# I Mechanical system with de Broglie-like features

#### AVentura

Summary
relativistic spinning cylinder resembles a wave function
Hello, I am curious if I have this correct and if it has a name.

A thin walled cylinder is spinning on its axis along its length in a closed system. It begins to draw itself in converting its invariant mass to kinetic energy. In polar coordinates $E=\gamma_\theta m c^2, L=\gamma_\theta m v_\theta r, \omega r = v_\theta$. Assuming* $v_\theta$ can approach $c$ , all of the mass is converted and $E \to L \omega$.

It is very long. An observer is at rest with its axis along $z$ in reference frame $S$. It translates at $v_z$ with respect $S'$. To the observer in $S'$ the cylinder is twisted. A line down the surface that is straight in $S$ is a helix in $S'$. The magnitude of the helix twist is given by $\frac{d\theta}{dz} = \gamma_z \omega v_z/c^2$. Sorry I don't have a proof for this but references state that $\omega$ is for the non translating frame. So the helix twists $2\pi$ radians in $\frac{2\pi L c^2}{\gamma_z E v_z}$ units of length. Note $E$ is the constant energy in $S$ and all of it contributes to the momentum in $S'$. So expressed in $p_z$ the "wavelength" is $\lambda=\frac{2\pi L}{p_z}$.

One problem that arises is the translating observer in $S'$ sees a slower angular velocity $\omega'=\omega/\gamma_z$ and that is not proportional to the total energy for that observer. But, the helix propagates wave crests at a velocity of $\frac{\lambda \omega'}{2\pi}=\frac{c^2}{v_z}-v_z$. These are not physically moving objects but coordinates moving along a moving object. So the straight addition to the velocity of the cylinder gives $c^2/v_z$, resembling the phase velocity of a wavefunction. The helix line can then look like a wave that is not translating, but just spinning at $\gamma_z \omega$ which is proportional to the energy in $S'$, by $L$.

Lastly both observers see the surface of the cylinder moving at $c$ and the observed energy is proportional to momentum for that observer. In $S'$ it is moving in a spiral. In $S'$ if you break this momentum into components of $p_\theta$ and $p_z$ their Pythagorean sum is the relativistic energy-momentum equation (taking advantage of $v_\theta' \gamma_z = c$).

*clearly all the energy cannot be converted due to atomic structure, but is there a problem before then? Weak energy condition perhaps?

Thanks for reading. Any comments or help pointing out mistakes are welcome.

Last edited:
Related Special and General Relativity News on Phys.org

#### Dale

Mentor
It begins to draw itself in converting its invariant mass to kinetic energy.
You don’t convert invariant mass to kinetic energy. Invariant mass is conserved. What you can do is convert potential energy into kinetic energy conserving total energy and invariant mass. Also, you can break a system into subsystems such that the sum of the masses of the subsystems is substantially less than the conserved mass of the whole system.

#### AVentura

First thing that comes to mind would be an matter-antimatter engine converting invariant mass to photons with opposite momentum. They could be captured in such a way that causes the cylinder to contract (picture two overlapping boxes with the reaction in the overlapping space). Conservation of angular momentum causes the angular velocity to increase.

Or simply over stretched springs that are allowed to relax inward. They draw the cylinder in doing work in the process. As a result they have less invariant mass.

Last edited:

#### PeterDonis

Mentor
As a result they have less invariant mass.
This is wrong. Both of the scenarios you describe do not involve any interaction with the environment--no radiation is emitted and no force is acting between the object and its environment, all forces are internal. Under those conditions, invariant mass is a constant. Internal interactions can redistribute energy among parts of the object, but they can't change its total invariant mass.

#### Dale

Mentor
First thing that comes to mind would be an matter-antimatter engine converting invariant mass to photons with opposite momentum.
The invariant mass of the initial electron positron system is the same as the invariant mass of the final two photon system. The invariant mass of a system is conserved, not converted.

In units where $c=1$ and energy, momentum, and mass are measured in $\text{keV}$, the electron has a four-momentum of $(511,0,0,0)\ \text{keV}$ and the positron also has a four-momentum of $(511,0,0,0)\ \text{keV}$ for a total system four-momentum of $(1022,0,0,0)\ \text{keV}$ and an invariant mass of $1022 \ \text{keV}$. After the anhillation the system consists of a pair of photons with four-momenta $(511,511,0,0)\ \text{keV}$ and $(511,-511,0,0)\ \text{keV}$ for a conserved total system four-momentum of $(1022,0,0,0)\ \text{keV}$ and a conserved system mass of $1022 \ \text{keV}$.

Last edited:

#### AVentura

It is my understanding that spinning objects do not have clearly defined invariant masses so perhaps this is the problem.

But, spinning objects can increase their tangential velocity in a closed system, like a figure skater. If the energy is given by $E=\gamma m c^2$ and is constant, then $m$ decreases as $\gamma$ increases. This is the mass I'm referring to.

