Photon Question

[Qfwfq]

If a field is moving with respect to an observer he views it as an electric field. If the observer is moving through the field he views it as magnetic.

I don't know where you get this idea. Did you try following the discussion here, of Lorentz covariance of the fields? If you found it difficult to understand, ask questions cuz else we can't know what you're missing.

[Little Bang]

If a charge is moving past me do I see the field as electric? Yes. If I am moving past the charge do I do I see the field as magnetic? Yes.

[CraigD]

When I say the word car most can imagine a picture of a car in their mind. When I say the word photon no one can conjure up a picture.

Since hearing about photons (possibly in the first runs of Star Trek, as they showed torpedoes nominally made of ‘em ;)) and eventually studying them in introductory modern physics lessons, I’ve had a mental picture of the photon and photons.

 

Like most figments of the imagination, it changes continually, gaining and losing details as different needs arise, but more or less, I imagine the photon as looking like a monochrome amoeba, that, when touched, snaps into a single sphere of the same color.

 

The amoeba shape represents the photons uncertain position, until it’s detected.

 

The size of the imaginary sphere corresponds to the wavelength of its color, scaled down to the greatest size that the holes in a conductive screen may be and act as a Faraday cage for photons of that frequency. Microwaves, thus, I picture as about the size of the little holes in the window of my microwave oven, visible light about 20 million times smaller.

 

This isn’t, of course, what a photon really looks like, but, for me, a useful mental image, serving to cement in my mind several significant physics concepts.

 

We can’t really see or feel an image of a photon, because we see and feel with photons. We can directly perceive the interaction of photons with tissues in our eyes – I’m doing it right now, watching the little letters appear on my computer’s display screen.

 

Our eyes and brains can’t usually perceive single photons, though there appear to be exceptions to this rule, such as the “closed eye light flashes” astronauts have reported since the 1960s, which are believed to be caused by single cosmic ray photons passing through their eyes, optic nerves, or vision regions of their brains (see ESA - Lessons online - Radiation and life).

[Little Bang]

Very interesting view of the photon. It is at least an attempt at an imaginary picture which I have been unable to do. I will ponder your idea.

[Little Bang]

Another thing we know about a photon. Drop a rock into a pond and the resulting wave propagates at ninety degrees to the direction of acceleration or in this case deceleration. An accelerated charged particle emits photons at ninety degrees to the direction of acceleration. Does this influence your picture at all?

[Qfwfq]

If a charge is moving past me do I see the field as electric? Yes. If I am moving past the charge do I do I see the field as magnetic? Yes.

Given the principle of relativity, there's no difference between these two cases; in each of them the field is described as both electric and magnetic.

 

Our eyes and brains can’t usually perceive single photons

Actually they can. I've never tried it but I've heard of ways to set up a situation in which one can see them, like little stars blinking at random. For instance, with a dilute flourescent solution under a microscope, arranged so there aren't too many of the molecules at one time in the field of view. Light the slide with UV and then switch out the ordinary light.

[CraigD]

Another thing we know about a photon. Drop a rock into a pond and the resulting wave propagates at ninety degrees to the direction of acceleration or in this case deceleration. An accelerated charged particle emits photons at ninety degrees to the direction of acceleration. Does this influence your picture at all?

:) Can you back up this claim (accelerated charged particiles emitting photons at 90° to their acceleration) with a link or reference, LB?

 

I believe you’re confusing QED’s explanation of electromagnetic force as carried by virtual photons with the emission of actual photons.

 

Charged particles are accelerated by exchanging virtual photons. When this acceleration results in a particle having less energy than before the interaction, it emits actual photons with energy equal to the difference in energy. A difference between virtual and actual photons is that the virtual ones are detectable only by their effect on the particles that exchange them, while actual photons can be detected by any particle that interacts with them. Actual photon emission is called radiation, while virtual photon emission isn’t – it might better be called “QED bookkeeping”. This “double role” of photons in QED has long struck me as one of the theory’s weirder features.

 

As for analogies between quantum mechanics and waves on the surface of a pond from rocks thrown into it, we should be careful drawing too much from them. Keep in mind that when a rock strikes a pond surface, it produces a lot of forces and waves traveling through the water in many directions, most of them longitudinal sound waves. The surface waves arise from a complicated series of forces between molecules of the rock, the water, neighboring water molecules, and gravitational force between the water and the Earth – terribly complicated stuff compared to a typical quantum effects like an electron in an atom changing energy level and radiating a single photon. Underlying each of the complicated fluid mechanical interactions of the rock thrown into a pond are exchanges of virtual particles (not only electrons, but gluons in atomic nuclei). Every mechanical effect is the result of such interactions, but not every mechanical effect is a good analogy for them.

