Movement of a guitar string at relativistic speeds

[Joda]

TL;DR Summary

I want to know if a string moving at relativistic speeds can be understood without delving into String Theory or Quantum Field Theory, and some tips would be appreciated. :)

Hey! I'm and undergrad in the third year of my applied physics program. I'm taking a course in Special Relativity, and due to Corona the exam has been replaced by a pretty free project where we delve deeply into a topic related to the course.

I'm interested in music, so my professor suggested I read about classical relativistic strings. As I started reading about it, I was hoping that this would not require me to take a crash course in String Theory, as I don't think I have all the requirements or time for it (don't know any general relativity and very limited group/representation theory). Problem is, as soon as I write the two words "string" and "relativity" into google, I get string theory, and as I'm unsure if I need this. I need some advice on where to start.

My question: Is there a theory that deals with something along the lines of how a massive, unbreakable string would move, if its constituent parts moved at relativistic speeds?

 

[dRic2]

Maybe the first chapter of this could help https://allisonmorgan.github.io/images/thesis.pdf

From that link:

The aim of this thesis is not to talk about string theory in all of its 10 dimensional glory. It is to investigate a relatively simple, mechanical system which behaves the same as a string. This model should be somewhat accessible to an advanced introductory or sophomore-level level physics student. In order to compare this system to a string, the equation of motion for a relativistic string will first be developed, starting from the fundamentals of special relativity. Once this equation has been derived, we will discuss the parameter choices under which our model behaves the same as a string. This model contains an arbitrarily large number of masses that are connected by springs and move with a speed close to the speed of light. Throughout my thesis, this model will be called a relativistic N-coupled oscillator or a string. Hopefully a reader will learn from this work that there is more than one interpretation of a relativistic string

See if it fits your needs.

 

[Joda]

Thanks! This looks pretty good. My professor also mentioned that Nambu-Goto Action would be worth looking into. I understand this has to do with String Theory? How would one differentiate between the string in String Theory and just a mechanical relativistic string? Is it simply that the string in String theory is a very small version of a mechanical relativistic string, and String Theorists study how these "mini-strings" influence each other? Or is the string in String Theory a completely different object, without a mass for example? Thanks for your reply.

 

[PeroK]

The best I can suggest is to watch as many old Jimi Hendrix videos as you can.

 

[Joda]

The best I can suggest is to watch as many old Jimi Hendrix videos as you can.

Good one Perhaps that will be my way to actually get some type of time off during the winter holidays! With two other exams immediately afterwards, it easily turns into a pair of extra study weeks! :)

 

[Dale]

Summary:: I want to know if a string moving at relativistic speeds can be understood without delving into String Theory or Quantum Field Theory, and some tips would be appreciated. :)

Problem is, as soon as I write the two words "string" and "relativity" into google, I get string theory, and as I'm unsure if I need this. I need some advice on where to start.

Maybe look for relativistic pendulum or relativistic oscillator or something similar.

 

[sysprog]

The best I can suggest is to watch as many old Jimi Hendrix videos as you can.

The amazingly talented Jimi Hendrix was naturally left-handed and he accordingly re-strung his Fender Stratocaster guitar upside-down.

 

[PeroK]

The amazingly talented Jimi Hendrix was naturally left-handed and he accordingly restrung his Fender Stratocaster guitar upside-down.

I've been to a gig in Jimi Hendrix's bedroom! Not that the man himself was there - this was only a couple of years ago.

 

[sysprog]

I've been to a gig in Jimi Hendrix's bedroom! Not that the man himself was there - this was only a couple of years ago.

Well I guess that could still make you Experienced.

 

[Halc]

My question: Is there a theory that deals with something along the lines of how a massive, unbreakable string would move, if its constituent parts moved at relativistic speeds?

String theory has nothing to do with vibration of guitar string.

Point 1: Any guitar can (and does) move at relativistic speeds, and per Galilean relativity, is unaffected by this. Relative to the frame in which it is moving at say .99c, the strings will vibrate at about a 7th of the rest rate, and will mass about 7 times more. This isn't specific to the guitar since any object moving like that experiences such time dilation and relativistic mass change.

Point 2: There's no such thing as an unbreakable (or unstretchable) string, even in principle. Such a string would have infinite speed of sound, which violates locality. If you accelerate continuously at say 1G dragging a string behind you, there's a finite length of string that can be thus dragged regardless of how strong it is or the lack of the string pulling anything except itself. For 1G, it is about a light year.

