# How does the apparent brightness of linear light sources fall with distance?

It is common knowledge that point-like light sources fall in brightness like $$1/r^2$$. For example, see this NASA website that describes why this is the case. It is well known that surface brightness, the luminosity per unit area, of a resolved surface is invariant with distance (ignoring cosmic dimming caused by the expansion of the universe).

How does the apparent brightness of a linear source (i.e. long and resolved in one direction, unresolved in the other) fall?

• I know this seems like a home-work like question, but I've recently had a professional astronomer reviewing a proposal I'd written who got this wrong, so I feel the fact needs more documentation. Aug 3, 2021 at 19:34
• As you speak of resolved objects it may bei useful to think about how this would be solved in photography? I.e. first, calculating the radiance, which is preserved and not decreasing with distance. Then multiplying it with the aperture of your lens and the solid angle under which a pixel sees the object. I'm not sure what to do if the pixel is only partly filled (first guess, just use the fraction that is filled, which of couse makes trouble in case of an ideal one-dimensional object...). Aug 3, 2021 at 23:03
• @CharlesTucker3 Basically, an object which is resolved in both directions (i.e. angular size >> resolution) and has constant radiance will follow the constant surface brightness law. One that is unresolved in both directions will follow the point-source law. One that is resolved in 1 dimension, unresolved in the other will follow the linear source law. What you describe is how you would need to handle the edge cases that are on the boundary between the main cases. Aug 4, 2021 at 4:39

## 1 Answer

The inverse square law can be derived from conservation of energy and symmetry concerns, the same way Gauss's law shows that electric fields fall of as $$1/r^2$$. A similar exercise with an infinite plane is what shows that surface brightness is invariant with distance.

Performing that same exercise on an infinite line shows that linear sources will have an apparent brightness that falls like $$1/r$$.

In more detail: place an imaginary cylinder coaxial with the linear light source that has radius $$r$$ and height $$h$$. By reflection symmetry, the circular ends will have an equal amount of light flowing in each direction through them, so the net energy flux is zero. All of the energy, therefore, flows out through the lateral side of the cylinder, which has area $$A=2\pi r h$$. The total amount of power enclosed in the cylindar is proportional to the length $$P = \lambda h$$, with $$\lambda$$ the linear brightness of the source (power per unit length).

Setting both powers equal gives: \begin{align} \lambda h &= A_{\mathrm{lateral}} F \\ &= 2\pi r h F. \end{align} Thus, the flux from the source is \begin{align} F & = \frac{\lambda}{2\pi r}, \end{align} which falls like $$r^{-1}$$, as expected.