any sort of limit.

In relativity the "infinite rest energy" comes from taking the limit $c \to \infty$ for a system which has no internal energy other than its rest mass, and which has a finite rest mass (see further comments below). In classical mechanics, such a system has zero energy when at rest, period.

In relativity this is not possible for a system with finite mass.

Yes, in relativity, all of the stuff we've been calling "internal energy" in classical mechanics--heat, specific energy of fuels, etc.--is part of the rest mass of the system. So postulating "infinite internal energy" in relativity would mean postulating a system with infinite rest mass (with $c$ having its usual finite value). In classical mechanics, the two things are separate, as noted above; a postulated system with infinite internal energy could still have a finite mass, since the two are separate, independent properties.

 

[DrStupid]

What do you think the theoretical limit on such forms of energy is in relativity?

E/m = c²

Yes, which is exactly what I said: "a system storing zero heat energy (or other internal energy)". Did you read what you quoted?

I highlighted "If" for a reason. The system has has zero energy at rest if it is storing zero heat energy (or other internal energy). This was to point out that your statement is limited to this condition.

No, it doesn't, because in classical mechanics you just postulate "internal energy" as a separate thing from the mass of the system or its motion.

Just like with $c \to \infty$. In this limit mass and energy become separate things due to

\mathop {\lim }\limits_{c \to \infty } \frac{{dm}}{{dE}} = 0

In relativity the "infinite rest energy" comes from taking the limit $c \to \infty$ for a system which has no internal energy other than its rest mass, and which has a finite rest mass (see further comments below). In classical mechanics, such a system has zero energy when at rest, period.

That makes no sense. In the next paragraph you say

Yes, in relativity, all of the stuff we've been calling "internal energy" in classical mechanics--heat, specific energy of fuels, etc.--is part of the rest mass of the system.

which is correct. But what makes you think, that a system with rest mass, including heat, specific energy of fuels, etc. has no energy in classical mechanics? If a system has stuff like heat, specific energy of fuels, etc. in relativity it has them in classical mechanics as well. Of course you may set the total energy to zero by definition - knowing that this results in a negative total energy when these energies are released. But nobody is forced to do that. Only the unknown parts of the internal energy are usually set to zero. And these parts have no theoretical limit in classical mechanics.

 

[PeterDonis]

E/m = c²

That doesn't answer the question; $E$ and $m$ as you're using them here are the same thing, just in different units. If you insist on having the question rephrased to avoid quibbles, what do you think the theoretical limits are in relativity on $m$ (i.e., rest mass)?

In this limit mass and energy become separate things

No, they don't. Rest mass = rest energy in relativity regardless of the value of $c$.

what makes you think, that a system with rest mass, including heat, specific energy of fuels, etc. has no energy in classical mechanics?

I never said it did. You are not reading my posts carefully. I said a system with rest mass but no heat, specific energy of fuels, etc., has zero energy in classical mechanics.

 

[DrStupid]

That doesn't answer the question; $E$ and $m$ as you're using them here are the same thing, just in different units.

They can be assumed to be the same thing if c doesn't change. To our current knowledge that's the case in reality but not what we are talking about. A special relativity with c different from the real speed of light doesn't describe reality anymore (which is by the way another similarity to classical mechanics).

If you insist on having the question rephrased to avoid quibbles, what do you think the theoretical limits are in relativity on $m$ (i.e., rest mass)?

The answer is still E/m = c². If you insist in always using a system of units with c=1 (or any other finite number) even if c changes compared to the known value than you would have no system of units at all if c goes infinite. That wouldn't be helpful.

I never said it did. You are not reading my posts carefully. I said a system with rest mass but no heat, specific energy of fuels, etc., has zero energy in classical mechanics.

You explicitely referenced to your further comments according to which heat, specific energy of fuels, etc. is included in the rest mass. But even if you ment a system with rest mass but no heat, specific energy of fuels, etc., you still can only define it's energy to be zero in classical mechanics. The internal energy cannot be determined and it is not limited by any known law of nature in classical mechanics. If you know a method to determine the internal energy or a law that limits it in classical mechanics than please post a corresponding reference. Just claiming its value to be equal to random numbers is not helpful.

