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[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 805 points806 points  (149 children)

The energy stored in an atom would be the mass-energy of the atom, found by E=mc2.

If you annihilate the atom with an antimatter atom, you could convert it all into energy in the form of EM radiation. It's technically possible, but you're more likely to get a whole bunch of other particles too.

[–]purpleoctopuppy 288 points289 points  (18 children)

To add some detail about the other particles, when you combine hydrogen with anti-hydrogen, the vast majority of the energy released will be in the form of pions, which will decay into photons (about a third of the total energy), muons (about half), and neutrinos (the rest). The muons will then decay into neutrinos and electrons/positrons.

Of course, this is just the dominant decay pathway, you get a lot more different particles at lower probability just from the sheer amount of energy you're dealing with.


[–]DarkFireRogue 32 points33 points  (8 children)

Why are certain particles preferred? Does that have to do with their entropy? Why wouldn't the energy quickly form new protons and electrons?

[–]dcnairb 27 points28 points  (5 children)

The interactions are described probabilistically as functions of the energy scales (mass and momenta of incident particle(s)) as well as depending on the process by which the interaction is carried out

[–]WiggleBooks 3 points4 points  (4 children)

Are they truly random as in there are no hidden variables? Or is that an open question?

[–]dwarfarchist9001 10 points11 points  (0 children)

Its kinda an open question. We have ruled out certain categories of possible hidden variables but so called "non-local hidden variables" are still a possibility.


[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 2 points3 points  (0 children)

There are no local hidden variables.

[–]dcnairb 2 points3 points  (0 children)

We've ruled out local hidden variables, in general I don't think many people subscribe to the idea of (some set of) hidden variables (being somewhere to govern physics). A probablistic interpretation of QM and QFT is standard.

[–]mfb-Particle Physics | High-Energy Physics 1 point2 points  (0 children)

There are deterministic interpretations, but even there it looks random to an observer in the universe.

[–]13esq 1 point2 points  (7 children)

I love learning about this stuff, please can you recommend a book for beginners who'd like to learn more?

[–]foofdawg 10 points11 points  (4 children)

A good starting place for all things math and science is the sixty symbols YouTube channel (their website might be a better place if you're interested in specific topics though). They started off explaining what different symbols are used for in math and science and how/why they are used practically and have expanded a lot since then. Planck lengths, neutrinos, relativity, etc.

There's also a great interview done with Richard Feynman called "fun to imagine" which is one of my favorite videos on the web. You can find it whole or in parts on YouTube.

Are you interested in specific learning about small particles or?

[–]DenSem 65 points66 points  (89 children)

you could convert it all into energy in the form of EM radiation

Hypothetically, lets say we convert it. How much work could we do with that much EM radiation from a single atom? i.e. "we could power a car for 10 miles", "we could power a ocean liner across the pacific", etc)

[–]Unearthed_Arsecano 395 points396 points  (39 children)

Hydrogen, the lightest atom, has a mass energy of 1.5 x 10-10 Joules. Uranium, the heaviest element that occurs naturally on Earth, has 3.6 x 10-8 Joules.

The energy in one hydrogen atom is enough to lift up a single grain of sand by the thickness of a sheet of paper. Uranium will let you lift a (dead) fruit fly about half an inch.

Keep in mind though that a grain of sand weighs more than 10,000,000,000,000,000,000,000 hydrogen atoms, so that's a pretty incredible amount of energy.

[–]11181514 176 points177 points  (19 children)

The energy in one hydrogen atom is enough to lift up a single grain of sand by the thickness of a sheet of paper. Uranium will let you lift a (dead) fruit fly about half an inch.

I've never seen it phrased like this before. Thanks for the visual, that really helped me grasp the concept.

[–]50StatePiss 20 points21 points  (18 children)

That's fascinating /u/Unearthed_Arsecano By those calculations we would only need 4% of a sand grain's worth of hydrogen to send a grain of sand to Alpha Centauri. Whoa

[–]throwaway48159 26 points27 points  (17 children)

Much, much less than that. Gravity gets weaker as you get further away.

[–]50StatePiss 3 points4 points  (16 children)

Oh yeah! Thanks. I wonder; then, how fast it could be accelerated to before it hits the star and how long it would take. I guess that would tell us what fraction of that 4% we'd actually need. I wouldn't know how to calculate that; my maths is limited to finance, not physics unfortunately.

[–]Unearthed_Arsecano 5 points6 points  (15 children)

To reach Alpha Centauri, you'd near enough need to entirely leave the gravity wells of both the Earth and the Sun (the second one is much harder). Assuming you're smart about it and use the Earth's orbital velocity to help you escape the solar system, you'd need a delta-v (change in velocity) of about 16.6 km/s, which for a 10 mg grain of sand is ~1,400 Joules of kinetic energy or 0.000000015 mg mass equivalent, which is 0.00000015% of the mass in question.

This is all very back of the envelope, bear in mind, but it should be correct to within an order of magnitude or two.

[–]DonQuixotel 10 points11 points  (0 children)

I'm within an order of magnitude or two of being pretty well off, so I'll allow it!

[–]SquidCap 2 points3 points  (13 children)

Grain of sand would experience quite a bit of air resistance, allthou the whole thought exercise is insane, we need to add a bit more to escape the atmosphere.. 16km/s means it still is subjected to the atmosphere for few seconds. Drag is ^2 after all (iirc). Maybe add magnitude of order or two more force.

[–]Unearthed_Arsecano 8 points9 points  (12 children)

A grain of sand at 16km/s would immediately vaporise in our atmosphere so trying to consider drag is rather futile. It would be energetically equivalent to just carry it out of the atmosphere on a big ladder and provide the remaining energy from there, so I propose we do that instead of you making me solve a rather tiresome differential equation at 11pm ;)

[–]were_z 13 points14 points  (0 children)

I just wanted to thank you for the imagery, this reply was where i grasped the idea best.

