Photon Question

[Little Bang]

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

[UncleAl]

Special Relativity combines electric and magnetic fields into a rank-2 electromagnetic tensor. Changing reference frames mixes these components.

 

Electromagnetic tensor - Wikipedia, the free encyclopedia

 

Special Relativity

How Relativity Connects Electric and Magnetic Fields

On the Electrodynamics of Moving Bodies

 

An observer cannot move past a photon. The photon is traveling at lightspeed.

 

Relativistic Doppler effect - Wikipedia, the free encyclopedia

Relativistic Doppler Effect

Relativistic Doppler Effect

http://http://spiff.rit.edu/classes/phys314/lectures/doppler/doppler.html

on the lab bench

 

Given any achievable velocities V1 and V2 and any finite lightspeed, Lorentz invariance requires the bound on the relative velocities of V1 and V2 as viewed by any inertial observer in 1 or 2 cannot exceed

 

(V1 + V2)/[1 +(V1)(V2)/c^2]

 

 

 

Photon vs. photon in 180-degree opposition is still lightspeedd. This is transformation of velocities parallel to the direction of motion. For velocities at an arbitrary angle theta,

 

u_parallel = (u'_parallel + v)/(1+(v dot u')/c^2)

u_perp = u'_perp/(gamma_v(1+(v dot u')/c^2))

 

...and while we're here, vote Uncle Al a big juicy 10!

UNDER SATAN'S LEFT FOOT

[CraigD]

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.

This statement is, in a subtle, linguistic way, incorrect.

 

Photons don’t have electromagnetic fields (which have electric and magnetic components) – they are the particle-like description electromagnetic fields.

 

Charged particles, such as electrons and protons, have electromagnetic fields.

 

One can accurately say electromagnetic fields are made of photons.

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.

An electromagnetic field is described entirely by its electric component when the charged particles are stationary relative to the observer. When the charged particles are moving relative to the observer, their electromagnetic field has both an electric and a magnetic component.

 

A good intuitive technique for visualizing this is to consider that the electric component of a EM field produce magnetic force between charged particles, while the magnetic component can move charged particles, which is also called inducing an electric current. If the magnetic component is zero, no current is induced, which matches the intuitive observation that a non-moving generator generates no current.

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

As described above, it’s not the velocities of photons relative to the observer that’s significant to electromagnetism, but the velocities of charged particles.

 

The photons that carry electromagnetic force (also called interaction) between charged particles are virtual, meaning that they are detectable only by the interaction they carry. Such interactions always involve the exchange of pairs of virtual photons – intuitively, just as classical mechanical forces are always equal and opposite, force carriers are always exchanged as equal virtual particles and opposite vectors.

[Little Bang]

Thanks for attempting to make me understand. It was, alas, a wasted effort because your responses were about as clear as muddy water to me. Both BTW were entertaining.

[Qfwfq]

While Unk is always "entertaining", I don't see what was entertaining about Craig's post. I found your query entertaining too, and less clear than their replies; I don't think they understood it, it needs to be translated into what you were probably trying to get at.

 

Suppose there are two observers, [imath]O[/imath] and [imath]O'[/imath] who's coordinates are related by a pure Lorentz boost with velocity [imath]v[/imath] in the [imath]x[/imath] direction. The [imath]x[/imath] components of fields will be the same according to both (at a same point but it won't have the same coordinates for both) while the [imath]y[/imath] and [imath]z[/imath] components will be related as follows:

 

[math]E_y'= E_y\cosh\phi - B_z\sinh\phi;\, E_z'= E_z\cosh\phi + B_y\sinh\phi[/math]

 

[math]B_y'= B_y\cosh\phi + E_z\sinh\phi;\, B_z'= B_z\cosh\phi - E_y\sinh\phi[/math]

 

with the signs of the [imath]\sinh[/imath] terms inverting according to which way along the [imath]x[/imath] assis velocity is. As you should know, light is a transverse wave, components are zero in the propagation direction. If this direction is the [imath]x[/imath] assis the polarization will be different for the two observers, otherwise they will also see it with a different propagation direction (Lorentz transform the coordinates) and the field components must change so as to still be tranverse.

