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January 8

COVID-19 particles

How big is a COVID-19 particle and can that particle just float around in the air by itself or MUST it be affixed to some sort of larger droplet in order for it to remain airborne? Essentially what I’m trying to approximate is what “size” is the airborne threat, as surely this is an important question in determining what PPE is appropriate etc. --Uhooep (talk) 03:17, 8 January 2022 (UTC)[reply]

The average diameters of most coronavirus virions fall in the range of 80 to 120 nm.^[1] For SARS-CoV-2, the coronavirus species that causes COVID-19, a diameter of around around 120 nm has been reported.^[2] A normal cloth face mask will not filter these out; quoting from the article: "One 2010 study found that 40–90% of particles in the 20–1000 nm range penetrated a cloth mask and other fabric materials.^[16]" However, an individual's immune system can easily beat a single particle. The main risk is inhaling a drop of fluid with a high viral load. Most airborne drops of fluid will be caught by a cloth mask, and the larger (and more risky when inhaled) they are, the more likely they are to be caught. Cloth masks do not offer the same protection as N95 masks, but definitely help, when used systematically, to lower the reproduction number, which needs to get and stay below 1 to end this pandemic. --Lambiam 10:42, 8 January 2022 (UTC)[reply]

There was a big controversy about whether covid was airborne, with the "establishment" saying no and upstarts saying yes. Now it looks like views are tilting towards yes. See covidisairborne.org for lots of links. 2602:24A:DE47:B8E0:1B43:29FD:A863:33CA (talk) 02:28, 13 January 2022 (UTC)[reply]

Field strength and inverse-square law

The first sentence of Antenna_(radio)#Arrays_and_reflectors begins, "The field strength from a transmitting antenna at a specified distance is reduced according to the inverse-square law..." Is there really a sense of "field strength" in which this can be correct, or is it a mistake? My understanding otherwise is that both radiant flux and signal strength decrease with inverse distance, not inverse-square [EDIT: in the context of EM radiation, I mean]. Do I have it right? -Amcbride (talk) 16:41, 8 January 2022 (UTC)[reply]

Yes, the electromagnetic fields fall off as 1/r, not 1/r^2 in case of electromagnetic radiation. The power that escapes to infinity is then finite, because the Pointing flux, which is proportional to the square to the fields, then falls off like 1/r^2, and multiplied with the area of a sphere of radius r, this then tends to a finite limit. In case of slowly varying fields, like if you move a charge and push another charge away, the energy that escapes to infinity will be zero, because Pointing flux times surface area tends to zero in the limit of infinite radius. The energy transfer from the first to the second charge comes with a feedback on the first charge. In case of radiation the energy moves away and can hen affect charges without these charges causing a feedback on the original charge. Count Iblis (talk) 16:52, 8 January 2022 (UTC)[reply]

Thanks! I'd edit the article, but there are several places it uses the term field strength, and for now I'm not sure which are correct and which should be intensity instead. -Amcbride (talk) 18:42, 8 January 2022 (UTC)[reply]

I have not looked at the article, but "electric field strength" normally refers to an electrostatic field, which does obey Coulomb's inverse-square law. This should be distinguished from the strength of the electromagnetic tensor. --Lambiam 08:18, 9 January 2022 (UTC)[reply]

I see what you mean, but the rest of the sentence I partially quoted is "...since that describes the geometrical divergence of the transmitted wave." So the context does seem to be radiation, not electrostatics. -Amcbride (talk) 16:11, 9 January 2022 (UTC)[reply]

The author of this sentence had rather a cloudy understanding of the basic physics. Ruslik_Zero 20:17, 9 January 2022 (UTC)[reply]

Another radiation question

Is there a word for the sum of the magnitudes of irradiance/intensity from all directions? That is, the power received by an ideal isotropic antenna divided by its effective aperture? (Or am I confused and these aren't the same idea? Or am I so confused that one or more of these ideas is not even coherent?) -Amcbride (talk) 19:36, 8 January 2022 (UTC)[reply]

Radiant intensity? Ruslik_Zero 20:21, 8 January 2022 (UTC)[reply]

That's per steradian, though... What I'm imagining I want seems like it would still be in units of power per area. Something like, surround the receiving point by a sphere, measure total radiant flux through that sphere, then divide by its area. (EDIT: Then take the limit as the radius of this sphere goes to zero. My intuition is that would be a well-defined limit, but I could be way off.) -Amcbride (talk) 20:42, 8 January 2022 (UTC)[reply]