[–]GWJYonder 20 points21 points  (9 children)

To extend this to a more human-conceivable bit of fuel, that grain of sand has a mass of .67 mg. Getting 100% of the mass energy of that grain of sound would net you 60.3 gigajoules. That is just over the amount of energy you get from 500 gallons of gasoline.

[–]somewhat_random 1 point2 points  (0 children)

500 gallons seemed low but I checked and you are correct - good mathing sir/mam!

[–]elnegativo 1 point2 points  (2 children)

Is it possible theoretically to convert 100 of any amount of mass?

[–]GWJYonder 4 points5 points  (1 child)

If you mean convert as in "actually get the anti-matter to come in contact with 100% of the matter" sure, but it seems like it would be very hard to practically do so with a fuel pellet of even a microscopic size. These conversions are so energetic that a grain of sand, for instance, would certainly be blown to bits far out of the reaction area long before the entire grain was converted.

That said, it doesn't matter, unless for some reason we are using some ridiculous crazy form of normal matter for these reactors/bombs/whatever, the limiting factor will be the anti-matter. As long as we are converting 100% of the anti-matter than even a a 1% matter conversion rate will be fine, we'll just go get more grains of sand.

Now if you mean "is it possible theoretically to get 100% of the energy out of this conversion as useful work" then no. First we're hit with the limit of the carnot cycle, the limit to how efficient work generation can be (tbf the temperatures of an antimatter reaction are so high that I imagine the carnot limit would be 1 for quite a few sig figs).

Aside from that though, is the "how much of that energy is theoretically capturable so that we can feed it into our carnot engine?" u/purpleoctopuppy posted a breakdown of the most likely decay pathway for a hydrogen and anti-hydrogen collision. I know basically nothing about Pions and Muons, maybe he or someone else could elaborate on the theoretical ways that those particles could be harnessed but any of the energy that is converted into neutrinos is gone for good, no way you are capturing that.

[–]elnegativo 1 point2 points  (0 children)

It was the second, thanks for the answer.

[–]c4mma 0 points1 point  (3 children)

There is some reason why (dead) fruit and not normal fruit? Why highlight (dead)?

[–]Unearthed_Arsecano 3 points4 points  (2 children)

A living fruit fly can fly, and thereby be lifted to an arbitrary height without expending the energy of our atom.

[–]007T 2 points3 points  (0 children)

and thereby be lifted to an arbitrary height

Careful though, if your arbitrary height is too great then we're back to dead fruit flies again.

[–]sxbennettComputational Materials Science 74 points75 points  (12 children)

Hardly anything from a single atom. One atom is unbelievably small, so the mass energy is around a billionth of a Joule. However, macroscopic objects contain unbelievably large numbers of atoms. 1 gram of matter contains enough energy (9x1013 Joules) to power the entire US for 20 minutes.

[–]CaptainCaspar 9 points10 points  (9 children)

When phrased like that, it really makes you think..

We're VERY bad at getting energy out of mass efficiently. We've got thousands of huge power plants that burn probably millions of tons of fuel in total every day, and yet, simply 72 grams of matter could give us the same energy if we could harness it efficiently.

[–]sxbennettComputational Materials Science 21 points22 points  (0 children)

It's not as much of an "efficiency" issue as it is an issue with the fact that there's probably no good way to convert 100% of matter into energy at scale that we can control. The energy you get out of a fission or fusion reaction might only be a fraction of the mass energy of the fuel, but most of that isn't wasted energy, it's just leftover mass. The real question is what are the byproducts of the reaction, how can those be used or recycled, and are you getting more energy out of the fuel than you put into mining, refining, overhead, etc.

[–]StrukkStar 15 points16 points  (1 child)

We're VERY bad at getting energy out of mass efficiently.

No, we're actually pretty good at it, as good as we can be if we consider modern physics to be correct. And we learned to do it without modern computing.

The energy just isn't accessible.

[–]a_trane13 4 points5 points  (2 children)

Well, by our underlying theories of physics we can't really harness it completely. Smashing an atom always produces other types of particles along with energy.

Hopefully someday someone can do it, but it would have to be under extremely foreign conditions to us.

[–]lynnamor 0 points1 point  (1 child)

Just getting out of the fractions of a percent harnessed would be a pretty huge leap, wouldn't it?

[–]a_trane13 2 points3 points  (0 children)

Sure, yes, but you have to accelerate particles to smash them together and break them. That takes a lot of energy. The particle accelerators like the LHC are huge energy consumers. You'd have to come up with a way to break atoms without using a lot of energy. Particles get very hard to accelerate when they're going very fast.

[–]corvus_curiosum 10 points11 points  (26 children)

E = mc2

For a single hydrogen atom m = 1.67e-27 kg The speed of light c = 3e8 m/s

E = 1.5e-10 J

For reference a gallon of gasoline contains about 1.22e8 J, so significantly more, but atoms are also really tiny. There are 6.02e23 hydrogen atoms in a gram.

If we use a full gram (of anything, it's all the same in this case)

E= 0.001 kg * (3e8)2

E= 9e13 J or about 700,000 gallons of gasoline, enough to drive a decently efficient car 14,000,000 miles, about halfway to Mars, which would take 20 years of driving all day and night at 75 mph.

TLDR a gram could fuel your car for your entire life.

[–]Futtermax 0 points1 point  (20 children)

Serious question- I'm definitely missing something.

How, if these atoms have such little energy, is a nuclear warhead able to produce such a large explosion?

Is it creating a chain reaction? Splitting up a lot of uranium at once? It doesn't make sense to me that something that can barely move a fruit fly can make a city explode.

Thanks in advance for your potential answer!

[–]space_keeper 7 points8 points  (1 child)

He's explained roughly how much mass-energy there is in a gram of stuff. Nuclear fission warheads contain several kilograms of uranium.

1 atom -> slightly moving a dead fruit fly

1 kilogram -> ~1023 atoms. Ten thousand billion billion fruit flies.