 

[vanhees71]

Hm, string theory... I'd rather look at relativistic continuum mechanics, particularly relativistic elasticity. A nice paper is

https://www.researchgate.net/profile/David_Williams220/publication/30830467_The_elastic_energy-momentum_tensor_in_special_relativity/links/5b9d155da6fdccd3cb596bbd/The-elastic-energy-momentum-tensor-in-special-relativity.pdf

Another great source is

https://books.google.de/books?hl=de...Y8X-hJp8CR7ww&redir_esc=y#v=onepage&q&f=false

 

[Joda]

String theory has nothing to do with vibration of guitar string.

Point 1: Any guitar can (and does) move at relativistic speeds, and per Galilean relativity, is unaffected by this. Relative to the frame in which it is moving at say .99c, the strings will vibrate at about a 7th of the rest rate, and will mass about 7 times more. This isn't specific to the guitar since any object moving like that experiences such time dilation and relativistic mass change.

Point 2: There's no such thing as an unbreakable (or unstretchable) string, even in principle. Such a string would have infinite speed of sound, which violates locality. If you accelerate continuously at say 1G dragging a string behind you, there's a finite length of string that can be thus dragged regardless of how strong it is or the lack of the string pulling anything except itself. For 1G, it is about a light year.

Great answer. Pretty much exactly what I needed, thanks! There's one thing I didn't quite understand though:
Why would an unbreakable string have infinite speed of sound?

 

[PeroK]

Point 1: Any guitar can (and does) move at relativistic speeds, and per Galilean relativity, is unaffected by this. Relative to the frame in which it is moving at say .99c, the strings will vibrate at about a 7th of the rest rate, and will mass about 7 times more. This isn't specific to the guitar since any object moving like that experiences such time dilation and relativistic mass change.

In what way does the guitar experience time dilation and relativistic mass change?

Note that the PF standard is not to use relativistic mass:

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[PeroK]

Great answer. Pretty much exactly what I needed, thanks! There's one thing I didn't quite understand though:
Why would an unbreakable string have infinite speed of sound?

I assumed you were talking about the guitar strings vibrating at relativistic speed relative to the frame of the guitar. Not that the whole guitar is moving at relativistic speed!

 

[PeterDonis]

Why would an unbreakable string have infinite speed of sound?

"Unbreakable" means that no finite amount of stress could break the string; in other words, the string's breaking strength is infinite. But the speed of sound in a material is a function of its breaking strength: if the breaking strength is infinite, the speed of sound must also be infinite.

The actual limit imposed by relativity on the breaking strength of materials is that the speed of sound in the material cannot exceed the speed of light. That translates into a finite breaking strength for the material. (This finite limit is many, many orders of magnitude higher than the breaking strength of any known material, so in a practical sense it is not something that ever has to be dealt with directly.)

 

[Sagittarius A-Star]

My professor also mentioned that Nambu-Goto Action would be worth looking into.

I found with Google:

2 The classical relativistic string 11
2.1 The Nambu-Goto action . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 The Polyakov action . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2.1 Action, symmetries, equations of motion . . . . . . . . . . 13
2.2.2 Relation to two-dimension field theory . . . . . . . . . . . 15
2.2.3 Polyakov action in conformal gauge . . . . . . . . . . . . 15
2.2.4 Lightcone coordinates . . . . . . . . . . . . . . . . . . . . 17
2.2.5 Momentum and angular momentum of the string . . . . . 19
2.2.6 Fourier expansion . . . . . . . . . . . . . . . . . . . . . . . 20

Source:
https://www.liverpool.ac.uk/~mohaupt/Strings07.pdf

I found with https://scholar.google.com

Abstract

The authors present a general solution of the equations of motion for a classical relativistic string in Minkowski space. For an 'open' string the solution is determined by a single real null curve which satisfies a quasiperiodicity condition. A formula is deduced which generates all such curves in terms of freely specifiable functions. For a 'closed' string two such curves are required. Their theory establishes a rigorous basis for the numerical investigation of 'cosmic strings', and also forms a starting point for a constraint-free Lorentz covariant analysis of the quantum relativistic theory.