 

The only thing that stumps me and I still have to figure better is when I tried to check if the energy density matches up with transformation of frequency, I immediately ran into a paradox that I have yet to sort out, so I might have made some blunder in the algebra. Check it.

[martillo]

One way I thought photons could be stored is inside a fiber optic ring.

There could be a little technical problem to solve in how to insert the photons in the ring and how to take them out when needed (I suppose the intention to store them is to use them at some time).

[UncleAl]

The photons are circulating at lightspeed/(refractive index). If there is an RI other than 1 there must be optical absorbance, by the Kramers-Kronig relationship. They won't store. For fused silica at RI(nD) = 1.4585, each second is 205,550,000 meters pathlength. You put them in and take them out with a touching prism closer than the evanescent wave distance above the totally internally reflecting surface.

 

Store photons with reflection from first surface dielectric mirrors in a long length of hard vacuum, e.g., LIGO with ~20 kilowatts of laser light in the cavity. This is still not viable for long term storage. Stars shine through a billion lightyears of vacuum... but you cannot get back those photons.

 

--

Uncle Al

UNDER SATAN'S LEFT FOOT

Vote a 10 for the experiments!

[modest]

Suppose there are two observers, [imath]O[/imath] and [imath]O'[/imath] who's coordinates are related by a pure Lorentz boost with velocity [imath]v[/imath] in the [imath]x[/imath] direction. The [imath]x[/imath] components of fields will be the same according to both (at a same point but it won't have the same coordinates for both) while the [imath]y[/imath] and [imath]z[/imath] components will be related as follows:

 

[math]E_y'=\cosh E_y - \sinh B_z;\, E_z'=\cosh E_z + \sinh B_y[/math]

 

[math]B_y'=\cosh B_y + \sinh E_z;\, B_z'=\cosh B_z - \sinh E_y[/math]

 

Are you sure you need the cosh terms functions? I ask only because the first source I find has:

 

[math]E' = \begin{pmatrix} E_x \\ E_y - \gamma v B_z \\ E_z + \gamma v B_y \end{pmatrix}, \quad \quad B'= \begin{pmatrix} B_x \\ B_y + \gamma v E_z \\ B_z - \gamma v E_y \end{pmatrix}[/math]

 

http://math.stanford.edu/~dbaskin/maxwell.pdf

 

At the top of page 4. :confused:

 

~modest

[TheBigDog]

One way I thought photons could be stored is inside a fiber optic ring.

There could be a little technical problem to solve in how to insert the photons in the ring and how to take them out when needed (I suppose the intention to store them is to use them at some time).

Why store photons? Turn on the light and photons happen.

 

Bill

[Qfwfq]

Certainly not much ponit in storing photons for a long time.

 

Are you sure you need the cosh terms functions?

I edited my post to the way I meant to make it yesterday (but forgot :doh:), with a more standard notation instead of the shorthand I had used in my scribblings, just to get through the algebra. Again I don't have the paper with me and no time to check it out. However, I find the Stanford result surprising but maybe you could try doing the matrix algebra too; note that the [imath]\gamma[/imath] and [imath]v\gamma[/imath] are the hyperbolic trig functions (all on the same angle, the tangent of which is [imath]v[/imath]).

[CraigD]

One way I thought photons could be stored is inside a fiber optic ring.

Uncle Al describes to the near impossibility of storing photons for long in any reflective container (of which a fiber optic ring, or loop, is a specific example), while TheBigDog notes the lack of practical need to store photons.

 

I’d like to throw out some approximate calculations of the energy density such a system would have, ignoring the above issues.

