<ol>
<li>It would be good to specify what "majority of radiation" means. If it means "radiative energy", then this premise is surely wrong. To find the energy emitted per wavelength, take the Planck function <span class="math-container">$B(T, \lambda)$</span> at a certain temperature <span class="math-container">$T$</span> and multiply with the wavelength <span class="math-container">$\lambda$</span>. You will find that the short-wavelength radiation always dominates for a black-body.</li>
<li>Wien's law gives you the peak of the Planck function. Radiation at all other wavelengths exists.</li>
</ol>

<p>Note that objects predominantly loose most of their energy, if photons and matter are in equilibrium. If this equilibrium is disturbed, or particle accelerations play a role (like in the presence of magnetic fields), a whole zoo of non-thermal processes can also lead to important radiative energy loss.<br>
Non-thermal processes in space can be radio emission from gyrating particles, line radiation, Masers, Bremsstrahlung, Compton radiation... Those processes then do not show a Planck spectrum, but other characteristic spectra.</p>

<hr>

<p>Edit: I've added a plot of the Planck function at two different temperatures. In a linear plot (on the left) it might look as if there's a cut-off, but in a logarithmic plot (on the right), one sees what mathematical analysis already betrays: The function is defined everywhere. It just becomes really small on the high-energy end, as a commentator pointed out correctly.</p>

<p><a href="https://i.sstatic.net/Q8J86.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Q8J86.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-08-16T14:59:33.180Z
I am confused about #2. You say that this is the peak of the Planck function, so how can there be anything emitted past the peak? Doesn’t the “peak wavelength” mean the cutoff, shortward of which there is zero emission?
John Smith
CC BY-SA 4.0
2019-08-16T15:03:41.733Z
@JohnSmith What, no? Why would it be? The Planck function is a continuous function, defined on the whole space of real numbers, there is no cutoff. Just plot it to see this.
AtmosphericPrisonEscape
CC BY-SA 4.0
2019-08-16T15:08:44.547Z
@JohnSmith it does drop off pretty fast on the short wavelength side of the peak. The reason it isn't a hard cutoff is that the distribution of thermal energy is somewhat random. While most atoms in a lump of iron at say 500K do not have enough energy to emit a photon of visible light, a few "lucky" ones will.
Steve Linton
CC BY-SA 4.0
2019-08-16T15:30:00.817Z
Wait, so a black body at ANY temperature will have nonzero emission at ALL wavelengths?
John Smith
CC BY-SA 4.0
2019-08-16T15:35:28.450Z
@JohnSmith At high frequencies quantization (shot noise) becomes an issue. The intensity may be so low that it amounts to less than one expected photon in the duration of your experiment, so you would expect to see no emmission at those wavelengths.
Steve Linton
CC BY-SA 4.0
2019-08-16T16:48:48.450Z
I see. But my question is whether Planck’s law predicts identically zero emission anywhere, especially past the Wien wavelength peak. I’m asking about the theory itself, not hypothetical lab measurements.
John Smith
CC BY-SA 4.0
2019-08-16T18:45:12.783Z
@JohnSmith the Planck function is widely available (e.g. Wikipedia). It isn't exactly zero at any frequency
ProfRob
CC BY-SA 4.0
2019-08-16T19:16:56.030Z
@JohnSmith: Your question has been answered. Is there still something you want to know?
AtmosphericPrisonEscape
CC BY-SA 4.0
2019-08-16T21:24:36.230Z
Ok, so what I’m understanding is that 1) not all objects emit the majority of their radiated energy in the radio (in fact, very few do, as they must be colder than 10 K for this to be the case) and what part of the spectrum they will emit most of their energy in will depend on their temperature (e.g., the Sun emits most of its energy in the visible, and much less of its energy in the radio), and 2) Planck’s law says any black body with a temperature above absolute zero has at least a VERY SMALL, but NONZERO amount of radiation at ALL WAVELENGTHS. Are these statements both accurate?
John Smith
CC BY-SA 4.0
2019-08-16T21:36:54.113Z
@JohnSmith Yes, both those statements are correct -- I think you've got it.
Peter Erwin
CC BY-SA 4.0
2019-08-16T21:38:47.480Z
@JohnSmith: Yes, they are accurate. If you are still studying, then I'd encourage you to think about Earth's Planck function (at ~300 K) and compare that as test case to the sun. Later, should you study atmospheric physics you can go and understand phenomena like global warming from there, and many other useful things.
AtmosphericPrisonEscape
CC BY-SA 4.0
2019-08-16T22:51:10.987Z
One point wasn't explained adequately. The OP thought "peak wavelength" referred to the highest wavelength. Instead, it refers to the wavelength at which the highest amount of energy is radiated.
Bit Chaser
CC BY-SA 4.0
2019-08-17T01:20:40.067Z
I agree with @bitchaser in his assessment. It sounds like OP misunderstood the meaning of the word 'peak'.
Jim421616
CC BY-SA 4.0
2019-08-16T14:54:30.867Z
2019-08-16T15:20:06.610Z
2019-08-16T15:20:06.610Z
AtmosphericPrisonEscape
<p>Two questions:</p>

<ol>
<li><p>I read online that all objects emit the majority of their thermal radiation in the radio, and that this emission increases with temperature. Does this mean that all glowing bodies emit the majority of their radiation in the radio, regardless of their temperature?</p></li>
<li><p>Wien’s law is used to find the peak wavelength of a black body. Does this mean that there is zero emission at any wavelength shorter than this peak wavelength?</p></li>
</ol>

<p>Thank you in advance for your help. </p>

CC BY-SA 4.0
2019-08-16T15:12:21.400Z
Only very cold objects (e.g., $T < 10$ K) would emit most of their radiation in the radio.
Peter Erwin
CC BY-SA 4.0
2019-08-17T22:53:22.350Z
`+1` for asking a clear question and following up with several comments. This is how Stack Exchange should be used. In the future if you find another passage that you'd like to understand e.g. "I read online that...", it would be better if you included a short quote.
uhoh
<radiation><thermodynamics><black-body>
Questions about Planck’s law
John Smith
# Questions about Planck’s law

(Asked by: username_0 on 2019-08-16)

Two questions:

1. I read online that all objects emit the majority of their thermal radiation in the radio, and that this emission increases with temperature. Does this mean that all glowing bodies emit the majority of their radiation in the radio, regardless of their temperature?
2. Wien’s law is used to find the peak wavelength of a black body. Does this mean that there is zero emission at any wavelength shorter than this peak wavelength?

Thank you in advance for your help.

**Tags:** radiation, thermodynamics, black-body
> **Comments:**
> **username_1**: Only very cold objects (e.g., $T < 10$ K) would emit most of their radiation in the radio.  
> **username_2**: `+1` for asking a clear question and following up with several comments. This is how Stack Exchange should be used. In the future if you find another passage that you'd like to understand e.g. "I read online that...", it would be better if you included a short quote.  

---

## 1 answers

### Answer by username_3 on 2019-08-16. Score: 3

1. It would be good to specify what "majority of radiation" means. If it means "radiative energy", then this premise is surely wrong. To find the energy emitted per wavelength, take the Planck function $B(T, \lambda)$ at a certain temperature $T$ and multiply with the wavelength $\lambda$. You will find that the short-wavelength radiation always dominates for a black-body.
2. Wien's law gives you the peak of the Planck function. Radiation at all other wavelengths exists.

Note that objects predominantly loose most of their energy, if photons and matter are in equilibrium. If this equilibrium is disturbed, or particle accelerations play a role (like in the presence of magnetic fields), a whole zoo of non-thermal processes can also lead to important radiative energy loss.  
Non-thermal processes in space can be radio emission from gyrating particles, line radiation, Masers, Bremsstrahlung, Compton radiation... Those processes then do not show a Planck spectrum, but other characteristic spectra.

---

Edit: I've added a plot of the Planck function at two different temperatures. In a linear plot (on the left) it might look as if there's a cut-off, but in a logarithmic plot (on the right), one sees what mathematical analysis already betrays: The function is defined everywhere. It just becomes really small on the high-energy end, as a commentator pointed out correctly.

[![enter image description here](https://i.sstatic.net/Q8J86.png)](https://i.sstatic.net/Q8J86.png)

> **Comments:**
> **username_0**: I am confused about #2. You say that this is the peak of the Planck function, so how can there be anything emitted past the peak? Doesn’t the “peak wavelength” mean the cutoff, shortward of which there is zero emission?  
> **username_3**: @JohnSmith What, no? Why would it be? The Planck function is a continuous function, defined on the whole space of real numbers, there is no cutoff. Just plot it to see this.  
> **username_4**: @JohnSmith it does drop off pretty fast on the short wavelength side of the peak. The reason it isn't a hard cutoff is that the distribution of thermal energy is somewhat random. While most atoms in a lump of iron at say 500K do not have enough energy to emit a photon of visible light, a few "lucky" ones will.  
> **username_0**: Wait, so a black body at ANY temperature will have nonzero emission at ALL wavelengths?  
> **username_4**: @JohnSmith At high frequencies quantization (shot noise) becomes an issue. The intensity may be so low that it amounts to less than one expected photon in the duration of your experiment, so you would expect to see no emmission at those wavelengths.  
> **username_0**: I see. But my question is whether Planck’s law predicts identically zero emission anywhere, especially past the Wien wavelength peak. I’m asking about the theory itself, not hypothetical lab measurements.  
> **username_5**: @JohnSmith the Planck function is widely available (e.g. Wikipedia). It isn't exactly zero at any frequency  
> **username_3**: @JohnSmith: Your question has been answered. Is there still something you want to know?  
> **username_0**: Ok, so what I’m understanding is that 1) not all objects emit the majority of their radiated energy in the radio (in fact, very few do, as they must be colder than 10 K for this to be the case) and what part of the spectrum they will emit most of their energy in will depend on their temperature (e.g., the Sun emits most of its energy in the visible, and much less of its energy in the radio), and 2) Planck’s law says any black body with a temperature above absolute zero has at least a VERY SMALL, but NONZERO amount of radiation at ALL WAVELENGTHS. Are these statements both accurate?  
> **username_1**: @JohnSmith Yes, both those statements are correct -- I think you've got it.  
> **username_3**: @JohnSmith: Yes, they are accurate. If you are still studying, then I'd encourage you to think about Earth's Planck function (at ~300 K) and compare that as test case to the sun. Later, should you study atmospheric physics you can go and understand phenomena like global warming from there, and many other useful things.  
> **username_6**: One point wasn't explained adequately. The OP thought "peak wavelength" referred to the highest wavelength. Instead, it refers to the wavelength at which the highest amount of energy is radiated.  
> **username_7**: I agree with @bitchaser in his assessment. It sounds like OP misunderstood the meaning of the word 'peak'.  

---