 

[Sorcerer]

[...]
You explicitely referenced to your further comments according to which heat, specific energy of fuels, etc. is included in the rest mass. But even if you ment a system with rest mass but no heat, specific energy of fuels, etc., you still can only define it's energy to be zero in classical mechanics. The internal energy cannot be determined and it is not limited by any known law of nature in classical mechanics. If you know a method to determine the internal energy or a law that limits it in classical mechanics than please post a corresponding reference. Just claiming its value to be equal to random numbers is not helpful.

Not to get nosy and interject here, but isn't it true that if you remove all of that noise and just focus on a point particle that has mass and is at rest, the only "rest energy" left in Newtonian physics would be potential energy due to position, which can be transformed to zero with a suitable reference frame choice? Conversely, in special relativity there is no choice of reference frame or origin that can remove all energy from a point particle at rest. In Newton, I can choose an origin so that mgh = 0, which means in this point particle case there would be no energy at all for that particle in that particular reference frame. If I do the same thing in relativity, I'm still stuck with $mc^2$. I guess if we bring gravity into it the ideal coordinate choice would be where the object is in free fall, and is therefore inertial. But at rest in its own local inertial frame, the $mc^2$ still doesn't go away, does it?

So correct me if I'm wrong, but whatever internal energy you decide upon for Newtonian physics, it's still of a different kind that what is found in the rest energy of special relativity. So the difference appears that in Newton you can choose a reference frame where the point particle has zero energy, but you cannot do the same in special relativity.What do you think? Is that a fair assessment?

 

[PeterDonis]

The answer is still E/m = c².

This doesn't answer the question because it doesn't tell me what the allowed range of values of $m$ is. What do you think it is?

You explicitely referenced to your further comments according to which heat, specific energy of fuels, etc. is included in the rest mass.

Not when I was describing in what circumstances the energy would be zero in classical mechanics. Again, you're not reading my posts carefully.

even if you ment a system with rest mass but no heat, specific energy of fuels, etc., you still can only define it's energy to be zero in classical mechanics.

No, you don't have to "define" its energy to be zero in classical mechanics for this case, because it's already zero since you've ruled out all possible kinds of energy being present by saying the system is at rest (no kinetic energy) and there is no heat, specific energy of fuels, etc. (no internal energy).

The internal energy cannot be determined and it is not limited by any known law of nature in classical mechanics.

That's true, but irrelevant to the case in which the internal energy is zero because we have specified there is no heat, specific energy of fuels, etc (and kinetic energy is zero because the object is at rest). And that is the only case in which I have said that the energy is zero in classical mechanics.

 

[DrStupid]

This doesn't answer the question because it doesn't tell me what the allowed range of values of $m$ is. What do you think it is?

m>0

Again, you're not reading my posts carefully.

It still reads:

In relativity the "infinite rest energy" comes from taking the limit $c \to \infty$ for a system which has no internal energy other than its rest mass, and which has a finite rest mass (see further comments below). In classical mechanics, such a system has zero energy when at rest, period.

And in the „further comments“ you still say:

Yes, in relativity, all of the stuff we've been calling "internal energy" in classical mechanics--heat, specific energy of fuels, etc.--is part of the rest mass of the system.[…]

And that still means that a system with no internal energy other than its rest mass (including heat, specific energy of fuels, etc.) has zero energy at rest in classical mechanics. And that is still wrong.

No, you don't have to "define" its energy to be zero in classical mechanics for this case, because it's already zero since you've ruled out all possible kinds of energy being present by saying the system is at rest (no kinetic energy) and there is no heat, specific energy of fuels, etc. (no internal energy).

By „saying the system is at rest (no kinetic energy) and there is no heat, specific energy of fuels, etc. (no internal energy)“ you actually define internal energy to be zero.

 

[DrStupid]

but isn't it true that if you remove all of that noise and just focus on a point particle that has mass and is at rest, the only "rest energy" left in Newtonian physics would be potential energy due to position, which can be transformed to zero with a suitable reference frame choice?

Internal energy cannot be determined in general. In special relativity we can use the rest energy instead. But that doesn't work anymore if we use Lorentz transformation with another invariant speed. With the speed limit going infinite or in classical mechanics there is no limit for internal energy at all. That applies to all systems, including point particles. In case of point particles you just don't need to care about internal energy because they are assumed to be isolated systems and internal energy is conserved. Therefore it doesn't matter how much internal energy the point particle contains. Zero is a convinient choice but you can also use any other value.

