Recently, my teacher just told us that intensity is not linearly dependent on temperature and that it's ONLY dependent on photons. But then, what about Boltzmann's law? Isn't intensity dependent on the fourth power of absolute temperature?

Even if you only consider temperature, we do observe that even though intensity is non-uniformly changed, increasing temperature does increase the MAXIMUM intensity. So.. technically, isn't intensity also non-linearly dependent upon temperature?

UPDATE: I asked my teacher this question and he explained that Stefan's law applies when we're restricted to Classical Physics, not Quantum Physics. So, when you're talking about the macro-level, AKA, a body emitting light as some of the answers here explain, Stefan's law applies. But when you're solely talking about a photon, Stefan's law doesn't apply and so.. intensity of a photon isn't dependent on temperature IN QUANTUM PHYSICS. Don't know how... that works.

UPDATE: I might have, sort of, figured it out.. fact of the matter is, when we're discussing BLACK BODY RADIATIONS, specifically, the intensity of light does increase with temperature, regardless of whether you're considering classical or quantum physics. Classical physics is correct in predicting the behavior of radiation emitted when the body was heated, but only to specific wavelengths. And, finally, Plank's Quantum Theory is correct for all wavelengths. So, yeah.. intensity of light coming from a BLACK BODY does increase, regardless of whether we're studying classical theories or quantum theories. Cheers, everyone :D! Thank you so much for giving me your time!

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    $\begingroup$ Intensity is energy per area per time. Thus, it certainly does not only depend on the number of photons. Just take the same number of photons, but change their individual energy (wavelength). Thus, I suppose you take your teachers comment out of the context. I believe you should go and talk to your teacher. Usually, teachers are happy if students show interest in the topic. $\endgroup$ – Semoi Dec 5 '20 at 16:06
  • $\begingroup$ Gotcha! Due to lockdown and everything, we only get to talk to the teachers like for ~30 minutes everyday and since we've a day off tomorrow, I can only talk to him on Monday. I just don't like waiting when I'm doubtful of a concept, haha $\endgroup$ – ihateelectricalphysics Dec 5 '20 at 16:08
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    $\begingroup$ Also good if you ask your teacher for additional literature. "Where can I read more about this, at a simple level and also at a more precise level?". Teachers are usually happy when students are curious and want to dig deeper into a topic. $\endgroup$ – pglpm Dec 7 '20 at 16:23
  • $\begingroup$ It's just that here in the Asian countries (Pakistan, to be more specific), the teachers aren't.. that good. At least at the high-school/college level, they have pretty average concepts and so we try to seek out other ways to understand stuff. Especially since we've JUST been introduced to Quantum Physics. However, I still try to seek help from them first. Cheers, everyone. $\endgroup$ – ihateelectricalphysics Dec 7 '20 at 16:29

Checking Wikipedia:

Several measures of light are commonly known as intensity:

Radiant intensity, a radiometric quantity measured in watts per steradian (W/sr)
Luminous intensity, a photometric quantity measured in lumens per steradian (lm/sr), or candela (cd)
Irradiance, a radiometric quantity, measured in watts per meter squared (W/m2) Intensity (physics), the name for irradiance used in other branches of physics (W/m2) Radiance, commonly called "intensity" in astronomy and astrophysics (W·sr−1·m−2)

Without any further qualification, I would take "intensity" to mean the first. Wattage is number of photons per unit of time times energy per photon, and energy per photon is proportional to frequency.

Recently, my teacher just told us that intensity is not linearly dependent on temperature and that it's ONLY dependent on photons.

If you claim that intensity is dependent on temperature, that's a rather unclear what that means. The term "temperature" doesn't apply to photons, at least not directly. One can establish a correspondence between frequency and temperature, for instance by taking the energy and dividing by Boltzmann's constant. But that's the temperature that corresponds to the photon, not the temperature of the photon. Given that interpretation, the intensity is proportional to the temperature.

Another interpretation of the statement is that it's referring to the temperature of an object emitting light. And one can further interpret that as meaning the intensity of the light emitted by an object emitting black-body radiation at that temperature. Note that when an object is emitting light, it's not necessarily emitting it as black body radiation. If it is being emitting as black body radiation, then Boltzmann's law applies.

  • $\begingroup$ Yes! We were specifically talking about Black Body radiations at the time, and this explanation suits it perfectly! The thing is, I asked my teacher this question and he explained that Stefan's law applies when we're restricted to Classical Physics, not Quantum Physics. So, when you're talking about the macro-level, AKA, a body emitting light as you said, Stefan's law applies. But when you're solely talking about a photon, Stefan's law doesn't apply. I guess, like you said, it's because you can't measure the temperature OF a photon. $\endgroup$ – ihateelectricalphysics Dec 7 '20 at 15:35

You are both at least somewhat right.

Photons are more or less particles of light. So the more particles, the more light. This is true at any temperature.

But making an object hot makes it emit more photons. Furthermore, it makes it emit photons of higher frequency and thus higher energy. This makes the energy go up faster than the number of photons.

There are ways of making photons other than heating an object. For example, a laser makes photons all of the same frequency. So there intensity is proportional to number of photons.

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    $\begingroup$ Yes. It increases the number of photons per second, and it makes the hot objects emit more energetic photons. The light energy per second from a hot object goes up faster than linearly with temperature. $\endgroup$ – mmesser314 Dec 5 '20 at 16:35
  • $\begingroup$ Ah, I see. But then, since we already have that out of the way, where does Boltzmann's equation come into the picture? How's it of any use to us? $\endgroup$ – ihateelectricalphysics Dec 5 '20 at 16:39
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    $\begingroup$ The Stefan-Boltzmann equation says what I said, but much better. It tells you exactly how fast energy goes up with temperature. $\endgroup$ – mmesser314 Dec 5 '20 at 16:42
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    $\begingroup$ This may help. hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html $\endgroup$ – mmesser314 Dec 5 '20 at 16:47
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    $\begingroup$ @ihateelectricalphysics - Thank you for the comment, but I am not sure it is entirely correct to say that this is strictly classical. HyperPhysics has a derivation that starts from the Planck radiation law. This assumes quantized radiation. $\endgroup$ – mmesser314 Dec 8 '20 at 0:44

Yes, but your instructer was refering to the intensity as measured by a detector independent of the source. In that case the intensity detected is just the number of photons. The temperature of a star will dictate the number of photons emitted. But the number detected gives the intensity of photons on a detector.


The problem is that the statement is a little imprecise, so it has some truth and some falsity.

A photon is, by definition, a quantum of energy exchanged between the elecromagnetic field (more precisely: a mode of the electromagnetic field) and something else. So by definition the energy exchanged is proportional to the number of photons, and therefore the light intensity is proportional to the number of photons exchanged through a surface, divided by the surface area and the time of the exchange. We speak of "quanta" of energy because the electromagnetic field cannot give or receive energy continuously, but only in multiples of a minimum amount – the quantum, the photon.

But this isn't the full story. The energy of a photon depends on the frequency $\nu$ of the electromagnetic field exchanging it: the energy is $h\nu$, where $h$ is Planck's constant. So if the same number of photons is exchanged in two different situations, through equal areas and in equal amounts of time, the intensity can still be different in the two situations.

How does the temperature enter in all this? Roughly speaking, in many situations we're actually uncertain about the number of photons exchanged by the electromagnetic field. We only have a probability that a particular number is exchanged. Typically this probability depends on the temperature $T$. For example (thermal state of light) we can have that the probability that $n$ photons of frequency $\nu$ are exchanged is $$P(n)=\frac{\exp\bigl(-n\,\frac{h\nu}{kT}\bigr)}{1-\exp\bigl(-\frac{h\nu}{kT}\bigr)} \ ,$$ where $k$ is Boltzmann's constant.

This means that the precise value of the intensity is also uncertain, because the number of photons is. We can calculate the average or expected value of the intensity: by definition this average is proportional to $$\sum_n n\ P(n) \equiv \sum_n n\ \frac{\exp\bigl(-n\,\frac{h\nu}{kT}\bigr)}{1-\exp\bigl(-\frac{h\nu}{kT}\bigr)} \ .$$

To summarize, the intensity is proportional to the number of photons, but the probability of the exact number of photons depends on the temperature. Therefore the average intensity depends, indirectly, on the temperature. It's also important to specify whether we're speaking about the temperature at one specific frequency (or of one specific electromagnetic mode), or through the whole spectrum, and so on.

It's good to emphasize that notion of temperature in electromagnetism is tricky. From a classical thermodynamic point of view the electromagnetic field does not have any temperature. Rather, we speak of the temperature of the matter in which it resides or with which it interacts. From a statistical-mechanical point of view we can speak of the (statistical) temperature $T$ of the electromagnetic field, in the sense that if we're uncertain about its state we then give each possible state a probability proportional to $\exp[E/(kT)]$ (or some other Gibbsian formula), where $E$ is the state's energy. This is almost a definition of the (statistical) temperature of the field. This definition goes over to the quantum description, and that's where the formula for the probability given above comes from. The two notions of temperature get connected only through the presence of matter: in equilibrium the statistical temperatures of the electromagnetic field and of the matter it interacts with must be the same; but the statistical temperature of matter is in turn connected with its thermodynamic temperature (in ways that are still under debate).

Good references about all this are Leonhardt's Measuring the Quantum State of Light (Cambridge U. Press 1997); Jackson's Classical Electrodynamics (Wiley 1999); Hutter, van de Ven, Ursescu's Electromagnetic Field Matter Interactions in Thermoelastic Solids and Viscous Fluids (Springer 2006); and also Biró's Is There a Temperature?: Conceptual Challenges at High Energy, Acceleration and Complexity (Springer 2011).

  • $\begingroup$ Thank you so much! I had one question, though. My teacher seems to explain that since Stefan's law is restricted to classical physics (at the macro level), therefore, it fails to explain the behavior of Photons when we start to study Quantum Physics (study at the micro-level.) And hence, while intensity IS proportional to the temperature, it's only so when we're talking about classical physics and not Quantum Physics. Do you agree? $\endgroup$ – ihateelectricalphysics Dec 7 '20 at 15:46
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    $\begingroup$ Not quite. The relation "intensity $\leftarrow$ energy $\leftarrow$ probability $\leftarrow$ temperature" is valid both in classical and quantum physics. Do you mean "is proportional to temperature", as opposed eg to "is inversely proportional to temperature"? Or do you mean "is dependent on temperature" as opposed eg to "is dependent on photon number"? $\endgroup$ – pglpm Dec 7 '20 at 16:00
  • $\begingroup$ I asked him about dependence. Like, if we increased the temperature of a black body emitting radiation (since you can't increase the temperature of a photon), would the intensity also increase? He says that when you only consider classical physics, then yes, it would. But, when we started to study photons at the micro-level, i.e in Quantum Physics, it was found that intensity IS NOT dependent on temperature. $\endgroup$ – ihateelectricalphysics Dec 7 '20 at 16:05
  • $\begingroup$ I don't think that's what your teacher meant. Intensity is dependent on photon number. Photon number is dependent on temperature. Hence intensity is (indirectly) dependent on temperature. $\endgroup$ – pglpm Dec 7 '20 at 16:07
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    $\begingroup$ @ihateelectricalphysics Yes for a blackbody source it should be directly proportional. One must make clear if one is speaking about expected intensity, intensity at a particular frequency or through the spectrum, etc. Maybe your teacher meant that there are additional corrections to the temperature dependence when one considers additional quantum effects – I don't know about this. It's also good if you ask your teacher for literature references, so you can go, examine, and check without time constraints. $\endgroup$ – pglpm Dec 7 '20 at 16:18

It depends how you defined intensity and what units you are measuring that in.

Lux, lumens, candela, watts/m^^2, cycles per second, photons per steradian, or what?

  • $\begingroup$ Ah, my bad! We were discussing it in the units of Watt/m^2 $\endgroup$ – ihateelectricalphysics Dec 7 '20 at 15:47

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