The Sun lives!

Our Sun, having spent the last year and a half in the doldrums, is
finally becoming slightly active. This heralds the arrival of the
much anticipated solar cycle 24.

In evidence thereof, I present a strip chart of a solar radio emission
event that I recorded last Wednesday.
http://jovearchive.gsfc.nasa.gov/rjd...0120193700.png

To find more of this stuff:
http://jovearchive.gsfc.nasa.gov/cgi...r/calendar.cgi

I know, I know... everyone wants to know how my antenna can be at 24
million K. Kinda warmish for copper, right?

To explain it, we're going to have to actually do some mathematics. If
math stimulates the glands that produce your eye-glaze juice, perhaps
you better skip the remainder of this message. If, on the other hand,
you want to learn more about how the universe works, read on.

In radio astronomy, there are two common ways of communicating signal
strength.

One way is by stating it in terms of flux density, or watts per square
meter per hertz, with the understanding that there exists a given
center frequency and a given bandwidth. The common unit is the jansky
(Jy), which is 10^-26 W/m^2/Hz.

The other way is to state it in terms of antenna temperature in
kelvin. This is the temperature that an imaginary black body
surrounding the antenna would have to possess in order to create the
signal observed in the antenna. This, like the flux calculation,
assumes a given center frequency and a given bandwidth.

What can we say about a solar radio burst producing a 24 MK
temperature in my antenna?

For a start, we can convert that to power.

P = Boltzmann's constant * Temperature * Receiver bandwidth

In my case

P = (1.38e-23 W*s/K)(2.4e7 K)(6 kHz)
P = 2e-12 W

What does 2 picowatts mean? That's the power at the antenna's feed
point.
But so what?

Well, if we know something about the antenna, we can now say something
about the signal's flux density, also known as S.

S = P / (Antenna effective area * receiver bandwidth)

But wait, "effective area"? What?

The effective area of an antenna is:

A = G (lambda^2) / (4 pi)

Great, so what's G? G is the gain of the antenna in terms of it's
ability to emit (and by symmetry collect) RF power better than an
isotropic radiator at the expense of angular coverage. An isotropic
radiator has a gain of 0.0 dBi. A Yagi antenna might have about 14
dBi at its design frequency in the direction of maximum gain. The
70-meter dishes of the Deep Space Network have a gain of about 80 dBi
in the Ka band (31 to 38 GHz).

In my case, I'm using a simple two-dipole phased array designed for
20.1 MHz. This arrangement has a maximum gain at the center of the
beam of just over 8 dBi. The arrangement has a -3 dB (half power)
beam width of about 100° in azimuth and 53° in altitude. The beam
center is fixed at 183° azimuth and 51° in altitude.

The Sun at the time of observation was at 214° azimuth and 32°
altitude, putting it at roughly the -2 dB area of the beam. As such,
for this observation, we can claim the antenna has a gain of (8 - 2) =
6 dBi.

Before we can use that, we have to convert it from decibels to a power
ratio. Decibels are just a convenient way to express the ratio of two
power measurements.

Since decibels = 10 * log ( P / P_0 ), we can say that gain G is P /
P_0
where

G = 10^(decibels/10)

or in my case

G = 10^(6/10) = ~ 4.

We can now find the antenna's effective area:

A = G (lambda^2) / (4 pi)

where lambda for 20.1 MHz = 15 meters

A = 4 * (15^2) / (4 pi) = 72 m^2

We can now solve for flux density.

S = P / (Antenna effective area * receiver bandwidth)

S = 2e-12 W / (72 m^2 * 6 kHz)

S = 4.6e-18 W/m^2/Hz = 460 million Jy.

It is also apparent from the fact that we used bandwidth twice that we
can get rid of it if we go straight to flux density from antenna
temperature--but then we would have lost the opportunity to discuss
power.

S = P / (area * BW)

P = Boltzmann * temp * BW

so,

S = ( Boltzmann * temperature ) / antenna area

S = 1.38e-23 W*s/K * 2.4e7 K / 72 m^2

S = 4.6e-18 W*s/m^2 = 4.6e-18 W/m^2/Hz = 460 MJy

Now, knowing the signal's flux density at my antenna, what can we do?

Well, if we *assume* the source radiated this energy
isotropically--i.e., in all directions evenly--then we can find out
how much power it was producing. This may *not* be a safe assumption
for astronomical sources; the RF energy is often emitted in a beam by
electrons interacting with magnetic fields--e.g., cyclotron and
synchrotron radiation. Nevertheless, for sake of argument, lets say
it was isotropic--because that's easier and I have to draw the line
somewhere.

The Earth--and hence my antenna--is 1 AU from the Sun. A sphere of
radius 1 AU has an area of 4piAU^2, or about 2.8e23 m^2.

My antenna saw a flux density of 4.6e-18 W/m^2/Hz. If we assume the
source on the Sun illuminated this whole 1 AU radius sphere evenly,
then the specific power is:
4.6e-18 W/m^2/Hz * 2.8e23 m^2 = 1.3 MW/Hz

My chart doesn't show it, but according to the NOAA Space Weather
Prediction Center, this particular emission event covered the
frequency range from 25 to 180 MHz.

Look for Event # 5910.
http://www.swpc.noaa.gov/ftpdir/indi...0120events.txt

However, the emission isn't flat across this 155 MHz bandwidth. This
was a Type III solar burst, which means it didn't emit all at once
over that 155 MHz bandwidth. These bursts start high, then slew
rapidly to lower frequencies at one or two dozen MHz per second.

At any one time--specifically, at the time the emission peak sweeps
past 20 MHz--the emission bandwidth is unknown. The available
spectrograms of the event just don't go to that level of detail.

For sake of argument, let's *assume* this burst was 10 MHz wide when I
recorded it.

To find the total power:
1.3 MW/Hz * 10 MHz = 13 Terawatts.

Now we need to make some corrections to that figure.

A dipole antenna a quarter wavelength off the ground, like mine, has a
radiation efficiency of only 0.9 or so. That gives us 13 TW / 0.9 =
14 TW

Then, add the fact that this was a mid-day observation, meaning the
ionosphere's F layer was thick as a Thule fog, which means it
reflected a lot of the incoming solar 20 MHz radiation back out into
space--in much the same way it would ahve reflected terrestrial
signals back to Earth. How much does that cut down the received
signal? Let's take a SWAG and say that the ionosphere reflected 80%
of the incoming signal.

To find the source power, we have
14 TW / 0.2 = 70 TW.

By comparison, the world's total installed electricity generating
capacity is only about 4 TW.
http://www.eia.doe.gov/pub/internati...06/table64.xls

This was a pretty strong burst as these things go. The strongest ones
are over ten times more powerful.

So much for the description. What causes these bursts to occur?

Best anyone knows, streams of fast (but not relativistic) electrons
are accelerated outward from a point low in the solar corona. As they
pass through the coronal plasma, they excite the plasma to radiate RF
energy. It is thought that the electrons set up Langmuir waves
(plasma oscillations) within the plasma, and that these oscillations
are somehow converted to RF energy--but, nobody's too sure just how
that happens--nor is anyone sure what generates the streams of
electrons in the first place.
--
Dave Typinski
AJ4CO Observatory