I E=mc^2 and acceleration

[cianfa72]

[Mentors' note: This thread was spun off from a thread about the applicability of $E=mc^2$ to massless particles]

Yes. The general equation is
$$
E^2 = p^2 c^2 + m^2 c^4
$$
where $E$ is the energy, $p$ is the momentum, and $m$ is the rest mass. For the case of a particle with nonzero rest mass at rest, we have $p = 0$, so we get $E = m c^2$.

In the realm of SR consider an accelerating massive particle (i.e. accelerating in the sense of proper acceleration). Does the above $E=mc^2$ apply to the MCRF (Momentarily Comoving Reference Frame) at any given event/point along the particle's worldline?

 

[Nugatory]

[Mentors' note: This thread was spun off from a thread about the applicability of $E=mc^2$ to massless particles]

In the realm of SR consider an accelerating massive particle (i.e. accelerating in the sense of proper acceleration). Does the above $E=mc^2$ apply to the MCRF (Momentarily Comoving Reference Frame) at any given event/point along the particle's worldline?

Yes, but only momentarily when $p=0$.

 

[cianfa72]

Yes, but only momentarily when $p=0$.

Yes, since the particle is supposed to proper accelerate, there is a different MCRF at any point along the particle's worldline (such that the particle's $p=0$ there).

I believe the same holds true in GR as well even though the MCRF is just local (i.e. any MCRF isn't a global inertial frame though).

 

[Nugatory]

By Jove! I think they've got it!

 

[cianfa72]

Take a body made of two or more particles and consider its Center of Mass (CoM). As whole the law $E=mc^2$ applies to any MCRF where the body's CoM is instantaneously/momentarily at rest. The body's rest mass (i.e. simply its mass) includes the body's internal energy evaluated in CoM's MCRF.

 

[PeroK]

Take a body made of two or more particles and consider its Center of Mass (CoM). As whole the law $E=mc^2$ applies to any MCRF where the body's CoM is instantaneously/momentarily at rest. The body's rest mass (i.e. simply its mass) includes the body's internal energy evaluated in CoM's MCRF.

Consider a system of particles, each with mass, energy and momentum $m_i, E_i, \vec {p_i}$ - as measured in some IRF. For each particle we have:$$m_i^2c^4 = E_i^2 - p_i^2c^2$$The quantity, $M$, defined by:
$$M^2c^4 = E^2 - P^2c^2 \equiv \bigg (\sum E_i \bigg )^2 - \bigg(\sum \vec {p_i} \bigg )^2c^2$$Is invariant and representes the invariant mass of the system.

 

[Herman Trivilino]

Take a body made of two or more particles and consider its Center of Mass (CoM). As whole the law $E=mc^2$ applies to any MCRF where the body's CoM is instantaneously/momentarily at rest. The body's rest mass (i.e. simply its mass) includes the body's internal energy evaluated in CoM's MCRF.

The way you have chosen to word things is clear, so I am really picking nits here. To me, the phrase "at rest" means to maintain a velocity of zero. Certainly an object can have a velocity of zero at a time, while spending zero time doing so. And I prefer to state it as having a velocity of zero at an instant rather than being at rest at an instant.

Of course we all know what you mean but to the uninitiated this can be a confusing affair. They will have the problem of understanding how the acceleration can be zero when the (coordinate) acceleration is nonzero.

This is why I reserve the phrase "at rest" to mean that the velocity remains zero for some nonzero amount of time. Thus they are confused by the phrases "momentarily at rest" or "instantaneously at rest".

 

[cianfa72]

Of course we all know what you mean but to the uninitiated this can be a confusing affair. They will have the problem of understanding how the acceleration can be zero when the (coordinate) acceleration is nonzero.

You are talking of a body/particle with zero proper acceleration. In any non (local) inertial frame (no MCRF), it results to have a nonzero coordinate acceleration.

Actually I was thinking of a (proper) accelerating body that therefore has a nonzero coordinate acceleration in the MCRF at any given point along the body's worldline.