This point about "invariant" is very well taken. Because the observer translating by must consider $\gamma_\theta m$ as the mass. Because it must be invariant to them.

I only need this $m$ to be the same one that's in the $L$ formula for this to work.

Last edited:

#### Dale

Mentor
It is my understanding that spinning objects do not have clearly defined invariant masses so perhaps this is where the disagreement stems from
Where did you hear that? It is incorrect. The invariant mass of a spinning object is well defined. The disagreement has nothing to do with rotating systems specifically.

If the energy is given by E=γmc2 and is constant, then m decreases as γ increases. This is the mass I'm referring to.
The relationship between energy and mass is $m^2 c^2=E^2/c^2-p^2$. Due to the conservation of four momentum, as discussed above, E, p, and m are all conserved.

In addition, m is invariant. Note that the concept of invariance is completely different from the concept of conservation. Conserved (E, p, and m) means that it is the same before and after some interaction. Invariant (m only) means that it is the same in different reference frames.

So as you look at different frames m is the same. Neither E nor $\gamma$ are invariant. So your analysis of $E=\gamma m c^2$ is off. As you go to faster moving reference frames E and $\gamma$ increase, while m stays the same since it is invariant.

#### Dale

Mentor
According to those folks my definitions are correct.
The point I am trying to communicate with you is closely related to this comment from that thread:
Very briefly, while in one inertial frame, energy is additive, invariant mass is not.
Although I am not convinced about the following sentence:
Further, the invariant mass of a rotating body as a whole is, in some sense, impossible to define.
Which seems like the source of your previous comment. That doesn’t seem right to me with regard to conserved quantities.

Last edited:

#### AVentura

Maybe we should restrict the conversation to the frame where the cylinder is only rotating. I hold that it can pull itself in, decreasing its radius and increasing its tangential velocity $v_\theta$. And that at any given time $E/L=\omega c^2/v_\theta^2$.

#### PeterDonis

Mentor
I hold that it can pull itself in, decreasing its radius and increasing its tangential velocity $v_\theta$.
As long as there is some internal interaction present that can cause this to happen, sure. For example, a skater can pull in their arms to increase their rotational speed, but this requires internal forces to be exerted between different parts of their body. It doesn't just happen by magic without some internal forces being involved. And those internal forces require some internal stored potential energy to act. See further comments below.

If the energy is given by $E=\gamma m c^2$ and is constant, then $m$ decreases as $\gamma$ increases.
The $E$ you are referring to here is the kinetic energy of a particular piece of the rotating object (say, for example, the skater's left hand). This $E$ is not constant as the radius decreases; it is increasing, because the rotational velocity is increasing and so the kinetic energy is increasing. But the invariant mass $m$ of the skater's hand does not change during any of this.

What I think is confusing you is that, if $E$ increases for, say, the skater's left hand as they pull in their arms, where does this extra energy come from? It comes from stored energy in the skater's body. The skater is able to move because their metabolic processes are burning various chemicals like glucose in oxygen and using the released energy to power muscle movements. So basically the skater pulling in their arms is an example of chemical energy being converted to rotational energy. The total energy of the skater remains constant.

But actually the skater is a bad example because they are not an isolated system. The skater is radiating heat to the environment all the time; in fact, about 90% of the chemical energy released by the skater's metabolism goes to heat, not muscle movements, so the energy balance I suggested above between chemical energy and rotational energy doesn't actually work--much more chemical energy is burned in the skater than comes out as rotational energy. The excess gets radiated to the environment. Also, the skater is interacting with the ground, so not even the mechanical energy will balance exactly.

For a truly isolated system, such as a spinning cylinder floating in free space, assumed to be completely isolated from its environment (not even radiating any heat), for it to decrease its radius requires some stored potential energy to be converted to rotational energy. The total energy remains constant.

#### AVentura

The $E$ you are referring to here is the kinetic energy of a particular piece of the rotating object (say, for example, the skater's left hand). This $E$ is not constant as the radius decreases; it is increasing, because the rotational velocity is increasing and so the kinetic energy is increasing. But the invariant mass $m$ of the skater's hand does not change during any of this.

...

For a truly isolated system, such as a spinning cylinder floating in free space, assumed to be completely isolated from its environment (not even radiating any heat), for it to decrease its radius requires some stored potential energy to be converted to rotational energy. The total energy remains constant.
I just want to point out that my $E$ there is total energy (in $S$), not kinetic. Which is constant as you say. Stored potential energy is needed. But, before it is used to increase the tangential velocity it already contributes to $E$ as matter. It contributes to the centrifugal "force" too.

#### Dale

Mentor
Maybe we should restrict the conversation to the frame where the cylinder is only rotating. I hold that it can pull itself in, decreasing its radius and increasing its tangential velocity $v_\theta$. And that at any given time $E/L=\omega c^2/v_\theta^2$.
No objections to that.

