Does Light Possess an Intrinsic Frequency?
Henri Wilson
Below is an observer O and a spatial wave, represented by 'Vs'.
O
XVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV\
1) Click on 'reply'.
2) Place the cursor between the X and the first V.
3) place one finger on the 'space bar' and another on the 'backspace'.
4) Move the wave forwards and backwards with these keys.
Questions:
1) If there are n 'Vs' per metre, does the structure possess an intrinsic
wavelength of 1/n?
2) Does the line of Vs oscillate or change in any way as it moves?
3) If the whole structure moves at 'c' metres/sec across the screen, does O
count (n x c) Vs going past per sec ?
4) Does the equation de/dt=de/dx.dx/dt describe the observed 'frequency of
events(e)' seen by O?
5) Does this model simulate 'frequency' as applied to light?
6) If so, what processes might account for the wavelike structure?
7) If there WERE an intrinsic oscillation in the wave itself (in its own
frame), how might this affect de/dt?
Henri Wilson.
The BIG BANG Theory = The creationists' attempt to hijack science!
But they didn't succeed!
See my animations at:
http://www.users.bigpond.com/HeWn/index.htm
Tom Roberts
> [...]
> 5) Does this model simulate 'frequency' as applied to light?
No.
Tom Roberts tjro...@lucent.com
HenriWilson
>Henri Wilson wrote:
>> [...]
>> 5) Does this model simulate 'frequency' as applied to light?
>
>No.
Well come on, give me a better model, Tom.
Where does the 'frequency' of light come from?
What property of light gives rise to the sensation of 'events per second'?
I'm talking 3D space here. That's how we observe light so that's how we must be
able to explain what we observe.
>
>
>Tom Roberts tjro...@lucent.com
S. Enterprize Company
Yes. The Smart Model CLEARLY shows how, why, where, & when at ALL levels of
infinity and this universe.
S. Enterprize Co. (Membership)
http://www.s-enterprize.com/
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
Tom Roberts
> On Tue, 08 Jul 2003 22:35:08 -0500, Tom Roberts <tjro...@lucent.com> wrote:
>>Henri Wilson wrote:
>>>[...]
>>>5) Does this model simulate 'frequency' as applied to light?
>>No.
>
> Well come on, give me a better model, Tom.
> Where does the 'frequency' of light come from?
The observed frequency of a light beam is NOT intrinsic to the light
beam, but is a relationship among the source of the light beam, the
trajectory of the light beam, the geometry along the trajectory, and the
receiver.
It is easier to think about the time interval between successive crests
of the light (i.e. its period), rather than its frequency. In the
instantaneously-comoving inertial frame of the source, the 4-vector
displacement between two successive crests is purely along the time
axis, and has a value corresponding to the period of the wave. This
4-vector is parallel propagated along the trajectory to the receiver by
the motion of the wave (this trajectory is inherently a null geodesic
path). In the instantaneously-comoving inertial frame of the receiver
(when the same pair of crests is received), to measure the time interval
between the crests the receiver projects the received signal onto the
time axis; this projection is mathematically equivalent to taking the
4-vector dot product of the parallel-propagated initial displacement
4-vector with the unit 4-vector of the receiver's time axis. In general
this last value will not have the same value as the magnitude of the
initial displacement 4-vector, and the difference is the redshift or
blueshift of the wave.
Note this description is in the context of GR; it is valid
for both "gravitational red/blueshift" and red/blueshift
due to relative motion of source and receiver, or any
combination.
Tom Roberts tjro...@lucent.com
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> What property of light gives rise to the sensation of 'events per second'?
If by light you mean visible light, i.e., electromagnetic radiation from
~400 nm to ~700nm, then the answer is no property of light gives rise to
the *sensation* of events per second.
I can definitely measure wavelength on the order of 400-700 nm. I can
definitely measure the propagation speed of this radiation in vacuum.
And from that I can infer a frequency as c divided by wavelength. But I
am not aware of any means to directly measure or sense in any fashion
whatever the frequency implied by measurement of wavelength and c for
visible light.
HenriWilson
Precisely. There IS NO frequency associated with light. Light has a
'wavelength' but there is no evidence of an intrinsic 'frequency'.
There is a frequency assciated with light's movement past an observer as
spatial regularities give rise to a periodic event.
Run a saw blade over a knife edge and see what you get.
Henri Wilson.
Why is the creative output of one SRian equal to the total produced by one million of them?
Henri Wilson
>Re: Does Light Possess an Intrinsic Frequency?
>
> Yes. The Smart Model CLEARLY shows how, why, where, & when at ALL levels of
>infinity and this universe.
I couldn't find anything relating to this question on your we site..
HenriWilson
>HenriWilson wrote:
>> On Tue, 08 Jul 2003 22:35:08 -0500, Tom Roberts <tjro...@lucent.com> wrote:
>>>Henri Wilson wrote:
>>>>[...]
>>>>5) Does this model simulate 'frequency' as applied to light?
>>>No.
>>
>> Well come on, give me a better model, Tom.
>> Where does the 'frequency' of light come from?
>
>The observed frequency of a light beam is NOT intrinsic to the light
>beam, but is a relationship among the source of the light beam, the
>trajectory of the light beam, the geometry along the trajectory, and the
>receiver.
No Tom, you are talking the observed wavelength.
There is no evidence of a 'frequency', at all.
>
>It is easier to think about the time interval between successive crests
>of the light (i.e. its period), rather than its frequency.
Are you not refering to the time interval between the ARRIVAL of successive
crests at a particular point, Tom?
>In the
>instantaneously-comoving inertial frame of the source, the 4-vector
>displacement between two successive crests is purely along the time
>axis, and has a value corresponding to the period of the wave. This
>4-vector is parallel propagated along the trajectory to the receiver by
>the motion of the wave (this trajectory is inherently a null geodesic
>path). In the instantaneously-comoving inertial frame of the receiver
>(when the same pair of crests is received), to measure the time interval
>between the crests the receiver projects the received signal onto the
>time axis; this projection is mathematically equivalent to taking the
>4-vector dot product of the parallel-propagated initial displacement
>4-vector with the unit 4-vector of the receiver's time axis. In general
>this last value will not have the same value as the magnitude of the
>initial displacement 4-vector, and the difference is the redshift or
>blueshift of the wave.
>
> Note this description is in the context of GR; it is valid
> for both "gravitational red/blueshift" and red/blueshift
> due to relative motion of source and receiver, or any
> combination.
Tom that is all very elegant but you are still talking about 'wavelength' and
not an intrinsic 'time periodic' property.
The only 'frequency' I can recognize is that which is associated with the bulk
movement of light past an observer.
>
>
>Tom Roberts tjro...@lucent.com
Henri Wilson.
Why is the creative output of one SRian equal to the total produced by one million of them?
See my animations at:
http://www.users.bigpond.com/HeWn/index.htm
kenseto
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:beilut$c...@netnews.proxy.lucent.com...
> HenriWilson wrote:
> > On Tue, 08 Jul 2003 22:35:08 -0500, Tom Roberts <tjro...@lucent.com>
wrote:
> >>Henri Wilson wrote:
> >>>[...]
> >>>5) Does this model simulate 'frequency' as applied to light?
> >>No.
> >
> > Well come on, give me a better model, Tom.
> > Where does the 'frequency' of light come from?
>
> The observed frequency of a light beam is NOT intrinsic to the light
> beam, but is a relationship among the source of the light beam, the
> trajectory of the light beam, the geometry along the trajectory, and the
> receiver.
>
> It is easier to think about the time interval between successive crests
> of the light (i.e. its period), rather than its frequency.
The time interval between two successive crests surely is dependent on the
motion of the receiver relative to these crests. So from that we can
conclude that red or blue shift is a function of the absolute motion of the
receiver.
Ken Seto
Chuck Simmons
>
> On Thu, 10 Jul 2003 05:38:09 GMT, Bill Rowe <bjr...@earthlink.net> wrote:
>
> >In article <a7ungvcise9e7smjp...@4ax.com>,
> > He...@the.edge(HenriWilson) wrote:
> >
> >> What property of light gives rise to the sensation of 'events per second'?
> >
> >If by light you mean visible light, i.e., electromagnetic radiation from
> >~400 nm to ~700nm, then the answer is no property of light gives rise to
> >the *sensation* of events per second.
> >
> >I can definitely measure wavelength on the order of 400-700 nm. I can
> >definitely measure the propagation speed of this radiation in vacuum.
> >And from that I can infer a frequency as c divided by wavelength. But I
> >am not aware of any means to directly measure or sense in any fashion
> >whatever the frequency implied by measurement of wavelength and c for
> >visible light.
>
> Precisely. There IS NO frequency associated with light. Light has a
> 'wavelength' but there is no evidence of an intrinsic 'frequency'.
>
> There is a frequency assciated with light's movement past an observer as
> spatial regularities give rise to a periodic event.
> Run a saw blade over a knife edge and see what you get.
This is a semantic boondoggle. For travelling waves in any medium,
frequency is defined as the speed of propagation divided by the
wavelength. In the case of lower frequencies, the time elapsed between
say crests of a wave may be measured. At higher frequencies yet, several
gigaHz and above, wavelength measurements are used and converted to
frequency. BTW, many wavelength measurements are made on standing waves.
That should give your rant even more ammunition.
Chuck
--
... The times have been,
That, when the brains were out,
the man would die. ... Macbeth
Chuck Simmons chr...@webaccess.net
Tom Roberts
> On Wed, 09 Jul 2003 22:27:46 -0500, Tom Roberts <tjro...@lucent.com> wrote:
>>HenriWilson wrote:
>>>Where does the 'frequency' of light come from?
>>The observed frequency of a light beam is NOT intrinsic to the light
>>beam, but is a relationship among the source of the light beam, the
>>trajectory of the light beam, the geometry along the trajectory, and the
>>receiver.
> No Tom, you are talking the observed wavelength.
Not true. I was discussing the PERIOD of the light, which is simply
1/frequency. Nowhere in my article did I mention or use wavelength in
any way, shape, or form.