 

The maximum power/cross-sectional area of typical commercial optical fiber is about 1000000000 W/m²

The speed of light in typical glass (refractive index about 1.5) is about 200000000 m/s

The density of typical fiber optic glass is about 2000 kg/m³

 

So the energy density of a fiber optic photon storage system is about 1000000000/200000000/2000 = 0.0025 J/kg

 

Compared to other common fuels and energy storage machines, this is very low. For example, gas, jet fuel, etc, has an energy density of about 46000000 J/kg, a wind-up clock spring about 300 J/kg, and a simple weight lifted 1 m against Earth’s surface gravity (such as in a grandfather or cuckoo clock), about 10 J/kg.

 

A fiber-optic photon storage ring, therefore, is on the order of 1/1000th as energy dense as a cuckoo clock. A cuckoo clock with a solar cell, a motor/generator and a light bulb would be a better photon storage system (with plenty of power to spare to make the little bird pop out and go “cuckoo”).

[martillo]

ja, ja, ja...

 

I have years in the forums and I never seen an acknowledgment for some good idea...

Actually I don't expect one.

The best answer would be silence, isn't it?...

 

I'm just curious about some (honest?) comment, actually missing, from "little bang" who asked for a possible way to store photons. He could answer for example in what application he was thinking to ask that.

But actually I don't expect anything.

[Qfwfq]

Martillo, I don't see where the OP:

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

was asking about how to store photons. Please avoid off topic replies.

 

Answering Modest, while I'm here, I got around to checking that computation and that Stanford .pdf has a computation error; the result is what it would be if the diagonal elements which are [imath]\gamma[/imath] were 1 instead. Like that, of course, it would not be a Lorentz boost. Anyway it was quite fishy that the result only has [imath]\beta[/imath] in it. Tell the Dean that someone there is not casting the best of light on the establishment.

[modest]

Answering Modest, while I'm here, I got around to checking that computation and that Stanford .pdf has a computation error; the result is what it would be if the diagonal elements which are [imath]\gamma[/imath] were 1 instead. Like that, of course, it would not be a Lorentz boost. Anyway it was quite fishy that the result only has [imath]\beta[/imath] in it. Tell the Dean that someone there is not casting the best of light on the establishment.

 

Thank you Q. I tried working out the matrix myself and fell way short. I suspected, like you say, that the paper was simply mistaken. And, this morning I found another source agreeing with you:

 

http://www.wbabin.net/hamdan/hamdan1.pdf

 

equation 7a and 7b

[math]E_x' = E_x \quad B'_y = \gamma(B_y + uE_z/c^2) \quad B'_z = \gamma(B_z - uE_y/c^2)[/math]

[math]E'_y = \gamma(E_y - uB_z) \quad E'_z = \gamma(E_z + uB_y)[/math]

 

~modest :thumbs_up

[UncleAl]

The other energy storage miracle would be a very large superconducting solenoid. Pump up the field during the night when demand is low. Bleed it out during the day when demand is high. However... A solenoid suffers Faraday force. The equation of state looks like the Ideal Gas Law with field strength as pressure. Strong electromagnets want to rip themselves apart. The fun in a supercon is thermal energy of shifting a winding is sufficient to quench it to normal conduction. CERN must repeatedly anneal each of its supercon magnets before they can be trusted to run at full strength. In an NMR magnet the volume of two fists the energy from a quench can be shunted to a (disposable) resistor bank dump. In a 100 megaton-equivalent supercon energy storage solenoid, you take a capital equipment writedown plus real estate bust.

 

--

Uncle Al

UNDER SATAN'S LEFT FOOT

Vote a 10 for the experiments!

 

 

.

[Little Bang]

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?

[Little Bang]

Ok, no description of a photon. So what is the only clue we have for the photon. It has both a magnetic and electric component at ninety degrees to each other. 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. So how does the photon propagating at C exhibit the property of moving past the observer and yet at the same time appear to the observer as if he is moving past the photon?