Linear polarizers: angle between the electric & magnetic fields

[narsep]

From wikipedia: "An electromagnetic wave such as light consists of a coupled oscillating electric field and magnetic field which are always perpendicular; by convention, the "polarization" of electromagnetic waves refers to the direction of the electric field."

A polarizer is an object that acts on electric field identically as on magnetic field with a 90 degrees orientation difference. Is there any object that behaves to these fields (electrical and magnetic) differently? e.g. Has the max transparency when the angle between these two fields is not 90 degrees?

 

[BvU]

Hello narsep,

Also from your link:

The electric and magnetic fields in EMR waves are always in phase and at 90 degrees to each other

so it doesn't happen !

 

[narsep]

Is there a theoretical explanation why this is so?

 

[BvU]

Google "why are b and e perpendicular" and pick one you like ...

 

[Ibix]

Is there a theoretical explanation why this is so?

The short answer is that non-perpendicular E and B fields don't satisfy Maxwell's equations, isn't it? (Edit: or, more precisely, a propagating wave with non-perpendicular E and B fields isn't a solution of Maxwell's equations.)

 

[narsep]

Thanks for your answers. However reading them, I realized that all of them are based on the relation that the cross product of EF and MF equals to zero and hence they are perpendicular. (Am I right?)
But what if EF and MF are not represented by cartesians coordinates (that enforce geometric perpendicularity) but by another coordinate system that "perpendicularity" (cross product=0) does not mean geometric perpendicularity?
This is related to the first answer in "https://www.quora.com/Why-are-electric-and-magnetic-fields-perpendicular-in-an-electromagnetic-wave" :
"

• Interesting point: If there had been one more entity consisting of three interacting charges (say electron, proton and XX, a tripole) then the same field, EF, MF would have another name in the plane (with reference to EF) affecting that TRIPOLE. say TRIPOLE field, TF.
• So this also answers the fifth question, the fields do not generate each other but are called different names in different planes."

Perpendicular in cartesian system or a different angle between planes (where cross product=0), in another coordinate system.

 

[Vanadium 50]

But what if EF and MF are not represented by cartesians coordinates

The physics is not affected by the coordinate system a human being chooses to describe that physics.

 

[narsep]

I agree with Vanadium 50.
So, I return to my first question: "Is there any object that behaves to these fields (electric and magnetic) differently? e.g. Has the max transparency when the angle between these two fields is not 90 degrees?"
If physics is not affected by the coordinate system a human chooses, then polarizers of 90 degrees are special case among different polarizers that have their max transparencies when the angle between these two fields is not 90 degrees. Otherwise it is as if cartesian coordinate has the God's blessing.

PS: From the above site: "However, in inhomogenous, non-linear, or isotropic media, the E and B fields may not be perpendicular, e.g. in a crystal (which is isotropic)."

 

[Nugatory]

Thanks for your answers. However reading them, I realized that all of them are based on the relation that the cross product of EF and MF equals to zero and hence they are perpendicular. (Am I right?)

You have it exactly backwards. They are perpendicular; therefore the cross product will be zero.

But what if EF and MF are not represented by cartesians coordinates (that enforce geometric perpendicularity) but by another coordinate system that "perpendicularity" (cross product=0) does not mean geometric perpendicularity?

The coordinates don't "enforce perpendicularity"; the magnetic and electric fields are what they are and are perpendicular to one another the same way no matter what coordinates we choose.

Perpendicular in cartesian system or a different angle between planes (where cross product=0), in another coordinate system.

Do remember that the formula for calculating the cross-product changes when you change coordinate systems; that's part of the coordinate transformation.

 

[narsep]

Let us try to be quite clear:

You have it exactly backwards. They are perpendicular; therefore the cross product will be zero.

They are physically perpendicular, that means their cross product is zero, but not necessarily geometrically perpendicular. We never detect the orientation of a field experimentally, we only deduce it from experiments. So we can not start an argument from the physical perpendicularity of the fields.