> There is no evidence of a 'frequency', at all.
While we cannot directly measure the frequency of a light beam, AFAIK,
we can do so for radio- and micro-waves. We have no reason to suspect
that the relationship among frequency, speed, and wavelength for them
does not hold for visible light. Certainly the measurements we do make
for visible light are consistent with that relationship being valid.
>>It is easier to think about the time interval between successive crests
>>of the light (i.e. its period), rather than its frequency.
> Are you not refering to the time interval between the ARRIVAL of successive
> crests at a particular point, Tom?
The concept applies to both the departure from the source and the
arrival at the detector.
> [...]
> Tom that is all very elegant but you are still talking about 'wavelength' and
> not an intrinsic 'time periodic' property.
Not true. I discussed the TIME INTERVAL between two successive wave
crests. To see the periodic nature of this, simply repeat as necessary
for successive pairs....
> The only 'frequency' I can recognize is that which is associated with the bulk
> movement of light past an observer.
I was discussing the measurement of the period of the light wave.
Period=1/frequency -- note there is no wavelength or c in that
relationship. So my discussion directly applies to frequency -- as I
said, it is easier to discuss the period rather than the frequency
(period can be described as a displacement 4-vector, frequency cannot).
Tom Roberts tjro...@lucent.com
Tom Roberts
> The time interval between two successive crests surely is dependent on the
> motion of the receiver relative to these crests.
No. The "motion of the receiver relative to these crests" is always c
for any receiver, so it is not useful in describing this. For a fixed
light trajectory without gravitation, it is motion of the receiver wrt
THE SOURCE that is important, as my discussion shows.
> So from that we can
> conclude that red or blue shift is a function of the absolute motion of the
> receiver.
Only when you "put it in by hand" implicitly by trying to discuss
"motion of the receiver relative to these crests".
Tom Roberts tjro...@lucent.com
Dirk Van de moortel
"kenseto" <ken...@erinet.com> wrote in message news:vgqmrja...@corp.supernews.com...
>
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:beilut$c...@netnews.proxy.lucent.com...
[snip]
> > The observed frequency of a light beam is NOT intrinsic to the light
> > beam, but is a relationship among the source of the light beam, the
> > trajectory of the light beam, the geometry along the trajectory, and the
> > receiver.
> >
> > It is easier to think about the time interval between successive crests
> > of the light (i.e. its period), rather than its frequency.
>
> The time interval between two successive crests surely is dependent on the
> motion of the receiver relative to these crests. So from that we can
> conclude that red or blue shift is a function of the absolute motion of the
> receiver.
"Absolutely Relative Seto Logic"
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/RelAbs.html
Dirk Vdm
kenseto
"Tom Roberts" <tjro...@Lucent.com> wrote in message
news:3F0D6BC7...@Lucent.com...
> On 7/10/2003 7:29 AM, kenseto wrote:
> > The time interval between two successive crests surely is dependent on
the
> > motion of the receiver relative to these crests.
>
> No. The "motion of the receiver relative to these crests" is always c
> for any receiver, so it is not useful in describing this. For a fixed
> light trajectory without gravitation, it is motion of the receiver wrt
> THE SOURCE that is important, as my discussion shows.
But the motion of the receiver wrt THE SOURCE is just the vector difference
of their absolute motions. So from that we can still conclude that red or
blue shift is a function of the absolute motions of the source and the
receiver.
Ken Seto
Bilge
>
>>Henri Wilson wrote:
>>> [...]
>>> 5) Does this model simulate 'frequency' as applied to light?
>>
>>No.
>
>Well come on, give me a better model, Tom.
>
>Where does the 'frequency' of light come from?
>
>What property of light gives rise to the sensation of 'events per second'?
>I'm talking 3D space here. That's how we observe light so that's how we
>must be able to explain what we observe.
This has been explained to you numerous times henry. The number of
"events per second" observed is a direct the result of the number of
"events per second" that produce the light at the source.
Tom Roberts
You assume some sort of absolute motion, and then conclude this is "a
function of the absolute motions" -- that's circular.
Moreover, your approach cannot handle the case of gravitational
redshift, where both source and receiver are motionless (wrt each other
and wrt the earth) but are at different altitudes.
Tom Roberts tjro...@lucent.com
HenriWilson
>On 7/10/2003 5:27 AM, HenriWilson wrote:
>> On Wed, 09 Jul 2003 22:27:46 -0500, Tom Roberts <tjro...@lucent.com> wrote:
>>>HenriWilson wrote:
>>>>Where does the 'frequency' of light come from?
>>>The observed frequency of a light beam is NOT intrinsic to the light
>>>beam, but is a relationship among the source of the light beam, the
>>>trajectory of the light beam, the geometry along the trajectory, and the
>>>receiver.
>> No Tom, you are talking the observed wavelength.
>
>Not true. I was discussing the PERIOD of the light, which is simply
>1/frequency. Nowhere in my article did I mention or use wavelength in
>any way, shape, or form.
Tom, your first statement began, "It is easier to think about the time interval
between successive crests of the light (i.e. its period), rather than its
frequency."
Tom, there are SPACES between successive wave crests, not TIME intervals.
'Frequency' means 'events per unit time'.
Certainly 'wave peaks' reach an observer at a certain frequency because the
wave is presumeably moving towards him; but that is not an intrinsic frequency.
>
>
>> There is no evidence of a 'frequency', at all.
>
>While we cannot directly measure the frequency of a light beam, AFAIK,
>we can do so for radio- and micro-waves. We have no reason to suspect
>that the relationship among frequency, speed, and wavelength for them
>does not hold for visible light. Certainly the measurements we do make
>for visible light are consistent with that relationship being valid.
>
I don't agree with this analogy. I regard a generated frequency as a wave
reflecting 'photon density variation' rather than just 'one long photon'.
>
>>>It is easier to think about the time interval between successive crests
>>>of the light (i.e. its period), rather than its frequency.
>> Are you not refering to the time interval between the ARRIVAL of successive
>> crests at a particular point, Tom?
>
>The concept applies to both the departure from the source and the
>arrival at the detector.
that is not important.
Frequency of light is implied by the equation f=c/lambda.
That equation should be f=c.n, where n is the number of wave peaks per unit
length.
The question is, "what is the real 3D nature of these 'wavepeaks?'.
Obviously photons do not resemble the line of 'Vs' that I drew in the original
post. There must be some kind of intrinsic process that sets up the Maxwellian
type wave form.
>
>
>> [...]
>> Tom that is all very elegant but you are still talking about 'wavelength' and
>> not an intrinsic 'time periodic' property.
>
>Not true. I discussed the TIME INTERVAL between two successive wave
>crests. To see the periodic nature of this, simply repeat as necessary
>for successive pairs....
Tom, I repeat, there is SPACE between successive wave crests, not TIME.
>
>
>> The only 'frequency' I can recognize is that which is associated with the bulk
>> movement of light past an observer.
>
>I was discussing the measurement of the period of the light wave.
>Period=1/frequency -- note there is no wavelength or c in that
>relationship. So my discussion directly applies to frequency -- as I
>said, it is easier to discuss the period rather than the frequency
>(period can be described as a displacement 4-vector, frequency cannot).
There is no need to go into 4 vectors for this. You were using that basically
to show doppler effects for different observers.
I want to know if there is any time periodicity associated with a light wave
other than the plain 'f=c/lambda', which obviously refers to the rate at which
wave crests are moving past a point.
SO WHAT EXACTLY ARE THESE WAVE CRESTS MADE OF?
HenriWilson
Ken is quite justified in holding that point of view. It certainly makes as
much sense as your 4D 'wise after the event' crap.
>
>Dirk Vdm
>
Henri Wilson.
Why is the creative output of one SRian equal to the total produced by one million of them?
See my animations at:
http://www.users.bigpond.com/HeWn/index.htm
HenriWilson
So let's speculate about the possible 3D form of a 'photon' which might allow
this "events per second" information to be conveyed to other 3D places.
The static 'line of Vs' I drew in my original post is one possible way this can
be done. But I'm sure you will agree that the 'waveform' of an individual
photon cannot be static. So what might it be?
Henri Wilson.
Why is the creative output of one SRian equal to the total produced by one million of them?
See my animations at:
http://www.users.bigpond.com/HeWn/index.htm
kenseto
"Tom Roberts" <tjro...@Lucent.com> wrote in message
> On 7/10/2003 1:46 PM, kenseto wrote:
> > "Tom Roberts" <tjro...@Lucent.com> wrote in message
> > news:3F0D6BC7...@Lucent.com...
> >>On 7/10/2003 7:29 AM, kenseto wrote:
> >>>The time interval between two successive crests surely is dependent on
> >>> the motion of the receiver relative to these crests.
> >>
> >>No. The "motion of the receiver relative to these crests" is always c
> >>for any receiver, so it is not useful in describing this. For a fixed
> >>light trajectory without gravitation, it is motion of the receiver wrt
> >>THE SOURCE that is important, as my discussion shows.
> >
> > But the motion of the receiver wrt THE SOURCE is just the vector
difference
> > of their absolute motions. So from that we can still conclude that red
or
> > blue shift is a function of the absolute motions of the source and the
> > receiver.
>
> You assume some sort of absolute motion, and then conclude this is "a
> function of the absolute motions" -- that's circular.
No...consider driving down the highway at a constant speed and you see other
cars are moving relative to you at different speeds. Now consider the
highway is invisible and all you see is the other cars are moving at
different speed relative to you. The relative speed of any specific car wrt
you is the vector difference of your speed wrt the highway and the other
car's speed relative to the highway.
>
> Moreover, your approach cannot handle the case of gravitational
> redshift, where both source and receiver are motionless (wrt each other
> and wrt the earth) but are at different altitudes.
It can indeed handle gravity. Consider the source and the receiver are
moving in concentric circular highways side by side. There is no relative
motion between them. However their motion (absolute motion) relative to the
highway (the aether) are different. Similarly when the source and the
receiver are at different altitudes as the earth rotates causes them to have
different states of absolute motion and thus the redshift.
