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I Linear polarizers: angle between the electric & magnetic fields

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  1. Nov 22, 2017 #1
    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?
     
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  3. Nov 22, 2017 #2

    BvU

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    Hello narsep, :welcome:

    Also from your link:
    so it doesn't happen !
     
  4. Nov 23, 2017 #3
    Is there a theoretical explanation why this is so?
     
  5. Nov 23, 2017 #4

    BvU

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    Google "why are b and e perpendicular" and pick one you like ...
     
  6. Nov 23, 2017 #5

    Ibix

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    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.)
     
    Last edited: Nov 23, 2017
  7. Nov 24, 2017 #6
    Thanks for your answers. However reading them, I realised 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.
     
  8. Nov 24, 2017 #7

    Vanadium 50

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    The physics is not affected by the coordinate system a human being chooses to describe that physics.
     
  9. Nov 25, 2017 #8
    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)."
     
    Last edited: Nov 25, 2017
  10. Nov 25, 2017 #9

    Nugatory

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    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.
     
  11. Nov 26, 2017 #10
    Let us try to be quite clear:
    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.
    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:
     
    Last edited: Nov 26, 2017
  12. Nov 26, 2017 #11

    Ibix

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    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.
     
  13. Nov 26, 2017 #12

    BvU

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    And here's me thinking the dot product is zero when vectors are perpendicular :smile:
     
  14. Nov 26, 2017 #13

    vanhees71

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    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##.
     
  15. Nov 26, 2017 #14

    Nugatory

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    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.
     
  16. Nov 26, 2017 #15
    Thanks vanhees71 and BvU for correcting us.
    Physically perpendicular means that their dot (NOT cross) product is zero.
    Geometrical perpendicular from Wikipedia "perpendicular"
    Ibix wrote:
    NO. I hope this is resolved after the correction.
     
  17. Nov 26, 2017 #16

    Ibix

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    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.
     
  18. Nov 26, 2017 #17

    vanhees71

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    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|##.
     
  19. Nov 26, 2017 #18
    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).
     
  20. Nov 26, 2017 #19

    Ibix

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    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.
     
    Last edited: Nov 26, 2017
  21. Nov 26, 2017 #20
    I ask for a short time out ... :smile:
     
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