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| Greg Hennessy |
 Posted: Wed Jan 19, 2005 9:41 am |
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In article <DVxHd.958$rc4.359@fe07.usenetserver.com>,
greywolf42 <mingstb@marssim-ss.com> wrote:
[quote:e704b010fc]Now, in the Joseph's case, above, we are talking about effects similar to
measuring paramecia with a meter stick. The COBE resolution is 1 part in
10,000 at any given intensity. But the absolute value of the reported
results are 1 part in 100,000 from the background blackbody curve.
[/quote:e704b010fc]
Well, I see greywolf is back to claiming one part in 10,000 for COBE,
when his reference claimed something else. Of course the value that
Lerner quoted was for the COBE FIRAS instrument, and the value of 1 in
100,000 was from the COBE DMR instrument, as has been explained to
greywolf multiple times. But he seems to find it convenient to ignore
data that conflicts with his worldview. |
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| Randy Poe |
| Posted: Wed Jan 19, 2005 2:30 pm |
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greywolf42 wrote:
[quote:0eb100d80d]If I make a series of such measurements that are
STATISTICALLY INDEPENDENT I can improve that accuracy to the limit
of
the systematic errors involved, by averaging multiple measurements.
1) Can you support this claim, instead of simply assert it?
[/quote:0eb100d80d]
This is just what is usually called "standard error of
the mean". The error in the mean of n measurements goes
as sqrt(n).
The theory is elementary. Suppose each measurement X1, X2,...
Xn has a variance of V (so a standard deviation of sqrt(V)).
If the measurements are independent, then the variance
of Xsum = (X1 + X2 + ... + Xn) = (V + V + ... + V) = n*V.
To find the variance of Xmean = Xsum/n, you need to know
that for any constant a and random variable X with variance
Vx, the variance of aX is a^2*Vx.
So the variance of Xmean = Xsum/n is var(Xsum)/n^2 =
n*V/n^2 = V/n.
The standard deviation of Xmean is sqrt(V)/sqrt(n). Take
100 measurements and you reduce the uncertainty in Xmean
by 10. Take 10000 measurements and you reduce it by 100.
- Randy |
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| greywolf42 |
| Posted: Tue Jan 25, 2005 5:04 pm |
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Greg Hennessy <greg.hennessy@tantalus.cox.net> wrote in message
news:csmd54$fi4$1@tantalus.no-ip.org...
[quote:f4611fb5ab]In article <DVxHd.958$rc4.359@fe07.usenetserver.com>,
greywolf42 <mingstb@marssim-ss.com> wrote:
Now, in the Joseph's case, above, we are talking about effects similar
to measuring paramecia with a meter stick. The COBE resolution is 1
part
in 10,000 at any given intensity. But the absolute value of the
reported
results are 1 part in 100,000 from the background blackbody curve.
Well, I see greywolf is back to claiming one part in 10,000 for COBE,
when his reference claimed something else. Of course the value that
Lerner quoted was for the COBE FIRAS instrument, and the value of 1 in
100,000 was from the COBE DMR instrument, as has been explained to
greywolf multiple times.
[/quote:f4611fb5ab]
LOL! What a pathetic lie. Just like last time, Greg.
[quote:f4611fb5ab]But he seems to find it convenient to ignore
data that conflicts with his worldview.
[/quote:f4611fb5ab]
LOL! So please provide the reference for the COBE DMR.
Bye in this thread, too.
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| greywolf42 |
| Posted: Tue Jan 25, 2005 5:04 pm |
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Randy Poe <poespam-trap@yahoo.com> wrote in message
news:1106163023.502161.111940@f14g2000cwb.googlegroups.com...
[quote:9e292579d1]
greywolf42 wrote:
If I make a series of such measurements that are
STATISTICALLY INDEPENDENT I can improve that accuracy to the limit
of the systematic errors involved, by averaging multiple measurements.
1) Can you support this claim, instead of simply assert it?
This is just what is usually called "standard error of
the mean". The error in the mean of n measurements goes
as sqrt(n).
The theory is elementary. Suppose each measurement X1, X2,...
Xn has a variance of V (so a standard deviation of sqrt(V)).
[/quote:9e292579d1]
But, in this case, there is no variance, within the known limits of
systematic error (the physical resolution of the device).
[quote:9e292579d1]If the measurements are independent,
[/quote:9e292579d1]
That *IS* the question, here.
[quote:9e292579d1]then the variance
of Xsum = (X1 + X2 + ... + Xn) = (V + V + ... + V) = n*V.
To find the variance of Xmean = Xsum/n, you need to know
that for any constant a and random variable X with variance
Vx, the variance of aX is a^2*Vx.
So the variance of Xmean = Xsum/n is var(Xsum)/n^2 =
n*V/n^2 = V/n.
The standard deviation of Xmean is sqrt(V)/sqrt(n). Take
100 measurements and you reduce the uncertainty in Xmean
by 10. Take 10000 measurements and you reduce it by 100.
