• B
• vijayantv
In summary, the conversation discusses the concept of spectral radiance and its relationship to the number of photons emitted per unit area and wavelength interval. It also touches on Planck's formula and the importance of using gauge-invariant quantities when making physical interpretations. The conclusion is that the "photon flux" and "energy per photon" are not actual quantities and should not be used in calculations or interpretations.
vijayantv
TL;DR Summary

1). It explains that the Spectral radiance measurement of 7000K temperature is the same as both 300 and 520 nm wavelength light.
See here both A and B shows 60.
Is my understanding correct?

2). Is spectral radiance the number of photons? If not, what exactly it is?

1) Yes.

2) Looking at the units, I'd say the area under the curve denotes a power emitted per area. That's a number of photons per second per area.

vijayantv said:
2). Is spectral radiance the number of photons? If not, what exactly it is?
No, it is the radiated energy per square meter, per unit wavelength interval, and per unit of solid angle. At optical wavelengths it is rather unusual to express a wavelength interval in picometers. But you can write ## 60 \rm W m^{-2} pm^{-1} sr^{-1} ## also as ## 60 \rm kW m^{-2} nm^{-1} sr^{-1} ##, or ## 6 \rm W cm^{-2} nm^{-1} sr^{-1} ##.

The photon flux is different at the two wavelengths. Because 520 nm photons have lower energy than 300 nm photons, there must be more of them if the radiance is the same.

vanhees71
WernerQH said:
No, it is the radiated energy per square meter, per unit wavelength interval, and per unit of solid angle. At optical wavelengths it is rather unusual to express a wavelength interval in picometers. But you can write ## 60 \rm W m^{-2} pm^{-1} sr^{-1} ## also as ## 60 \rm kW m^{-2} nm^{-1} sr^{-1} ##, or ## 6 \rm W cm^{-2} nm^{-1} sr^{-1} ##.

The photon flux is different at the two wavelengths. Because 520 nm photons have lower energy than 300 nm photons, there must be more of them if the radiance is the same.

When compare with 400 nm and 300 nm we get Spectral-radiance 70 and 60 respectively.

Here, the 300 nm photon has more energy than the 400 nm photon. how about the number of photons, in this scenario?

vijayan_t said:
how about the number of photons, in this scenario?
You simply divide the spectral radiance by ## h \nu (= hc / \lambda) ##, the energy per photon, to get the flux of photons. You get a different curve because the factor is different for different wavelengths.

You have seen Planck's famous formula, haven't you? You can work out all numbers yourself using $$I_\lambda = \frac {2 h c^2} {\lambda^5} \left(e^{hc/\lambda kT} - 1\right)^{-1}$$ for the spectral radiance, and $$\Phi_\lambda = \frac {2 c^2} {\lambda^4} \left(e^{hc/\lambda kT} - 1\right)^{-1}$$ for the photon flux.

vijayan_t
WernerQH said:
You simply divide the spectral radiance by ## h \nu (= hc / \lambda) ##, the energy per photon, to get the flux of photons. You get a different curve because the factor is different for different wavelengths.

You have seen Planck's famous formula, haven't you? You can work out all numbers yourself using $$I_\lambda = \frac {2 h c^2} {\lambda^5} \left(e^{hc/\lambda kT} - 1\right)^{-1}$$ for the spectral radiance, and $$\Phi_\lambda = \frac {2 c^2} {\lambda^4} \left(e^{hc/\lambda kT} - 1\right)^{-1}$$ for the photon flux.
Got it. very clear ! .
Thanks .

WernerQH said:
You simply divide the spectral radiance by ## h \nu (= hc / \lambda) ##, the energy per photon, to get the flux of photons. You get a different curve because the factor is different for different wavelengths.