[Little Bang]

Craig, I have been unable to find the link that I read several years ago but have found two that suggest the the same thing here.

 

http://arxiv.org/PS_cache/physics/pdf/0506/0506049v5.pdf

 

http://www.cv.nrao.edu/course/astr534/PDFnewfiles/LarmorRad.pdf

[TheBigDog]

When I say the word car most can imagine a picture of a car in their mind. When I say the word photon no one can conjure up a picture. Most of us have seen a mathematical graphic simulation of an electromagnetic wave but not a photon. So what does a photon look like?

I've always imagined photons as singular points riding on waves. I think that photons are problematic when thought of in non quantum terms. I think they are neither energy or matter, and both energy and matter. When you measure a photon you determine its state. That is how they seem to violate SR since having mass would prevent them from traveling at light speed, but they are light itself to the observer. Being in a dual state, or a neither state allows them to behave both as matter (measurable impacts) and as energy traveling in waves at the speed of light.

 

Of course I could be wrong, but in simplest terms that is how I understand photons from my understanding of physics.

 

Bill

[Qfwfq]

The maximum in the angular distribution of intensity is at right angles to the acceleration, this can be found in any half decent 2nd year ph. textbook. In terms of photons it means that this goes for the probability distribution. It doesn't mean that each photon goes out at right angles, of course.

[Farsight]

The photon has both an electric and magnetic component. The determination of whether a field is electric or magnetic is dependent on the observer moving through the field or the field moving past the observer.

You missed something there. As CraigD said a couple of weeks ago, the photon doesn't "have" an electric and a magnetic field, the photon is an electromagnetic field variation.

 

That said, how does the photon exhibit the property of moving past the observer and also of the observer moving past the photon?

If you were standing in an electric field, you could play around with charged particles to detect that electric field. If you were travelling through it, you could play around with a compass to detect the magnetic field it now appears to be. But it isn't like this for a photon. It doesn't have a fixed electric field. And the moot point is that you experience a magnetic field when an electric field is changing, not just when you move through a static electric field.

 

I need to get down to the fundamentals to answer your question. It's very simple, and it goes back a long way, to Maxwell and his electric elasticity equation E=1/ε D , but Minkowski's wrench is my favourite reference. For some odd reason nobody seems to know about it. Take a look at Space and Time, two pages from the end. You can maybe see it on this book search. Here's the interesting bit:

 

"Then in the description of the field produced by the electron we see that the separation of the field into electric and magnetic force is a relative one with regard to the underlying time axis; the most perspicious way of describing the two forces together is on a certain analogy with the wrench in mechanics, though the analogy is not complete".

 

Note that there's only one field, but two forces. That's why it's the electromagnetic field, even though we all talk about the electric field and the magnetic field, because we tend to think about the vector-field "what it does" and tend to forget about the geometrical "what it is". As for what it is, the word wrench is the key. It's associated with a screw thread, and is an analogy that says the electric field is a "twist" field. It's twisted space. If you move through it, or it moves through you, you see it as a magnetic field, which is "turn" field. This is why we have generators and dynamos, and the right-hand rule. See Right-hand rule - Wikipedia, the free encyclopedia and look at this picture:

 

 

Now find a drill-bit or a reamer. It's got a twist to it. Grip it in your right hand, put your left thumb on the bottom, and push. It turns. In similar vein if you have a current going up the wire, you have a moving electric field, and so you have a magnetic field too. Back to the photon, take a look at the wiki picture below:

 

 

A photon conveys energy, and the dimensionality of energy is pressure x volume, so go with the flow and think about a "twist field". Start from the zero line just to the right of the M, and work left. The electric field rises to a peak, which is where the twist in space is at a maximum. It then reduces to zero, then descends to a minimum, which is where space is twisted in the opposite direction. Then it goes back to zero, and you've finished. In essence the sine wave is telling you the slope of twisted space, and you can draw a sinusoidal "hump" where the slope starts at zero, rises to a maximum midway up the hump, goes to zero at the top of the hump, then slopes the other way, then back to zero again. It's basically a hump of spacewarp, a pressure pulse, like an ocean wave, and travelling at a velocity dictated by the properties of the space it's travelling in. If this was moving past you through a cubic lattice, you wouldn't see a particle, all you'd see is the squares twisting out of true. That's the electric field variation. But as they did, you'd see the lines turning. That's the magnetic field. It's best to look at the reamer to understand the orthogonality, rather than looking at the cubic lattice on either side of the photon.

