a-cosmology-group/acg#274

Well played Han, your simulation seems quite impressive and prescient!

As Dr. Smethurst says, the data is analyzed with “AI trained on what black holes might look like”, trained on the results of general relativity. It’s not surprising then that they get results “consistent with the predictions of general relativity”!

That the “ring” has the size of the point spread function is a very strong argument from the Miyoshi team. It’s known in optics that there is no unambiguous way to deconvolve data that has noise and fluctuates in time, so for all we know both claimed “images of a black hole” might be wrong…

LIGO is not science, it’s just mathematicism running amok in a delusional frenzy of model-fitting and data-massaging.

@budrap00 writes: >LIGO is not science [ .. ]

Read this: How to measure a distance of one thousandth of the proton diameter? The detection of gravitational waves This one too: Gravitational Waves and LIGO

I’m very familiar with the LIGO song and dance (or is it smoke and mirrors?) Here’s the only question that matters: The LIGO equipment is entirely composed of atoms (scale=10^-10 meters). The claimed detection is of a signal at 10^-18 meters. That means the signal is 100 million times smaller than the chaotically vibrating atoms that make up the all of the system including the mirrors. Consequently, at the atomic scale there is no physical meaning to the concept of a measurable fixed length separation between the two mirror sets of the interferometer.

The nature and intent of an interferometer - to measure a variance in the separation between two mirrors has no physical meaning at the atomic scale - because there is no meaningful fixed separation at that scale. The physical concept of an interferometer does not work at the atomic scale, let alone at a scale 100 million times smaller. So how then does LIGO manage to detect a signal that does not exist in terms of its equipment?

The LIGO song and dance team claims that the thermal noise can be subtracted from the data to reveal the signal. But that’s just mathematicism - the equipment cannot pick up a signal it is not sensitive to. It’s like claiming you can measure an atom with a micrometer just because you can do it with some math. The LIGO claims do not make physical sense.

This doesn’t even address the peculiar fraction-of-a-second completion of the BH infall while according to standard BH theory a distant observer can never observe the infall of matter into a BH because of time dilation. That’s the beauty of mathematicism though; the physics doesn’t have to make sense as long as there are handy assumptions of mathematical convenience available to smooth out the rough edges of physical reality and basic logic. Apparently 1BH + 1BH = 0 time dilation. Sure, why not? BHs are just BS anyway.

Not only this. Suppose that an ant is crawling around in the neighborhood of a LIGO. Wouldn’ t that give rise to a giant earthquake when compared with the impact of a gravitational wave?

You should support that ‘ant earthquake’ statement with a calculation Han @HanDeBruijn ! What’s the change in gravitational acceleration when an ant moves by 1mm, at the position of a mirror 10m away, and assuming the ant moves away from the mirror (i.e. from 10.000m to 10.001m)? How much displacement of the LIGO mirror will this acceleration cause in 0.1 seconds? (longest response time of the system)

@RedshiftDrift . I cannot remember that I’ve made a statement. I have been asking a question ! (Maybe a bit suggestive ;-)

@budrap00 could you justify your statement “The physical concept of an interferometer does not work at the atomic scale”?

Please read a paper by myself and colleagues (attached) where a laser is shown to be stabilized by a Fabry-Perot interferometer at a level of $5\times 10^{-16}$ in 3000 seconds averaging. The interferometer is 25cm long, so we detect length variations of $5\times 10^{-16} \ \times \ 25$ cm $= 1.3\times 10^{-16}$ m. We’re only one of several labs around the world doing this routinely, with a budget far smaller than LIGO’s, and we’re already well below the atomic scale by a factor of one million.

A length-variation measurement with an interferometer is possible to a high sensitivity. But what LIGO detects is the energy of high-energy particles absorbed by the mirrors (not gravitational waves).

The Fabry-Perot interferometer we characterize in the paper. It is shown inside the vacuum chamber, during a test phase. One mirror is visible on the right-hand side of the glass block.