Ken Seto
Tom Roberts
> Tom, your first statement began, "It is easier to think about the time interval
> between successive crests of the light (i.e. its period), rather than its
> frequency."
>
> Tom, there are SPACES between successive wave crests, not TIME intervals.
Not when one sits at one place and observes the wave as a time-series,
as I implied.
> 'Frequency' means 'events per unit time'.
Right. Each event is the observation of a crest at the position of the
detector. No wavelength there at all. By such an observation one cannot
tell that this is a "traveling wave" at all.
> Certainly 'wave peaks' reach an observer at a certain frequency because the
> wave is presumeably moving towards him; but that is not an intrinsic frequency.
Right. That's my point.
Tom Roberts tjro...@lucent.com
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> On Thu, 10 Jul 2003 05:38:09 GMT, Bill Rowe <bjr...@earthlink.net> wrote:
> >If by light you mean visible light, i.e., electromagnetic radiation from
> >~400 nm to ~700nm, then the answer is no property of light gives rise to
> >the *sensation* of events per second.
> >I can definitely measure wavelength on the order of 400-700 nm. I can
> >definitely measure the propagation speed of this radiation in vacuum.
> >And from that I can infer a frequency as c divided by wavelength. But I
> >am not aware of any means to directly measure or sense in any fashion
> >whatever the frequency implied by measurement of wavelength and c for
> >visible light.
> Precisely. There IS NO frequency associated with light. Light has a
> 'wavelength' but there is no evidence of an intrinsic 'frequency'.
Not exactly. There is clearly a frequency asscociated with
electromagnetic energy in say the 1 meter band. I can easily measure
this directly on my oscilloscope. And all available experimental
evidence indicates visible light is simply the same as 1 meter
electromagnetic radiation with a shorter wavelength. And that is
certainly evidence there is a frequency associated with visible light
whether or not I can directly measure it or sense it.
dlzc@aol.com (formerly)
"Bill Rowe" <bjr...@earthlink.net> wrote in message
news:bjrowe-DA7245....@nnrp02.earthlink.net...
...
> Not exactly. There is clearly a frequency asscociated with
> electromagnetic energy in say the 1 meter band. I can easily measure
> this directly on my oscilloscope. And all available experimental
> evidence indicates visible light is simply the same as 1 meter
> electromagnetic radiation with a shorter wavelength. And that is
> certainly evidence there is a frequency associated with visible light
> whether or not I can directly measure it or sense it.
What you measure of the photon; energy, momentum, frequency, or wavelength
is *not* intrinsic to the photon stream. It is as much a measure of your
velocity relative to the source, as it is the action of the source that
emitted it.
By adjusting your speed, you can make your 1m wavelength anything you
choose, visible light, gamma, or 100,000 km wavelength.
Sorry to disagree with a fine point.
David A. Smith
YBM
[...]
>>You assume some sort of absolute motion, and then conclude this is "a
>>function of the absolute motions" -- that's circular.
>
>
> No...consider driving down the highway at a constant speed and you see other
> cars are moving relative to you at different speeds. Now consider the
> highway is invisible and all you see is the other cars are moving at
> different speed relative to you. The relative speed of any specific car wrt
> you is the vector difference of your speed wrt the highway and the other
> car's speed relative to the highway.
The relative speed of any specific car wrt you is as well the
vetor differerence of you speed wrt anything else (say a third car) and
the other car's speed relative to this third thing.
So there is an infinity of absolute frames. Good.
Is the vertical directition isotropic in all of them ? (arf)
>
>>Moreover, your approach cannot handle the case of gravitational
>>redshift, where both source and receiver are motionless (wrt each other
>>and wrt the earth) but are at different altitudes.
>
>
> It can indeed handle gravity. Consider the source and the receiver are
> moving in concentric circular highways side by side. There is no relative
> motion between them. However their motion (absolute motion) relative to the
> highway (the aether) are different. Similarly when the source and the
> receiver are at different altitudes as the earth rotates causes them to have
> different states of absolute motion and thus the redshift.
I suppose that you are about to introduce the isotropy of the circular
direction, aren' you ?
Dirk Van de moortel
"YBM" <ybm...@nooos.fr> wrote in message news:3f0e6a2c$0$16257$79c1...@nan-newsreader-02.noos.net...
> kenseto wrote:
[snip]
> > It can indeed handle gravity. Consider the source and the receiver are
> > moving in concentric circular highways side by side. There is no relative
> > motion between them.
hehe ;-)
> > However their motion (absolute motion) relative to the
> > highway (the aether) are different. Similarly when the source and the
> > receiver are at different altitudes as the earth rotates causes them to have
> > different states of absolute motion and thus the redshift.
>
> I suppose that you are about to introduce the isotropy of the circular
> direction, aren' you ?
Ask him whether the moon and the earth have no relative
motion either, after all, they are moving more or less in
concentric circles as well.
Wanna bet he won't even get the point?
Please try asking - he won't reply to me since he pretends
having killfiled me ;-)
Enjoy...
Dirk Vdm
YBM
[...]
>>I suppose that you are about to introduce the isotropy of the circular
>>direction, aren' you ?
>
>
> Ask him whether the moon and the earth have no relative
> motion either, after all, they are moving more or less in
> concentric circles as well.
> Wanna bet he won't even get the point?
> Please try asking - he won't reply to me since he pretends
> having killfiled me ;-)
> Enjoy...
I'd like to, but he pretend to have killfiled me as well. It seems
he killfiled everyone who asked him annoying questions...
HenriWilson
>HenriWilson wrote:
>> Tom, your first statement began, "It is easier to think about the time interval
>> between successive crests of the light (i.e. its period), rather than its
>> frequency."
>>
>> Tom, there are SPACES between successive wave crests, not TIME intervals.
>
>Not when one sits at one place and observes the wave as a time-series,
>as I implied.
That's the point I was making.
>
>
>> 'Frequency' means 'events per unit time'.
>
>Right. Each event is the observation of a crest at the position of the
>detector. No wavelength there at all. By such an observation one cannot
>tell that this is a "traveling wave" at all.
>
>
>> Certainly 'wave peaks' reach an observer at a certain frequency because the
>> wave is presumeably moving towards him; but that is not an intrinsic frequency.
>
>Right. That's my point.
Good. We are in agreement.
What I want to know is what kind of 'wave' can be static as far as its 'wave
peaks' are concerned yet dynamic in accordance with, say, Maxwell's concept.
What I mean is, those 'V's I drew must be effectively moving as one unit to
give light a 'frequency' yet they must be driven by another intrinsic
(standing) wave' - or something of that nature.
HenriWilson
I don't think RF is mae of one single photon. I think generated RF etc is a
'photon density wave'. It still moves at c but it is a wave signifying some
property of a whole bunch of photons.
However, as with single photons, the 'frequency' received is given by the rate
at which wave crests arrive.
I want to know what property of all these photons is manipulated in order to
produce that traveling wave.
Henri Wilson.
Why is the creative output of one SRian the same as that produced by one million of them?
HenriWilson
(Bilge) wrote:
> HenriWilson:
> >
> >So let's speculate about the possible 3D form of a 'photon'
>
> No, let's not.
>
>
>
Obviously that would be beyond you.
Henri Wilson.
Why is the creative output of one SRian the same as that produced by one million of them?
Tom Roberts
> What I want to know is what kind of 'wave' can be static as far as its 'wave
> peaks' are concerned yet dynamic in accordance with, say, Maxwell's concept.
No electromagnetic wave is static at all.
> What I mean is, those 'V's I drew must be effectively moving as one unit to
> give light a 'frequency' yet they must be driven by another intrinsic
> (standing) wave' - or something of that nature.
No. Your 'V's don't correspond to an EM wave in any sensible way -- your
'V's have intrinsic lengths and EM waves don't; your 'V's have an
intrinsic rest frame and EM waves don't. These are precisely the
properties you are confused about, probably because you insist on trying
to force a model corresponding to your 'V's apply to EM waves -- that
will not work. Nature simply does not work that way. <shrug>
We have enough experimental evidence and theoretical
interpretation of it that I can make that last statement
with reasonable confidence.
Tom Roberts tjro...@lucent.com
HenriWilson
>HenriWilson wrote:
>> What I want to know is what kind of 'wave' can be static as far as its 'wave
>> peaks' are concerned yet dynamic in accordance with, say, Maxwell's concept.
>
>No electromagnetic wave is static at all.
>
>
>> What I mean is, those 'V's I drew must be effectively moving as one unit to
>> give light a 'frequency' yet they must be driven by another intrinsic
>> (standing) wave' - or something of that nature.
>
>No. Your 'V's don't correspond to an EM wave in any sensible way -- your
>'V's have intrinsic lengths and EM waves don't; your 'V's have an
>intrinsic rest frame and EM waves don't. These are precisely the
>properties you are confused about, probably because you insist on trying
>to force a model corresponding to your 'V's apply to EM waves -- that
>will not work. Nature simply does not work that way. <shrug>
You can shrug all you like. If light DOES NOT have 'static wavepeaks' then it
will not exhibit a frequency=c/lambda.
It's as simple as that.
Don't argue against yourself Tom.
> We have enough experimental evidence and theoretical
> interpretation of it that I can make that last statement
> with reasonable confidence.
>
>
>Tom Roberts tjro...@lucent.com
Henri Wilson.
Why is the creative output of one SRian the same as that produced by one million of them?
Bill Rowe
"dl...@aol.com \(formerly\)" <dlzc1.cox@net> wrote:
> Dear Bill Rowe:
>
> "Bill Rowe" <bjr...@earthlink.net> wrote in message
> news:bjrowe-DA7245....@nnrp02.earthlink.net...
> ...
> > Not exactly. There is clearly a frequency asscociated with
> > electromagnetic energy in say the 1 meter band. I can easily measure
> > this directly on my oscilloscope. And all available experimental
> > evidence indicates visible light is simply the same as 1 meter
> > electromagnetic radiation with a shorter wavelength. And that is
> > certainly evidence there is a frequency associated with visible light
> > whether or not I can directly measure it or sense it.