 

[cianfa72]

Consider a system of particles, each with mass, energy and momentum $m_i, E_i, \vec {p_i}$ - as measured in some IRF. For each particle we have:$$m_i^2c^4 = E_i^2 - p_i^2c^2$$The quantity, $M$, defined by:
$$M^2c^4 = E^2 - P^2c^2 \equiv \bigg (\sum E_i \bigg )^2 - \bigg(\sum \vec {p_i} \bigg )^2c^2$$Is invariant and representes the invariant mass of the system.

In the system's CoM rest IRF (or better in the MCRF the system's CoM has zero velocity at a given instant as @Mister T pointed out) by definition $\vec P=0$ hence here $M^2c^4 =E^2=\bigg (\sum E_i\bigg )^2$.

Where is the internal energy contribution/term into it ?

 

[PeroK]

In the system's CoM rest IRF (or better in the MCRF the system's CoM has zero velocity at a given instant as @Mister T pointed out) by definition $\vec P=0$ hence here $M^2c^4 =E^2=\bigg (\sum E_i\bigg )^2$.

Where is the internal energy contribution/term into it ?

I don't now if I totally understand your question, but (in the CoM frame):
$$Mc^2 = \sum E_i \ge \sum m_ic^2$$With equality iff all particles are at rest in the CoM frame.

 

[cianfa72]

I don't now if I totally understand your question, but:
$$Mc^2 = \sum E_i \ge \sum m_ic^2$$With equality iff all particles are at rest in the CoM frame.

Ok, my question is: suppose there are internal forces acting between system's particles. Can we adopt a sort of potential energy to evaluate the system's internal energy w.r.t. the system's CoM frame ?

 

[PeroK]

Ok, my question is: suppose there are internal forces acting between system's particles. Can we adopt a sort of potential energy to evaluate the system's internal energy w.r.t. the system's CoM frame ?

That's another layer of complexity.

 

[cianfa72]

That's another layer of complexity.

Are you saying that the sum $\sum E_i = Mc^2$ evaluated in the system's CoM rest frame (or in the MCRF at each point along the CoM worldline whether the latter is proper accelerating) assumes the particles composing the system as non-interacting each other?

 

[PeroK]

Are you saying that the sum $\sum E_i = Mc^2$ evaluated in the system's CoM rest frame (or in the MCRF at each point along the CoM worldline whether the latter is proper accelerating) assumes the particles composing the system as non-interacting each other?

I assumed an inertial CoM frame.

 

[jbriggs444]

Are you saying that the sum $\sum E_i = Mc^2$ evaluated in the system's CoM rest frame (or in the MCRF at each point along the CoM worldline whether the latter is proper accelerating) assumes the particles composing the system as non-interacting each other?

As I understand the thrust of this question, it is about binding energy.

If, for instance, we had a loose cluster of two protons and two neutrons, all slowly moving, this system would have a rest energy given by the sum of the rest energies of the individual particles. The total for this system would be:$$6.695098782 \times 10^{-27}\text{ kg}$$
If we arranged for these particles to fuse into an alpha particle the resulting system would have a mass of:$$6.644657230 \times 10^{-27}\text{ kg}$$
So yes, an assumption going into the formula given by @PeroK is that there are no inter-particle interactions that result in a significant net positive or negative unlocalized potential energy.


There are two different inequalities in play here.

One is that the rest energy of a collection of non-interacting particles will always be greater than or equal to the sum of the rest energies of the individual particles. The discrepancy, if any, traces to the kinetic energies of the individual particles in the CoM frame.

The other is that the rest energy of a collection of interacting particles can be greater than, less than or equal the sum of the rest energies of the individual particles. The discrepancy is in the binding energy associated with the interactions. And also due to any kinetic energies.

 

[PeterDonis]

The other is that the rest energy of a collection of interacting particles can be greater than, less than or equal the sum of the rest energies of the individual particles. The discrepancy is in the binding energy associated with the interactions. And also due to any kinetic energies.

