Why do Photons Attract? Physics Q&A

[vince_jt]

My physics professor assigned an extra credit question that asks, why do two photons attract one another even though they are massless and have no charge. He said it involved quantum physics, but I never have taken quantum physics I have no clue where to start. If some one would point me in the correct direction I would appreciate it.

vince

 

[vince_jt]

could it be something to do with the duel properites. Light being a particle and a wave?

 

[lethe]

My physics professor assigned an extra credit question that asks, why do two photons attract one another even though they are massless and have no charge. He said it involved quantum physics, but I never have taken quantum physics I have no clue where to start. If some one would point me in the correct direction I would appreciate it.

vince

classically, two electromagnetic beams (in flat space) do not attract each other. but quantum corrections do indeed give them an interaction. for each photon can turn into a virtual pair of electrons which annihilate back into photons. this is the box diagram.

 

[turin]

How can they be said to attract each other if they don't have a position? Or am I mistaken? I thought that a photon had Δp = 0, and therefore Δx -> infinity. Is the question really asking about why two beams of light attract each other? I'm confused.

 

[Janitor]

Could he be talking about the tendency of identical bosons to gather into identical states? Bose-Einstein statistics and all that?

 

[LURCH]

How can they be said to attract each other if they don't have a position? Or am I mistaken? I thought that a photon had Δp = 0, and therefore Δx -> infinity. Is the question really asking about why two beams of light attract each other? I'm confused.

You think you're confused; I thought it had been experimentally verified that wto beams of light don't attract each other! Is there experimental data that says they do? Is this actually an observed phenomanon?

 

[Haelfix]

This is actually a very neat result...

2 photons traveling parallel to one another do not attract or know about each other gravitionally.

2 photons traveling antiparalel attract.

From General relativity, light has a contribution to the stress energy content, therefore a photon will feal another photon. But it doesn't give us the actual dynamics, but rather a large scale hint.

Quantum field theory gives us the dynamics, via graviton propagators.

In practise, electromagnetic effects will be much larger.. Mostly coming from one loop contributions.

 

[vince_jt]

not sure

Honestly, I don't know what my professor is talking about using quantum mechanics. Remember I am just in a Phyics 298 course, (basic physics). 2 photons do attract according to my professor. lethe, when you said "for each photon can turn into a virtual pair of electrons which annihilate back into photons. this is the box diagram." I looked up information about the "box diagram" and did not find anything on it. I understand that they are massless implying that there is no gravitational force. And I understand they have no electrical charge in classical physics. Haelfix when you say the "feal (feel?) one another" what is the name of that method? I am interested in learning this material if someone would just point me in the right direction

 

[Haelfix]

I'm going to talk about the electromagnetic interaction here, and use what's known as a feynman diagram.. This is what Lethe is talking about. Think of it roughly as a vacuum process (the quantum vacuum constantly replaces mass and energy and viceversa.. energy becomes mass.. mass becomes energy)

In this main interaction, two photons come close to one another, they exchange a virtual photon carrying momentum, and two photons leave. Now, why do they feel each other? Its partly because, to second order, each of those initial photons, decomposes (as a vacuum process) into an electron and a positron.. those electron and positrons annihilate and become the photon again.. Now, what happens in the intermediate step, if one of those electrons happens to reach over, and find its partners positron and viceversa. All of a sudden you have a charged interaction (you can think of it almost like a van der waal force), even though its smallish in effect, it does contribute to the net probability of the interaction (and remember you can never say which event happens, you can only measure the probability.. the square of the amplitude.. so in principle every possible event has to be calculated and added to the net amplitude). So this vacuum process outputs a net tugging.. And it turns out, if you keep going like this (thinking of all possible vacuum processes and summing over them all), you will see that the amplitude MUST output an attractive force..

 

[vince_jt]

thanks for the info

vince

 

[TeV]

This is actually a very neat result...

2 photons traveling parallel to one another do not attract or know about each other gravitionally.

2 photons traveling antiparalel attract.

.

