What is the actual size of a photon?

[Neo_Anderson]

A high-energy photon needs a smaller aperture to get down to its limit of resolution. Conversely, a radio wave needs an enormous aperture if it is to go through it (an RF waveguide, eg, is like 3"x7").
So what exactly is the size of the photon? Since they're created in atoms, I'd think they're about the size of an atom, maybe smaller. Interesting, since an atom can generate an RF wave that can't even fit in something smaller than a waveguide, whereas another atom can generate a photon that has no problem entering a 600 nm aperture...
A ridiculous question, I know, but valid nonetheless.

 

[hamster143]

How large is a wave on the pond?

 

[ansgar]

Photons are pointparticles in QM, they have no spatial extension, just as electrons, quarks and force carrier bosons.

 

[dmtr]

Schwarzschild radius?

 

[ansgar]

Schwarzschild radius?

mass = 0...

 

[dmtr]

mass = 0...

Energy != 0.

 

[ansgar]

Energy != 0.

scharchild radius is not related to "physical" radius (whatever that is...)

 

[dmtr]

scharchild radius is not related to "physical" radius (whatever that is...)

How comes? Schwarzschild radius is a real radius in the real space.

 

[ansgar]

How comes? Schwarzschild radius is a real radius in the real space.

Can you explain to me what the Schwarzschild radius is? The Earth has a radius differnt than it's density distribution radius (which is what we commonly call "size")

 

[dmtr]

Can you explain to me what the Schwarzschild radius is? The Earth has a radius differnt than it's density distribution radius (which is what we commonly call "size")

Er.. http://en.wikipedia.org/wiki/Schwarzschild_radius
For an electron it is a lot smaller compared to the Plank's length. Still, it is better, compared to the point!

 

[dmtr]

The original question was "what exactly is the size of the photon", it can be reformulated - "in how small volume can you 'squeeze' the photon", well, to do that you only need to throw it in the black hole. The resulting volume wouldn't be zero.

 

[ansgar]

Er.. http://en.wikipedia.org/wiki/Schwarzschild_radius
For an electron it is a lot smaller compared to the Plank's length. Still, it is better, compared to the point!

I know what it is, but you didn't seem to know it so I asked you to explain it.

One can also use the "Classical electron radius" (http://en.wikipedia.org/wiki/Classical_electron_radius) if one just wants a number, but the figure of interest is the radius for the matter distribution, not "Schwarzschild radius" etc.. and the radius asked for should be understood as the matter radius...

 

[dmtr]

I know what it is, but you didn't seem to know it so I asked you to explain it.

One can also use the "Classical electron radius" (http://en.wikipedia.org/wiki/Classical_electron_radius) if one just wants a number, but the figure of interest is the radius for the matter distribution, not "Schwarzschild radius" etc.. and the radius asked for should be understood as the matter radius...

AFAIK Schwarzschild radius sets a fundamental limit of how much mass/energy (information) you can pack in space. And it do set the lower limit on the size of the photon and electron.

 

[ansgar]

AFAIK Schwarzschild radius sets a fundamental limit of how much mass/energy (information) you can pack in space. And it do set the lower limit on the size of the photon and electron.

Can you say what the smallest size of a photon then, which has zero mass...

 

[Dmitry67]

In QM, it is a point particle.
All speculations about the plank length, Schwarzschild radius etc require the Quantum Gravity theory which is not ready yet.

 

[ansgar]

In QM, it is a point particle.
All speculations about the plank length, Schwarzschild radius etc require the Quantum Gravity theory which is not ready yet.

that was supposed to be my next move! :D

 

[dmtr]

Can you say what the smallest size of a photon then, which has zero mass...

R_{s} = \frac{2Gh\nu}{c^{4}}

And just for the fun of it:
A = 4\pi{R_{s}}^{2},

S = \frac{c^{3}A}{4 \hbar G} bits

 

[ansgar]

R_{s} = \frac{2Gh\nu}{c^{4}}

And just for the fun of it:
A = 4\pi{R_{s}}^{2},

S = \frac{c^{3}A}{4 \hbar G} bits

derivation?...

 

[dmtr]

In QM, it is a point particle.
All speculations about the plank length, Schwarzschild radius etc require the Quantum Gravity theory which is not ready yet.

Well, you are right of course. But I see nothing wrong in throwing an electron or a photon into a black hole and considering 'how small can it get there'.

 

[dmtr]

derivation?...

E = mc^{2} = h\nu

R_{s} = \frac{2Gm}{c^{2}}

R_{s} = \frac{2Gh\nu}{c^{4}}

 

[hamster143]

The original question was "what exactly is the size of the photon", it can be reformulated - "in how small volume can you 'squeeze' the photon", well, to do that you only need to throw it in the black hole. The resulting volume wouldn't be zero.

That is incorrect.

