I LIGO and light speed constancy

[timmdeeg]

Move inertially is defined as moving along a particular trajectory which, in this case, is staying at rest relative to the transverse traceless coordinates. Most of the atoms in the bar are moving along an inertial trajectory. There are changing interatomic forces but they net to 0 for most atoms in the bar. This change in the interatomic force arrises from the stress field induced by the GW. It nets to zero because this stress field has zero divergence. Geometrically the mechanical stress is due to the underlying geometry (or distance between inertial points) changing with time. If you find this confusing, you're normal.

Anyhow thanks for your patience. I don't know the meaning of the underlined terms, but suspect it might be difficult to explain it in ordinary language. Please consider my layman level.

Hopefully some questions will show what I'm missing.

Why are "Most of the atoms in the bar moving along an inertial trajectory."? I understand that most of the interatomic forces cancel within the bar. But any two neighboring atoms feel these forces, so how can one tell that they are inertial though?

Supposed the GW just induces stretching are all interatomic distances stretched so that this sums up to stretch the whole bar?
Or does the interatomic stretching cancel so that the bar isn't stretched?

Do you agree that the change of interatomic distances in a bar which is falling radially towards a black hole is so tiny that its length increase is negligible compared to the case were there was magically no electrostatic bonding between the atoms of the bar?

 

[timmdeeg]

Can you explain how this is important?

No. I just mentioned atomic vibrations but wasn't sure if they are of any relevance regarding this discussion.

 

[Paul Colby]

I don't know the meaning of the underlined terms, but suspect it might be difficult to explain it in ordinary language.

Transverse traceless coordinates are those coordinates in which particles at rest WRT the coordinates remain at rest WRT the coordinates unless acted upon by a force. It can rightly be viewed as an inertial system in this case which is the weak GW limit. It's existence and construction isn't easy or simple IMO. I choose TT coordinates because Newton's laws work as advertised. As I've said before, continuum mechanics is hard even without GR to help.

Zero divergence? Well, that's vector calculus which I know how to use but not explain in simple terms so let me botch an explanation for you. Mechanical stress is like pressure (actually pressure is one type of stress). A gas at a uniform pressure doesn't move because the forces all cancel. It's only when there is a spatial pressure change or gradient that things start moving. The divergence is the sum of the gradient components. You have to have a non-zero gradient sum for there to be a net force at a point according to the equations of motion. To show this I'd have to use vector calculus. Safe to say I probably don't understand it either .

 

[PeterDonis]

I understand that most of the interatomic forces cancel within the bar. But any two neighboring atoms feel these forces

If the forces on a particular atom cancel, then the atom feels no force, and is therefore inertial ("inertial" means "feels no force"). That's what "cancelling" means. The term "zero divergence" can be thought of as a fancy way of saying "the forces cancel".

 

[timmdeeg]

If the forces on a particular atom cancel, then the atom feels no force, and is therefore inertial ("inertial" means "feels no force"). That's what "cancelling" means. The term "zero divergence" can be thought of as a fancy way of saying "the forces cancel".

Yes, understand, thanks for clarifying this point.

 

[PeterDonis]

Do you agree that the change of interatomic distances in a bar which is falling radially towards a black hole is so tiny that its length increase is negligible compared to the case were there was magically no electrostatic bonding between the atoms of the bar?

This is a very different case from the case of a GW passing through a bar. In the GW case, the change in the metric coefficients is small (at least for all the cases under discussion in this thread) and periodic--which means it doesn't build up over time, it just oscillates. So the induced stress in the bar stays well within the elastic limit of the bar--i.e., it never permanently deforms its structure.

In the case of a bar falling radially into a black hole, the change in the metric coefficients does not oscillate--it builds up over time, getting larger and larger. That means it will eventually stretch any material whatsoever beyond its elastic limit and start permanently deforming it (and ultimately tearing it apart). That is because relativity imposes a finite limit on the tensile strength of materials (roughly, that the speed of sound in the material can't exceed the speed of light), whereas there is no limit on the stretching that can be induced by tidal gravity on a bar oriented radially when falling into a black hole (the stretching increases without bound as the singularity inside the hole is approached). So this case is very different from the case of a GW, and should be discussed in a separate thread if you want to know more about it.

 

[timmdeeg]

This is a very different case from the case of a GW passing through a bar. In the GW case, the change in the metric coefficients is small (at least for all the cases under discussion in this thread) and periodic--which means it doesn't build up over time, it just oscillates. So the induced stress in the bar stays well within the elastic limit of the bar--i.e., it never permanently deforms its structure.

