# Understanding Planck's law

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1. Jan 27, 2016

### Alettix

Hello!

I have a little trouble with understanding Planck's law of radiation, and wondered if you could help me with it. :)
The equation is:
$\frac{dI}{d\lambda} = \frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)}$ (1)
where T is the temperature, k Boltzmann's constant, h Planck's constant, $\lambda$ the wavelenght, c the speed of light in vacuum and I the radiated intensity per unit area.

What confuses me is the differential form on the left-hand side of the equation. As I have understood, the total intensity radiated in a small wavelenght interval from $\lambda$ to $\lambda + \Delta\lambda$ is given by:
$I = \frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)} \Delta\lambda$ (2)
and the total intensity of all wavelenghts:
$I = \int_{0}^{\infty} \frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)} d\lambda = \sigma T^4$ (3)
which is just the Stefan-Boltzmann law.

What I wonder about is how we can find the radiated intensity for one wavelenght only. That would mean $\Delta\lambda \rightarrow 0$ which would give $I \rightarrow 0$ with (2) if I am not wrong. But this doesn't seem logical, thus it would mean that no single wavelenght radiates any energy, wouldn't it? When trying to find information, I came accross writings which said that the radiation intensity at one wavelenght is simply given by:
$I = \frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)}$ (4)
But to me this seems wrong as well. If $\Delta\lambda$ is small (which it is assumed to be), we get that (4)>(2) . This means that a wavelenght interval is radiating less energy than one single wavelenght in that interval, which just can't be right, can it?

I would be really glad if you could help me sort this out. :)
Thank you!

2. Jan 27, 2016

### IAN 25

The radiation is measured over a small wavelength interval in practice which gives rise to the use of Δλ (or Δf if a frequency range is used). Clearly a single photon of a given frequency carries energy, represented by a single λ as you say. However, if we take a small interval Δλ one gets intensity ΔI. If you replace I in (2) by ΔI, then integrating over all wavelengths gives (3). Does this help?

3. Jan 27, 2016

### BvU

You are absolutely correct and (4) is wrong (unless they mean something else with their I).

If you divide your (1) by (3) you get a probability distribution. And you'll readily accept that if you narrow the range ($\Delta \lambda$), the probability for that range decreases proportional to the width.

As Ian said:
Compare the spectrograph or spectrophotometer: to pick out a single wavelength you need to narrow the slit to zero -- and get nothing reaching the detector.

4. Jan 27, 2016

### Alettix

Thank you, the probability example was very useful.

So if I understand you correctly, my equation (2) is right about $\Delta I \rightarrow 0$ (using Ian's notation) when $\Delta\lambda \rightarrow 0$ because the fraction between the radiated intensity of the single wavelenght λ and the total radiated intensity approaches zero ($\frac{\Delta I}{I} \rightarrow 0$)? But the single wavelenght λ still carries the energy $E = hf$?

5. Jan 27, 2016

### Ray Vickson

This is no different from densities in other fields, such a mass density, charge density, current density, energy density, etc. In the Planck case the function on the right is just the radiation density. The real issue is for you to understand the concept of density in general; it has not much at all to do with the details of black-body radiation as such.

6. Jan 27, 2016

### Alettix

I thought I had an understanding of density, but it seems I don't know how to apply it here. I can see that the units on the y-axis become $\frac{\frac{J}{m^2}}{m} = \frac{J}{m^3}$, which looks like energy density and makes me think about something like an amount of energy distributed in a volume. But I how should I picture radiation density? Could you please help me with it Sir?

7. Jan 27, 2016

### Ray Vickson

The function $f(\lambda)$ on the right of your original post is just the "intensity per unit of wavelength"; that is, the total intensity falling in the range $(\lambda, \lambda + \Delta \lambda)$ is just $f(\lambda) \Delta \lambda$, and the total intensity falling in the range $(\lambda_1, \lambda_2)$ is
$$\text{Intensity}(\lambda_1 \to \lambda_2) = \int_{\lambda_1}^{\lambda_2} f(\lambda) \, d \lambda \;\;\ (\star)$$
Basically, eq. (*) tells you exactly what $f(\lambda)$ it means.

It may seem odd to define a "density" per unit of wavelength (rather than, say, per unit of length or area or volume), but that is how the people who invented the subject chose to present things. You just have to live with it.

8. Jan 29, 2016

### IAN 25

I agree with Ray. Also the Intensity is really a flux density, which is the amount of energy carried by the radiation (Or flux of photons) per unit wavelength, per unit area: ie in Joules λ-1 m-2, where here density means amount passing through per square metre. It is an area density rather than your more familiar volume density.

9. Feb 1, 2016

### Alettix

I think I start to understand this, but when thinking about the denisty as "per wavelength" I feel like the density should be multiplied with the wavelength:
$I= \frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)} \cdot \lambda$
But that's wrong, isn't it?

10. Feb 1, 2016

### BvU

Yes. It is $$dI= \frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)} \ d\lambda$$(exactly what you wrote as (1) in post #1). For $I$ you need to integrate, e.g. as in (3) in post #1 where you integrated over all wavelengths. If you only want the intensity from red light you need to define a range of wavelength that you consider 'red' and integrate over that range.

11. Feb 6, 2016

### Alettix

Thank you! I think I understand it now. So if we want to approximate the I for a small wavelenght Δλ we can use:
$$I= \frac{2\pi hc^2}{\lambda^5(e^{hc/\lambda kT}-1)} \Delta\lambda$$
can't we?

12. Feb 6, 2016

### Ray Vickson

Yes, exactly as was written in #7.