# Black body radiation: can you help me understand it (better)?

1. Jul 26, 2009

### nietzsche

My apologies if I am posting this in the wrong forum, and if this question has already been asked. This is my first post.

First off: I'm a first-year undergraduate student. My program is Philosophy of Physics.

I am trying to understand (more fully) black body radiation. I am having trouble understanding the experiment involving a cavity with a small hole in it.

From what I understand, a cavity with a small hole in it can be approximated as a black body. Photons enter the the hole and become trapped in the cavity, and all their energy is absorbed. The cavity will eventually reach "thermal equilibrium" with the radiation, which means that the amount of radiation absorbed is equal to the amount of radiation emitted. (Please, correct me if I'm wrong.)

1) The Ultraviolet Catastrophe: I don't understand this at all, to be honest. Here is what I think: EM radiation with a very high frequency will have a very high energy. When this kind of radiation enters the cavity, such as ultraviolet light, the amount of energy absorbed will be very high and consequently the amount of energy emitted will be infinite. Like I said, I'm not sure if what I'm saying makes sense. Also, what does this have to do with "the number of modes that can fit inside the cavity"? What does that even mean?

2) Thermal Radiation: The energy that is emitted is in the form of EM radiation. But what does this resulting energy have to do with the original energy absorbed in the form of photons? If, for example, a "red" photon entered the cavity, what would happen to the energy it was carrying? Does it necessarily have to be re-emitted as a red photon?

3) What is the relationship between energy and radiation? For example, let's take a microwave. Why makes the microwaves "intense" enough to heat the food? Is it because there are MANY photons? Because we are exposed to micro waves all the time, but we aren't cooked. What makes a microwave oven different?

4) What does it mean for EM radiation to have a "temperature"? I know that heat can be transferred through a vacuum as radiation. But the actual radiation does not really have a temperature, does it? It is only when it hits something that is able to detect it that we can measure its temperature. Is that correct?

Thanks in advance for your help. I'm trying my best to understand this stuff, but it's very confusing to me.

2. Jul 26, 2009

### JazzFusion

First off from me: Welcome to Physics Forums!! Glad to have you here, and I hope you will visit often and continue posting as you progress in your studies. You will be surprised how quickly you will go from posting to ask for help, to posting to give help.

You can never have infinite energy emitted, and Conservation of Energy shows that you can never have more energy emitted than what went inside to begin with. The best theory developed at that time (prior to the development of the theory of Quantum Mechanics) predicted that you would get that infinite amount of energy out. This was called the 'Ultraviolet Catastrophe', because the scientists could see that the prediction did not make any sense. It showed that the current theory had to be wrong, but no one at the time knew how to 'fix' it.

No. It is emitted as energy, but not necessarily at the same wavelength.

Here's a simple example: leave your car parked out in the sun in Miami, FL in July. Visible light (400 - 600 nm wavelength or so) will enter through the glass. The interior of your car will absorb this energy. The car seats and dash will try to re-emit it as thermal energy in the infrared part of the spectrum (8 - 15 microns wavelength or so). Your eyes can't see radiation at 8 - 15 microns, so you can't 'see' that your car is hot, but you sure can feel all that thermal energy when you climb in your car.

Others may answer this one more clearly, but the short version is that there are two things at work here: FIRST: the wavelength of the radiation (microwave, infrared, visible, ultraviolet, X-ray, etc) tells something about the energy of the individual particle. SECOND: You are right, you can have few or many particles hitting at any one time.

I would not describe EM radiation as 'having' a temperature. There is, however, a point in the spectrum of emitted energy where radiation 'peaks', and this depends on the temperature of the object. So, for instance, you can see the sun because it is hot enough to emit lots of radiation in the visible portion of the spectrum. If you went out in space and looked at the Earth, however, you could not see light emitted from the night side (you would only see light reflected from the sun from the daylight side). That is because the Earth is much cooler than the sun (lucky for us!), so it emits primarily in the IR part of the spectrum.

If you look up "http://en.wikipedia.org/wiki/Black_body_radiation" [Broken]" in Wikipedia, there is a very nice graph on the right-hand side part way down that shows how the peak wavelength changes with the temperature of the emitting object.

Never stop learning!!!! Best to you, and hope this helps.

Last edited by a moderator: May 4, 2017
3. Jul 27, 2009

### nietzsche

Thanks very much JazzFusion. You cleared up a lot of points that were confusing me. I had already been browsing Wikipedia but I still needed some clarification. Very much appreciated.

4. Jul 27, 2009

### Nabeshin

To expand upon this, the theory was called the Raleigh-Jeans theory which said essentially:
$$I \propto \frac{1}{\lambda^5}$$
So you can easily see that as lambda gets small (into the ultraviolet, say), the intensity of radiation at that particular wavelength gets arbitrarily large. This was derived entirely from "classical" (non-quantum, non-relativistic) physics, such as thermodynamics and statistical mechanics. It works quite well at large wavelengths, where the actual form converges to something very similar to this, but failed at high energies because of quantum concerns.

Last edited: Jul 27, 2009
5. Jul 27, 2009

### alexepascual

I think you meant ".. gets arbitrarily large.." ?

6. Jul 27, 2009

### lstellyl

I believe what is actually happening that makes the microwaves "intense" is taking advantage of fact that their wavelength resonates with water (which happens to be in most of the foods that we eat... making them "microwavable") which excites or vibrates the water molecules effectively heating up the food.

I think (not positive) that dishes and such that are not "microwave safe" have molecules that similarly resonate with the waves.

Keep in mind, however, that microwaves are quite common... and lower energy microwaves (some microwave ovens exert over 1000w) will not have near the same "energizing" effect on food or water because the waves will not have the same amplitude

someone feel free to correct me if anything i said is incorrect, this is just from what i understand.

7. Jul 28, 2009

### JazzFusion

You are correct in why microwave ovens can cook food. However, you are also correct that there are many sources of microwave radiation that are more or less powerful.

Microwaves themselves, due to their wavelength, carry a certain amount of energy. This is regardless of their source (your oven, your cell phone, or a tracking radar ). The total energy has to do with intensity - which drops off as an inverse square with distance, just like sound or light.

8. Jul 28, 2009

### mgb_phys

Microwave ovens aren't in a resonance with water, they do use the fact that water is dielctric (has different charges on different ends) to cause the water molecules to move but the bond energy is very different.

Microwave safe generaly means they wont melt (although the dishes aren't directly heated they are in contact with hot foods) and don't contain chemicals that will leak out at high temperatutes.

There is a difference between power and energy. Shorter wavelengths have more energy, so a single photon from a 1800Mhz cell phone will have more energy than an 2.4Ghz photon from a microwave. But a cell photon only puts out a fraction of a watt of power while a microwave generates 1000W = a lot more photons.

9. Jul 28, 2009