Why is a resonant frequency assumed in blackbody radiation?

[kelly0303]

Hello! In the descriptions of the BB radiation that I read, I see that we assume we have a cavity at a fixed temperature in equilibrium, we make a hole in it and we look at the intensity of different frequencies emitted through that hole. As far as I understand, the intensity dependence on the frequency (and temperature) doesn't depend on the shape of the cavity. I am not sure I understand why. For a given shape of the cavity, certain frequencies are not allowed (for example for a box, only wavelength that are a divisor of the length of the box work). So how can we talk about the intensity at a given frequency (and temperature), if that frequency is not allowed at all in the cavity? Shouldn't it be zero? And doesn't this mean that the shape of the cavity matters? What am i missing? Thank you!

 

[Mentz114]

Hello! In the descriptions of the BB radiation that I read, I see that we assume we have a cavity at a fixed temperature in equilibrium, we make a hole in it and we look at the intensity of different frequencies emitted through that hole. As far as I understand, the intensity dependence on the frequency (and temperature) doesn't depend on the shape of the cavity. I am not sure I understand why. For a given shape of the cavity, certain frequencies are not allowed (for example for a box, only wavelength that are a divisor of the length of the box work). So how can we talk about the intensity at a given frequency (and temperature), if that frequency is not allowed at all in the cavity? Shouldn't it be zero? And doesn't this mean that the shape of the cavity matters? What am i missing? Thank you!

I think you are confusing two different things. In a resonant cavity experiment which must take place in cold conditions the energy must be resonant with the cavity dimensions to be there at all. In the BB experiment a large cavity is heated and its material radiates thermally. There is no resonance and the shape is pretty much irrelevant.

 

[kelly0303]

I think you are confusing two different things. In a resonant cavity experiment which must take place in cold conditions the energy must be resonant with the cavity dimensions to be there at all. In the BB experiment a large cavity is heated and its material radiates thermally. There is no resonance and the shape is pretty much irrelevant.

Thank you for your reply. So the book I am reading is this: https://link.springer.com/book/10.1007/978-1-4419-1302-9. In chapter 2.2 he derives the formula for a blackbody. However, he uses a rectangular cavity and makes explicit use of the fact that only certain modes can be exist inside the cavity. Yet at the end he says that the cavity shape doesn't matter. I am just a bit confused about why is he using the allowed modes in the cavity if they don't matter.

 

[Mentz114]

Thank you for your reply. So the book I am reading is this: https://link.springer.com/book/10.1007/978-1-4419-1302-9. In chapter 2.2 he derives the formula for a blackbody. However, he uses a rectangular cavity and makes explicit use of the fact that only certain modes can be exist inside the cavity. Yet at the end he says that the cavity shape doesn't matter. I am just a bit confused about why is he using the allowed modes in the cavity if they don't matter.

In that case I'm guessing that perhaps if the BB cavity is large enough and hot enough then the absent modes would contribute very little to the total radiation.
I hope someone can set me right.

 

[kelly0303]

In that case I'm guessing that perhaps if the BB cavity is large enough and hot enough then the absent modes would contribute very little to the total radiation.
I hope someone can set me right.

But still, the frequency must be given by the Planck formula. Yet, if a mode can't exist inside the cavity, its intensity would be zero, regardless of the Planck formula. So something is definitely wrong with my understanding.

 

[Nugatory]

For a given shape of the cavity, certain frequencies are not allowed (for example for a box, only wavelength that are a divisor of the length of the box work).

That is true only if the walls of the box are idealized perfect conductors and are optically flat. It is not true, (but is a good approximation) across a wide range of frequencies if the walls of the box are realistic not-quite-perfect conductors finished to as close to optical flatness as possible. It is just plain not true and not even close if the walls are non-conductors.

However, all of that is just a red herring. Assume for the sake of argument that we have a rectangular cavity machined to perfect optical flatness out of ideally conductive high-grade unobtainium. With these assumptions it really is the case that the only possible modes will be those with wavelength $n\pi/L$ where $n$ is an integer and $L$ is the width of the cavity, something on the order of ten centimeters.

Now try calculating for yourself just what the energy gap is between two adjacent modes in the far infrared and shorter wavelengths. How bad is the approximation that we have a continuous spectrum?

 

[kelly0303]

That is true only if the walls of the box are idealized perfect conductors and are optically flat. It is not true, (but is a good approximation) across a wide range of frequencies if the walls of the box are realistic not-quite-perfect conductors finished to as close to optical flatness as possible. It is just plain not true and not even close if the walls are non-conductors.

However, all of that is just a red herring. Assume for the sake of argument that we have a rectangular cavity machined to perfect optical flatness out of ideally conductive high-grade unobtainium. With these assumptions it really is the case that the only possible modes will be those with wavelength $n\pi/L$ where $n$ is an integer and $L$ is the width of the cavity, something on the order of ten centimeters.

Now try calculating for yourself just what the energy gap is between two adjacent modes in the far infrared and shorter wavelengths. How bad is the approximation that we have a continuous spectrum?

Thank you so much! So the idea is that even if, in this idealize case, some frequencies are not allowed, the allowed ones are so close to each other that one can approximate it as a continuum. Is this right? However I am still confused about some stuff. I agree that for an order of few centimeters that holds. However, in the book I am reading they say that the size of the cavity doesn't matter. So if the cavity is few micro meters or 1 km instead, the result should be the same. However at a point the continuity approximation of the frequencies would fail, wouldn't it? Also just looking at the formula itself, the value of the intensity at a given frequency is not zero i.e. if I plug the $\nu$ in the Planck formula, for some $\nu$ not allowed by the cavity, I won't get zero. Yet the real intensity of that mode is zero (again in an idealized case, but also Planck formula is an idealized case, as no perfect black body exists, right?).

 

[Heikki Tuuri]

Suppose that the cavity has a diameter of 1 meter. An oscillator at one wall emits a photon whose wavelength is roughly one micrometer. The photon is absorbed by another atom at an opposing wall, or the photon escapes through the hole to the surrounding space.

I do not see any reason why the photon should be of a resonant frequency for the cavity.

A resonant frequency is needed if we want the photon to bounce many times from the walls of the cavity. In that case we may say that the empty space of the cavity has "absorbed" the photon.

 

[kelly0303]

Suppose that the cavity has a diameter of 1 meter. An oscillator at one wall emits a photon whose wavelength is roughly one micrometer. The photon is absorbed by another atom at an opposing wall, or the photon escapes through the hole to the surrounding space.

I do not see any reason why the photon should be of a resonant frequency for the cavity.

A resonant frequency is needed if we want the photon to bounce many times from the walls of the cavity. In that case we may say that the empty space of the cavity has "absorbed" the photon.

Well I am not sure why it should be resonant, but the author of that book definitely assumes so. I am just trying to understand why.