Microwave Power
Chalky
frequency of CMB radiation, is inversely proportional to 1 + z, where
z is its redshift.
However, I am not so sure if this also true for the received power per
unit area of detector surface.
On balance, I think it probably would, because I would expect the
received power and temperature of the radiation to be inearly related.
Nevertheless, I have still some doubts because, outside of GR, one
would expect the received power of an EM radiator to be proportional
to the square of its Doppler shift in frequency.
Can anyone put me straight on this?
Igor Khavkine
> If I understand correctly, the temperature and (more obviously)
> frequency of CMB radiation, is inversely proportional to 1 + z, where
> z is its redshift.
> However, I am not so sure if this also true for the received power per
> unit area of detector surface.
> On balance, I think it probably would, because I would expect the
If I look at the Sun from 1AU or from 1 light year, the temperature of
the Sun is the same (~6000K). On the other hand, the power received by
a detector at 1AU is much larger than that by a detector at 1 light
year (larger by the square of the ratios of the distances, ~2.5e-10).
The total power of the radiation and the temperature of its source are
not related.
The temperature is imprinted in the spectrum. If the radiation is
thermal, once the total power is normalized away, the distribution of
radiated power among different frequencies should follow the Planck
law.
> Nevertheless, I have still some doubts because, outside of GR, one
> would expect the received power of an EM radiator to be proportional
> to the square of its Doppler shift in frequency.
I think you are still off in your reasoning here. The amplitude and
frequency of electromagnetic radiation are in general not in a fixed
relationship with respect to each other.
Hope this helps.
Ulf Torkelsson
> If I understand correctly, the temperature and (more obviously)
> frequency of CMB radiation, is inversely proportional to 1 + z, where
> z is its redshift.
Yes.
> However, I am not so sure if this also true for the received power per
> unit area of detector surface.
No, it is not. The energy density of the blackbody radiation field
is proportional to temperature to the fourth power, so it is
proportional to (1+z)^{-4}. Received power is proportional to energy
density times c times a geometric factor.
Ulf Torkelsson
FrediFizzx
news:798679d0-b802-4140...@z41g2000yqz.googlegroups.com...
I have been wondering about what you say for quite some time. It seems
to me that the amplitude of EM radiation can only be dependent on the
number of photons and the frequency of said such photons. Of course,
relative to an observer. Consider an ideal case of EM radiation
composed of identical in phase photons. I obtain for the electric field
strength squared which would be the energy density,
E^2 = (hbar w^4 n)/(4pi^2 c^3)
Where w is omega, angular frequency, and n is the number of photons. Of
course this this is an expectation value since we are considering the
number of photons. AFAICT, plugging that into Maxwell's equations gives
the correct answer for the electric field strength at a certain distance
from the source of EM radiation of this ideal case according to what
Jackson says in Classical Electrodynamics. So if this is true, then we
do have a fixed relationship of some sort between frequency and
amplitude. I have never really seen that expression before, so if true,
then this is something possibly new. Or.... what other variable am I
missing?
Best,
Fred Diether
Moderator sci.physics.foundations
Peter
Let me add, have a look at the Stefan-Boltzmann law
Peter
Chalky
> On Dec 14, 11:02 am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > If I understand correctly, the temperature and (more obviously)
> > frequency of CMB radiation, is inversely proportional to 1 + z, where
> > z is its redshift.
> > However, I am not so sure if this also true for the received power per
> > unit area of detector surface.
> > On balance, I think it probably would, because I would expect the
> > received power and temperature of the radiation to be linearly related.
>
> If I look at the Sun from 1AU or from 1 light year, the temperature of
> the Sun is the same (~6000K).
I actually meant the temperature of the radiation relative to the
detector, not the emitter (otherwise the question would clearly be
completely stupid).
> On the other hand, the power received by
> a detector at 1AU is much larger than that by a detector at 1 light
> year (larger by the square of the ratios of the distances, ~2.5e-10).
