Thermal Radiation of human body

I've read that all bodies emit electromagnetic radiation of all wavelengths. Does this mean that humans emit gamma rays too (excluding the radiation from radionuclides present in humans)?

Does this mean that humans emit gamma rays

No.

The problem is right here:

all bodies emit electromagnetic radiation of all wavelengths

You have not read that. You have read that all bodies above absolute zero emit electromagnetic radiation. You did not read "of all wavelenghts".

You can find an image of the emissions from a hot body here. The curve shape remains like this, but the peak and the total area under the curve changes with temperature. This is the black-body radiation curve.

Notice that there is a sharp cutoff at short wavelengths, in this case on the left side of the curve. For a human, that cutoff is in the infrared. We release very little above the IR spectrum.

UPDATE: I've been trying to find a good explanation of the high-frequency cut-off, but so far everything's been super-technical. Basically, above a certain frequency (below a certain wavelength) any incoming energy is down-converted to a lower frequency. This is why the earth is bright in infra-red in spite of visible light shining on it.

• I read it here learningweather.psu.edu/node/18 – Hark Apr 17 '18 at 12:19
• Great. I suggest sending them the links above. – Maury Markowitz Apr 17 '18 at 12:22
• The high frequency cut off is technical. It's called "quantum mechanics". See "Ultraviolet Catastrophe". – JEB Apr 17 '18 at 13:41
• But it's not a "hard cut off". It is an exponential decay. More explanation required. – Rob Jeffries Apr 17 '18 at 14:48
• It is a hard cut off. The body will not release a photon with more energy than the body contains. Yet that would be required to meet the condition "all". There are lots of practical limits far below that, and they are technical, but somewhat inside-baseball for the level of this question. – Maury Markowitz Apr 18 '18 at 13:06

A body of temperature $T$ radiates according to Planck's law: $$B(\nu)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_B T}-1}.$$ This can be integrated across all frequencies to get Stefan-Boltzman's law, but we are interested in how much energy there is for frequencies between $\nu_{min}$ and $\infty.$ This is $$E=\left(\frac{2\pi h}{c^2}\right)\left(\frac{k_B T}{h}\right)^4 \int_{x_0}^\infty \frac{x^3 dx}{e^x-1}$$ where $x_0=h\nu_{min}/kT$.

If $T=310$ K and $\nu_{min}=10^{19}$ Hz. So $x_0=1.5481\times 10^{6}$. Numerically, the integral is very, very small. Using the trick in this page, we can turn it into $$\int_{x_0}^\infty \frac{x^3 dx}{e^x-1} = \sum_{n=1}^\infty \int_{x_0}^{\infty} x^3 e^{-nx} dx.$$ $$\int_{x_0}^{\infty} x^3 e^{-nx} dx = (1/n^4)[-e^{-nx}(n^3x^3+3n^2x^2+6nx+6)]_{x_0}^{\infty} = e^{-nx_0}(n^3x_0^3+3n^2x_0^2+6nx_0+6)/n^4 \approx x_0 e^{-nx_0}/n.$$ So the first term in the sum will be $x_0e^{-x_0}\approx 10^6e^{-10^6}\approx 10^{-434289}$. The second term will be about the square of that, and so on. So it is not a bad approximation to set the sum and hence the integral to this value.

The rest of the energy factor is 80.6398 Watt. So the conclusion is that the energy found in gamma rays is vanishingly small. Given that human metabolism is $\approx 100$ Watt and gamma rays have energies $h\nu>6.6261\times 10^{-15}$ joule there will be less than $(100/h\nu)10^{-434289}\approx 10^{ -434273}$ gamma photons per second. It is unlikely that you will ever emit one in your lifetime... even if you live until proton decay in $10^{36}$ years.

• I have used the approximation that humans have 1 square meter surface area. The actual value is 1.5-2 square meters. – Anders Sandberg Apr 17 '18 at 16:23