- #1

- 9

- 0

And a related question, if there is, does that mean there is a difference between the gravitational field and a gravity wave (graviton?), or the electric field and an electric wave?

Thanks

You should upgrade or use an alternative browser.

- I
- Thread starter josh777
- Start date

- #1

- 9

- 0

And a related question, if there is, does that mean there is a difference between the gravitational field and a gravity wave (graviton?), or the electric field and an electric wave?

Thanks

- #2

- 5,432

- 292

The quantum theory of electrodynamics (QED) describes the interaction between electrons and light quanta (photons ?) and this is where 'photons' can be found. There is no such idea in Maxwells equations.

The electric field is linear mathematically because ##\vec{E}##s may be added. This not the case for gravity and GWs can be found only in the linearised version of the general theory of relativity.

- #3

- 9

- 0

1. So if for example, I've got current flowing past in a wire and thus have created a magnetic field around that wire, would that magnetic field would be somehow completely different from the "light quanta" that are generated when electrons drop energy levels? Is this magnetic field also quantizable?

2. Could I, idk, replicate the photoelectric effect with a strong enough magnetic field?

3. The way I understand the "frequency" of an EM wave is I imagine the electron that produced it having oscillated positions either faster or slower, faster means higher frequency and slower means lower frequency (again, I could be completely wrong). What would the "frequency" be of an EM field generated by a wire, or is there no such thing?

4. (This one sums them up) Is there a difference between a magnetic field and a magnetic wave?

Thank you!

- #4

- 5,432

- 292

This mixes the classical and quantum theories. Not sure what you mean.Hey, @Mentz114 thanks so much for your answer. So I've got a couple questions:

1. So if for example, I've got current flowing past in a wire and thus have created a magnetic field around that wire, would that magnetic field would be somehow completely different from the "light quanta" that are generated when electrons drop energy levels? Is this magnetic field also quantizable?

I doubt it.2. Could I, idk, replicate the photoelectric effect with a strong enough magnetic field?

The frequency of light quanta is proportional to their energy.3. The way I understand the "frequency" of an EM wave is I imagine the electron that produced it having oscillated positions either faster or slower, faster means higher frequency and slower means lower frequency (again, I could be completely wrong). What would the "frequency" be of an EM field generated by a wire, or is there no such thing?

What is a magnetic wave ?4. (This one sums them up) Is there a difference between a magnetic field and a magnetic wave?

Thank you!

- #5

- #6

PeterDonis

Mentor

- 37,793

- 15,543

You appear to be thinking of these two models as different possible ways that electromagnetism could be. That's not a good way to think about them. They are just different ways of describing and making predictions about electromagnetic phenomena. Neither of them are necessarily the way electromagnetism "really is"; they're just models that we have constructed to describe the phenomena and make predictions about them.

- #7

- 20,129

- 10,868

No, that's not what is "revealed" by relativity.

The quantum theory of electrodynamics (QED) describes the interaction between electrons and light quanta (photons ?) and this is where 'photons' can be found. There is no such idea in Maxwells equations.

The electric field is linear mathematically because ##\vec{E}##s may be added. This not the case for gravity and GWs can be found only in the linearised version of the general theory of relativity.

The electromagnetic field consists of six field-degrees of freedom, we are used to split into an electric field ##\vec{E}## and a magnetic field ##\vec{B}##. There are different ways to treat the transformation properties of the field under Lorentz transformations. The most common one is to arrange the six components into an antisymmetric 2nd-rank tensor with components ##F_{\mu \nu}=-F_{\nu \mu}##. The three space-time components are given (with the appropriate signs, I'm to lazy to figure out now) by the electric and the space-space components by the magnetic components. Under Lorentz transformations it transforms according to the rules of 2nd-rank tensor components, i.e.,

$$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x),$$

where

$$x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho}.$$

Now photons are certain states of the quantized free (sic!) electromagnetic field, socalled single-photon Fock states. I don't think that the formalism can adequately be given in a physics-forums posting, but let me stress that on the popular-science level it is well more save to think of all electromagnetic fields, quantized or not, rather from a field-point-of view than a particle-point-of view. Photons are as far from classical point particles as anything can be. You cannot even define a position observable for them. They thus cannot be localized like a massive point particle in any way.