If the system is bound, the rest energy of the system will always be less than the sum of the rest energies of the individual particles. The (negative) binding energy will always outweigh the (positive) kinetic energies of the particles in the CoM frame. If that were not the case, the system wouldn't be bound.

If you want to consider a more general case where the system might not be bound, the term "interaction energy" rather than "binding energy" would be more appropriate. The interaction energy might be negative or positive, depending on the details of the interaction (for example, the interaction energy between oppositely charged particles is negative, but between same-charged particles it's positive).

 

[cianfa72]

If the system is bound, the rest energy of the system will always be less than the sum of the rest energies of the individual particles. The (negative) binding energy will always outweigh the (positive) kinetic energies of the particles in the CoM frame. If that were not the case, the system wouldn't be bound.

The system's rest energy includes the sum $\sum E_i$ where the terms $E_i$ are evaluated w.r.t. the system CoM's inertial rest frame (or w.r.t. the MCRF at any point along the system CoM's worldline whether the latter is proper accelerating). On the other hand $$E_i^2 = m_i^2c^4 + p_i^2c^2$$ therefore to get a total system's rest energy less than the sum $\sum m_ic^2$ a negative binding energy term, as pointed out by @PeterDonis , will always outweigh the positive contribute due to $p_i^2c^2$ terms (i.e. basically to the kinetic energies of particles w.r.t. the CoM rest frame).

If you want to consider a more general case where the system might not be bound, the term "interaction energy" rather than "binding energy" would be more appropriate. The interaction energy might be negative or positive, depending on the details of the interaction (for example, the interaction energy between oppositely charged particles is negative, but between same-charged particles it's positive).

In this case the "interaction energy" evaluated in the system CoM's rest frame is somehow related to a sort of potential energy describing the forces/interactions between the charged particles.

 

[Herman Trivilino]

In this case the "interaction energy" evaluated in the system CoM's rest frame is somehow related to a sort of potential energy describing the forces/interactions between the charged particles.

I think it is the potential energy of the composite body. When that interaction energy is negative we call it the binding energy.

 

[PeterDonis]

In this case the "interaction energy" evaluated in the system CoM's rest frame is somehow related to a sort of potential energy describing the forces/interactions between the charged particles.

It is the potential energy due to the interactions between the particles.

When that interaction energy is negative we call it the binding energy.

No, when the system is bound we call it the binding energy. The interaction energy can be negative even if the system is not bound--for example, consider an electron and a proton scattering off each other when their kinetic energies are high enough that they aren't bound. The interaction energy is still negative even though the total energy is greater than the sum of the rest energies.

 

[cianfa72]

It is the potential energy due to the interactions between the particles.

I wonder whether it makes sense in relativity to define a system's potential energy since there is no action at distance at all.

No, when the system is bound we call it the binding energy. The interaction energy can be negative even if the system is not bound--for example, consider an electron and a proton scattering off each other when their kinetic energies are high enough that they aren't bound. The interaction energy is still negative even though the total energy is greater than the sum of the rest energies.

Ok.

 

[PeroK]

I wonder whether it makes sense in relativity to define a system's potential energy since there is no action at distance at all.

https://en.wikipedia.org/wiki/Electromagnetic_four-potential

 

[Sagittarius A-Star]

If the system is bound, the rest energy of the system will always be less than the sum of the rest energies of the individual particles.

Isn't the proton a counter-example?

The mass of a proton is about 80–100 times greater than the sum of the rest masses of its three valence quarks, while the gluons have zero rest mass.

Source:
https://en.wikipedia.org/wiki/Proton#Quarks_and_the_mass_of_a_proton

 

[PeterDonis]

I wonder whether it makes sense in relativity to define a system's potential energy since there is no action at distance at all.

Potential energy in relativity is not defined in terms of "action at a distance". It's defined in terms of fields, which are local. For example, the interaction energy between charged particles is really the energy of the electromagnetic field. The particles are sources of the field, but changes in the field propagate at the speed of light; they aren't instantaneous, and there is no "action at a distance" anywhere. But if you consider a composite system, the field energy makes a contribution that works the same as the "potential energy" you're used to seeing in non-relativistic mechanics.

