What Happens to Light Energy?

It radiates outwards into the universe for ever. Unless it is stopped by an opaque medium of some sort*...

Of course the intesity diminishes rapidy as the energy is spread out over a greater and greater wave front. Think about ripples on a pond, and extrapolate it into 3 dimensions.

* E.g. The living room curtains

So if you point a torch (flash light) up to the night sky and flick it ON the OFF that pulse of light will travel outwards. If you switch the torch on for 1/10 of a second.... the pulse of light will be about 3x10⁷ metres long .

Del

Of course the intensity diminishes rapidly as the energy is spread out over a greater and greater wave front. Think about ripples on a pond, and extrapolate it into 3 dimensions.

Uhhh....aren't the ripples in a pond already occurring in three dimensions?

It leaves really, really fast!

Del's right - it either diffuses or is absorbed (transfers its energy) by an opaque object.

Hypothetical next question: "But then where doe it go, ultimately?"

When it is absorbed by matter, it increases the temperature of that matter causing it to then be re-radiated at a longer wavelength where it then travels until it is absorbed by a colder body that warms and re-radiates it at a longer wavelength, and so on and so on down the arrow of entropy until all energy is uniformly distributed throughout the universe and then it gets really dark and the devil comes and plucks out your eyes.

My God I love it when science and insanity run smack into each other! Now I'll have nightmares about very long wave lengths and the devil plucking my eyes out!!! Thank you very much!!!

Freddie from design

Are we talking about coherent light or incoherent light?

White light or incoherent light spreads out into different directions and dilutes itself to the point that as soon as you secure the source it dissipates into nothing.

Coherent light on the other hand will travel in a straight line until it is absorbed.

Just think, it takes eight minutes for the light from the Sun to reach us.

We all seem to miss the possibility that light could be a form of matter such as photons. These photons contribute to the changes that take place in solid materials when they are hit by the light such as weathering, discoulouring and photo sensitive transformations. These processes all must take up some amount of energy as well so that the light, when reflected or refracted, will have less energy left after the event.

Eventually though it is all dispersed and diluted which is a form of redistribution if you will. The energy increase the object may have suffered due to the impacting light will eventually be sucked into a black hole or a giant gas star and burn again, just to be repeated over and over again. No energy ever gets lost.

Wait a minute. Are you all saying that light is energy, and not just light? I thought that the energy input to an illumination devise dissipates for the most part as heat, which is energy. Please clarify for me.

Yes. Light is energy. It is part of the spectrum of electromagnetic radiation. Heat is also energy; thermal radiation. Heat output + Light output = Electrical input. That is the very basis of the First Law of Thermodynamics.

JMM and CSM,

Thanks for that information--all of it including the last part about different kinds of light producers. People speak of "visible" light, as if there were invisible light, but I learned (if I may use that term) somewhere that in actuality visible light is not actually visible in the sense that it can be seen; it can only be observed as a light source or light reflector as you pointed out. For this reason I was told that a visible beam of light, say from a flashlight, can only be seen if there is dust or moisture in the air. But a couple of other things still puzzle me.

CSM's equation, energy in = heat energy out + light energy out, suggests to me that the light energy contains no heat and will add no warming to anything it strikes. Is this correct? If so, then what is it besides light that causes warming of the earth by the sun during daytime. Would it be some part of spectrum, like infrared only, other than the light portion of the spectrum? Thanks.

The "light" that hits the objects, are not only the visible kind. All the other forms of emitting radiations that the source produces, is also hitting the object. Most of the heat will come from these other forms of radiation as they are better at exiting the atoms of the solid object, which is exactly what heat is. Heat is nothing more than a frantic movement, or vibration, of the atoms. If any light makes the atoms vibrate, which we call excitation, the solid item heats up.

This also explains why we always loose some efficiency in transformation of energy from one form to another. The movement of atoms, the heat, is wasted kinetic energy as you can rarely get this back or perform another task. The moment the heat disperses into the space around the object, it is lost as waste but that does not mean lost to the overall picture. It only means our intended process lost it.

As for your first point, you can see the source but you cannot see the beam unless it is hitting some dust. This is correct but it also misses an important point. When your eyes sees the light, what exactly does it see?

It sees light particles, either direct from source or reflected, that enter your eye and which is converted by your brain into an image. You can see the light as that is the image.

Thank you all and it's becoming quite clearer so you all give each other some good answer points, please. As I now think I understand, any heating that might occur by excitation by visible light would depend on what was hit by the visible light and also the other radiations that accompany the visible light. Looking at the previous answers, it seems to me also that light source brightness and color would be kind of an opposite of light reflector. If a brighter, whiter light emitter is hotter, then a reflecting object that is visibly brighter and whiter must be cooler than a dark object because it is reflecting and not absorbing the light that is striking it? This part still has me stumped a little. I know that Venus can be very bright but its very warm compared to other planets farther from the sun.