Myself installing it in position (my hands were very steady 22 years ago) for the final time before closing the vacuum chamber.

One of our papers: Dubeetal2009Anarrowlinewidth.pdf

@RedshiftDrift . >The interferometer is 25cm long, so we detect length variations of $5 \times 10^{-16}\times 25 \mbox{ cm}$ [ .. ]

That’s weird. So the error in length measurement is directly proportional to the length of the interferometer. Why then does LIGO have such lengthy arms? Looks like a disadvantage ..

we’re already well below the atomic scale by a factor of one million.

No. According to a very simple argument in The Limited Validity of Special Relativity at my website It is impossible to sychronize clocks within intervals smaller than the Compton time and It is impossible to compare lengths within intervals smaller than the Compton (wave)length. Where Compton length and Compton time are defined respectively by

$$ \frac{h}{m_e c} \approx 2.4 \times 10^{-12} m \quad ; \quad \frac{h}{m_e c^2} \approx 8 \times 10^{-21} s $$

where $h=$ Planck’s constant, $c=$ speed of light, $m_e=$ electron (rest) mass. It’s a great technological achievement that ticks of modern Atomic clocks are only one or two orders of magnitude remote from that maximally achievable accuracy in seconds. But there is clearly an isue with the shortest length that can be measured with optical means. Contrary to

A length-variation measurement with an interferometer is possible to a high sensitivity.

At last I quote from the “Read this” above (page 6 of the PDF):

The Michelson Interferometer detects differential effects in arms.

Contrary to

But what LIGO detects is the energy of high-energy particles absorbed by the mirrors (not gravitational waves).

It’s the other way around. A gravitational wave, if such thing exists, produces a relative space distortion

$$ \epsilon = \frac{\Delta L}{L} $$

so the measured displacement $\Delta L = \epsilon L$ is proportional to the strength of the space distortion $\epsilon$ and the length $L$. Hence it is advantageous to have a long interferometer.

Your simple argument is for detection of a single quantum. The LIGO measurement is done with a large number of photons reflecting off a large number of atoms on the surface of the mirrors, averaging out and improving on the quantum limit you calculated. With unlimited sources of funding, LIGO has laser beams of about 12cm diameter, and that beam is shared between both arms, eliminating common noise.

Page 6 of that slide presentation is the usual story, but the signal also depends on an additional sensitivity of the mirrors to the energy deposited in them by particles. Read this paper: 2008_Yamamoto et al._Effect of energy deposited by cosmic-ray particles.pdf

But what LIGO detects is the energy of high-energy particles absorbed by the mirrors (not gravitational waves).

Me too, I’d like to see the continuation of this story. Sounds like the official interpretation is scandalously wrong (because they should know its problems)…

The LIGO measurement is done with a large number of photons reflecting off a large number of atoms on the surface of the mirrors, averaging out and improving on the quantum limit you calculated.

Averaging out, yes. Improving on the quantum limit? Let’s see.

The definition of averaging out a number of $N$ distances $x_k$ is

$$ \overline{x} = \frac{1}{N} \sum_{k=1}^N x_k $$

A measure for the error in the average is

$$ s_x = \sqrt{\frac{1}{N-1} \sum_{k=1}^{N} (x_k-\overline{x})^2} $$

Let’s double the (immensely large) number of atoms $N$ that we are talking about.

$$ \overline{x}’ = \frac{1}{2N} \sum_{k=1}^{2N} x_k \approx \overline{x} $$

Because: it is evident that the latter mean value will hardly be different from the first one. Likewise for the error measure.

$$ s_x’ = \sqrt{\frac{1}{2N-1} \sum_{k=1}^{2N} (x_k-\overline{x})^2} $$

$$ s_x’ = \sqrt{\frac{N-1}{2N-1}\left[\frac{1}{N-1}\sum_{k=1}^{N} (x_k-\overline{x})^2 + \frac{1}{N-1}\sum_{k=N+1}^{2N} (x_k-\overline{x})^2\right]} $$

with

$$ \frac{1}{N-1}\sum_{k=N+1}^{2N} (x_k-\overline{x})^2 \approx \frac{1}{N-1}\sum_{k=1}^{N} (x_k-\overline{x})^2 $$

