How much should a nuclear explosion be slowed for its energy to be safely converted to electrical energy?

Harvesting explosions by containing them in a time bubble.

Here's an interesting sci-fi concept, which involves generating a time bubble in order to slow a nuclear explosion to the point where the energy that leaks out of the bubble can be safely harvested. Over a sufficiently long period of time 100% of the explosion would be converted to electrical energy. A spherical chamber could encircle the time bubble to collect light, heat and radiation emitted by the explosion. In a certain way, this is a scaled down version of a Dyson sphere or a fusion reactor but we'll get to that later.

The time dilation would turn the light into infrared, which can be harvested using thermal energy conversion. Same system as nuclear reactors. BUT! I plan on using an H-bomb (hydrogen bomb), which fuses hydrogen into helium upon detonation, giving us extra energy compared to an ordinary atomic bomb. Like a nuclear/fusion reactor. All this energy being released instantly means I'll need to know how much time would need to be dilated to not overheat and destroy the thermal power plant surrounding the explosion.

By how much should time be slowed for this to be safe and efficient?

• I think the commercial implementation of this is called a nuclear reactor....
– L.Dutch
Feb 21 at 15:38
• For that matter, the output could be shifted in wavelength so how do you want to absorb it? Feb 21 at 17:27
• Won't the light be red-shifted to such an absurd degree that it will be useless for photovoltaics? And I have no idea how you harvest fast neutrons, or that they'd survive once time-dilated... they can only exist freely for about 20 minutes before they decay, is that slowed too? Feb 21 at 17:42
• And what is a “second” when you’re changing the length of a second? It’s a paradox that can only be answered after you define the physics of your time dilation. For example, say “Time dilation costs 50 MJ per 1% dilation” would give something to put into a calculator. You’ve given us nothing for an efficiency calculation as it is. Feb 21 at 19:47
• Interesting concept indeed, but I have to put one note about your H-bomb plan... "I plan on using an H-bomb (hydrogen bomb), which fuses hydrogen into helium" that will complicate your time warping considerably ! The H-bomb is detonated by a fission bomb inside. You'll have to allow that bomb to explode, before you'll have actual fusion. In some way, containment of that initial fission explosion is needed, else you loose your device.. en.wikipedia.org/wiki/Thermonuclear_weapon#Design_variations Feb 21 at 21:45

Step 1: Handwave away all those pesky things that science tells us

There are not very many ways in which one can even theoretically cause a significant amount of time-dilation, and those that do exist involve costs that would make your time reactor unbelievably more expensive to build and operate that you could get out of it... that said, many authors introduce cheap time dilation that they can just blame it on some unique property of Unobtanium or Clarke Tech and move on. The other thing to handwave away is 100% efficiency. 70-90% efficiency is far more realistic; so, to account for this you will either need to adjust your final figures or again go the Clarke Tech route for collecting all that energy.

The smallest hydrogen bomb ever tested is North Koria's 30kt hydrogen bomb. It is widely believed that this test was actually faked, making the smallest confirmed H-bomb the 100kt W76 warhead, but either way, you don't want to go with the smallest H-bomb if you have cheap time-dilation. You want to go with the biggest possible H-bomb. The reason for this is that the most expensive part of an H-bomb is actually the fission bomb that you use to get it started. So, assuming you get a positive energy return at all out of this invention, it means you get more bang for you buck going with a bigger bomb. So instead we will want to pick something more like the 50,000kt Tsar bomb which translates to a power output of about 210,000,000,000 MJ

Step 3: Pick your power output.

Since this is basically a nuclear reactor, you should look at the power output of a single reactor in a nuclear power plant. You should not be looking at the output of non-thermal reactors, or total output of a multi-reactor plant since this is going to be a single thermal reactor per bomb. The highest output per individual thermal reactor in the world are about about 1300 MW each. So to convert a Tzar Bomb's blast into an output of just 1300 MW, you would need time to slow down to an average of about 1/160000000th normal time giving you an explosion that would last for just a bit over 5 years. Note that I say average for a reason since, as Dan's answer points out, it will not be a uniform flow of energy you need to slow down.