The coordinates don't "enforce perpendicularity"; the magnetic and electric fields are what they are and are perpendicular to one another the same way no matter what coordinates we choose.

Indeed, the coordinates don't "enforce physical perpendicularity". The magnetic and electric fields are what they are and are physically perpendicular to one another the same way no matter what coordinates we choose. Their geometric perpendicularity depends on the used coordinate system, obeying to the coordinate transformation. Only in the cartesian coordinate system physical and geometric perpendicularity means the same. This is what I meant by "enforce of perpendicularity" by the cartesian system; physically perpendicular enforce to be also geometric perpendicular.
PS. Do remember the starting quest:

Is there any object that behaves to these fields (electric and magnetic) differently? e.g. Has the max transparency when the (geometric) angle between these two fields is not 90 degrees?

 

[Ibix]

They are physically perpendicular, that means their cross product is zero, but not necessarily geometrically perpendicular.

Do you have a reference for what "geometrically perpendicular" means? It's not a term I'm familiar with, and the only meaning I can impute to it from your usage here is that you are defining (1,0) and (0,1) to be "geometrically perpendicular" regardless of basis. But that seems to me to be a total misapplication of both words, since the vectors are not, in general, perpendicular (their inner product is not necessarily zero) and this definition is to do with coordinates, not geometry.

 

[BvU]

They are physically perpendicular, that means their cross product is zero,

And here's me thinking the dot product is zero when vectors are perpendicular

 

[vanhees71]

You have it exactly backwards. They are perpendicular; therefore the cross product will be zero.
The coordinates don't "enforce perpendicularity"; the magnetic and electric fields are what they are and are perpendicular to one another the same way no matter what coordinates we choose.
Do remember that the formula for calculating the cross-product changes when you change coordinate systems; that's part of the coordinate transformation.

I'm a bit puzzled about this.

If two vectors $\vec{a}$ and $\vec{b}$ are perpendicular, then the dot product vanishes, $\vec{a} \cdot \vec{b}=0$, but the cross product $\vec{c}=\vec{a} \times \vec{b}$ gives a third vector perpendicular to the other two.

It's easy to see, because $\vec{a} \cdot \vec{b}=|\vec{a}| |\vec{b}| \cos \phi$ and $|\vec{a} \times \vec{b}|=|vec{a}| |\vec{b}| \sin \phi$, where $\phi$ is the angle between the vectors. For $\phi=\pi/2$ you have $\cos(\pi/2)=0$ and $\sin(\pi/2)=0$. The cross product vanishes if the vectors are linear dependent (i.e., if they are parallel to each other), because then $\phi=0$ or $\phi=\pi$.

 

[Nugatory]

I'm a bit puzzled about this.

If two vectors $\vec{a}$ and $\vec{b}$ are perpendicular, then the dot product vanishes, $\vec{a} \cdot \vec{b}=0$, but the cross product $\vec{c}=\vec{a} \times \vec{b}$ gives a third vector perpendicular to the other two.

It's easy to see, because $\vec{a} \cdot \vec{b}=|\vec{a}| |\vec{b}| \cos \phi$ and $|\vec{a} \times \vec{b}|=|vec{a}| |\vec{b}| \sin \phi$, where $\phi$ is the angle between the vectors. For $\phi=\pi/2$ you have $\cos(\pi/2)=0$ and $\sin(\pi/2)=0$. The cross product vanishes if the vectors are linear dependent (i.e., if they are parallel to each other), because then $\phi=0$ or $\phi=\pi$.

I'm sorry, of course you're right. I was so fixated on the coordinate-independence (which I thought in is the sense of the OP's misunderstanding) that I didn't even notice which vector operation we were talking about.

 

[narsep]

Thanks vanhees71 and BvU for correcting us.
Physically perpendicular means that their dot (NOT cross) product is zero.
Geometrical perpendicular from Wikipedia "perpendicular"

In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). The property extends to other related geometric objects. ...

