Remember, the change in intensity of light is proportional to

If the change in distance is very small compared to the initial distance, this ratio is one, and the intensity will remain essentially unchanged. Let's have another look at a 1/R² graph light curve which is intended to depict a person walking away from a wall while he or she shines a flashlight on the wall. The graph represents the intensity of the light on the wall as the person walks away. Let's suppose the time to get to the midpoint on the graph is about 5 seconds.

Notice the graph now has a scale and the time to walk to the midpoint of the graph (where the intensity is greatly reduced) is just 5 seconds. Certainly over this timespan the effect of the light reduction and therefore the curvature of the 1/R² graph can be seen.

But lets now consider what must happen for this effect to be seen with the intensity of M31. M31 is traveling very fast no doubt, but it has been traveling fast for a very, very long time. When you begin to measure the intensity of it from its lightcurve, which is certainly possible, it is starting at an enourmous distance from the earth (and you the observer). Even at a tremendous speed the galaxy covers a distance which is trivial when compared to the distance it was from you alreadly. M31 is at a distance of approximately 300,000 parsecs (pc) away from the earth. And.... a parsec is approximately equal to 200,000 astronomical units (AU) or in brief it is about 3 x 10⁵ radii to the sun! Maybe to make some sense of this we should look at the equations regarding this again.

The final intensity proportional to the initial intensity and inversely proportional to the distance to M31 squared. But what happens to the equation when Dr is very, very small as compared to r? Here's a little spreadsheet which shows the results.

So...now you know why the light curve for M31 looks flat!

Click here to return to the beginning and try another experimental approach.