It's not that simple.

A nuclear explosion is not a simple thing. It comes in phases. Those phases are, of course, separated by very small amounts of time.

One phase is the release of gamma rays. You can see the effect of this in the so-called "rope trick." The image is from the Tumbler-Snapper test, one of the above ground tests in Nevada. It shows the cables holding the shot cab being heated to vaporization temperature ahead of the shock wave of the blast.

(Note: Thank you to Vikki for pointing out I had mis-identified the source of the image.)

The gammas come out in the first few micro-seconds. The shock wave requires milli-seconds to get as far as it has in the image. So the cables had time to begin their own little explosions before the main explosion shock front got there. You can even see the ground under the explosion begining to glow due to heating.

The doubling time of the rate of energy release in the explosion is something on the order of 1 microsecond. It depends on the design. The gammas will carry some few percent of the total energy of the explosion. And they will come out at the speed of light, nearly $$3\times10^8$$ m/s. So you get this pulse of gammas with a few percent of the energy coming out within a few micro-seconds.

So you will need to slow that part of the explosion by a huge factor or you will get nicely fried if you are anywhere near the explosion. If it was a 1 MT explosion, then probably 20 kT worth is in the gamma pulse. And you need to spread that roughly 5 microsecond pulse out to long enough that it does not fry you.

The next thing to come out will be the neutrons. They come out over many milli-seconds. This is because they come out with a wide range of energy, so a wide range of speed. The high end ones are doing about 10 MeV, say something like 0.005 the speed of light. The slowest can be arbitrarily slow. This will be followed by a soup of other radiation, alphas, betas, protons, etc. This pulse will also be some few percent of the total energy. And will come out in a liesurely few milliseconds. And will continue to be released at diminishing rates anywhere there are fission fragments.

Following will come the fission fragments, and the heated up non-reacting components of the bomb. These will be coming at speeds round about a few 100 km/s. This last relatively slower pulse contains something over 90% of the total energy.

So your tuning of the time-slower device needs to be extremely fine, and have at least three settings. Otherwise you wind up either getting totally cooked by one of the pulses, or waiting months for the useful energy.

• As an aside, the time of a neutron generation (the shake of a lamb's tail) is roughly 10 nanoseconds, the 'nuclear' part of the primary is over in about half a microsecond. See, for example, B. Cameron Reed's article on the Serber plot (American Journal of Physics 88, 565 (2020), doi: 10.1119/10.0001206). Feb 22 at 14:55
• It gets even worse if you're using an H-bomb like OP suggests, since that involves (if I'm not mistaken) first setting off a fission reaction, with all the stages you describe, in order to get to the more energetic fusion reaction that happens in yet another later stage. Feb 22 at 17:01
• Neutron generation time depends on bomb design. I know much less about fission-fusion reactions, but it would probably add at least one more pulse, possibly two. That also would depend on the design. A tritium-deuturium burn, for example, will produce a whack-load of 14 MeV neutrons that might even out-pace the ones from the fission reaction.
– Dan
Feb 22 at 20:37
• That image isn't from Trinity, it's from one of the Operation Snapper tower shots in 1952. Feb 22 at 21:47

Over a sufficiently long period of time 100% of the explosion would be converted to electrical energy

About 5% of the energy will be lost as neutrinos. There's no realistic way to recapture this energy... doing so might even be less realistic than magical time-bubble reactors.

You're using a thermal power plant, and that means you're limited by the efficiency of the Carnot cycle: the theoretical maximum possible efficiency is $$\eta = 1 - {T_C \over T_H}$$ where $$T_C$$ is the temperature of the heat sink, and $$T_H$$ is the temperature of the heat source. If the temperature of the hot end was 1000K and the cold end was ambient, you get about 72% efficiency. If you can run your heat engine at 3000K (and no, you probably can't) you reach over 90% theoretical maximum efficiency, and much above that you'll find that everything starts melting.

Anyway, you can't reclaim 100% of that energy.