 

[PeterDonis]

Isn't the proton a counter-example?

No, just a much more complicated interaction, meaning that the common heuristic picture of a proton as being "made of" three quarks isn't correct, and isn't even a good approximation. (A proton does have three "valence quarks", but they aren't anything like, say, the electron and proton in a hydrogen atom as far as the discussion in this thread is concerned.) In fact, the concept of "binding energy" that we've been using in this thread isn't even applicable to a proton, because that concept assumes that the individual particles making up the system can be separated and each one taken to infinity as a free particle. You can't do that with quarks.

 

[cianfa72]

But if you consider a composite system, the field energy makes a contribution that works the same as the "potential energy" you're used to seeing in non-relativistic mechanics.

Ok, so for a composite system in the system CoM's MCRF one is allowed to write $$E= \sum E_i + F$$ where $E_i = \sqrt {m_i^2c^4 + p_i^2c^2}$ for the $i$-particle and $F$ is the "interaction energy" or "field energy" term that can be either positive or negative (negative for sure in bound systems).

Note that the sources of the field are the system's particles alone (e.g. charges), i.e. the term $F$ doesn't include any external field energy.

 

[Sagittarius A-Star]

The other is that the rest energy of a collection of interacting particles can be greater than, less than or equal the sum of the rest energies of the individual particles. The discrepancy is in the binding energy associated with the interactions. And also due to any kinetic energies.

For a bound system this is true, if with particle a sub-system, consisting of nucleons, is meant. If such a sub-system gets isolated, it has it's own mass-defect.

Source:
book "Introduction to Special Relativity, 2nd edition", chapter 27, page 75 (W. Rindler)
https://www.amazon.com/Introduction...-Publications/dp/0198539525?tag=pfamazon01-20

 

[A.T.]

Yes. The general equation is
$$
E^2 = p^2 c^2 + m^2 c^4
$$
where $E$ is the energy, $p$ is the momentum, and $m$ is the rest mass.

In geometrized units where $c = 1$ the above becomes
$$
E^2 = p^2 + m^2
$$which looks like the Pythagorean theorem, similar to this relationship
$$
dt^2 = dx^2 + d\tau^2
$$where $t$ is the coordinate time, $x$ is spatial distance and $\tau$ is proper time of the moving object. When interpreted geometrically, those two Pythagorean equations indeed describe the same shaped (similar) right triangle.

This is visualized here (just with different names "dynamic / coordinate mass $m$" is used for energy $E$, "proper mass $\mu$" is used for rest mass $m$)
https://archive.org/details/L.EpsteinRelativityVisualizedelemTxt1994Insight/page/n125/mode/2up
https://archive.org/details/L.EpsteinRelativityVisualizedelemTxt1994Insight/page/n127/mode/2up

The linked book also does a good job of differentiating between two key types of relativistic acceleration processes on these pages:
https://archive.org/details/L.EpsteinRelativityVisualizedelemTxt1994Insight/page/n139/mode/2up

1) (page 127) Acceleration with constant total energy $E$, and decreasing rest mass $m$.

Here, propulsive energy is stored on board and consumed, thus decreasing the rest mass according to mass-energy-equivalence (not due to ejecting burned fuel). This applies to a wind-up-car, or a car with matter-antimatter-annihilation drive. Here the rest mass is effectively converted into momentum, which in the limiting case of annihilating all of the rest-mass lets you approach light speed (dynamic-mass arrow approaches horizontal), while total energy $E$ (length of the dynamic-mass arrow) remains constant.

2) (page 128) Acceleration with constant rest mass $m$, and increasing total energy $E$.

Here, propulsive energy is supplied to the object from outside. Like for an electric train, or linear particle accelerator. The geometry in figure 8-7 shows, that for non-zero rest mass $m$, approaching light speed (dynamic-mass arrow approaches horizontal) makes the energy $E$ (length of the dynamic-mass arrow) tend to infinity.