The process in a fission warhead isn't totally efficient (not even close), but it does release a large portion of that energy in a very short amount of time (unlike the moderated process in a nuclear reactor, which releases the energy over a longer period).

[–]ayemossum 2 points3 points  (1 child)

Well the energies we're talking about are a single hydrogen atom. A uranium bomb is dealing with a MINIMUM of 33 pounds of U235, which is roughly 3.8 * 1025 atoms. It's a matter of scale. If I have the amount of black powder in a toy cap gun (just enough to make a loud bang) it's unimpressive. If I have a 10 pounds of the stuff... It's a little bit different.

[–]SirButcher 2 points3 points  (1 child)

A nuclear warhead contains multiple kilogramms of material. The above example was for ONE ATOM. 1 kg plutonium is about 4.1 moles - this means one kg plutonium contains 2.408.856.600. piece of atom. Now compromise /u/corvus_curiosum example with this incredible huge number!

Each atom itself contains little energy - on our scale. But even the tiniest thing contains an incredibly huge amount of atom, and this small amount of little energy quickly add up. A thermonuclear device barely utilizes around 1% of the available energy. If you would be able to acquire a 1kg of antimatter that - when it meets with "normal" matter and they annihilate each other it would create explosion huge enough to destroy the planet by releasing most of the contained energy in an atom.

[–]FarleyFinster 2 points3 points  (0 children)

I'm definitely missing something.

That would probably be the bit about "one atom of hydrogen". Your typical nuclear warheads tends to split quite a few uranium atoms at the same time. Around 1.26×1026* of 'em in your in the typical nuke. So despite not getting anywhere remotely near the rest-mass energy out of fission, even a truly insignificant portion of what you're talking about is more than sufficient to deal with your fruit fly problem.


* To wrap your head around this number, a trillion (1012) is roughly the number of grains of fine sand the largest road-legal dump truck can carry. But this is exponents here; we're nowhere near halfway there yet. Got your dump truck full of sand? Good. You need another trillion of them. That's one truck full of sand for each grain of sand in the first truck.

Got all those trucks together? Excellent. Collect the same amount another 99 times. Now you've got 1026 grains of sand. You only need another 20% or so more to include that ".26" behind the decimal.

[–]PM_ME_GLUTE_SPREAD 1 point2 points  (6 children)

You aren't just splitting a single atom when you detonate a warhead. You start the reaction by splitting a bunch, which release neutrons(?) that go and split more, which continues until the fuel is spent and there isn't enough of a concentration to continue detonating atoms.

This happens on the order of nanoseconds and releases all this energy, essentially, at once which creates the huge explosion.

[–]chumswithcum 0 points1 point  (4 children)

It should be added that most of the fuel in nuclear device isn't even used, the device blows itself apart (thus stopping the reaction) before all the fuel can react. However, adding more fuel means there is more available to react in the very, very short time that the device is intact, and you get a bigger boom.

[–]PM_ME_GLUTE_SPREAD 0 points1 point  (3 children)

Do you know if the yield of a nuclear device has diminishing returns? Like, does the force of the explosion end up pushing the material away faster with more material you add?

[–]Peter5930 0 points1 point  (0 children)

The opposite, the more material you add the more efficient it becomes and the more material is able to fuse before it blows apart. If you add an entire Sun's worth of material it doesn't blow apart at all and instead the electromagnetic forces from photon pressure trying to blow everything apart end up in a stable balance with the gravitational force trying to crush everything together that creates the conditions for a self-regulating and slow fusion process that can last for billions of years.

[–]chumswithcum 0 points1 point  (0 children)

I don't know that particular bit. Here's an interesting factoid, though - by the time the casing of the warhead splits, the reaction is done. The giant flash and subsequent fireball are all emitted after the bomb is done being a bomb. It's a little insane how much energy is emitted in so short a time.

[–]quantasmm 0 points1 point  (0 children)

No, it does not have diminishing returns, it is the opposite, assuming similar engineering can be used to solve the problems of a larger bomb. A nuclear bomb goes off in about 500 nanoseconds. 99% of the energy is released in the last 50 nanoseconds. Its so fast, its own inertia limits how quickly it can expand. Assume for a second the core/tamper is going from rest to a million meters per second during one 10 nanosecond cycle; that's still only about half a centimeter of expansion. In this time, a fission bomb can double its energy output. The fusion parts of a bomb are even faster. Finally the core expands 20 centimeters or so and it starts on a path to equilibrium but not before an immense amount of energy is stored in an area smaller than a basketball, so much so that it reaches stellar temperatures.

Nuclear bombs of normal size have enough plutonium to reach "two critical masses" (im referring to k = 2, but this is a decent analogy). A much larger bomb would require that we cleverly space out "several critical masses" of plutonium that could reach much larger values of k. This could cause the bomb not to merely double every 10 ns, but something like 5x every 10 ns. (depends on solution of course) And the tamper will be much larger and have much more inertia. IMHO more energy would be released and it might even be more efficient. Edit: And the fusion part, its speed scales with temperature, so the fusion is faster as well.

We stopped building larger bombs not because we don't know how, but because we dont need a larger boom. We can already build a bomb with as many stages as we want, the "boom" is only limited to the raw materials we decide to put into it.

[–]corvus_curiosum 0 points1 point  (0 children)

Yes, it's a chain reaction with lots of atoms, ideally all of them, splitting very quickly. Uranium will ocassionally release neutrons, and when one splits it releases several neutrons that may hit others and cause them to split as well. Fissile material are either subcritical, critical, or super critical masses. Subcritical masses decay exponentially, there aren't enough atoms nearby to maintain the reaction, and most neutrons exit the material without striking another nucleus causing the reaction to fizzle out. In a critical mass there are enough atoms around for each split to cause any leading to a steady reaction. Super critical masses are ones were each split causes more than one other nucleus to split leading to an exponential increase. By doing things to increase the rate of this growth, using more pure fuel with less non fissionable atoms taking up space/absorbing neutrons, adding neutron reflector, and using conventional explosives to quickly turn subcritical masses critical by either joining two separate subcritical masses or compressing a single mass, it's possible to increase the rate of the reaction to the point that it releases a large amount of its energy before it destroys itself.