And so

$$ s_x’ \approx \sqrt{\frac{N-1}{2N-1}2\left[\frac{1}{N-1}\sum_{k=1}^{N} (x_k-\overline{x})^2\right]} = \sqrt{\frac{2N-2}{2N-1}} s_x \approx s_x $$

where the $\approx$ is very near common equality $=$ . I do not see any improvement on my quantum limit. Where does that fairy tale come from ??

On the other hand:

• Erste Filmaufnahme eines Elektrons

• A Boy And His Atom: The World’s Smallest Movie

it is evident that the latter mean value will hardly be different from the first one.

@HanDeBruijn “1” is smaller than “2” by a factor of two. By assuming $1 \approx 2$ and extrapolating to any $N$, you make the mistake of arguing $1 \approx N$.

The signal to noise ratio decreases as the square-root of the number of samples, as you correctly identified. From one atom to $10^{20}$ reflecting the laser beam on the LIGO mirrors, there is, in principle, an improvement by a factor $10^{10}$ in position measurement. Other limitations also have to be considered, such as the number of photons detected in the measurement period. However, the quantum limit of a single photon measurement on a single atom is easy to beat.

@AndriiKyryl

But what LIGO detects is the energy of high-energy particles absorbed by the mirrors (not gravitational waves).

Me too, I’d like to see the continuation of this story. Sounds like the official interpretation is scandalously wrong (because they should know its problems)…

When it was announced that the first gravitational wave had supposedly been detected, only a few days after they turned the detector on, everybody was questionning the analysis procedure. I think they have shown that the analysis is correct, especially when later, signals were detected in coincidence by independent detectors separated by a large distance. LIGO and Virgo have the same mirror size, 35cm/40 kg, KAGRA mirrors are a bit smaller 22cm/?kg.

However, I did ask questions along with other people, insisting on particle energy. They all referred to this same paper 2008_Yamamoto et al._Effect of energy deposited by cosmic-ray particles.pdf. There is no other paper for this calculation that is key to the interpretation of LIGO signals.

That paper only considers the mechanical longitudinal-modes of the mirrors, but it totally ignores bending modes which change mirror curvature and also affects the optical modes of the interferometer by a larger amount. (We’ve had to condider the curvature problem even for our 25cm-interferometer!)

There is no continuation of the story, unfortunately. When asked, Yamamoto el al. never responded, they moved on to some other project (I think Yamamoto himself was a graduate student; no longer with the group now). But who would publish a paper saying that they forgot to include a much bigger effect that voids the meaning of gravitational measurements? Other theorists working on gravitational wave detection are specialized in relativity, they have no clue how an optical interferometer works.

@AndriiKyryl perhaps you could write a paper about this. The calculation is not that difficult, the particle energies are given in Yamamoto’s paper and the optics equations for mirror curvature are available in optics books.

@RedshiftDrift writes: >you make the mistake

No, I didn’t make a mistake. >there is, in principle, an improvement by a factor $10^{10}$ in position measurement.

There is no improvement at all. >the quantum limit of a single photon measurement on a single atom is easy to beat.

The quantum limit of a single photon measurement cannot be beaten.

Here is a little (Pascal) program that proves it.

program ACG;
var
  k,N : integer;
  x : array of double;
  mean,error : double;
begin
  N := 1000;
{ N := 1000000; }
  SetLength(x,N);
  for k := 0 to N-1 do
  begin
    x[k] := Random-0.5;
  end;
  mean := 0;
  for k := 0 to N-1 do
  begin
    mean := mean + x[k]/N;
  end;
  Writeln(mean);
  error := 0;
  for k := 0 to N-1 do
  begin
    error := error + sqr(x[k]-mean)/N;
  end;
  error := sqrt(error);
  Writeln(error);
end.

