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A “strange” unit radio astronomy
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I'm reading up on radio astronomy, and I came across this paper from 1964. At the bottom of page 193, the author uses a unit that I've not seen before in discussing radio power emission from stars:
Now the outbursts on the Sun give an intensity on Earth of $10^{19}$ to $10^{20}$ $wm^{-2}(c/s)^{-1}$
I'm guessing it's "Watts per square meter per something per second", but I'm not sure what the something is.
A similar unit appears in this paper on the first line on page 364:
The comparison band in the radiometer, being separated approximately 3.25 Mc from the signal band, never encounters the hydrogen range of frequencies.
Again, this looks to me like megasomething. Can anyone shed some light on this?
On page 362 of the second paper, the unit appears as $(Watts/M^2
)/(C/S)$ as a unit of flux. There, the $C$ looks like coulombs, but that makes the $3.25 Mc$ in the second quote seem weird.
radio-astronomy units
asked 4 hours ago
Jim421616Jim421616
569211
$endgroup$
add a comment |
$begingroup$
I'm reading up on radio astronomy, and I came across this paper from 1964. At the bottom of page 193, the author uses a unit that I've not seen before in discussing radio power emission from stars:
Now the outbursts on the Sun give an intensity on Earth of $10^{19}$ to $10^{20}$ $wm^{-2}(c/s)^{-1}$
I'm guessing it's "Watts per square meter per something per second", but I'm not sure what the something is.
A similar unit appears in this paper on the first line on page 364:
The comparison band in the radiometer, being separated approximately 3.25 Mc from the signal band, never encounters the hydrogen range of frequencies.
Again, this looks to me like megasomething. Can anyone shed some light on this?
On page 362 of the second paper, the unit appears as $(Watts/M^2
)/(C/S)$ as a unit of flux. There, the $C$ looks like coulombs, but that makes the $3.25 Mc$ in the second quote seem weird.
radio-astronomy units
asked 4 hours ago
Jim421616Jim421616
569211
$endgroup$
add a comment |
$begingroup$
I'm reading up on radio astronomy, and I came across this paper from 1964. At the bottom of page 193, the author uses a unit that I've not seen before in discussing radio power emission from stars:
Now the outbursts on the Sun give an intensity on Earth of $10^{19}$ to $10^{20}$ $wm^{-2}(c/s)^{-1}$
I'm guessing it's "Watts per square meter per something per second", but I'm not sure what the something is.
A similar unit appears in this paper on the first line on page 364:
The comparison band in the radiometer, being separated approximately 3.25 Mc from the signal band, never encounters the hydrogen range of frequencies.
Again, this looks to me like megasomething. Can anyone shed some light on this?
On page 362 of the second paper, the unit appears as $(Watts/M^2
)/(C/S)$ as a unit of flux. There, the $C$ looks like coulombs, but that makes the $3.25 Mc$ in the second quote seem weird.
radio-astronomy units
asked 4 hours ago
Jim421616Jim421616
569211
$endgroup$
I'm reading up on radio astronomy, and I came across this paper from 1964. At the bottom of page 193, the author uses a unit that I've not seen before in discussing radio power emission from stars:
Now the outbursts on the Sun give an intensity on Earth of $10^{19}$ to $10^{20}$ $wm^{-2}(c/s)^{-1}$
I'm guessing it's "Watts per square meter per something per second", but I'm not sure what the something is.
A similar unit appears in this paper on the first line on page 364:
The comparison band in the radiometer, being separated approximately 3.25 Mc from the signal band, never encounters the hydrogen range of frequencies.
Again, this looks to me like megasomething. Can anyone shed some light on this?
