# I Sticky beads and a simple wave

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1. Jul 31, 2016

### Ibix

A while ago, robphy posted a link to a paper by Saulson describing a "step" gravitational wave described by the metric $ds^2=-dt^2+[1+h(t)]dx^2+[1-h(t)]dy^2$, where $h(t)=h_0H(t-\tau)$ is a step function of (constant) height $h_0$, with the step occuring at $t=\tau$. Presumably, $\tau$ is a function of time and the suppressed $z$ dimension.

I was wondering about Feynman's sticky beads in such a metric. As I understand it, if I aligned the rod parallel to the x direction then the distance between two beads jumps from $l$ to $l\sqrt { 1+h_0}$ as the gravitational wave comes through (more precisely, the round trip time for light pulses travelling from one bead to the other and back goes up by a factor of $\sqrt{1+h_0}$, which I choose to interpret as a distance change). Similarly, the rod's length increases by the same factor - but it has structural strength, so it immediately begins to relax. Thus, the way I'm describing things, in this case the heating effect comes from the contracting rod moving through the (initially stationary) beads.

Normally Feynman's beads are described in terms of the beads moving on a rigid rod. Am I right in thinking that there's no sensible way of using this description in this scenario because there's no way to define a rigid unstressed rod when the gravitational wave is passing through? However, more realistic waves (like those detected by LIGO) are periodic (ish) on a timescale of order 0.1s. Since we would expect stresses in a metal rod (length 1m, speed of sound ~1km/s) to relax away in a few milliseconds, it can be treated as rigid. The length changes as a result of the metric are relaxed away as fast as they accumulate. In that case it makes sense to think of the beads moving along the rod because the changes in the rod length get relaxed away while the changes in the bead separation continue to accumulate.

Is that all correct?

2. Jul 31, 2016

### Staff: Mentor

If this were true, it would also be true in Feynman's original scenario; a "step" GW is not the only possible kind of GW, and any kind of GW gives rise to the same question.

The answer is that, of course an unstressed rod will not be rigid when a GW passes through, but it's perfectly possible for a stressed rod to be rigid, at least to a good enough approximation, when a GW passes through. The "step" GW model won't show this because it assumes an (unphysical) instantaneous discontinuity in the curvature of spacetime. Any real GW will involve spacetime curvature changing at a finite rate, and there will be a wide range of possible GWs where physically possible materials would be able to respond to the change fast enough to remain rigid to a good approximation. Feynman's original scenario basically assumed this kind of setup.

The stresses don't "relax away"; they don't disappear, at least not as long as the GW is passing through. If the stresses disappeared, that would mean the rod was not staying rigid. The stresses have to be there to keep the rod rigid while the spacetime curvature in its vicinity is changing. What is true is that the stresses will change as the GW passes through, fast enough to keep the rod rigid despite the changes in spacetime curvature. But they won't relax to zero until the GW has passed and the spacetime curvature in the rod's vicinity returns to what it was before the GW passed (assuming that that's what happens--and strictly speaking, we're also assuming that that "original" spacetime geometry was flat on the scale of the rod, otherwise the rod would not have been unstressed at the start).

3. Aug 1, 2016

### GeorgeDishman

Curious, I would have thought as Ibix implied, that the stress would appear instantly and would relax thereafter as an exponentially decaying cosine (assuming the metal rod "rings" a bit) settling back to the same 'length' as before the step change occurred.
The wave period was around 0.007s at peak amplitude and 0.004s thereafter during ringdown. There's a handy Factsheet here but the numbers are rounded a bit.
If I was trying to calculate the effect, I'd assume a slight delay so that there was a tiny strain, enough to induce just sufficient stress to accelerate the mass of the bar at the rate needed to keep the 'length' nearly constant. Is that the right approach?

Last edited: Aug 1, 2016
4. Aug 1, 2016

### Staff: Mentor

No, it won't. What is going on here is that there are two effects working against each other:

(1) The atoms in the rod are trying to maintain a constant separation;

(2) The tidal gravity of the GW is trying to change their separation, by making them follow geodesics that diverge or converge.

In the presence of the GW, (1) and (2) are incompatible--the atoms in the rod can't both maintain a constant separation and follow the geodesic paths that the tidal gravity of the GW is trying to make them follow. The result is stress in the rod.