Hm,maybe I'm missinterpreting you here but from the stadpoint of stationary observer in GR , two parallel concentrated beams of light should produce gravitational field and attract each other regardless of their propagating directions.Refering just to non-quantum interaction in framework of GR I believe the amount of gravitational interaction could be even calculated in weak field aproximation (?).Never saw sombody did it though..

 

[lethe]

In this main interaction, two photons come close to one another, they exchange a virtual photon carrying momentum, and two photons leave.

photons cannot exchange virtual photons, they can only exchange virtual fermions.

 

[lethe]

lethe, when you said "for each photon can turn into a virtual pair of electrons which annihilate back into photons. this is the box diagram." I looked up information about the "box diagram" and did not find anything on it.

i scoured the web with google, and found a picture

this is a purely electrodynamical effect. no gravity necessary.

I understand that they are massless implying that there is no gravitational force.

No, according to Einstein, anything with energy gravitates, and photons have energy, even though they have no rest mass.

And I understand they have no electrical charge in classical physics.

right. so this is a second order quantum effect. a classical electromagnetic beam will not attract another one classically via an electromagnetic effect (although it will do so gravitationally)

Haelfix when you say the "feal (feel?) one another" what is the name of that method? I am interested in learning this material if someone would just point me in the right direction

Look up photon-photon scattering. it has been observed in the laboratory. i assume that the gravitational interaction between photons is many many orders of magnitude too small to be seen, so while it exists in principle, the box diagram will drown any signal out.

 

[Haelfix]

Yes, apologies, ignore what I wrote above... Silly brainfart at 6am in the morning. The box diagram is what is important for the interaction. And again, you should see intuitively why this is so, since the fields are being decomposed into charged pairs that should in principle know about one another electromagnetically (which is what I really meant and was describing above). One caveat though, this effect is highly suppressed for low energy photons energetically.

In the case of graviton exchange, what I wrote a few posts up is correct however. Parrelel photons don't feel one another, anti parralel photons do. The nature of this quantum mechanically is a little technical, see A Zee's book quantum field theory in a nutshell for a good introduction to this effect. You can derive this result using GR as well, if you write down appropriate worldlines.. I believe one of Einstein's students discovered this first, look up Tolman-Ehrenfest-Podolsky.

 

[TeV]

.
You can derive this result using GR as well, if you write down appropriate worldlines.. I believe one of Einstein's students discovered this first, look up Tolman-Ehrenfest-Podolsky.

Interesting.Intuivtively,I would say that long parralel beams would gravitationaly interact with each other too.They should induce gravitational field force perpendicular to the stationary observer.Hence,my thinking was their mutual attraction as well.

 

[ranyart]

Interesting.Intuivtively,I would say that long parralel beams would gravitationaly interact with each other too.They should induce gravitational field force perpendicular to the stationary observer.Hence,my thinking was their mutual attraction as well.

The distance of two parralel photons is what gives observational results. For instance if one was to be directly behind the photons(at source) then the distance of travel for the photons moving away will result in the observer 'seeing' the two parralel beams converging at the maximum observational horizon?

In simplistic terms we can imagine two ships (A+B) leaving a shore, whilst mainting a constant distance from each other, say 100metres. Now if there was no curvature of the Earth, and imagine that the ocean extends to infinity, any observer maintaining observation of A +B would see the ships collide at the maximum distance of observation.

Now if we could travel along overhead with the Ships, say at 100metres above (Angel-One), and could relay our 'constant observation' of the event below, to the shoreline observer (Angel-Two), there would be a paradoxical dis-agreement of observations at the distance of Maximum Horizon! The passenger ,Angel-One would have to convince Angel two that what he is seeing does not conform to Local Reality?..as observed by Angel-One.


The local Reality of near field experimentations within QM's, conflicts with Local Relative Observational Relativity, this is what Relativity means.

P.S I've left out obvious Lorentz Transformations for the two-ships and Angel-One, these are scale dependent and cannot be observed at Near or Far away location (Angel-One would be contracting to scale with the two Ships and would not notice anything, Angel-Two would be beyond an horizon at a far off location, and so is out of the Equation alltogether).

 

[TeV]

The local Reality of near field experimentations within QM's, conflicts with Local Relative Observational Relativity, this is what Relativity means.

