[yor_on]

Ron, why not take it to the other thread instead?

I'm already 'arguing' as good as I know there, which doesn't say much )
And I'm sure you will get better responses to your equation there as I hope the 'heavy artillery' will roll out to define what that elusive light does and does not

I want answers !!!

Heh.
And ahem

[yor_on]

Sim have you seen this one?
Measuring the Recoil of Photons. That one seems to say that there is a measurable recoil?

But if there is no acceleration? You could use it as a proof of that a photon have a 'source' that it 'propagates' from, if correct too, it seems? But how does something without a intrinsic 'time frame' leave a recoil? That one I'm not sure how to see at all.
==

And yeah, Ron, I agree, so much for discussing 'photons' at the 'correct' thread.
Ah well

[syhprum]

I always understood that a photon gun radiating 300MW had a recoil of 1 Newton, hence the photon drive spaceships beloved by sci fi authours.

[yor_on]

That is when 'bouncing' isn't it?
I never thought of it in any other way when reading about Enterprises warpdrive?
Hmm shouldn't the correct sentence be?

  "Beam me up Scotty, but hey, watch out for that dam*d recoil please.."

That way it's acceptable for me, but if we consider one photon, just 'materializing'?

[JP]

But if there is no acceleration? You could use it as a proof of that a photon have a 'source' that it 'propagates' from, if correct too, it seems? But how does something without a intrinsic 'time frame' leave a recoil? That one I'm not sure how to see at all.

I think the problem is that you're trying to apply classical mental pictures to non-classical processes.  With photons you're dealing with particles (quantum mechanics) that are moving fast (relativity), so classical intuition might not hold.  In this case, it doesn't because the photon has no rest frame (relativity), plus the photon at some point is created by some process (quantum mechanics).  This all together means you don't have to accelerate the photon because it moves at the speed of light always, including when it gets created.  There is no acceleration.

Even quantum mechanics has to obey some familiar rules, though.  You can use conservation of momentum to determine recoil if you know the momentum of the photon.

[yor_on]

So the momentum will be 'ring/sphere' formed at its 'materialization' then?
Or how it is thought to work, I always thought of momentum as a 'force' well, at least pointing in the direction of the photons velocity, but it seems hard to state that the recoil lies before the photon?

Maybe that's more correct way to look at it thinking of it, as something evenly distributed? But to talk about a recoil from something just 'coming into existence' should then also mean that what we call that recoil take place everywhere at that photon it seems. The easiest way i can imagine a recoil in this way is if there would be some sort of 'sticky elasticity' involved in its materialization, like boundaries letting go. Does that make sense?

[Ron Hughes]

If the equation above is correct ( and we know it is ) then what ever emits the photon must have a recoil.

[JP]

So the momentum will be 'ring/sphere' formed at its 'materialization' then?
Or how it is thought to work, I always thought of momentum as a 'force' well, at least pointing in the direction of the photons velocity, but it seems hard to state that the recoil lies before the photon?

Momentum is a vector, so its like an arrow pointing in the direction of the photon's motion.  Momentum of the photon source is a vector pointing in the direction of its motion.  If the source was initially stationary, its initial arrow was zero length.   Since momentum is conserved, after emitting a photon the arrow of the photon plus the arrow of the source have to add up to zero length (meaning that if you align them tip-to-tail and follow them, you end up where you started after traversing both arrows). 

In Newtonian mechanics, change in momentum means that something changed its velocity.  This means acceleration, which by Newton's second law means there was a force.  However, the creation of photons isn't Newtonian mechanics, so the photon doesn't have to experience a force or accelerate.

[yor_on]

"If the source was initially stationary, its initial arrow was zero length. Since momentum is conserved, after emitting a photon the arrow of the photon plus the arrow of the source have to add up to zero length (meaning that if you align them tip-to-tail and follow them, you end up where you started after traversing both arrows)." 

Makes perfect sense, except for one thing. A photon can't be seen as 'stationary', can it?
Or do we allow it a 'instant' inside our arrow, where it is 'stationary' before it starts to move, without accelerating?

[Ron Hughes]

A photon or wave ( http://www.thenakedscientists.com/forum/index.php?topic=34333.0 ) can never be stationary, anytime an electron or proton is moved the change in it's field position will propagate away at C. A photon/wave is only created by the movement of charged particles. If there is another way please tell me.

[JP]

"If the source was initially stationary, its initial arrow was zero length. Since momentum is conserved, after emitting a photon the arrow of the photon plus the arrow of the source have to add up to zero length (meaning that if you align them tip-to-tail and follow them, you end up where you started after traversing both arrows)." 

Makes perfect sense, except for one thing. A photon can't be seen as 'stationary', can it?
Or do we allow it a 'instant' inside our arrow, where it is 'stationary' before it starts to move, without accelerating?

You're perfectly right.  A photon's momentum is never zero, so its momentum can always be drawn as an arrow of non-zero length.  What's key here is that the photon didn't exist initially, so the total momentum was zero initially.  When the photon comes out of the source, in order for the total momentum to stay the same, the source has to recoil in the opposite direction of the photon. 

[Pmb]

I can't get the actual concept or meaning of various forms of energies . Actually what is energy ? My textbook defines energy as capacity to do work.Most of the websites define energy in the same way. But i can't understand any thing from that definition. I also want to know what is work.
Actually, why do we need these quantities?

Thanks in advance.

[MOD EDIT - PLEASE PHRASE YOUR POST TITLES AS QUESTIONS. THANKS. CHRIS]

We've discussed this subject in the recent past so you might want to look up that debate. I created a web page on my website at

http://home.comcast.net/~peter.m.brown/mech/what_is_energy.htm


to support the answer that nobody knows what energy is. There is an e-book online which discusses it too. See "What is Energy?, by Dave Broyles. Basically it' what Richard Feynman said (See my web page for reference for this quote and Feynman's reasoning for it)

It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and we add it all together it gives “28” -  always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

In the former discussion I presented some arguments which I believe render the "Energy is the ability to do work" meaningless.

[Ron Hughes]

It's funny how someone can post a logical irrefutable fact and it means nothing to the discussion underway.

[Pmb]

It's funny how someone can post a logical irrefutable fact and it means nothing to the discussion underway.

I posted a response to the comment what is energy?What are you talking about?

[Ron Hughes]

Pmb, that wasn't pointed at you. Energies lowest common denominator is radiation.Any other explanation requires the use of matter's momentum which is used every day to do work and matter is composed of radiation, E = mC^2.

[lightarrow]

Recoil reduces energy of a photon. Your laser gun has no recoil. Therefore energy of your photons is not reduced.

Ok, but what this has to do with what pushes away a photon from light source?

[lightarrow]

yor, take a photon with E = fh and calculate it's mass from m = fh/C^2. The equation is derived from  fh = E = mC^2. As you can see we could claim mathematically that all photons have mass.

Photons are massless, and that's all, no possibility of opinions here.

[lightarrow]

and matter is composed of radiation, E = mC^2.

Wrong. That's only your idea.

[Ron Hughes]

Then the equation must be wrong? I agree that photons are massless, the equation only shows the energy required to produce that photon.

[lightarrow]

Then the equation must be wrong? I agree that photons are massless, the equation only shows the energy required to produce that photon.

That equation is wrong for photons. The correct one, valid for all particles and regardless if stationary or moving (in SR) is this:

E² = (mc²)² + (cp)²

p = momentum.

For light, in classical EM, E = cp. If you apply it to photons (certainly you will agree on the fact this is allowed) you find:

(cp)² = (mc²)² + (cp)²

=> 0 = (mc²)²

so m = 0.