[/quote:9e292579d1]
I see you missed the point. I'm well aware of the basic math.
Now, can you point me to any reference that states that you can use this
math for regions in which a measuring device cannot operate (i.e. below the
physical resolution of the device)?
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| Greg Hennessy |
| Posted: Tue Jan 25, 2005 5:58 pm |
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In article <MnzJd.6298$VA5.5695@fe07.usenetserver.com>,
greywolf42 <mingstb@marssim-ss.com> wrote:
[quote:0e6647b7a3]But he seems to find it convenient to ignore
data that conflicts with his worldview.
LOL! So please provide the reference for the COBE DMR.
Bye in this thread, too.
[/quote:0e6647b7a3]
Do you want me to provide references, or are you gone? |
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| Bjoern Feuerbacher |
| Posted: Wed Jan 26, 2005 3:49 am |
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greywolf42 wrote:
[quote:42b583c220]Greg Hennessy <greg.hennessy@tantalus.cox.net> wrote in message
news:csjn4d$rec$1@tantalus.no-ip.org...
In article <wgcHd.810$rc4.391@fe07.usenetserver.com>,
greywolf42 <mingstb@marssim-ss.com> wrote:
[/quote:42b583c220]
[snip]
[quote:42b583c220]and quoted you the paper that shows
that the resolution of the instrument in question is 7 degrees, not a
number of microK.
But Roberts and Joseph and Bjoern claim a few microK.
[/quote:42b583c220]
Well, I simply used your term and did not bother to
correct you. But Greg is right - "resolution" is simply
the wrong term here. The resolution of an instrument
refers to its ability to distinguish things which are
spatially close to one another.
Look it up at www.m-w.org. The relevant definition would
be 1h here:
"the process or capability of making distinguishable the individual
parts of an object, closely adjacent optical images, or sources of light"
OTOH, the "sensitivity" of an instrument tells us how
small the signal is which it can still measure (and
distinguish from noise).
[quote:42b583c220]The units of the sensitivity of the instrument is
Kelvin,
I believe that you are mistaken. The device is simply a multiple channel
intensity recorder, with 100 channels. Each channel "sensitive" to 0.1%.
[/quote:42b583c220]
And where did you get this from?
[snip]
Bye,
Bjoern |
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| Dirk Van de moortel |
| Posted: Wed Jan 26, 2005 12:20 pm |
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"Greg Hennessy" <greg.hennessy@tantalus.cox.net> wrote in message news:ct74hf$k5d$1@tantalus.no-ip.org...
[quote:8b601cbdad]In article <MnzJd.6298$VA5.5695@fe07.usenetserver.com>,
greywolf42 <mingstb@marssim-ss.com> wrote:
But he seems to find it convenient to ignore
data that conflicts with his worldview.
LOL! So please provide the reference for the COBE DMR.
Bye in this thread, too.
Do you want me to provide references, or are you gone?
[/quote:8b601cbdad]
He says "Bye in this thread" and then he continues:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GoodBye.html
He's a bit off, if you ask me.
Dirk Vdm |
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| Bill Rowe |
| Posted: Sat Jan 29, 2005 12:49 am |
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In article <IXwKd.6769$VA5.2919@fe07.usenetserver.com>,
"greywolf42" <mingstb@marssim-ss.com> wrote:
[quote:b4db61a04f]Tom Roberts <tjroberts@lucent.com> wrote in message
news:ctdikk$du6@netnews.proxy.lucent.com...
Bill Rowe wrote:
greywolf42 said:
If you know that there is a systematic error, then you
redo the experiment.
[...]
When things are done using best experimental technique, systematics
errors can be eliminated.
These claims are naive.
There are classes of systematic errors that cannot be eliminated.
LOL! What are these classes, Tom?
[/quote:b4db61a04f]
My comment was far too strong and careless as Tom Roberts correctly
points out.
For a simple example of systematic error that cannot be eliminated
consider a temperature measurement. For the measurement to take place
there must be heat transfer from the object whose temperature you are
trying to measure to the instrument you are using. That will clearly
cool the thing being measured, an unavoidable systematic error.
--
To reply via email subtract one hundred nine |
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| Androcles |
| Posted: Sat Jan 29, 2005 5:08 am |
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"Bill Rowe" <readnewscix@earthlink.net.invalid> wrote in message
news:readnewscix-A21103.21481428012005@news1.west.earthlink.net...
[quote:1dd6f4097d]My comment was far too strong and careless as Tom Roberts correctly
points out.
For a simple example of systematic error that cannot be eliminated
consider a temperature measurement. For the measurement to take place
there must be heat transfer from the object whose temperature you are
trying to measure to the instrument you are using. That will clearly
cool the thing being measured, an unavoidable systematic error.
[/quote:1dd6f4097d]
How do you measure the temperate of a star, and how does your
measurement
affect the star?
Why is the act of looking at the star "clearly" cooling it?
Or look at a human being through night vision IR devices, perhaps....
It isn't "clear" to me at all, since the source was radiating anyway.
