# Does a photon have physical volume or geometrical size?

### #1

Posted 12 November 2007 - 11:47 PM

A college student asked the question if a photon has physical volume or geometrical size. Does anyone have a good answer?

From the particle point of view, it should have volume/size; from wave point of view, it should not have volume/size.

A photon travels from a remote star to the earth (Billion years), does not fall apart during the long journey, it should have certain volume/size.

### #2

Posted 13 November 2007 - 12:07 AM

### #3

Posted 13 November 2007 - 06:41 AM

**wave**, or as having a 2d surface that projects at right angles to the

**direction**of travel or movement. There are 2 degrees of freedom for a photon, along this traveling 2d surface (let´s say it looks like a little circle), and there are 2

**components**, an electrical and a magnetic component, which ´resolve´ into a momentum. The wave is said to

**collapse**when this happens. But this is a model, an idea of what a photon being absorbed or emitted ´looks like´.

A 2-d surface can´t have volume so it´s meaningless in such a space, but area isn´t. The

**radius**for photons in this (mathematical) space is never

**more than**the

**same**constant value (but the two components vary sinusoidally about a

**zero point**, so that the area is also cycled this way, from zero to a constant value, and the cycle time, or frequency, determines the

**energy**of a particular photon), this (maximum constant amplitude) appears to be related to its apparent

**velocity**, somehow.

The energy is not related to the distance traveled, unless the photon interacts with another photon (or an electrical or magnetic field, or collides with an electron or other charged bit of matter). In other words you could say that the energy in a photon is bounded by (integrable over) a single period of its cycle (or something similar), like a packetised bit of energy, rather than the integral of all the periods it has cycled through on its journey.

Ex

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### #4

Posted 13 November 2007 - 10:12 AM

Both of these are simply interactions with the atoms in the table, and in fact we never make comntact between the atoms in our fingers and the atoms in the table - they are repelled by electromagnetism before we get there.

So, 'size' is difficult to define. Does it mean the distance at which you start to interact with the object? In that case, and electron for example has infinite size because its electromagnetic field extends forever (though getting very small). A photon is neutral so generates no such electromagnetic charge, but in principle if you get really close you will start to distinguish it converting into electron-positron pairs and back again.

The electrons and positrons are charged but very close together. So it still looks neutral from a distance, but if you get up really close you will start to feel an electromagnetic force.

In other words, the photon does have a size, since you start to feel it pushing back against you if you go to close. This happens at about half a fermi, or roughly m.

### #5

Posted 13 November 2007 - 01:54 PM

I agree with you! And that push back is because of the distance from the photon. It will not push back any more you are maintaining a distance of m (at least they told us this at school:doh: ), because this distance is said to be neutral. And if the distance is larger then the force should normally be attraction!In other words, the photon does have a size, since you start to feel it pushing back against you if you go to close. This happens at about half a fermi, or roughly m.

### #6

Posted 13 November 2007 - 03:22 PM

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### #7

Posted 13 November 2007 - 09:31 PM

I thought that could mean thatA 2-d surface can´t have volume so it´s

meaninglessin such a space

**volume**is the same thing as ´mass´ for a photon. I could however be incorrect (along with a few other guys, like Mr. Einstein, say).

Hang about... If a photon moves through a vacuum and has 2-DOF along a traveling surface, and the two components are like changing vectors, the momentum at any point in time is some area on this 2-d surface. It must be a constant, rather than cyclic area, because it represents the photon´s momentum -the sum of both moments: the electric and magnetic moments. Otherwise how is a photon´s momentum transferred, or conserved? The area described by both moments or vectors must be constant, and depend on frequency -could that make the momentum look like a volume after all... there´s the ´area´ described by the two perpendicular wave fronts, and the frequency could look like a ´depth´, so a shorter cylindrical ´volume´ would be a more energetic photon -the height of the cylinder is inversely proportional to the momentum...? At least when the wave collapses or whatever.

Ex

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### #8

Posted 14 November 2007 - 12:38 PM

The energy and momentum of a photon depend

only on its frequency νor, equivalently, its wavelength λ

and consequently the magnitude of the momentum is

**k**is the directional wave(number) vector =.

Doesn't look like there's any room left for

**volume**. The Wikipedia entry for photons goes on to describe how the momentum is the result of the two moments, or components, that have an angular frequency (just the e and m components). The spin angular momentum is

**independent**of the frequency -it is related to the way a photon propagates (as a wave of EM energy, and the fact it has no mass, but has a

**mass equivalent**like Einstein says).

Wikipedia again:

So the momentum is due solely to the frequency, which means that the area (of the 2-d "wavefront") isTo illustrate the significance of these formulae, the annihilation of a particle with its antiparticle

mustresult in the creation of at least two photons for the following reason[:]

In the center of mass frame, the colliding antiparticles have no net momentum, whereas a single photon always has momentum. Hence, conservation of momentum requires that at least two photons are created, with zero net momentum. The energy of the two photons—or, equivalently, their frequency—may be determined from conservation of four-momentum. Seen another way, the photon can be considered as its own antiparticle. The reverse process, pair production, is the dominant mechanism by which high-energy photons such as gamma rays lose energy while passing through matter.

The classical formulae for the energy and momentum of electromagnetic radiation can be re-expressed in terms of photon events. For example, the pressure of electromagnetic radiation on an object derives from the transfer of photon momentum per unit time and unit area to that object, since pressure is force per unit area and force is the change in momentum per unit time.

**constant**--at least it must be when the photon collapses, or if it varies with time, then somehow it "becomes" its actual value...?

