4 ms·
Sure, 60 GHz may be a resonant frequency of O₂ molecules and signals at that frequency may transfer energy to them and thus heat them up. However, remember tha
by jdiez17 5y ago
Sure, 60 GHz may be a resonant frequency of O₂ molecules and signals at that frequency may transfer energy to them and thus heat them up.
However, remember that radio signals lose energy to free space proportional to the (distance and frequency) _squared_ [1]. A 60GHz signal is attenuated by 88dB at 10 meters. That means that if it is transmitted with 100 W (20dBW), the RF power that arrives at your position is -68dBW, or 0.000158 milliwatts. That isn't going to heat up much.
[1] https://en.wikipedia.org/wiki/Free-space_path_loss https://en.wikipedia.org/wiki/Free-space_path_loss
- mlyle 5y agoNote that this math is misleading for calculating how much things will heat up, because effective antenna aperture of the thing that you'd heat up increases with the square of frequency. So the square terms of frequency cancel each other out. It's still small, but the best simple approximation is just to take the 100W and divide by the proportion of the antenna pattern your target occupies. A 100W source on 2GHz or 60GHz at 10 meters distance is going to dump about the same amount of energy into you. If the transmit antenna has 3dB of gain, that's going to be about 50mW.
- jdiez17 5y agoThat's interesting. I'm not quite sure how that works but I'll look into it. If you have some resources, I'd appreciate it!
- mlyle 5y agoSee the derivation section in your link. It's assuming an isotropic antenna at the receiver, whose effective area gets smaller as the frequency increases. But people do not get smaller as the frequency increases: they occupy the same fraction of the 10m radius sphere around the antenna.