> What you measure of the photon; energy, momentum, frequency, or wavelength
> is *not* intrinsic to the photon stream. It is as much a measure of your
> velocity relative to the source, as it is the action of the source that
> emitted it.
Quite true
> Sorry to disagree with a fine point.
I am puzzled. What in my comments leads you to think I disagree?
dlzc@aol.com (formerly)
"Bill Rowe" <bjr...@earthlink.net> wrote in message
news:bjrowe-5416E0....@nnrp04.earthlink.net...
> In article <%hrPa.1337$u51.776@fed1read05>,
> "dl...@aol.com \(formerly\)" <dlzc1.cox@net> wrote:
...
> > Sorry to disagree with a fine point.
>
> I am puzzled. What in my comments leads you to think I disagree?
Something is the way you said:
>There is clearly a frequency asscociated with
> electromagnetic energy in say the 1 meter band. I can easily measure
> this directly on my oscilloscope.
To Henri Wilson's:
>> Precisely. There IS NO frequency associated with light. Light has a
>> 'wavelength' but there is no evidence of an intrinsic 'frequency'.
It is now clear to me that you were simply describing a particular
measurement, and not directly responding to his statement. He words things
so oddly...
David A. Smith
FrediFizzx
news:p5fvgv0q2qgldhgmm...@4ax.com...
| On Fri, 11 Jul 2003 21:55:40 -0500, Tom Roberts <tjro...@lucent.com>
wrote:
|
| >HenriWilson wrote:
| >> What I want to know is what kind of 'wave' can be static as far as its
'wave
| >> peaks' are concerned yet dynamic in accordance with, say, Maxwell's
concept.
| >
| >No electromagnetic wave is static at all.
| >
| >
| >> What I mean is, those 'V's I drew must be effectively moving as one
unit to
| >> give light a 'frequency' yet they must be driven by another intrinsic
| >> (standing) wave' - or something of that nature.
| >
| >No. Your 'V's don't correspond to an EM wave in any sensible way -- your
| >'V's have intrinsic lengths and EM waves don't; your 'V's have an
| >intrinsic rest frame and EM waves don't. These are precisely the
| >properties you are confused about, probably because you insist on trying
| >to force a model corresponding to your 'V's apply to EM waves -- that
| >will not work. Nature simply does not work that way. <shrug>
|
| You can shrug all you like. If light DOES NOT have 'static wavepeaks' then
it
| will not exhibit a frequency=c/lambda.
Well, a photon would only be static in its own "frame". I think there may
be something seriously wrong with the modern concept of a photon. It is not
hard to show that a free space photon will not fit in a volume smaller than
lambda^3/2pi. This makes the volume for radio wave photons huge. I guess
this would be interpreted as it being harder to localize low energy photons.
However, it could be interpreted as its energy density is all spread out in
that volume. If this is the case, then I predict that a photon is two
wavelengths long and has a frontal radius of lambda/2pi. Its spin is what
gives the impression of having wave crests, etc. Really it is just a chunk
of EM energy.
FrediFizzx
http://www.flashrock.com/upload/photong/photong.html
kenseto
> HenriWilson wrote:
> > What I want to know is what kind of 'wave' can be static as far as its
'wave
> > peaks' are concerned yet dynamic in accordance with, say, Maxwell's
concept.
>
> No electromagnetic wave is static at all.
>
>
> > What I mean is, those 'V's I drew must be effectively moving as one unit
to
> > give light a 'frequency' yet they must be driven by another intrinsic
> > (standing) wave' - or something of that nature.
>
> No. Your 'V's don't correspond to an EM wave in any sensible way -- your
> 'V's have intrinsic lengths and EM waves don't; your 'V's have an
> intrinsic rest frame and EM waves don't. These are precisely the
> properties you are confused about, probably because you insist on trying
> to force a model corresponding to your 'V's apply to EM waves -- that
> will not work. Nature simply does not work that way. <shrug>
If a photon does not have intrinsic frequency as you seem to imply then how
come in the equation for the enery of a photon shows that it is a function
of frequency?
I endorse the idea that a photon is a wave packet. A model of a photon is
shown in my website under the section "Past Experiments Detecting Absolute
Motion"---the Photoelectric Experiment.
http://www.erinet.com/kenseto/book.html
Ken Seto
HenriWilson
wrote:
>Dear Bill Rowe:
>
>"Bill Rowe" <bjr...@earthlink.net> wrote in message
>news:bjrowe-5416E0....@nnrp04.earthlink.net...
>> In article <%hrPa.1337$u51.776@fed1read05>,
>> "dl...@aol.com \(formerly\)" <dlzc1.cox@net> wrote:
>...
>> > Sorry to disagree with a fine point.
>>
>> I am puzzled. What in my comments leads you to think I disagree?
>
>Something is the way you said:
>>There is clearly a frequency asscociated with
>> electromagnetic energy in say the 1 meter band. I can easily measure
>> this directly on my oscilloscope.
>
>To Henri Wilson's:
>>> Precisely. There IS NO frequency associated with light. Light has a
>>> 'wavelength' but there is no evidence of an intrinsic 'frequency'.
>
>It is now clear to me that you were simply describing a particular
>measurement, and not directly responding to his statement. He words things
>so oddly...
>
>David A. Smith
>
Well come on! Tell us all about light's intrinsic frequency.
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> Tell us all about light's intrinsic frequency.
Light does not have an *intrinsic* frequency. An intrinsic property of a
thing is by definition a property of that thing. An intrinsic property
of a thing cannot be changed without changing the thing.
Measurements of wavelength for light change when you change your motion
with respect to the source. That means wavelength cannot be an intrinsic
property of light. Since wavelength times frequency is a constant, it is
clear frequency must also change and cannot be an intrinsic property of
light.
In fact, both wavelength and frequency are properties of the
relationship between the observer and the light source.
Bill Rowe
"kenseto" <ken...@erinet.com> wrote:
> If a photon does not have intrinsic frequency as you seem to imply then how
> come in the equation for the enery of a photon shows that it is a function
> of frequency?
The answer should be obvious. Photons do not have an intrinsic energy
either.
Jeff Krimmel
If I had a shiny nickel for every time Ken couldn't understand something
that should be obvious...
Jeff
FrediFizzx
What is generating that relationship? OK, lets say the the observer is in
the rest frame of the source. What then? The frequency is generated by the
source and the same frequency should end up at the observer. So the photon
has something that carries this. I thought it would be the relationship of
spin and momentum.
FrediFizzx
Tom Roberts
> If a photon does not have intrinsic frequency as you seem to imply then how
> come in the equation for the enery of a photon shows that it is a function
> of frequency?
The energy of a photon is not intrinsic to the photon, either. No kind
of energy whatsoever is intrinsic to the object which "has" it. Energy
is a relationship, not a thing (hence the quotes on "has", because there
is a pun on that word).
The energy of an object is a relationship between the object's
intrinsic 4-momentum and the coordinates of the observer
measuring the energy: energy is the 0-th component of that
4-momentum projected onto the observer's coordinates. For
continuous structures things are more complicated, but the
local energy density is the 0,0-th component of the
stress-energy tensor.
I am, of course, using the context of modern physics, SR/GR.
Tom Roberts tjro...@lucent.com
Tom Roberts
> "Bill Rowe" <bjr...@earthlink.net> wrote in message
> news:bjrowe-D54BFE....@nnrp02.earthlink.net...
> | In article <ngc3hvclci9kvv3ui...@4ax.com>,
> | He...@the.edge(HenriWilson) wrote:
> |
> | > Tell us all about light's intrinsic frequency.
> |
> | Light does not have an *intrinsic* frequency. An intrinsic property of a
> | thing is by definition a property of that thing. An intrinsic property
> | of a thing cannot be changed without changing the thing.
> |
> | Measurements of wavelength for light change when you change your motion
> | with respect to the source. That means wavelength cannot be an intrinsic
> | property of light. Since wavelength times frequency is a constant, it is
> | clear frequency must also change and cannot be an intrinsic property of
> | light.
> |
> | In fact, both wavelength and frequency are properties of the
> | relationship between the observer and the light source.
>
> What is generating that relationship?
Let me discuss a light beam, and use classical electrodynamics and GR to
describe it.
Unlike what Bill Rowe said, the basic relationship is between the light
beam and the observer. But there is a fixed relationship between the
light beam and the source generating it, so Bill's statement is not
wrong, it's merely incomplete. The basic problem with Bill's statement
is that all physics is local, but his statement is not -- there MUST be
more going on.
A light ray is a physical object, and therefore exists independent of
any human coordinate system one might use to describe it. This means we
must use a coordinate-independent quantity to represent it. That
eliminates all of these quantities: energy, 3-momentum, frequency,
wavelength (all are inherently dependent on coordinates). As the light
ray has a direction, clearly a 4-vector is the most appropriate object
to use to model it.
For a particle of mass m, the relevant 4-vector is 4-momentum. When
projected onto Cartesian locally-inertial coordinates it has components
(E,Px,Py,Pz). With Planck's work in mind, the obvious corresponding
4-vector components for a light ray are proportional to (f,Kx,Ky,Kz),
where f is its frequency wrt these coordinates, and K is its wave
3-vector (=n/w where n is the unit 3-vector in the direction of
propagation and w is its wavelength). This is indeed a 4-vector[#], and
we call it the wave 4-vector (in the standard definition, neither
Planck's constant nor intensity are included in it).
[#] see MTW page 573.
So the frequency of a light wave in a given set of coordinates is merely
the time component of the wave 4-vector, and its wavelength can be
obtained from the spatial components of the wave 4-vector.
The wave 4-vector is intrinsic to the light beam. It is determined by
the source. The receiver projects the wave 4-vector of the beam onto the
locally-inertial coordinates of the detector, separating it into
frequency and a 3-vector related to its wavelength and propagation
direction. Now we have a local description of the relationship between
the wave and the detector, and one that is independent of coordinates.
