# Stargazing A new telescope can detect a candle across the Atlantic

1. Sep 12, 2009

### map19

Publicity often says that a new telescope can detect a candle across the Atlantic, etc.
Well, a candlepower is defined as 1/683 Watts at 540 X 10^9Hz.
This gives photon energy at 2.2 eV.
One candlepower produces 4.2 x 10^15 photons/s radiated isotropically.
At 3,000km the area of the sphere is about 10^14m^2
So the photon density at the telescope is 4.2 x 10^15/10^14 or about 4 photons per sq meter per second.
Two things to note:
The associated electromagnetic field would be very small, but not zero.
A second is a long time in quantum matters.
My questions are: How do you demonstrate the area covered by a photon at the telescope mirror ?
How long is the photon wavetrain ?
(I can’t believe I’m asking this) does the field collapse to the photon particle ?
Is this like the photoelectric effect where a classical calc shows that 10 minutes or so would be required to dislodge the electron, where in fact it occurs in about 10^-9s.

2. Sep 12, 2009

### f95toli

Re: photons

The "practical" answer is that you can "pretend" that your photon is a classical EM wave up until it hits the actual detector (which in real-life is always a farily small device; usually just microns in size); i.e the mirror/antenna or whatever you are using (depends on the wavelength) to focus the radiation is just designed using standard optics/engineering and only the bolometer (or whatever type of detector is used) needs to be designed using QM.

3. Sep 13, 2009

### map19

Re: photons

No need to 'pretend' you can measure the electric and magnetic fields to show it is an EM field.
Bohm has demonstrated that there is no 'collapse' and his ontological description of boson fields does not imply boson particles, such as photons.
for an electron to be raised to an excited state by the field by interaction, the whole system - field plus atom - enters a new state in which the quantum of energy has gone into the atom. Somehow the quantum of energy has been gleaned from the whole field.
The field has continuous coverage and you can not specify that it is in packets before it reaches the atom.
That's why I gave the example of the photoelectric effect.

4. Sep 13, 2009

### f95toli

Re: photons

That is not what I meant.

My point was that even for a single photons the classical EM equations seems to work well when designing antennas and optics. Note that this is not a "trivial" result; a full QM (QED)description of e.g. a waveguide will give you a result that is in general NOT identical to what you expect from classical physics.
However, for a number state it turns out that the classical equations and QM agree (although there might be some higher order corrections; I don't remember the details), which is why we can use our normal "toolbox" of techniques and equations for just about everything except the detector itself.

What interpretation you use is irrelevant; that the classical equations are a good approximation (or in some cases identical) to the full QM model is a mathematical -not a philosophical- result.

5. Sep 13, 2009

### map19

Re: photons

Agreed. But this still leaves me frustrated.
I want to know why, let's use the photoelectric effect, energy from the field can appear within the orbit of an atom and trigger excitation within a nanosecond, when the classical calc says that 10 minutes of continuous exposure would be required at the quoted field strength.
And, in general, if you quote photons only, see QED by Feynman, how long is the wavetrain. it has to be at least a few cycles.