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Comment by tomxor | Hacker News Reader

> If you have a 1/r^2 decreasement from a point source, the if you have an infinite wall it becomes a constant value that does not decrease with the distance.

It will still decrease with distance. If you represent the wall as a collection of point sources distributed over a plane, as the distance to the plane diminishes, the distance to individual points do not diminish to and from the same distance as each other, and do not diminish at the same rate. However they still all diminish, and thus the sum intensity must diminish. Though the appropriate function describing the sum of the contributions from each point is no longer 1/d^2.

There will be some continuous equation to describe this... someone more familiar with radiance feel free to chime in!

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They’re referencing a classic Electricity & Magnetism proof. Given an infinite sheet of uniform charge density, the electric field is constant at all distances from the sheet, which is deeply unintuitive.

That is unintuitive, but then is my mistake only in representing the wall as an infinite plane? Would my parent comment still be true for a finite plane, or is that merely technically true but missing the point of the function shape being almost constant up to some threshold (depending on the limits of the finite plane).

I feel like I want to plot this now to build an intuition.

The problem is that this wall does not exist in a vacuum - the OPs comment is otherwise valid. The air is absorbing the radiation (very effectively).

Yes, I understand that. I'm wondering separately (in a vacuum if you will), what the falloff looks like with a finite plane, and how that relates to the infinite version. Obviously we cannot actually represent the wall as an infinite plane, from my current location it's accurate enough to represent it as a point source... so there is some threshold for a particular finite plane after which the "constant" nature is clearly not true.

If you're looking down at an infinite plane, then the plane occupies half your field of view (assuming you have a 360 degree field of view) no matter how far you are from it. Given a 2d object of uniform density, the component of the gravitational force towards that object in the direction perpendicular to the plane it lies in is proportional to the solid angle of your field of view it occupies. For an infinite plane, the components in other directions cancel out. I think this is related to how a sphere acts equivalently to a point at its center - an infinite plane is just a sphere of infinite radius.

The situation is similar for an opaque light source (apart from the lack of atmosphere in the way, I don't think the sun would be much brighter if you were much closer to it and looking through a pinhole), but I'm not sure it applies to gamma radiation which is emitted uniformly in all directions from each point and treats most stuff as pretty transparent - unless there's some weird interference you don't get to cancel out the parts coming from different directions.

Yes, I actually kind of get the infinite part, but thanks for the explanation it's interesting. I was wondering what the intensity looks like for a distance field around a finite plane and how that relates. This might be harder to describe.

What I'm suspecting is that under a certain distance for a given finite plane it's almost constant similar to the infinite version, but outside of that distance there must be some non uniform falloff function.

For a finite circle, where you're on a line perpendicular to it going through its center, the gravitational force you experience towards it is exactly proportional to the amount of your field of view it occupies - if I'm right, that should be proportional to `1-cos(atan(r/d))` where r is the radius of the circle and d is the distance between you and the center of the circle.

You are correct. For a short distance it's a almost a constant 1 like an infinite wall with the same charge density. For a long distance is almost 1/r^2 like punctual source. For an intermediate distance, it's a mess and you hope nobody ask for an exact solution.

A rectangle is a mess. It's easier with a circle. The approximations at short and long distance are the same, but a circle has an "easy" formula in between. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.h... 1-x/sqrt(x^2+R^2) where R is the radio of the circle. Let's use R=10 to keep it simple.

For a short distance 0<x<R/10, then it's almost a constant https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%...

For a long distance, x>4R it's almost like R^2/(2*x^2) https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%...

There are some trick to add more term to the approximations to reduce the middle part where both approximations are bad, like using A+Bx or A+Bx+Cx^3 for short distance and D/x^2+E/x^4 or D/x^2+E/x^4+F/x^4 for long distance. It depends on how much you care about the precision and how many calculations you want to do.

In some particular cases like a circle or the shell of a sphere there are closed formulas for the intermediate distances. In other cases there is no nice formula.

I don‘t think that‘s unintuitive. What goes into a source-free volume element has to go out of that volume element (radiation or field).

It’s certainly unintuitive to me in that I wouldn’t expect the field an inch and a trillion miles from the sheet to be identical

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