Great. This clarifies what I was missing. Thanks a lot.

 

[pervect]

Can you explain how this is important? For a mass to move non-inertially one has to have a net force applied. The geometry of space is changed by the GW in such a way that the net force on most of the bar matter is zero. In fact the only (net) forces applied by the GW are the to the bar end surfaces[1].

I don't think this can be right. If we have a bar in a slowly varying GW, so that the internal forces of the bar act fast enough to keep the particles of the bar the essentially at the same proper separation, only the center of the bar should have zero proper acceleration. This implies that there IS a net force on any particle not at the center of the bar. I did some more detailed calculations once upon a time of the necessary congruence to keep particles at a constant separation in a 2d plane, I'd have to dig it up, I'm not sure there is the interest. But basically, if all the particles of the bar were force-free, their separation from their neighbor would be changing.

 

[Paul Colby]

I don't think this can be right. If we have a bar in a slowly varying GW, so that the internal forces of the bar act fast enough to keep the particles of the bar the essentially at the same proper separation, only the center of the bar should have zero proper acceleration. This implies that there IS a net force on any particle not at the center of the bar. I did some more detailed calculations once upon a time of the necessary congruence to keep particles at a constant separation in a 2d plane, I'd have to dig it up, I'm not sure there is the interest. But basically, if all the particles of the bar were force-free, their separation from their neighbor would be changing.

You're 100% correct[1] in the limit you are discussing. The post you quote is the tail end of a long discussion in which different limiting cases were discussed. One thing that I think is an important takeaway is a deeper understanding of the interaction of GW with matter so you might be incline to read them. The post you quoted is referring to a very short duration GW pulse which is not slowly varying relative to the speed of sound in the bar.

[1] Well, the force due to the GW is applied just to the ends if the bar is made of isotropic materials. This is true in all cases.

 

[PeterDonis]

the force due to the GW

It might be worth clarifying the usage of the term "force" in this context. Strictly speaking, as I believe I said in a post a while back, a GW does not apply a force to anything--objects moving solely under the influence of a GW, subject to no other interactions, (such as isolated dust particles) move inertially and feel no force. The force felt by atoms at either end of the bar is due to unbalanced forces from nearby atoms. Those forces come into being because the GW changes the metric coefficients, which changes the physical distance between nearby atoms; so the GW does cause the forces, but only indirectly.

 

[timmdeeg]

You distinguished between two cases in #63.

Well, there are two limiting cases both non-resonant, both quite different. In case 1) the frequency of GW is well below the lowest mechanical resonance, while in case 2) the GW frequency is well above the lowest mechanical resonance. By well above and well below let's say 3 orders of magnitude in each case. I was referring to case 2, not case 1, in the quoted text. In case 1) the ends of the bar can and will move to compensate for spatial distance changes due to the slowly varying metric. In case 2 the bar ends simply can't move fast enough to compensate for the distance change due to the GW.

In case 2 the atoms move inertially except those very close to the surface. So almost all atoms follow the change of the metric coefficients in free fall within their elastic limit if the GW passes by. Why then "can't the bar ends move fast enough to compensate for the distance change due to the GW" ? Because of the few atoms which are non-inertial?

 

[Paul Colby]

You distinguished between two cases in #63.

In case 2 the atoms move inertially except those very close to the surface. So almost all atoms follow the change of the metric coefficients in free fall within their elastic limit if the GW passes by. Why then "can't the bar ends move fast enough to compensate for the distance change due to the GW" ? Because of the few atoms which are non-inertial?

Case 2 was intended for the extreme limiting case in which the GW is much shorter duration than the transit time of sound in the bar. In case 2 we are discussing what happens to the atoms over a time scale on the order of the GW pulse duration.

The length of the bar before the GW hits is $L$ with every atom (bit of bar) at rest. In the instant after the GW hits every atom is at exactly the same coordinate it was at prior to being hit by the GW, however, the distance between points has changed so the bar length is now, $h_{xx}(t)L$. The bar is either compressed in length or stretched in an instant[1] depending on the GWs sign even though none of the atoms coordinates have changed. This GW induced length change suddenly puts the bar into uniform stress. Uniform stress in classical mechanics produces 0 force on points interior to the bar but gives rise to forces on the boundary. These (very tiny) boundary forces begin accelerating[2] the bar material at the boundary first setting up an acoustic wave in the bar which propagates inward at the speed of sound in the bar.

[1] Again this is a limiting case where the GW is an arbitrarily sharp square wave hitting the bar broadside.