On page 362 of the second paper, the unit appears as $(Watts/M^2
)/(C/S)$ as a unit of flux. There, the $C$ looks like coulombs, but that makes the $3.25 Mc$ in the second quote seem weird.
radio-astronomy units
radio-astronomy units
asked 4 hours ago
Jim421616Jim421616
569211
asked 4 hours ago
Jim421616Jim421616
569211
edited 4 hours ago
Jim421616
asked 4 hours ago
Jim421616Jim421616
569211
asked 4 hours ago
Jim421616Jim421616
569211
asked 4 hours ago
Jim421616Jim421616
569211
569211
add a comment |
add a comment |
1 Answer
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I would expect the authors to be talking about the signal in terms of janskys, the now-commonly-used units of flux density. The typical definition is
$$1text{ Jansky}=10^{-26}text{ Watts meters}^{-2}text{ Hertz}^{-1}$$
One Hertz is one cycle per second, which makes me suspect that the "c" stands for cycle. It does seem curious that the authors choose to use cycles/second instead of Hertz, but that could be simply a convention of the time.
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
$endgroup$
1
$begingroup$
The fact that they are from older papers makes me agree with you, that it's an old convention. Jansky is consistent with flux. Thanks!
$endgroup$
– Jim421616
4 hours ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I would expect the authors to be talking about the signal in terms of janskys, the now-commonly-used units of flux density. The typical definition is
$$1text{ Jansky}=10^{-26}text{ Watts meters}^{-2}text{ Hertz}^{-1}$$
One Hertz is one cycle per second, which makes me suspect that the "c" stands for cycle. It does seem curious that the authors choose to use cycles/second instead of Hertz, but that could be simply a convention of the time.
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
$endgroup$
1
$begingroup$
The fact that they are from older papers makes me agree with you, that it's an old convention. Jansky is consistent with flux. Thanks!
$endgroup$
– Jim421616
4 hours ago
add a comment |
$begingroup$
I would expect the authors to be talking about the signal in terms of janskys, the now-commonly-used units of flux density. The typical definition is
$$1text{ Jansky}=10^{-26}text{ Watts meters}^{-2}text{ Hertz}^{-1}$$
One Hertz is one cycle per second, which makes me suspect that the "c" stands for cycle. It does seem curious that the authors choose to use cycles/second instead of Hertz, but that could be simply a convention of the time.
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
$endgroup$
1
$begingroup$
The fact that they are from older papers makes me agree with you, that it's an old convention. Jansky is consistent with flux. Thanks!
$endgroup$
– Jim421616
4 hours ago
add a comment |
$begingroup$
I would expect the authors to be talking about the signal in terms of janskys, the now-commonly-used units of flux density. The typical definition is
$$1text{ Jansky}=10^{-26}text{ Watts meters}^{-2}text{ Hertz}^{-1}$$
One Hertz is one cycle per second, which makes me suspect that the "c" stands for cycle. It does seem curious that the authors choose to use cycles/second instead of Hertz, but that could be simply a convention of the time.
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
$endgroup$
I would expect the authors to be talking about the signal in terms of janskys, the now-commonly-used units of flux density. The typical definition is
$$1text{ Jansky}=10^{-26}text{ Watts meters}^{-2}text{ Hertz}^{-1}$$
One Hertz is one cycle per second, which makes me suspect that the "c" stands for cycle. It does seem curious that the authors choose to use cycles/second instead of Hertz, but that could be simply a convention of the time.
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
answered 4 hours ago
HDE 226868♦HDE 226868
20k265125
20k265125
1
$begingroup$
The fact that they are from older papers makes me agree with you, that it's an old convention. Jansky is consistent with flux. Thanks!
$endgroup$
– Jim421616
4 hours ago
add a comment |
1
$begingroup$
The fact that they are from older papers makes me agree with you, that it's an old convention. Jansky is consistent with flux. Thanks!
$endgroup$
– Jim421616
4 hours ago
1
1
$begingroup$
The fact that they are from older papers makes me agree with you, that it's an old convention. Jansky is consistent with flux. Thanks!
$endgroup$
– Jim421616
4 hours ago
$begingroup$
The fact that they are from older papers makes me agree with you, that it's an old convention. Jansky is consistent with flux. Thanks!
$endgroup$
– Jim421616
4 hours ago
add a comment |
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