Re: An Analysis of the Resolution of the Michelson-Morley Experiment
From: Tom Roberts (tjroberts_at_lucent.com)
Date: 01/25/05
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Date: Tue, 25 Jan 2005 13:04:23 -0600
Nomenclature: "precision" refers to how well a given value can be
represented, and says nothing about how close the value is to the actual
value being measured. That is _not_ what I am discussing. The usual
words we use are: "errors", "errorbars", and "resolution"; "accuracy" is
not used because we do not know the true value. I did not use "error"
because around here that is certain to be misconstrued. I deliberately
chose to use the word "resolution, AND I TOLD THE READER WHAT I MEAN BY
THAT WORD. Regardless of what greywolf42 says, this _IS_ how these words
are used today by working physicists.
I choose to ignore (and snip) all of greywolf42's insults and vacuous
statements.
greywolf42 wrote:
> Tom Roberts <tjroberts@lucent.com> wrote in message
> news:x1TId.8807$ZV4.7443@newssvr31.news.prodigy.com...
>> with an error analysis of the measurement, it can be determined
>> whether or not the measurement is consistent with a theory that
>> predicts a null result.
>
> And error analysis can also determine whether or not the measurement is
> consistent with a theory that predicts a *non-null* result.
Sure. But please look at my title -- I am discussing their RESOLUTION,
not their results.
> But in the experiment, we don't really care what various theories claim.
> The observations (data) stand on their own.
No, they don't. One must know how accurate they are. One cannot know
that without knowing the resolution of the underlying measurements.
Hence my article.
And you are quite wrong on a different level: the theory Michelson
implicitly used gives a wildly different value for "speed relative to
the ether" than does Cahill's. As is well known, a theory is required to
interpret experimental data -- otherwise one has just a set of numbers
without meaning.
> If there is a 'signal'
> that shows above pure noise or the physical resolution of the instrument, it
> exists. Even if we don't (yet) have a theory that predicts that
> experimental result.
Hmmmm. You cannot know that without knowing the noise level and/or
resolution of the instrument. THAT is what I am discussing here. And the
conclusion is clear: the variations seen in their plot on page 340 are
much smaller than the errorbars that should be drawn on the data points.
>>I have identified three approaches:
>> 1. Look into a modern Michelson interferometer and estimate the
>> measurement resolution.
>
> What about duplicating the MMX interferometer. Why make this a *different*
> interferometer?
A. OHSA would never permit the construction or use of such an instrument
today (gallons of mercury in the open -- you've got to be kidding!).
B. No competent physicst today would use such an instrument to make
this sort of measurement -- VASTLY better techniques are avilable
today. We live in 2005, not 1887. <shrug>
C. The details of the interferometer are not very important -- what is
at issue is the actual appearance of interference fringes, and the
ability of the human eyeball to resolve and locate them. That has
not changed significantly since 1887, AFAIK.
D. I have neither the resources not interest to construct such an
apparatus. But in college I did have the opportunity to use a
version of a Michelson interferometer, as have many generations of
physicists.
>> 3. Use the original authors' data in a statistical analysis of the
>> resolution displayed by the actual data.
>
> #3 has nothing to do with the physical resolution of the instrument -- which
> doesn't change regardless of how many readings we take.
You have to READ my article and UNDERSTAND the analysis I am
performing. The analysis I perform does indeed relate directly to the
physical resolution of the instrument. <shrug>
>>So I will assume that
>>the errors in the individual measurements are uncorrelated, and normally
>>distributed. While such an assumption is undesirable, the available data
>>essentially force it -- a competent modern repetition of this experiment
>>would take pains to accurately measure the actual resolutions.
>
> Tom is calling Michelson incompetent!
Yes, in 2005 a paper like this would be rightly rejected due to its
author's incompetence. But in 1887 it was state-of-the-art. Perhaps you
should look around you and notice that science has progressed since
1887. <shrug>
>>Fortunately, the presence of a rather large systematic error in the data
>>implies that this statistical independence is reasonably likely[#].
>
> I am presuming that Tom is referring to the apparent drift in the readings
> (2 to 5 divisions per minute) as a "large systematic error".
Yes. Over the course of measuring one row in their table this totaled as
much as 30 divisions (0.6 fringe width), or almost 30 times the largest
excursion from zero in the figure on page 340 of their paper. I think
that qualifies as "large". That it is a systematic error is clear -- the
value at mark 16 is not the same before and after a turn.
>>Note, however, that this analysis technique _forces_ the data to be
>>cyclical. That is, the above subtraction ensures that at the beginning
>>and end of each turn the value will be exactly zero; any non-zero
>>measurement in between will naturally appear to be "cyclical".