[–]R01ne 0 points1 point  (3 children)

How much to boil all water on earth? Me and some friends tried to figure it out once, and came up with an approximate amount of 50 kg, but we did it hung over in the back of a car with no paper. I'd love to know the actual amount.

[–]corvus_curiosum 1 point2 points  (2 children)

Rest mass calculations a child's play compared to that question, but I think I can do it within an order of magnitude.

There are approximately 1e21 kg of water on Earth. Assuming it is at an average of 0°C, all those deep dark oceans, it will take 418600 J per kg to raise it to 100°C and another 2257000 J to boil. So let's say 1e27 J all together. (1e27J)/(3e8)2 = m

About 1e10 kg, so a bit more than 50. Which seems a bit high to me, but there is a lot of water on Earth, and it really doesn't like boiling.

[–]dfryer 1 point2 points  (0 children)

It seems like a lot, but that's still only 1kg matter+antimatter per 100 billion kg of water! It just goes to show how big of a doomsday device you'd need if you really want to destroy the world...

[–]bigscience87 7 points8 points  (6 children)

Let's assume a uranium atom annihilates with an anti-uranium anti-atom, since those are nice and heavy and therefore have lots of energy.

Using E=mc², you get... About 71 nanojoules. Enough energy to run a 100W light bulb for 0.71 nanoseconds.

[–]torontonorth 0 points1 point  (0 children)

This has finally made the co-relation for me. I studied all this stuff but never really knew why and now I know.

[–]CthulhuChild 1 point2 points  (0 children)

Just to tie this back to your original context (re: fission), nuclear weapons lose a lot of energy due to incomplete consumption of their fissionable material. The explosion spreads the material out, at which point it ceases to be supercritical. I think the bombs dropped on Japan used something like 2% of the reaction mass. The rest just becomes a low-density hot mist of radioactive material.

[–]Azurealy 1 point2 points  (3 children)

You're missing your momentum term unless this atom is staying perfectly still.

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 1 point2 points  (2 children)

The context of OP's question implies that they're asking about some sort of "fuel" efficiency, so I assumed it was the non-relativistic case.

[–]Azurealy 0 points1 point  (1 child)

Even still, on the scale of an atom, it would still have momentum wouldnt it? Though i suppose youre right because it would be so small we wouldn't care much right?

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 1 point2 points  (0 children)

The momentum term in the relativistic energy equation is related to the kinetic energy of the particle, and the kinetic energy of a particle due to thermal vibrations is tiny compared to its rest energy.

[–]guidedhand 0 points1 point  (2 children)

Didn't veritasium do a good video showing that most of the mass of particles doesn't come from theass part of Eisenstein's equation, but from the energy stored in binding the particles together? Like the mass /energy equivalency was only a smaller part

[–]SenorTron 1 point2 points  (0 children)

That energy binding the particles together is converted to Mass, or vice versa.

The moment I learned that the mass of a proton, neutron, and electron counted individually is different to the mass of a hydrogen atom made of all three combined was mind blowing.

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 0 points1 point  (0 children)

I think you're remembering his video wrong. He was probably saying most of the mass doesn't come from the the interaction with the Higgs field but from the binding energy.

[–]Minguseyes 0 points1 point  (1 child)

When calculating anti-matter anhilation energy, do you use m (the mass of the normal matter) or 2m (the combined mass of normal and anti-matter) ?

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 0 points1 point  (0 children)

You use 2m. But the antiparticle has mass too, which means since it's part of your original "fuel", the average energy released per particle is still mc2.

[–]flacidturtle1 0 points1 point  (2 children)

So how do we turn these other particles into energy?

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 0 points1 point  (1 child)

A lot of those other particles are actually particle-antiparticle pairs, which will eventually decay into EM radiation too.

The other (non-matter-antimatter) particles eventually decay into stable particles, after which you can only extract energy from them by annihilation again with their antiparticles.

[–]flacidturtle1 0 points1 point  (0 children)

So a machine designed to shoot particles at each other at fast enough velocities could technically shoot multiple particles( even those that break off) and bounce particles back into the cross fire to produce more energy?

[–]LaplaceMonster 0 points1 point  (4 children)

I have a question, I wish I knew more about these things. The conversion of that energy, strictly speaking, is that 100% efficient? Is the E=mc2 energy completely equal in value to the EM radiation energy or is there some loss attributed to this due to entropy increase?

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 0 points1 point  (3 children)

If it gets converted completely to EM radiation, 100% of the energy goes into the radiation.

[–]Canbot 0 points1 point  (2 children)

there are no units of measure in e=mc2. How many 100w lightbulbs could a hydrogen atom power for an hour?

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 1 point2 points  (1 child)

there are no units of measure in e=mc2

The units are in SI, so joules, kilograms, and meters per second.

How many 100w lightbulbs could a hydrogen atom power for an hour?

The speed of light is ~3 x 108m/s, a hydrogen atom is ~10-27kg, so you get ~10-10 J of energy. It's enough to power 10-12 100 W bulbs for a second, which is one picobulb for a second.

[–]Canbot 0 points1 point  (0 children)

Thanks, I guess it would have been more useful to ask how many bulbs a gram of hydrogen could power.

[–]Panzerkatzen 0 points1 point  (1 child)

If I may ask a related question for something I never fully understood:

Does that mean anti-matter and matter meeting will annihilate each other and leave nothing, or that they will annihilate each other with the full power of a nuclear explosion from each, or just a massive EMP?

[–]Dubanx 98 points99 points  (49 children)

As mentioned, the energy stored is equal to E=MC2 or Energy is equal to mass times the Speed of light squared.

As for releasing 100% of the energy of matter, that can accomplished with antimatter. That isn't done, though, because

A) Antimatter is incredibly difficult to contain as it explodes violently whenever it contacts regular matter.


