Movement of a guitar string at relativistic speeds

In summary: If you stretch a string too much, it will break. If you apply a force perpendicular to the string's axis, the string will recoil and the force will be perpendicular to the string's new length.
  • #36
Sagittarius A-Star said:
Why shall the radius contract?

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

Sagittarius A-Star said:
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.
 
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  • #37
PeterDonis said:
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)?
 
  • #38
Sagittarius A-Star said:
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.
 
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  • #39
PeterDonis said:
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.)
 
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  • #40
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.
 
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  • #41
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.
 
  • #42
Googling I found another paper, according to which ##R > R_0##:
paper said:
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
 
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  • #43
Halc said:
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? :)
 
  • #44
Joda said:
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.
 
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  • #45
PeterDonis said:
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?
 
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  • #46
Redbelly98 said:
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.

Redbelly98 said:
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).
 
  • #47
PeterDonis said:
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 [itex]v=(T / \mu)^{1/2}[/itex]. (v, T, and [itex]\mu[/itex] 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.
 
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  • #48
Redbelly98 said:
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.

Redbelly98 said:
In the relation I'm familiar with there are no coefficients, just [itex]v=(T / \mu)^{1/2}[/itex].

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.)
 
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  • #49
In case you have to go to string theory, it looks like ~50 pages in Zwiebach (chaps 4-6)

85E47789-FBDF-4EC5-A9AD-60511EDBB042.jpeg
 
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  • #50
PeterDonis said:
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.
 
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  • #51
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
 
  • #52
vanhees71 said:
I'm a bit puzzled.

See post #48.
 
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  • #53
caz said:
A rigid body

There is no such thing as a perfectly rigid body in relativity.
 
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  • #54
PeterDonis said:
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”?
 
  • #55
caz said:
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).

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

It can't. Who said it could?

caz said:
a non-relativisitic supersonic flow problem

Has nothing to do with the topic of this thread.
 
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  • #56
I see. I call theoretical strength (commonly approximated as E/10) what you call breaking/ultimate strength. I read your post #48 too quickly.
 
  • #58
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
 
  • #59
caz said:
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##.
 
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  • #60
T
vanhees71 said:
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?
 
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  • #61
If anyone sees this, I'd like to ask another question:

How do we know that the Nambu-Goto action is the one we want to minimize to get the correct equation of motion for the string?
 
  • #62
Joda said:
If anyone sees this, I'd like to ask another question
Maybe it can be seen better, if you start for the new question also a new thread with a headline, which asks specifically for the Nambu-Goto action.

Eventually you could ask a moderator, if the new thread should be issued here in the relativity forum or better in the quantum theory forum. I think, formally it belongs to the relativity form, because it is in literature discussed regarding classical (non-quantized) relativistic strings. But eventually, you find in the quantum theory forum more users, who have better knowledge about it, because of the D-dimensional Minkowski space and also the additional dimension of the particle.

Joda said:
How do we know that the Nambu-Goto action is the one we want to minimize to get the correct equation of motion for the string?
I don't know. I can only provide additional links:
https://www.damtp.cam.ac.uk/user/tong/string/string.pdf
https://en.wikipedia.org/wiki/Nambu–Goto_action
 
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