>
> A linear assumption in no way forces a cyclical variation onto data.
This linear SUBTRACTION most definitely does. It forces the signal to
be zero at the beginning and end of a turn.
> And the correct way of addressing the issue is to do a chi squared fit on a
> sinusoidal pattern.
I am discussing their RESOLUTION, not their results.
> Tom has also championed 0.01 fringe
> for Cahill -- simply because Tom "likes" Cahill's conclusion.
You sure cannot read at all. I disagree completely with Cahill's
approach, and have said so several times. And I have _NEVER_ "championed
0.01 fringe" for any manual optical measurement like this. Quite the
contrary, I have repeatedly pointed out that such a claim is far too small.
>>3. A statistical analysis of the resolution displayed in the data
>>-----------------------------------------------------------------
>>The key to doing this is to find instances in the data where they
>>measured the same value multiple times; then a histogram of the multiple
>>measurements will give a distribution of the errors, and the resolution
>>can be obtained from the distribution.
>
> The physical resolution has already been obtained, above.
Yes, two estimates of it. I am determining multiple estimates so they
can be compared. And estimate #3 is the one most relevant to be used
in fits to their data, becuase it COMES FROM their data.
> Variations in actual measurements are merely
> statistical error.
You simply do not understand the analysis here.
If one makes multiple measurements of a single quantity, and then
histograms the results, the sigma of that histogram gives an estimate of
the resolution of the underlying measurement. That is what I am doing
here.
>>[In fact for perpendicular arms there is an additional
>>90-degree symmetry, unexploited by all authors including
>>me.]
>
> Symmetry is irrelevant in determining statistical variations (precision).
The symmetries are used to determine which values are multiple
measurements of the same quantity, so the differences between such
measurements can be histogrammed to determine the resolution of the
underlying measurement.
I chose not to use the 90-degree symmery, because it is more complicated
than the 180-degree one, and because the arms are not exactly perpendicular.
>>of the original measurements is
>>sqrt(6) times this value.
>
> Nice try, but no. You cannot determine the variation in the orginal data
> from the average value.
I am not looking at "the average value", I am looking at the difference
between two average values that measure the same real signal. With the
assumption of independent errors one can determine the resolution of the
original meausurement from this (as described above).
The main reason I relate this back to the resolution of the original
position measurement is to compare to the other estimates of that. As
you point out, this does not affect the application of the error
estimate to the data below.
>>To compare to the original authors' data table, each row is an average
>>of 6 turns,
>>and so should have a resolution of 0.14/sqrt(6) = 0.057 fringe width.
>
> You can't determine the precision of the average measurement from the
> absolute value of the measurement!
I did not do that. I took the estimate of their resolution and used it
to determine the resolution of the average. Given my assumptions this is
valid. As you pointed out, this cancels the early multiplication by sqrt(6).
> As I noted in the other thread, Tom assumed a zero result, and calcuated the
> "sigmas" from zero.
That is BLATANTLY false. I determined the sigma of their measurement
from the histogram of the DIFFERENCES between measurements 180 dgrees
apart. As any real signal must be the same for orientations 180 degrees
apart, this is a valid way to measure their actual resolution because
_any_ real signal is removed. You really should READ the article you are
attempting to criticise.
I repeat: I am discussing their RESOLUTION, not their result.
> But he did not determine a chi-squared fit to a
> sinusoidal curve, and then compare the two -- to see which theory was the
> 'better' fit.
Right. I am discussing their RESOLUTIONS, not their result. In order to
perform such a fit, one must know their actual resolution.
> Tom is now claiming that the MMX is both "consistent" with SR (null result)
> and "equivocal" with a non-null result.
Let me jump ahead 1 paragraph:
>>Interestingly, when one histograms the data with the assumed-linear
>>systematic subtracted, the deviations from zero are roughly Gaussian
>>distributed with a mean of -0.01 fringe and a sigma for individual
>>measurements of 0.1 fringe. While this is _not_ an error plot [...]
Yes, here and nowhere else I am briefly discussing ther result, and not
their resolution. And I said so ("While this is _not_ an error plot [...]").
This histogram of their data has a sigma of 0.1 fringe. And the
estimates for their resolution are 0.11 or 0.15 fringe. That directly
implies that if one computes the chi-square for the fit to a 0-parameter
straight line one will obtain a good fit and a high-confidence
chi-square. That's what I meant when I said
>> it solidly demonstrates that
>>the measurements are consistent with the hypothesis of a truly null
>>result.
>
> The same would be true of a pure sine wave, Tom.
