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If a blackbody has a temperature such that its peak frequency was well within our audible range, for example $1\ \mathrm{kHz}$, what would that sound like if we used Planck's law to plot its spectral curve in the frequency domain and performed a transformation (like an inverse FFT) to obtain a waveform?

Planck's law tells us the peak wavelength and electromagnetic spectral emission curve of an ideal radiator an some temperature $T$.

If we know the peak frequency, $f$, then we can work backwards to figure its peak wavelength, $\lambda$. For our example, if $f=1\ \mathrm{kHz}$, then $\lambda\approx170\,471\ \mathrm m$, so $T\approx17\ \mathrm{nK}$.

If we plot the power spectral density of a 17 nanokelvin blackbody as a function of frequency and performed an inverse FFT on that curve, what would the resulting waveform sound like?

According to this Wikipedia article, blackbody radiation is just thermal noise (Johnson–Nyquist noise); if that's what I'm looking for, what does it sound like? Just to clarify, I'm looking for a waveform, maybe a WAV file, rather than a verbal description.

ayane_m
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    NB for anyone who attempts this: You can't just inverse Fourier transform the spectrum to get the time domain signal. You have to respect the fact that noisy time domain signals give noisy frequency domain signals. The amplitude and phases of your frequency domain points must be drawn from the proper probability distributions in order for the resulting time domain signal to have the right statistical properties. This is a commonly missed aspect of signal processing. – DanielSank Nov 03 '15 at 06:54
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    What kind of answer can possibly be given to this question? Suppose I were to construct the sound wave using my computer and a speaker. Can I post an answer describing the sound verbally? Can I post a link-only answer with a recording for everyone else to hear? I really like this question, but I'm wondering if/how it can be answered. – DanielSank Nov 03 '15 at 07:19
  • @DanielSank I think the answer should be objective rather than subjective (i.e. a description); perhaps you could upload a small wav file? I'm curious to analyze its spectrum using different window functions. – ayane_m Nov 03 '15 at 07:27
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    It's funny for me to inverse Fourier transform the Plank spectrum, and for you to then analyze it with a window function. Why not just use the windowed spectrum to produce the time domain signal in the first place? ^^ – DanielSank Nov 03 '15 at 07:28
  • yuki96, "blackbody radiation is just thermal noise". It is thermal noise in general but if the temperature is 10 nK, because this was your initial question, the spectrum of the black body radiation has a bell shape and is in the audible domain of frequencies between 0.5 kHz and 2.5 kHz. – Energizer777 Nov 03 '15 at 08:54
  • DanielSank, The bell shape of a spectral density means that the time domain signal is quite nice and looks like this. It is clear for me that you and yuki96 do not have a clear understanding of the time domain <-> frequency domain correspondence. – Energizer777 Nov 03 '15 at 10:00
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    Peak wavelength is not $c/f_{peak}$. – ProfRob Nov 04 '15 at 22:22
  • @RobJeffries Thank you for pointing that out. I corrected the data with the proper ones. – ayane_m Nov 05 '15 at 00:23
  • In order to know "what it sounds like", the phases are not significant, as our hearing is practically insensitive to phase. The cochlea does a kind of mechanical Fourier transform, mapping the frequencies to different regions. Some timing information from the hair cells is then used in binaural direction estimation. –  Apr 29 '17 at 11:46
  • @DanielSank I saw that this was your favorite question but I do not understand it, I mean the question as asked. My retina can see some part of the spectrum of the BB radiation but cannot hear any of it. How can anybody? Instead, one could ask how do phonons of an elastic body, a crystal, or one's eardrum, react when immersed in BB radiation. I do not know the equilibrium physics or the transfer function of such reaction but the answers/comments seem to overlook this issue. Am I misunderstanding this question? – hyportnex Jan 24 '23 at 02:23
  • @hyportnex The idea is to take the spectrum and scale it down to audio frequencies. You can hear that. – DanielSank Jan 25 '23 at 20:10
  • @DanielSank sea-dog legend has it that sailors could hear the pulse rate of a high power radar around PRF=1KHz. The explanation I was told that somehow the inner ear (cochlear fluid?) could rectify the 3GHz pulses whose length was may be 1 microsec repeated at 1msec intervals and this was felt by the brain. Is this the sort of thing you think the ear could hear when immersed in bb radiation at peak around 1kHz? – hyportnex Jan 25 '23 at 20:29
  • @hyportnex That is not what I had in mind. I had in mind that blackbody radiation has a certain spectral density. We can choose a temperature so that a significant amount of power in the spectral density is in the audio range. Then we can produce sample waveforms from that spectral density and play those waveforms from a speaker. – DanielSank Jan 26 '23 at 04:47
  • @DanielSank then I just do not understand the importance of the question. I mean the "publications world" is full of faux spectra shown on charts where the various intensities are represented by an associated color chart. You could play an associated audio, too for deeper experience. How is this different, what am I missing? But I do think the equilibrium interaction of acoustic waves and BB radiation, if there is any at a low enough temperature, may be an interesting question, I am just not competent enough to ask it any more intelligently. – hyportnex Jan 26 '23 at 10:35
  • @hyportnex Nobody guarantees that your experience reading questions and answers here will be life-changing. – DanielSank Jan 26 '23 at 16:46

3 Answers3

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This problem can be solved with noise-shaping. Since the shape of the spectrum is known, it can be used as a base for the power spectral density:

$$ P(f,T)=\frac{ 2 h f^3}{c^2} \frac{1}{e^\frac{h f}{k_\mathrm{B}T} - 1} $$

where $k_\mathrm{B}$ is the Boltzmann constant, $h$ is the Planck constant, and $c$ is the speed of light. This outputs the relative power of each band as a continuous function of frequency, $f$, and temperature, $T$. Since the output quantity must be expressed in decibels (dBr) to be meaningful for audio, we simply use a log scale and add an offset (a gain) to normalize the peak to 0. The equation of the EQ curve is:

$ E(f,T) = 10 \log{ P(f,T) } + G_{t}(T) $

where $G_{t}(T)$ is the gain required to normalize the peak to 0 dB. The required gain depends on the inverse cube of the temperature plus a constant, $G$ (187 dB); thus, $ G_{t}(T) = G - 10 \log T^3 $. The leading coefficient $10$ converts bels to decibels. Simplifying gives us:

$$ E(f,T) = 10 \log{ \left( \frac{ 2 h f^3}{c^2 T^3} \frac{1}{e^\frac{h f}{k_\mathrm{B}T} - 1} \right)} + G $$

tl;dr:

We obtain our waveform by applying an EQ to gaussian white noise from AudioCheck.net.


Examples:

  1. 17 nanokelvins is the temperature at which black noise has a peak frequency of 1 KHz. Its passband is limited to 1 Hz to 12 KHz.

  2. 30 nanokelvins is the lowest temperature at which black noise has a passband that spans the entire hearing range.

  3. 55 nanokelvins is the temperature at which black noise has a peak frequency of approximately 3 KHz, the most sensitve frequency of human ears.

  4. 340 nanokelvins is the temperature at which black noise has a peak frequency of just under 20 KHz, which is the limit of human hearing. Most of the audible spectrum is a linear upward ramp, which is very similar to violet noise. At higher temperatures, the frequency domain will be almost identical to violet noise.

All EQ filter parameters are in the descriptions of the tracks on SoundCloud.

ayane_m
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  • Feel free to edit my answer or give feedback. I did all of this in Audacity. – ayane_m Nov 04 '15 at 20:35
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    This is probably my favorite Stack Exchange post ever. Great job! It would be really nice to post the code so that others can check it, but if you did it in a GUI application that may not be possible. It would also be really nice to post the original white noise for comparison, and perhaps a few more temperatures. – DanielSank Nov 04 '15 at 20:38
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    What is $B$ in the equation? It's good to define all symbols, even if they're defined in the external link. To clarify, it would be nice to upload the white noise to soundcloud so people don't have to download it etc. etc. – DanielSank Nov 04 '15 at 20:42
  • @I've updated my answer; I'll upload an XML EQ preset containing the parameters I used soon. FOSS FTW ;) – ayane_m Nov 04 '15 at 21:40
  • Clicking the equation will show a page with it graphed. – ayane_m Nov 04 '15 at 21:57
  • Yeah, I was wondering about using the log scale like that. It's probably clearer to just write the power, i.e. get rid of the dB scale. However, with the new sentence explaining the level it's much clearer already. – DanielSank Nov 04 '15 at 21:59
  • Basically yuki96 took a white noise in the band 0 - 20 kHz and applied to it a band pass filter having the shape of Plank's curve. It is self evident that he obtained an ~white noise in a narrower band. However, this has nothing to do with the sound emitted by a black body at 10 - 20 nK, see yuki96's initial question. Yuki96 have not demonstrated anything as long as he assumed from the beginning that the sound emitted by a black body at 10 - 20 nK is a white noise. – Energizer777 Nov 04 '15 at 23:05
  • @Energizer777 This is not band-limited white noise. White noise contains equal energy at all bands in a linear scale. By your logic, pink, violet, Brownian, etc noise are all white noise, which they aren't (and I did not assume anything along those lines either). A bandpass filter is flat; mine is not. My question was asking what the waveform would be if we know the frequency domain (which is thermal noise). At higher temperatures, this would be violet noise. Also, not everyone on the internet is a guy. – ayane_m Nov 04 '15 at 23:34
  • yuki96, Do the inverse Fourier transform of Plank's curve. That will be the sound generated by a black body at 10 - 20 nK. If you do not want to do it then demonstrate mathematically than a white noise, filtered with Plank's curve, is the same thing as the inverse Fourier transform of Plank's curve. – Energizer777 Nov 05 '15 at 00:13
  • @Energizer777 It would sound like a chirp, as you've hinted in your answer. However, a chirp is finite-time and is not a musical note. Plus, a finite-time signal is not self-similar: an FT with any window size on any given interval of such a signal will not be identical or near-identical to another FT in another interval. (I.e., a chirp will have a bell-curve FT in finite time [using a window size that spans the whole chirp], but will be dynamic in continuous time; a moving-window FT with a size less than the full tone's size will not retain the same frequency curve at every interval.) – ayane_m Nov 05 '15 at 01:10
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    Sounds just like the tinnitus in my ears. :) – anna v Nov 05 '15 at 04:36
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    @DanielSank I added links for noise at two more temperatures. I also cleaned up the answer and refined the formula a bit. – ayane_m Nov 06 '15 at 05:24
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    @yuki96 I think this answer would be more clear if the first equation were to give just the power spectral density. Then, in a second equation, you can give the equalization curve, noting that audio processing is typically done with logarithmic scales. It's confusing (to me, and probably other physicists), to see the Plank law written as an equalizer curve expressed on a log scale. Please note that I'm only bugging you because I love this question and answer so much :D – DanielSank Nov 06 '15 at 07:33
  • @DanielSank Thank you, I'm flattered! I really appreciate the feedback; I'll change what you noted. I'm all ears for more! – ayane_m Nov 06 '15 at 07:58
  • Just noticed something maybe worth mentioning: I think the spectrum used for this noise is for the 3D case. For what it's worth, the spectrum is different in 1D or 2D because the density of states is different. Love this Q&A ^^ – DanielSank Feb 25 '16 at 22:43
  • Reminds me of the sound of a waterfall. – Virgo Aug 31 '19 at 23:23
5

If you are cooling your object that you wish to hear, then the exact sound will depend on the exact temperature (as given by yuki96's answer at 17nK).

However, any temperature above the nanoKelvin temperature scale will sound the same, but the volume will increase with temperature (according to the Stefan-Boltzmann law).

The sound of a warm blackbody (such as what you would get at room temperature) would sound like a violet or purple noise. You can listen to a sample of purple noise here.

-5

What does a blackbody sound like?

It will sound like a musical note. Any spectrum looking like a bell is a musical note with a lot of harmonics. The narrower the bell the purer the musical note. If the bell is wide the ear perceives the sound more like a pop because the wider the spectrum the shorter its time image gets.

Planck's law

Planck's law

To be more precise, the inverse Fourier transform (the signal as a function of time) of a spectrum having the shape of a bell (like Plank's diagram) looks like in the image below, row no. 4 from the top.

Continuous and discrete spectra

Continuous and discrete spectra

Because I am seeing there are people who do not understand, the spectrum of a finite duration signal (a real musical note for instance) is continuous. Only repetitive signals have discrete spectra. If the musical note is played continuously its spectrum is discrete if not, and this is the case in practice, the spectrum is continuous.

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    Harmonics are multiples of a fundamental tone. How can harmonics give a continuous bell shape? – DanielSank Nov 03 '15 at 07:05
  • Harmonics would be discrete, not continuous, in the frequency domain. An example is a square wave. – ayane_m Nov 03 '15 at 07:10
  • If the number of harmonics of a musical note is big the spectrum of the note appears continuous. See the diagrams from here. – Energizer777 Nov 03 '15 at 07:20
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    @Energizer777 That has nothing to do with harmonics. The continuous spectrum there looks like it is due to the finite time duration of the measured sound wave causing spectral leakage in the DFT. It's a signal processing effect. – DanielSank Nov 03 '15 at 07:24
  • @Energizer777 A band-limited signal has a finite number of harmonics, and harmonics are integer multiples of the fundamental. As DanielSank said, low-order windowing function tends to show wider "lobes". – ayane_m Nov 03 '15 at 07:35
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    yuki96: "Harmonics would be discrete, not continuous, in the frequency domain. An example is a square wave." That square wave has to last indefinitely to have a discrete spectrum. If it last a finite number of periods its spectrum will be continuous. – Energizer777 Nov 03 '15 at 08:18
  • Since there's confusion here, let's sort it out in chat. – DanielSank Nov 03 '15 at 08:53
  • @yuki96 pinging you to point you to the chat link. – DanielSank Nov 03 '15 at 09:20
  • @Energizer777 "That square wave has to last indefinitely to have a discrete spectrum. If it last a finite number of periods its spectrum will be continuous." That is correct; I said the square wave is band-limited but not time-limited. – ayane_m Nov 03 '15 at 21:49
  • Hey Energizer777, note in @yuki96's answer the sound is nothing at all like a musical tone. It's noise. The spectral density of a noise process is not the same thing as the Fourier transform of a deterministic signal. – DanielSank Nov 04 '15 at 20:43
  • Yuki96 has to sample the Plank's curve and perform an ifft. That will be the sound generated by a black body. Yuki96 did something completely different, he filtered a random signal with Plank's curve. If yuki96 had filtered a melody with Plank's curve he would have obtained the same song a bit distorted but still intelligible. – Energizer777 Nov 04 '15 at 23:24
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    @Energizer777 If I applied the EQ curve anything other than white noise, I won't preserve the energy contour of the EQ curve, because the white noise energy spectrum is more or less constant and independent of the time domain. Even assuming the melody is a steady-state wave, it will not have equal energy like white noise in the frequency domain. Please note that a blackbody is a continuous time phenomenon and not a pulse like a chirp. – ayane_m Nov 04 '15 at 23:43