A run with $N=1,000$ gives:

F:\DSchijf\Delphi\2025jul>ACG
-5.98569136764854E-0003
 2.87279621001668E-0001 # error

And a run with $N=1,000,000$ gives:

F:\DSchijf\Delphi\2025jul>ACG
-1.45917460046704E-0004
 2.88436296324332E-0001 # error

No significant difference can be observed between error(1,000) and error(1,000,000), as I’ve argued. Again, I don’t see where the “improvement” fairy tale comes from.

@HanDeBruijn ok, you didn’t make that mistake, you make another mistake. Your program calculates the sample standard deviation $s$. What is measured is the mean, and the standard deviation of the mean is $S = \sqrt{s^2/N}$, smaller than $s$ by $\sqrt N$. See https://physics.stackexchange.com/a/704608

From the reference, titled What is the ‘correct’ way of determining uncertainty of an average value from multiple measurements? >I am very confused as to what the correct way is to calculate the uncertainty of the average of values ($x_{avg}$) in a data set of measurements $(x_1…x_N)$. I have found at least four different ways of doing it around the internet [ .. ]

Not uncommon with that “discipline” called Statistics. But okay, let’s proceed. >the standard deviation of the mean is

$$ S = \sqrt{\frac{s^2}{N}} \quad \mbox{with} \quad s = \sqrt{\frac{\sum_{k=1}^N(x_k-\overline{x})^2}{N-1}} \quad \Longrightarrow \quad S = \sqrt{\frac{\sum_{k=1}^N(x_k-\overline{x})^2}{N(N-1)}} $$

Now take a look at the motivation in the accepted answer.

you intuitively want a measure that will decrease as $N$ increases, because the more (independent) measurements you make, the smaller you expect your uncertainty to be.

Sure, physics intuition as a “scientific” argument. And you expect ? Speak for yourself, please ! Different methods are suggested in the other answers. So the answer as quoted above may be just wishful thinking, a self-fulfilling prophecy.

Actually there is no best way.

Therefore I see no reason (yet) to share your conclusions, @RedshiftDrift .

I was expecting you would look at the equations, not the confused comments of the confused person asking the question. Haven’t you learned to do your own research? You are even more literal than I am… I have to feed you pre-digested answers for every single detail of my arguments!

Since I can’t find my old book on statistics, perhaps one of those is sufficiently rigorous for a mathematician: Last equation of https://online.stat.psu.edu/stat414/lesson/24/24.4 First equation of https://en.wikipedia.org/wiki/Standard_error#Standard_error_of_the_sample_mean Eq. 7.1.2 in Introductory_Statistics_with_Probability_(CUNY)

I was expecting you would look at the equations, not the confused comments of the confused person asking the question.

Actually the confused comments of the confused person are identical with mine.

It’s very simple, Louis. I categorically refuse to understand things that cannot reasonably be understood. With the words of @budrap00 : the concept of a   $Var(\overline{X})$   is pure mathematicism without any physical counterpart. There is no mean that is meaner than the mean, if you know what I mean. But okay, that first result happens to be standard. The second one is different: there is no variance of the sample mean beyond the variance that we already have for samples. With other words: there is no smoother mirror beyond the physical mirror with its molecules jiggling around all the time.

Haven’t you learned to do your own research? You are even more literal than I am…

I have spent the whole evening to make sense of that $\sigma^2$ divided by $n$ but did not succeed in doing so. Draw any conclusion that you want. Just remember that I am never impressed by the siren voices of authorities.

A bit back on topic. The Bohr radius of an atom like hydrogen is

$$ a_0 = \frac{\hbar}{m_e c.\alpha} \approx 137 \times \mbox{Compton length} $$

We have seen that it is indeed possible to make a picture of separate (carbon monoxide) atoms, which is because of the $(137 \times)$ .

