How Does Light Remain After Polarization Despite Changes in Its Electric Field?

In summary, the topic being discussed is polarization of light and the confusion about the electric and magnetic fields. The intensity of the electric field is used to track polarization, but the form of the fields (oscillation with mutually orthogonal electric and magnetic components) is still maintained. A polarizer will typically pass the electric field according to the cosine of the angle. In a general free electromagnetic field, the vector potential in radiation gauge can be written as a Fourier integral. The electric and magnetic fields are given by the derivatives of the vector potential. The type of polarizer being discussed consists of metal rods forming a grid, and the wave must have an E-field perpendicular to the grids and a magnetic field along the rods to pass through.
  • #1
sinus
13
1
I'm studying optics in this semester and one of the topic is polarization of light.

Please help me.
For ease of understanding we only consider the electric field E, right? I'm confused, if light passes through a polarizer then there will only be one field whose direction of vibration is only in one plane, say E whose direction of vibration is up-down. Then why is there still light after being polarized? Isn't light an E field and a B field that are perpendicular to each other? if one of the fields is missing (unable to pass through the polarizer) then it is no longer light right after being polarized?
I would be very thankful if someone explains this as clearly as possible.
 
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  • #2
The intensity of the E field is simply chosen as the way to keep track of the polarization because it is usually the strongest interacting component in our description of light. It is understood that the form of the fields (oscillation with B and E mutually orthogonal and orthogonal to propagation direction) is maintained except perhaps in a very small region A polarizer will typically pass E field according to cos of the angle (and the energy is quadratic in he fields hence cos2)). The quantum mechanical description is mostly compatible with this as is the passage through "linear" media where polarization is cleverly included.
Of course there are books dedicated to this subject
 
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  • #3
An electromagnetic wave has in any frame of reference always electric and magnetic components. It's most easily seen when expanding the free em. field in plane waves.

To that end we use the radiation gauge for the potentials, i.e., ##\Phi=0## and ##\vec{\nabla} \cdot \vec{A}=0##. One can show that this fixes the gauge freedom completely and can always be done for free fields, i.e., we get the general free em. field when applying these gauge constraints.

Now we write the vector potential as a Fourier integral,
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} [\vec{a}(t,\vec{k}) \exp(\mathrm{i} \vec{k} \cdot \vec{x}) + \vec{a}^*(t,\vec{k}) \exp(-\mathrm{i} \vec{k} \cdot \vec{x})].$$
From the gauge condition, ##\vec{\nabla} \cdot \vec{A}=0## you get
$$\vec{k} \cdot \vec{a}(t,\vec{k})=0,$$
i.e., for each ##\vec{k}## the ##\vec{a}(t,\vec{k})## are orthogonal to ##\vec{k}##, i.e., we have transverse waves. Now defining (for simplicity real) vectors ##\vec{\epsilon}_j(\vec{k})## with ##\vec{\epsilon}_{1}(\vec{k}) \cdot \vec{\epsilon}_2(\vec{k})=\vec{k} \cdot \vec{\epsilon}_j(\vec{k})=0## and ##\vec{\epsilon}_{j}^2(\vec{k})=1## such that ##\vec{\epsilon}_1(\vec{k}) \times \vec{\epsilon}_2(\vec{k})=\vec{k}/\|vec{k}|##, you can write
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \sum_{j=1}^2 \vec{\epsilon}_j(\vec{k}) [a_j(t,\vec{k}) \exp(\mathrm{i} \vec{k} \cdot \vec{x}) + a_j^*(t,\vec{k}) \exp(-\mathrm{i} \vec{k} \cdot \vec{x})].$$
Further the Maxwell equations tell you that
$$\Box \vec{A}=\left (\frac{1}{c^2} \partial_t^2 -\Delta \right) \vec{A}=0,$$
and this implies that
$$\left (\frac{1}{c^2} \partial_t^2 + \vec{k}^2 \right) a_j(t,\vec{k})=0 \; \Rightarrow \; a_j(t)=a_j)\vec{k} \exp[-\mathrm{i} \omega(\vec{k}) t], \quad \omega(\vec{k})=c |\vec{k}|.$$
So finally you get for a general free em. field the vector potential in radiation gauge
$$\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \sum_{j=1}^2 \vec{\epsilon}_j(\vec{k}) [a_j(\vec{k}) \exp(-\mathrm{i} \omega(\vec{k}) t + \mathrm{i} \vec{k} \cdot \vec{x}) + a_j^*(\vec{k}) \exp(\mathrm{i} \omega t -\mathrm{i} \vec{k} \cdot \vec{x})].$$
Now the electric and magnetic fields are given by
$$\vec{E}=-\partial_t/c \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
For each plane-wave mode you have transversal electric and magnetic fields, which together with ##\vec{k}## build a orthonormal dreibein: ##\vec{E} \times \vec{B} \propto \vec{k}##.
 
