Which is easier to make: light or sound?

Originally posted by sonhouse
I took that quote to mean you are in a dark room and how much energy does it take to cause a visible flash of light. There wasn't much in the way of citations though. I'll have to see where I saw that one. I couldn't find the 2E-17 reference but did find this one, first talks about light in terms of electron volts then has a conversion chart at the end.
ht ...[text shortened]... e packet and inverting that you get 3.1E18 as the # of such packets to = one joule of energy.

1 eV = 1.60217653(14) E-19 Joules - you have to divide by the charge on an electron. To get the energy of a photon in electon Volts we have E = (hc/e)/lambda, Since we´ve got lambda in nm we should multiply by a billion. So replace (hc) with (hc*1E+09/e) = 1239.84 which is how he gets his 1240 figure. So a 635nm photon has energy 1239.84/635 = 1.95 eV.

Direct sunlight has an intensity of 120 W/m² so assuming the pupil is about 12 mm² we´ve got about 10 microwatts hitting the retina. Based on your figure of 3.1E18 photons for a joule this gives us of the order of 10^13 photons per second in direct sunlight. There are of the order of 150,000,000 receptors (wiki page on retina) so that gives us a few tens of thousands of photons per receptor per second in direct sunlight.

Originally posted by DeepThought
1 eV = 1.60217653(14) E-19 Joules - you have to divide by the charge on an electron. To get the energy of a photon in electon Volts we have E = (hc/e)/lambda, Since we´ve got lambda in nm we should multiply by a billion. So replace (hc) with (hc*1E+09/e) = 1239.84 which is how he gets his 1240 figure. So a 635nm photon has energy 1239.84/635 = 1.95 e ...[text shortened]... so that gives us a few tens of thousands of photons per receptor per second in direct sunlight.

Not to quibble but sunlight hitting the top of the atmosphere has an intensity of about 1300 w/m^2 and maybe 1/3 of that hitting the ground makes it more like 450 w/M^2 so there are about three times the amount of photons hitting as you suggest on average. So I must be wrong about that 2E-17 Joules it takes to cause a visible flash. I saw one reference to 3 photons needed to produce a flash of light. Of course that depends on which receptor it hits, because of the striking difference in sensitivity of rods and cones. Maybe they were talking about the amount of energy needed to make a colored flash of light, orders of magnitude higher.

Here is an article about a gamma ray source about 3 billion LY away emitting one of the strongest gamma rays ever seen, 150 billion EV, seems to work out for each packet=2.4E-8 joule. I think an erg is one ten millionths of a joule so that makes each packet = .24 erg. Quite a wallop for one rf packet!
http://www.sciencedaily.com/releases/2009/03/090305150921.htm

Originally posted by sonhouse
Not to quibble but sunlight hitting the top of the atmosphere has an intensity of about 1300 w/m^2 and maybe 1/3 of that hitting the ground makes it more like 450 w/M^2 so there are about three times the amount of photons hitting as you suggest on average. So I must be wrong about that 2E-17 Joules it takes to cause a visible flash. I saw one reference to ...[text shortened]... about the amount of energy needed to make a colored flash of light, orders of magnitude higher.

The Wikipedia page on sulight helpfully puts that 120 W/m² figure in the introduction so I latched onto it. Closer inspection reveals that it is the minimum illumination for light reaching the ground to count as sunlight - you´d have thought that coming from the sun would a sufficient criterion, but there you are.

The other problem is that that the 1,300 figure includes all frequencies (I assume) so we´d need to find the proportion of light in the visible range (not that hard, but enough of a pain of a calculation I don´t want to do it) to get a good figure. Also, the eye is kind of lossy, so you have to discount some fraction of the incoming light as it is just going to be absorbed by non-photosensitive cells.

I don´t know where I first heard that rod cells were sensitive to single photons. It could be one of those scientific urban myths like the notion that the Corriolis force is responsible for water in your sink always going anti-clockwise (it doesn´t - other factors dominate). Someone writes it the first time and it just gets uncritically copied.

What I think is true is that it takes less energy to get a response out of the eye than the ear, just because of the amount of material you have to move around to detect sounds.

The littlest quantum of light is one photon. But a photon's energy can be arbitrary small ( yet > 0 ) depending of wavelength, right?

The littlest quantum of sound does not exist. (Is there such a quantum, called a sonon?) It can be arbitrary small ( yet > 0 ). Sound is a kinetic vibration. Even the smallest atom vibrates. You have to cool it down to zero Kelvin to still an atom and then other effects kicks in.

So both the energy of light and sound can be arbitrary small ( yet > 0 ) so, according to me, the question has no meaning.

Am I right? (I might be.) If not, please tell me where I reason in a wrong direction.

Originally posted by FabianFnas
The littlest quantum of light is one photon. But a photon's energy can be arbitrary small ( yet > 0 ) depending of wavelength, right?

The littlest quantum of sound does not exist. (Is there such a quantum, called a sonon?) It can be arbitrary small ( yet > 0 ). Sound is a kinetic vibration. Even the smallest atom vibrates. You have to cool it down to z ...[text shortened]... aning.

Am I right? (I might be.) If not, please tell me where I reason in a wrong direction.

It's not how little energy phonons or photons can have, it's how much energy does it take to produce a notable reaction to the eyes or ears.

Originally posted by sonhouse
It's not how little energy phonons or photons can have, it's how much energy does it take to produce a notable reaction to the eyes or ears.

In the original question "Which is easier to make: light or sound?", nothing was said about "a notable reaction to the eyes or ears".
Sorry to say, I misunderstood the critieria that we have to 'produce' the light or sound - which is easiest to produce?

However, the preacher was wrong, whatever he said. 😉

Originally posted by sonhouse
The energy of one of those famous 635 jobs is about 1E-31 joules, so do the math.

I found where I got the 635nm figure from. It corresponds to a transition of Silicon at 635.5nm found in the absorbtion spectrum of type Ia supernovae, but absent from type Ib and Ic supernovae. Strange that the figure stuck in my mind, but the context was completely lost.

http://en.wikipedia.org/wiki/Type_Ib_and_Ic_supernovae