The discussion in MTW referenced above shows another property
intrinsic to the light beam: the phase of the wave at each
point in the manifold (this is clearly a scalar function on the
manifold, and the wave 4-vector is merely its exterior
derivative[%]). This directly displays the important geometrical
aspect of a light beam.
[%] in physics we always have a metric available, so the
distinction between vector and covector is not very important
(at least at this level).
Tom Roberts tjro...@lucent.com
FrediFizzx
Thanks, Tom. I have been working on a string model for the photon and was
scratching my head about this very problem. So the wave 4-vector is
intrinsic to the light beam.
FrediFizzx
The answer comes to you shortly after you give up hope of finding it.
HenriWilson
>In article <ngc3hvclci9kvv3ui...@4ax.com>,
> He...@the.edge(HenriWilson) wrote:
>
>> Tell us all about light's intrinsic frequency.
>
>Light does not have an *intrinsic* frequency. An intrinsic property of a
>thing is by definition a property of that thing. An intrinsic property
>of a thing cannot be changed without changing the thing.
>
>Measurements of wavelength for light change when you change your motion
>with respect to the source. That means wavelength cannot be an intrinsic
>property of light. Since wavelength times frequency is a constant, it is
>clear frequency must also change and cannot be an intrinsic property of
>light.
Typical SRian logic, "Wavelength is velocity dependent, therefore wavelength is
dependent ONLY on velocity and on nothing else."
How stupid ARE you people?
>
>In fact, both wavelength and frequency are properties of the
>relationship between the observer and the light source.
So bloody what. That is not what CAUSES wavelength.
HenriWilson
That's great Tom. It describes the end result perfectly. But it doesn't tell us
anything about the possible structure of photons. Some intrinsic physical
attribute is responsible for giving an observer the impreesion of 'wavelength'
and by inference, a frequency. Now what might that property be?
>
>The wave 4-vector is intrinsic to the light beam. It is determined by
>the source. The receiver projects the wave 4-vector of the beam onto the
>locally-inertial coordinates of the detector, separating it into
>frequency and a 3-vector related to its wavelength and propagation
>direction. Now we have a local description of the relationship between
>the wave and the detector, and one that is independent of coordinates.
Now translate that into a 3D picture of whatever is responsible for
'wavelength'.
>
> The discussion in MTW referenced above shows another property
> intrinsic to the light beam: the phase of the wave at each
> point in the manifold (this is clearly a scalar function on the
> manifold, and the wave 4-vector is merely its exterior
> derivative[%]). This directly displays the important geometrical
> aspect of a light beam.
>
> [%] in physics we always have a metric available, so the
> distinction between vector and covector is not very important
> (at least at this level).
>
>
>Tom Roberts tjro...@lucent.com
kenseto
> kenseto wrote:
> > If a photon does not have intrinsic frequency as you seem to imply then
how
> > come in the equation for the enery of a photon shows that it is a
function
> > of frequency?
>
> The energy of a photon is not intrinsic to the photon, either. No kind
> of energy whatsoever is intrinsic to the object which "has" it. Energy
> is a relationship, not a thing (hence the quotes on "has", because there
> is a pun on that word).
The other interpretation is: A photon is a wave packet in a stationary and
structured aether and thus the energy of a photon is intrinsic. The
intrinsic frequency of a photon (wave packet) as it exists in the aether is
dependent on the absolute motion of the source. Different observers will
each measure a different energy for the same photon due to the different
states of absolute motion of these observers in the aether.
Ken Seto
kenseto
Question: Is the speed of light an intrinsic property of a photon?
Ken Seto
Tom Roberts
> > [my discussion]
> That's great Tom. It describes the end result perfectly. But it doesn't tell us
> anything about the possible structure of photons. Some intrinsic physical
> attribute is responsible for giving an observer the impreesion of 'wavelength'
> and by inference, a frequency. Now what might that property be?
I alluded to this in my post:
>> The discussion in MTW referenced above shows another property
>> intrinsic to the light beam: the phase of the wave at each
>> point in the manifold (this is clearly a scalar function on the
>> manifold, and the wave 4-vector is merely its exterior
>> derivative[%]). This directly displays the important geometrical
>> aspect of a light beam.
The phase of the light beam is a scalar function on the manifold, and is
probably the most fundamental property of the light beam. For example,
from it one can deduce the wave 4-vector I discussed earlier. It is
determined by properties of the source, by properties of EM waves in
general, and by the underlying geometrical structure of the manifold.
From the phase function it is natural and easy to understand why
different observers measure different frequencies and wavelengths for a
given light beam, and why each observer will measure the value c for its
local speed. This model can account for all of the classical aspects of
light beams, qualitatively and quantitatively, for both geometrical
optics and wave optics approximations, as well as all the details of
Maxwell's equations.
The quantum aspects of light are far more complicated, and there's no
point in trying to discuss them here (analogy: you need to master
arithmetic before attempting calculus; there is a reason why schools
teach classical physics before quantum mechanics).
Tom Roberts tjro...@lucent.com
Tom Roberts
> Question: Is the speed of light an intrinsic property of a photon?
No. The speed of any object is never an intrinsic aspect of the object
itself. Speed is always a relationship between the object in question
and some observer's coordinate system.
Note "speed of a photon" is really not well defined; I am
answering for a light pulse, not a photon. The quantum
complexities are simply too great....
The fact that the local speed of light is independent of the observer's
coordinates is an aspect of the underlying geometry and the relationship
between EM radiation and the geometry.
Only invariant quantities can be intrinsic to an object, but
invariants need not be intrinsic to some object.
Tom Roberts tjro...@lucent.com
HenriWilson
>kenseto wrote:
>> If a photon does not have intrinsic frequency as you seem to imply then how
>> come in the equation for the enery of a photon shows that it is a function
>> of frequency?
>
>The energy of a photon is not intrinsic to the photon, either. No kind
>of energy whatsoever is intrinsic to the object which "has" it. Energy
>is a relationship, not a thing (hence the quotes on "has", because there
>is a pun on that word).
That claim has been made before. It seems to apply to KE and PE but what about
heat energy?
Again, the dependence of light energy on observer frame is just a result of the
postulate that its speed is constant for all observers. One cannot apply
ballistic principles so one must assume 'frequency' changes.
>
> The energy of an object is a relationship between the object's
> intrinsic 4-momentum and the coordinates of the observer
> measuring the energy: energy is the 0-th component of that
> 4-momentum projected onto the observer's coordinates. For
> continuous structures things are more complicated, but the
> local energy density is the 0,0-th component of the
> stress-energy tensor.
>
>I am, of course, using the context of modern physics, SR/GR.
That's all very well, Tom. That tells us how light energy varies with observer.
We are trying to establish the exact nature of 'light frequency'. Where is
evidence of any 'time periodicity' associated with light other than that which
I have suggested (with the 'moving line of Vs')?
I say the 'light frequency' concept is completely bogus.
>
>
>Tom Roberts tjro...@lucent.com
Henri Wilson.
Why is the creative output of one SRian the same as that produced by one million of them?
HenriWilson
The equation should be E=hc.n, where n is the number of wavecrests per meter.
E=h.nu is wrong.
There is no 'frequency' directly associated with EM. A generated EM based wave
like RF uses variations in combined properties of many 'photons'.
Proof: RF does not change frequency when transmitted down a gravity well. Light
does.
Henri Wilson.
Why is the creative output of one SRian the same as that produced by one million of them?
HenriWilson
wrote:
Dead right Freddy.
This 'photon' concept is all a bit of mystery.
One thing is certain though. Energy CAN BE transmitted across vast distances of
3D space by something. Tom's maths tells us the end result.
I want to know the instantaneous state of affairs.
What constitutes a 'free photon'?
Spin could provide the answer to the 'wavecrest' question.
>
>FrediFizzx
>
>http://www.flashrock.com/upload/photong/photong.html
HenriWilson
Quite right Ken.
And according to my H-aether theory, that energy will vary through space
according to local Haether density and turbulence.
>
>Ken Seto
HenriWilson
But Tom, nothing you are saying here points to evidence of any 'time
periodicity' associated with light.
HenriWilson
>On 7/14/2003 8:06 AM, kenseto wrote:
>> Question: Is the speed of light an intrinsic property of a photon?
>
>No. The speed of any object is never an intrinsic aspect of the object
>itself. Speed is always a relationship between the object in question
>and some observer's coordinate system.
>
> Note "speed of a photon" is really not well defined; I am
> answering for a light pulse, not a photon. The quantum
> complexities are simply too great....
Why don't you admit it Tom. We simply don't know much about the propagation of
light at all.
>
>
>The fact that the local speed of light is independent of the observer's
>coordinates is an aspect of the underlying geometry and the relationship
>between EM radiation and the geometry.
There is no evidence that the OW light speed is independent of source velocity
over short ranges.
If you run my short demo www.users.bigpond.com/hewn/photons.exe
you will see that your argument is impossible.
>
> Only invariant quantities can be intrinsic to an object, but
> invariants need not be intrinsic to some object.
>
>
>Tom Roberts tjro...@lucent.com
Tom Roberts
> The equation should be E=hc.n, where n is the number of wavecrests per meter.
> E=h.nu is wrong.
No. This is quite clear once one understands that the wave 4-vector is
the exterior derivative of the phase function on the manifold, and its
relationship to 4-momentum.
> There is no 'frequency' directly associated with EM.
Right. But there's no wavelength intrinsic to an EM wave, either.
> Proof: RF does not change frequency when transmitted down a gravity well. Light
> does.
What do you do, just make up statements to bolster your argument and
provide whatever "proof" you desire? You are quite simply wrong, and any
research into the issue at all would have shown you that this is wrong.
All EM waves experience gravitational red/blueshift in the same way.
Tom Roberts tjro...@lucent.com
FrediFizzx
What don't you understand about the following that Tom said?