Here's the real problem though:

The time dilation would turn the light into infrared

A gamma ray photon emitted from a nuclear reaction has a lot of energy. An infrared photon does not. Where's that energy gone? Well, energy isn't necessarily conserved under general relativity, but momentum certainly appears to be, and photons have momentum too. So where'd it go?

Either:

1. All the gamma ray photons get redshifted into more tractable wavelengths, but the energy and momentum associated with them has poofed away into the ether. This results in a colossal loss of energy... those 1 MeV gamma photons have now become 1 eV near-IR photons, so that's a millionfold loss.
2. The gamma ray photons are still gamma ray photons when they leave the stasis field... maybe the power of the system is lower than a nuclear bomb because the deadly radiation is trickling out over a longer timescale, but the stuff shooting out of the system is definitely not infrared
3. The interior of the stasis bubble is optically thick, such that you can reasonably expect all short wavelength light from the nuclear reactions to be absorbed by whatever is in the bubble, and as the resulting plasma cools and recombines it emits nice tractable black-body radiation

If you want to extract useful amounts of energy from your bomb, it can't be (1). No redshifting for you.

(3) is also awkward, because the interior of the bubble needs to be large enough to allow the fireball to develop and expand without breaching the walls of the bubble, because due to the no-redshift thing you'll end up with an actual nuclear fireball impinging on your heat engine, and that sort of thing is what engineers usually refer to as A Bad Thing. How big it would have to be I don't know, but you're going to want it to be at least a couple of hundred metres across.

So what you've ended up with is not a magical nuke-driven heat lamp, but instead a giant incandescent radioactive fireball, which you've hopefully slowed down enough to capture all the heat and light from without accidentally nuking yourself. Unlike a regular nuclear reactor, the fallout won't be nicely contained in fuel rods either... lots of free gas and dust instead. Fun to confine and clean up.

I don't want to think about the problem of the bubble of hot, high-pressure gas wanting to rise up... maybe you can fix this by having a tile-dilation bubble shaped like an egg standing on its pointier end, but suddenly your reactor chamber is hundreds of metres across and a kilometre high and you have to wonder whether old fashioned fission reactors were really that bad after all.

Maybe it is easier just to have a merely large chamber, lined with something highly refractory (like tungsten) that should be able to catch and thermalize the short-wavelength photons and high-speed particle radiation without melting itself. Then you just have the problem of cooling it fast enough that the interior layer doesn't vaporize, because repairs will be awkward.

Now it is just a question of how effectively you can cool your blast chamber. Due to Carnot efficiencies, the quoted power of real world thermal plans is actually a lot lower than the actual amount of heat coming out of the source. The world's biggest nuclear plants today are hitting about 8GW of electric output, so they can presumably soak up more like 11-12GW of heat.

Releasing 2.5-3 tonnes-of-tnt-equivalent worth of energy over a second generates about that much power. A Tsar Bomba at its maximum yield of 100 megatonnes would run such a plant for a little under 404 days, though you might find the multi-kilometre-wide fireball a bit awkward for option (3).

If your heat-extraction capacity is lower, your time-dilator needs to be better, or you get nuked. If your time-dilator isn't free to operate, you're going to face diminishing returns as the size of the power plants decrease... your system seems like it might scale up, but it won't do a good job of scaling down.

• hehe "About 5% of the energy will be lost as neutrinos. There's no realistic way to recapture this energy... doing so might even be less realistic than magical time-bubble reactors." well magic can do anything.. and recently they did measure the mass of these little buggers.. 0.8 eV/c2. You won't get much out of that (despite the speed) but maybe it can be done, some day! nature.com/articles/s41567-021-01463-1 Feb 21 at 21:28
• @Goodies: The problem with neutrinos is that they barely interact with anything physical. An enormous quantity of them are streaming out of the sun at all times, and of those that reach the Earth, the vast majority pass all the way through the planet. If your magic is powerful enough to pull energy from neutrinos, then you can probably find a better power source than a slowed-down nuclear bomb. Feb 22 at 8:19
• You can imagine that stasis field generator starts producing energy instead of consuming it, much like the recuperative electric motor, when it slows down gamma rays to IR. This will also enable reclaiming of most of neutrino power - they'll just come out much slower (still near-C). Feb 22 at 8:39
• @MichaelSeifert if I had a cent for every time someone pointed that out I'd have... about 6 cents, maybe? Clearly there are too few nuke questions on this place. Feb 22 at 19:07
• "A Tsar Bomba would run such a plant for nearly 6 minutes" <- you may want to double check your math on this one. A 50 megaton blast is ~210,000,000 GJ. Even at 12 GW, this is ~203 days. Feb 22 at 20:40