Source:
https://iopscience.iop.org/article/10.1088/0264-9381/5/3/001/meta

I think with "classical" they mean "not quantized".

 

[Halc]

Why would an unbreakable string have infinite speed of sound?

PeterDonis's reply is excellent. Yes, I was talking about the whole guitar moving at relativistic speed since due to the very finite speed of motion of any known material, no string vibrates (traverse or longitudinal) at any rate where relativity comes into play.

If there are two strings that are almost parallel moving perpendicular to their length, the point at which they appear to cross each other from a certain point of view can move at far faster than light speed. Hence a Moire pattern has no apparent limit to the speed at which the patterns propagate, but this is not cause and effect. Just mentioning it as something related to fast moving relationships between strings.

In what way does the guitar experience time dilation and relativistic mass change?

ex·pe·ri·ence
verb
encounter or undergo (an event or occurrence)

A fast moving object undergoes length contraction.
I wasn't using the philosophical definition, if that's what you're asking.

 

[PeroK]

A fast moving object undergoes length contraction.

This is not right at all. According to the principle of relativity, there is no such thing as an absolute state of motion; hence no such thing as a fast moving object.

An object moving relative to a frame of reference is measured to be length contracted in that frame; but in no sense does the object itself experience or undergo length contraction.

 

[Halc]

This is not right at all. According to the principle of relativity, there is no such thing as an absolute state of motion; hence no such thing as a fast moving object.

I know that the speed of an object is relative. I did not assert otherwise. I even referenced in my first post a guitar that is moving fast relative to a frame in which it has high speed. Yes, I know it only is contracted relative to such a frame and not objectively.

but in no sense does the object itself experience or undergo length contraction.

That's a matter of how one looks at things. For instance, a fast spinning (spin is objective, not relative) ring really does decrease in circumference (and radius) as it gets to relativistic speeds, and your wording suggests that such contraction isn't real. You didn't say that, but how is it not the object itself being changed here? All abstract of course since in practice, no material is strong enough to spin at a rate where such contraction can be measured, for the same reason we can't have an infinitely strong guitar string. Any real ring would expand as you spin it up due to strain of the material.

 

[PeterDonis]

a fast spinning (spin is objective, not relative) ring really does decrease in circumference (and radius) as it gets to relativistic speeds

Careful. There are a lot of complexities lurking here, of which I will only mention two (plus there is a third which you mentioned yourself--see the end of this post):

(1) What is this "decrease" determined relative to? The most common comparison to make is between a spinning disk and a non-spinning disk with the same radius--but you're saying the radius changes as well, which leaves no way at all to compare the spinning disk with anything, and therefore no way to say what is "decreased".

(2) If we make the comparison holding the radius fixed, as mentioned in #1 above, then it appears that what is changing is the ratio of the circumference to the radius--i.e., the geometry of the disk is no longer Euclidean, but non-Euclidean, with a circumference smaller than $2 \pi$ times the radius. But this "geometry of the disk" has to be defined very carefully; it is not, for example, the same as any spacelike slice at a particular "time" taken out of the disk's "world tube".

how is it not the object itself being changed here?

In the case of the spinning disk, there will be internal stresses in the material of the disk. A non-spinning disk does not have them. So the physical differences between the two can be attributed to those internal stresses, which in turn can be attributed to the fact that, except for the very center of the disk, the parts of the disk are not moving inertially.

However, in the case of linear relative motion, none of those things are true. The object that is said to be "length contracted" has no internal stresses, and all of its parts are moving inertially. So there is no sense in which the "moving" object is "changed" at all. The length contraction is purely a matter of perspective--of one's choice of reference frame.

Any real ring would expand as you spin it up due to strain of the material.

To put this more generally, it is impossible to spin up a disk while maintaining the same relative distances between all of its parts. (A more technical way of stating this conclusion is that it is impossible to spin up a disk with a Born rigid motion. This is one consequence of a theorem called the Herglotz-Noether theorem.) This is the third complexity that I mentioned above: there is no well-defined way to compare the spinning disk with a non-spinning disk, because there is no distance-preserving transformation from one to the other.

 

[Halc]

Careful. There are a lot of complexities lurking here, of which I will only mention two (plus there is a third which you mentioned yourself

I agree, there are complexities. I’m assuming a Born-rigid ring, which of course just shatters if you put any stress on it.

(1) What is this "decrease" determined relative to?

