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Recently I've started to be really intrigued with the electromagnetic spectrum and bumped into this problem:

According to the wave theory of light (or any electromagnetic wave, really), the magnetic field is perpendicular to the direction of the wave, and the electric field is perpendicular to both the direction of the wave and the magnetic field. So the electric field is vertical (at least according to the pictures I found up) and the magnetic field is horizontal. So far so good.

But is the electric field always vertical? And vertical with respect to what? Gravity? Are light waves, when viewed from the direction of the wave looking like this '+'? What happens if I rotate the source of light? Will the waves be compiled like this 'x' ? Or is the space orientation of the magnetic and electric waves purely arbitrary? Do they undergo spinning across the vector of the direction of the wave?

Hope I made my point clear. Thanks!

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The direction of the electric field is called polarization. The direction of the electrical field in the free space lies in the plan perpendicular to the direction of propagation, and if this direction is unique for all the beam, it is said to be linear polarization. So, for linear polarization, the electric field can point in whatever direction that lies in the plan perpendicular to the direction of propagation. (As a side remark, in some cases there exist also longitudinal e.m. waves, e.g. such waves can be obtained in a resonant cavity.)

But there exist also elliptical polarization, in particular, circular polarization. In this case the polarization is not fixed, it rotates during the time, look in Wikipedia, http://en.wikipedia.org/wiki/Circular_polarization. You will see an animation. There is right polarization, in which case the electrical vector rotates counter-clockwise, and left polarization, in which it rotates clockwise.

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  • $\begingroup$ You said 'the electric field can point in whatever direction [] perpendicular to the direction of propagation'. Ok, I understand that, but which direction is it? Is it like + or x? Or something in between? There is an infinite number of possible arrangements, just by rotating the wave line around. $\endgroup$ – Roll Dec 10 '14 at 22:49
  • $\begingroup$ @Roll : Let's distinguish between linear polarization and circular (elliptical is a combination of these two, s.t. let's keep simple). Let's admit that a beam of light comes straight towards you. And let's take this direction as direction z. Now, in the plane x, y take an axis x horizontal, and an axis y vertical. For linear polarization, take an axis that makes some arbitrary angle with x. This axis, let's call it x', is a perfectly valid axis for the linear polarization. $\endgroup$ – Sofia Dec 10 '14 at 23:16
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Vertical with respect to whatever arbitrary set of coordinates you devise. If most of the pictures you see have the E-field pointing "up", that's just some kind of cultural bias. The E-field can point in any direction at all.

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There is a problem with diagrams like this

enter image description here

Yes if that was a physical object and you looked at it end-on it would look like a plus-symbol. However it isn't intended to be interpreted physically like that.

These diagrams shouldn't be interpreted as showing a vertical or horizontal displacement

What is being shown is field strength and direction at a single point over time.

Here's a diagram of circularly polarised light

enter image description here

What you'd see looking into this end-on would be a point (or dot).

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As has already been said: the orientation of the field drawn in your book is arbitrary. Once you have decided that the wave is say going into the page, then all that is required is that the E-field be drawn on the page (i.e. at right angles to the direction of wave propagation). You can put it up, down, left, right, diagonally, it doesn't matter, it would be a valid representation of an electromagnetic wave. The direction of the B-field would then not be arbitrary. It would have to be drawn in the page (so also at right angles to the wave propagation), at right angles to the E-field, and it must also be the case that the vector product $\vec{E} \times \vec{B}$ points in the direction of wave propagation.

The degrees of freedom for the E-field direction then allow other forms of polarisation. The situation I described is linear polarisation, but so long as the E-field vector stays on the page, it can vary with time, it can rotate, it can get bigger or smaller in amplitude etc.

What controls this behaviour is the way that the wave is produced and how it interacts (interferes) with other electromagnetic waves or with interfaces between different materials. For instance if you produce EM waves from a simple linear, oscillating electric dipole, this will produce linearly polarised light where the E-field oscillates up and down along a single line drawn on the page. On the other hand if I then mix this radiation with light with a perpendicular polarisation that oscillates in phase, I will produce light polarised at some angle in between the two (45 degrees if they are of equal amplitude). On the other hand if one of the two waves I mix lags the other in phase by $\pi/2$ I would produce a rotating electric field vector - circularly polarised light. If it leads by $\pi/2$ the electric field vector rotates the other way.

You can also get unpolarised light where the light is an incoherent mixture of different random polariation states. You could turn (some of) this into polarised light by passing it through a polarising filter or by reflecting the polarised light off a block of glass at the Brewster angle. There are lots of interesting ways to mess about with the polarisation of light!

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  • $\begingroup$ Ok if the orientation of the field in my book is arbitrary, what's the reality like? Let's say I take a laser pointer and I aim it in vacuum at some spot, let's say, 10meters away from the pointer. I will have this nice and neat line of light - but what about the polarisation? Will all the light beams be polarized the same? Will they be polarized randomly? Or will they undergo that circular thing? I know it may get all tricky when subject to external influence, but what happens in this neat ideal-like example? $\endgroup$ – Roll Dec 11 '14 at 0:25
  • $\begingroup$ @Roll It's not neat or ideal. Some lasers produce linearly polarised light, some don't; it depends on their mode of operation and output optics. You certainly couldn't say which way any polarisation was oriented without knowing how the laser was put together, or by measuring it. $\endgroup$ – Rob Jeffries Dec 11 '14 at 0:36
  • $\begingroup$ Ok so what about a candle light? Let's say I lit a tea candle; how will the waves be perpendicular to the wave of propagation? This is natural light so it shouldn't be subject to any mode of operation or output optics. Will this kind of light have circular waves or just straight waves? $\endgroup$ – Roll Dec 11 '14 at 2:02
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    $\begingroup$ @Roll Unpolarised: random, rapidly changing E-field vectors perpendicular to the direction of wave propagation. The light is the result of many unrelated, incoherent emission events, each of which last of order $10^{-7}$ s. I should also have said in response to your OP, that gravity has no effect on defining the polarisation state. $\endgroup$ – Rob Jeffries Dec 11 '14 at 7:33

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