Obviously, but, again, I took this as implied (or I had hoped it would
be). So I was querying the effect of the (1+z) factor, after these
things had been factored in.
> The total power of the radiation and the temperature of its source are
> not related.
Of course not, because of the above considerations, which I had hoped
would be obvious. In this context, perhaps now would be an appropriate
time to quote Robert Zimmerman "If you don't underestimate me, I won't
underestimate you".
> The temperature is imprinted in the spectrum. If the radiation is
> thermal, once the total power is normalized away, the distribution of
> radiated power among different frequencies should follow the Planck
> law.
I don't know what you mean here by "normalised away". What I mean,
physically, by CMB temperature, is what you would get from a (ideally
black) thermometer located far away from other sources of radiation.
Phillip Helbig---undress to reply
<f21abcec-4134-4624...@m25g2000yqc.googlegroups.com>,
Chalky <chalk...@bleachboys.co.uk> writes:
The wavelength of each photon goes up by (1+z), or the frequency down by
the same factor. One can apply the same argument to the individual
photons as one does to the crests of the waves. This gives another
factor of (1+z). Power is energy per time, so the power will drop by
two factors of (1+z).
On a related note: luminosity distance is defined via bolometric
luminosity. These two powers of (1+z) in the luminosity give us 1
factor of (1+z) in the distance (luminosity goes down with the square of
the distance) compered to the situation in a non-expanding universe. In
the triangle defining the luminosity distance, the apex is at the source
and the defining solid surface is at the observer, i.e. the value NOW
matters. In the case of the angular-size distance, the defining apex is
at the observer and the defining length (or surface) is at the object.
Thus, the distance at the time the light was emitted is important.
These differ by a factor of (1+z). Add to that the other factor of
(1+z) discussed above and one obtains the famous result that the
luminosity distance is two factors of (1+z) greater than the angular
size distance, independent of the cosmological parameters.
This means that surface brightness goes down like (1+z) to the -4! This
is for bolometric luminosity. Since an observation in a given
wavelength range corresponds to a wavelength range in the frame in which
the light was emitted which is (1+z) times smaller, realistic
observations in a finite band have the surface brightness go down with
(1+z) to the -5. (For the experts: this assumes that this bandwidth
effect is the only source of the K-correction.) For Poisson photon
statistics, this means that my signal-to-noise goes like (1+z) to the
-10. No wonder it is difficult to observe high-redshift objects.
Chalky
undress to reply) wrote:
> The wavelength of each photon goes up by (1+z), or the frequency down by
> the same factor. One can apply the same argument to the individual
> photons as one does to the crests of the waves. This gives another
> factor of (1+z). Power is energy per time, so the power will drop by
> two factors of (1+z).
I would understand your argument here (I think), if we were talking
about a discrete and hard edged "billiard ball" type radiator at a
large distance and, thus, redshift.
However, this was not the scenario that triggered my further musings
on this subject.
Instead, I was attempting to calculate the total background power of
EM radiation relative to us, by integrating the contributions of
subsequent shells of matter of constant "lookback time" thickness, in
a "bucket of dust" type model of a homogeneous and isotropic universe.
>From this point of view, your argument for a decrease in number of
photons per unit time at higher redshift, is then wiped out by the
fact that the density of radiant matter increases in this constant
thickness direction (at least approximately) in proportion to (1+z).
>From this perspective it doesn't really matter which shell we are
discussing, but I pitched the question at around z = 1000
because such shells look particularly promising for prodding with a
theoretical probe.
Incidentally, does anyone have a reference (ideally online) for the
original paper (was it by Rutherford? [I forget]) deriving the total
background energy / temperature of starlight? Because I suspect this
may mean that the calculation I am trying was originally performed
long ago.
Chalky
> the triangle defining the luminosity distance, the apex is at the source
> and the defining solid surface is at the observer, i.e. the value NOW
> matters.