It also doesn't make any sense to say "the electromagnetic field consists of photons". Photons are a special sort of quantum states of the quantized electromagnetic field (single-photon states) is the correct statement. If you have prepared the electromagnetic field in such a state you have a single photon. This can be done by a process called parametric downconversion. You should a laser pulse into certain sorts of berefringent crystals enabling you to filter out true single-photon states. The process involved is a non-linear-optics process, where the laser fields interacts such with the crystal that a laser mode of energy ##\hbar \omega## is converted into an (entangled) pair of photons, each of energy ##\hbar \omega/2##. Then you can use one of these photons to "herald" the other photon, then being sure to have a single photon.

This single photon then interacts with your optical apparatus you use to investigate its properties. And FAPP you can think about the photon as a electromagnetic wave (not a particle!) but the intensity of the field providing the probability to register this photon at a place where the photodetector is located. There's no other adequate picture to describe what a photon is than this probabilistic one. A full understanding is only possible by studying the full math of quantum electrodynamics.

- #8

- 20,129

- 10,868

Sigh :-((. Everything is correct in this posting, but not the semisentence "which are particle-like". Just erase this semi-sentence! Then the posting becomes perfectly right!modelsof electromagnetism. There is the classical EM model, using electric and magnetic fields and based on Maxwell's Equations. And there is the quantum model, based on QED and using the photon--more precisely, using a quantum field, certain states of which are particle-like and are referred to by the term "photon". The quantum model is the more fundamental of the two; the classical model is an approximation to the quantum model that works well for many phenomena.

You appear to be thinking of these two models as different possible ways that electromagnetism could be. That's not a good way to think about them. They are just different ways of describing and making predictions about electromagnetic phenomena. Neither of them are necessarily the way electromagnetism "really is"; they're just models that we have constructed to describe the phenomena and make predictions about them.

Photons are the least particle like of all single-quantum Fock states possible! This is due to the fact that they are massless quanta with spin 1 and thus do not admit the definition a proper position observable!

If there's one most important particle property, it should be the possibility to prepare states where the object described by this state is localized in the sense that its position is determined within an appropriately small spatial region. Photons do not even have a position observable, and physically it's thus impossible to make sense of a spatially localized single-photon state!

- #9

- 5,432

- 292

I presume you are objecting to the phrase 'the ##\vec{E}## and ##\vec{B}## fields transform like space and time under Lorentz transformations'.No, that's not what is "revealed" by relativity..

This comes from Itzyksen & Zuber (1985) where they split the Lorentz transformation into two separate transformations, one for ##x\rightarrow x'## and one for ##t\rightarrow t'## ( equations 1.23, page 5). These equations obviously mix t and ##\vec{x}##. When these equations are applied to ##\vec{E},\ \vec{B}## then this shows that they mix like ##t## and ##x## (equations 1.35, page 9). I may have misinterpreted this but I think my statememt above is true, if uninteresting.

With ##F=\pmatrix{0 & Ex & Ey & Ez\cr -Ex & 0 & 0 & 0\cr -Ey & 0 & 0 & 0\cr -Ez & 0 & 0 & 0}## and ##\Lambda=\pmatrix{\frac{1}{\sqrt{1-{\beta}^{2}}} & \frac{\beta}{\sqrt{1-{\beta}^{2}}} & 0 & 0\cr \frac{\beta}{\sqrt{1-{\beta}^{2}}} & \frac{1}{\sqrt{1-{\beta}^{2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1}