This is a great sci-fi idea because almost any solution for getting electricity would be better - but few would be cooler

Having said that, this question falls into my "how many angels can dance on the head of a pin?" rule. Answer: "as many as wanting." You're seeking (I assume) a factual answer to a fictional scenario. One doesn't exist without making whomping assumptions. To wit:

• I'm using solar panels using today's reasonable-best efficiency standard of converting 20% of light to electricity for 2kwh per-panel in a 10-hour daylight period, suggesting I could get 4.8 kwh if the panel were exposed to continuous light and each panel is one square meter in size.

• I'm going to assume that 100% of the energy released by the nuclear blast is photonic in the light spectrum acceptable to the solar panel. Note: This is an assumption so wild, so insane, so incredibly wrong that it caused angels to weep. But it really simplifies the analysis.

• I'm going to wrap my panel around the explosion in all three dimensions and assume no equipment gets in the way. This is also a cause-angels-to-weep assumption. In reality (hah) you would at least have the equipment generating the time dilation field that the panel could not cover. Energy from the explosion that penetrates the field in the direction of this equipment is lost (and could damage the equipment), but like I said, we're ignoring all that.

• The total solar panel inner surface is 1.4706 square meters. This will make more sense later. It has a lot to do with me being lazy.

• I'm going to assume that I'm going to slow down the bomb's detonation to ensure that no more energy impacts the solar panel than would sunlight.

• We want energy for a long time, so we're going to detonate the Tsar Bomba — about 225 peta-joules.

• I'm ignoring entirely the fact that, insofar as we understand it, as one increases the need to slow down time, one also increases the energy required to do it.1 In other words, this can't work based on our understanding of time and physics. But it's your world. Throw us all the proverbial vulgar hand gesture and do it anyway.

All those wild-and-crazy assumptions make the calculation fairly straightforward.

One watt = 1 joule per second. And from here we learn that an explosion requires give-or take .0000008 seconds. If you read that response on Quora, it'll point out that this is a fairly arbitrary statement as the actual number will depend on a boat-load of variables that we're completely ignoring.

So, I'm looking to get (for completely arbitrary purposes that have to do with me being lazy), 48 kwh (10x4.8kwh) over a 24-hour period. From here we learn that watts = (kWh × 1,000) ÷ hrs. So, 48 * 1000 / 24 = 2000 watts. Watts are joules-per-second, so to get 200 watts we need to accommodate efficiency (20%) getting us to needing 5,000 joules to impact the surface of the panel every second.

Now, NASA says Earth gets 1,360 watts/square meter from the Sun (now you know why the inner surface of the solar panel 1.4706 square meters. 10X(200/1360)=1.4706).

So, all that's left is slowing down the detonation. This is going to be so unrealistic that's it's a good thing the idea is uber cool.

Basic scalar math:

$$\frac{225 Pj}{.0000008 S} = \frac{5000 j}{X}$$

Which, when you do all the acrobatics tells you that you need the following time dilation ratio:

1:56.25x1018

That's not a little slow-down. That's a lotta slow-down. If you're a fan of metric prefixes, that's almost2 0.2 zepto-seconds of explosion for every second of unadjusted time. The ratio gets better as you increase the efficiency, capability, and surface area of the solar panels... but not by what I'd consider a practical amount.

And remember all those weeping angels. We made some outrageous assumptions to get to a number that you don't really need because are you really going to start playing around with the size, type, quality, and efficiency of the nuclear device, the size, type, quality, and efficiency of the solar panels (aka energy converters or accumulators), the distance between them, the size, placement, and occlusion of your time dilation generator, etc., etc., etc. This is why many scifi authors don't explain the gory details. There's always some group of geeks who will check every number just to see if you really know your science or not.

Just tell your readers that a time dilation field is used to slow the detonation of a nuclear bomb so that accumulators can convert the detonation energy to electricity. If Star Trek can make gobs of money powering Romulan warbirds with a contained "quantum singularity" (aka, a small black hole), then you're good to go!

1A fun-to-watch explanation for this comes from Doctor Who's "The Pirate Planet" where the Doctor explains why the time dams keeping Xanxia's body alive can't work. Tom Baker. Doctor Who. What more do you need?

2As if the minor rounding error would matter here.

• Nice answer, thank you! Rest assured, it's not just the angels who are weeping at the idea. Feb 22 at 7:25
• I don't even think the Warbirds' singularity engines are the most egregious dodge. I reserve that for the Heisenberg Compensator. A physics journal once asked a Star Trek writer how the Heisenberg Compensator worked. "Very well, thank you! Next question!" Feb 22 at 23:59
• " 'how many angels can dance on the head of a pin?'... Answer: 'as many as wanting.' " Not so. Sandberg (2001), "Quantum Gravity Treatment of the Angel Density Problem", establishes an upper bound of $8.6766 \times 10^{49} \text{angels}/(4/3\; \pi\; \text{angstrom}^3)$. (This assumes electrically neutral angels, but he argues that charged angels would exceed the material tolerances of the pin at far smaller numbers)
– Ray
Feb 24 at 22:18
• @Ray I unofficially but enthusiastically declare Anders Sandberg an honorary 100,000+ rep ranking on Worldbuilding and gladly bow to such audacious mathematics and analysis. L'audace, l'audace, toujours l'audace!
– JBH
Feb 25 at 4:09

There is a similar real-world plan that I have read years ago, so the idea is more or less viable:

1. Make an underground nuclear explosion, deep enough to be contained.
2. Use the trapped heat underground heat just like a natural geothermal source.

Rinse (not really), repeat, fix the thermal wells when needed.

• a company in Germany patented a design in the 70s where they heat water to supercritical temperatures using a high-yield nuclear bomb and then vent the supercritical water over time (because of decompression it would generate large amounts of water vapour). I think, tbh, that the idea of OP is more complicated than real designs already imagined Feb 23 at 12:23

1 "kiloton" is 1.16 gigawatt-hours.

As nuclear power stations designs were refined into the 1970s, economics favored a fairly narrow range of sizes - most newer plants are 1.0 to 1.5 gigawatts. So if we claim "the average semi-modern nuclear power station is 1.16 gigawatts", we're pretty darn close.

1 hour of their runtime is thus 1.16 gigawatt-hours. Or 1 kiloton.

"That was easy"

A 22 kiloton nuke = 22 hours of run-time of our "average" nuclear reactor.

A megaton nuke = 1000 hours of run-time.

A lot, but there can be no net energy gain in this scenario

As others have said, 100% energy capture is not physically possible with real physics. I'll skip the Carnot efficiency calculation as you're not asking for it and someone already provided it. No process gives 100% useful work. That's the easy part.

Next is the calculation of time dilation. That isn't particularly hard but to reach an actual number you have to know certain parameters going in. So I will give the formula then go through the work for you.

To safely extract energy from an H-bomb by time dilation, a time dilated second s$$_1$$ passes inside your time bubble in a number of real seconds according to the formula: $$s_1 = \frac{\textbf{H}_0}{\textbf{H}_U \textbf{H}_F(1-\textbf{M}_S)}s$$

Where:

• S$$_1$$ is the number of real-world seconds will pass every time one second passes inside the time bubble
• $$\textbf{H}_0$$ is the power density of the H-bomb at the surface of the time bubble
• $$\textbf{H}_F$$ is the power density that will cause your weakest part of the energy harvesting device to fail
• $$\textbf{H}_U$$ is the power density that will will be converted into useful work by your harvesting device (based on the device efficiency)
• $$\textbf{M}_S$$ is your design safety margin, or, the percentage of "slop" you want to allow your design so it will not fail

For reasons explained below your $$\textbf{H}_0$$ power density will have many components, but ultimately every form of energy coming from the detonation can be converted into Watts/m$$^2$$ and plugged into this formula. This is the number you need to find based on the yield of your bomb and the diameter of the sphere.