B) It can only be made in exceedingly small quantities by using a particle accelerator.

So it takes as much energy to create antimatter as it releases (to get a nuclear sized explosion you would have to spend a nuclear bomb's worth of energy to create that much antimatter), and the difficulty in containment means any sort of failure in containment will cause it to detonate. Though, it would probably be impossible to store that much antimatter in one place in the first place.

[–]W1D0WM4K3R 12 points13 points  (10 children)

Question. Does antimatter explode with all regular matter, or just it's regular matter counterpart?

[–]Dubanx 37 points38 points  (2 children)

Its counterpart.

If you were to toss a positron at a proton they would repel each other because of the like charges, rather than merge and annihilate.

[–]mikelywhiplash 19 points20 points  (1 child)

Right - and even if we could overcome the electric repulsion, you'd run into another problem: if two positively-charged particles are destroyed, charge is not conserved.

[–]314159265358979326 13 points14 points  (0 children)

Note that it's not atoms matching atoms in antimatter annihilation, it's electrons annihilation with positrons and anti-protons annihilating with protons.

[–]Unearthed_Arsecano 6 points7 points  (2 children)

An electron will only annihilate with a positron (an anti-electron), a proton will only annihilate with an anti-proton, but on a larger scale the constituent particles of say, carbon and anti-nitrogen would still annihilate, it wouldn't matter that they were in a different atomic configuration (though you'd have a few particles left over from the anti-nitrogen).

[–]theWyzzerd 15 points16 points  (1 child)

Huge missed opportunity for the physics community to call anti-protons negatrons if you ask me.

[–]I_Cant_LogoffChemical Physics | Atomic and Optical Physics 1 point2 points  (0 children)

The particle physics community actually attempted to rename the electron the negatron back when the positron was discovered, but the name didn't really catch on. Naming the antiproton the negatron would just cause even more confusion.

[–]Sedu 6 points7 points  (2 children)

Only its counterpart... but keep in mind that:

a) particles and their anti-particles are strongly attracted to one another


b) All matter which we are familiar with on a day to day basis is made (effectively) of nothing but electrons, protons, and neutrons.

[–]Unearthed_Arsecano 6 points7 points  (1 child)

Neutrons are not strongly attracted to anti-neutrons, because both have no net charge. Any interaction between the two would occur when they are already extremely close to each other.

[–]Sedu 1 point2 points  (0 children)

True enough! Although they should most typically be found in antimatter atoms’ nuclei, which would attract due to the protons/positrons.

[–]jcy 2 points3 points  (1 child)

does antimatter occur naturally in the universe at all?

[–]reallydarkcloud 1 point2 points  (0 children)

Positrons (anti-electrons) are emitted as part of the radioactive decay of some isotopes, including K-40, which is a naturally occurring (and the source of radiation emitted by bananas.

There are apparently other sources, but they're outside my range, so you might want to try wikipedia.

[–]ShadoWolf 5 points6 points  (1 child)

From my understand antimatter isn't the only way you could go about extracting energy from matter. You could focus a whole ton of laser light to generate a kugelblitz black hole. if you could feed said micro blackhole.. or generate said black hole with some feed matter at the focal point you would get some approaching antimatter for matter to energy conversion

[–]liamguy165[S] 2 points3 points  (29 children)

Okay, that makes sense. But I do not understand why anti-matter annihilation with matter is the only way to convert a significant (even 100%) percentage of mass to energy. Is there any other known ways or theorized ways?

[–]Gigazwiebel 26 points27 points  (25 children)

In principle, you can compress the matter to a black hole and wait until it evaporates by Hawking radiation. That achieves the same.

[–]liamguy165[S] 3 points4 points  (24 children)

Okay, thank you. It seems that E=mc2, while true, hides the difficulty of the process in that equals sign lol. But also, since (so I’ve heard) fission/fusion releases 1% of the energy, why can’t we achieve even 30% mass to energy release?

[–]GuitarCFD 10 points11 points  (9 children)

But also, since (so I’ve heard) fission/fusion releases 1% of the energy, why can’t we achieve even 30% mass to energy release?

Think about what happens in a fission reaction. A single neutron is added to a U-235 nucleus making it a U-236. That nucleus becomes unstable and splits into Kr-92 and Ba-141. Fission releases such a small amount as actual energy because quite a bit of the energy remains intact as actual matter. Someone with a better handle on the math than I can show you how you take E=MC2 and evaluate the Kr-92 and Ba-141 to find the remainder is the energy that was actually released during fission. (3 neutrons worth right? Someone tell me if i'm wrong there).

You want to talk about how inefficient Nuclear Fission is as a process. How about how we gather energy and convert it to electricity. Nuclear Power Plants are basically high tech boilers. They use the heat in the reaction to heat water and turn a turbine. The energy released is small compared to what's held in a U-236 nucleus, but what is released comes out in the form of heat and electromagnetic radiation. We only have the technology currently to use the heat.

[–]liamguy165[S] 1 point2 points  (7 children)

Ah I see, thank you. I’m guessing if we could pull apart atoms to the extent that it “breaks” and releases it’s mass as energy, it would of course just take the exact or less amount of energy to pull it apart as is released. Thanks again!

[–]GuitarCFD 6 points7 points  (6 children)

it would of course just take the exact or less amount of energy to pull it apart as is released.

That's probably left for someone with an actual degree, but I can tell you that what makes Fission a net gain energy wise is that we don't have to spend energy to get energy. (yeah, yeah enrichment aside). We are taking advantage of a natural process of nuclear decay. Certain isotopes of elements on the periodic table just have unstable nuclei and break apart in nature. Some very smart people realized that if you gathered enough of that isotope in a dense enough object you'd reach what we call "critical mass" and the decay would become a chain reaction. Fission bombs are technically 2 halves of critical mass that get slammed together and then...well "boom".