[Qfwfq]

I don't find it appropriate to describe the EM field as being "twisted space", especially considering that the equivalence principle cannot be valid.

 

Minkowski's wrench is my favourite reference. For some odd reason nobody seems to know about it.

From briefly reading that book search on wrench, he is talking about what is now quite well known, even if not commonly called the wrench. He also says the analogy is not complete.

 

In any case, we were talking about it here. After some hesitation I had decided not to talk about the four-potential that is mentioned in there, from which the [imath]F[/imath] tensor can be defined:

 

[math]F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}[/math]

 

Which makes [imath]F[/imath] manifestly antisymmetric in the two indices. Hence it has the six independent components which, upon Lorentz transformations including parity, notoriously behave as the components of a polar 3-vector and an axial one.

 

[math]E_{\alpha}=F_{0\alpha}[/math] and [math]B_{\alpha}=\epsilon_{\alpha\beta\gamma}F^{\beta\gamma}[/math]

 

Edit: Note that your lower image is incorrect and not consistent with your own text. You have the zeroes and maxima of the two fields coinciding.

[Farsight]

I didn't understand what you were saying about the equivalence principle cannot be valid. Please can you restate? The geometrical picture is rather unfamiliar, but it is supported. Think of the polar 3-vector as being caused by an isotropic distortion of space akin to a hairball with Fibonacci-spiral hairs as opposed to straight hairs. In similar vein think of the axial 3-vector as being caused by motion of a series of such distortions past you, as per the current in the wire. Also see Maxwell re Heaviside replacing Maxwell's quaternions with vectors. Or better still, consider the magnetic field-variation of a light wave. See Magnetic field and note the literal truth of magnetic fields are produced by electric currents. There's a magnetic field variation in a light wave, which means there's a current, which means a photon is a quantum of alternating current, which is why impedance Z0 = √(μ0/ε0) applies. Then note that the D in Maxwell's E=1/ε D is displacement current. Take a displacement as the cause of the vector-field effect, and the photon pressure-pulse "spacewarp" visualisation works rather well.

 

Can you also elaborate on the lower image being incorrect and not consistent with the text? It's a wikipedia "light-wave" image rather than my own, and the maxima are usually shown as coincident as per these further examples:

 

Encyclopaedia of occupational health ... - Google Books

 

Food Analysis: Principles and ... - Google Books

 

Principles of Physical Chemistry - Google Books

[Ben]

After some hesitation I had decided not to talk about the four-potential that is mentioned in there, from which the

[imath]F[/imath] tensor can be defined:

 

[math]F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}[/math]

If you won't talk about this, then I shall extemporize for mine own amusement:

 

Look at this: [math]\partial_{\tau} F_{\mu \nu} =(\frac{\partial^2}{\partial^{\tau}\partial^{\mu}}A_{\nu}+\frac{\partial^2}{\partial^{\tau}\partial^{\nu}}A_{\mu}) -(\frac{\partial^2}{\partial^{\mu}\partial^{\tau}}A_{\mu}+\frac{\partial^2}{\partial^{\nu}\partial^{\tau}}A_{\nu})[/math]

 

[math]=(\frac{\partial^2}{\partial^{\tau}\partial^{\mu}}A_{\mu}-\frac{\partial^2}{\partial^{\mu}\partial^{\tau}}A_{\mu})

+(\frac{\partial^2}{\partial^{\nu}\partial^{\tau}}A_{\nu}-\frac{\partial^2}{\partial^{\tau}\partial^{\nu}}A_{\nu})=0+0[/math],since mixed partial second derivatives commute.

 

Because of this and also because on antisymmetry, and noting the implied factor of 2 in the above, I can define the 2-form [math]\mathbf{F} = \frac{1}{2}F_{\mu\nu} dx^{\mu}\wedge dx^{\nu}[/math], whence [math] d\mathbf{F} = 0 = d(d\mathbf{A})[/math] (where now [math]\mathbf{A}[/math] is a 1-form).

 

From which is a quick(ish) task to write the Maxwell equations in terms if this identity.

 

Which makes [imath]F[/imath] manifestly antisymmetric in the two indices. Hence it has the six independent components which, upon Lorentz transformations including parity, notoriously behave as the components of a polar 3-vector and an axial one.