"With Planck's work in mind, the obvious corresponding
4-vector components for a light ray are proportional to (f,Kx,Ky,Kz),
where f is its frequency wrt these coordinates, and K is its wave
3-vector (=n/w where n is the unit 3-vector in the direction of
propagation and w is its wavelength). This is indeed a 4-vector[#], and
we call it the wave 4-vector (in the standard definition, neither
Planck's constant nor intensity are included in it)."
FrediFizzx
"The path to truth is thinner than a razor's edge."
FrediFizzx
news:70i6hv47mg23p8k50...@4ax.com...
What don't you understand about the following that Tom said?
"With Planck's work in mind, the obvious corresponding
4-vector components for a light ray are proportional to (f,Kx,Ky,Kz),
where f is its frequency wrt these coordinates, and K is its wave
3-vector (=n/w where n is the unit 3-vector in the direction of
propagation and w is its wavelength). This is indeed a 4-vector[#], and
we call it the wave 4-vector (in the standard definition, neither
Planck's constant nor intensity are included in it)."
FrediFizzx
Tom Roberts
> But Tom, nothing you are saying here points to evidence of any 'time
> periodicity' associated with light.
Sure it does -- a detector measures the wave's phase function along its
local time axis, and observes a signal of definite frequency.
Real detectors clearly exhibit such time periodicity for all EM waves
they detect.
Visually, let me suppress X and Y and draw a wave traveling along the +Z
axis, with the Z axis horizontal and the T axis vertical. I'll use "*"
to draw the wave crests, and "+" to indicate a detector at rest in these
coordinates and '<' to indicate a detector moving in the -Z direction at
0.5 c (use a fixed-width font):
T
^
| * * * * * * * *+ * * * * * * < * * * The angle
|* * * * * * * * +* * * * * * * <* * * of the lines
| * * * * * * * + * * * * * * < * * * of *s is the
| * * * * * * * *+ * * * * * * * <* * * speed of
|* * * * * * * * +* * * * * * * < * * light.
| * * * * * * * + * * * * * * * <* * *
| * * * * * * * *+ * * * * * * * < * * Each detector
|* * * * * * * * +* * * * * * * * <* * must move
| * * * * * * * + * * * * * * * < * * slower than
| * * * * * * * *+ * * * * * * * * <* * the speed
|* * * * * * * * +* * * * * * * * < * of light.
| * * * * * * * + * * * * * * * * <*
-----------------------+-------------------------<--> Z
Clearly both detectors measure a time-periodic signal from the wave. And
the < detector measures a higher frequency than the + detector (it
intersects 7 *-lines while the other intersects 4.5; time dilation
applies but is a smaller effect than this difference in number of lines
crossed).
Tom Roberts tjro...@lucent.com
kenseto
But the direction will change with different observers. So that too,
according to you, cannot be an intrinsic property of the light ray.
> The discussion in MTW referenced above shows another property
> intrinsic to the light beam: the phase of the wave at each
> point in the manifold (this is clearly a scalar function on the
> manifold, and the wave 4-vector is merely its exterior
> derivative[%]). This directly displays the important geometrical
> aspect of a light beam.
Here you assume that the light ray is composed of waves and the phase of the
wave is an intrinsic property. So why isn't the number of wave per unit time
(frequency) an intrinsic property of the light ray?
Ken Seto
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:bet4u2$l...@netnews.proxy.lucent.com...
>>any human coordinate system one might use to describe it. This means we
>>must use a coordinate-independent quantity to represent it. That
>>eliminates all of these quantities: energy, 3-momentum, frequency,
>>wavelength (all are inherently dependent on coordinates). As the light
>>ray has a direction, clearly a 4-vector is the most appropriate object
>>to use to model it.
>
> But the direction will change with different observers. So that too,
> according to you, cannot be an intrinsic property of the light ray.
The direction is NOT intrinsic to the light ray, the wave 4-vector is.
Differently-moving observers project it onto different spatial
coordinates, and obtain different wavelengths and directions for the
observation.
Here I used direction in the 3-vector sense, as your question
implied. In the 4-vector sense, of course, the direction of
the ray is indeed intrinsic and invariant.
> Here you assume that the light ray is composed of waves and the phase of the
> wave is an intrinsic property. So why isn't the number of wave per unit time
> (frequency) an intrinsic property of the light ray?
The wave 4-vector is an intrinsic property of the wave.
Differently-moving observers project its time component onto different
time coordinates, and obtain different frequencies for the observation.
The key to understanding the basis of SR and GR is to learn that they
are GEOMETRICAL theories, and the usual acts of detection and
observation are represented mathematically by PROJECTION of the
underlying physical system onto the observer's coordinates.
Tom Roberts tjro...@lucent.com
HenriWilson
wrote:
I feel it is just an unnecessarily complicated way of expressing in 4D what I
said in my original post in 3D + time.
>
>FrediFizzx
>
>"The path to truth is thinner than a razor's edge."
HenriWilson
Tom, what you say here is not in dispute. Of course differently moving
observers will intersect a different number of wavecrests per unit time.
You don't appear to be grasping the point I am trying to make.
The above model assumes that light (photons) are rigid objects which are
characterised by a 'spatial wave pattern' along their 'axes'. As this wave
passes an observer, that observer interprets the intersection of arriving
wavecrests as a 'frequency.
This is consistent with my moving line of 'Vs' in the original post. It is also
consistent with my 'sawtooth' or 'serated bullet' model of photons.
Whether you analyse this process in 3D or 4D matters not.
The question is, light cannot be a rigid 'line of wavecrests'. So what is it?
Some intrinsic property must be responsible for the wave shape. What might that
be? Some process maintains a series of 'fixed wave crests' along a photon as it
traverses space.
Think 3D. Forget your 4-vector stuff.
No 'frequency' has ever been directly associated with individual photons.
'E=h.nu' in inferred from the correct equation: E=hc/lambda.
Radio signals are alternating macroscopic effects involving many photons. If
the latter DO possess some kind of intrinsic periodicity, there is no evidence
that this plays any part in the broadcast frequency.
HenriWilson
>HenriWilson wrote:
>> The equation should be E=hc.n, where n is the number of wavecrests per meter.
>> E=h.nu is wrong.
>
>No. This is quite clear once one understands that the wave 4-vector is
>the exterior derivative of the phase function on the manifold, and its
>relationship to 4-momentum.
You will have to explain that in simple language to me Tom.
>
>
>> There is no 'frequency' directly associated with EM.
>
>Right. But there's no wavelength intrinsic to an EM wave, either.
But there must be. What you are really saying is that the size of that
wavelength is frame dependent.
There has to be an intrinsic property that gives an impresion of 'wavelength'.
>
>
>> Proof: RF does not change frequency when transmitted down a gravity well. Light
>> does.
>
>What do you do, just make up statements to bolster your argument and
>provide whatever "proof" you desire? You are quite simply wrong, and any
>research into the issue at all would have shown you that this is wrong.
>All EM waves experience gravitational red/blueshift in the same way.
Where is your evidence of that Tom?
>
>
>Tom Roberts tjro...@lucent.com
HenriWilson
How do you determine the zero of your 'spatial axes' here? Are you not assuming
some kind of spatial absolutivity?
>
>
>The key to understanding the basis of SR and GR is to learn that they
>are GEOMETRICAL theories, and the usual acts of detection and
>observation are represented mathematically by PROJECTION of the
>underlying physical system onto the observer's coordinates.
>
>
>Tom Roberts tjro...@lucent.com
kenseto
This is a round about way to explain the frequency shift of a light ray. It
is designed to avoid the implication that the absolute motions of the source
and the detector are responsible for the observed red or blue shift.
Ken Seto
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> On Sun, 13 Jul 2003 19:45:27 GMT, Bill Rowe <bjr...@earthlink.net> wrote:
> >Measurements of wavelength for light change when you change your motion
> >with respect to the source. That means wavelength cannot be an intrinsic
> >property of light. Since wavelength times frequency is a constant, it is
> >clear frequency must also change and cannot be an intrinsic property of
> >light.
> Typical SRian logic, "Wavelength is velocity dependent, therefore
> wavelength is dependent ONLY on velocity and on nothing else."
Looks to me like you are describing *your* problem with logic. It is
your words that say "therefore wavelength is dependent ONLY..." not mine.
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> The equation should be E=hc.n, where n is the number of wavecrests per meter.
But since hc.n does not have units of energy it is clear this equation
cannot be correct. And
> E=h.nu is wrong.
clearly is correct as it matches experimental observation.
> There is no 'frequency' directly associated with EM.
There clearly is. I can measure it directly on an oscilloscope as long
as the frequncy is less than ~100GHz or so.
> Proof: RF does not change frequency when transmitted down a gravity
> well. Light does.
Whatever gives you the idea RF does not change in the same manner as
light does in a gravity field?
There is no experimental evidence whatever to show RF is any different
than light save wavelenght and the consequences due to different
wavelengths.
S. Enterprize Company
>
It only works in special limited cases.
The original E = hf delt with blackbodies as far as I know.
.
Smart's Last Will And Testament Song
http://smart1234.s-enterprize.com/BeefmeatBlues.MP3
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
FrediFizzx
Correction: The *average* radius is lambda/2pi.
| Dead right Freddy.
| This 'photon' concept is all a bit of mystery.
I think I might have it all worked out and will present it in about a month.
Think about "natural" free space boundary conditions.
| One thing is certain though. Energy CAN BE transmitted across vast
distances of
| 3D space by something. Tom's maths tells us the end result.
| I want to know the instantaneous state of affairs.
|
| What constitutes a 'free photon'?
A free photon is one that is not still connected to a real particle.
| Spin could provide the answer to the 'wavecrest' question.
Maybe so.
FrediFizzx
The answer comes to you shortly after you give up hope of finding it.
HenriWilson
>In article <0rd6hv8r99ban2fud...@4ax.com>,
> He...@the.edge(HenriWilson) wrote:
>
>> The equation should be E=hc.n, where n is the number of wavecrests per meter.