To put it another way: consider the rod's state before the GW passes. By hypothesis, spacetime is flat before the GW passes (at least to a good enough approximation in the vicinity of the rod). That means the rod's atoms can both maintain an constant separation and follow geodesic paths, because the geodesic paths in flat spacetime don't diverge or converge. So the reason the rod is unstressed before the GW passes is not that it always "relaxes" to being unstressed, but that it is in a flat region of spacetime, and its equilibrium state in a flat region of spacetime is unstressed, because all of the rod's atoms can follow geodesics and still maintain constant separation.

While the GW is passing, spacetime in the vicnity of the rod is no longer flat, so the rod's equilibrium state is no longer unstressed. Instead, its equilibrium state will be that state in which the stresses in the rod are just sufficient to counteract the geodesic deviation due to tidal gravity, and keep the rod's atoms at a constant separation.

5. Aug 2, 2016

### GeorgeDishman

All you say is correct but bear in mind that Ibix was asking about a particular form of change and I was responding to that case:
Prior to $t=\tau$, I assume the rod has relaxed and is sitting at rest in flat space. After $t=\tau$, space is again flat but the rod has the "wrong" length so there is internal stress. Thinking about a small element of the rod length $dL$ of mass $dM$ at the end, the stress applies a force but immediately after $t=\tau$, it still has zero momentum. The stress therefore causes it to accelerate from rest towards its new relaxed position. When the rod reaches that length, assuming the motion is under-damped, $dM$ still has a non-zero velocity and will overshoot. The whole thing is just classic damped harmonic oscillation in response to a step change input. So is what I said not correct?
In the case of real waves, you have a damped harmonic oscillator with a driving term which is the full chirp waveform in the case of the LIGO detections.

6. Aug 2, 2016

### Ibix

Thanks, Peter. I am definitely happy to consider a free-falling rod in flat spacetime (plus gravitational wave).

Right - so as long as there is geodesic deviation there is an induced strain and a resulting stress holding the rod together and approximately rigid. It's only when the geodesic deviation goes to zero that the rod relaxes.

I was wondering about the limit of the rigidity approximation. As I understand it, for a rod perpendicular to the travel direction of a plane wave, the strain induced is constant along the rod (analogous to Hubble's law, as Saulson points out). Hooke says the stress is also constant. So the middle region of a very long rod is stressed but in equilibrium as the gravitational wave comes through, and won't move with respect to beads on it. This is not true, however, for regions near the ends of the rod which get the message (travelling at the speed of sound) that the ends are free to move. Right?

None of this affects Feynman's argument, of course, even if I'm right. Just chop up the long rod and you get a flock of the transducers he was talking about.

7. Aug 2, 2016

### Ibix

I was contrasting that with a more typical gravitational wave - it was the behaviour of extended bodies in the latter that I was trying to understand. I'm beginning to think that this was not a helpful approach, as it lead me wrong and, I think, is generating a misunderstanding between you and Peter here.

8. Aug 2, 2016

### GeorgeDishman

It's helpful to me :-) Either what I said above is right which is useful confirmation or its wrong and hopefully someone will tell me why I need to change the way I am thinking about the situation. It's win-win for me.

Also, it's made me realise that a change in the metric can, to put it crudely, "leave a bit of matter behind". I had the impresion that a geodesic kept a small object at the same place in the metric, like "comoving galaxies" in cosmology but clearly that's a naive mistake. I think it's a mental remnant of Newtonian "inertia".
I think it's still valid. For a rod and slow sinusoidal waves, it seems to me that the material will respond by shrinking and stretching but to do that requires a small internal stress so there must be a matching strain (analogy: like the error signal in a servo) hence there will be a small lag between the wave and the response of the material.

For a larger object like the spherical "Earth frame" in the previous thread, I think that could be sliced into thin discs in the plane of the wavefronts and each would undergo stretching and shrinking plus maybe some other distortions depending on the polarisation (thinking of the polar format), but then there is also shear between adjacent discs to consider if the object's z extent is comparable to the wavelength.

9. Aug 2, 2016

### Staff: Mentor

This makes sense if you are saying the rod has the "wrong" length because it has been stretched instantaneously by the passing GW wave front, and after $t = \tau$ it just returns to its original unstressed length. But it doesn't seem like that's what you're saying; see below.

This is not really consistent. If you're idealizing the applied force as instantaneous, you have to also idealize the impulse, i.e., momentum change, as instantaneous. Otherwise you have no effect at all. To put it another way, in reality the force can only change the rod's momentum if it is applied for a finite time, however short. So idealizing the force to be applied for zero time idealizes away any momentum change--unless you adopt a corresponding idealization for momentum change.