[2] This is an I level thread so I assume people know basic mechanics. Accelerating from 0 velocity changes the velocity and position of a mass continuously. It takes a finite period of time to change position and velocity, by $F=ma$ and all. The time scale over which the bar length changes is not so constrained in GR. The length is dictated by metric coefficients and such. Yes, no true square wave GWs are likely produced in nature, however, their sharpness or rate of change is not limited by the physics of GW propagation so we can discuss these limits in meaningful ways.

 

[timmdeeg]

The length of the bar before the GW hits is $L$ with every atom (bit of bar) at rest. In the instant after the GW hits every atom is at exactly the same coordinate it was at prior to being hit by the GW, however, the distance between points has changed so the bar length is now, $h_{xx}(t)L$. The bar is either compressed in length or stretched in an instant[1] depending on the GWs sign even though none of the atoms coordinates have changed.

Thanks. My problem is to understand that as you said "the bar ends simply can't move fast enough to compensate for the distance change due to the GW" even though the bar length "follows" the changes of the metric coefficients. I was naively thinking that if the bar length changes fast then its ends have to move as fast. Or at least almost as fast if one takes the forces on the boundary into account. And I'm not really sure why this reasoning is wrong.

 

[Paul Colby]

I was naively thinking that if the bar length changes fast then its ends have to move as fast. Or at least almost as fast if one takes the forces on the boundary into account. And I'm not really sure why this reasoning is wrong.

One really needs to define what "move" means. It's always defined as changing distance with respect to some well defined location. The bar ends don't initially move with respect to the coordinates (which are used to mark off positions and times). But, the coordinates as defined are moving relative to each other by virtue of the metric coefficients. Hard core relativity folks usually prefer to use coordinate free descriptions because of just this type of confusing situation. For the description of continuums mechanics, I simply don't know a coordinate free way of saying what needs to be said. Sorry if this is hard to understand.

 

[pervect]

You're 100% correct[1] in the limit you are discussing. The post you quote is the tail end of a long discussion in which different limiting cases were discussed. One thing that I think is an important takeaway is a deeper understanding of the interaction of GW with matter so you might be incline to read them. The post you quoted is referring to a very short duration GW pulse which is not slowly varying relative to the speed of sound in the bar.

[1] Well, the force due to the GW is applied just to the ends if the bar is made of isotropic materials. This is true in all cases.

Sorry, I should read these long threads more carefully, but - there's just too much of them, I tend not to follow them closely as I ideally would.

From my perspective, the response of a steel bar to a GW and the response of steel bar to an external perturbing force are basically the same for a bar with the dimensions of Ligo's interferometer.

You don't view gravity as a "perturbing force", but for the case under consideration the model works just fine, and I think it will be more familiar to a lot of readers.

Let's talk about the response of a bar to a perturbing force.

We can model the distributed bar as the limit of a lumped spring-mass system. There are two general cases When the springs are strong, we can more or less ignore the effect of the mass, and if the spring is really really stiff, the spring doesn't change length much under perturbing force and we have the classic rigid bar.

The other limit is when the springs are really weak. Then the bar acts like a bunch of mostly disconnected masses, though they aren't perfectly disconnected, because the spring is still there, it's just weak enough to be ignored in the first approximation.

We know how disconnected masses move. They follow geodesics in space-time. Which is what the Ligo test masses do. So the non-rigid bar acts mostly like the test masses, and the rigid bar doesn't. And as an aside, the TT coordinates are ideally suited to this case (the weak-spring case).

So that's the physics of the response of the bar. I'll sketch out my argument about why the "perturbing force" model works at all. It's basically a matter of scale. If we can create a nearly born-rigid congruence of worldlines, there is no issue with changing coordinates away from the TT gauge coordinates, and instead using the Born rigid congruence. The mathematical techniques needed to do this are rather sophisticated, alas, but the results are easy to talk about. When the approach works, it means that you can think of gravity as a perturbing force in a flat space, and if you can also ignore the time dilation issues, you have a flat, essentially Newtonian, space-time with a perturbing force. In the Born coordinate system, a point with a constant cooordinate isn't following a free-fall trajectory, (a space-time goedesic), it's following the trajectory of one of the worldines in the Born congruence.

It's really just a coordinate change, but for many applications I find it makes things simpler than using the TT gauge coordinates. There's no really need to view the TT coordinates as what's "really going on", they're just coordinates, we can use whichever coordinates we like.