Do you agree that total energy, momentum, and invariant mass are all conserved during such a process?

#### AVentura

No objections to that.

Do you agree that total energy, momentum, and invariant mass are all conserved during such a process?
Energy is conserved because it is a closed system. Total momentum is conserved because it is a loop that increases in tangential speed (and it is a closed system). For invariant mass I can just say that if this cylinder was rotating in a box and an outside observer was measuring its contents, he would always get the same invariant mass regardless of the angular velocity.

#### Dale

Mentor
For invariant mass I can just say that if this cylinder was rotating in a box and an outside observer was measuring its contents, he would always get the same invariant mass regardless of the angular velocity.
Yes. Since $E$ is conserved and $p$ is conserved then $m$ is necessarily conserved since $m^2 c^2 = E^2/c^2-p^2$

#### AVentura

Note the initial tangential velocity could have been zero and the radius infinity. I can claim all that energy as mass at rest from the start. But it doesn't matter for the sake of this question.

$E\to L \omega$ and now we have a cylinder that twists to translating observers. And I believe the "wavelength" approaches $2\pi L/p_z$.

#### Dale

Mentor
I can claim all that energy as invariant mass
You cannot claim energy as invariant mass because energy is not invariant.

#### AVentura

Ok matter at complete rest. Then it's not. In a closed system.

Mentor
Ok. That works.

#### PeterDonis

Mentor
matter at complete rest.
But in a rotating object, the matter is not at complete rest. The invariant mass of a very small piece of the object is the same as the mass of that piece in its rest frame; but its rest frame is not the same as the rest frame of the object as a whole.

Considering the object as a whole, the invariant mass is not just the sum of the invariant masses of all of its parts. The invariant mass of the object is a whole also includes the kinetic energy of the individual parts in the rest frame of the object as a whole (i.e., its center of mass frame)--or, to put it another way, it also includes the rotational energy of the object, as well as the sum of the invariant (rest) masses of all the parts. And if there is any stored potential energy in the object as a whole--things like compressed springs or batteries that are charged--the stored potential energy contributes to the object's invariant mass as a whole. That is what @PAllen was referring to in the other thread you linked to, when he said invariant mass is not additive.

#### PeterDonis

Mentor
the initial tangential velocity could have been zero and the radius infinity. I can claim all that energy as mass at rest from the start.
Only if you are going to consider some of that rest mass as actually stored potential energy, because you are going to convert some of it to energy (via matter-antimatter annihilation reactions, say) in order to start the object spinning as you decrease its radius.

#### Dale

Mentor
Only if you are going to consider some of that rest mass as actually stored potential energy, because you are going to convert some of it to energy (via matter-antimatter annihilation reactions, say) in order to start the object spinning as you decrease its radius.
This is exactly the sort of terminology I was objecting to. Mass is not converted to energy in annihilation. The mass and energy are both the same before and after the annihilation. It is electrons and positrons that get converted to photons, not mass that gets converted to energy.

#### PeterDonis

Mentor
This is exactly the sort of terminology I was objecting to. Mass is not converted to energy in annihilation.
I understand your objection, but we have to have some way of talking about the fact that if an electron-positron pair annihilates into a pair of photons, we go from a system with constituents that have nonzero invariant mass by themselves, to a system with constituents that have zero invariant mass by themselves, while the invariant mass of the system as a whole stays the same. That's what "mass is converted to energy" is trying to describe. How would you describe that without using those words and without using the unwieldy language I used?

Also, the post of @AVentura that I was responding to talks about starting with a "cylinder" with an infinite radius whose constituents are all at rest, so its invariant mass is in fact the sum of the invariant masses of all its constituents. Then he wants to start decreasing the radius while increasing the rotational speed. This will require reducing the invariant masses of the constituents, since the rotational kinetic energy has to come from somewhere. (There is also a question about binding energy lurking here that we haven't even addressed.) How would you describe this process without using the phrase "mass is converted to energy"?

The mass and energy are both the same before and after the annihilation.
The invariant mass of the system, and its total energy in any given frame, are the same before and after. But the invariant masses of the constituents of the system are not the same.

#### Dale

Mentor
How would you describe that without using those words and without using the unwieldy language I used?
”Internal PE is converted to KE”.

What is described as mass being converted to energy is always actually some form of energy, whether it is the energy stored in an elastic band, a chemical, or an electron. I would even accept the term “mass energy” if there is some reason to distinguish it from other types of PE.

#### PeterDonis

Mentor
I would even accept the term “mass energy” if there is some reason to distinguish it from other types of PE.
Considering rest mass as part of the potential energy is indeed a common thing to do in a relativistic context.

"Mechanical system with de Broglie-like features"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving