Estimating Radius for Half of Total Luminosity: Luminosity Integral Help

Click For Summary
To estimate the radius corresponding to half of the total luminosity from the integral L_T = ∫_0^{∞} e^{-(r/a)^{1/4}} r^2 dr, one must solve for r_{1/2} in the equation ∫_0^{r_{1/2}} e^{-(r/a)^{1/4}} r^2 dr = L_T/2. The challenge lies in the difficulty of directly solving this integral analytically. However, numerical methods or computational tools can effectively approximate r_{1/2}. Utilizing these computational resources is recommended for accurate estimation.
Logarythmic
Messages
277
Reaction score
0
If the total luminosity is given by

L_T = \int_0^{\infty} e^{-(r/a)^{1/4}} r^2 dr

estimate the radius r(a)[/itex] corresponding to half of the total luminosity.<br /> <br /> This would be to integrate from zero to r and get a function r(a,L) but this is impossible so anyone got an idea on how to estimate this?
 
Physics news on Phys.org
Well the primitive of the integrand exists, you knew that right?

The difficult part is solving for r_{1/2} in \int_0^{r_{1/2}} e^{-(r/a)^{1/4}} r^2 dr=L_T/2. But computers are good with this.
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
11K
  • · Replies 32 ·
2
Replies
32
Views
7K
  • · Replies 1 ·
Replies
1
Views
956
  • · Replies 21 ·
Replies
21
Views
2K