You have seen Planck's famous formula, haven't you? You can work out all numbers yourself using $$I_\lambda = \frac {2 h c^2} {\lambda^5} \left(e^{hc/\lambda kT} - 1\right)^{-1}$$ for the spectral radiance, and $$\Phi_\lambda = \frac {2 c^2} {\lambda^4} \left(e^{hc/\lambda kT} - 1\right)^{-1}$$ for the photon flux.
That's an important point, because that's indeed the only way to make physical sense of something like an average "photon number". One should be aware that what's physically measurable must be related to gauge independent local observables, and in the case of photons the measure related to intensity is energy density, which is both gauge invariant and a local observable, i.e., obeying the microcausality principle. That's not the case for a naive photon number density using the annihilation and creation operators of the mode decomposition of the free vector potential, because this is a gauge-dependent quantity, which does not fulfill the microcausality condition (in the radiation gauge).

vanhees71 said:
That's an important point, because that's indeed the only way to make physical sense of something like an average "photon number". One should be aware that what's physically measurable must be related to gauge independent local observables, and in the case of photons the measure related to intensity is energy density, which is both gauge invariant and a local observable, i.e., obeying the microcausality principle. That's not the case for a naive photon number density using the annihilation and creation operators of the mode decomposition of the free vector potential, because this is a gauge-dependent quantity, which does not fulfill the microcausality condition (in the radiation gauge).

Please explain in detail what is not correct in his explanation.

Thanks

vijayantv said:
see the attached image
Where is this image from? Please give a reference.

vijayan_t said:
Please explain in detail what is not correct in his explanation.
The "photon flux" in post #5 is not an actual number of photons. The state of the quantum electromagnetic field in this scenario is not even an eigenstate of photon number, so there is no definite "number of photons" at any frequency. At best the "photon flux" can be thought of as an expectation value of photon number for a given frequency.

Similar remarks apply to the "energy per photon" at a given frequency; this can't possibly be an actual energy per photon since there isn't even a definite number of photons present.

vanhees71
vijayan_t said:
What question of yours has not been answered?

vanhees71
Only gauge-invariant quantities from local observables have a physical interpretation. In the case of the electromagnetic field the photon-number operators do not fufill this criterion, because the field operators ##\hat{\vec{A}}## in the radiation gauge do not fulfill the microcausality condition, because of the constraint ##\vec{\nabla} \cdot \hat{\vec{A}}=0##.

Observable are indeed intensities, i.e., the components of the energy-momentum tensor (particularly energy density and energy-current density/momentum density aka "Poynting vector"), because they derive from the physical field components ##\vec{E}## and ##\vec{B}## which fulfill microcausality conditions and are gauge invariant. So indeed "the flux", derived from the energy-current density of the em. field (Poynting vector) is an observable, but that must not be confused with a "photon-number current", which is not an observable. What you can define of course is the energy-current density divided by the energy and call this "photon-number current".

In the above quoted book by Zombeck it's of course done correctly, using the energy density ("radiance") as it must be.

Last edited by a moderator:
gentzen
Everybody understands "photon flux" as I said, namely as a count rate which is really measured. It's indeed defined via the Poynting vector and not some non-physical photon number, which indeed is a purely theoretical construct without any physically interpretible meaning.

## 1. What is Black Body Radiation?

Black body radiation is the electromagnetic radiation emitted by an object due to its temperature. It is a continuous spectrum of radiation that is dependent on the temperature of the object and follows a specific distribution known as the Planck curve.

## 2. Why is it called "Black Body" radiation?

The term "black body" refers to an idealized object that absorbs all radiation that falls on it and emits radiation at all wavelengths. This means that a black body does not reflect or transmit any radiation, and therefore appears black. It is used as a theoretical model to study the behavior of radiation in different systems.

## 3. How is Black Body Radiation related to the color of an object?

The color of an object is determined by the wavelengths of light it reflects. A black body, by definition, does not reflect any light, so it appears black. However, as the temperature of a black body increases, it emits more radiation at shorter wavelengths, which can give the appearance of a different color (e.g. red hot metal). This is why the color of an object changes as it is heated.

## 4. What is the significance of Black Body Radiation in physics?

Black body radiation has been a fundamental concept in physics for over a century. It has been used to explain various phenomena, such as the thermal radiation from stars and the cosmic microwave background radiation. It also played a crucial role in the development of quantum mechanics and the understanding of the atomic structure.

## 5. How is Black Body Radiation measured?

Black body radiation can be measured using a spectrometer, which separates the radiation into its different wavelengths. The resulting spectrum can then be compared to the theoretical Planck curve to determine the temperature of the object. Alternatively, the total amount of radiation emitted by an object can be measured using a radiometer.

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