Ibix wrote:

... the only meaning I can impute to it from your usage here is that you are defining (1,0) and (0,1) to be "geometrically perpendicular" regardless of basis.

NO. I hope this is resolved after the correction.

 

[Ibix]

Physically perpendicular means that their dot (NOT cross) product is zero.
Geometrical perpendicular from Wikipedia "perpendicular"

Given that definition, these two things are the same. Having an inner product of zero means that the two vectors meet at a 90° angle, as you can see from vanhees71's definition. So I don't see the distinction you are trying to make.

 

[vanhees71]

Well, I'm not sure whether I understand the question, but if it's about plane em. waves, we just have to use what's always answers all questions about em. waves as long as we have no quantum effects to take into consideration, i.e., the good old Maxwell equations. Here we look for the plane-wave solutions (which never occur in nature but you can build all free-field soloutions of Maxwell's equations from them via Fourier integrals). The equations are (using Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0,$$
$$\vec{\nabla} \cdot \vec{E}=0, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0.$$
We start with the ansatz
$$\vec{E}=\vec{E}_0 \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x}),$$
$$\vec{B}=\vec{B}_0 \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x}),$$
where it is understood that the real parts of these expressions are the physical fields. The constant coefficients $\vec{E}_0$ and $\vec{B}_0$ are in $\mathbb{C}^3$.

Now we go into the equations. The vanishing divergences of both the electric and the magnetic components means that
$$\vec{k} \cdot \vec{E}_0=\vec{k} \cdot \vec{B}_0,$$
i.e., both electric and magnetic field components are transverse to the direction of wave propagation, given by the real wave vector $\vec{k}$.

Next we consider the 2nd equation (Faraday's Law), leading to
$$-\mathrm{i} \vec{E}_0 \times \vec{k}-\mathrm{i} \frac{\omega}{c} \vec{B}_0=0 \; \Rightarrow \vec{k} \times \vec{E}_0=\frac{\omega}{c} \vec{B}_0,$$
i.e., $\vec{E}_0$ and $\vec{B}_0$ are orthogonal to each other, and $\vec{E}_0$, $\vec{B}_0$, and $\vec{k}$ in these order form a positively oriented orthogonal system of vectors.

Then the 4th equation (Ampere-Maxwell Law) gives
$$-\mathrm{i} \vec{B}_0 \times \vec{k} +\frac{\mathrm{i} \omega}{c} \vec{E}_0=0 \; \Rightarrow \vec{B}_0 \times \vec{k}=\frac{\omega}{c} \vec{E}_0.$$
Now we use the equation we got before to get rid of $\vec{B}_0$
$$\frac{c}{\omega} (\vec{k} \times \vec{E}_0) \times \vec{k}=\frac{\omega}{c} \vec{E}_0$$
or, using $\vec{E}_0 \cdot \vec{k}=0$
$$\frac{c}{\omega} [\vec{E}_0 \vec{k}^2-\vec{k} (\vec{E}_0 \cdot \vec{k})]=\frac{c}{\omega} \vec{k}^2 \vec{E}_0=\frac{\omega}{c} \vec{E}_0.$$
Since we want to have $\vec{E}_0 \neq 0$, we must have
$$\vec{k}^2=\frac{\omega^2}{c^2},$$
which is the dispersion relation for electromagnetic waves.

This also implies that
$$\vec{B}_0 \times \frac{\vec{k}}{|\vec{k}|}=\frac{\omega}{c|\vec{k}|} \vec{E}_0=\vec{E}_0.$$
Since $\vec{k} \cdot \vec{B}_0=0$, you can easily prove from this that $|\vec{E}_0|=|\vec{B}_0|$.

 

[narsep]

vanhees71 wrore: "If two vectors are perpendicular, then the dot product vanishes,... It's easy to see, because cos(π/2)=0"
This is true ONLY if the two vectors are collinear to the basis vectors. This happens in cartesian system but NOT generally. Generally, we have to project the two vectors to the axis of the basis vectors that results to a dot product different from zero as cos(φ)≠0 (where φ is the angle between basis vectors that is not necessarily π/2).

