Is light visible ?

[timo]

to understand why light cannot "bounce off" light you must think of it the light as a wave...... as waves have no mass, hence no surface area, there is nothing for the light to "bounce" off.

The reason is that photons interact with electric charge. Since photons are electrically uncharged, they don´t directly interact with other photons.

[bombus]

"But please ask a real physicist, because I'm probably lying to you."

 

 

You fraud.

 

You will be hearing from my solicitors.

 

But in the meantime, have another question :

 

The thought experiment continues. You're on your spaceship returning to Earth, when you accidentally hit the "Planetary Destruct" button instead of the retro-rockets, thereby wiping out all life on Earth.

 

Assume no life exists elsewhere in the universe.

 

And sooner or later, you die too.

 

(Sorry, this isn't the most fun thought experiment in town, is it ?)

 

At the precise moment of your expiry, what happens ?

 

There are no longer any eyes for photons to go into.

 

So, does the scene change ?

 

Does the universe become invisible ? In an instant ?

 

In what sense could it be visible, if there are no eyes ?

 

What do you reckon, Cap'n ?

 

Ant.

 

As far as an individual is concerned, once you die a whole universe dies with you. When I die, you will all disappear from existence too, and the universe will end - and you won't be able to prove me wrong! Unless there is an afterlife of course...

[grifter]

The reason is that photons interact with electric charge. Since photons are electrically uncharged, they don´t directly interact with other photons.

 

I was just trying to "dumb it down" for Mr Jones, by relating photons interacting with charge upon a particle to the photons "bouncing" of of particles, it was probabally still a bit wide of the mark even for a "dumbed down" aproach, never mind, eh.

[swansont]

The reason is that photons interact with electric charge. Since photons are electrically uncharged, they don´t directly interact with other photons.

 

But there is an indirect interaction. According to QED photons should have a lower-order interaction with other photons, but this would be very weak and only become noticable when the photon density is very high. IIRC it involves each photon becoming a virtual electron/positron pair and interacting, so the probability scales with photon energy.

[Sisyphus]

I was just trying to "dumb it down" for Mr Jones,

 

Sure you were.

[grifter]

is you calin me dum loike ?

 

No really... I was attempting to make it a bit simpler, since this Mr Jones finds it difficult to grasp the basic principals of reflection I thought introducing photon energy and particle interaction might fry his brain.

 

but, i myself would like some explanation on how virtual electron/positron pairs interact, as I have only a basic understanding... I guess its cool to ask about that in this thread or would it be better to start another thread...

[GutZ]

well, the dictionary does occasionally list the scientific meaning along with other nonscientific meanings. the scientific definition of sound is a mechanical vibration in an elastic medium. the sensory part is not a soundwave therefore should not be included in the definition. just as we do not describe photoresistors in our definition of light because they can detect light.

 

and as klay said, this is the physics forum.

 

The original question is ill-formed then. The part about a human being there or not is irrelevant by all physical means, therefore not really a question, but I dont think that the intent.

 

So to me answering yes would be incorrect in other ways.

[timo]

but, i myself would like some explanation on how virtual electron/positron pairs interact, as I have only a basic understanding... I guess its cool to ask about that in this thread or would it be better to start another thread...

I´m not sure what you mean by "how". The photons interact with a charged field creating a particle/antiparticle pair of charged particles that can then interact with the other photons involved. Schematically, a feynman diagram could look like the one I´ve attached, where the outer wavy lines are the photons (you can pick for yourself which two lines are the incoming and which two are the outgoing) and the inner square is some charged particle (electron, myon, tauon, quark, W) running in a loop (or equivalently two particle/antiparticle pairs if you consider going in opposite direction of the arrow as an anti-particle).

 

[grifter]

Thanks, after your explanation that makes sense, (excuse my poor question).

[Severian]

Essentially.

 

Light does not reflect off of light because photons are massless, so there's really nothing to reflect off of.

 

But please ask a real physicist, because I'm probably lying to you.

 

You are indeed lying. Photons do interact with one another via virtual charged particle loops. This is known as light-by-light scattering. It is very small but it exists.

 

So even in the most literal sense of the word invisible (can't scatter light), light is not invisible.

 

Edit: Bugger. I should read the whole post before replying. Atheist even gave a piccy of it. Bah humbug!

[Cap'n Refsmmat]

Well, that makes me feel better about the disclaimer, then.

[grifter]

we should enforce a mandatory disclaimer policy on all posts, I'm going to add one to my signature right now

 

 

The above post is not necessarily representative of Grifter's views, any views expressed in this message may be inaccurate and/or incorrect.

The above mentioned is not responsible for any actions that may occur as a direct or indirect result of reading the above post

[imp]

This thread has stopped making sense; it's getting too hypothetical for my little abilities. So, here is my input on the original question:

 

Is light visible? NO.

 

Light is a certain wavelength region of Electromagnetic Radiation. That particular region "excites" the surface of objects which we "see", by virtue of those objects reflecting light back to our eyes. If we see it as "white", it's reflecting many different wavelengths, if black it's reflecting very few.

 

Simplistic, perhaps, but easiest way for me to "see" it. imp