For the safety margin, assume you have some coupling in the collector that will be destroyed with exposure to 100,000W/m$$^2$$ over 5 seconds. If you want a 5% safety margin, the max power density you can allow to "safely" (per the question) operate this device would be 5kW/m$$^2$$. You would dilate time enough to make s$$_1$$ long enough that power doesn't exceed this number.

In the end, unless you are dilating time for free, you will not pull a net energy from this device, however you will make energy useful where it normally is not. Now the calculation:

You will be taking normally destructive energy and making it useful, but at a very low efficiency. Note also, that a thermonuclear device is an uncontrolled chain reaction with more energy yield as it grows larger. Someone calculated you will need a 200m diameter sphere, but the problem isn't really asking for a fixed yield so I will only touch on the why of this argument. If you bottle the bomb up into a small bubble, you will have an exponentially smaller yield compared to the uncontrolled bomb in open air. No matter, you didn't want max yield.

Disregard the thermonuclear bomb for now and just say you have a radiating body of any kind. In your case, it radiates electromagnetic waves, gamma rays, neutrons, protons, neutrinos, etc., etc. You have also introduced the novel physics that time dilation alters the wavelength of light when it is measured outside the bubble in normal time. This should not happen in your case of special relativity (we are talking about time dilation, after all) because it violates laws of conservation. An infrared photon has less energy than a red photon, for example. Where did the extra energy go?

Your question is very simple and the qualifier "safely" means we only need to find the most likely item to fail, and assume the machine fails as a whole soon afterward. That would be the safety margin. If I make this radiation come out of the bubble more slowly, smaller doses of the radiation will reach the parts of your machine per second of real-time. So you simply need to determine how many Watts per second will destroy your energy harvesting machinery. The destructive elements are many, but the calculation is very simple, and can only be given a number after knowing what your harvesting machine can withstand:

Electricity

If you have copper wiring anywhere, for example, you need to keep the temperature at least below the melting point of 1,984°F, but preferably much lower than that if you want your wires to be efficient at conducting electricity. You can have better conductors with gold or platinum, but you will also need to keep the temperature lower. You will heat up any copper parts from 25° to 1,984° ($$\Delta T=1,959$$°) by exposing it to your radiation. This reduces the efficiency of the electrical conductor. But how much can it take? That answer depends completely on a review of your engineering design.

We have to assume you want to harvest energy rather than destroy it. A particle leaving the blast will be moving through time more slowly but have the same energy. All you have done is change the definition of a second within that bubble. When the hydrogen's proton (or any particle) that has enough energy to undergo fusion reaches the perimeter of the bubble, unless you have caused the energy to leave the universe magically, it is still in the particle. The particle's momentum is conserved. So the proton will be traveling at the same speed as it did if time were never affected. What will change is the number of these protons reaching the perimeter per normal second. So this reduces to a simple calculation of what dosage of proton radiation will destroy your machine. As stated, the design of the machine completely determines that number. We will treat the weakest part of your machine as a black body which is absorbing heat energy (we won't use a gray body calculation because anything transmitted or reflected eventually ends up in your machine somewhere else). You need to have some heat removal method to offset the heat transfer going in.

The radiation energy per unit time from a black body is proportional to the fourth power of the absolute temperature and can be expressed with Stefan-Boltzmann Law as: $$q=\sigma T^4A$$ where:

• q = heat transfer per unit time (W)
• σ = The Stefan-Boltzmann Constant = $$5.6703e-10^{-8} \text{(W/m}^2 \text K^4$$)
• T = absolute temperature in kelvin (K)
• A = area of the emitting body (m$$^2$$)

Now, conveniently, you are asking about a thermonuclear reaction. The sun is doing exactly what you are asking for here, and so is any fusion reactor. So let's look at the sun as an example to gauge how much energy is radiated in real time. From that, you can slow down time enough that the the heat transfer $$q$$ will not destroy your most delicate component.

