I Two blackbodies at two foci inside an ellipsoidal shell

AI Thread Summary
Two spherical blackbodies within an ellipsoidal heat-reflecting shell can emit the same power despite having different radii and temperatures, as their surface power flux density is inversely proportional to their radius. If their temperatures were to equalize, they would then emit different powers due to the relationship between power emission and surface area. The discussion raises the question of whether these temperatures will converge over time. It is generally accepted that two bodies in thermal equilibrium will reach the same temperature. Therefore, it is expected that the temperatures of the two blackbodies will eventually equilibrate.
particlezoo
Messages
111
Reaction score
4
Let's consider two spherical blackbodies at two foci inside an ellipsoidal heat-reflecting shell. Consider the situation that they both have different radii and that their temperatures are such that they emit the same power. Thus, the surface power flux density of each is inversely proportional to the square of their radius. Therefore, when these spherical blackbodies are emitting the same power, they are at different temperatures.

Conversely, if their blackbody temperatures were to become the same, they would be emitting different powers, as the power emitted would be proportional to the surface of the object.

So should I expect these temperatures to come together, or not?
 
Physics news on Phys.org
particlezoo said:
Let's consider two spherical blackbodies at two foci inside an ellipsoidal heat-reflecting shell. Consider the situation that they both have different radii and that their temperatures are such that they emit the same power. Thus, the surface power flux density of each is inversely proportional to the square of their radius. Therefore, when these spherical blackbodies are emitting the same power, they are at different temperatures.

Conversely, if their blackbody temperatures were to become the same, they would be emitting different powers, as the power emitted would be proportional to the surface of the object.

So should I expect these temperatures to come together, or not?
Good question.
My vote is that the two bodies will come to thermal equilibrium.

( Would the two blackbodies have all of their emission transmitted to the other m even if they are situated at the loci? )
 
When the two balls are emitting the same total Wattage, their temperatures are different as you say. If left to evolve from this initial state, the temperatures of the black balls would eventually become equal.

What reason is there to think the temperatures would not equilibrate?
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Hello! I am generating electrons from a 3D gaussian source. The electrons all have the same energy, but the direction is isotropic. The electron source is in between 2 plates that act as a capacitor, and one of them acts as a time of flight (tof) detector. I know the voltage on the plates very well, and I want to extract the center of the gaussian distribution (in one direction only), by measuring the tof of many electrons. So the uncertainty on the position is given by the tof uncertainty...
Back
Top