Maxwell’s equations appear to have no limitation as to the length of an electromagnetic wave that can be produced by an accelerating electric charge. So in theory, if I use a charged rod to oscillate a charged pith ball at, say $1\, \mathrm{Hz}$, the wavelength of this wave is $3\times10^8\,\mathrm{m}$! Although the energy of this wave is very weak, it is an electromagnetic wave. I have two questions:

How would someone go about to detect such a weak wave, if possible?

Has anyone ever detected electromagnetic waves in nature (I am thinking outer space) with a comparable wavelength?

  • $\begingroup$ Although not quite as low frequency as you are asking about (though close), Extremely Low Frequency radio has been used by militaries to communicate with submerged submarines. It is also produced by various natural sources. Wikipedia lists 1 Hz in the "Tremendously Low Frequency" band, saying it's "Natural and man-made electromagnetic noise" but doesn't say any more than that. $\endgroup$ – Michael Brown Jun 19 '13 at 16:35

While there does not exist currently any standardized technology for detecting such long-wavelength radiation, the technology does exist. In theory, one could build an extremely large antenna out of smaller antennae placed between here and the Moon (it would need to be that large to accommodate the long wavelength) and tune the receiver to that frequency. Or one could set up an omni-directional antenna that resonates at $1Hz$. However, the practicality of it is minimal. A long wavelength like that might easily pass though some objects, but the wave would have such low energy that massive amplification would be required for any transmitting or receiving apparatus. This would put the power consumption at some ridiculous level and any benefit of using a $1Hz$ signal would be lost.

As for detecting this frequency from nature, no. Everything emits blackbody radiation. The warmer something is, the higher the peak frequency and the reverse for colder things. The coldest something could get in nature (meaning a system in equilibrium with its surroundings) is around $2.7K$. The Cosmic Microwave Background radiation is emitted at approximately this temperature. That means the CMB should have one of the lowest peak frequencies of all natural bodies. The low-end peak frequency for the CMB is still, however, ~$160.2GHz$, a wavelength of ~$1.873 mm$.

That being said, all objects emit radiation at lower frequencies, including at the $1Hz$ range. However, when we observe these objects, we tend to look for frequencies that will have enough energy to be easily visible and using well known technology. Currently, we use Radio waves as the lowest range (Yes, I see the tautology there), specifically the mid to high-end radio waves. The reason being manifold; as wavelength goes up, your antenna dish needs to be on the same scale as it or larger. This means for even poor resolution, a $1Hz$ telescope would need to span almost to the Moon (as I mentioned earlier). Also, $1Hz$ is very far from the peak emitted frequency of any object, which means that the power at that frequency would be extremely low. In most cases, it is so low that one could approximate the amount of radiation at $1Hz$ as zero.

In summary, $1Hz$ is a very natural frequency, in fact cosmologists often talk about frequencies much lower; with wavelength on the order of the size of the visible universe. The technology does exists to be able to measure such a wave, but it is (in almost all cases) impractical to do so and thus, there is no standard equipment existing that can do it. All objects emit $1Hz$ radiation, however this is nowhere near the peak frequency of any natural body, so detecting it from a natural object would be pointless and require massive amounts of power.

  • $\begingroup$ Wow! It never occurred to me that the scale of a measuring apparatus would have to be so large, but of course, it makes sense after you explained it Jim. I like the way you brought in the CMB into the picture. Thanks. $\endgroup$ – Carlos Jun 19 '13 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.