Photon "density"

[kirkster501]

So me and you are looking at a quasar a billion light years away say.  You are standing next to me with your scope and you can see the quasar just like I can.  So that means photons emiited by the quasar have descibed a sphere of an almost unimaginable surface area and some of the photons of that sphere hit your eye and some hit mine.  But that means that the photons must have spread out on that journey.  Spread out from only a few hundred million miles.  So what must the "density" of the photons have been like when they were emitted?  Does that make sense?

[ronin]

Isn't it just the normal (1/r) squared law?

r= however many meters or Km or whatever, just need to count how many you get here per unit area. :confused:

Expect a lot of 0's. :grin:

[Kropster]

Yes... the inverse square law.

The surface area of a sphere is proportional to the radius squared.

So, the amount of photons per unit area is proportional to 1 divided by the radius squared.

Doubling the radius gives a quarter the amount of photons etc...

[Tiki]

3C 273 has a magnitude of 12.9 and an absolute magnitude of -26.7. ie. It would be as bright as the sun if viewed from a distance of 32.6 light years.

[Macavity]

Yes (as above!) I think you want some sort of (classical) "Photon flux density" (Watts/m²) ...

rather than (a more obscure) quantum mechanical "density of states" wibbley thing!

If you know the photon Energy (Joules) from Planks Constant x Frequency (Hz)

Remembering Watts are Joules / second... and dividing stuff... You could get:

number of photons per metre squared per second  as a "meaningful number".

As ever, you need to "get the units right" (laddie!) lol

See: http://en.wikipedia.org/wiki/Planck_constant

An exercise for the reader, as they say... 

[Knight of Clear Skies]

One thing to bear in mind is that Quasars do not emit light evenly, most the light is radiated in a relatively narrow jet. So if you performed the calculation above you'd get an overestimate of the Quasar's rate of energy production.

[Kropster]

Getting the units right is important..... I can remember getting things like square seconds at school!

[Tiki]

The sun emits about 10^45 photons/sec. 3C273 is about 10^14 times more luminous than the sun and therefore emits about 10^59 photons/sec.

3C273 is less than a ly in diameter so supposing it to have a radius of 1/2 a ly we can write that its surface area is just pi ly^2.

pi ly^2 = pi*(4.5*10^15) m^2 = 6.4*10^31 m^2

The average number of photons emitted per sqare metre of 3C273's surface is therefore 10^59/6.4*10^31 = 1.6*10^17 photons/sec/m^2. 

These thin out to about 8,500 photons/sec/mm^2  (note the unit change) by the time they travel to earth, with about 100,00 entering the well orientated Mk.1 eyeball every second!!!!

(6mm pupil diameter assumed)