Do you mean:
1) the distance from the emitter at the time of emission, to the
detector at the time of detection, relative to the detector?
Or
2) the distance from the emitter at the time of detection, to the
detector at the time of detection (aka co-moving radial distance,
iiuc)?
If the latter, doesn't this strike you as a bit bonkers?
It would require premonition of the subsequent history of the emitter,
over the light travel time between emission and detection.
p.ki...@ic.ac.uk
> Consider an ideal case of EM radiation
> composed of identical in phase photons. I obtain for the electric field
> strength squared which would be the energy density,
> E^2 = (hbar w^4 n)/(4pi^2 c^3)
> Where w is omega, angular frequency, and n is the number of photons. Of
> course this this is an expectation value since we are considering the
> number of photons. AFAICT, plugging that into Maxwell's equations gives
> the correct answer for the electric field strength at a certain distance
> from the source of EM radiation of this ideal case according to what
> Jackson says in Classical Electrodynamics. So if this is true, then we
> do have a fixed relationship of some sort between frequency and
> amplitude.
> [...]
Field "amplitude" is in your equation in _two_ places: in "E^2",
and in "n": to get a larger field I have to add photons (at least
in suitably simplistic cases, e.g. for coherent states). For a fixed
frequency, n and E^2 are directly proportional.
If you want to fix n, then the only way to get a higher energy density
is to increase the frequency of your chosen n photons (an easy mathematical
manipulation, but not so easy to do physically).
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
Phillip Helbig---undress to reply
<2782c5bb-92f6-42b8...@m3g2000yqf.googlegroups.com>,
Chalky <chalk...@bleachboys.co.uk> writes:
> > In
> > the triangle defining the luminosity distance, the apex is at the source
> > and the defining solid surface is at the observer, i.e. the value NOW
> > matters.
>
> Do you mean:
> 1) the distance from the emitter at the time of emission, to the
> detector at the time of detection, relative to the detector?
> Or
> 2) the distance from the emitter at the time of detection, to the
> detector at the time of detection (aka co-moving radial distance,
> iiuc)?
The latter. Yes, without the extra factors due to the redshifting of
the photons, it is the co-moving radial distance.
> If the latter, doesn't this strike you as a bit bonkers?
>
> It would require premonition of the subsequent history of the emitter,
> over the light travel time between emission and detection.
Remember, these definitions apply in a universe which can be
approximating as expanding homogeneously and isotropically.
Once the emitter emits the light I detect now, it can vanish. Once this
light is on the way to us, we can calculate what happens to it.
Or perhaps I don't understand your objection.
glird
> If I understand correctly, the temperature and (more obviously)
> frequency of CMB radiation, is inversely proportional to 1 + z, where
> z is its redshift.
Please give us an example using the known value of z; and give us
the direction indicated by z.
> However, I am not so sure if this also true for the received power per
> unit area of detector surface.
> On balance, I think it probably would, because I would expect the
> received power and temperature of the radiation to be linearly related.
> Nevertheless, I have still some doubts because, outside of GR, one
> would expect the received power of an EM radiator to be proportional
> to the square of its Doppler shift in frequency.
>
> Can anyone put me straight on this?
Not if you stay within GR; because a "straight line" is curved,
there.
FrediFizzx
news:51esv6-...@ph-kinsle.qols.ph.ic.ac.uk...
> FrediFizzx <fredi...@hotmail.com> wrote:
>> Consider an ideal case of EM radiation
>> composed of identical in phase photons. I obtain for the electric
>> field
>> strength squared which would be the energy density,
>
>> E^2 = (hbar w^4 n)/(4pi^2 c^3)
>
>> Where w is omega, angular frequency, and n is the number of photons.
>> Of
>> course this this is an expectation value since we are considering the
>> number of photons. AFAICT, plugging that into Maxwell's equations
>> gives
>> the correct answer for the electric field strength at a certain
>> distance
>> from the source of EM radiation of this ideal case according to what
>> Jackson says in Classical Electrodynamics. So if this is true, then
>> we
>> do have a fixed relationship of some sort between frequency and
>> amplitude.