Seemingly the inertial frame of the center of gravity of the ring. Given no deformation to the thin ring (any finite thickness results in deformation, which is why we don’t use a disk), a spinning ring should in theory be able to pass through an identical one not spinning.

This cannot be done with linear motion. A ring becomes an ellipse relative to a frame in which it is not stationary, but no manipulation of speeds will allow one ring to pass through another barn-paradox style.

The most common comparison to make is between a spinning disk and a non-spinning disk with the same radius--but you're saying the radius changes as well, which leaves no way at all to compare the spinning disk with anything, and therefore no way to say what is "decreased".

The radius can be measured by a series of spokes with length markings on them, and which pass through holes in the unattached rings. The spokes (and holes) contract in width leaving a sort of ship’s steering wheel shape spinning. The same radius can be measured by a stationary stick held next to the spinning ring.

(2) If we make the comparison holding the radius fixed

Not doing that. My example was of a spinning rigid thin ring. You can paint length marks on the ring, and so long as it doesn’t deform, an observer on the surface of the spinning ring will measure the same circumference, but an observer in the inertial frame will measure a smaller one.

In the case of the spinning disk, there will be internal stresses in the material of the disk.

Which is why I didn’t use a disk. A Born-rigid disk will be infinitely brittle and cannot spin at all without breaking. It can accelerate linearly, but only if force is applied uniformly to all points of the object. You can’t gently push it from one point. This gets more obvious with large objects or high acceleration rates.

In theory, one can have a spinning rigid disk if it was manufactured that way, no torque ever applied. Stopping it would break it, or if it was thin enough, I suppose make an interesting negatively curved non-Euclidean surface.

However, in the case of linear relative motion, none of those things are true. The object that is said to be "length contracted" has no internal stresses, and all of its parts are moving inertially. So there is no sense in which the "moving" object is "changed" at all.

Agree. It’s proper dimensions are not a function of its motion, as things must be, else it would be a violation of the principle of relativity.

A further complexity that you didn’t mention is the communication of forces between parts of the ring. This thing is spinning fast and the forces in balance must in theory communicate to (‘cause the acceleration of’) other parts of the ring which are not local. This must be done instantly (frame dependent on the inertial frame of the portion of the ring in question, which is a different frame for every portion) for a rigid ring, which of course is why no material can be rigid like that. A true rigid material violates relativity, and a non-rigid material is going to deform if under any kind of stress.

How might we get a real ring to contract? The acceleration cannot be the result of tension in the ring. If there is no tension, only external centripetal acceleration (applied say by some kind of LHC-style accelerator), then a non-rigid material can be spun up and contracted without deformation of the thin material. It is left as an exercise to the engineer how to get the force-applying machine to adjust to the decreasing radius of the object being spun.

 

[PeterDonis]

I’m assuming a Born-rigid ring, which of course just shatters if you put any stress on it.

No, that's not correct. Born rigidity is not a property of a material or an object. It's a property of a motion. A ring that comes apart if you put any stress on it is not a Born rigid ring; it's a ring with vanishing breaking strength. And it is perfectly possible to have an object undergoing Born rigid motion with internal stresses present; a ring or disk spinning with constant angular velocity is just such an object.

a spinning ring should in theory be able to pass through an identical one not spinning

I see that I misread "ring" for "disk" in your previous post. So some of my previous post is actually not relevant to the case you're discussing; sorry about that.

For a ring, there are actually multiple possibilities for how the ring can be spun up, but yes, one of those possibilities will reduce the ring's radius so that it can pass through a similar ring that was not spun up. See further comments below.

You can paint length marks on the ring, and so long as it doesn’t deform, an observer on the surface of the spinning ring will measure the same circumference

By "so long as it doesn't deform", you mean "so long as it doesn't deform in the tangential direction". It's not possible for the ring to not deform at all; that would imply that there was a Born rigid motion from the ring not spinning to the ring spinning, and there isn't. You have simply specified that the ring is spun up so that all the deformation is in the radial direction, and none in the tangential direction.

Note that you can't idealize this away by saying that the ring has zero extension in the radial direction; that just means the internal radial stress of the ring is no longer well-defined mathematically, so your model is not consistent. Idealizing the ring as having zero radial extension works if all you're interested in is kinematics, but as soon as you start considering things like deformation, you're no longer interested in just kinematics.