>
>But since hc.n does not have units of energy it is clear this equation
>cannot be correct. And
What are you talking about?
E=h.nu=hc/lambda
Lambda=wavelength (metres)
n=1/wavelength=1/L
E=hc.n has units of energy.
>
>> E=h.nu is wrong.
>
>clearly is correct as it matches experimental observation.
>
>> There is no 'frequency' directly associated with EM.
>
>There clearly is. I can measure it directly on an oscilloscope as long
>as the frequncy is less than ~100GHz or so.
That is not associated with the frequency of a single photon. That is a
generated wave involving a multitude of photons.
>
>> Proof: RF does not change frequency when transmitted down a gravity
>> well. Light does.
>
>Whatever gives you the idea RF does not change in the same manner as
>light does in a gravity field?
>
>There is no experimental evidence whatever to show RF is any different
>than light save wavelenght and the consequences due to different
>wavelengths.
Wrong. There is no experimental evidence to show that it DOES behave like
light.
Henri Wilson.
HenriWilson
Well this 'wavelength' thing doesn't just appear from nowhere.
First we had the 'tick fairies' now we have 'wavelength fairies'!
brew...@ecn.ab.ca
>I want to know what property of all these photons is manipulated in order to
>produce that traveling wave.
It's like trying to define light as no mass with the speed of
light. It seems to me that "photon" is just a convenient way to express
light as a particle, and then giv it properties of a wave, too. It might
be useful to define it as the cycle of a light wave, in which case it
is also wavelength.
brew...@ecn.ab.ca
(on trying to define the form of a photon)
Obviously that would be beyond you.
I would also refuse to define the form of a photon. It's not a
shape unless you polarize it, and even then, you hav to realize that both
your unit of time and your unit of length are arbitrary standards, which
leaves us with more than enough to establish.
It's like a word that Lewis Carroll's Humpty Dumpty used to mean
a paragraph. Nobody remembers the word.
I believe that light has an intrinsic frequency, but I also know
that this frequency rises when the emitter is approaching you.
brew...@ecn.ab.ca
do a lot of experimentation. He did explain why there doesn't have to be a
planet (Vulcan) between Mercury and the Sun, which Newtonian mechanics and
observation predicted there would be. His theory also jibes with
observations that light doesn't go faster when it travels in the same
direction as the Earth, and that nuclear energy is orders of magnitude
greater than any chemical reaction we know.
So if there's a quack in this thread trying to tell us that light
shifts into the red by induction with some kind of aether or slows down to
"local speed" (whatever that is), maybe he should tell us another way to
explain why the planet Vulcan wasn't.
brew...@ecn.ab.ca
(...)
>>> Typical SRian logic, "Wavelength is velocity dependent, therefore
>>> wavelength is dependent ONLY on velocity and on nothing else."
That's not what anyone said. And this discussion nudges the definition of
intrinsic, which makes me bored as hell. Previously, I think he tried to
undefine frequency.
Dirk Van de moortel
"HenriWilson" <He...@the.edge> wrote in message news:5rf4hv476mupor36u...@4ax.com...
[snip]
> Typical SRian logic, "Wavelength is velocity dependent, therefore wavelength is
> dependent ONLY on velocity and on nothing else."
>
> How stupid ARE you people?
Not as stupid as this:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LogicBull.html
Dirk Vdm
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> On Wed, 16 Jul 2003 03:26:21 GMT, Bill Rowe <bjr...@earthlink.net> wrote:
> >But since hc.n does not have units of energy it is clear this equation
> >cannot be correct. And
> What are you talking about?
My mistake. I failed to notice you defined n as 1/wavelength. I thought
you defined n as the number of wavecrests, i.e., a dimensionless
parameter.
> >There is no experimental evidence whatever to show RF is any different
> >than light save wavelenght and the consequences due to different
> >wavelengths.
> Wrong. There is no experimental evidence to show that it DOES behave like
> light.
You must be kidding. Refer to any good text. Maxwells equations describe
light equally well as they describe RF energy.
S. Enterprize Company
I will answer this based on what I think you mean. Does light exist only at
a certain frequency or just at visual frequency range? No light exists within
the whole electromagnetic spectrum of frequencies with in frequencies NOT only
in the visual EM spectrum range.
Example:
We can observe the line spectra in the visual range for hydrogen, but this
doesn't mean this is the only area where the line spectra exists for hydrogen.
The line spectra also exists outside of the visual range, too.
Smart's First Will And Testament Song
HenriWilson
>HenriWilson (He...@the.edge), in article nntp:/<jfeugv4lol2poa24a...@4ax.com> , wrote:
>
>(on trying to define the form of a photon)
>Obviously that would be beyond you.
One has to start somewhere.
>
> I would also refuse to define the form of a photon. It's not a
>shape unless you polarize it, and even then, you hav to realize that both
>your unit of time and your unit of length are arbitrary standards, which
>leaves us with more than enough to establish.
Time and space don't have to be absolute in irder than the form a photon can be
determined.
>
> It's like a word that Lewis Carroll's Humpty Dumpty used to mean
>a paragraph. Nobody remembers the word.
>
> I believe that light has an intrinsic frequency, but I also know
>that this frequency rises when the emitter is approaching you.
So bloody what. Everyone knows that.
What causes that 'intrinsic frequency' then?
HenriWilson
Newtonian Mechanics needs to be upgraded. That's all.
HenriWilson
One thing is certain. Empty space carrying EM is different from empty space
devoid of EM.
WHY?
HenriWilson
I don't doubt that RF is made of photons in some kind of waving structure. But
I will continue to argue that RF frequency doesn't change in a gravity well
like photon 'c/lambda' does.
There is no proof either way except maybe for the Pioneer anomaly.
tj Frazir
?
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> I don't doubt that RF is made of photons in some kind of waving structure. But
> I will continue to argue that RF frequency doesn't change in a gravity well
> like photon 'c/lambda' does.
Clearly, you can argue as you like. But since there is no evidence
there is any difference between photons associated with light and
photons associated with RF, your arguement above is rather pointless.
HenriWilson
No it isn't. Photons accelerate down a gravity well like anything else but
their combined wave configuration retains its emission frequency.
See www.users.bigpond.com/hewn/falling wave. (I haven't finished this yet but
it will explain what I mean)
Wavelength increases as the velocity of photons increases but the arrival
frequency of the wavecrests does not alter.
brew...@ecn.ab.ca
>One thing is certain. Empty space carrying EM is different from empty space
>devoid of EM.
>WHY?
IMHO, because empty space devoid of EM is much like a vakyuum or
absolute zero, more of a theoretical posibility than a practical one.
Consider that radiant heat is in the infra-red band, so eliminating it
would require temperatures that are either extremely low or absolute zero.
...It _might_ exist within a cavity in a superconductor, but I'm at a loss
to guess whether that cavity would also hav to be evacuated.
brew...@ecn.ab.ca
(...)
>Newtonian Mechanics needs to be upgraded. That's all.
Isn't an upgrade what you hav in a theory that simplifies to
Newtonian Mechanics when force and speed is insignificant with respect to
light.
brew...@ecn.ab.ca
(...)
>Photons accelerate down a gravity well like anything else but
>their combined wave configuration retains its emission frequency.
Assume that the Sun is a gravity well. That would be a curious
effect, and certainly not true within the gravity of our solar system or
you'd aim flashlights like you aim canons. I've heard that black holes
attract light, but I haven't read any details, much less observed it.
Also, I don't believe it's established that you can accelerate
photons without, INSTEAD, changing their frequency. The same concept comes
with some kinds of electrical coils where to induce electrons would seem
to be also to change their speed -- they don't actually change in speed,
but their voltage changes. So if black holes have in fact been observed to
bend rays of light, they should also bring their frequency up.
>See www.users.bigpond.com/hewn/falling wave. (I haven't finished this yet but
>it will explain what I mean)
>Wavelength increases as the velocity of photons increases but the arrival
>frequency of the wavecrests does not alter.
If you take that statement, and assume that photons can't change
in speed, you'll require that time be variable to explain a wavelength
that doesn't correspond with the frequency at a constant rate.
brew...@ecn.ab.ca
(...)
>Time and space don't have to be absolute in irder than the form a photon can be
>determined.
I think the only definition of a photon's form is its polarization.
>>
>> It's like a word that Lewis Carroll's Humpty Dumpty used to mean
>>a paragraph. Nobody remembers the word.
>>
>> I believe that light has an intrinsic frequency, but I also know
>>that this frequency rises when the emitter is approaching you.
>So bloody what. Everyone knows that.
>What causes that 'intrinsic frequency' then?
Fire is the shortest answer. Many things are the initial cause of a
frequency. They might be the distance from an electron's free state to a
more stable state in an ionized gas or heated metal-- a question with
answers ranging from (the outermost shell) to the maximum ionization level
that the medium exhibits. They might be electrons hitting the phosphors of
your cathode ray tube. They might be short-wave photons being converted
into visible light within fluorescent light phosphors.
If you meant why is it intrinsic, it is simply that we haven't found a
reason to suspect otherwise.
Paul B. Andersen
"HenriWilson" <He...@the.edge> skrev i melding news:b9i6hv0dc79pmt6ev...@4ax.com...
> On Mon, 14 Jul 2003 09:43:20 -0500, Tom Roberts <tjro...@Lucent.com> wrote:
> >The fact that the local speed of light is independent of the observer's
> >coordinates is an aspect of the underlying geometry and the relationship
> >between EM radiation and the geometry.
>
> There is no evidence that the OW light speed is independent of source velocity
> over short ranges.
> If you run my short demo www.users.bigpond.com/hewn/photons.exe
> you will see that your argument is impossible.
And in that "short demo" we find the following demonstration
of Wilsonian logic at its very best:
| Based on the second postulate, SR has derived a formula
| showing that velocities must be added relativistically
| using the formula w=(u+v)/(1+(uv/c^2)), where w is the
| velocity of object A in the observer frame, v is the velocity
| of B in the observer frame and u the velocity of A in B's frame.
| If we apply this formula to a photon (A) emitted at velocity c
| from a moving source (B), we first assume that the photon's speed
| must be added to that of the source.
| In other words, it is source dependent.
