I Movement of a guitar string at relativistic speeds

[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.

 

[PeterDonis]

Why shall the radius contract?

It will contract if the ring is spun up in such a way as to make it contract.

the direction of the radius is perpendicular to the direction of motion and therefore not lenght-contracted

You are confusing length contraction as a frame-dependent effect, which has nothing to do with dynamics, with the dynamics of spinning up the ring. The dynamics of spinning up the ring must deform the ring, and there is a continuous range of possible solutions for how the ring deforms. One endpoint of that range is all radial deformation--the radius of the ring contracts enough to keep the tangential separation between atoms in the ring, in the instantaneous rest frame of each atom, constant. See my post #24.

 

[Sagittarius A-Star]

and there is a continuous range of possible solutions for how the ring deforms.

Shouldn't there be, after the spin-up is over, a unique equilibrium-state, given the number of atoms along the circumference (contracted in the non-rotating frame)?

 

[PeterDonis]

Shouldn't there be, after the spin-up is over, a unique equilibrium-state

Given a particular process of spin-up, there will be a unique end state for that process. But there is not one unique process of spin-up. There is a continuous range of them, and so a continuous range of end states.

It's possible that at least some of that continuous range of end states are unstable, in the sense that there are stresses out of balance. But if that is the case, one would expect that a stable end state would have a balance between radial and tangential stresses, not stresses that were all radial or all tangential. The end state you are describing, where the radius is "length contracted" by the amount you state, is a state with all radial stress, and so would not be expected to be in balance by the criterion I have just described.

This once again illustrates the difference between "length contraction" as a frame-dependent "perspective" effect and actual dynamics. As I noted in an earlier post, an object moving in a straight line at a constant speed that looks length contracted in a frame in which it is moving has no internal stresses; it has not been subjected to any dynamics that would change its shape. But any rotating object will have internal stresses, so there can't be any unique correspondence between a rotating object and "the same object but not rotating", since a non-rotating object will not have internal stresses (assuming that in both cases the center of mass of the object is moving inertially) which will affect its shape.

 

[PeterDonis]

It's possible that at least some of that continuous range of end states are unstable, in the sense that there are stresses out of balance. But if that is the case, one would expect that a stable end state would have a balance between radial and tangential stresses, not stresses that were all radial or all tangential.

Greg Egan has an analysis of relativistic rotating rings and hoops which seems to bear this out:

https://www.gregegan.net/SCIENCE/Rings/Rings.html

The "hoop" case (basically the limit of a ring as the inner and outer radius approach the same value, which is the case we have been discussing) shows the hoop initially expanding, then shrinking as angular velocity increases. (Note that his plots are of equilibrium solutions at various angular velocities; he does not analyze the spin-up process itself.) This can be seen as a result of two effects: centrifugal force, which tends to make the hoop expand (as in the Newtonian case), and the relativistic limitation that the individual atoms of the hoop cannot move faster than light, which tends to make the hoop shrink when the angular velocity gets high enough (since that limit equates to $\omega r < 1$, where $\omega$ is the angular velocity and $r$ is the hoop radius). The latter limitation can be viewed as a sort of "length contraction", but it's worth noting that, until the angular velocity gets fairly high, the expanding due to centrifugal force outweighs the shrinking due to the relativistic effect, and the hoop's radius is actually larger than its starting radius in the non-rotating state.

(One caveat to all of this is that Egan's model is highly idealized, to the point of being "unphysical" in certain regimes. He discusses that towards the end of the article.)

 

[pervect]

Greg Egan did some non-peer reviewed work on relativistic hoops - not rigid ones, but using a "hyperleastic" material model. See for instance https://www.gregegan.net/SCIENCE/Rings/Rings.html

While it isn't peer reviewed, it's an honest effort to look at the problem, and he quotes some peer reviewed papers, especially when he's deriving his material model.

Pre-requisites for understanding Egan's attempt at an approach are some familiarity with the stress-energy tensor $T^{ab}$, and the continuity equations it must satisfy $\nabla_a T^{ab} = 0$. A Lagrangian field approach would also work, as $T^{ab}$ can be derived from the assumptions of a Lagrangian.

Some takeaways from my time looking at the issue. It is unwise to assume either existence or uniqueness of any proposed models - both must be demonstrated formally.

One should also avoid any model where the speed of sound in the material is greater than the speed of light. This is not an automatic feature of a generalized Lagrangian model, but must be imposed.

