Timeline for Diffraction in Electromagnetic Waves
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| S May 26, 2016 at 13:26 | history | suggested | YakovL | CC BY-SA 3.0 |
removed two-storey fractions with tiny numerators
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| May 26, 2016 at 13:25 | review | Suggested edits | |||
| S May 26, 2016 at 13:26 | |||||
| Jan 27, 2014 at 1:08 | comment | added | user24082 | Oh I think I see. The field is just the solution, and a wave front is what you get when you send something along that solution. | |
| Jan 27, 2014 at 1:03 | comment | added | user24082 | So the field is an amalgamation of every possible solution? I'm still confused- I know it's trivial but I think I should be able to understand this. I think of a wave as something that moves as an imaginary exponential- is the field every possible solution? | |
| Jan 27, 2014 at 0:12 | comment | added | Selene Routley | @Anthony ....the same shape but the $f(z-c\,t)$ solution will move at constant speed $c$ in the $+z$ direction. | |
| Jan 27, 2014 at 0:11 | comment | added | Selene Routley | @Anthony I'm thinking of the solution to Maxwell's equations that fits the boundary conditions of the problem at hand. I guess you may be thinking of a wavefront: a contour of constant phase, which moves through the field. For example, all six Cartesian electromagnetic components fulfill $(\nabla^2 - c^{-2} \,\partial_t^2) f(x,y,z,t) = 0$: in one dimension this equation becomes $(\partial_z^2 - c^{-2} \,\partial_t^2) f(z,t) = 0$ and has the basic solution of $f(z\pm c\,t)$ for any twice differentiable function $f$. So, if $f$ is a bump at $t=0$, thereafter the wave will be a "bump" of .... | |
| Jan 26, 2014 at 23:30 | comment | added | user24082 | When you say field, are you thinking of some ambient, large, field? | |
| Jan 26, 2014 at 23:17 | comment | added | Selene Routley | @Anthony Either "Field" and "Wave" will do here I think, for this discussion they have the same meaning: a solution to e.g. Helmholtz's equation or the D'Alembert wave equation $(\nabla^2 - c^{-2} \,\partial_t^2)\psi = 0$. Any number of solutions to these equations can be added together to get another solution (interference id thus simply a linear superposition phenomenon), and a wave with its edges clipped off by a slit cannot be one plane wave: you have to add many plane waves together to match the boundary conditions. | |
| Jan 26, 2014 at 20:15 | comment | added | user24082 | Thank you for all your help, but I just don't get the last bit. When you say a field, are you talking about an ambient field? I always think of light as just moving as it's own wave. | |
| Jan 23, 2014 at 8:00 | comment | added | Selene Routley | @Anthony ... transforming is equivalent to diagonalising a matrix. We apply the propagation transformation to th eigenfunctions because this is the easiest way for analysis. | |
| Jan 23, 2014 at 7:59 | comment | added | Selene Routley | The key thing to understand is that the aperture is of finite width. So there are Fourier components with nonzero spatial frequency. If the aperture is very wide, these spatial frequencies get smaller and smaller: in the limit of an infinitely wide aperture, we truly do have only a plane wave and there is NO diffraction. This is why laser beams diverge. Why do we use the Fourier transform? Because plane waves are the eigenfunctions of freespace: these are the fields that propagate by taking on a phase but their form stays the same. So Fourier transforming - phase delay - inverse Fourier ... | |
| Jan 23, 2014 at 4:49 | comment | added | user24082 | Or are you looking at the surface perpendicular to propagation? And we choose to Fourier transform this because of Huygens principle? | |
| Jan 23, 2014 at 4:46 | comment | added | user24082 | I don't follow, I'm afraid. The wave front is moving forward through the aperture, isn't this a single plane wave? | |
| Jan 23, 2014 at 4:37 | comment | added | Selene Routley | Because a plane wave in one direction makes a purely sinusoidal variation across a transverse plane as it cuts through it. If it is travelling precisely in the direction of propagation, the phase front aligns with the transverse plane and there is no transverse variation. So any field with a non sinusoidal variation across a tranverse plane must have more than one plane wave Fourier component. Paraxial fields are fields comprising plane waves running at shallow angles relative to some "mean" propagation direction. | |
| Jan 23, 2014 at 4:34 | comment | added | user24082 | Why does the field have components in all directions, not just the direction of propagation? | |
| Jul 9, 2013 at 7:20 | history | edited | Selene Routley | CC BY-SA 3.0 |
Corrected ground state energy in footnote
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| Jul 9, 2013 at 3:39 | vote | accept | CommunityBot | ||
| Jul 9, 2013 at 2:04 | history | answered | Selene Routley | CC BY-SA 3.0 |