P.S. And in case of, sorry: never try to be funny in a language that is not your own :-(

@HanDeBruijn How do you explain this: On Monday, I go to the lab and measure the frequency of the laser that is proportional to the length of the 25 cm interferometer, and on Tuesday I remeasure the frequency of that laser, and do so for every day of the week. Comparing the five numbers, they all agree to $4\times 10^{-15}$, corresponding to a length change smaller than $10^{-15}$ m.

@RedshiftDrift .

Comparing the five numbers, they all agree to $4\times 10^{-15}$ [ .. ]

Insufficient information. $4\times 10^{-15}$ as what frequency ? Please keep in mind that not everyone of us is an optics professor. Things obvious to you may be not obvious for people out of the know.

corresponding to a length change smaller than $10^{-15} m$ .

How ? Or did you already use the variance of the sample mean in your calculations ?

@HanDeBruijn The measurement and analysis go as follows: The frequency on a given day is obtained by reading numbers $f_k$ on a device that takes a laser frequency measurement every second. (The reference frequency is from the spectroscopy of a strontium ion.) We get about $N_d = 30000$ numbers each day and calculate the average frequency with your equation

$$ \overline f = \frac{1}{N_d} \sum_{k=1}^{N_d} f_k. $$

For each day $i$, an average frequency $x_i \equiv \overline f_i$ is obtained:

Monday: $x_1 =$ 444 779 044 095 485.2 Hz Tuesday: $x_2 =$ 444 779 044 095 482.9 Hz Wednesday: $x_3 =$ 444 779 044 095 484.3 Hz Thursday: $x_4 =$ 444 779 044 095 485.4 Hz Friday: $x_5 =$ 444 779 044 095 481.2 Hz

These values are independent from each other (laser and frequency counters are turned off every night) except for the interferometer which is the same one every day.

The standard deviation of these values, as per your equation, is

$$ s_x = \sqrt{ \frac{1}{5-1} \sum_{k = 1}^5 \left(x_k − \overline x \right)^2 } $$

is $s_x = 1.7$ Hz $\equiv \Delta \nu$. This gives a relative uncertainty $\Delta \nu/\nu = 1.7/\overline x = 1.7/444779044095483.8 = 4\times 10^{-15}$. Since the laser is servo-locked to the interferometer, the laser frequency is proportional to the inverse of the length of the interferometer

$$ \nu = q\frac{c}{2L}, $$

where $q$ is a constant integer, characteristic of the experimental system ($q$ is the number of half-wavelengths inside the interferometer), and $L$ is the length of the interferometer. Relative variations in the frequency of the laser $\Delta \nu/\nu$ are equal in magnitude (and of opposite sign) to relative variations in the length of the interferometer $\Delta L/L$ since, from the above equation

$$ \Delta \nu = -q\frac{c}{2L^2} \Delta L, $$

and

$$ \left| \frac{\Delta \nu}{\nu} \right| = \left| -\frac{\Delta L}{L} \right|. $$

Therefore,

$$ \frac{s_x}{445 THz} = 4\times 10^{-15} = \frac{\Delta L}{ 0.25\mbox{ m}}. $$

From day to day, the length of the interferometer is stable to $\Delta L = 10^{-15}\mbox{ m}$.

I have used your statistics equations for the average and the standard deviation to show that the average length of the interferometer does not change by more than the size of a proton from day to day. That is, even if each molecule at the surface of the mirrors is vibrating by a million times that size.

The signal stability is already evident in $x_1 - x_5$ By dividing by 30,000 the results compare the (average) measures per day, and there would be no indication of ‘atom jitter’ in those results.

@RedshiftDrift do you have any analysis on the spread of the original $f_k$ which might give some indication of stability? Not sure how this would be interpreted though, since there is no way to correlate a reading to an atom’s position. It might just be interesting to see how much it varies.

Even for individual samples, a dose of averaging could already be built into your measuring instruments too, where some measure of $V_{rms}$ may be present, and it also uses ‘atoms’ to receive the signal. Again, even if you could reduce the smoothing, trying to correlate an error signal to atom jitter is an unlikely task.