BTW, you guys seem like very good teachers. Are you teachers?

Guest,

I am not a teacher by profession or job title, but I find that every day, when I work with people, some teaching and some learning both happen. Your post mentions many topics.

You ask: "If a brighter, whiter light emitter is hotter, then a reflecting object that is visibly brighter and whiter must be cooler than a dark object because it is reflecting and not absorbing the light that is striking it?" Let me lead you to a better answer. Take two objects that are identical except for the color, with one being white while the other is black. While you are holding them and carrying them to a place in the sun you will notice that they feel the same--neither one is hotter or cooler. Leave them in the sun a few minutes and then feel them. The black one will be much warmer to touch than the white one. Reason: the black one is absorbing nearly all of the light which reaches it, while the white one is reflecting most of it.

Physics people have an equation: E=hν. In this, E is energy, h is a constant value named Plank's constant because he first figured this equation out, and ν is actually the Greek letter "nu" (our letter equal to this is "n") which is the frequency of the radiation. This equation describes the energy content to be found in all types of electromagnetic radiation. This is true of the portion we see as visible light, and the portion we feel as heat, and the portion we describe as X-rays, and the portion we describe as radio waves, and all other portions.

Venus is much closer to the sun, so it gets more energy from the sun the same way you would get more energy from a fire by being closer to it. For many years scientists assumed that the thick cloud cover over Venus kept it from being very hot. They were very surprised when the first satellite to reach Venus melted in seconds. After a number of attempts, Russian and USA scientists got satellites to survive longer, and measured a surface temperature of about 470 deg-C. They also found a very high concentration of carbon dioxide in its atmosphere (over 100x what is on the Earth). Hmm.

Got to go--JMM

"Reason: the black one is absorbing nearly all of the light which reaches it, while the white one is reflecting most of it."

True, true, but what of the light energy that does not strike either one of them? Since photons from the light source (whatever it may be) emanate spherically in all directions what then? In other words, a very small part of the Sun's EM energy is impacted by the local planets, asteroids, etc., but "What happens to Light Energy?" as originally posted, by the majority of EM that goes off into never-never land, i.e., the void of deep space?

Then taking it one step further, what about the EM that collides with EM from other light sources (stars)? Surely they will multiply is they collide "in-phase" and the reverse, etc. But, again, what happens to this energy? Does it just travel around the universe contributing to the total energy contained therein?

Johnjohn,

Profound questions. Now, If light travels in paths that are bent by gravity, is it possible that some of it can curve around so a very distant star may be our own?

Perhaps, except for that nagging one-way arrow of time.

On the question of the energy of EM (electromagnetic spectrum) radiation source, we extrapolate its total energy based on the tiny sample that reaches our detectors, but usually detectors measure a specific designed in frequency, and, in general, ignore all the rest (frequencies).

The original question was regarding Light Energy, so do we conclude that to mean only "visible" light? Your previous example, for instance, applied to heat energy (light and dark objects), which is, of course, invisible to (naked) human eyes, i.e., no thermal imaging in effect. Not to be confused with mirages, etc.

So, what happens to EM energy in the 400nm to 700nm range?

-John

Addendum to #29:

I, like probably a lot of us, used to think about black holes like the Sci-fi writers portrayed them; like a huge, yawning funnel sucking up everything around the funnel's "mouth". However, in an earlier thread, vermin expanded my (usually) limited way of thinking by pointing out that a black hole is spherical by necessity, i.e., the event horizon exists in all directions around it.

I'm just relating this story as it pertains to the spherical radiation of EM- it goes in ALL directions, not just where we happen to "observe" it.

There will indeed be some light of our own sun (who owns it by the way?) that comes back after several bends in space.

But will we see it as a star?

I don't think so, it disappears in the back radiation, the noise.

Yes, even when absorbed and re-radiated at a longer wavelength it must continue on, even if only in the lowly form of heat. Where else could it go, Detroit?

I know, it keeps re-radiating at longer wave lengths until it reaches audio and that's that hum you hear when your eyes are closed: "OM".

After that, it finally becomes the sound of one hand clapping and you are allowed to leave the temple.

LMAO! That's probably the best explanation yet. Can't be Detroit though. That's our closest known event horizon and thus nothing could be re-radiated.

But how could you not hear it grasshopper? -Master Po

It goes to the Fridge.

the energy gets absorbed equal to

E=hf.

h=Planck's constant

f=frequency.

This is first year second semester physics.