High School Why does a LIGO arm stretch and not move in unison?

[Henri Garcia]

... it would only make the background worse.

Apparently you have never seen a spherical chicken. :-) :-) And thanks for the answer. Can you (and/or Peter) recommend reading material (text?) for a PhD in physics who wants to understand GW and LIGO?

 

[Justin Hunt]

LIGO splits a beam of photons down two perpendicular 4km arms bounce back and fourth then join back up again at the detector and the difference in their arrival is detected. With no noise or GW they would arrive simultaneously.

Now, Gravity waves are disturbances to the very fabric of space time itself just like gravity. Similar to how gravity effects the trajectory of photons, gravity waves also effect the photons traveling down the arms by distorting space time. The effect is so subtle though that they need such long arms to detect it. The noise is only relevant when the photon interacts with something like a mirror or detector, during its travel down the arms it is in a vacuum and distortions to space time is all that can effect it.

@Henry the symmetry I was talking about is the kind of symmetry we find in things such as humans. Your right side is symmetric to your right. The plane of symmetry of LIGO would be 45 degrees between the two arms. The light takes the path of two sides of a 45 45 90 degree triangle (not the hypotenuse) so the line or plane of symmetry would cut the hypotenuse in half. My question earlier was about whether this symmetry matters for GW. I did and still don’t understand the mechanics of GW themselves, but it would be feasible that gravity waves with propagation rays that lie on the plane of symmetry would have identical effects on the photons traveling down the arms and thus have a zero net effect on the detector.

 

[Ibix]

Similar to how gravity effects the trajectory of photons, gravity waves also effect the photons traveling down the arms by distorting space time.

That's not really what LIGO detects. What it detects is a difference in flight times of light down the two arms. Usually this is interpreted as a changing difference in the length of the arms. In principle it's doing nothing more complex than racing light pulses down the two arms and watching for anything other than a draw. The practice is rather more complex because the signals are so weak. (Edit: there's more than one way of describing what LIGO does, but that's one way).

I did and still don’t understand the mechanics of GW themselves, but it would be feasible that gravity waves with propagation rays that lie on the plane of symmetry would have identical effects on the photons traveling down the arms and thus have a zero net effect on the detector.

Gravitational waves (not gravity waves, by the way - those are a kind of surface wave on water) cause changes in distances perpendicular to the direction of propagation of the wave. Say the wave is propagating in the z direction. In one direction (call it x) distances stretch and then squish; in the other (y) direction distances squish then stretch. You are correct that if a wave happens to impinge on LIGO with its x and y directions exactly at 45° to the arms the wave will not be detected. That's one of the reasons for building multiple detectors with different orientations, because each detector is blind to gravitational waves from some parts of the sky. You also need multiple detectors for triangulation and speed estimates and probably other things.

 

[mfb]

You are cotrect that if a wave happens to impinge on LIGO with its x and y directions exactly at 45° to the arms the wave will not be detected.

As an example, Virgo was close to this orientation for the observed neutron star merger, and had a very small signal as result.

 

[Marcus Parker-Rhodes]

Because the wave doesn't move them in unison. It's a wave of tidal gravity; it alternately stretches and squeezes the arm.

As far as I understand a gravitational wave flexes the whole of space/time spreading across the firmament
like a ripple on a pond, but it seems to me that would not only stretch and squash the machine and the scientists operating it, but also include time and so the apparent speed of light.

 

[Ibix]

As far as I understand a gravitational wave flexes the whole of space/time spreading across the firmament
like a ripple on a pond, but it seems to me that would not only stretch and squash the machine and the scientists operating it, but also include time and so the apparent speed of light.

No. You can pick a notion of time so that the stretch-and-squish effect is perpendicular to this, so the gravitational wave has no effect on time. That isn't necessarily the most convenient coordinate system for describing the LIGO instrument, but if you can make the case that the instrument works in one coordinate system, it must work in all coordinate systems, even if the description is more complex.

 

[Marcus Parker-Rhodes]

Yet gravity does effect time. Even in wave form without the usual accompanying heavy stuff. Should two gravitational waves cross each other, at the intersection, might not a black hole form, where time stops altogether? forgive me, It’s getting very late and the mind wanders. good night!

 

[mfb]

With implausibly high intensities two intersecting gravitational waves could form a black hole. Same as everything else that can put sufficient energy in a sufficiently small volume.

We have gravitational waves from billions of different sources going through us all the time. They are just too weak to be relevant.

 

[PeterDonis]

might not a black hole form, where time stops altogether?

Time does not stop in a black hole.

 

[Marcus Parker-Rhodes]

Sorry, I should have added "to an outside observer."

 

[PeterDonis]

Sorry, I should have added "to an outside observer."

That doesn't help; the statement is still not correct.

 

[pervect]

Sorry, I should have added "to an outside observer."

A more correct statement is that if one takes the limit as one approach the event horizon, time (in Schwarzschild coordinates, coordiinates which are frequently represented as being associated with an "outside observer"), "time" slows down towards zero.

This can be misunderstood, because the meaning of "time slowing down" isn't really clear. The idea is commonly used in popularizations though, so I won't belabor its limitations. My goal here is more modest, it's to simply point out what's wrong with taking this idea and extending it to saying that "time stops at the horizon".