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  • #4
sinus said:
I'm studying optics in this semester and one of the topic is polarization of light.

Please help me.
For ease of understanding we only consider the electric field E, right? I'm confused, if light passes through a polarizer then there will only be one field whose direction of vibration is only in one plane, say E whose direction of vibration is up-down. Then why is there still light after being polarized? Isn't light an E field and a B field that are perpendicular to each other? if one of the fields is missing (unable to pass through the polarizer) then it is no longer light right after being polarized?
I would be very thankful if someone explains this as clearly as possible.
Consider the type of polariser which consists of metal rods forming a grid. The wave must not have an E-field acting along the metal rods, which would cause it to be reflected back as for a sheet of metal, so it should be at right angles to the grids in order to pass through. On the other hand, the magnetic field must be acting along the rods so as not to induce currents - the flux is not cutting the conductor.
 
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  • #5
tech99 said:
Consider the type of polariser which consists of metal rods forming a grid. The wave must not have an E-field acting along the metal rods, which would cause it to be reflected back as for a sheet of metal, so it should be at right angles to the grids in order to pass through. On the other hand, the magnetic field must be acting along the rods so as not to induce currents - the flux is not cutting the conductor.
Getting a suitable mental picture in one"s head is quite demanding; not so much with plane waves with basic H or V polarisation but picturing what's going on when a wave of arbitrary polarisation arrives from an arbitrary direction on a grid of parallel wires, at another arbitrary angle. Getting familiar with The Normal is the only way forward. Search for loads of images and find a variety of presentations and explanations of polarisation.
 
  • #6
If we place the polariser normal to a beam of unpolarised light this is how to analyse what gets through. The source in this case can be considered as two sources having orthogonal (opposite) polarisation. One polarisation (as defined by the E-field) is lined up with the slits and the other is at right angles. I will call these co- and cross-polarised. As the source is unpolarised, we drive each of our sources by a separate noise generator. Now we can see that one polarisation passes unscathed through the polariser whilst the other is stopped. So we can see that the power transmitted is one half. For angles other than normal there is no alternative to some geometry to find the magnitude of the component of each polarisation lining up with the slits. The power of each is then proportional to the magnitude squared. The two contributions may then be added on a power basis.
If the polariser is other than linear polarised, we should select two source polarisations which are co-and cross polarised to it. (Sorry to use that word so often)!
 
  • #7
tech99 said:
The source in this case can be considered as two sources having orthogonal (opposite) polarisation.
I always think that putting it that way can sound too arbitrary and it confused me in the middle of the last century. It helps to make the point that, resolving in orthogonal directions, gives components which are independent of each other. If, instead of right angles, you chose x and y axes not at right angles, you would get values of x which contained a component of y advice versa (a sort of crosstalk effect). It's also quite arbitrary to choose horizontal and vertical; it often but not always just makes the sums more straightforward.
tech99 said:
The two contributions may then be added on a power basis.
Hmm - yebbut. The two orthogonal components add vector-wise. Energy Conservation says that the powers must add up right in the same way that Pythagoras's theorem works; sum of the squares of the sides is the square of the hypotenuse. Adding vectors is less open to error than adding powers. You always know where you are, imo and adding posers won't tell you the direction of the resultant.
 
  • #8
sophiecentaur said:
Hmm - yebbut. The two orthogonal components add vector-wise. Energy Conservation says that the powers must add up right in the same way that Pythagoras's theorem works; sum of the squares of the sides is the square of the hypotenuse. Adding vectors is less open to error than adding powers. You always know where you are, imo and adding posers won't tell you the direction of the resultant.
For an unpolarised source, such as the Sun, the vectors of any two orthogonal polarisations are incoherent, so we cannot add them. We cannot add vectors which are varying in amplitude in a random way, but we can add the two powers.
 
  • #9
tech99 said:
For an unpolarised source, such as the Sun, the vectors of any two orthogonal polarisations are incoherent, so we cannot add them. We cannot add vectors which are varying in amplitude in a random way, but we can add the two powers.
I'm not disagreeing with your Science. But, imo, you have pulled the level of the thread way up. We had got as far as a simple linearly polarised wave (nicely modelled in terms of microwaves. Going into sunlight in one bound is introducing much more sophistication. We very easily get distracted into higher level technical dogfights (mea culpa) and can lose an OP.
 
  • #10
Agree.
 
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  • #11
tech99 said:
Agree.
The terms Polarisation, Bandwidth and Coherence are tightly bound together but seldom discussed in a helpful way, That's why I like the Radio Transmitter as a model source; it's as coherent as necessary. Ionised atoms, otoh. . . . . . .
 

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