<p>First, let's get a bit of background. "<a href="https://en.wikipedia.org/wiki/Absolute_hot" rel="nofollow noreferrer">Absolute hot</a>" is the concept that a maximum temperature <strong>exists</strong>. It describes your question exactly. For a long time, no one was able to figure out whether an absolute hot exists, but in the 20th century major revolutions in theoretical physics gave us answers.</p>

<p>"Absolute hot" isn't the name of the hottest temperature, but it's the concept that one exists. So what is our answer — is there a highest possible temperature that fits the concept of an "absolute hot"? Well, <strong>yes</strong>. But first we must talk about heat and temperature.</p>

<p>Photons act both like particles and waves. As we add more energy, the photon's wavelength decreases, and its frequency increases. Thus, the wavelength of the light emitted by hot objects generally depends on the temperature of the object. Hotter objects will emit more energetic photons with shorter wavelengths.</p>

<p>According to modern physics, a Planck length is the smallest meaningful length in the Universe (the reason for that is an entirely different question). If we keep raising the temperature of our object, nonstop, the wavelength of the photons it emits will decrease until it reaches a Planck length. Our object would reach a temperature called <strong><a href="https://en.wikipedia.org/wiki/Planck_temperature" rel="nofollow noreferrer">Planck temperature</a></strong>. That's about $1.417×10^{32} \; \mathrm{K}$.</p>

<p>Planck temperature is theoretically the hottest possible temperature. At that point, the object could not theoretically get any hotter, as the photon's wavelength cannot decrease beyond a Planck length.</p>

<p>But what happens if we were to add <em>more</em> energy to the object? Would it go beyond Planck temperature? Well...God knows. At that point, we wouldn't even call it temperature anymore. Nobody knows what exactly would happen, but it certainly wouldn't match the definition of temperature.</p>

CC BY-SA 3.0
2016-06-12T14:40:05.797Z
Thank you for your amazing skill of making something that is complicated to something even I could understand.
HotSaucey
CC BY-SA 3.0
2016-06-12T14:41:51.680Z
Let's say that this theoretical photon got even more energy. Why would it not match the definition of temperature?
HotSaucey
CC BY-SA 3.0
2016-06-12T14:46:35.890Z
@Mobal Well, the photon would become hotter than what is physically allowed. I like your question; theoretically, there might not be a limit to how much energy we could add to the photon. But we have no idea what would happen if it went beyond the Planck temperature, as the wavelength shouldn't be able to decrease anymore. It really wouldn't follow the concepts that we've established already, like the Planck length being the smallest possible length.
Sir Cumference
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2016-06-12T14:50:16.310Z
If we found a way to add enough energy to something to heat it up past the Plank temperature, wouldn't that be indirect evidence of the disproving of the Planck length being the shortest possible length?
HotSaucey
CC BY-SA 3.0
2016-06-12T14:56:15.810Z
@Mobal I think that's an unsolved problem in physics, even though according to our models, the Planck length must be the shortest possible length (you could read more about it here: https://goo.gl/4VBO8T). But ultimately, we have no way of testing that. Temperatures that hot would actually form a black hole (called a Kugelblitz). In Einstein's theory of General Relativity, energy is a source of gravity too.
Sir Cumference
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2016-06-12T15:01:01.557Z
I am not a whiz at physics but is energy a source of gravity because of E = mc^2
HotSaucey
CC BY-SA 3.0
2016-06-12T15:02:21.450Z
@Mobal I haven't studied General Relativity too thoroughly. But that equation probably has something to do with it. Other things, like pressure, mass, and momentum are sources of gravity too. Somehow we got from temperature to GR, haha.
Sir Cumference
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2016-06-12T15:17:27.923Z
I swear GR is connected to everything astronomy
HotSaucey
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2016-06-12T18:55:10.830Z
Thanks for the great answer. It makes me wonder about temperature inside a black hole.  What about "merely" inside neutron or quark stars: are we getting close to these temperatures, or, not even close??
Fattie
CC BY-SA 3.0
2016-06-12T19:32:35.523Z
@JoeBlow The maximum temperature of a neutron star is 10^12. So Planck temperature is hundreds of millions of trillions times that amount.
Sir Cumference
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2016-06-12T22:30:06.787Z
uh, perhaps i'm just naive, but i don't think temperature is a well-defined quantity when the energy of a system is fixed (like a single photon, for instance)...
randomquestionasker77
CC BY-SA 3.0
2016-06-12T22:39:53.370Z
@colnegn I'm sorry? Why not?
Sir Cumference
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2016-06-12T22:58:47.023Z
well i'm no expert on the subject, but I believe that E and T are conjugate variables (they are related by a Fourier transform), and therefore have an associated uncertainty. 

similarly, in a system with a fixed volume: the pressure is ill-defined.
randomquestionasker77
CC BY-SA 3.0
2016-06-13T01:50:24.960Z
In reality, wouldn't a photon basically turn into particles well before it reached Planck energy?   I understand the answer was theoretical, but when asked "what if you gave it more energy", in reality, long before the photon reached that energy it would give up some energy in the form of particle creation.  (or am I mistaken?).  I don't think I'm mistaken.
userLTK
CC BY-SA 3.0
2016-06-13T01:59:18.393Z
@userLTK Yes, pair production would probably happen well before it reached Planck temperature. This is just a theoretical explanation for Planck temperature. Still, in the history of the Universe, matter *has* Planck temperature before — at 10^-42 seconds before the Big Bang, the entire Universe was at that temperature. That's just about 200 Planck times after the Big Bang (you can read up on Planck time, it's the shortest amount of time possible)!
Sir Cumference
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2016-06-13T02:21:26.493Z
@SirCumference so, it's theoretically possible that there were no photons at or prior to 10^-42 seconds after the big bang because it was too hot for photons?   or am I reading too much into this.
userLTK
CC BY-SA 3.0
2016-06-13T02:40:25.770Z
@userLTK This all took place during the Planck epoch, the earliest period after the Big Bang. There is no currently available physical theory to describe such short times. As you get closer and closer to t(0), our current theories begin breaking down. I'm gonna guess some aspects of GR and QM break down more than others, so physicists could determine the temperature at those times. Still, for the most part, we have no idea what went on back then.
Sir Cumference
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2016-06-13T02:57:31.600Z
@userLTK Also, I made a typo earlier on, when I said "at 10^-42 seconds before the Big Bang". I meant *after* the Big Bang XD
Sir Cumference
CC BY-SA 3.0
2016-11-21T05:57:18.433Z
@userLTK   so when photon gains more energy it starts populating the space ?
Sanket Deshpande
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2016-11-21T05:59:15.157Z
@SanketDeshpande What do you mean, "populating the space"?
Sir Cumference
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2016-11-21T06:02:09.297Z
@SirCumference maybe give away heat by creating more matter ? not sure.
Sanket Deshpande
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2016-11-21T06:05:27.130Z
@SanketDeshpande Still unsure what you mean by "populating the space", unless you're thinking of particles classically as spheres.
Sir Cumference
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2016-11-21T06:05:37.277Z
@SanketDeshpande  I think he's asking in relation to pair production, which is a super-high energy situation.   Also, I learned in another thread, a single photon can't produce a pair of particles in and of itself, but if it hits something it can.   High enough temperatures would probably start creating pairs of particles, but a single photon, without hitting something, no.  But the energy this requires is enormously high, only created by very violent events like black hole accretion disks or inside particle accelerators.
userLTK
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2016-11-21T06:08:58.577Z
@userLTK exactly, thats what i mean, but was unsure about single photon doing it
Sanket Deshpande
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2016-11-21T06:32:56.387Z
@SanketDeshpande  I asked a somewhat related question here.  The answer to my question is also an answer to yours if you want more detail.    http://physics.stackexchange.com/questions/266507/photon-pair-production-at-relativistic-speeds
userLTK
CC BY-SA 3.0
2016-11-21T14:41:48.917Z
interesting, so if we take an isolated system with 2 photon and continuously heat it what would be the outcome? Will it produce matter and consume all the given space? any thoughts?
Sanket Deshpande
CC BY-SA 3.0
2016-06-12T14:04:43.250Z
2016-11-21T03:12:29.220Z
2016-11-21T03:12:29.220Z
Sir Cumference
<p>As I stated in my comment in Sir Cumference's answer, I might just be naive on the subject, but I believe that:</p>

<p>Assuming that the density of states (the number of states per possible energy) for a given system can be taken to be effectively continuous, one can define the inverse temperature:</p>

<p>$$ \frac{1}{k_BT} = \beta = \frac{dS}{dE}$$</p>

<p>Where E is the system's energy, and S is the system's entropy ($ S= \ln(\Omega(E) $), where $\Omega(E)$ is the density of states (a function of the energy).</p>

<p><strong>This means that for a given density of states of a system: if the derivative of its natural log has any extrema (local maxima <em>or</em> minima), then the system will have <em>infinite</em> temperature at those energies (at the locations of the extrema--remember that $\Omega$ is a function of the energy).</strong></p>

<p>Interestingly, by this definition: at energies where the slope of a system's entropy, S, is negative, the system is defined to have a negative energy (which ironically is more energetic than states with positive temperature!)... although to be honest: I am not totally convinced that systems at negative temperatures are actually at equilibrium, which (I would think) indicates that temperature is ill-defined... but definitely a curiosity worth some consideration.</p>

CC BY-SA 3.0
2016-06-12T22:52:49.057Z
2016-06-12T22:52:49.057Z
randomquestionasker77
<p>Temperature is a measure of the average kinetic energy of particles in a given place (correct me if I am wrong) so there must be a definite limit to heating like there is an absolute zero to cooling. And what process could achieve this temperature?</p>

CC BY-SA 3.0
2016-11-20T23:46:47.923Z
I'm voting to close this question as off-topic because it should be on Physics SE.
ProfRob
CC BY-SA 3.0
2016-11-20T23:52:44.307Z
See for example http://physics.stackexchange.com/questions/231354/hotter-than-the-absolute-hot http://physics.stackexchange.com/questions/1775/why-is-there-no-absolute-maximum-temperature?noredirect=1&lq=1
ProfRob
<temperature><heat>
Is there a limit to how hot an object an get?
HotSaucey
# Is there a limit to how hot an object an get?

(Asked by: username_0 on 2016-06-12)

Temperature is a measure of the average kinetic energy of particles in a given place (correct me if I am wrong) so there must be a definite limit to heating like there is an absolute zero to cooling. And what process could achieve this temperature?

**Tags:** temperature, heat
> **Comments:**
> **username_1**: I'm voting to close this question as off-topic because it should be on Physics SE.  
> **username_1**: See for example http://physics.stackexchange.com/questions/231354/hotter-than-the-absolute-hot http://physics.stackexchange.com/questions/1775/why-is-there-no-absolute-maximum-temperature?noredirect=1&lq=1  

---

## 2 answers

### Answer by username_2 on 2016-06-12. Score: 5 | Accepted answer

First, let's get a bit of background. "[Absolute hot](https://en.wikipedia.org/wiki/Absolute_hot)" is the concept that a maximum temperature **exists**. It describes your question exactly. For a long time, no one was able to figure out whether an absolute hot exists, but in the 20th century major revolutions in theoretical physics gave us answers.

"Absolute hot" isn't the name of the hottest temperature, but it's the concept that one exists. So what is our answer — is there a highest possible temperature that fits the concept of an "absolute hot"? Well, **yes**. But first we must talk about heat and temperature.

Photons act both like particles and waves. As we add more energy, the photon's wavelength decreases, and its frequency increases. Thus, the wavelength of the light emitted by hot objects generally depends on the temperature of the object. Hotter objects will emit more energetic photons with shorter wavelengths.

According to modern physics, a Planck length is the smallest meaningful length in the Universe (the reason for that is an entirely different question). If we keep raising the temperature of our object, nonstop, the wavelength of the photons it emits will decrease until it reaches a Planck length. Our object would reach a temperature called **[Planck temperature](https://en.wikipedia.org/wiki/Planck_temperature)**. That's about $1.417×10^{32} \; \mathrm{K}$.

Planck temperature is theoretically the hottest possible temperature. At that point, the object could not theoretically get any hotter, as the photon's wavelength cannot decrease beyond a Planck length.

But what happens if we were to add *more* energy to the object? Would it go beyond Planck temperature? Well...God knows. At that point, we wouldn't even call it temperature anymore. Nobody knows what exactly would happen, but it certainly wouldn't match the definition of temperature.

> **Comments:**
> **username_0**: Thank you for your amazing skill of making something that is complicated to something even I could understand.  
> **username_0**: Let's say that this theoretical photon got even more energy. Why would it not match the definition of temperature?  
> **username_2**: @Mobal Well, the photon would become hotter than what is physically allowed. I like your question; theoretically, there might not be a limit to how much energy we could add to the photon. But we have no idea what would happen if it went beyond the Planck temperature, as the wavelength shouldn't be able to decrease anymore. It really wouldn't follow the concepts that we've established already, like the Planck length being the smallest possible length.  
> **username_0**: If we found a way to add enough energy to something to heat it up past the Plank temperature, wouldn't that be indirect evidence of the disproving of the Planck length being the shortest possible length?  
> **username_2**: @Mobal I think that's an unsolved problem in physics, even though according to our models, the Planck length must be the shortest possible length (you could read more about it here: https://goo.gl/4VBO8T). But ultimately, we have no way of testing that. Temperatures that hot would actually form a black hole (called a Kugelblitz). In Einstein's theory of General Relativity, energy is a source of gravity too.  
> **username_0**: I am not a whiz at physics but is energy a source of gravity because of E = mc^2  
> **username_2**: @Mobal I haven't studied General Relativity too thoroughly. But that equation probably has something to do with it. Other things, like pressure, mass, and momentum are sources of gravity too. Somehow we got from temperature to GR, haha.  
> **username_0**: I swear GR is connected to everything astronomy  
> **username_3**: Thanks for the great answer. It makes me wonder about temperature inside a black hole.  What about "merely" inside neutron or quark stars: are we getting close to these temperatures, or, not even close??  
> **username_2**: @JoeBlow The maximum temperature of a neutron star is 10^12. So Planck temperature is hundreds of millions of trillions times that amount.  
> **username_4**: uh, perhaps i'm just naive, but i don't think temperature is a well-defined quantity when the energy of a system is fixed (like a single photon, for instance)...  
> **username_2**: @colnegn I'm sorry? Why not?  
> **username_4**: well i'm no expert on the subject, but I believe that E and T are conjugate variables (they are related by a Fourier transform), and therefore have an associated uncertainty.   similarly, in a system with a fixed volume: the pressure is ill-defined.  
> **username_5**: In reality, wouldn't a photon basically turn into particles well before it reached Planck energy?   I understand the answer was theoretical, but when asked "what if you gave it more energy", in reality, long before the photon reached that energy it would give up some energy in the form of particle creation.  (or am I mistaken?).  I don't think I'm mistaken.  
> **username_2**: @userLTK Yes, pair production would probably happen well before it reached Planck temperature. This is just a theoretical explanation for Planck temperature. Still, in the history of the Universe, matter *has* Planck temperature before — at 10^-42 seconds before the Big Bang, the entire Universe was at that temperature. That's just about 200 Planck times after the Big Bang (you can read up on Planck time, it's the shortest amount of time possible)!  
> **username_5**: @SirCumference so, it's theoretically possible that there were no photons at or prior to 10^-42 seconds after the big bang because it was too hot for photons?   or am I reading too much into this.  
> **username_2**: @userLTK This all took place during the Planck epoch, the earliest period after the Big Bang. There is no currently available physical theory to describe such short times. As you get closer and closer to t(0), our current theories begin breaking down. I'm gonna guess some aspects of GR and QM break down more than others, so physicists could determine the temperature at those times. Still, for the most part, we have no idea what went on back then.  
> **username_2**: @userLTK Also, I made a typo earlier on, when I said "at 10^-42 seconds before the Big Bang". I meant *after* the Big Bang XD  
> **username_6**: @userLTK   so when photon gains more energy it starts populating the space ?  
> **username_2**: @SanketDeshpande What do you mean, "populating the space"?  
> **username_6**: @SirCumference maybe give away heat by creating more matter ? not sure.  
> **username_2**: @SanketDeshpande Still unsure what you mean by "populating the space", unless you're thinking of particles classically as spheres.  
> **username_5**: @SanketDeshpande  I think he's asking in relation to pair production, which is a super-high energy situation.   Also, I learned in another thread, a single photon can't produce a pair of particles in and of itself, but if it hits something it can.   High enough temperatures would probably start creating pairs of particles, but a single photon, without hitting something, no.  But the energy this requires is enormously high, only created by very violent events like black hole accretion disks or inside particle accelerators.  
> **username_6**: @userLTK exactly, thats what i mean, but was unsure about single photon doing it  
> **username_5**: @SanketDeshpande  I asked a somewhat related question here.  The answer to my question is also an answer to yours if you want more detail.    http://physics.stackexchange.com/questions/266507/photon-pair-production-at-relativistic-speeds  
> **username_6**: interesting, so if we take an isolated system with 2 photon and continuously heat it what would be the outcome? Will it produce matter and consume all the given space? any thoughts?  

---

### Answer by username_4 on 2016-06-12. Score: 2

As I stated in my comment in Sir Cumference's answer, I might just be naive on the subject, but I believe that:

Assuming that the density of states (the number of states per possible energy) for a given system can be taken to be effectively continuous, one can define the inverse temperature:

$$ \frac{1}{k\_BT} = \beta = \frac{dS}{dE}$$

Where E is the system's energy, and S is the system's entropy ($ S= \ln(\Omega(E) $), where $\Omega(E)$ is the density of states (a function of the energy).

**This means that for a given density of states of a system: if the derivative of its natural log has any extrema (local maxima *or* minima), then the system will have *infinite* temperature at those energies (at the locations of the extrema--remember that $\Omega$ is a function of the energy).**

Interestingly, by this definition: at energies where the slope of a system's entropy, S, is negative, the system is defined to have a negative energy (which ironically is more energetic than states with positive temperature!)... although to be honest: I am not totally convinced that systems at negative temperatures are actually at equilibrium, which (I would think) indicates that temperature is ill-defined... but definitely a curiosity worth some consideration.


---