Congratulations, Henry.
You have demonstrated that the fact that the speed of light
is independent of the velocity of its source prove that the speed
of light depend on the velocity of its source.
That should settle the matter.
SR must be wrong.
Paul
Bill Rowe
He...@the.edge(HenriWilson) wrote:
> Wavelength increases as the velocity of photons increases
But experimental evidence shows photons travel at c and experience
neither an increase nor decrease in speed. In fact, if you are using SI
units as your measurement system, you can never measure the speed of
photons to be anything other than c due to the way SI units are defined.
> See my animations at:
> http://www.users.bigpond.com/HeWn/index.htm
And what would be the point? If you animation supports your remark "the
velocity of photons increases" then it has no relevance to valid physics.
Tom Roberts
> There has to be an intrinsic property that gives an impresion of 'wavelength'.
The intrinsic property RELATED to wavelength and frequency is the phase
function on the manifold. Different observers observe that function
differently, and obtain different values for frequency and wavelength of
a given wave. Neither wavelength nor frequency are intrinsic to the
wave, as they are relationships between the wave itself the observer.
<shrug>
> On Mon, 14 Jul 2003 22:44:51 -0500, Tom Roberts <tjro...@lucent.com> wrote:
>>Visually, let me suppress X and Y and draw a wave traveling along the +Z
>>axis, with the Z axis horizontal and the T axis vertical. I'll use "*"
>>to draw the wave crests, and "+" to indicate a detector at rest in these
>>coordinates and '<' to indicate a detector moving in the -Z direction at
>>0.5 c (use a fixed-width font):
>>
>>T
>>^
>>| * * * * * * * *+ * * * * * * < * * * The angle
>>|* * * * * * * * +* * * * * * * <* * * of the lines
>>| * * * * * * * + * * * * * * < * * * of *s is the
>>| * * * * * * * *+ * * * * * * * <* * * speed of
>>|* * * * * * * * +* * * * * * * < * * light.
>>| * * * * * * * + * * * * * * * <* * *
>>| * * * * * * * *+ * * * * * * * < * * Each detector
>>|* * * * * * * * +* * * * * * * * <* * must move
>>| * * * * * * * + * * * * * * * < * * slower than
>>| * * * * * * * *+ * * * * * * * * <* * the speed
>>|* * * * * * * * +* * * * * * * * < * of light.
>>| * * * * * * * + * * * * * * * * <*
>> -----------------------+-------------------------<--> Z
>>
>>Clearly both detectors measure a time-periodic signal from the wave. And
>>the < detector measures a higher frequency than the + detector (it
>>intersects 7 *-lines while the other intersects 4.5; time dilation
>>applies but is a smaller effect than this difference in number of lines
>>crossed).
>>
>>
>>Tom Roberts tjro...@lucent.co
>
> The above model assumes that light (photons) are rigid objects which are
> characterised by a 'spatial wave pattern' along their 'axes'.
No. I make no attempt to discuss photons at all. I am discussing
electromagnetic waves in the wave optics approximation. The quantum
aspects of photons are FAR too complicated to permit into this discussion.
The phase of the wave is a function on the manifold. This is independent
of observer (or frame or coordinates). In some sense this is "rigid",
but I would not apply that word to it, as it has too many invalid
connotations.
E.g. every "rigid" object has a rest frame, but this EM
wave has none. Every "rigid" object can be pushed or pulled,
not this EM wave. The only sense of "rigid" that applies to an
EM wave is that each portion of the wave has a fixed spatio-
temporal relationship to every other portion.
> As this wave
> passes an observer, that observer interprets the intersection of arriving
> wavecrests as a 'frequency.
Yes.
> This is consistent with my moving line of 'Vs' in the original post.
No. Your 'Vs' have an intrinsic rest frame, and an intrinsic length. The
electromagnetic wave I discuss has neither.
> The question is, light cannot be a rigid 'line of wavecrests'.
Hmmmm. I think you are confusing yourself with the invalid connotations
of the word "rigid".
> Some intrinsic property must be responsible for the wave shape.
Yes. The wave function on the manifold.
> No 'frequency' has ever been directly associated with individual photons.
True. When it is possible to detect individual photons, they act as if
they were particle-like. YOU ARE CONFUSING YOURSELF BY TRYING TO DISCUSS
THINGS YOU KNOW NOTHING ABOUT. stop trying to write about such stuff and
go LEARN about it.
I suggest you start with: Feynman, _QED_.
Tom Roberts tjro...@lucent.com
HenriWilson
You are very confused.
Nobody has performed a OWLS experiment so don't try to make out you are an
expert on the subject.
The frequency at which photon 'wavecrests' arrive at an observer DOES change
with its varying speed in a gravity well.
However, the frequency of an RF signal DOES NOT change when falling down a
gravity well because its 'wavecrests' become more spatially separated as the
group of photons that make up the signal accelerate.
HenriWilson
>In article <f71fhvg0msjd374v4...@4ax.com>,
> He...@the.edge(HenriWilson) wrote:
>
>> Wavelength increases as the velocity of photons increases
>
>But experimental evidence shows photons travel at c and experience
>neither an increase nor decrease in speed. In fact, if you are using SI
>units as your measurement system, you can never measure the speed of
>photons to be anything other than c due to the way SI units are defined.
Bill, there is absolutely NO experimental evidence to prove that the one way
speed of light is constant - so don't try to make out you know anything about
the subject.
>
>> See my animations at:
>> http://www.users.bigpond.com/HeWn/index.htm
>
>And what would be the point? If you animation supports your remark "the
>velocity of photons increases" then it has no relevance to valid physics.
According to the Pound-Rebka expt, - which is the only one even remotely
believable - light accelerates when falling just like anything else.
Henri Wilson.
HenriWilson
wrote:
Welcome back Paul. Your education has suffered markedly because of your recent
absence from this NG.
Yes, SR's little maths trick once again disguises the fact that SR is just
another aether theory.
By combining source dependency with its velocity addition equation, SR manages
to prove that light speed is absolute even without an aether.
Thus, Einstein again made the aether irrelevant whether it exists or not - just
as he did with his clock synch definition.
HenriWilson
>HenriWilson wrote:
>
>> There has to be an intrinsic property that gives an impresion of 'wavelength'.
>
>The intrinsic property RELATED to wavelength and frequency is the phase
>function on the manifold. Different observers observe that function
>differently, and obtain different values for frequency and wavelength of
>a given wave. Neither wavelength nor frequency are intrinsic to the
>wave, as they are relationships between the wave itself the observer.
><shrug>
TOM, you are still refering to the 'number of wavecrests passing per second'.
We all know that this is observer dependent.
I am asking "what , in the 3D sense, might be responsible for a photon even
POSSESSING a property that gives rise to an observable 'wavelength' .
Put it this way. As a photon goes past, something about it causes the observer
to 'hear' boom!boom!boom!boom!boom!.............................
That is suggestive of a fixed spatial periodicity as in the case of say a
moving saw blade.
So is a photon a discreet quanta that has intrinsic properties or is it just a
traveling wave in a medium?
They are probably irrelevant anyway.
>
>The phase of the wave is a function on the manifold. This is independent
>of observer (or frame or coordinates). In some sense this is "rigid",
>but I would not apply that word to it, as it has too many invalid
>connotations.
>
> E.g. every "rigid" object has a rest frame, but this EM
> wave has none. Every "rigid" object can be pushed or pulled,
> not this EM wave. The only sense of "rigid" that applies to an
> EM wave is that each portion of the wave has a fixed spatio-
> temporal relationship to every other portion.
>
>
>> As this wave
>> passes an observer, that observer interprets the intersection of arriving
>> wavecrests as a 'frequency.
>
>Yes.
>
>> This is consistent with my moving line of 'Vs' in the original post.
>
>No. Your 'Vs' have an intrinsic rest frame, and an intrinsic length. The
>electromagnetic wave I discuss has neither.
Tom, everything you are saying seems to agree with my claim that photon
'frequency' is merely inferred from the fact that some kind of spatial
irregularity in moving past a particular observer.
I am asking the question, "if (in the 3D picture) light behaves as though it
consists of a 'moving line of V's' but obviously cannot be of that form, what
intrinsic properties might it possess which can explain this phenomenon?
>
>
>> The question is, light cannot be a rigid 'line of wavecrests'.
>
>Hmmmm. I think you are confusing yourself with the invalid connotations
>of the word "rigid".
I am thinking in terms of a 3 model. After all, our observations about light
are ll made in 3D so we should be able to construct such a model.
>
>
>> Some intrinsic property must be responsible for the wave shape.
>
>Yes. The wave function on the manifold.
I think you need a medium for that. Please explain.
>
>
>> No 'frequency' has ever been directly associated with individual photons.
>
>True. When it is possible to detect individual photons, they act as if
>they were particle-like. YOU ARE CONFUSING YOURSELF BY TRYING TO DISCUSS
>THINGS YOU KNOW NOTHING ABOUT. stop trying to write about such stuff and
>go LEARN about it.
Tom I don't think anyone knows much about light. Certainly nobody knows
anything about one-way light speed.
So why do you and others here try to make out you do?
I repeat, the frequency normally associated with photons is merely inferred as
being c.n where n is the number of 'wavecrests' passing per unit time due to
photon movement. You apparently agree with this.
If you know so much , please tell me what, in the 3D sense, might constitute a
'photon wavecrest'?
>
> I suggest you start with: Feynman, _QED_.
>
>
>Tom Roberts tjro...@lucent.com
HenriWilson
I wasn't trying to eliminate EM or even suggest that there might exist parts of
space that are completely devoid of it.
HenriWilson
You just don't get it do you "?
We know the answer.
A frequency is easily associated with light. It is just f=c/lambda.
But no such frequency has ever been measured. It is only inferred.
So what intrinsic property of a light quanta constitutes 'lambda'.