# Difference between wavenumber and wave mode?

1. Jul 7, 2014

### sawer

Are these two, wave number and wave mode, same thing?

If not, what is the difference?

Thanks...

2. Jul 7, 2014

### Simon Bridge

Have you checked the definitions in, say, a dictionary, to see?
What did you find out?

3. Jul 7, 2014

### sawer

Yes I checked.

Linear wave number is L/(λ/2) for a specific frequency. That gives antinode number in a standing wave. So these two must be same thing.
http://web.utk.edu/~cnattras/Physics221Spring2013/modules/m10/images/stand.gif [Broken]
I'm trying to understand blackbody radiation. But the documents that explain blackbody radiation use "wave mode" term confusingly.

For exampe according to Rayleigh–Jeans every mode of a standing wave for a particular frequency has energy of kt. but what does that mean?

Does it, every antinode has an energy of kt? (So wave mode and wave number and antinode numbers are all same thing???)

Last edited by a moderator: May 6, 2017
4. Jul 7, 2014

### WannabeNewton

A wave mode refers to a normal mode of the electromagnetic radiation or equivalently a Fourier mode of the Fourier decomposition of the associated electromagnetic field. Recall that the electromagnetic field associated with the radiation in the cavity can be expressed as a superposition of independent harmonic oscillator solutions. The normal modes of the radiation then refer to these independent harmonic oscillator solutions. The normal modes therefore have associated with them a wave number $k$ corresponding to the frequency of oscillation so you can use wave mode and $k$ interchangeably. But nodes are different from modes.

Each mode of the electromagnetic field of the radiation in the cavity has an energy $k_B T$ by application of the equipartition theorem to the electromagnetic energy density.

5. Jul 7, 2014

### sawer

So if we rewrite this sentence: each "antinode" of a standing wave has an energy of $k_B T$ . Right?

So for that diagram:

http://web.utk.edu/~cnattras/Physics221Spring2013/modules/m10/images/stand.gif [Broken]

Total energy is: $3 * k_B T$ . Because wave number is 3. Righ?

And all the calculations of density of wave for blackbody is basically counting the number of antinodes. And all of them are same: number of antinode = number of waves = wave mode

Thanks WannabeNewton

Last edited by a moderator: May 6, 2017
6. Jul 7, 2014

### WannabeNewton

No. Each mode has an energy $k_B T$ from the equipartition theorem. A mode is not the same thing as a node or antinode.

Each mode of the radiation field has associated with it a certain number of nodes. The exact details of how many nodes each mode of the radiation field in the blackbody has are easy to work out so try it yourself.

7. Jul 7, 2014

### sawer

Why not?

If a standing wave has 3 antinodes and we say its mode number is 3. So why is mode not the same thing as antinode?

http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Here it says: "number of modes" per unit frequency per unit volume.

For the first picture: there are 3 standing waves and 5 waves. Now which number is the number of modes? 3 or 5?

It confuses me here, they say: for blackbody radiation, we must calculate the "number of modes".

Number of what? Standing waves, or antinode number of standing wave?

Thanks for help WannabeNewton

8. Jul 7, 2014

### WannabeNewton

Yes the number of antinodes of the standing wave is the same as the mode number but a mode itself refers to one of the independent harmonic oscillator solutions in the Fourier decomposition of the radiation field. The mode has an associated wave number $k$ and with periodic boundary conditions we get standing waves with $k \propto n$ so we can use the mode number $n$ to label a mode and the mode number also reflects the number of antinodes of the standing wave but the mode number $n$ does not mean we have $n$ modes it just means we are in the mode corresponding to $n$. Therefore each antinode does not have an energy $k_B T$; it is only each mode that has an energy $k_B T$ again as per the equipartition theorem.

Yes this represents the density of modes in the resonant cavity.

Which picture exactly? There are only standing waves of the radiation field inside the resonant cavity so I don't understand the distinction you are making between standing waves and waves. A single mode corresponds to a single standing wave.

9. Jul 7, 2014

### sawer

So wave mode refers to one particular frequency of a standing wave and wave number is the number of antinodes of that wave. So wave mode number indicates wave numbers but mode itself indicates frequency. Right? There are many different frequencies of waves(different wave modes), but, one type of mode indicates one frequency. Right? This was what I understood at first when I started to study on blackbody radiation.

But please look at this expression on the picture:

"For higher frequencies you can fit more modes into the cavity"

If we replace mode and frequency and rewrite the expression:

"For higher frequencies you can fit more high frequency of wave into the cavity"

It doesn't make sense.

So I began to think for wave mode as a number of antinodes, not as a standing wave of a particular frequency. Because increasing frequency means increasing the antinode number. Now we rewrite the expression again:

"For higher frequencies you can fit more antinode of wave into the cavity" It made sense.

So multiplying the "number of modes per unit frequency per unit volume" by $k_B T$ gives the energy of one standing wave of a particular frequency. so that meant every antinode of a standing wave must have an energy of $k_B T$.

I think I can explain my confusion now. can you please check again?

What does that mean: "For higher frequencies you can fit more modes into the cavity".

What does "more modes" mean? Doesn't one frequency indicate one mode?

Sorry, I meant the cavity which is on the right side of the picture. There are 3 standing waves and 2 of them has 2 antinodes one of them just 1.

10. Jul 7, 2014

### WannabeNewton

Yes this is all correct.

Yes.

Yes it doesn't. The article you linked is just worded terribly.

I see your confusion and I can see how it can be engendered by the bad wording in that article. But a standing wave of a particular frequency is exactly a normal mode of the radiation field.

Define $g(\omega)d\omega$ as the number of normal modes of the radiation field in the resonant cavity with frequency in between $\omega$ and $\omega + d\omega$. Let the cavity by a cube of side length $L$ so that its volume is $V = L^3$. From the boundary conditions imposed on the independent harmonic oscillator solutions corresponding to the normal modes obtained from Maxwell's equations inside the cavity, we find $\omega_{n_1n_2n_3} = \frac{c\pi}{L}\sqrt{n_1^2 + n_2^2 + n_3^2}$. Then $g(\omega)d\omega = \frac{V\omega^2 d\omega}{\pi^2 c^3}$-see if you can derive this from a basic argument; we can divide by $V$ if we wish to get the density of modes per unit frequency per unit volume. So $g(\omega) = \frac{V\omega^2}{\pi^2 c^3}$ by itself represents the density of modes per unit frequency which is why the article you linked is just worded very poorly; what the article should have involved is the quantity $g(\omega)d\omega$ which by definition is the number of normal modes in the frequency range $\omega$ to $\omega + d\omega$.

I hope that clears things up.

11. Jul 7, 2014

### Simon Bridge

Ah OK.
My dictionary says that the wave number is the number of wavelengths per unit length while the mode is the excitation number of a standing wave ... i.e. on a string.

This a mode is something that standing waves have while wave nuber is quite a different property (check the dimensions) of any periodic wave.

But you are trying to understand Blackbody radiation and you have already had a lot of assistance there so I won't confuse you further by butting in.