A true rigid material violates relativity, and a non-rigid material is going to deform if under any kind of stress.

Yes, agreed. This is related to why the speed of sound in a material cannot exceed the speed of light, and why relativity imposes a finite limit to the breaking strength of materials, as was discussed earlier in the thread.

Note, however, that the constraint imposed by the Herglotz-Noether theorem on Born rigid motions is even stronger than this. For the case of linear acceleration, it actually is possible in principle to induce a Born rigid motion by a "conspiracy of forces"--basically, attaching a tiny rocket to every atom of the object and pre-programming the rockets to fire in exactly the right way to accelerate the atoms so they maintain constant proper distance from each other. But for the case of angular acceleration, even that is not possible; there is literally no possible motion, kinematically, that takes a non-spinning ring to a spinning ring while maintaining constant proper distance between all of the atoms of the ring.

 

[Halc]

Born rigidity is not a property of a material or an object. It's a property of a motion. A ring that comes apart if you put any stress on it is not a Born rigid ring; it's a ring with vanishing breaking strength. And it is perfectly possible to have an object undergoing Born rigid motion with internal stresses present; a ring or disk spinning with constant angular velocity is just such an object.

OK, I see the difference, and I acknowledge that stress can be present with rigid motion, but change of stress cannot. Such change must be accompanied by change of strain, which then no longer qualifies as Born rigid motion.

I see that I misread "ring" for "disk" in your previous post. So some of my previous post is actually not relevant to the case you're discussing; sorry about that.

We can still discuss Luigi spinning the raw pizza crust. Does it contract? Probably not, which is why Luigi is doing it in the first place.

You can paint length marks on the ring, and so long as it doesn’t deform, an observer on the surface of the spinning ring will measure the same circumference

By "so long as it doesn't deform", you mean "so long as it doesn't deform in the tangential direction".

By ‘doesn’t deform’ I meant no change to the strain on the material compared to the non-spinning state. That looks locally like nothing is different to a person on the ring, who locally measures no change in dimensions but is in a highly accelerating and rotating point of view, making it less than trivial to assess the proper circumference of the entire ring.

It's not possible for the ring to not deform at all; that would imply that there was a Born rigid motion from the ring not spinning to the ring spinning, and there isn't.

Assuming a single-point-particle thickness ring, it gets smaller, but nothing else. I suppose that disqualifies it as Born rigid motion since the near and far side are not the same distance apart as they were before? Is that considered ‘deformation’? It’s definitely still circular, at least in the frame where the CoG of the ring is stationary. Your comment seems to contradict your statement at the bottom where length contraction doesn’t violate Born rigidity, so maybe I’m misunderstanding you.

You have simply specified that the ring is spun up so that all the deformation is in the radial direction, and none in the tangential direction.

OK. I guess I hadn’t considered relativistic length contraction to be deformation. For instance, if I take a rod and accelerate it to .8c, it will contract (relative to the original rest frame) to 0.6 of its proper length. You say that is ‘deformation’ despite the fact that the rod is unstressed and still its full proper length in its new frame. Not sure if @PeroK would agree with that who in post 18 said “in no sense does the object itself [ ] undergo length contraction“.

Note that you can't idealize this away by saying that the ring has zero extension in the radial direction

Understood. The dimensions of the ring are objectively different, and that makes is not Born-rigid motion.

For the case of linear acceleration, it actually is possible in principle to induce a Born rigid motion by a "conspiracy of forces"--basically, attaching a tiny rocket to every atom of the object and pre-programming the rockets to fire in exactly the right way to accelerate the atoms so they maintain constant proper distance from each other.

I had opened a thread (elsewhere) on exactly this topic, using a linear accelerator instead of tiny preprogrammed rockets to maintain the rigidity of the accelerating rod. The idea was to compute the minimum time to move a distance of one light hour, without applying any local strain, a 100 light-year object, as measured from the frame where the object is stationary before and after the exercise. I computed 55.3 days with a rather trivial calculation.

But for the case of angular acceleration, even that is not possible; there is literally no possible motion, kinematically, that takes a non-spinning ring to a spinning ring while maintaining constant proper distance between all of the atoms of the ring.