No. There is no "new relaxed position". Spacetime is flat after the GW passes, and the rod can only have one equilibrium state in flat spacetime. So its relaxed position must be the same after the GW passes as before. The only possible difference is that after the GW passes, the rod has nonzero momentum (see above), so it moves away from equilibrium and stresses are set up in the rod to restore that equilbrium.

10. Aug 2, 2016

### Staff: Mentor

No. "Matter" means "nonzero stress-energy tensor", and a GW, by hypothesis, is source-free, i.e., the SET is zero. You cannot have zero SET produce nonzero SET; that would require the SET to have a nonzero covariant divergence, which is prohibited.

11. Aug 3, 2016

### GeorgeDishman

There are two ways of looking at it which in this case I think are just a change of coordinates, i.e. no change to the physics. In another situation, I think there is a difference but that is for another day. I've taken a screen grab of the Wikipedia animation and added rulers (I don't have time to animate it at the moment, I'm on a lunch break).

On Wikipedia, the ring of particles oscillates while the rulers are static. The alternative is that the ring is motionless and the rulers get stretched, one expanding while the other shrinks. I got the impression that these two visualisations were what Ibix was comparing:
If you think of the rod in the discussion as having ruler marking on it, the usual picture is the beads moving along a rigid ruler, but the alternative is to consider the beads as unmoving and the rod shrinking and stretching within them.
No, what I am thinking was proposed is that there was an instantaneous change from a Minkowski Metric $ds^2=-dt^2+dx^2+dy^2$ to a Minkowski Metric $ds^2=-dt^2+(1+h_0) dx^2+(1-h_0) dy^2$. In the second view above (fixed particles, stretchy rulers), the rod is relaxed and unmoving before the step change and remains the same size on the screen immediately after the change of metric, and also remains unmoving at that instant. In the first view above (fixed rulers, moving particles), the ends of the ruler jump instantaneously to new locations (i.e. like teleporting, not passing through points between). Again that results in the instantaneous appearance of stress which starts the ends accelerating.
The ends have no momentum at the moment of the change but there is an instantaneous change from unstressed to a non-zero stress in the metal. That starts accelerating the ends so the momentum starts to build. This is a graph of the stress and acceleration, the velocity and momentum would be the integral of this and would die away to zero as the rod relaxes (the scales are arbitrary). I think the stress should be an invariant other than maybe a scaling by $1+h_0$, is that right?

In the first view, the rod magically jumps to a new (proper?) length instantly then relaxes back to its usual value. In the second view, the coordinate length stays the same when the step changes in the metric occurs but it then relaxes to a new coordinate length (back to the original proper length).

The problem is that the verbal descriptions depend on the choice of coordinates even though the physics being described is the same all the time.

12. Aug 3, 2016

### Staff: Mentor

Ah, I see; the GW wave front is basically a "domain wall" between two flat regions with different coordinate scalings. The domain wall passes through the plane at a given value of $z$ at a time $\tau$, where the simplest assumption would be that $\tau = t - z$, i.e., the wave front is traveling at the speed of light in the positive $z$ direction, and passes through the plane $z = 0$ at coordinate time $t = 0$.

I think this is one possible view, but not the only one. See below.

This would be consistent with the above--basically you are idealizing the GW's effect as an instantaneous stretch or shrink of the rod to a new proper length. This causes an instantaneous nonzero stress in the rod, and the effect of the stress is to drive the rod back to its original proper length.

However, I think there's an alternative assumption we could make, which is that the GW's effect is an instantaneous impulse applied to the particles in the rod, so that what changes discontinuously is not the rod's proper length but the momenta of the particles in it. This would then cause the rod to start stretching or shrinking--but the stretch or shrink would cause nonzero stress in the rod, which would drive it back to its original proper length.

The reason I prefer the second alternative is that it keeps the worldlines of the particles in the rod continuous. The first alternative, a discontinuous jump in the proper length of the rod, makes the worldlines of all the particles in the rod discontinuous. In the second alternative, the worldlines are continuous but their derivatives are not (because of the instantaneous jump in momentum and hence in 4-velocity, i.e., in the tangents to the worldlines). The latter seems more tractable mathematically and more reasonable physically.

