# The Draper point and why can't we see human radiation

• jaumzaum

#### jaumzaum

I was wondering how many visible photons humans emit in a day. I did all the calculations with the Planck's Law (it's quite easy, because in the visible spectrum and for temperatures above 0 degrees the exponential term is much larger than one, and the "-1" can be taken out of the integral, leaving us with an integral of the form):

$$P=\int_{F_{min}}^{F_{max}} A f^3 e^{-B f} \,df$$
$$P=-A e^{-B f} (f^3/B+3 f^2/B^2 + 6 f/B^3 + 6/B^4)$$

Where:
$$A= \frac {\epsilon 2hS \Omega} {c^2} = 7,06 \cdot 10^{-55} (SI units)$$
$$B =\frac{h}{ k_B T} =1,55 \cdot 10^{ -13} (SI units)$$

Where:
epsilon = emissivity of human body (considered to be 0,96)
P = Power emitted in the visible spectrum
h = Planck constant
c = light speed
kb = Boltzmann constant
f = frequency
T = temperature of human body (considered 36°C)
Fmin = minimum visible frequency assumed to be 400THz
Fmax = maximum visible frequency assumed to be 800THz
S = superficial area of human body (assumed to be 2 meters squared)
Omega = solid angle of our 2 pupils (assumed to have 4 mm each one), assuming a distance of 2 meters to a point source

Integrating I get:
$$P = 3,1 \cdot 10^{-23} W$$
If we consider the average visible photon to have 600 THz = 4E-20 J:
$$n = 7,8 \cdot 10^{-7} photons/second$$

With this calculation we find that in a day an observer in a distance of 2 meters to an average person will perceive one visible photon once in a 2-week period (very low).
But this is the perceived photons (the one that arrives in the pupils). If we take the solid angle to be 4pi, the emitted visible radiation of a human body is found to be 0,4 photons/second, which is 33600 photons a day.

Ok, this is more interesting. The real question is: how many photons per second can wee actually see?
In this experiment:
https://math.ucr.edu/home/baez/physics/Quantum/see_a_photon.html
They find that in a completely dark room, 60% of the people will answer that they indeed saw something when a flash of 510 nm and about 90 photons arrives in their eyes. It's considered that only 10% (9 photons) get to the retina. So we need to have 90 photons in each synapse period, which is about 1ms from the 2 way round trip. If we substitute that in the calculations (and change our pupil size to 8 mm in the completely dark, and the emissivity to be one) we find that we could see objects with a temperature of 225°C.

But this does not seem to be true. Actually far from true. The Draper point in defined as the point where objects begins to glow, and it is 525°C.
It may seem that this gives some margins to the assumptions (i.e. size of the pupils, emissivity, synapse time...) but I would accept a 50°C change, not a 300°C. If we input this temperature in the calculations we find 317 million visible photons getting in the eye each millisecond, 3 million times more photons than we needed, an error of 300000000%.

I can't wee why the Draper point is much higher than the one I calculated, can someone help me?

I'm just wondering, but aren't you assuming in your calculations that all photons are emitted radially away from the object? In reality every small surface area will radiate evenly into a hemisphere. This would mean that there are much less photons directed towards your eye.

It is very difficult to follow this somewhat stream-of-consciousness message. But if your question is why we don't see an object putting out a photon every 3 seconds or so when the eye's threshold is 9, it's because it takes half a minute to get to 9. The eye doesn't "remember" that it saw a photon half a minute ago.

• sophiecentaur and jaumzaum
epsilon = emissivity of human body (considered to be 0,96)
In the infrared. I expect it to be lower for visible light.

• sophiecentaur and jaumzaum
I'm just wondering, but aren't you assuming in your calculations that all photons are emitted radially away from the object? In reality every small surface area will radiate evenly into a hemisphere. This would mean that there are much less photons directed towards your eye.
That’s why I used the solid angle for the retina. I considered our retina to be 4 mm, so it’s area will be 2pi*4^2=100 mm2. That gives as a solid angle of 2,5E-5 setereoradians.

In the infrared. I expect it to be lower for visible light.
I could not find that information on internet, but even though it’s lower, it’s not 300 million times lower. I cannot spot the assumption (or calculation mistake) that made the outcome be 300 million times smaller than the real result.

It is very difficult to follow this somewhat stream-of-consciousness message. But if your question is why we don't see an object putting out a photon every 3 seconds or so when the eye's threshold is 9, it's because it takes half a minute to get to 9. The eye doesn't "remember" that it saw a photon half a minute ago.

Actually the real question was about the Draper point. In my calculation (assuming the time between two messages to the brain is 1ms, which is twice the value of an average synapse) we find that a point object with emissivity one at 225 Celsius degrees emits 90 visible photons per milisecond to the two 8 mm radius pupils of a person that is 2 meters from the object (could you confirm if this calculation is correct? ). Considering that we can see 90
photons per synapse, then we would be able to see this object. So why we consider the visible temperature (Draper point) to be 300 Celsius degrees above this?
Thanks

I presume the eye has its own noise temperature, which must be at least 300K, so an object, to be visible, would need to be somewhat higher than this.

I have remebered that on a very dark night outdoors, away from starlight in a forest, I can see a sparkling background of noise, like a low light camera. I wonder if others can see this?

I do not wish to reproduce your calculation. It does have the following flaw: the human body is not a point source and so the emission from it does not focus to a spot on the retina (as would the image of a star) The photon threshold you quote is, I believe, for a localized photoreceptor cell and so must correspond to a localized source focused on the retina. Somebody's fat... um...forehead 1 meter away does not qualify (by a lot). So your calculation otherwise may be correct...there are 100 million photoreceptive cells in the retina.

There you go...good question.

I do not wish to reproduce your calculation. It does have the following flaw: the human body is not a point source and so the emission from it does not focus to a spot on the retina (as would the image of a star) The photon threshold you quote is, I believe, for a localized photoreceptor cell and so must correspond to a localized source focused on the retina. Somebody's fat... um...forehead 1 meter away does not qualify (by a lot). So your calculation otherwise may be correct...there are 100 million photoreceptive cells in the retina.

There you go...good question.
Thanks! That is indeed a very good explanation. So, you think the 9 photons must reach the same photoreceptors (or photoreceptors near apart) to be visible? That makes sense, and would indeed explain the 300 million error in the result.

• hutchphd
Thanks! That is indeed a very good explanation. So, you think the 9 photons must reach the same photoreceptors (or photoreceptors near apart) to be visible? That makes sense, and would indeed explain the 300 million error in the result.
So, for an object extremely small (a coin for example) you think 225 Celsius degrees would be visible?

Actually I think the object needs to be even smaller than this. Much smaller. I will do some calculations.

Humans may not be able to 'see' each other but I know of snakes that can 'see' their prey with specially adapted 'pits'. So perhaps the OP could feel that the idea is not actually that unworkable (I'm suspecting that a SciFi plot could be somewhere in here).
I'd bet that a snake couldn't read a newspaper under those night time conditions.

• jaumzaum
I'd bet that a snake couldn't read a newspaper under those night time conditions

Even though they are so good at it at noon?

• • Ibix, sophiecentaur and nasu
Even though they are so good at it at noon?
How can you be so judgemental about reptiles?

I was hinting at the sort of resolution the snake's system might have. The prey's temperature would be not much higher than the background and the sensor is more or less just a 'tube'. I would bet that distinguishing food from background would be based on movement characteristics of a little blob, rather than optical recognition. A warm piece of dung could look more or less the same as a mouse. (Remember Jurassic Park - would that be a good enough citation I wonder?)

• jaumzaum