To answer your question: The crux of the talking point or “theory” does center around the energy released by nuclear fission, and not necessarily the “heat” caused by that activity.
The heat is simply the moving of molecules that have been accelerated by the kinetic energy obtained from radiation and fast moving particles that are ejected from the nucleus of an atom. Those fast moving particles approach the speed of light, as your link to the “blue light” points out.
It is the PROCESS and the AFTERMATH of that process that I am trying to explain.

We could compare the fission of atoms to triggering a shotgun, the shot flies off at high speed, although only reaching a small fraction of the speed of light.
We could use the chemical energy and the flying shot from many shotguns to heat a room that could heat water that in turn would turn turbines to generate electricity. We could make the room walls heavy enough to withstand the heat and the damage that the flying shot would cause. If we did generate energy in such a way it would be far more benign than the nuclear process. The speeding shot from the shotgun will only cause the molecules of the shot and the walls, to increase their activity, once such extra thermal energy is radiated away the molecules return to their original energy level. Such is not the case in the nuclear process, the produced radioactivity “heats” the environment for thousands of years.
I am curious about your statement of: “Even if they all simultaneously had a meltdown, it wouldn’t raise global, let alone local temperatures a lick.” Did you actually figure out the amount of energy that would be released in such an event and how long the “produced” radioactive atoms would be “heating” the atmosphere?

The information, from the link that you gave, is that there are 436 active nuclear power plants, producing 370,494,000,000 watt hours of electricity, remember that this quantity is only a small fraction of the total energy that is being created, even if this quantity of energy was created by burning fossil fuel, it still would represent only about 30% of the total energy that was being produced, the rest is wasted heat. When fossil fuel is used to produce electricity, the fuel is depleted and the resulting heat is in the form of molecular activity, the molecular activity and atomic excitement is raised to the level where infrared energy is released. When the fuel supply terminates, the infrared and molecular excitement also cease within a relatively short time.
The nuclear reaction is different, once you set that activity into motion it does not terminate for many thousands of years, and during that time it “heats” the surroundings.

You can get very detailed information about nuclear reactors from this site.

http://www.nr.titech.ac.jp/coe21/eng/events/NuclReactorTheoryTextbook.pdf

The rest of my explanations dwell on the extrapolation of what will happen IF the Earth’s atmosphere is “heated” for a long time with high energy nuclear activity.
Basically the changes would consist of the following:

If the atmosphere gets hotter it expands. (obvious physics)
If the upper atmosphere gets very hot, the molecules will escape the Earth’s gravitational field. (thermosphere area)
If atmospheric molecules escape into space, then the atmospheric pressure will decrease.
If the atmospheric pressure decreases, then there will be a variety of changes in the global environment.
The severity of the changes will depend on the severity of the atmospheric loss.

If so, I think you’re underestimating two things.

One, the amount of nuclear reactors on the planet. There’s actually not that many in existence, in this day and age or any other (currently, there’s around 436 active reactors worldwide). Even if they all simultaneously had a meltdown, it wouldn’t raise global, let alone local temperatures a lick.

Which brings me to point two: insulation ability. Thermal energy is basically moving matter. The faster it’s moving around on an atomic scale, the hotter it is (except for “wind chill” which still makes no sense to me, but hey). Most materials in general are generally good at preventing heat from transferring from nearby sources; thermal conductors aren’t that common in nature last time I checked. The stuff they layer reactors with? Lead and concrete, pure and simple. It’s dense, not very thermally-conductive, and the more you pile on, the less likely anything’s going to affect you (If it wasn’t that effective, reactor workers would be dead from radiation before they knew it). Notwithstanding they vast amounts of water they use to cool the reactors, the blanket of air that covers the planet probably also makes a pretty good thermal insulator.

And if you are feeling slightly warmer around a reactor, chances are you’re probably seeing blue light too, in which case, you’re pretty much screwed.

I could be wrong here in my thoughts on what you’re saying here though. If I misinterpreted anything, let me know.

Pressure = Force / Area, so I suppose that, Pressure * Area = Force
In my example the area was one square mile measured in sq. in. and pressure was 14.7 lb. per sq. in; the difference or force was 29,510,250,000 lb. (per sq. in.)
Or did the sq. in. disappear, and if so what happened to them.

How come my 4000 lb car can be supported by 30 lb. per sq. inch of air pressure.

Want to go over and do politics for a while? I’m tired.

Since normal air pressure is 14.5 pounds per square inch, that means that a column of atmosphere that is one inch square on the surface of the earth and extends to outer space weighs 14.5 pounds. The density of granite is
approximately 0.1 pounds per square cubic inch, so a column of rock that is one inch square would have to be 145 inches tall to weigh 14.5 pounds, or 12 feet tall. So to answer your question, in a pressure differential situation, where the weight of the atmosphere on one inch were halved, you’d need an additional six feet of rock to offset that change.

A very shallow quake might be on the order of a mile down. Most are much deeper. A column of rock that is one inch square and a mile tall would weigh 6336 pounds. In other words, in order to affect an effective vertical pressure on a fault at that depth, the air pressure would have to be 3 tons per square inch. I’ll leave it to you to figure out how thick the atmosphere would have to be to weigh the same as 3 tons.

In structural geology calculations, where we derive the differential stress on a given plane, the weight of the atmosphere is so small it is thrown out of the calculations.

Suppose the seismic activity happens in a specific area of one square mile.
One square mile contains 4.015*10^9 square inches, there is about 14.7 pounds of air pressure per square inch on that surface area.
The total area then is under the pressure of 14.7 * 4.015*10^9 which equals 59,020,500,000 lb.
Suppose that there is a gas bubble underground with similar counter pressure, now suppose that the surface pressure is reduced by 1/2, then the total pressure difference is 29,510,250,000 lb.
How strong or thick must rock be on top of the gas bubble to hold it so that nothing moves between the pressure differences?

Years ago now, after many previous years of asking you to do the statistics, I did them for you. I clearly showed with data availiable to you from the USGS that there was no measurable difference in worldwide seismicity. I even graphed it up for you so that you’d see it with your own eyes.

All to no avail. I don’t feel inclined to repeat the exercise in futility

Please understand, J, that atmospheric forces are so small compared to tectonic forces that they have no influnce on earthquakes. None. You could strip this planet of its atmosphere and the forces of plate tectonics would not even notice the difference. It’s a dead end, mate, give it up.

From the following web site:

http://www2.seismosoc.org/FMPro?-db=Abstract_Submission_12&-sortfield=PresDay&-sortorder=ascending&-sortfield=Special+Session+Name+Calc&-sortorder=ascending&-sortfield=PresTimeSort&-sortorder=ascending&-op=gt&PresStatus=0&-lop=and&-token.1=ShowSession&-token.2=ShowHeading&-recid=224&-format=/meetings/2012/abstracts/sessionabstractdetail.html&-lay=MtgList&-find

They say that:

“A naturally-occurring rate change of this magnitude is unprecedented outside of volcanic settings or in the absence of a main shock, of which there were neither in this region.”

(paragraph tags added by moderator to clean up the subject line)