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Wikipedia says "Since rays in space can be parameterized by three coordinates, x, y, and z and two angles $\theta$ and $\phi$, as shown at left, it is a five-dimensional function"

I'm not understanding why $\theta$ and $\phi$ are necessary here, or why this needs to be a five-dimensional function at all. If you change the angle values of $\theta$ or $\phi$ doesn't that rotate the vector in space at its endpoints and thus change the x,y,z values? Can't you parameterize the ray in space with only x,y,z or only $\theta$ and $\phi$?

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You need to know two things about a ray:

  • its direction, which can be specified by two angles, $\theta$ and $\phi$;
  • and a point through which it passes, which needs three coordinates, $x,y,z$ (or equivalent in another coordinate system).

You can see that you need all of these by thinking about the case where you only have one or the other set:

  • if you only have $\theta$ and $\phi$, then obviously there is a whole family of rays which you can construct simply by starting with one and dragging it around, like filling space with a bundle of dried spaghetti;
  • if you only have $x, y$ & $z$, then you have a point in space, through which you can pass a whole family of rays in all different directions.

So you need both things.

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  • $\begingroup$ So I was assuming that all the rays are traveling "outward" from the origin of the field. In that case, their direction would be from the origin to the point specified by x,y,z. But it sounds like the rays can pass through directions/along lines that do not pass through the origin? How would a ray come from the origin in a direction that isn't related to the x,y,z? $\endgroup$ – Dante Feb 23 '16 at 20:40
  • $\begingroup$ Oh, xyz is the origin in the picture, and then the angle specifies where the ray is passing through the xyz point $\endgroup$ – Dante Feb 23 '16 at 20:46
  • $\begingroup$ @Dante Exactly! you need a point (the origin, if you like), and a direction. $\endgroup$ – tfb Feb 23 '16 at 20:48
  • $\begingroup$ So that means the plenoptic function accounts for rays that hit surfaces and reflect back in different directions? Not a theoretical field that isn't interacting with things? $\endgroup$ – Dante Feb 23 '16 at 20:49
  • $\begingroup$ Oh I see, it accounts for multiple light sources in the field, not necessarily reflective surfaces $\endgroup$ – Dante Feb 23 '16 at 20:50
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The point is that x,y,z are fixed and the angles can be independently varied. So since there are an infinite number of rays/lines that can go through point x,y,z we need a way to differentiate them, and the angles they make with the coordinate system works to do that.

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Just from reading the Wikipedia article, this "plenoptic function" is supposed to give the radiant intensity going through the point $(x,y,z)$ in the direction given by $(\theta,\phi)$.

You cannot omit any of those without losing information. If you omit $(\theta,\phi)$, then you can only speak about the radiant flux through $(x,y,z)$, but lose the directional information. If you omit the $(x,y,z)$, then you get the total radiant intensity in the direction $(\theta,\phi)$, but lose the information about where you are measuring this intensity.

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The angle parameters are there to indicate the direction the light ray takes as it leaves the position $(x,y,z)$. You need the three position parameters to indicate one point the ray passes, and the two direction parameters to indicate in which direction it does so.

Perhaps it's somewhat easier to think of these parameters as specifying two points that the ray passes, $(x,y,z)$ and $(x',y',z')$, as those will specify it uniquely. However, that information is redundant, since choosing the second point twice as far away (e.g. choosing $(x,y,z)=(0,0,0)$ and $(x',y',z')=(1,1,1)$ vs. $(x',y',z')=(2,2,2)$) results in the same ray, so you cull one degree of freedom in $\mathbf r'$ by reducing it to two angles.

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