In Newton, I can choose an origin so that mgh = 0, which means in this point particle case there would be no energy at all for that particle in that particular reference frame.

The outer energy is zero. The total energy is zero if the internal energy is definied to be zero.

but whatever internal energy you decide upon for Newtonian physics, it's still of a different kind that what is found in the rest energy of special relativity.

I'm not really sure what you mean but many parts of the internal energy are of the same kind and have a similar amount in classical mechanics and in special relativity.

A usual example is thermal energy. In case of an ideal gas it is the sum of the kinetic energies of it's particles. Under standard conditions the difference between the classical and relativistic kinetic energy of the particles is negligible because the speed of the particles is non-relativistic. That means that the thermal energy is usually almost identical in classical mechanics and relativity.

Another example is chemical energy. It is determined by quantum mechanics which in chemistry usually doesn't care about relativity or not. There are very vew examples where the difference matters. One of them is mercury. With increasing invariant speed it would most likely turn into the solid that it is expected to be according to non-relativistic quantum mechanics. That would indeed imply a significant change of the internal energy (mainly due to the missing melting heat). But assuming chemical energy to be equal in kind and amount in classical mechanics and relativity is usually a very good approximation.

I case of nuclear energy I would indeed expect a relevant difference and in case of annihilation of matter and antimatter I totally agree with you. That would need to be of completely different kind in classical mechanics (e.g. due to conservation of mass).

 

[PeterDonis]

that still means that a system with no internal energy other than its rest mass (including heat, specific energy of fuels, etc.) has zero energy at rest in classical mechanics. And that is still wrong.

It is? How?

 

[PeterDonis]

Internal energy cannot be determined in general.

Why not?

In special relativity we can use the rest energy instead. But that doesn't work anymore if we use Lorentz transformation with another invariant speed.

Why not? (I assume by "another invariant speed" you mean a finite value of $c$, not the limit $c \to \infty$.)

 

[DrStupid]

It is? How?

Because heat, specific energy of fuels, etc., which are included in the rest mass in special relativity, exist both in classical mechanics and in relativity.

Why not?

Because internal energy cannot be measured directly. Only changes of the internal energy can be measured in the form of work or heat. In classical mechanics you never know if the system has released all internal energy or not. No matter how much energy it releases, there can always be an unknown amount left.

Why not? (I assume by "another invariant speed" you mean a finite value of $c$, not the limit $c \to \infty$.)

In special relativity the maximum energy that can be released by a system with the mass m is m·c² where c is the speed of plane light waves in vacuum. We know that because the system will lose the mass m during this process (no matter what kind of process it is) and will therefore not exist anymore. As it makes sense to assume the internal energy of a non existing system to be zero, the rest energy m·c² can be used instead in relativiy.

If we use another invariant speed c' > c in lorenz transformation the resulting prediction for rest energy would be m·c'². But the maximum energy that can be release by the system is still m·c² because the use of another invariant speed only changes the theoretical predictions but not reality. With the value m·c'² we just have an upper limit for the internal energy. If c' goes infinite we don't even have this limit anymore and internal energy becomes completely unpredictable like in classical mechanics.

 

[PeterDonis]

Because heat, specific energy of fuels, etc., which are included in the rest mass in special relativity, exist both in classical mechanics and in relativity.

You are simply not responding to my point here. I don't know how else to get you to respond to it.

In classical mechanics you never know if the system has released all internal energy or not.

You're confusing model with reality. In the model, we say what forms of internal energy might be present and what it means for them to be present. Then we check the model against reality to see if it makes correct predictions. If a model of a particular system that says it has zero internal energy makes correct predictions, then that particular system in reality has zero internal energy for all practical purposes. And such a model in classical mechanics says that such a system has zero energy when at rest.

It is true that classical mechanics is not correct, so a classical mechanics model of such a system is not, strictly speaking, correct. But within its domain of validity, it makes accurate enough predictions for all practical purposes.

In special relativity the maximum energy that can be released by a system with the mass m is m·c² where c is the speed of plane light waves in vacuum.