Please explain.
Androcles |
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| Greg Hennessy |
| Posted: Sat Jan 29, 2005 8:20 am |
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In article <7gJKd.10374$B5.3390@fe1.news.blueyonder.co.uk>,
Androcles <dummy@dummy.net> wrote:
[quote:d2df4dd4e0]Why is the act of looking at the star "clearly" cooling it?
Or look at a human being through night vision IR devices, perhaps....
It isn't "clear" to me at all, since the source was radiating anyway.
[/quote:d2df4dd4e0]
Temperature is defined for a source in equilibrium. Pedantically
speaking, if the source is radiating, it does not have a temperature. |
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| Greg Hennessy |
| Posted: Sat Jan 29, 2005 9:54 am |
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In article <DDRKd.3919$g_3.2788@trndny08>,
Tom Capizzi <etianshrldu@verizon.net> wrote:
[quote:c6cd121bb5]Surely you are exaggerating.
[/quote:c6cd121bb5]
Surely I was making a point. Although I should have said "net
radiating" rather than radiating. |
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| Androcles |
| Posted: Sat Jan 29, 2005 1:52 pm |
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"Greg Hennessy" <greg.hennessy@tantalus.cox.net> wrote in message
news:ctgk56$vgb$1@tantalus.no-ip.org...
[quote:e1580fa068]In article <7gJKd.10374$B5.3390@fe1.news.blueyonder.co.uk>,
Androcles <dummy@dummy.net> wrote:
Why is the act of looking at the star "clearly" cooling it?
Or look at a human being through night vision IR devices, perhaps....
It isn't "clear" to me at all, since the source was radiating anyway.
Temperature is defined for a source in equilibrium. Pedantically
speaking, if the source is radiating, it does not have a temperature.
[/quote:e1580fa068]
Ah... I see. Stars do not have a temperature. Got it.
Androcles |
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| N:dlzc D:aol T:com (dlzc) |
| Posted: Sat Jan 29, 2005 1:54 pm |
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Dear Greg Hennessy:
"Greg Hennessy" <greg.hennessy@tantalus.cox.net> wrote in message
news:ctgk56$vgb$1@tantalus.no-ip.org...
[quote:7c7f882b51]In article <7gJKd.10374$B5.3390@fe1.news.blueyonder.co.uk>,
Androcles <dummy@dummy.net> wrote:
Why is the act of looking at the star "clearly" cooling it?
Or look at a human being through night vision IR devices, perhaps....
It isn't "clear" to me at all, since the source was radiating anyway.
Temperature is defined for a source in equilibrium. Pedantically
speaking, if the source is radiating, it does not have a temperature.
[/quote:7c7f882b51]
The spectrum radiated by one surface, to a Universe at the same
temperature, is radiating still. The incident radiation = the leaving
radiation (neglecting emissivity/absorptivity issues). It would be
expected that radiation from the left equals radiation from the right, for
any arbitrary section though a body. A source radiates if it is not at
absolute zero, and if binding energies (solids and liquids comply) provide
enough broadening of quantum energy levels to provide some sort of
continuity.
David A. Smith |
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| Tom Capizzi |
| Posted: Sat Jan 29, 2005 2:39 pm |
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"Greg Hennessy" <greg.hennessy@tantalus.cox.net> wrote in message
news:ctgk56$vgb$1@tantalus.no-ip.org...
[quote:4731c7230f]In article <7gJKd.10374$B5.3390@fe1.news.blueyonder.co.uk>,
Androcles <dummy@dummy.net> wrote:
Why is the act of looking at the star "clearly" cooling it?
Or look at a human being through night vision IR devices, perhaps....
It isn't "clear" to me at all, since the source was radiating anyway.
Temperature is defined for a source in equilibrium. Pedantically
speaking, if the source is radiating, it does not have a temperature.
[/quote:4731c7230f]
Surely you are exaggerating. Specifically, the Stefan-Boltzman law
relates the total amount of radiation emitted by a body to the 4th
power of its temperature. The relation is not restricted to bodies
at the same temperature as the surroundings. If you want to be
technical, even a body in equilibrium with its surroundings is
radiating. It is just absorbing at the same rate. |
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| Tom Roberts |
| Posted: Sat Jan 29, 2005 6:55 pm |
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Greg Hennessy wrote:
[quote:49e2971ede]Temperature is defined for a source in equilibrium. Pedantically
speaking, if the source is radiating, it does not have a temperature.
[/quote:49e2971ede]
Not true.
There can be a heat source inside an object, and the object reaches
equilibrium via radiation cooling. In fact, that is how one arrives at
the formula for blackbody radiation, and the ability to measure an
object's temperature from its emitted radiation.
Or an object could have a large volume-to-surface ratio and/or large
heat capacity, and for all practical purposes it is in equilibrium, as
the heat radiated over a typical observation time is tiny fraction of
its total.
The sun is an example of the first. The earth is a combination of both.
Tom Roberts tjroberts@lucent.com |
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