[–]liamguy165[S] 1 point2 points  (4 children)

Right, thank you. I see that we take advantage of the fact that the combination becomes unstable, and the released energy is just the subtracted mass basically to form into new particles. However, is there no process by which an atom is so unstable that the next most favorable configuration is to convert to energy? Say for example, Hydrogen, as it has nothing to destabilize into?

[–]rivenwyrm[🍰] 7 points8 points  (2 children)

Think of it like this: The amount of energy stored in an atom (due to the strong and weak forces) is stored there because that is the most stable repository for that energy. If it weren't, it would be somewhere else by now, billions of years into the universe. Naturally decaying atoms are unstable, so they shed energy until they reach a state of stability. But atoms that do not decay are already in the most stable form they can reach without 'outside intervention'. And there are a very limited number of ways you can 'intervene' inside an atom, simply due to physics.

Obviously, isotopes are unstable because the extra neutrons are throwing off the balance in the atom, but the best solution for the atom is to just get rid of the neutrons.

[–]liamguy165[S] 1 point2 points  (1 child)

Okay, that makes sense. So the equation + what we see everyday implies mass is more stable than energy?

[–]RobusEtCeleritasNuclear Structure | Nuclear Reactions 3 points4 points  (4 children)

Because no process exists that would do that.

[–]Mac223 1 point2 points  (7 children)

It seems that E=mc2, while true, hides the difficulty of the process in that equals sign

The equals sign doesn't imply a process.

One part of the equality is that if you do convert matter to energy, or vice versa, it tells you that c2 is the conversion factor.

A second point, which isn't clear unless you know the full context of the equality, is that energy is mass, and mass is energy. A ball of energy has a mass and gravitational attraction, just like matter.

[–]liamguy165[S] 0 points1 point  (4 children)

Do we have physical examples of such balls of energy (Quarks?)

[–]Mac223 3 points4 points  (2 children)

Photons is the example that comes to mind, but the point is more general than balls of energy you can point to. For instance, a spinning ball weighs more than a ball that is not spinning. The Earth itself is more massive due to its daily rotation.

[–]liamguy165[S] 0 points1 point  (1 child)

Wow. And we have measured this and confirmed this too?!

[–]Mac223 3 points4 points  (0 children)

I'm not aware of any direct measurement.

In general it's extremely difficult to observe the mass-energy equivalence outside of nuclear reactions, because to add, say, one gram worth of rotational energy to a spinning ball you'd have to add enough energy to lay waste to Hiroshima.

[–]Unearthed_Arsecano 0 points1 point  (0 children)

Quarks are matter. The comment above was a little loose in their use of "mass". Energy distribution is equivalent for purposes of both gravitation and inertia to a mass energy distribution. So, for example, the thermal energy in a hot cup of coffee will make it's effective mass as you would measure it a tiny fraction greater than the same cup of coffee at room temperature. Similarly, light (photons) carries energy, and so like an object of equivalent mass will both be affected by, and generate, a gravitational field.

[–]liamguy165[S] 0 points1 point  (1 child)

So can it be converted, or are they the same thing? If mass is equal to energy, then really isn’t everything one or the other?

[–]ikefalcon 1 point2 points  (0 children)

Matter can't be converted to energy without some mechanism for the conversion. The key is the law of conservation of energy. The mass-energy of the reactants must equal the mass-energy of the products.

Matter anti-matter annihilation is the only mechanism that results in most of the reactants being destroyed, so all of the mass-energy of the reactants is converted to energy.

In other types of energy-releasing reactions, there are massive (meaning containing mass) products. For instance nuclear fission takes a Uranium 235 atom and a neutron and results in a Krypton 92 atom, a Barium 141 atom, and three neutrons. The mass-energy of the products is less than the mass-energy of the reactants, so energy will be released, but there is still a tremendous amount of mass-energy in the products.

Even though only a small percentage of the potential energy is released from a nuclear fission reaction, it is still quite significant because a very small amount of mass converts to a tremendous amount of energy as we see in the E=mc2 equation.

[–]EuphonicSounds 0 points1 point  (0 children)

It's maybe worth emphasizing that the "conversion" in question here is a conversion of potential energy to kinetic energy.

Because protons are all positively charged, they repel each other electrostatically. The reason a nucleus stays together is that the (residual) strong force attracts the nucleons to each other, and this attraction is, well, stronger than the electrostatic repulsion. But the repulsive electrostatic force is still there, and associated with it is an electrostatic potential energy.

To facilitate fission, we arrange for the repulsive electrostatic force to overcome the attractive residual strong force. When this happens, bits of the nucleus fly apart from each other: the potential energy associated with their mutual electrostatic repulsion is converted into their kinetic energy.

[–]bermudi86 0 points1 point  (0 children)

Because fission and fusion release energy but you are left with some matter that wasn't converted into energy (the leftover has the remaining energy) whereas using antimatter you get only light (energy) and no leftovers.

[–]BearlyMoovin 0 points1 point  (1 child)

This is the basic concept between warp core reactors in Star Trek, correct?

[–]Dubanx 4 points5 points  (0 children)

I doubt it, at least not in any scientific manner. The energies in star trek are laughably absurd. They clearly added a bunch of zeros to all of their numbers blindly without even thinking about what those numbers meant. Like, energy outputs could literally be measured in planets per second absurd.

[–]onjens 21 points22 points  (4 children)

The energy released from fission is actually quite different from the energy "stored" in the atom in the form of mass. Binding subatomic particles into large atoms requires energy, some of which is then released when the atom is split into smaller atoms.

The energy required to bind subatomic particles together is tiny when compared to the energy required to create the same particles from thin air. Hence, the energy that is released in fission is tiny when compared to the energy that is released in antimatter annihilation.

[–]alex_snp 2 points3 points  (2 children)

Isnt the 99% of the mass of the proton due to binding energy?

[–]mumblerfishString Theory | Flux Compactification 11 points12 points  (2 children)

There are several factors to take into account. One would be the constituents for example. Say we study a fusion process, a very common one would be H+H -> He + rest, or more accurately you would use one of the isotopes, D or T. So, lets take a look at the process D+T -> He + n, as you can find on Wikipedia [1].