If someone asks "what is the size of dmtr", we don't start looking for the answer by squeezing dmtr with a hydraulic press.

The natural size of dmtr is the size of the smallest hole through which we can shoot him without significantly distorting the form of his wave packet. It is probably between 1 and 2.5 meters - many orders of magnitude greater than Schwartzschild radius of dmtr.

Similarly, the natural size of a photon with wavelength \lambda is approximately \lambda.

 

[dmtr]

That is incorrect.

If someone asks "what is the size of dmtr", we don't start looking for the answer by squeezing dmtr with a hydraulic press.

The natural size of dmtr is the size of the smallest hole through which we can shoot him without significantly distorting the form of his wave packet. It is probably between 1 and 2.5 meters - many orders of magnitude greater than Schwartzschild radius of dmtr.

Similarly, the natural size of a photon with wavelength \lambda is approximately \lambda.

My protest was to the notion of 'point particles' and the associated infinities. As to the size, I think it is very natural to measure sizes in bits. For example, imagine a photon with a wavelength comparable with the radius of the universe. All you can tell about that photon, is whether it is present, or not. Nothing else. That would be exactly one bit. So very naturally the size of that photon would be one bit. ;)

 

[jtbell]

The question of "size of a photon" comes up regularly here on PF. You might like to browse through some old threads:

http://www.google.com/#hl=en&q=photon+size+site:physicsforums.com

 

[ansgar]

E = mc^{2} = h\nu

Does not apply to a photon...

 

[hamster143]

My protest was to the notion of 'point particles' and the associated infinities.

Well, let's just agree that the protest was worse than the notion.

The other problem I have with Schwarzschild radius is that it's just a severely overhyped number. There's nothing magical about Schwarzschild radius. If you're falling into a black hole, you don't "hit the surface", so to speak, at \frac{2Gm}{c^{2}}. You may not even notice that anything has happened. The black hole itself is pointlike (or, rather, has the diameter on the order of Planck length).

 

[dmtr]

That is incorrect.
The natural size of dmtr is the size of the smallest hole through which we can shoot him without significantly distorting the form of his wave packet. It is probably between 1 and 2.5 meters - many orders of magnitude greater than Schwartzschild radius of dmtr.

I agree with the "The natural size of a photon is the size of the smallest hole through which we can shoot it". But you don't need to add 'without significantly distorting'. You can either shoot a photon through that hole or not. If the hole radius is smaller than the Schwartzschild radius for that photon, you wouldn't be able to shoot that photon through. It won't fit.

Note that the size of the hole have a units of area, in the natural units it is measured in bits (Plank's area is equal to one bit). As a result the natural size of a photon should be measured in bits. (The smallest photon would be the photon with the largest wavelength [radius of the universe], and AFAIK, if you calculate the size of it, it would come out as one bit ;)

 

[dmtr]

Nah. It doesn't come out as one bit. Not even close. It looks like:

R_{s} = \frac{2 G m}{c^{2}}

E = m c^2 = h \nu = \frac{h c}{\lambda}

A = 4 \pi {R_{s}}^{2}

A_{bit} = \frac{h G}{2 \pi c^{3}}

S_{bits} = \frac{A}{A_{bit}}= \frac{16 \pi^{2} h G c}{\lambda^{2}}

where {\lambda} is the photons wavelength.

 

[ansgar]

E = m c^2 = h \nu = \frac{h c}{\lambda}

Not valid, did you fail your special relativity class? :)

 

[dmtr]

Not valid, did you fail your special relativity class? :)

I'm a CS grad :). Why? Is there any other way to calculate the equivalent increase of the mass of the black hole, if you throw a photon into it?

 

[ansgar]

I'm a CS grad :). Why?

E = mc^2 is not valid for a photon...

E^2 = (mc^2)^2 + (pc)^2

or did you mean that m = m_0 x Gamma ?

then Gamma = infinity for photons...

sorry dmtr, there is something strange going on here :(

and what is a CS grad?

 

[dmtr]

E = mc^2 is not valid for a photon...

E^2 = (mc^2)^2 + (pc)^2

or did you mean that m = m_0 x Gamma ?

then Gamma = infinity for photons...

sorry dmtr, there is something strange going on here :(

and what is a CS grad?

Computer science graduate. Read "complete failure in physics".

Yes. E^2 = (mc^2)^2 + (pc)^2. But m = 0. So E = pc.
Now I'm trying to calculate the equivalent increase of the mass of the black hole (assuming I throw that photon into it), AFAIK I can use the mass-energy equivalence principle - E = mc^2 for that.

 

[ZapperZ]

The OP appeared to have not participated in this thread beyond the first post. As has been pointed out, this has been discussed already many times before, and a link to one of them has been provided.

How this thread has somehow meandered into a black hole is beyond me.

Zz.