Weak field aproximation is needed as said.I never did it ,but the Einstein's student who did the calculation first was (looks like) J.A.Wheeler.
Without taking QM effects under the consideration,two long light beams interact gravitationaly always.But only in the case of PERFECTLY parallel ones do not attract.Gravitomagnetic 'force' of the beams in this case exactly cancels out gravitoelectric 'force'.Light is EM wave,and direct Lorentz transform method applied may easily misslead (as in my case).
See:
http://arxiv.org/PS_cache/gr-qc/pdf/9811/9811052.pdf

regards

 

[ranyart]

Hi Tev, Please forgive me if it looks like I was making a direct response to your post, I was simplifying for perspective.

I first come across some reference to: John (baldy) Wheeler about two years ago. I was amazed at some of his work, needless to say I have the utmost respect for him, although I have been waiting for some 'change' of Idea's he was currently complimenting ( as stated in the article I came across).

Thanks for the link, I will add the paper (looks interesting!) to my collection, which stands at about 400+ at present!

 

[gnome]

Can't this question be approached much more simplistically?

I just read the section in my text about gravitational effects on light.

Using the effective inertial mass of the photon m_i = \frac{p}{c} = \frac{hf}{c^2} = \frac{h}{c\lambda} and assuming that the photon has a gravitational mass equal to its inertial mass, this value is used to determine the increase or decrease in energy (shift in wavelength) of a photon under the influence of a gravitational field, which has been confirmed by experiment.

So if we accept that photons have an effective gravitational mass of \frac{hf}{c^2}, wouldn't it be reasonable to expect them to attract each other gravitationally (even if the effect is much too small to measure)?

Let's see. Take two photons of wavelength 300nm, 1 cm apart:
Gmm/r^2 = \frac{Gh^2}{rc^2\lambda^2} = \frac{6.67\times10^{-11}\times(6.63\times10^{-34})^2}{.0001(3\times10^8)^2(300\times10^{-9})^2} = 3.62\times10^{-81}, I think.

Pretty small.

 

[gnome]

So, wuddaya all say about my reply (above)? Could that be what the prof was talking about? Is there any merit to it, or is it just nonsense?

 

[TeV]

So if we accept that photons have an effective gravitational mass of \frac{hf}{c^2}, wouldn't it be reasonable to expect them to attract each other gravitationally (even if the effect is much too small to measure)?

It would.They always gravitationaly interact .And attract in all the cases except in a case of perfect 'parralelism'.Problem is in defining word parralel. Parralel in a flat Euclidian space isn't equivalent to the meaning of the word parralel in a Riemann space.And odd consequence of Riemann geometry describing gravitational interaction of EM beams is a possibility of their traveling along nonconverging geodesics of mutually formed gravitational field.
This is not obvious at all. It's not nonsense you wondered about these things;GR has it's peculiar results.And it doesn't theoretically matter if the effect is too small to measure or not.For change you may take beams of powerfull gamma rays or something ,but there will be always one case of comptelly unattracting beams (which we can call address as parralel)regardless of gravity field strenght they induce in world.
The fact that this effect is first noticed in 1930's (some 15 years after publishing GR theory equations) shows the 'simplistic' approach is not enough here.Check the link few posts up.

 

[TeV]

Let's see. Take two photons of wavelength 300nm, 1 cm apart:
Gmm/r^2 = \frac{Gh^2}{rc^2\lambda^2} = \frac{6.67\times10^{-11}\times(6.63\times10^{-34})^2}{.0001(3\times10^8)^2(300\times10^{-9})^2} = 3.62\times10^{-81}, I think.

To oversimply again,one might ask can we somehow use the formula above (and receipe of inserting photon inertial mass in Newton's gravitational force formula and treating the field sources in a pointy fashion) to calculate photon-photon gravitational "attraction" ?Answer is yes.But the multiplying factors are needed.Particulary:In the case of antiparralel propagation the factor is 4.And in the case of a perfect parralelism (parralel propagation in the same direction) the multiplying factor is 0 (ie. no attraction).A bit of joking doesn't hurt here