Use a surface temperature of the sun at 5800 K, and assume that the sun can be regarded as a black body. The radiation energy per unit area can be expressed by modifying (1) to

q / A = σ T$$^4$$

= $$(5.6703e-10^{-8}) W/m^2K^4) (5800 K)^4$$

= $$6.42e-10^7$$ (W/m$$^2$$)

As you can see, the absolute temperature of your thermonuclear device—at the perimeter of the time bubble—must be known before any meaningful numbers can be drawn. That of course is determined by the diameter of your bubble, and the yield of the radiation source being considered. That can be the subject of another question, perhaps. No matter, since Watts are a unit measured against time, you are simply turning the second into a variable in this formula. You need to modify the value of the second so that the black body radiation finally coming out does not destroy your weakest link.

Let's say you have a little fancy active cooling. Now, let's say you have some thermoelectric device harvesting energy, and that device is destroyed (or becomes useless) above 1000K. That's pretty hot, but say you've designed this. Let's also say you have a small-ish time bubble and the radiation reaching the perimeter is "normally" at 5,800K, just like the sun. All you need to do is change the value of s in the Stefan-Boltzmann Law such that you are not putting more heat in than the device can dissipate.

Let's assume you can naturally dissipate heat by cooling fans at the rate of 10kW per second per meter$$^2$$. How long does a time-dilated second need to be to allow your cooling system to prevent overheating?

Solve the solution above ($$6.42e-10^7$$ (W/m$$^2$$)) for s by rewriting it:

• Recall: 1 Watt - 1J/s, $$\therefore$$ 1 s = 1J/W

$$6.42e-10^7 W/m^2 \\ = 6.42e-10^7 \frac{J}{s m^2} \\ \therefore s= \frac{1J}{6.42e-10^7 m^2}$$ But you need to know what the slowed second is ($$s_1$$) that will just introduce enough heat to destroy your weakest component, at $$10 \frac{kW}{s m^2}$$. So we rewrite the above solution for $$s_1$$ as follows: $$s_1= \frac{1J}{1e-10^4 m^2}$$ now we have an exact ratio for time dilation by dividing the unknown second by the normal one, $$\frac{s_1}{s}$$ as follows:

$$\frac{s_1}{s}=\frac{\frac{1J}{1e-10^4 m^2}}{\frac{1J}{6.42e-10^7 m^2}} \\ s_1= \frac{6.42e-10^7}{1e-10^4}s \\ S_1 = 6,420s$$

A:

Given these conditions for harvesting a thermonuclear detonation which has the same energy as the surface of the sun, you will need to slow time such that one second equals $$1.78\overline3$$ hours, or slow time by 642,000%

I made many assumptions but in simplest terms, design your energy collection device first, learn where its weakest link is, treat the time bubble as a black body radiator of a certain surface area, then modify the Stefan-Boltzmann equation to solve for your new value of a second.

• You are thoroughly mistaken with "The high yield comes from the hydrogen in the air that it fuses." Feb 21 at 23:14

The Hiroshima bomb had a yield of approximately 15kt. This is roughly equivalent to 62759999999991.5 joules (let's say 62.76 * 10^12).

The Three Gorges dam, the largest power station in the world at this writing, has about 22.5MW of capacity.

So, the energy from one Hiroshima-sized (15kt) explosion would provide 22MW for 2.8 * 10^6 seconds. This is just a little over a month.

If you don't need Three Gorges capacity, you can stretch it for proportionately longer. For all I know, though, you'll need at least that much to keep the time bubble running!

• Downvoted for the supposedly humorous conversion of approximately 15 kilotons (two significant digits) to a "rough" value with 15 significant digits and then "rounding" it to a value with four significant digits. Public service announcement: approximately 15 kilotons equivalent TNT is about 63 terajoules. Would downvote a second time for mindlessly ignoring the obvious effects of time dilation and not even bothering to tell the querent what they need to handwave. Feb 21 at 19:20
• Accepted yield for Little Boy was 12-18, so honestly 60 TJ is probably a better way to show it. Feb 24 at 4:23