## we get

##F'=\Lambda F \Lambda = \pmatrix{0 & Ex & \frac{Ey}{\sqrt{1-{\beta}^{2}}} & \frac{Ez}{\sqrt{1-{\beta}^{2}}}\cr -Ex & 0 & \frac{Ey\,\beta}{\sqrt{1-{\beta}^{2}}} & \frac{Ez\,\beta}{\sqrt{1-{\beta}^{2}}}\cr -\frac{Ey}{\sqrt{1-{\beta}^{2}}} & -\frac{Ey\,\beta}{\sqrt{1-{\beta}^{2}}} & 0 & 0\cr -\frac{Ez}{\sqrt{1-{\beta}^{2}}} & -\frac{Ez\,\beta}{\sqrt{1-{\beta}^{2}}} & 0 & 0}##

and the magnetic field components ( ##By=\frac{Ez}{\sqrt{1-{\beta}^{2}}}## and ##Bz=\frac{Ey}{\sqrt{1-{\beta}^{2}}}##) that arise used to be part of an electric field component. Mixing.

Last edited:

- #10

- 20,129

- 10,868

Another often more simple to handle way to describe the Lorentz-transformation properties is to introduce the Rieman-Silberstein vector, ##\vec{F}=\vec{E}+\mathrm{i} \vec{B}##. Then application of the transformation laws to ##\vec{E}## and ##\vec{B}## you get from the tensor formalism shows that this transformation can be represented by the group ##\mathrm{SO}(3,\mathbb{C})##, i.e., by the special orthogonal complex (sic!) rotations. The rotations are given by the subgroup ##\mathbb{SO}(3,\mathbb{R})##, and the boosts are given by rotations around an arbitrary axis with a purely imaginary rotation angle. Again, this clearly shows that ##\vec{E}## and ##\vec{B}## do NOT transform like time-like and spacelike components of four-vectors.

- #11

- 5,432

- 292

OK, that's me told.[cut to save space]

. Of course you formulae are incomplete since ##F_{\mu \nu}## also includes the magnetic field components to begin with.

Would you accept that ##\vec{E}## and ##\vec{B}## mix under Lorentz reansformation like space and time in a 4-vector ? This is what I'm trying to say. Obviously the rank-2 tensor needs two applications of ##\Lambda## but component-wise there is similar mixing.

(Welcome back )

Last edited:

- #12

- 20,129

- 10,868

- #13

- 5,432

- 292

Some misunderstanding here. Any linear transformation 'mixes' the components of whatever it operates on. In the case of the EM field the components happen to be the ##\vec{E}## and ##\vec{B}## fields. I appreciate your erudition but we do not use the word 'mix' in the same way.

[edit]

Just to clear up a point - the potential ##A^\mu=( A_0, A_x(t), A_y(t), A_z(t))## gives only electric fields in

##F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu##. Is this an impossible potential ?

- #14

- 20,129

- 10,868

Of course your example for a potential is not a priori wrong. Let's see what it is. I guess ##A_0=\text{const}##. Then ##\partial_{\mu} A^{\mu}=0##, i.e., you work in the Lorenz gauge. The electromagnetic field is given by

$$\vec{E}=-\partial_t \vec{A} - \vec{\nabla} A^0=-\partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}=0.$$

Now the homogeneous Maxwell equations are fulfilled due to the derivation of the electromagnetic field from a four-vector potential.

Now we have to fulfill the inhomogeneous equations as well. Gauß's Law reads

$$\vec{\nabla} \cdot \vec{E}=0,$$

because ##\vec{A}## is only a function of ##t##. I.e., there are no charges anywhere. Because of ##\vec{B}=0## the Ampere-Maxwell Law reads

$$\frac{1}{c} \partial_t \vec{E}=-\frac{1}{c} \vec{j}.$$

From Gauß's Law it's clear that ##\vec{\nabla} \cdot \vec{j}=0##. From Faraday's Law and ##\vec{B}=0## we have

$$\vec{\nabla} \times \vec{E}=0,$$

and thus also ##\vec{\nabla} \times \vec{j}=0##. This implies

$$\partial_t \vec{E}=0 \; \Rightarrow \; \vec{E}=\text{const},$$

i.e., you have a homogeneous electric field in the entire space, which doesn't exist in nature. So ##\vec{E}=0##.

Share: