# Does color temperature limit how much a laser of a given wavelength can heat a target?

The Sun has a peak wavelength of around 500 nm and an effective surface temperature of 5770 K, and sunlight cannot be focused to make something hotter than the Sun, because this would be heat flowing from a colder object to a warmer object.

My question is how (and whether) this applies to lasers that emit a single frequency. For example, does this mean that a 500 nm laser would also be incapable of heating any object beyond 5770 K, regardless of how many watts you pump the laser with, how well you insulate the object, or how many lasers you use?

• No, this does not apply to lasers because a laser is a coherent radiation source and not incoherent black body radiation. Commented Jul 17 at 2:46
• @FlatterMann Correct me if I'm wrong but is it really about power intensity at the emitter? i.e. the Sun has a certain W/m^2 at its surface which determines its blackbody spectrum, and a 500 nm laser can be pumped to a (much?) higher W/m^2 than that. (How high can it be pumped?) So then intensity at the emitter determines a limit on intensity at the target; target intensity cannot be greater than source intensity. And intensity at the target determines equilibrium temperature of the target, equilibrium reached when the target is radiating as much W/m^2 as it is receiving. Commented Jul 17 at 3:01
• Coherence is part of the story because spatial coherence is what allows the whole output of a laser to be focused onto an extraordinarily small target. IDK what role (if any) is played by the temporal coherence (i.e., by the extraordinarily narrow spectrum) of laser light. Commented Jul 17 at 4:28
• You are correct correct. The power density limitation for black bodies does not apply to laser sources. Commented Jul 17 at 4:54
• @stackoverblown Yes, you can make something burn. The point is you can't make it hotter than the sun's surface itself.
– bdsl
Commented Jul 17 at 17:27

The argument for heating by the Sun is that if the target object is hotter than the Sun it would heat the Sun rather than the Sun heating it. We could try to apply this argument to a laser, but how would we estimate the temperature of the laser?

The Sun is a black body, and for a black body there is a simple relationship between its temperature and the spectrum of the light it emits. However a laser produces light from a population inversion of excited states and this is not even close to being in a thermal equilibrium.

If we take the helium-neon laser as a common type of laser that many of us will have used, then the population inversion is generated by the transition of helium atoms from the ground state to excited states that have an energy of about $$20~\mathrm{eV}$$. These are generated by a beam of electrons not from thermal excitation, but we could guesstimate a temperature by equating that $$20~\mathrm{eV}$$ energy to $$kT$$ and we would get a temperature of about $$2 \times 10^5$$K or about $$40$$ times hotter than the surface of the Sun.

If we accept this somewhat arm waving approach it predicts an object would have to be hotter than $$200~\mathrm{kK}$$ before it could start "heating" the laser rather than the laser heating it.

• You are showing your age (like me - saw my first laser, a He-Ne, at a local college back in the early 1970s)! I don't think my 20-something kids have seen a He-Ne laser in their lifetime. All laser diodes now out in the world at large... Commented Jul 17 at 13:15
• I have to object to the Energy to temperature conversion: That's simply not how temperature is defined. While it's true that temperature increases linearly with energy content in many cases, the actual definition is via the entropy change in relation to an energy change. And with that definition, we find that the lasing medium is always at a temperature that tends towads $T = -\infty$. Commented Jul 18 at 19:29
• That is the consequence of the state inversion, and also precisely the reason why this state inversion is required: It means that the lasing medium can increase its entropy by getting rid of energy, and is thus happy with adding energy to the entropy free laser beam. Commented Jul 18 at 19:37

Neither the Sun nor a laser are objects in thermodynamic equilibrium
The second law of thermodynamics, which prohibits the flow of heat from a colder to a warmer object, applies to objects that have reached internally thermodynamic equilibrium. If this objects are brought in contact or otherwise made to interact, they would evolved towards the common thermodynamic equilibrium with a temperature between the initial temperatures of the two objects - that is the warmer object cools, whereas the colder one warms up.

Neither the Sun nor a laser are objects in thermodynamic equilibrium, but at best objects in a steady state: both constantly lose energy by radiating, but these losses are replenished by thermonuclear reactions in the Sun and the power supply of the laser. Being macroscopic objects they can be described by statistical distributions that are quite similar to the equilibrium ones, and attributed an effective temperature, but many of the conclusions of the equilibrium thermodynamics do not apply.

The true temperatures of the Sun and a laser
Given its quasi-equilibrium state, the radiation emanating from the Sun surface can be described as a black body of temperature about 6000K (see How does radiation become black-body radiation?), however this surface temperature is not the same as the temperature inside the Sun, which rises to millions of Kelvin in its core (another obvious manifestation that Sun is not in thermodynamic equilibrium.)

The wave-length of the maximum of black body radiation determines the rest of energy distribution and uniquely related to its temperature. This is manifestly not true for a laser, with a radiation spectrum very far from that of a black body. In fact, temperature is not meaningful for a laser, with the photon occupation numbers in the active mode much higher than those of lower energy modes. At best we can assign to a laser infinite temperature (although in laser physics one sometimes resorts to introducing negative temperature for laser radiation and active environment.) The constraints of the second law of thermodynamics clearly do not apply here.

Assuming that the target is a black body. The real critical factor affecting heat transfer is emit energy per unit area described by Stefan–Boltzmann law $$M=\sigma T^4$$. The temperature increase occurs when the received power density is higher than the radiated power density. And as you can see, this is not necessarily correlated with the wavelength of the center of the object's radiation (the correlation only holds for black-body light sources).

Further, lasers are not thermally equilateral systems and there is no corresponding relationship between their center wavelength and power density. Therefore, the temperature to which an object can be heated is also not related to the wavelength of the laser.

### UPDATE

I totally disagree with the conversion of laser photons to temperature in other answers. A black body can absorb all incident electromagnetic radiation, regardless of wavelength of incidence. What the relationship between 1 eV = 11, 600 K means? This is to say that the average kinetic energy possessed by a gas molecule/ particle at a temperature of 11, 600 K is about 1 eV. But this does not limit a molecule/atom to taking energy from only one photon, which means that the target is perfectly capable of being heated beyond the temperature corresponding to the energy of a single photon. In strong-field physics, there are a variety of multiphoton processes, the most important of which is multiphoton ionisation, which can strip even heavy atoms of all their electrons, and stronger lasers can even directly induce nuclear reactions. So it is the intensity of the laser that is important, not the wavelength of the laser.

To illustrate further, we can take a closer-to-life example: microwave ovens use 2.4 GHz electromagnetic waves to heat food, and obviously this can easily heat objects up to a few hundred kelvins. But if we convert to the energy of a single microwave photon, $$h\nu\approx 9.93\times10^{-6}$$ eV, implying $$\sim 0.115$$ K.

So you can see that the estimation of the photon energy doesn't give enough information, the more important value is the intensity of the laser. You can compare the intensity of the laser with $$\sigma T^4$$, where $$\sigma$$ is the Stefan–Boltzmann constant. This tells us the temperature a laser can heat a black body up to: $$T=\sqrt[4]{I/\sigma}$$. Here, $$I$$ is intensity of the laser.

• this does not limit a molecule/atom to taking energy from only one photon, which means that the target is perfectly capable of being heated beyond the temperature corresponding to the energy of a single photon. - a single atom cannot he characterized by temperature, which is a thermodynamic quantity, meaningful only in the limit $N\rightarrow\infty$. For a large number of atoms, there will atoms absorbing and emitting photons, with Boltzmann distribution of a given temperature, characterizing how many of them are excited. Commented Jul 19 at 21:11
• @RogerV. Thank you for your comment, it is true that individual atoms do not have a concept of temperature, and I am not characterising individual atoms in terms of temperature here. What characterises a single atom is its translational/vibrational/rotational kinetic energy. The target here refers to a number of atoms assembled into an ideal blackbody. As a side note, the Boltzmann distribution also only describes the collection of atoms in thermal equilibrium. Commented Jul 21 at 10:59

No, your reasoning does not apply to a laser. You can use Wien’s law to relate the peak emission wavelength to the temperature of a black body emitter (a perfect absorber/emitter across all wavelengths). However, this description cannot be applied to the coherent stimulated emission from a laser, which relies on a population inversion (i.e. is not at thermal equilibrium) and has much stronger emission/absorption at certain wavelengths (defined by atomic transitions). You could work out the temperature that a laser could heat a body to by considering a balance between absorption at the laser wavelength and thermal emission across all wavelengths. Choosing body emissivity $$\epsilon ( \lambda ) = 1$$ for $$\lambda = \lambda_{laser}$$ and $$0$$ otherwise would give a theoretical maximum.

Alternatively you could note that NIF lasers heat targets to millions of Kelvin using 351 nm light (coresponding to 8300 K according to Wien’s displacement law).

No, it is totally unrelated. Look at the National Ignition Facility (at LLNL). It has a laser, I do not know what its frequency is (probably a small multiple of $$c/1064\,$$nm, but it heats a pellet to the point that fusion occurs, which is many eV, while 1 eV is already

$$1\,{\rm eV} = 1\,{\rm V} \times \frac e k \approx 11,600{\, K}$$

which is twice the Sun's peak..not enough for fusion.

Edit (in progress): based on comments, my answer is not clear. I'm just saying the laser is not at all a thermal spectrum. Nevertheless, in the radiometer world, ppl talk about "brightness temperature, $$T_B$$, not actual temperature $$T$$. It includes emissivity as a frequency dependent fudge factor.

At a frequency $$\nu$$, a blackbody has a radiance:

$$B(\nu, T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}}$$

i.e., the well known Planck formula. Now if you observe a radiance of $$B$$ at $$\nu$$, the emissivity, $$\epsilon$$, satisfies:

$$B= \epsilon B(\nu, T) \equiv B(\nu, T_B)$$

Given the NIF configuration, I'll look at the energy density per unit volume:

$$u(\nu, T) = \frac{8\pi h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}}$$

and go with approximate numbers (since I don't know them).

Let's say:

Laser energy (from the news):

$$E = 1\,{\rm MJ}$$

in time (from the last YAG I used):

$$t = 1\,{\rm ns}$$

with frequency (frequency tripled YAG, rounded up):

$$\nu = 10^{15}\,{\rm Hz}$$

and band width (from $$\Delta\lambda=8$$ nm, from google):

$$\Delta\nu = \nu \frac{\Delta\lambda}{\lambda} \approx 10^{13}\,{\rm Hz}$$

in the volume defined by a pellet with:

$$R = 1\,{\rm mm}$$

[stand by]

• At LLNL, the laser heats the surface of a pellet enough to ablate the outer layer. This accelerates the outer layer radially outward. The reaction force crushes the pellet. The compression does the heating. In effect, the laser must be hot enough to light the fuse. Commented Jul 17 at 13:04
• but it's not a blackbody spectrum (duh). At the wavelength, if we say $\epsilon=1$, the brightness temperature is insanely high--hot enough for fusion, but IR-UV is so far below the Wein's Law peak frequency..so idk. I just don't think of a laser a thermal radiation --so the answer could be yes: look at NIF, it gets really hot.
– JEB
Commented Jul 17 at 14:34
• @mmesser314 I'm addressing your concern now, but I have to go work. (which is in radiometry from space....I hope my colleagues don't see my first answer..).
– JEB
Commented Jul 17 at 15:05