Here is where the explanation falls apart for me:

L is the length of the interferometer

That term has no physical meaning at the atomic scale of 10^-10 meters - at that scale a uniform, fixed or variable separation between the mirrors does not exist. The mirrors do not have the smooth evenly separated surfaces they have at the classical scale at or below the atomic scale.

Scale is the operative word here. Classical scale math does not apply to atomic and subatomic scales if it makes assumptions (like a smooth surface) that do not apply at that scale. This situation is exactly analogous to the case of classical Solar system gravitational math not working at galactic and higher scales. Scale matters.

It cannot be assumed that some perfectly reasonable math that works on a particular, defined and carefully constructed or constrained system has universal significance across all physical scales. Physical reality is not structured like that. Different scales have different operational properties that one-size-fits-all math does not accommodate.

Thanks, @RedshiftDrift , for your very elaborate answer. What happens here is exactly as I have predicted and repeated. Namely, what you are doing is taking the variance of sample means - i.e. not the variance of the samples themselves. As is obvious from your definition of the five $x_i$ out of 30000 samples each and then calculating the variance of these means. I have argued that this is mathematicism. It produces the illusion of an extraordinary precision that does not exist in the physics. If I am right, it is the only flaw in your derivation. Sadly the flaw is standard math. And the chain is as weak as its weakest link.

@Francois-Zinserling

a dose of averaging could already be built into your measuring instruments

Averaging - is - built into the instrument since we need to eliminate contributions from random thermal fluctions. The laser beam averages over about $10^{14}$ atoms on the mirrors, and the frequency meter gives a reading every second, averaging frequency fluctuations faster than one second.

do you have any analysis on the spread of the original $f_k$

The Allan deviation does exactly that! See graph below from the paper I posted above

What it tells us is the more you average, the lower the uncertainty on the measurement is. Eventually it reaches a minimum due to instabilities in the measuring system or the interferometer itself. In our case, we “beat the thermal noise” by measuring longer than one hour. Our instrument was optimized for stability since it is to be used as an atomic clock. From the graph, you could ‘in principle’ calculate the thermal noise of our interferometer, but since some noise also comes from electronics and laser noise, it would be hard to ‘extract’ the value of thermal vibrations from this measurement. (These are old results, the system does much better now but I no longer work with the group.) Also, not every types of ‘noises’ average as $1/\sqrt N$, there is a whole field of study on that.

The difficult thing is to average as much as possible over the shortest time. With a 1000x bigger budget and 100x the number of scientists, the stability limit can be lowered to get accurate measurements at a few milliseconds, like LIGO does (second figure below) by neglecting long term stability (LIGO would not be a good clock!) Their best stability is at around 10ms (100 Hz), the time interval required to measure “gravitational waves”. Note that at 100Hz, their biggest problem is thermal noise (red line).

Conclusion: small length variations can be detected in a few milliseconds, but the alternative cause for the LIGO signals is the detection of energy deposited by particles and bending of the mirrors (no length change), that give signals that look like GW-detections. Since you understand what’s going on @Francois-Zinserling , you can try to explain this to @HanDeBruijn and @budrap00. I do not have the time to teach online courses on measurement physics, statistics interpretation in physics, noise analysis, optical physics, servo-loop feedback, etc.

@RedshiftDrift , @Francois-Zinserling , @budrap00 . >Averaging - is - built into the instrument

An average of averages is an average, so that doesn’t harm. >What it tells us is the more you average, the lower the uncertainty on the measurement is.

The simple answer is: NO. Uncertainty remains the same (I hope that $\sigma$’s are not hardwired too). >there is a whole field of study on that.

Sure. There is also a whole field of study on Set Theory. And I don’t need it. >you can try to explain this to @HanDeBruijn and @budrap00 [ .. ]

Louis, be my hero and make a movie of the quarks inside a proton protons inside an Uranium nucleus. Then we will be on speaking terms again. >I do not have the time to teach [ impressive list deleted ]

A bon entendeur demi mot suffit. I rest my case.