Which is easily seen by inspection of this tensor in component form:

 

[math]F_{\mu \nu} =\begin{pmatrix}0&E_x &E_y& E_z\\-E_x&0& -B_z & B_y\\-E_y & B_z & 0 &-B_x\\-E_z & -B_y & B_x &0 \end{pmatrix}[/math].

 

So if the term "twisted space" is to have any meaning (which I rather doubt), we will have our signs in this matrix jumping around all over the place.

[Qfwfq]

After having had time to figure it out, I must recognize that the phases in that image are ok for a propagating wave with linear polarization. It would not be correct for a linear polarized standing wave, in which the maxima of one field are at the same [imath]x[/imath] as the zeroes of the other. Gosh these things are tricky, I was reckoning that the wave from an oscillating dipole would have a [imath]\frac{\pi}{2}[/imath] phase difference and it seemed convincing for energy density to be uniform, as it is in circular polarization. Foggy memories, foggy memories, it was only with the Poynting vector that I was able to sort it out completely.

 

I didn't understand what you were saying about the equivalence principle cannot be valid. Please can you restate?

It is valid for gravitation, because the charge is so precisely proportional to inertial mass that we can consider them as being the same thing.

 

It is not valid for electromagnetism; not all bodies have the same charge-mass ratio. Further, they are both gauge fields but in EM the connection is between phases of the charged field, not between the values of coordinate tensors. It makes no whatsoever sense to regard the em field as a geometric effect. Also, the displacement current is not D but instead it's time derivative, otherwise the field equations would be wrong.

 

In any case folks, let's ttry not to get toooooo far off topic. The OP wasn't asking for quite so much jazz.

[Farsight]

After having had time to figure it out, I must recognize that the phases in that image are ok for a propagating wave with linear polarization...

Thanks.

 

It would not be correct for a linear polarized standing wave, in which the maxima of one field are at the same [imath]x[/imath] as the zeroes of the other. Gosh these things are tricky, I was reckoning that the wave from an oscillating dipole would have a [imath]\frac{\pi}{2}[/imath] phase difference and it seemed convincing for energy density to be uniform, as it is in circular polarization. Foggy memories, foggy memories, it was only with the Poynting vector that I was able to sort it out completely.

Interesting that you should mention a standing wave. The "near field" of a radio transmitter is something I'm looking into: See Near and far field - Wikipedia, the free encyclopedia and Evanescent wave - Wikipedia, the free encyclopedia. But I digress.

 

It is valid for gravitation, because the charge is so precisely proportional to inertial mass that we can consider them as being the same thing.

I must point out that gravity is caused by energy, not by mass per se. A "massive" body only causes gravity because of its energy content. Hence I'd say that thinking of active gravitational mass as "gravitational charge" isn't ideal.

 

Also, the displacement current is not D but instead it's time derivative, otherwise the field equations would be wrong.

Sorry, slip of the tongue, see Electric displacement field - Wikipedia, the free encyclopedia.

 

It is not valid for electromagnetism; not all bodies have the same charge-mass ratio. Further, they are both gauge fields but in EM the connection is between phases of the charged field, not between the values of coordinate tensors. It makes no whatsoever sense to regard the em field as a geometric effect.

Unfamiliar as it might seem, it makes good sense. A positron has spin 1/2. Think of a simple geometrical object that fits the bill:

 

 

Humour me and make a moebius strip. Note what you have to do. Now think about a proton. Different mass, same spin. It's all horribly simple, but you have to take a step back from vector fields to appreciate it. A vector field is an effect, caused by a dynamical rotating entity in non-uniform space. The trick is to think of pair production and see an electron as a self-trapped photon, and then remember that light travels in straight lines. When it doesn't, we say space itself is curved or distorted in some fashion.

 

Vector field - Wikipedia, the free encyclopedia

 

In any case folks, let's ttry not to get toooooo far off topic. The OP wasn't asking for quite so much jazz.

Maybe a separate thread then? I did read the whole of this thread, and only felt moved to offer a contribution because the answers didn't seem to be getting to the bottom of it.

[Little Bang]

Unc, as usual you are correct and have admirably demonstrated that fact in your post. My use of the the word acceleration is obviously wrong. The term I should have used is, " A changing frame of reference ". My next post will be a thought experiment that I think will link not only gravity and inertia to a changing frame of reference but also Doppler shift.