No. You must include a constant term as well (except for an orientation
along marker 16 which was forced to have a zero signal by the
subtraction of the assumed-linear systematic). But yes, such a fit could
be equally good as long as the amplitude is small enough to fit inside
the error bars. <shrug>
>>When Consoli and Costanzo [2] display a graph of the July 9 PM data,
>>they drew error bars approximately 0.005 fringe -- more than a factor of
>>ten too small.
>
> No, they simply started with a value that was a factor of 10 smaller than
> you started with, Tom.
Again you do not understand my analysis. I have shown, that the value
they used is a factor of ~10 too small. I did not "start with" any
value, my estimates #2 and #3 come directly from the data; while that
phrase could apply to my estimate #1, it does not affect the value used.
And estimate #1 is not invalid or arbitrary, it only seems so to someone
who has never actually looked into a Michelson interferometer. <shrug>
>>They give no indication whatsoever how they arrived at
>>this value;
>
> In this, they are no different than you were, Tom.
Then you indeed cannot read. I have described in detail how I obtained
estimates of their resolution of 0.1-0.2, 0.07, and 0.15 fringe widths.
They give no indication whatsoever how they arrived at the value they
used. Anybody who can actually read can easily verify this.
> You simply started with
> a bald assertion that the physcial resolution *was* between 0.1 and 0.2
> fringe.
Go back and actually READ my original article. I described in detail
how I obtained 3 different estimates of their resolution.
> Tom is afraid to do the chi squared fit for a sinusoid versus a straight
> line.
I am not "afraid" at all. It's just that my subject here does not
include it -- I am discussing their RESOLUTIONS and not their results.
Yes, a fit to their data for small-amplitude sinusoids will have good
chi-squares. So what? I am discussing their RESOLUTION. <shrug>
> Because he can see the wave just as well as anyone else.
The "wave" is much smaller in amplitude than the errorbars on the
points. I cannot help it if Michelson and Morley did not actually draw
those errorbars -- in their day the necessity of doing so was not known.
But YOU are not living in 1887, and YOU have no such excuse.
Neither do Consoli and Costanzo; I cannot help it if they pulled
errorbars out of the air that are a factor of ~10 too small. I have
provided an analysis of the experiemnt that gives three estimates of
what the errorbars actually are.
>>It is clear that with the above error estimate a
>>zero-parameter flat line fits the data as well as their 10-parameter
>>Fourier decomposition.
>
> But then, there is no reason for them to use Tom's estimate of error.
Why is that? You think they are not using Michelson and Morley's data?
Internal relationships among those data imply they have an underlying
resolution of 0.15 fringe (estimate #3), and the author's statements
imply a resolution of 0.07 fringe (estimate #2). And the difference
between these estimates can be qualitatively explained by their large
systematic error, present in the former (#3) but almost absent in the
latter (#2). Applied to C&C's plot, these estimates yield errorbars ~10
time larger than they used, and which are larger than their entire plot.
> For
> Tom used his own (quite arbitrary) values at the beginning of his effort.
Please go back and READ the article. I used data from the Michelson and
Morley paper to obtain estimates #2 and #3 of their resolution. These
are not "quite arbitrary" at all.
> (Tom has in
> the past supported a 0.01 fringe physical resolution for Cahill.)
Blatantly false. You _REALLY_ need to learn how to read.
>>Conclusion
>>----------
>>The recent attempts to "re-analyze" the Michelson Morley
>>experiment[1][2] are woefully incomplete, and do not include an accurate
>>consideration of the experiment's actual resolution.
>
> Translation: They disagree with Tom.
You REALLY need to learn how to read. Neither of those papers "include
an accurate consideration of the experiment's actual resolution", which
anybody who can actually read can confirm quite easily.
>>In any case, the experiment is indeed solidly consistent with the
>>prediction of SR -- a null result.
>
> Only if you first fudge the numbers, then avoid doing any chi squared
> fits
I fudged no numbers. And I did do enough analysis of their result to
know that a chi-squared fit to the 0-parameter flat line will have an
excellent chi-squared (see above). But this article is focusing on their
RESOLUTION, not their result.
In another article greywolf42 wrote in reply to this (discussing
estimate #2):
>> Note, however, that such measurements were surely carried out over a
>> period much shorter than 36 minutes, or even 6 minutes,
>
> Michelson explicitly stated that each turn of the apparatus took 6 minutes.
Yes. But the "such measurements" here are their measurements of a fringe
width, presumably made while the interferometer is not rotating. Surely
this took much less than 6 minutes per measurement, as I said. Perhaps
you should READ the article you are attempting to criticise.
In summary, greywolf42 again proves he does not bother to actually read
the article he is trying to criticise. As that will likely happen for
this article, too, don't expect me to respond to him.
Tom Roberts tjroberts@lucent.com
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