Erratum

An average of averages is an average, so that doesn’t harm.

Really? Then read this: Why is an average of an average usually incorrect?

I’ve wondered where that 10^15m number came from and it turns out to be just some sloppy physics. Basically, a dimensionless ratio of the frequency variance of the system is converted to a dimensional (length) value without the benefit of converting from a frequency basis to a wavelength basis. If you do that conversion then the sensitivity of the system is to 10^-7 meters (essentially the wavelength of the laser) - a perfectly reasonable claim to which doubtless neither @HanDeBruijn nor I would object.

In addition to @budrap00’s reply.

Assuming that the formula below is mathematically correct for the physical uncertainty of the samples $\sigma$ and the pure mathemacist $S$ of the samples mean, we can calculate the other way around, knowing that the first and main average is over about $N = 10^{14}$ atoms on the mirrors. Resulting in a sensitivity

$$ \sigma = S \sqrt{N}= \sqrt{10^{14}} \times 10^{-15} m = 10^{-8} m $$

Update

First we ask AI (Google) about speed of light divided by frequency, giving wavelength. >A light wave with a frequency of 445 THz has a wavelength of approximately 674 nanometers (nm).

Second. From this reply we have: >in 3000 seconds averaging.

Third. A next reply says: >We get about $N_d=30000$ numbers each day and calculate the average

Let’s take 3600 seconds for convenience, that is an hour of averaging each second. Then our calculation “refines” into

$$ \sigma = \sqrt{3600} \times 10^{-8} m = 6 \times 10^{-7} m = 600 \times 10^{-9} m\approx 674\mbox{ }nm $$

At last it is read: >From one atom to $10^{20}$ reflecting the laser beam on the LIGO mirrors, there is, in principle, an improvement by a factor $10^{10}$ in position measurement.

Resulting in the following wayback to reality for LIGO.

$$ \sigma = 10^{-18} \times 10^{10} m = 10^{-8} m $$

Roughly the same as with the first estimate for the equipment in the laboratory of Louis. Which was to be expected.

In physics, a measurable quantity is defined by the operational procedure used to measure it. What I call “length of the cavity” is measured by using the instruments in the laboratory configured in a way that is sensitive to a quantity that has units of length.

Whichever way you see it, if someone walks by the interferometer and produces a force that changes its “length” by the size of a proton, the numbers on the counters will change in an unambiguous way. You can’t do that with sloppy physics.

A frequency counter counts Hertz (cycles per second) and the unit of Hz is inverse time (t^-1). To convert frequency to a unit of length in basic electromagnetic theory you have to use the standard formula fw=c which turns a frequency measurement into a wavelength, the unit of which is meters. Converting your 10^15Hz frequency measurement yields a 10^-7 meter measurement. You can’t just slap a meter unit on a frequency measurement without converting it - to do so is sloppy physics. It doesn’t matter matter how sophisticated your equipment is. If you can’t keep the measurement units straight you wind up making ridiculous claims - like a 10^-15 length measurement.

how would that account for the near-simultaneous signal occurrence at both detectors? - @budrap00

Catastrophic events, such as supernovae or colliding stars, emit a lot of high-energy particles. After travelling to Earth, some of these particles are simultaneously absorbed by every voluminous mirror that contains enough absorbing material. The deposited energy triggers a mechanical distortion in the large mirrors of every interferometer, producing the near-simultaneous detections of an oscillating signal. The 4 km arms of LIGO (3 km for Virgo and KAGRA) makes the device extremely sensitive to bending of the mirrors. It is not a change of length of the interferometer that is detected, it is a change of the mirror curvature due to bending that is detected, with a much greater sensitivity.