Maybe this is the part I’m not getting. I don’t know what you mean by ‘constant proper distance’ since I’m not sure how proper distance is measured in such a situation.
In the inertial frame of the ring as a whole, the atoms (a band of point-atoms a single atom thick) are getting close together, but they’re moving, so that’s not proper distance. In the local inertial frame of any atom, the proper distance between it and the immediate neighbor is maintained, altered by the fact that the neighbor is now rotating around it. In the rotating frame of the ring, the atoms are definitely getting closer to each other.

 

[PeterDonis]

stress can be present with rigid motion, but change of stress cannot. Such change must be accompanied by change of strain, which then no longer qualifies as Born rigid motion.

Yes.

Assuming a single-point-particle thickness ring

Which you can't, for the reasons I gave in my previous post: the radial stress is not well-defined.

I suppose that disqualifies it as Born rigid motion since the near and far side are not the same distance apart as they were before? Is that considered ‘deformation’?

Yes. More precisely, once you've corrected your model to give the ring a finite radial thickness so it's mathematically well-defined, any change in the radial distance from one side of the ring to the other is deformation.

I guess I hadn’t considered relativistic length contraction to be deformation.

It isn't. See below.

if I take a rod and accelerate it linearly in a Born rigid manner to .8c, it will contract (relative to the original rest frame) to 0.6 of its proper length. You say that is ‘deformation’

No, that's not what I'm saying. See the phrase in bold that I added in the quote above? It is critical. If you accelerate the rod in exactly the manner described in that bolded phrase (which, as I noted in a previous post, requires an exact "conspiracy of forces" applied to each atom of the rod and is not something that can be done practically at all), then the rod remains unstressed throughout the process (because by construction we are keeping each of its atoms at exactly the same proper distance from all other atoms) and is not deformed.

However, your scenario involves angular acceleration, not linear acceleration. You're not just taking an object and accelerating it in a straight line. You're spinning it up, which means each atom is subjected to both a tangential acceleration and a radial acceleration, i.e., an acceleration in two directions. (And, relative to an inertial frame, which direction is "radial" and which is "tangential" changes as an atom goes around.) And, as noted, there is no way to do this in a Born rigid manner, not even with an exact "conspiracy of forces". And so there is no way to avoid deforming the ring as you spin it up. But, as I have noted, you can spin it up in such a way that all of the deformation is radial--the atoms remain at the same tangential distance from each other at all times, but the radial distance between them gets smaller. Or you can spin it up in such a way that all of the deformation is tangential--in which case the radial distance between atoms stays the same but the tangential distance between atoms gets larger. Or you can do something in between, where there is deformation both radially and tangentially. There is a continuum of possibilities between those two extremes.

I don’t know what you mean by ‘constant proper distance’ since I’m not sure how proper distance is measured in such a situation.

By the stresses that are present! If there is zero stress, then proper distances are being kept the same. If there is nonzero stress (measured, say, with a strain gauge), then proper distances are changing.

Or you can use your local inertial frame definition...

In the local inertial frame of any atom, the proper distance between it and the immediate neighbor is maintained

...but then you have to understand that this statement is false. [1] It is impossible to spin up a ring and have it be true. In the local inertial frame of any atom in the ring, the proper distances between it and its immediate neighbors will change. The only freedom you have is to decide how those distances will change, as above. You don't have the option of having them not change at all. That's what the unavoidable presence of stresses in the ring as it is spun up is telling you, since the two definitions are physically consistent with each other; they both give the same answer (proper distances are or are not being kept the same) in any scenario.

[1] - But see my next post for a further note on this.

 

[PeterDonis]

the atoms (a band of point-atoms a single atom thick)

You can't model the atoms as point particles, for the reasons already given. But I suppose you could model the ring as a band of atoms of finite size, one atom thick radially. In that case you would find that the atoms themselves would have to deform: if you kept their tangential separation constant (which is what it seems like you are imagining), then they will deform radially, in the sense that, for example, the spatial configurations of their electron orbitals will be extended further radially than they are tangentially. And this would have an effect on the inter-atomic forces (something like a hoop stress would have to appear, so even though the atoms had the same center-to-center separation tangentially, they would still be under strain because of the internal deformation).

 

[PeterDonis]

In the rotating frame of the ring, the atoms are definitely getting closer to each other.

Be very, very careful here. I suggest actually trying to construct, mathematically, this "rotating frame of the ring", taking into account the ring's varying angular velocity relative to an inertial frame. It is an instructive exercise.