An ideal isotropic antenna is useful as a theoretical reference but it is impossible to construct. For field measurement purposes it may be approximated by three orthogonal antennas or sensing devices with a radiation pattern of the omnidirectional type

\sin(\theta )

, such as short dipoles or small loop antennas. However these are resonators whose response varies with frequency and cannot be so well matched to the impedance of free space as to capture all incoming radiated power at all frequencies. Only an ideal black body absorbs all incident electromagnetic radiation which it converts to heat. Philvoids (talk) 00:40, 9 January 2022 (UTC)[reply]

Thanks. I see what you mean about measuring the whole spectrum, and I'd never thought of a black body that way. But really I was thinking per-frequency or over some convenient frequency band, but I forgot to say that part. -Amcbride (talk) 16:43, 9 January 2022 (UTC)[reply]

Radiant flux? The concept does not require the radiant energy to be isotropic; it applies to any distribution. The solid angle subtended by a full sphere is 1 sp = 4π sr. So, assuming that the radiant intensity is isotropic, the radiant flux equals 4π times the radiant intensity. --Lambiam 08:04, 9 January 2022 (UTC)[reply]

Well, I understand that you can integrate irradiance (power per area) over the whole sphere and get total radiant flux (power). My question is, what would you call it when you divide that integral by the surface area of the sphere, getting units of power per area again? And am I right to think that doing so would lead to a quantity with a finite limit as the radius of the sphere goes to zero? Because if the answer to that last question is "yes", then it would seem to be a sensible way to characterize some vague idea of, "How much radiation (over some frequency band) is there at this point in space, regardless of what direction it's coming from?" -Amcbride (talk) 16:43, 9 January 2022 (UTC)[reply]

If the total power P remains constant while the radius of the sphere and thereby the surface area A shrinks to zero, the power per area, P/A (which is the radiant intensity), goes to infinity. --Lambiam 17:18, 9 January 2022 (UTC)[reply]

Wait, why would the power remain constant as the radius of the sphere shrinks? Maybe this is the source of my confusion. -Amcbride (talk) 17:40, 9 January 2022 (UTC)[reply]

If it does not, then we need to understand how the power is related to the size of the sphere. I see your spherical antenna is that of a receiver, not a transmitter, so then it seems reasonable to assume the amount of energy transferred from the field to the antenna is actually proportional to its area. That means we're back to radiant intensity. --Lambiam 23:47, 9 January 2022 (UTC)[reply]

That's what I've been assuming too, that the energy transferred would be proportional to the sphere's area. That's why it strikes me as a reasonable idea to divide total power by area to get a quantity characteristic of the sphere's location. The result would seem to be a quantity with the units of intensity (power per area) but unlike intensity it would be a scalar, not a vector. By contrast, radiant intensity has units of power per solid angle. -Amcbride (talk) 16:19, 10 January 2022 (UTC)[reply]

It is not obviously useful to hide the difference between a Scalar field (examples: temperature or pressure distributions) and a Vector field (examples: magnetic or gravitational forces, or radiant power) by specifying a new unit as the power impacting some arbitrary "standard" size of sphere. Philvoids (talk) 21:29, 10 January 2022 (UTC)[reply]

Irradiance is already watts per square meter. I suppose one could say that an antenna with a 1m^2 aperture is therefore in some sense (not a very noteworthy sense) the "standard" for measuring irradiance. A sphere with area 1m^2 would be, in just the same sense, a "standard size" for this summed-intensity-from-all-directions that I'm trying to talk about. So it doesn't seem to me that the idea of introducing a standard area is what makes the difference between the vector and scalar quantities. It's just whether you're summing over all directions or not. Let me put it this way. Say you've got one lamp to the north of a point, and one lamp to the west of a point, and no other light sources. From the perspective of this point, say the intensity coming from the northern lamp is three W/m^2 pointed southward, and the intensity coming from the western lamp is three W/m^2 pointed eastward. All I want is to talk about the quantity that we have *six* of at that point. Its units would have to be W/m^2, but it's not intensity because it ignores direction. Perhaps it doesn't have a name because no one in their right mind would ever refer to such a thing. -Amcbride (talk) 22:54, 10 January 2022 (UTC)[reply]

In case anyone is interested, the answer to my question turns out to be "scalar irradiance." There doesn't seem to be anything about it on Wikipedia, but here are some example links: [3], [4], [5] -Amcbride (talk) 04:01, 11 January 2022 (UTC)[reply]