First off, you are treating this as a Black Body radiation problem. When dealing with gas or plasma, radiative power is as much a function of density as surface area. Think about a star. The Chromosphere and Corona are much hotter than the Photosphere, but the radiative power of these outer layers of a star is negligible because the plasma is so tenuous.
(One of the subtleties of bomb pumped X-Ray lasers is that the lasing rods have to remain intimately connected to the actual bomb so that after the lasing media is pumped by the absorption of gamma rays from the explosion, the shockwave that propagates along the length of the still dense plasma that had once been the lasing rod compresses the plasma to stimulate lasing in the X-Ray spectrum.)
Secondly, you are ignoring the fact that the vast majority of the explosive energy is in the form of kinetic energy of the bomb residue. As the plasma cloud expands, the velocity of the particles becomes very uniform. In the absence of atmosphere for the residue to collide with, the effective, radiative temperature is an order of magnitude lower than you assume.
Thirdly, radiation in the visible spectrum requires electrons to have actually been recaptured by nuclei so that they can undergo the transitions in energy state that correspond to emissions in the visible portion of spectrum. By the time the plasma cloud has cooled sufficiently for electrons to be recaptured, it has expanded to such an extent that the rate of electron recapture becomes very low. Think of how we can see the gas cloud remnants from supernovae thousands of years after the event.
The bottom line is that the nuclear explosions from a space battle are going to be very visible, but only if you have an X-Ray or Gamma Ray telescope.
SWM wrote:Now for a closer look at a nuclear explosion. I've done some more reading. The tricky part is extracting the atmospheric effects from the descriptions.
When the bomb first explodes, the first thing that comes out is a burst of gamma rays, followed shortly by x-rays. This, of course, is what laserheads take advantage of to produce their x-ray lasers. The x-rays uniformly heat up the residue of the bomb.
In the first microsecond, the temperature of the residue is around 60 million degrees C or more. The residue radiates a nice blackbody spectrum, of course. This residue expands very rapidly. In an atmosphere, this would turn into blast and thermal effects, but not in space. 80% of the energy yield initially comes out in photons, and without an atmosphere, it is not converted to other effects.
Since there is no atmosphere, there is no fireball. But the residue radiates. As it expands, it cools rapidly. At a few microseconds into the explosion, the explosion is a few meters across and a temperature of tens of millions of degrees. By the time the explosion is 13 meters across, the temperature is down to 300,000 degrees. In an atmosphere, this would happen 100 microseconds into the explosion, but it will happen much faster without an atmosphere. Since the thermal radiation goes as the fourth power of the temperature, the total power being emitted has dropped by a factor of more than a thousand.
It is easy to see that the great majority of the energy is emitted in the first 100 microseconds. By 1 millisecond, the power emitted is a tiny fraction of what was emitted initially. It is fair to say that the flash occurs in the first millisecond or less. In an atmosphere, the radiated power is spread out over a much longer period, because it is absorbed and reradiated by the atmosphere, and much of the radiation comes from the hot shock wave. So I would agree that in an atmosphere, a flash of 1 second is reasonable. But without an atmosphere, even 1 millisecond is probably too long.
If we use a time of 1 millisecond and say that 80% of the yield is in a black-body spectrum, and assuming a 1 Megaton bomb, we get an effective power of 3.4e18 W.
Nameless is correct that much of this will be in non-visible wavelengths. It is a bit hard to guesstimate how much of the power would be in the visible range, especially with the temperature dropping so precipitously in the first millisecond. So let us assume that only 0.1% of the energy is emitted in the visible spectrum. Thus, the power in the visible spectrum is 3.4e15 W.
The luminosity of the sun is 3.8e26 W, and an absolute magnitude of 5. Thus, the sun is about 1.1e11 times brighter than the bomb. So we go through the calculations I demonstrated before. A factor of 10^11 is a difference of 28 magnitudes. So the bomb has an absolute magnitude of 33. Absolute magnitude is the apparent magnitude at 10 parsecs. The explosions in HoTQ took place at 150 million km. 10 parsecs is 3.1e14 km, or 2.0e6 times further, so the explosions in HoTQ will be 4.0e12 times brighter than magnitude 33. 4.0e12 is more than 31 magnitudes. That makes the explosions magnitude 2.
The explosions are quite easily visible on Grayson, even if we assume only a 1 Megaton bomb, and only 1/1000 of the radiated energy is in the visible spectrum. A 15 megaton bomb would be about 3 magnitudes brighter, or brighter than Sirius.
