# Conversion of mass into energy with 100% efficiency

Suppose you are given a mass of 1 kg. Einstein's theory of relativity says that all the mass can be converted into energy as per the equation $$E = mc^2$$.

So if you want to convert this whole mass of 1 kg into energy, what will you do to that mass?

• Converted to photons: ie. light and heat (BANG!) – slebetman Dec 12 '19 at 4:28
• with anti-matter? – Shing Dec 12 '19 at 16:05

Although pop sci sources often phrase it in terms of “converting” mass into energy, it is incorrect. $$E=mc^2$$ says that if you have a stationary mass $$m$$ then it already has an amount of energy $$E=mc^2$$. No conversion is necessary. The energy is already there otherwise energy would not be conserved. Also, an amount of stationary energy $$E$$ has a mass $$m=E/c^2$$, again no conversion necessary.

What you can do is to convert matter at rest with total mass $$m$$ into photons with total energy $$E=mc^2$$. In principle you could do that by using antimatter, so if your mass consisted of 0.5 kg of electrons and 0.5 kg of positrons then it would work.

• Or a black hole, although you might need to be prepared to wait a while. – Taemyr Dec 11 '19 at 13:25
• at least your black hole isn't as efficient as OP claims .. the top rate observed and calculated is about 42...43% . the remaining mass "falls into" the black hole – eagle275 Dec 11 '19 at 14:42
• @eagle275 You scoop up the remaining mass as Hawking radiation and repeat the process. That's where the waiting a while bit comes into play. – John Dvorak Dec 11 '19 at 15:12
• @eagle275 - I think what is meant is the black hole evaporation by Hawking radiation: note the comment "might need to be prepared to wait a while" :) – The_Sympathizer Dec 11 '19 at 15:21
• That is why I suggested 0.5 kg of electrons and 0.5 kg of positrons. Those decay almost (99.99%) exclusively to photons. Protons and anti-protons are messy. – Dale Dec 11 '19 at 17:53

One way to convert a kg of matter to other forms of energy is to bring along a 1 kg mass of antimatter and have them annihilate. However I have always felt that this is a bad method, because it is well known that antimatter is very rare, and very difficult to produce. So as an answer it merely shifts the problem into "how do I obtain or manufacture 1 kg of anti-matter?" Most methods to make antimatter will be very inefficient in practice.

So I prefer the much cleaner gravitational methods. The following is a beautiful piece of gravitational physics which deserves to be more widely known.

We dangle a mass using a massless string. According to General Relativity, for a static source, if you stand at $$B$$ and lower an object to $$A$$ using a massless string, then the energy extracted at $$B$$ is given by $$W = \left( 1 - \frac{e^{\Phi(A)/c^2}}{e^{\Phi(B)/c^2}} \right) E_B$$ where $$\Phi$$ is gravitational potential and $$E_B$$ is the energy the object has when at $$B$$. For someone at $$B$$ this would be the rest energy of the object.

Now we take $$B$$ to be at infinity, and use the Schwarzschild metric. Then $$\Phi(r) = (1/2) c^2 \ln(1 - r_s / r)$$ and we find $$W = \left( 1 - \sqrt{1 - r_s/r} \right) m_0 c^2.$$ To repeat, this is the energy (work) gathered at $$B$$ when someone stands there lowering the object using a massless string. Now lower the object all the way to the event horizon of a black hole: the place where $$r = r_s$$. In that case $$W = m_0 c^2$$. The entire rest energy of the object has been seamlessly and smoothly converted into work. The string then breaks and the lowered object enters the black hole. And here's the kicker: the mass of the black hole does not then change! The black hole doesn't get any of the rest energy of the lowered object; it all stays at $$B$$.

So that is how I would convert a kg with 100% efficiency.

Processes somewhat like this one occur in the accretion discs (i.e. orbiting material) around black holes, and the energy goes to the form of kinetic energy of the material in the disc. In other words, the disc gets very hot. There is no 'string' of course, but the collisions as matter slowly falls on a spiral orbit perform the function of such a string.

Finally, a thought which follows from the above. For any given object, including all the ordinary objects sitting around you as you read this, their rest energy can be regarded as the energy that would have had to be provided to raise them from the horizon of a black hole to the place where they are now located.

This is an equation that only tells half the story - For a more full picture of the energy we should use the equation; $$E= mc^2 + \hbar \lambda$$

For anything with mass; the last term can be approximated to 0. For anything without mass, the first term is 0.

This means that should you take some mass; and turn it into "not mass"; that energy must be converted into the last term - that last term is the energy of light..

This is how power stations etc work; (from a 10,000ft view) they're able to convert a small percentage of the fuel into light; and that light is then absorbed by water which heats up and turns to steam.

• (1) Your equation is wrong. You can use $E = \sqrt{m^2 c^4 + p^2 c^2}$ if you like, but $E=m c^2$ remains correct as a statement about rest energy, which is precisely what it is. (2) Power stations don't convert fuel into light, they convert the energy stored in molecular bonds into kinetic energy of molecules. – Andrew Steane Dec 13 '19 at 9:01
• @AndrewSteane sorry - should've said "nuclear" power stations. my mind was thinking what my fingers weren't typeing – UKMonkey Dec 13 '19 at 10:38
• OK; for nuclear power stations they convert the nuclear stored energy mainly into kinetic energy of neutrons and kinetic energy of the product nuclei; only a little goes to gamma rays I think. – Andrew Steane Dec 13 '19 at 16:26