 

[Algr]

A fast moving object undergoes length contraction. [relative to a stationary observer]

Well that brings up something odd. Supposing I have a strange guitar with two strings that are identical, but mounted perpendicular to each other. When I strike them, they both play the same note. Assuming the vibrations are in phase, the two strings would always cross at the same point. And it would make no difference if I attached the strings together at that point with a tiny thread.

But then Joda walks by at .8c, moving parallel to one of the strings. Joda would see the parallel string length contracted, and the perpendicular one not. So Joda would hear them playing different notes? What would happen to the tiny thread?

Edit: It seems to me that there should be some reason why the strings should continue to play the same note, even from Joda's perspective. It can't be time dilation, that would be the same for both strings. Perhaps the "shorter" string vibrates more slowly than its length would dictate because it is denser?

 

[Halc]

Well that brings up something odd. Supposing I have a strange guitar with two strings that are identical, but mounted perpendicular to each other. When I strike them, they both play the same note. Assuming the vibrations are in phase, the two strings would always cross at the same point. And it would make no difference if I attached the strings together at that point with a tiny thread.

So string 1 is strung along x, another along y, and both are plucked to attain vibration along the z axis. Yes, they can be effectively attached by a string.

But then Joda walks by at .8c, moving parallel to one of the strings. Joda would see the parallel string length contracted, and the perpendicular one not. So Joda would hear them playing different notes? What would happen to the tiny thread?

Both strings would continue to move together. That's a frame independent fact. What he'd hear is mostly due to Doppler effect, but both strings would be pitched higher at first and decreasing as he passed them, assuming the sound comes only from the one point on the string, not its whole length. In actuality the string along which he is passing will vibrate with less symmetry and he'd hear the near end of the string sooner than the far end. In reality he'd be moving too fast to hear it at all, so it's more what he sees as he watches the strings than what he hears.

Edit: It seems to me that there should be some reason why the strings should continue to play the same note, even from Joda's perspective. It can't be time dilation, that would be the same for both strings. Perhaps the "shorter" string vibrates more slowly than its length would dictate because it is denser?

Most simple explanation is simple time dilation which effects both equally. Both strings get a boost from relativistic mass gain. The short string retains its full cross section thickness, whereas the sideways string becomes thinner, so the density of both is higher but still identical.

 

[Algr]

Oops, that's right: the perpendicular string would be thinner, so the two strings would not have different densities. Thanks Halc.

But that kills my speculation about why the short string isn't vibrating faster. Normally, a short string would naturally vibrate at a faster rate then a long string. What is preventing that from happening in Joda's POV? (Maybe I should make it a xylophone so we can ignore the tension of the strings.)

 

[Halc]

Disclaimer: I don't know the mathematics of string vibration, so I'm just working through some guesses. Oh, and sorry for the hijack of your topic with the spinning ring diversion above.

The density of the string for purposes of the note played is probably not mass/volume, but rather mass/length. I suspect that two strings with the same mass/length, same length and tension will play the same note even if one string is twice as fat as the other.

So we have a short string that is fat, twice the mass per length. It will vibrate at the same rate as the same string split lengthwise. Connecting them doesn't change the note any more than connecting two balls makes them fall faster from the tower of Pisa.
So the short string is equivalently two strings each having a similar profile (length density) as the long string, but each taking half the total tension. That probably accounts for the two perpendicular strings playing the same note.

 

[PeroK]

For instance, if I take a rod and accelerate it to .8c, it will contract (relative to the original rest frame) to 0.6 of its proper length. You say that is ‘deformation’ despite the fact that the rod is unstressed and still its full proper length in its new frame. Not sure if @PeroK would agree with that who in post 18 said “in no sense does the object itself [ ] undergo length contraction“.

Put simply, the reason that we cannot talk about a rod experiencing length contraction is that the rod has a different length in the infinite possible reference frames. Which length contraction does it experience?

This is the very basis of the theory of SR. If the rod is moving inertially, then it has no sense that it is "really" moving at $0.8c$ and "really" length contracted by a factor of $0.6$.

In the above scenario, you could equally well analyse things from a frame in which the rod begins with a speed of $0.8c$ and decelerates to rest. In that frame, rather than being "contracted" by an acceleration, the rod is "stretched" by a deceleration.