13. Aug 4, 2016

### GeorgeDishman

Almost, $t=\tau$ is the time at which the wall passes through $z=0$. This is how Saulson describes the approach in the paper Ibix cited:
• "To analyze what goes on in an interferometer, it is simplest if we imagine that the gravitational wave has the form of a step function $h(t)=h_0 H(t-\tau)$, where $h_0$ is the (distressingly small, perhaps 10-21) amplitude of the wave, and $H(t-\tau)$ is the unit step function at time $\tau$. Although gravitational waveforms should come in many varieties, there is a class of them that will in fact have a net dc shift in $h$. [And we lose no generality by considering only a pure step function, since we can always approximate an arbitrary wave form $h(t)$ by a suitable succession of positive and negative step functions.]"
His "succession of positive and negative step functions" would have different values of $\tau$. In section IV he also says:
• "Next, consider what happens at $t=\tau$, when the gravitational wave arrives. Suddenly, all distances between freely falling masses along the x axis are increased by a factor of $(1+{\frac 1 2} h_0)$."
Exactly, and I think that is what Ibix envisaged too. Would that understanding affect your earlier replies?
I understand your proposal but I don't see how you get that from Saulson's metric or his text.
Right, if you think of that in coordinates matched to proper length, the rod ends jump to new locations then return to the original positions as the stress relaxes. I was contrasting that with coordinates where nearby free-fall masses are unmoved by the wave, the rod remains unaffected as the wave passes but then relaxes to a new coordinate length (which is the original proper length) hence it removes all the discontinuities. I got the impression from Pervect's posts last week that this might be related to "transverse-traceless" though the details of that are beyond me.

Last edited: Aug 4, 2016
14. Aug 4, 2016

### Staff: Mentor

This only works if all the equations are only supposed to apply in the plane $z = 0$.

Yes, but again, this only applies in the plane $z = 0$; basically this is saying that we have different possible domain walls that pass through the plane $z = 0$ at different times. In my more general formulation, each step function would be of the form $\tau = t - z + \tau_0$, where $\tau_0$ is a constant that labels the step function (by labeling either the value of $z$ where that particular domain wall is at $t = 0$, or by labeling the value of $t$ at which that particular domain wall passes through the plane $z = 0$--either way of viewing it is consistent with the math). In other words, in general, if we look throughout the spacetime instead of just at the plane $z = 0$, the argument of the step function cannot just be a function of $t$; it must be a function of $t - z$, because that ensures that each domain wall, i.e., each step function wave front, moves at the speed of light, i.e., that each spacetime domain wall is null.

Possibly. See below.

I haven't read all of Saulson's paper, but what you've quoted from it says nothing at all about how the worldlines of individual particles in the rod behave as the wave front passes. It only talks about the metric. The metric in itself does not constrain the rod's particles to move on particular worldlines; it limits the possible worldlines, but not enough to uniquely define them.

What I'm saying is independent of any choice of coordinates. I'm saying there are two distinct possibilities for how the worldlines of the particles in the rod behave when the wave front passes, both of which are consistent with all the math I've seen thus far:

(1) The particles in the rod "jump" discontinuously when the wave front passes, i.e., their worldlines are disconnected at the "domain wall" surface. This corresponds to the rod's proper length discontinuously changing immediately at the instant the wave front passes, and then returning to the original proper length.

(2) The particles in the rod move continuously through the wave front, but their momentum changes discontinuously, i.e., their worldlines are connected at the "domain wall" surface but the tangent vectors to those worldlines change discontinuously at that surface. This corresponds to the rod's proper length not changing immediately at the instant the wave front passes, but changing afterwards as a result of the discontinuous change in momentum, and then returning to the original proper length.

Whether or not the worldlines of the rod's particles are continuous at the domain wall surface is an invariant, independent of any choice of coordinates.

Not if the first possibility above is true; if that possibility is true, then in the chart you are describing here, the coordinate "grid lines" themselves are discontinuous at the "domain wall" surface, so the fact that the $x$ coordinate of each particle in the rod is the same after the wave passes does not mean the worldlines of the particles are continuous, so it does not mean the rod is "unaffected" by the wave. Whether or not the rod is "affected" should also be an invariant, independent of coordinates.

15. Aug 5, 2016

### Ibix

What other information do you need? I thought the metric and its derivatives were enough to describe all unaccelerated motion. Or is this a feature of the discontinuity in the metric?

16. Aug 5, 2016

### Staff: Mentor

First, "all unaccelerated motion" is certainly not a unique set of worldlines.

Second, the motion of the rod in the scenario under discussion is not unaccelerated from the instant the GW passes it, until it has relaxed back to its original proper length.

The discontinuity adds a third factor: whether or not the worldlines of the particles in the rod are also discontinuous at the GW wave front. The metric itself can't tell you that; it's consistent with both possibilities.