Graph of inverse square law for radiation intensity

AI Thread Summary
The discussion centers on the mathematical representation of the inverse square law for radiation intensity using logarithms. The equation lnI = ln(S/4π) - 2lnr is confirmed as correct, illustrating the properties of logarithms. Participants clarify the relationship between this equation and the linear form y = mx + c, identifying y as lnI, x as ln r, with a slope m of -2 and an intercept c of ln(S/4π). A correction is made regarding the notation for the slope, emphasizing that it should be m = -2, not -m = -2. Overall, the mathematical relationships and corrections are accurately discussed.
Luke1121
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If I=s/4pir2 would It be correct to write this in terms of logs like this:. lnI=(lns/4pi)-2.lnr Also how could this relate to y=mx+c? I think it's y=lnI. X=lnr. -m= -2 and c= lns/4pi. Is this correct? Thank you
 
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Luke1121 said:
If I=s/4pir2 would It be correct to write this in terms of logs like this:. lnI=(lns/4pi)-2.lnr

If you meant \ln I = \ln(S/4\pi) - 2 \ln r, then yes, this is correct, because

ln(abc) = ln(a) + ln(bc) (the log of a product equals the sum of the logs of the individual factors in the product).

and

ln(bc) = cln(b)

Luke1121 said:
Also how could this relate to y=mx+c? I think it's y=lnI. X=lnr. -m= -2 and c= lns/4pi. Is this correct? Thank you

That looks right. I mean, y is the dependent variable (in this case log of intensity), x is the independent variable (in this case log of radial distance) . The slope m is the constant factor that multiplies the independent variable. The intercept c is what you get when x = 0.

EDIT: It should be m = -2, NOT -m = -2.
 
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You wrote -m=-2 which results m=2, a positive slope of the ln(I)-lnr graph. Is it right? ehild
 
Ah of course it's not -m, seems like I confused myself. Thank you
 
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