P.S. About the zero-mass of photons: the photon is without intrinsic mass (it doesn't have

**any**at all, how sad). Since it doesn't have

**mass**, isn't saying "a photon's mass doesn't exist" a bit pointless? Especially since energy (which is photons) is

*equivalent*to mass? I think it kind of is, myself...

Ex

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### #9

Posted 14 November 2007 - 02:04 PM

Here's what Wikipedia says about photons and momentum:

kis the directional wave(number) vector =.

Doesn't look like there's any room left forvolume. The Wikipedia entry for photons goes on to describe how the momentum is the result of the two moments, or components, that have an angular frequency (just the e and m components). The spin angular momentum isindependentof the frequency -it is related to the way a photon propagates (as a wave of EM energy, and the fact it has no mass, but has amass equivalentlike Einstein says).

Wikipedia again:

So the momentum is due solely to the frequency, which means that the area (of the 2-d "wavefront") isconstant--at least it must be when the photon collapses, or if it varies with time, then somehow it "becomes" its actual value...?

P.S. About the zero-mass of photons: the photon is without intrinsic mass (it doesn't haveanyat all, how sad). Since it doesn't havemass, isn't saying "a photon's mass doesn't exist" a bit pointless? Especially since energy (which is photons) isequivalentto mass? I think it kind of is, myself...

I don't see how any of this is directly relevant to a spatial extent of a photon.

Transmission and diffraction obviously have a spatial dependence, so I don't see how one concludes the wavefront area is constant.

*Minutus cantorum, minutus balorum, minutus carborata descendum pantorum* **To go to the fortress of ultimate darkness, click the up arrow ^**

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### #10

Posted 14 November 2007 - 11:20 PM

### #11

Posted 14 November 2007 - 11:47 PM

A photon is definied as a volumeless, dimensionless particle having little or no mass.

No mass, they are massless, this is important.... and their spacial dimentions are undeffined IIRC, as swansont says if you try and put a photon through a subwavelength conductor it don't work...

Of course you can make subwavelength slit arrays in conductors and find some interesting effects...

### #12

Posted 15 November 2007 - 04:31 AM

Isn't 'important', except that it maybe doesn't exist, like mass and change in amplitude are non-existent, perhaps..a spatial extent of a photon.

What I want to nail down is how the momentum is "transferred", if both vectors are varying about a zero point, or does the "packet" get delivered during an oscillation (of both components), regardless of what the instantaneous values are for either (so the delivery occurs over time, or over a wavelength, or something)?

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### #13

Posted 15 November 2007 - 10:59 AM

Isn't 'important', except that it maybe doesn't exist, like mass and change in amplitude are non-existent, perhaps..

What I want to nail down is how the momentum is "transferred", if both vectors are varying about a zero point, or does the "packet" get delivered during an oscillation (of both components), regardless of what the instantaneous values are for either (so the delivery occurs over time, or over a wavelength, or something)?

Then please start a new thread rather than hijack an existing one. This is a different question than in the OP

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### #14

Posted 17 November 2007 - 07:00 AM

Thank you all for your nice replies to my thread. I got more clear picture about the "volume" of photons.

Sisyphus indicated that "AFAIK, the short answer is "no." That's why you can't hit a photon with a photon. However, the wave function does mean there is a finite (though not rigidly bounded) region where the wave's magnitude is non-negligible. So in a certain sense it does have a volume, but not in the way we're used to thinking about it."

The wave is in a "finite" region, such region is the "volmue", we should say that it is a "soft" volume or "transparant" volume, I mean that other objects can occupy this region also. So if you hit a photon with a photon, the two photons will occupy the same region when they meet and then pass through each other.

### #15

Posted 17 November 2007 - 11:01 AM

(Check the links listed here)

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### #16

Posted 22 November 2007 - 02:51 PM

*"Two-photon scattering is observed at such facilities when the approaching particles both emit photons, which then interact with each other via a charged fermion/antifermion pair".*

That seems to say the photons do not scatter off one another.

### #17

Posted 22 November 2007 - 03:35 PM

When I follow the link I saw this:

"Two-photon scattering is observed at such facilities when the approaching particles both emit photons, which then interact with each other via a charged fermion/antifermion pair".

That seems to say the photons do not scatter off one another.

Yes, that's the virtual particle pair in QED, and how the interaction is modeled (i.e. the Feynman diagram); one of the links points out "A photon can fluctuate into any pair of charged fermions" — virtual particles are to what "fluctuates" refers. It's photon-photon scattering; that's what goes in and comes out. Otherwise you have to argue that electrons really don't scatter off each other because of the virtual photon that mediates the interaction. But that would just be semantics.

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### #18

Posted 22 November 2007 - 05:11 PM

photon-photon use virtual fermions.

fermion-fermion use virtual photon.

See what I mean ?

### #19

Posted 22 November 2007 - 07:07 PM

Isn't it a little bit circular ???

photon-photon use virtual fermions.

fermion-fermion use virtual photon.

See what I mean ?

Yes, it is. But there is a reason for it - the only point interaction in QED is a photon interacting with an electron.

### #20

Posted 25 November 2007 - 08:35 AM

*mass*and momentum due to its mass. A photon doesn't so its momentum has to be explained in terms of its rotating vectors --the electric and magnetic moments which also explain

*charge*-- angular momentum.

(Also I think I should review or repair my earlier statement about the 2d surface, and the two rotating vectors. The area would not vary, but instead each vector is meant to change orthogonally, I recall, like sine and cosine, so the 'area' is constant -then it's the frequency or the angular momentum that gets transferred... And I think I can remember reading something once about how an electron doesn't change state instantaneously like I was told at Chem. lectures, but takes a few tenths of a nanosecond, maybe, or does that mean we can pin down a photon's escape to this interval even?)

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