 

[vanhees71]

Googling I found the following paper, which looks interesting:

https://doi.org/10.1007/BF02710244

There's also a short treatment in Eddington's book "The mathematical theory of gravitation", quoted in the above paper, but it's using the unphysical assumption that the proper density of the disk doesn't change, i.e., it's assuming an infinite Young modulus and thus and infinite speed of sound. I'd rather say, the "most rigid body" in relativity is one with a speed of sound of $c$, and that's discussed in the paper.

 

[Sagittarius A-Star]

Googling I found another paper, according to which $R > R_0$:

Abstract
We rederive the equations of motion for relativistic strings, that is, one-dimensional elastic bodies whose internal energy depends only on their stretching, and use them to study circular string loops rotating in the equatorial plane of flat and black hole spacetimes. We start by obtaining the conditions for equilibrium, and find that:
(i) if the string’s longitudinal speed of sound does not exceed the speed of light then its radius when rotating in Minkowski’s spacetime is always larger than its radius when at rest;
(ii) in Minkowski’s spacetime, equilibria are linearly stable for rotation speeds below a certain threshold, higher than the string’s longitudinal speed of sound, and linearly unstable for some rotationspeeds above it
...
Theorem 3.1.
In Minkowski’s spacetime, the radius $R$ of a rotating string loop which admits a relaxed configuration of radius $R_0$, satisfies the weak energy condition and possesses a well defined longitudinal speed of sound not exceeding the speed of light always satisfies $R > R_0$.

This is something that one would naively expect due to the balance between the centrifugal and elastic forces, but is not obvious in view of the length contraction term in equation (58).

Source:
https://arxiv.org/pdf/1712.05416.pdf

 

[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.

Hey Halc, do you have a source you'd recommend that covers your first point? :)

 

[Halc]

Hey Halc, do you have a source you'd recommend that covers your first point? :)

Depends on what part of the first point you're referring to. That time and mass (and length) are affected by a factor of 7 when moving at .99c relative to the frame in which those properties are measured is a straight consequence of SR theory, 1905. That the guitar is moving at that speed relative to say a muon is simple Galilean relativity, many centuries old, which says simply if x is moving at velocity v relative to y, then y is moving at velocity -v relative to x.

The guitar being unaffected by this means that the guy playing it (who's also moving at the same velocty as the guitar) has no way of noticing any difference. Hence the lack of a test for absolute motion under special relativity.

This was not true before SR. Under Newtonian physics, the guitar could move at say 1.1c and the guy playing it would notice this because light would only come from one side and he could not see in the direction he had been since he's outrunning light coming from that direction, all very similar to how a supersonic aircraft cannot hear anything approaching it from behind.

 

[Redbelly98]

Joda: "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.)

I'm having trouble understanding this, for two reasons. (1) Isn't the speed of sound a function of Young's modulus rather than the breaking strength? And (2) for vibrating strings, the speed of transverse waves depends instead on the string's tension and mass-per-unit-length, right? Or am I missing something?

 

[PeterDonis]

Isn't the speed of sound a function of Young's modulus rather than the breaking strength?

Breaking strength is also a function of Young's modulus, so you can re-express the speed of sound as a function of breaking strength by simply inverting the function that gives breaking strength in terms of Young's modulus.

for vibrating strings, the speed of transverse waves depends instead on the string's tension and mass-per-unit-length, right?

Yes, but it also depends on the material properties of the string (those are contained in the coefficients that relate the wave speed to the tension and mass per unit length).

 

[Redbelly98]

Breaking strength is also a function of Young's modulus, so you can re-express the speed of sound as a function of breaking strength by simply inverting the function that gives breaking strength in terms of Young's modulus.

Interesting, I was unaware of such a dependence. Is there an online reference that shows it? I couldn't find one in an (admittedly quick) online search.

Redbelly98: for vibrating strings, the speed of transverse waves depends instead on the string's tension and mass-per-unit-length, right?

PeterDonis: Yes, but it also depends on the material properties of the string (those are contained in the coefficients that relate the wave speed to the tension and mass per unit length).

In the relation I'm familiar with there are no coefficients, just v=(T / \mu)^{1/2}. (v, T, and \mu are the usual wave speed, string tension, and mass-per-length, respectively.) What am I missing? The only material property that comes into play is the density of the string material.

 

[PeterDonis]

Interesting, I was unaware of such a dependence. Is there an online reference that shows it? I couldn't find one in an (admittedly quick) online search.

I was being sloppy. It would be more correct to say that macroscopic properties like the Young's modulus and the breaking strength and the sound speed are all related, since they all ultimately depend on microscopic properties of the atoms in the material and the forces between them. The key point for this discussion is that infinite breaking strength would also imply infinite sound speed, since both would ultimately have to arise from inter-atomic forces propagating instantaneously throughout the material.