If the system is at rest, yes, this is true.

If we use another invariant speed c' > c in lorenz transformation the resulting prediction for rest energy would be m·c'². But the maximum energy that can be release by the system is still m·c² because the use of another invariant speed only changes the theoretical predictions but not reality.

You're confusing model with reality again. The only reason to use a model with an invariant speed of c' instead of c would be if the invariant speed actually were c' instead of c in reality. (I'm assuming here that there are no changes of units.) It makes no sense to talk about a model with invariant speed c' if the invariant speed in reality is actually c.

If your response to the above is that we might want to use a model with an invariant speed c' > c as an approximation, the only reason we would do this is if for some reason the model with c' was easier to use than the one with c, but still made predictions that were accurate enough in the domain under consideration. As a matter of fact, that's only true for a model in the limit $c \to \infty$, not for a model with a finite c' > c. So I don't see the point of even discussing the latter case.

 

[DrStupid]

You are simply not responding to my point here. I don't know how else to get you to respond to it.

I did respond to your point and I don't know how else to get you to accept it. There are three major possibilities: 1. I don't anderstand you. 2. You don't understand me. 3. Both. It would require a more detailed discussion to identify and solve the problem. But I do not know if this is a) worthwhile and b) possible without escalation.

If a model of a particular system that says it has zero internal energy makes correct predictions, then that particular system in reality has zero internal energy for all practical purposes.

What models are you talking about? For major parts of the internal energy there is not even a model in classical mechanics that could predict changes of internal energy. Chemical reactions for example have always been known to be accompanied by release or absorption of energy. But there never was a model in classical mechanics that could predict the corresponding change of internal energy. All values for standard enthalpy of formation result from experimental observations only (Try to make an educated guess why there are no values for standard enthalpy.) And you are going to tell me that there are models that not only predict changes of internal energy but imply information about the value of internal energy itself? I hope this is just a misunderstanding. If yes, please explain in detail what you actually mean.

If your response to the above is that we might want to use a model with an invariant speed c' > c as an approximation, the only reason we would do this is if for some reason the model with c' was easier to use than the one with c, but still made predictions that were accurate enough in the domain under consideration. As a matter of fact, that's only true for a model in the limit $c \to \infty$, not for a model with a finite c' > c. So I don't see the point of even discussing the latter case.

Please demonstrate how to get the limit for $c’ \to \infty$ without considering $c' > c$.

 

[sweet springs]

Say N photons of frequency $\nu$ are packed into a vessel of ideally no mass. The packed vessel has photon gas energy $Nh\nu$ and thus show inertial mass $\frac{Nh\nu}{c^2}$ when put into force. This is a case that mass is explicitly coming from energy.

 

[DrStupid]

This is a case that mass is explicitly coming from energy.

In relativity mass is always coming from energy. In classical mechanics it never comes from energy. With c'>c we are somewhere in the middle.

 

[sweet springs]

You can measure photo gas energy by measuring electromagneric field in the container.
You can measure inertial mass by putting force on the container.
You will find the relation E=mc^2.
I am not sure it is in the category of classical mechanics you are saying, but it is a fact we can observe.

 

[DrStupid]

You will find the relation E=mc^2.

Yes, but what does that contribute to the topic? In relativity we know that E is the total energy inside the container. How do you prove that in classical mechanics?

 

[sweet springs]

In relativity we know that E is the total energy inside the container. How do you prove that in classical mechanics?

We now know E=mc^2 as a fact that classical mechanics or saying more accurately classical non relativistic mechanics does not anticipate, need and prove. I prefer the fact to a theory.　Don't put new wine into old bottles.

 

[DrStupid]

We now know E=mc^2 as a fact that classical mechanics or saying more accurately classical non relativistic mechanics does not anticipate, need and prove.

I already wrote above that internal energy cannot be determinied in classical mechanics. That includes that classical mechanics doesn’t anticipate, need and prove E=mc². So I still do not see a new argument. Do I miss something?

 

[sweet springs]

Now you know that non relativistic mechanics failed. I do not think you miss anything.

 

[DrStupid]

Now you know that non relativistic mechanics failed.

I was well aware of it from the beginning (I even mentioned it in #18). You are quite off-topic if that was all you are trying to tell me.