I Mathematical disagreement with inverse square law?

[enslay]

Looks OK.

Note that for the plane point on the x-axis you get the usual 1/r² dependency when changing R (here R = r).

But when keeping R constant and moving along the plane in yz-directions you get 1/r³ dependency due to the changing angle of incidence (AOI) combined with 1/r².

And it gets even more complex when the plane point is not on the x-axis but some non-zero fixed yz-coordinates, and you change R. For R=0 you get 0 intensity, because the AOI is 90° and r is non-zero. Then, with increasing R, intensity reaches some maximum value, and for R->∞ it goes to zero again, approaching the 1/r² dependency, because the AOI approaches 0° and is not changing much anymore.

Sweet! And this is certainly easier to derive than using conditional probabilities. It's neat that I got the same answer though! To deal with the vanishing intensity in $R \to \infty$, I essentially considered the ratio of the intensities at a location $(y,z)$ for the $x=2R$ and $x=R$ planes. That ratio gives the expected 1/4. But thank goodness for that missing angle incidence term as everything works out exactly now! You're my hero!

Of course not: I think you may be relying too much on equations here and abandoning common sense. Do you think the amount of radiation detected depends only on the surface area of the director? What if you take a square detector and turn it through 90 degrees so it no longer faces the source?

You're right. I do have a tendency to think of the equations too much. A lot of what I did was very mechanical with the math without a lot of thought of the big picture.

Although, the planar detector here is infinite. You couldn't make it face away from the source. But you could cause some numerical problems by having the source exactly on the plane!

When I first derived the planar detector, I was obsessed with making it a proper probability density function (PDF). Both for the infinite plane and for a finite-sized detector. But this turned out to be a bad idea since, in the case of the infinite plane, only 1/2 of the "photons" hit the infinite plane detector. Well, that is unless you erroneously let "photons" also travel negative distances (which I initially did). So the infinite plane PDF really ought to integrate to 1/2 (which isn't a PDF anymore) reflecting that. And similarly for the finite-sized detector, lots of "photons" will never hit the finite-sized detector. So all my extra cumulative distribution function derivation for normalizing the density was a complete waste of time. The integral of the density function (for the infinite planar detector) over the finite plane then tells you what fraction of "photons" that actually hit that detector. Similar story for the cylinder detector.

 

[pbuk]

You're right. I do have a tendency to think of the equations too much. A lot of what I did was very mechanical with the math without a lot of thought of the big picture.

Although, the planar detector here is infinite. You couldn't make it face away from the source. But you could cause some numerical problems by having the source exactly on the plane!

You have made exactly the same mistake again - your inappropriate mathematical model has introduced a singularity whereas inspection of the "big picture" should tell you that with true linear dispertion the number of particles that hit a surface parallel to their motion is exactly zero.

The integral of the density function (for the infinite planar detector) over the finite plane then tells you what fraction of "photons" that actually hit that detector.

If you want to know the answer for something finite it is usually better to start with something infinitessimal and work up rather than something infinite and work down.

 

[enslay]

You have made exactly the same mistake again - your inappropriate mathematical model has introduced a singularity whereas inspection of the "big picture" should tell you that with true linear dispertion the number of particles that hit a surface parallel to their motion is exactly zero.

What's inappropriate about starting with a point source emitting "photons" in uniform random 3D directions and transforming that distribution into a distribution of "photons" hitting a parameterized surface? It's not like I chose to have an infinite plane before hand. The infinite support emerges as a consequence of the original distribution... which would actually span an infinite plane (or infinite cylinder), wouldn't it?

If you want to know the answer for something finite it is usually better to start with something infinitessimal and work up rather than something infinite and work down.

I started with this distribution
$$
1 = \frac{1}{4 \pi} \int_{-\pi}^{\pi} \int_0^{\pi} \sin u du dt
$$
Which transforms into this distribution with infinite support
$$
1 = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{|R - x_0|}{\left ( (R - x_0)^2 + (y-y_0)^2 + (z-z_0)^2 \right )^{\frac{3}{2}}} dy dz
$$
Directly w.r.t. these conditions
$$
\begin{align*}
R = x &= x_0 + r \sin u \cos t \\
y &= y_0 + r \sin u \sin t \\
z &= z_0 + r \cos u
\end{align*}
$$
where $u, t$ are sampled from the first distribution mentioned above and $r$ is the distance that "photon" travels to the plane $x = R$.

Now we can integrate the Cauchy PDF and confirm that, say, point source at $(x_0,y_0,z_0) = (R,0,0)$ does not even hit a finite portion of the plane spanned by $y \in (5,10),\ z \in (5,10)$.
$$
0 = \frac{1}{2 \pi} \int_5^{10} \int_5^{10} \frac{|R-R|}{\left( (R-R)^2 + y^2 + z^2 \right )^{\frac{3}{2}}} dy dz
$$
In the limiting case for the "singularity", the Cauchy PDF, being a PDF, will still integrate to unity over its infinite support.

My issues span from fixating on the math of transforming one PDF into another and missing, firstly, that only half the photons hit an infinite plane (so it should integrate to 1/2). And secondly, an integration over a finite portion of the plane should not be normalized to integrate 1. I was fixated on keeping things as PDFs. It's more useful to be able to integrate the PDF on a finite portion of the plane and know the proportion of "photons" that will hit that finite portion of the plane.

EDIT: Eek, swap du and dt! And now let's substitute $y_0 = 0, z_0 = 0$ too!

 

[PeterDonis]

The probability density functions characterizing where photons strike the detector surface always seems to have a 1/r^3 in it.

This claim requires a valid reference. And any such reference should have the math to back up the claim. That is the sort of basis we would need to have a useful discussion of the topic.

In the absence of such a basis, this thread is closed. @enslay if you can find a valid reference, PM me and it can be reviewed.