>> [...]
>
> Field "amplitude" is in your equation in _two_ places: in "E^2",
> and in "n": to get a larger field I have to add photons (at least
> in suitably simplistic cases, e.g. for coherent states). For a fixed
> frequency, n and E^2 are directly proportional.
Yes, that is rather obvious.
> If you want to fix n, then the only way to get a higher energy density
> is to increase the frequency of your chosen n photons (an easy
> mathematical
> manipulation, but not so easy to do physically).
Why not easy to do physically? Tunable RF transmitters are not
possible? :-) Well, probably both n and w would be changing in that
case anywise. One would probably have to carefully adjust transmitter
power to keep n somewhat constant. But consider the case of a single
photon; then n = 1 and the "amplitude" then does have a fixed
relationship to the frequency only.
Best,
Fred Diether
moderator sci.physics.foundations
Chalky
> But consider the case of a single
> photon; then n = 1 and the "amplitude" then does have a fixed
> relationship to the frequency only.
Aha!
But what is that fixed relationship?
I have often pondered this in a vaguely wooly manner in the past, but
not come up with the answer (perhaps because I am mathematically lazy).
Chalky
> On Dec 14, 5:02 am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > If I understand correctly, the temperature and (more obviously)
> > frequency of CMB radiation, is inversely proportional to 1 + z, where
> > z is its redshift.
>
> Please give us an example using the known value of z; and give us
> the direction indicated by z.
Your request doesn't make much sense. If classical bb theory is
correct, that z depends on when the observer is located. Eg an
observer located 13.7 billion years in our past would see that
radiation at a much higher frequency. Consequently, in this context,
there is no such thing as "the known value of z", only its inferred
value now.
The indicated direction of the source is, likewise, the past. As you
should know, the cmb arrives from all directions in space.
On Dec 22, 7:19 pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
undress to reply) wrote:
> Or perhaps I don't understand your objection.
You did understand it and addressed it succinctly.
FrediFizzx
news:9347a0a8-f5a1-41f4...@n38g2000yqf.googlegroups.com...
> On Dec 23, 5:46 pm, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>
>> But consider the case of a single
>> photon; then n = 1 and the "amplitude" then does have a fixed
>> relationship to the frequency only.
>
> Aha!
> But what is that fixed relationship?
E^2 = (hbar w^4)/(4pi^2 c^3)
For the case of a single photon n = 1. w is omega, angular frequency.
It is the only variable in the above expression. In units of hbar = c =
1, the "static" E field amplitude of a photon (if we can even talk of
such a thing) would be,
E_0 = +,- w^2/2pi
> I have often pondered this in a vaguely wooly manner in the past, but
> not come up with the answer (perhaps because I am mathematically
> lazy).
Well, perhaps that is the answer above. :-) I'm wondering why no one
has asked how I derived it. I think it could be useful in some
approximations. Also if one might be curious about approximately how
many photons there might be in a given E field (for the expression with
n in it).
Chalky
> E^2 = (hbar w^4)/(4pi^2 c^3)
>
> For the case of a single photon n = 1. �w is omega, angular frequency.
> It is the only variable in the above expression. �In units of hbar = c =
> 1, the "static" E field amplitude of a photon (if we can even talk of
> such a thing) would be,
>
> E_0 = +,- w^2/2pi
[...]
> Well, perhaps that is the answer above. :-) �I'm wondering why no one
> has asked how I derived it.
Well, yes, that would be nice too (I had hoped that anyone answering
this question would also explain why the answer was what it was)
I am also curious about how you appear to have managed to sidestep the
issue of how close to monochromatic the source is. From Fourier
theory, the more monochromatic the source, the more cycles are needed
within each photon wavepacket, which will obviously impact on the
amplitude of the wave.