Black Body Radiation: Exploring Radiated & Perpendicular Power

Click For Summary
SUMMARY

The discussion focuses on the concept of black body radiation, specifically addressing the relationship between radiated power and perpendicular power. It clarifies that radiated power is greater than perpendicular power due to the need to account for the angle of incidence, which requires the use of the cosine function (cos(theta)) to determine the perpendicular component. Additionally, the averaging of power over various angles is necessary to accurately represent the total energy transmitted through a surface, rather than simply summing powers from different angles.

PREREQUISITES
  • Understanding of black body radiation principles
  • Familiarity with the cosine function and its application in physics
  • Knowledge of power transmission concepts
  • Basic grasp of angular measurements in physics
NEXT STEPS
  • Research the Stefan-Boltzmann Law and its application in black body radiation
  • Learn about the Planck's Law of black body radiation
  • Explore the concept of angular distribution of radiation
  • Study the implications of radiative heat transfer in engineering applications
USEFUL FOR

Physics students, educators, and professionals in engineering or thermal dynamics who seek to deepen their understanding of black body radiation and its mathematical foundations.

dsdsuster
Messages
30
Reaction score
0
Can someone help explain what's going on here?
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radpow.html#c1

Specifically the part about how radiated power is greater than perpendicular power.
I also don't understand why the power is being averaged over the various angles. Doesn't it make more sense to add the powers from light arriving at different angles to get the total energy transmitted?
 
Science news on Phys.org
I can't follow this very well, nothing is defined and I can't work out the overall goal.

I think it's saying that the radiation going through the specified surface from some angle is not perpendicular, so you need to use cos(theta) to find the perpendicular component of the vector.
 

Similar threads

  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
17K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
6K