'faint' photons and photon energy

[nytecam]

Perhaps someone can answer this for me? I appreciate the Xray photons have a higher energy than visible light and radio wavelength photons have less energy.

However In a recent SGL post I wrote faint photons that I quickly corrected for scarce photons eg do photons, at a given wavelength, have a common and equal 'energy' ? I assume that when taking a DSO image we expose for longer to capture photons that arrive at irregular or infrequent intervals and not because the photon stream is constant but weaker

[username]

Aren't you just referring to the flux of photons. You expose any image for longer if the amount of photons per unit time at the area of detection is small compared to if it's large.

The stream of photons will always be constant if the conditions are the same. Just less of them.

[michael.h.f.wilkinson]

For a given wavelength a photon has a given energy. When we talk about light intensity we are really talking about the average number of photons emitted or received per unit of time. Since emission and detection of photons are random processes, the actual number detected in a unit of time fluctuates randomly around that average (this is what we call photon noise).

[username]

Sorry I missed the first bit, at constant wavelength they do have same energy. You expose for longer due to the mount of them per time per area. As a DSO is very far away and radiates for arguments sake isotropically the amount of photons reduces by the inverse of the square of the distance but the flow of photons to your detector may be infrequent but will be regular.

[andrew s]

The energy of a photon is directly proportional to its frequency and so related to it's wavelength via speed © = wave length x frequency. For a faint object the average rate of arrival of photons of a given energy is low and subject to poission statistics (for visible light photons) so for an average rate of n photon per second you get a standard deviation of sqrt(n) in arrival rate.

Hope this helps Andrew

[michael.h.f.wilkinson]

The energy of a photon is directly proportional to its frequency and so related to it's wavelength via speed © = wave length x frequency. For a faint object the average rate of arrival of photons of a given energy is low and subject to poission statistics (for visible light photons) so for an average rate of n photon per second you get a standard deviation of sqrt(n) in arrival rate.

Hope this helps Andrew

Not just visible photons, all photons have Poisson noise. (technical point: If you collect a sufficient number the Poisson distribution starts to look like a Gaussian, so in radio astronomy, people often assume the noise is Gaussian, which is close enough)

[andrew s]

Not just visible photons, all photons have Poisson noise. (technical point: If you collect a sufficient number the Poisson distribution starts to look like a Gaussian, so in radio astronomy, people often assume the noise is Gaussian, which is close enough)

That is not quite right. Photons are in fact Bosons and conform to Bose-Einstein statistics which at visible frequencies approximates to Poisson statistics. At low enough energies you get correlations in the arrival times of the photons which leads to them bunching together and arriving in bursts.

Regards Andrew

[nytecam]

That is not quite right. Photons are in fact Bosons and conform to Bose-Einstein statistics which at visible frequencies approximates to Poisson statistics. At low enough energies you get correlations in the arrival times of the photons which leads to them bunching together and arriving in bursts. Regards Andrew

On recollection that was my thought too

[J_M_Franklin]

Perhaps someone can answer this for me? I appreciate the Xray photons have a higher energy than visible light and radio wavelength photons have less energy.

However In a recent SGL post I wrote faint photons that I quickly corrected for scarce photons eg do photons, at a given wavelength, have a common and equal 'energy' ? I assume that when taking a DSO image we expose for longer to capture photons that arrive at irregular or infrequent intervals and not because the photon stream is constant but weaker

There is a simple explanation. All electro-magnetic radiation spreads out from the source as a result of the inverse-square rule. Thus for every doubling of the distance the flux (number of photons) hitting a given point drops by a factor of 4.

Example:

A source emits photons that hit a target 1km away at a rate of 100,000,000 photons per cm² per second

Double this distance to 2km and the rate drops to 25,000,000 photons per cm² per second

At 4km the rate drops to 6,250,000 photons per cm² per second

At 8km the rate drops to 1,562,500 photons per cm² per second

At 16km the rate drops to 390,625 photons per cm² per second

Do you see? So to receive a given number of photons per cm² you need to expose your receiver for a time proportional to the number of photons you will receive at your detector.

Thus using the above figures, to receive the original number of 100,000,000 photons per cm² and your detector is 16km from the source, you need to expose it for 100,000,000 / 390,625 = 256. Thus the detector needs to be exposed for 256 seconds to receive the same number of photons it would receive at 1km per second.

The energy of the photons is totally irrelevant to the detection time, only the detection method. The inverse square rule works for radio, microwave, Ir, Visible, UV, X-ray and gamma ray photons and the above example is true to for all.

Hope this helps.