The energy released in this process, that will be transferred to momentum of the final constituents, is the difference in mass. So, lets use the numbers we can find on Wikipedia for those masses and see what the difference is:

(m_D + m_T) - (m_He + m_n) = (2.01410178+3.0160492)-(1.00866491588+4.002602) [u] = 0.018884064119999877 [u]

So, as people have been saying, use E=mc2 to find this in energy if you like, but to answer your question we only need to figure out how much mass is gone (m_lost/m_before = m_lost c2 / (m_before c2 ) = E_lost/E_before). That is:

((m_D + m_T) - (m_He + m_n)) /(m_D + m_T) = 0.018884064119999877/(2.01410178+3.0160492) = 0.0037541744164505926

that is about 0.374% of the original mass. So at least the "not even 1%" number is in the right ballpark. This differs for which processes you use.

Note also that the full (2.01410178+3.0160492) [u] = 5.03015098 [u] of before-mass is not easily accessible. Fusion and fission works so that you have some starting constituents and some final ones, D+T and He+n respectively above, and you use the difference in those masses to extract energy.

So if you are looking for "How much energy [...] is it technically possible to 'extract'[?]" then you would have to: Take some starting particles (above D+T), find a series of interactions (above one fusion reaction) that leads to a final state which has the least amount of mass possible. Basically, in some oversimplified terms, it is the binding energy of the constituents of the atoms themselves (in D for example, and proton and a neutron) that you access.

[1] https://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_candidates_for_terrestrial_reactions

[–]liamguy165[S] 4 points5 points  (1 child)

Ah I see, that makes sense to me. Fission and fusion access the binding energy and not so much the mass-energy equivalence. It seems unlikely to me though that there is no natural process of “destroying” mass to create energy that we can take advantage of.

[–]mumblerfishString Theory | Flux Compactification 4 points5 points  (0 children)

It seems unlikely to me though that there is no natural process of “destroying” mass to create energy that we can take advantage of.

So what is energy then? In my description above, energy will be stored as the kinetic energy of the final constituents, which you can extract (into heat for example). Say there is a hypothetical process that results in excess energy, but no mass; where would it go? The only way to store the energy somewhere would be to use massless particles: the photons. So then I would refer you to the other answers and discussion threads under this here question regarding particles--anti-particles. But to summarise the problem there: anti-particles are not easily accessible; not easy to create in large amounts.

The strength of fusion (and fission) is that you have an abundance of, say, hydrogen, and even though you only use ~0.3%, when repeating it a lot (say realistic amounts, ~10{23} or so [1]) you would get a lot of energy in total. A single particle--anti-particle annihilation would not be much energy, so you'd have to repeat it a lot, which would be the hard part. You'd only get a factor 100 increase by using a single particle--anti-particle pair [2], so you'd have to create ~10{21} (23-2 = 21) anti-particles to even be able to start competing.

[1] Assuming processes effectively use the order of grams worth of starting masses https://en.wikipedia.org/wiki/Avogadro_constant

[2] Assuming comparable masses, which is not a good assumption, since building whole atoms of anti-matter is even harder.

[–]JustAGuyFromGermany 7 points8 points  (1 child)

Relevant minutephysics video about "efficiently" converting mass into energy: https://www.youtube.com/watch?v=t-O-Qdh7VvQ

tl;dw Antimatter is way too impractical. Black holes are the way to go and have a efficiency of up to ~40% provided you have some sci-fi tech to catch gamma rays emitted from a near by black hole. Also: Cats are totally a unit of measurement!

[–]lantech 1 point2 points  (18 children)

Everyone keeps saying e=mc2 but what does that actually mean? Mass times the speed of light, so the mass of a single atom times 186,000 miles per hour second? What the hell is that?

What is the amount of energy in a meaningful measurement (like Kilotons of TNT, or joules maybe) in a single hydrogen atom for example?

What atoms have the most mass btw?

*edit: thank you!

[–]mumblerfishString Theory | Flux Compactification 3 points4 points  (0 children)

mass of a single atom times 186,000 miles per hour? What the hell is that?

So there is a conversion of units in there:

Joule = kg m2 / s2

so using SI units for mass and speed (kilograms, and meter/second) you get, by using E = m c2, SI units for energy back: Joule.


Edit to add: Mass increase further down you go in the periodic table (higher "mass number"). A quick look on Wikipedia [2] tells us:

Oganesson has the highest atomic number and highest atomic mass of all known elements.

[2] https://en.wikipedia.org/wiki/Oganesson

[–]sxbennettComputational Materials Science 2 points3 points  (0 children)

It's mass times the speed of light squared, which has units of energy. The mass energy of a single hydrogen atom is 1.5x10-10 Joules, which isn't much but that's a very very small amount of mass. One gram of mass is equivalent to over 20 kilotons of TNT. The heaviest naturally occurring element is Uranium 238.

[–]--Squidoo-- 2 points3 points  (0 children)

What is the amount of energy in a meaningful measurement (like Kilotons of TNT, or joules maybe) in a single hydrogen atom for example?

Here's one I've always liked, from Richard Rhodes's magisterial The Making of the Atomic Bomb: "the energy from each bursting uranium nucleus would be sufficient to make a visible grain of sand visibly jump".

[–]Kalan_arkais 1 point2 points  (0 children)

Using SI units, if you multiply mass in kilograms by the square of meters per second, you get an energy in Joules.

[–]etrnloptimist 1 point2 points  (1 child)

That's actually miles per second.

The speed of light in a familiar unit of speed is 670,615,200 miles per hour.

So now, the units are familiar, but the value is nonsensical in terms of everyday life. Alas.

[–]Oznog99 0 points1 point  (0 children)

It means if you had 0.5g of antimatter and 0.5g of matter and combined them, they would annihilate each other and produce e=1g*c2 joules= 89875518MJ of gamma rays.