Any proper deformation of the rod is due to the forces applied to it, not to any absolute velocity that you can ascribe to the rod. If we take a practical view, whether a rod is really stretched or contracted depends on whether we pull it or push it. Imagine a large spring: if we accelerate it by pulling it, it will stretch; and, if we push it, it will contract. This has nothing to do with the length contraction of SR.

 

[vanhees71]

Note, however, that the constraint imposed by the Herglotz-Noether theorem on Born rigid motions is even stronger than this. For the case of linear acceleration, it actually is possible in principle to induce a Born rigid motion by a "conspiracy of forces"--basically, attaching a tiny rocket to every atom of the object and pre-programming the rockets to fire in exactly the right way to accelerate the atoms so they maintain constant proper distance from each other. But for the case of angular acceleration, even that is not possible; there is literally no possible motion, kinematically, that takes a non-spinning ring to a spinning ring while maintaining constant proper distance between all of the atoms of the ring.

Indeed, and that's the solution of the disk paradox. You can have a disk rigidly rotating with constant angular velocity but it's impossible to bring it from rest to this "state" keeping its Born-rigid-body properties for the time of angular acceleration. A much more realistic relativistic model for what we call a rigid body in Newtonian mechanics is the (quasi)rigid elastic body, as described, e.g., here

https://arxiv.org/abs/1912.08221

 

[Halc]

Put simply, the reason that we cannot talk about a rod experiencing length contraction is that the rod has a different length in the infinite possible reference frames. Which length contraction does it experience?

Again you assume I made an absolute statement. I did not, having specified a specific reference frame relative to which my statement was meaningful.

In the above scenario, you could equally well analyse things from a frame in which the rod begins with a speed of $0.8c$ and decelerates to rest.

But my statement explicitly said "relative to the original rest frame". My statement of .99c in post 10 said "Relative to the frame in which it is moving at say .99c". I know that velocity figures require a reference.

Any proper deformation of the rod is due to the forces applied to it

I never suggested proper deformation of a object due to linear motion.

The ring does. I had made it thin to minimize the contraction effect between the higher speed outside radius and the lower speed inside radius. I thought all the stress was due only to that, but Peter showed otherwise. Suppose I have a thick ring of radius 3 (inside) and 4 (outside) and we ignore the difference from a length contraction perspective. The material has a thickness of 1. The inside of the stationary ring is 75% of the material of the outside ring. Now we spin it up and contract the radius by half, ignoring the fact that the inside is slower and contracts less. Now the ring is 2 (outside) and either 1 or 1.5 inside. It is 1 if we maintain constant radial thickness (since there's no motion in that direction) and 1.5 if we maintain identical 75% material ratio. The two numbers not matching is additional strain on top of the fact that the inside is slower and not contracted as much. I did not see that additional strain until it was pointed out, so I stand corrected that a thin ring cannot change angular velocity and still maintain rigid motion.

But you're treating me like a total amateur that still thinks in terms of absolute motion despite my pointing out my frame references. It's beginning to be insulting.

Imagine a large spring: if we accelerate it by pulling it, it will stretch; and, if we push it, it will contract. This has nothing to do with the length contraction of SR.

When doing relativity calculations, I typically assume a force applied in proportion to all parts of the spring so it can be accelerated without any proper deformation at all. Notice that I don't say uniform acceleration since the acceleration at the rear of the object is greater than at the front.

Another problem with the ring is that there's no front and rear to its motion when under torque.

 

[PeroK]

I'm sorry if I've offended you. My original statement was simple:

In no way does an object experience time dilation, length contraction or relativistic mass increase (*). To say otherwise, IMO, is to get off on the wrong foot altogether with SR.

(*) Not least because every object is always time dilated, length contracted (in any direction) and has increased mass - all to any possible degree, depending on the frame or reference in which it is studied. It seems odd to say that it experiences all these things simultaneously!

 

[Sagittarius A-Star]

Now we spin it up and contract the radius by half, ignoring the fact that the inside is slower and contracts less.

Why shall the radius contract? If the radius of a rotating ring is measured as R in the non-rotating frame, then it is also R in the rotating frame, because the direction of the radius is perpendicular to the direction of motion and therefore not lenght-contracted. The circumference is then $U=2\pi R$ in the non-rotating frame and $U'=2\pi \gamma R$ in the rotating frame.