Ethics of Photoediting

[Maris]

A photograph, made with film, is a sample of the real world? What a bizarre notion!

Actually yes. Do the sums. A 8x10 sheet of 100 ISO film gets about 10 ^ -23 Kilograms heavier at the moment of exposure. All that extra mass came from the subject matter and nowhere else. That's why camera are made light-tight.

 

[keithwms]

LOL

Gosh I love physics humour.

 

[CGW]

Well I think you're making this about an entirely different subject. Sure, we compose and that does have the essence of 'pose' in it- we select what we want to have in the photogrpah and thereby influence how the scene is perceived. fine.

But manipulation of the sort described in the article is at a very different level

Manipulation to alter perception is, well, still manipulation. It's really not a matter of degree or kind. Composing, cropping, and posing are all manipulation. What about the elephant that Roger Fenton cropped out of his famous Crimea shots of the Valley of the Shadow of Death? Could have been one there, right? Still think you're also being willfully naive about the degree of manipulation in pre-Adobe photography.

 

[cliveh]

Actually yes. Do the sums. A 8x10 sheet of 100 ISO film gets about 10 ^ -23 Kilograms heavier at the moment of exposure. All that extra mass came from the subject matter and nowhere else. That's why camera are made light-tight.

Can you reveal the source of this calculated evidence?

 

[Maris]

Can you reveal the source of this calculated evidence?

I'm the source. The calculation is not particularly difficult although it requires some knowledge of basic physics. The number is of an "order of magnitude" accuracy because it involves recasting photometric quantities into energy units and assumes some values for a "typical" film. Annoyingly, I can't find a reference to anyone else having done the sums. I'd appreciate it if you or someone else on APUG would run the numbers and check my answer.

 

[keithwms]

I'm the source. The calculation is not particularly difficult although it requires some knowledge of basic physics. The number is of an "order of magnitude" accuracy because it involves recasting photometric quantities into energy units and assumes some values for a "typical" film. Annoyingly, I can't find a reference to anyone else having done the sums. I'd appreciate it if you or someone else on APUG would run the numbers and check my answer.

Haha sounds like something I'd inflict on my poor students!

First of all, let's be clear: photons don't have mass so the argument that they deposit mass would be incorrect.

However, photons do carry a momentum, namely p=E/c, and so you could pick ~532 nm as your center wavelength and that is, what, 2.3 eV energy, roughly.

So then the argument would be that the impulse F*t equals the change in momentum imparted by the photons. Each photon contributes a momentum of roughly 2.3 eV/c or 4x10^(-19) kg m^2/s^2 divided by 3x10^8 m/s, so let's say ~10^(-27) kg m/s . Then in one second you get an effective force of, what, 10^(-27) Newtons. You set that "weight" equal to m*g and use g= 10 m/s^2 to get a mass of something like 10^(-26) kg. Then you say you get a thousand photons deposited on your film in a second, and you arrive at 10^(-23) kg.

 

[michaelbsc]

I dare say that a thousand photons is a bit shy of reality in a typical exposure.

I didn't check your math, though I have no reason to doubt the accuracy.

But I would expect the number of photons to be closer to 10^7. That's just a wild guess on my part.

 

[nhemann]

Now this is just getting amusing - what were you all talking about?

Another option would be to caclutale the total change in energy of the the atomic bonds on a pre and post exposure film and then relate that to its mass/energy equivalent.

Just don't forget to include the weight of the soul you captured.......
n

 

[cliveh]

I'm the source. The calculation is not particularly difficult although it requires some knowledge of basic physics. The number is of an "order of magnitude" accuracy because it involves recasting photometric quantities into energy units and assumes some values for a "typical" film. Annoyingly, I can't find a reference to anyone else having done the sums. I'd appreciate it if you or someone else on APUG would run the numbers and check my answer.

Did you do it on your bathroom scales?

 

[keithwms]

Oh I don't know how many photons hit the film offhand, Michael, I was just using whatever I remember. To be more correct, one would need to specify the light source intensity, aperture and bellows factor etc.

But the estimate is just a light-hearted joke (I hope); there is no mass gained by exposure to photons. Whatever mass change actually results is a loss of mass during development.