The problem with extrapolating this to saying that "time stops at the event horizion" is that the Schwarzschild coordinates are singular there. So the issue is one of omitting the notion of taking a limit. This is important, because to justify the idea of taking the limit, one needs to make sure that the limiting process makes sense. In this case it doesn't, not really, as there is an implied belief that everythign is just fine at the horizon other than time stopping there. But it isn't.

The easy solution to the whole issue is to not insist on using Schwarzschild coordinates, and use coordinates which are well behaved at the horizon. There are multiple choices, including but not limited to Kruskal coordinates or Painleve coordinates, that do not have this issue.

Coordinate charts are basically a kind of "map" of spacetime, and the issue we have here is that the Schwarzschild coordinate map is confusing". THe best solution is to use a different map. But people cling to the confusing map, rather than try and use a different one. I suspect there may be some deeper philosophical reasons for this clinging to confusion, but it'd take this post too far afield to go into them. My goal here is relatively modest, simply to point out the issue with the "time stops at the horizon" idea.

 

[PeterDonis]

This can be misunderstood, because the meaning of "time slowing down" isn't really clear.

Not as you state it, but looking at the math makes it clear whose "time" is being referred to: the proper time of a static observer, "hovering" at a constant altitude above the horizon, as compared to the proper time of an observer very, very far away from the hole, where the two times are compared by, for example, the observers exchanging round-trip light signals and seeing how long the round trip takes by their respective clocks.

The reason this can't be extrapolated to the horizon itself is not that there is a coordinate singularity there in Schwarzschild coordinates; it is that there are no static observers at the horizon. It's physically impossible to "hover" there. So even if we take limits as the horizon is approached, to avoid the coordinate singularity issue, we don't obtain any physically meaningful quantity that can be interpreted as "time stops".

 

[Marcus Parker-Rhodes]

Protected from Spaghettification with the assistance of the Flying Spaghetti Monster, I drop into a black hole.
I notice no sign of time slowing down or, disappointingly, of any sign of Schwarzschild’s event horizon. I continue to fall inwards towards the singularity, but just as I am about to reach it, I find myself back in the void, which is empty because the universe ended some time ago, and the black hole has finally evaporated, releasing me. My watch continues to behave as it should, but I had got it cheap on the internet. Perhaps I should have paid a bit more and got one with an eternity hand.
http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q15.html

 

[PeterDonis]

I drop into a black hole.
I notice no sign of time slowing down or, disappointingly, of any sign of Schwarzschild’s event horizon.

Ok so far.

continue to fall inwards towards the singularity, but just as I am about to reach it, I find myself back in the void, which is empty because the universe ended some time ago, and the black hole has finally evaporated, releasing me.

Wrong. You will hit the singularity long before the hole evaporates. (And according to our best current model, the universe will never end.)

http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q15.html

This is a pop science site, not a textbook or peer-reviewed paper, so it's not a valid source. Its claim about comparing times on the infalling clock and the distant clock is not correct; there is no well-defined way to compare those times. What the distant observer can say is that light signals from the infalling clock take longer and longer to reach him as the infalling clock gets closer and closer to the horizon.

In the purely classical picture of an eternal black hole (i.e., ignoring Hawking radiation and black hole evaporation), the distant observer never sees light signals emitted by the infalling clock at or below the horizon.

In the simplest model in which Hawking radiation and black hole evaporation are included (the one Hawking originally used), the distant observer sees light emitted at the horizon by the infalling clock at the same time he sees the light from the final evaporation of the hole. But the infalling clock still falls in and hits the singularity and is destroyed long before that.

Quantum physicists object to the model I just described because it destroys quantum information: the quantum states in the infalling clock get destroyed in the singularity, which according to standard QM is impossible. However, there is no well established alternate model (although some quantum physicists, like Susskind, like to claim there is.) This is referred to as the "black hole information paradox", and remains unresolved (although, again, some like to claim it has been resolved in favor of their preferred models).

 

[Nugatory]

Protected from Spaghettification with the assistance of the Flying Spaghetti Monster,

You don't need the FSM, you just need to pick a very massive black hole to fall into. The larger the black hole, the smaller the tidal forces between your head and your feet, and the weaker the spaghettification of an object of any given length.

but just as I am about to reach it, I find myself back in the void, which is empty because the universe ended some time ago, and the black hole has finally evaporated, releasing me. My watch continues to behave as it should, but I had got it cheap on the internet. Perhaps I should have paid a bit more and got one with an eternity hand.

That's a common misconception and you might want to give this paper a try: https://arxiv.org/abs/0804.3619

With less math, there are two interesting moments for the outside observer:
1) The last time that they can send a message to you and expect a reply, which is to say the last time that a message they send will reach you before you've crossed the horizon.
2) The last time that they can send a message to you and you will receive it. Any light signal sent after that moment won't catch up with you before you reach the singularity, so can only follow you into it.

Both of these moments pass in a matter of seconds or minutes (depending on the size of the black hole), long long before any hypothetical end of the universe.

 

[mfb]

We can also define similar points for the infalling observer.

1) The last time they can send a reply out and expect a response. This is somewhere outside the event horizon.
2) The last time they can send a reply out and know it will be received. This is just before crossing the event horizon.
3) If the observer outside sends a clock signal, the last timestamp the infalling observer will receive before reaching the singularity. This is microseconds to hours ahead of to their own (previously synchronized) clock, depending on the size of the black hole.