If we orient the bar so that it's length is transverse to the direction of propagation of the GW, the critical dimension for the existence of nearly Born rigid coordinates is not the length of the bar, but it's height, the height being oriented to point along the direction of propaation of the GW. This dimension isn't critical at all to the operation of Ligo. Unfortuantely, this isn't a textbook result - it's something I worked out. So I suppose it needs more scrutiny. I should note that I worked out the 2d case first (suppressing the dimesnion in the direction of the GW propagation). Working out the full 3d case and finding the "height" limit came as a bit of a relevation as to the limits of the approach.

 

[Paul Colby]

We can model the distributed bar as the limit of a lumped spring-mass system. There are two general cases When the springs are strong, we can more or less ignore the effect of the mass, and if the spring is really really stiff, the spring doesn't change length much under perturbing force and we have the classic rigid bar.

"Strong" and "weak" spring limits are better understood in terms of fast and slow sound speed IMO. The Earth as a big material sphere are well into the "weak" spring limit as far as LIGO bands are concerned so the dimension on the "bar" under consideration matters as well. This is an important consideration in the few seismic GW detection experiments that people have tried.

 

[PeterDonis]

If we can create a nearly born-rigid congruence of worldlines, there is no issue with changing coordinates away from the TT gauge coordinates, and instead using the Born rigid congruence.

This brings up an issue that I think is worth commenting on. As I understand the usual Newtonian formulation of continuum mechanics, constant coordinate distance is equated with constant physical distance, in the same sense that, to take an example from non-continuum mechanics, a spring whose coordinate extension is equal to its equilibrium extension is assumed to be unstressed. This is what allows a law like Hooke's law to assume its usual simple form: $F = - k x$ assumes that equilibrium--zero force--corresponds to $x = 0$.

If we imagine that spring being affected by a gravitational wave, however, and we choose the TT gauge coordinates, the above statement is no longer true: if the spring starts out unstressed, so its coordinate extension is equal to its equilibrium extension, then as the GW passes, if we are in the high frequency GW limit, the spring's coordinate extension will stay the same (because in this limit its ends move inertially), but it will not remain unstressed. By contrast, if we use Born rigid coordinates, then the change in the spring from unstressed to stressed, induced by the GW passing, will show up as a change in its coordinate extension. Similar remarks would, I would think, apply to continuum mechanics when discussing things like stress-strain relationships.

I mention this because it was mentioned earlier in this thread that TT gauge coordinates are the ones usually used to analyze GW detectors (which AFAIK is the case), but also that in these coordinates the Newtonian formulation of continuum mechanics works as usual. In the light of the above remarks, I'm not entirely sure whether that is true--although it might still be a good enough approximation for weak GWs.

 

[Paul Colby]

I mention this because it was mentioned earlier in this thread that TT gauge coordinates are the ones usually used to analyze GW detectors (which AFAIK is the case), but also that in these coordinates the Newtonian formulation of continuum mechanics works as usual. In the light of the above remarks, I'm not entirely sure whether that is true--although it might still be a good enough approximation for weak GWs.

It's a worthwhile question and one I've thought about some. Seems to me the underlying assumption in the TT-gauge-Newton-works-as-usual approach is that the interatomic forces remain the same (as a function of distance) in a curved space-time. While this now seems a reasonable assumption to me it's far from clear it is the correct one to me. I believe (i.e. I don't really know) that the same assumption is being made in the Born ridged coordinates.

 

[PeterDonis]

Seems to me the underlying assumption in the TT-gauge-Newton-works-as-usual approach is that the interatomic forces remain the same (as a function of distance) in a curved space-time.

Yes, but only if "distance" means proper distance, not TT gauge coordinate distance. If "distance" means TT gauge coordinate distance, this seems obviously false for the case of a GW passing.

I believe (i.e. I don't really know) that the same assumption is being made in the Born ridged coordinates.

If you mean "distance" as proper distance, then the assumption is not dependent on any choice of coordinates--it's a physical assumption about how inter-atomic forces work. If you mean "distance" as coordinate distance, then yes, Born rigid coordinates satisfy the assumption--but, as above, I think TT gauge coordinates do not.

The key property of Born rigid coordinates, which is what makes me say they satisfy the assumption in terms of coordinate distance, is that coordinate distance does equal proper distance in these coordinates (whereas it doesn't always do so in TT gauge coordinates). That means the spatial metric coefficients cannot vary within the patch of spacetime covered by the coordinates. Using Born rigid coordinates to describe a GW would mean that all of the variation in spacetime curvature due to the GW would show up as variation in the time-time (or time-space, if the metric is not diagonal) metric coefficients, whereas in TT gauge coordinates all of that variation shows up in the space-space metric coefficients.