 

[Ibix]

vanhees71 wrore: "If two vectors are perpendicular, then the dot product vanishes,... It's easy to see, because cos(π/2)=0"
This is true ONLY if the two vectors are collinear to the basis vectors. This happens in cartesian system but NOT generally. Generally, we have to project the two vectors to the axis of the basis vectors that results to a dot product different from zero as cos(φ)≠0 (where φ is the angle between basis vectors that is not necessarily π/2).

This is exactly what I said you were doing back in #11 and you denied in #15. You are presuming that the dot product of two vectors is the sum of the products of their components ($\vec a\cdot\vec b=a_1b_1+a_2b_2+...$) regardless of the basis. This is not correct in general. The dot product is the product of the moduli of the vectors times the cosine of the angle between them, or (in component notation with an assumed sum over repeated indices) $g_{ij}a^ib^j$. It's just that in cartesian coordinates, $g_{ij}$ is zero everywhere except where $i=j$ (i.e., is the Kronecker delta), and $g_{ij}a^ib^j$ simplifies to the sum of component products.

In other words, the inner product of two vectors is a coordinate independent value. If you are getting coordinate dependence then you are calculating something other than what physicists mean by the dot product.

 

[narsep]

I ask for a short time out ...

 

[Delta2]

Well, I'm not sure whether I understand the question, but if it's about plane em. waves, we just have to use what's always answers all questions about em. waves as long as we have no quantum effects to take into consideration, i.e., the good old Maxwell equations. Here we look for the plane-wave solutions (which never occur in nature but you can build all free-field soloutions of Maxwell's equations from them via Fourier integrals).

I may deviate from the main subject of this thread but wanted to ask, can spherical waves be build from Fourier integrals of plane waves? My main intuitive difficulty in understanding that is that in spherical waves the wave vector is not constant but changes from point to point (I know it always points in the radial direction though). In contrast all plane waves have constant wave vectors, so I believe any Fourier integral of plane waves will have a constant wave vector as well.

 

[vanhees71]

Yes, you can! You can expand any em. wave in terms of vector spherical harmonics (aka multipole expansion; equivalent to the energy-angular momentum single-photon states) as well as plane waves (equivalent to the energy-momentum single-photon states in QED).

 

[vanhees71]

vanhees71 wrore: "If two vectors are perpendicular, then the dot product vanishes,... It's easy to see, because cos(π/2)=0"
This is true ONLY if the two vectors are collinear to the basis vectors. This happens in cartesian system but NOT generally. Generally, we have to project the two vectors to the axis of the basis vectors that results to a dot product different from zero as cos(φ)≠0 (where φ is the angle between basis vectors that is not necessarily π/2).

As the name suggests the scalar product maps two vectors to a scalar, which is independent of any basis. Also vectors are independent of the choice of bases you might use to decompose them into components. In a Euclidean vector space the fact that two vectors are prependicular to each other, i.e., their scalar product vanishes, is independent of the choice of basis.

 

[vanhees71]

Given that definition, these two things are the same. Having an inner product of zero means that the two vectors meet at a 90° angle, as you can see from vanhees71's definition. So I don't see the distinction you are trying to make.

Well, Euclidean geometry can be defined by the axioms of a Euclidean affine manifold. That's way more convenient for the purposes of physics than the original way Euclid defined it in ancient times. Of course, it's still the same mathematical thing.

 

[narsep]

Well, I was wrong about dot product and the perpendicularity staff. Thanks everybody for your help.
Is the quote: "However, in inhomogenous, non-linear, or isotropic media, the E and B fields may not be perpendicular, e.g. in a crystal (which is isotropic)." right? Apart from the "isotropic" (that should be "anisotropic").

 

[vanhees71]

That's correct.