That's enough gamma to boil 35M liters of water from 20C.

When you cool down water in a closed system, you actually DO lose mass- it's just infinitesimally small. Each water molecule loses weight due to lower energy. It is not a specific particle leaving.

[–]tpdmech 0 points1 point  (0 children)

With regards to fission, when a fresh fuel bundle is put into the reactor, its composition is mostly uranium 238 (99.3%) and some uranium 235 (0.7%) for Candu (Canadian) reactors. (Note, many US reactors use enriched uranium with higher % of 235). When the U238 is bombarded with neutrons, it will fission into many different elements, including plutonium 239. Only U235 and Pu239 will sustain the reaction because they have low absorption cross section and high fission cross section (you want the neutrons to cause fission, not be absorbed). Along with these good fission products, you get a ton of "bad" elements. These elements will absorb the neutrons and the reaction will eventually stop. So after the fuel bundle has been in the reactor for months, there will be too many bad elements built up in the fuel and they have to be removed. This occurs when only about 1-2% of the U238 has fissioned. So in theory, the fuel bundle is hardly used, but unless you could remove the bad absorbing elements from within the bundle, it's NFG. Fyi, there's is research looking into doing this so we can "recycle" all our used fuel, estentially extending the fuels life by a huge factor.

[–]shiningPate 0 points1 point  (0 children)

The mass converted to energy atomic explosions comes from the particles that make up the carriers of the strong force that hold the protons and neutrons in the nucleus together --i.e. there are not any protons, electrons or neutrons directly converted from mass into energy. Instead it comes from more basic sub atomic particles. You can see this if you total up the weights of the fission mother nucleus and fission product daughter nucleons. Take U235 for example, the full atomic weight of U235 is 235.0439299 atomic units. The primary fission products are Barium 144, atomic weight 143.922953 and Krypton 89, atomic weight 88.91763. There are also 2 neutrons produced at 1.00866491588 atomic units each (plus two massless gamma ray photons). The grand total atomic weight of these fission products is 234.85781283 atomic units, leaving 0.186 atomic weight units of mass "missing" or converted to energy. This represents only about 0.079% of the mass energy equivalent of the U235 atom, but that is not typically how the efficiency of a nuclear weapon is measured. Rather, that quantity of mass converted to energy would equate to a 100% efficient nuclear explosion. Numbers I've seen have indicated the efficiency of the hiroshima little boy bomb was about 2%-3% efficient. You can run some sample calculations. See if you can get a conversion of kilotonnes of TNT to joules. Compare the 15 KT of the hiroshima explosion to what 50 grams of mass energy conversion would have produced.

[–]Hypothesis_Null 0 points1 point  (0 children)

The amount of energy in an atom is described by its mass E=MC2 , but that is not typically what people are referring to when they say '1%'.

In nuclear bombs, while a core for a critical mass might be several kg of Uranium or Plutonium, only a few grams of the material will actually fission before the core is blown apart and becomes sub-critical. The Hiroshima bomb had about 10kg of highly enriched Uranium for its core, but i think the ammount that actually fissioned weighed something on the order of a dollar bill.

In nuclear power, fuel rods are made of unenriched or lightly enriched Uranium oxide. In PWRs, 3%U235 fuel is kept in the reactir until most of the U235 is consumed (0.7% remaining). The remaining 97% of the Uranium is U238. If this were bred into Pu239, it could all be fissioned as well. Actually about 3% of the U238 does get bred into Pu239/240/241 and half of that gets fissioned, contributing 35% of the released energy and leaving about 1.5%Pu in the spent fuel. In general, a rid of 'spent' nuclear fuel has ~24x as much energy left in it.

These are the contexts for that "only a few percent" claim. Nuclear bombs and nuclear reactors both only operate on a few percent efficiency (in their utilization of fuel), if that. They're not referring to the actual 100% mass to energy conversion. Which you could technically achieve by annihilation with anti-matter, but making antimatter is a rediculosly energy inefficient process - not worth it.

As for energy releasable by fission, technically energy from fissioning an atom is released all the way down to Iron. But we can only fission Uranium and Plutonium isotopes efficiently via chain-reactions. Their fusion products release secondary amounts of energy since they tend to be radioactive isotopes due to excessive neutrons. Of the energy that is available from fissioning, we can actually get most of it in a proplery built nuclear reactor. We just currently don't due to economic and saftey condiderations. Plus politicized fear that hamstrings experimentation and research in the area.

[–]canadave_nyc 0 points1 point  (2 children)

This discussion is fascinating, particularly the bits about the amount of energy "locked up" in a single atom. Given the enormous amounts of energy that is in a single atom, is it true to say the universe, as a whole, has an untold enormity of "potential energy" (sort of like a coiled spring, although obviously not the same analogy)? What would that imply (i.e. philosophically, cosmologically) if it were true to say that?

[–]liamguy165[S] 0 points1 point  (1 child)

Indeed, that’s a part of what I’m asking sort of. If a baseball’s mass in equivalent energy could annihilate much of the world, then what about a planet? The amount of energy that could be potentially converted is enormous, and seems to point towards an idea about the origin of the universe and the Big Bang.

[–]canadave_nyc 0 points1 point  (0 children)

Another thought that occurs to me--there's X amount of energy "floating around the universe" right now, and Y amount of mass. But of course there's no reason this has to be so, right? Our universe could theoretically be 99.99999% energy and 0.000001% mass, or 23.67% energy and 72.33% mass, or some other proportion. Why is the mass/energy balance the way it is in our universe right now? And is it possible our universe has X amount of energy (in whatever form, whether it be pure energy or its equivalent in mass), whereas other universes have more total energy (or less)?

[–]bremidon 0 points1 point  (1 child)

Related question: when matter and antimatter collide, the mass is converted into energy, so does this mean that mass (rest mass) is no longer conserved? I've gotten into arguments about this; IIRC, the main gist of the argument was that for an outside observer, the rest mass of the entire system is conserved. Is this true?