In the relation I'm familiar with there are no coefficients, just v=(T / \mu)^{1/2}.

Ah, yes, you're right, as long as the string is in the elastic regime (where strain is a linear function of stress), the specific material properties of the string don't matter. The limitation imposed by the specific string material is how much tension it can withstand and still remain in the elastic regime.

(Strictly speaking, the specific material properties do have some effects--for example, a guitar with steel strings sounds different from a guitar with nylon strings, because the higher harmonics are different. But that doesn't affect the transverse wave speed.)

 

[Frabjous]

In case you have to go to string theory, it looks like ~50 pages in Zwiebach (chaps 4-6)

 

[vanhees71]

Breaking strength is also a function of Young's modulus, so you can re-express the speed of sound as a function of breaking strength by simply inverting the function that gives breaking strength in terms of Young's modulus.

I'm a bit puzzled. Doesn't Young's modulus only refer to an elastic body in its "linear regime", i.e., for small strains, where the stress can be considered proportional to the strain? For a real-world material the response should become non-linear before the string breaks, i.e., you are outside of the linear range, where Hooke's Law is valid.

 

[Frabjous]

I think there is some confusion about sound speed. A rigid body has infinite sound speed because information about what happens at a single point is known instananeously by the entire body. For bodies that can deform, this information is transferred via material deformation with a sound speed of (relevant modulus/density)^0.5

 

[PeterDonis]

I'm a bit puzzled.

See post #48.

 

[PeterDonis]

A rigid body

There is no such thing as a perfectly rigid body in relativity.

 

[Frabjous]

There is no such thing as a perfectly rigid body in relativity.

I do not know what you mean by breaking strength. I hear it and think of an elastic-failure or like @vanhees71 an elastic-plastic-failure criterion.

I do not understand how a non-rigid body can have an infinite wave speed. I’ll grant that my knowledge of high speed flows is incomplete, but I am not aware of a non-relativisitic supersonic flow problem that has an infinite sound speed solution, although if there is one I would be really interested in it. Is it coming from theoretical strength calculations? I would also be interested in the details of that calculation. If you are driving E/10 to infinity, why isn’t that a “rigid body”?

 

[PeterDonis]

I do not know what you mean by breaking strength.

It's also called "ultimate strength"--the maximum stress the material can sustain without breaking apart. This is different from (and greater than) the elastic limit, which is the maximum stress the material can sustain and still remain in the elastic regime (i.e., if a force causing stress is removed the material will return to its original shape).

I do not understand how a non-rigid body can have an infinite wave speed.

It can't. Who said it could?

a non-relativisitic supersonic flow problem

Has nothing to do with the topic of this thread.

 

[Frabjous]

I see. I call theoretical strength (commonly approximated as E/10) what you call breaking/ultimate strength. I read your post #48 too quickly.

 

[pervect]

I have only skimmed it, but https://arxiv.org/abs/gr-qc/0605025 is a PHD thesis recommended by Greg Egan on the topic of relativistic elasticity and dynamics. There's quite a lot of material there.

 

[Sagittarius A-Star]

Greg Egan's page on Relativistic Elasticity, in which the above mentioned PHD thesis is recommended for further reading:
http://www.gregegan.net/SCIENCE/Rindler/SimpleElasticity.html

Gron: Covariant formulation of Hooke's law:
https://www.researchgate.net/profile/Oyvind_Gron2/publication/252618282_Covariant_formulation_of_Hooke%27s_law/links/58a34b75458515d15fd98f25/Covariant-formulation-of-Hookes-law.pdf

 

[vanhees71]

I do not understand how a non-rigid body can have an infinite wave speed.

No body can have an infinite speed of sound, because that would mean you could use a sound wave to propagate information instantaneously and that violates relativistic causality. The conclusion is the opposite: There cannot be a rigid body in relativity and there cannot be an elastic body with a speed of sound larger than $c$. The material closest to a rigid body would be a hypothetical material where the speed of sound is $c$.

 

[Frabjous]

T

No body can have an infinite speed of sound, because that would mean you could use a sound wave to propagate information instantaneously and that violates relativistic causality. The conclusion is the opposite: There cannot be a rigid body in relativity and there cannot be an elastic body with a speed of sound larger than $c$. The material closest to a rigid body would be a hypothetical material where the speed of sound is $c$.

I understand that. I was tring to come to terms with the infinite breaking strength/sound speed idea. Out of curiosity, what is the sound speed of a string in string theory? Is it c or something else?