From the timing of the detected events, gravitational-wave-astronomers use the speed of light (the speed they think “gravitational waves” are travelling at) to calculate the direction of the event. This works because high energy particles almost travel at the speed of light. I believe there was at least one confirmed optical counterpart to a “gravitational wave event”. But what LIGO saw was the particles emitted by that ‘catastrophic event’, which also emitted a lot of light.

In reality, “GW”-detectors detect and identify the direction of a source of high-energy particles with their mirrors, which are sensitive particle-energy detectors.

Maybe, but it seems a bit of a reach. Most cosmic rays don’t make it to the surface of the Earth. That the signal LIGO massages out of the data is of extraterrestrial origin is entirely speculative - where it came from cannot be determined from the data. LIGO is primarily an exercise in mathematicist self-delusion. Hell of a laser though.

@budrap00

it seems a bit of a reach

Can you quantify this with a number? This is not “a reach”. The paper cited above gives the energy deposited by high-energy particles in the mirrors and the frequency at which this happens. Perhaps you could read the paper before guessing that the numbers are too low?

Most cosmic rays don’t make it to the surface of the Earth

Read the paper, enough cosmic rays make it to the surface of Earth. There are many observatories that routinely detect high-energy and indirect showers of particles: https://www.scientificamerican.com/article/high-energy-cosmic-rays-come-from-outside-our-galaxy/ Enough muons can penetrate through massive amounts of material and produce an image: https://www.sciencefocus.com/future-technology/how-scientists-are-using-cosmic-radiation-to-peek-inside-the-pyramids If a small muon detector can detect cosmic particles going through pyramids, a sensitive LIGO mirror can detect the few high-energy particles that reach the ground.

that the signal LIGO massages out of the data is of extraterrestrial origin is entirely speculative

Not when the LIGO signal is correlated with a Virgo signal, and an optical counterpart is detected in another galaxy I found one paper here.

I read the linked paper did you? From the conclusion:

The arrival frequency of high-energy cosmic-ray muons that generate enough large showers inside the mirrors of LCGT and Advanced LIGO is one per a millennium. If a shower that occurs near a mirror brings more than a thousand particles to the mirror (an original particle of the shower has an energy that is more than 1 TeV), the vibration will be observed in LCGT and Advanced LIGO interferometers.

The first sentence, despite employing incoherent terminology (cosmic-rays are not muons), seems to “quantify” my point that most cosmic-rays do not penetrate all the way to the Earth’s surface. The second sentence discusses a muon shower that occurs “near” a mirror. Your assumption that similar nearby events would occur simultaneously at the remote interferometers is what I would call a reach.

If a small muon detector can detect cosmic particles going through pyramids, a sensitive LIGO mirror can detect the few high-energy particles that reach the ground.

The problem with the LIGO setup is that it is hypersensitive. It picks up more signals than it can account for. Sabine Hossenfelder did a good write up on this some years back that also points out the dubious nature of the one-time-only “multi-messenger” event detection you referred to: https://backreaction.blogspot.com/2019/09/whats-up-with-ligo.html

BTW, you can build a muon detector for $100 so testing your theory would be cheap. Just install a few in the immediate vicinity of the LIGO detectors and if you detect a spike in both locations at the same time LIGO claims a BH merger event - you nailed it.

From What’s up with LIGO? according to bud rap. >Then there is the rather physically preposterous claim that a classical scale interferometer with 4-kilometer arms is somehow capable of measuring a signal that is 100 million times smaller than the atoms of the mirrors that supposedly detect the signal. Such a claim is physically inexplicable, but with a little creative math, I guess you can come up with a pretty picture of such a signal.

Where the little creative math is identified as the variance of the sample mean - i.e. not the variance of the samples themselves. The variance of the sample mean is made up in such a way that it is inversely proportional to the number of samples. And this mathematicism is thought to be of physical significance. That’s how they arrive at the 100 million times smaller than the atoms of the mirrors.

and also a reason I did not mention, that all of those atoms are vibrating chaotically.

Story continuing in Cosmological surveys .