# Can matter be annihilated to form kinetic energy

I don't know much about conversion of energy and all, but I had a simple question: Can the energy released by converting matter into energy be used to set other particles in motion (increase their kinetic energy)? If yes, does this imply that particles can travel faster than light? $$E=E_k$$ $$mc^2=\frac12mv^2$$ $$c^2=\frac12v^2$$ [assuming mass destroyed is equal to mass not destroyed (which will receive the kinetic energy)] $$2c^2=v^2$$ $$\sqrt{2c^2}=v$$ $$v=\sqrt2c$$ $$v>c$$

• Your formula for kinetic energy doesn't take relativistic effects into account. – raptortech97 Dec 6 '14 at 17:03
• How can the mass that has been annihilated, still be around to receive the kinetic energy? The mass of the lightest particle after the annihilation is many times greater than the mass annihilated. – LDC3 Dec 6 '14 at 17:06
• Possible duplicates: physics.stackexchange.com/q/91501/2451 and links therein. – Qmechanic Dec 6 '14 at 17:24
• Kinetic energy isn't a "stuff", it is a property of an object or system. I know, I know. We talk about energy using language that treats it as a thing, but you have to keep in mind that it is not something that you can isolate and say "Ta da! I have a beaker of energy!". – dmckee --- ex-moderator kitten Dec 6 '14 at 17:44

Yes, and this happens all the time in beta decay.

In beta decay a neutron turns into a proton, an electron and an anti-neutrino. If you add up the masses of the proton, electron and anti-neutrino you get a total mass slightly smaller than the mass of a neutron. This missing mass has been turned into the kinetic energy of the electron and anti-neutrino.

For more on this have a look at my answer to What keeps mass from turning into energy?.

However your suggestion about accelerating a particle to FTL speeds is flawed because you're using the wrong expression for the kinetic energy. The expression $KE = \tfrac{1}{2}mv^2$ is a low energy approximation, and if you're considering speeds approaching the speed of light you need to use the relativistic equation for the total energy (including the rest energy):

$$E^2 = p^2c^2 + m^2c^4 \tag{1}$$

where $p$ is the relativistic momentum and is given by:

$$p = \frac{mv}{\sqrt{1 - v^2/c^2}} \tag{2}$$

The total energy, $E$, is equal to $mc^2$ plus the energy from the amount of matter you're proposing to convert to energy, $M$, so:

$$E = mc^2 + Mc^2$$

In this case you're suggesting we annihilate an amount of matter equal to the mass of our particle so $M = m$ and therefore $E = 2mc^2$. Putting this in for the energy in equation (1) we get:

$$4m^2c^4 = p^2c^2 + m^2c^4$$

and this simplifies to:

$$p = mc\sqrt{3}$$

If substitute equation (2) for $p$ we get:

$$\frac{mv}{\sqrt{1 - v^2/c^2}} = mc\sqrt{3}$$

And this gives:

$$v = c\sqrt{\tfrac{3}{4}} \approx 0.866c$$

I won't prove it here, but you'll find it takes an infinite amount of energy to get your particle to the speed of light, so however big you make the amount of matter, $M$, you're annihilating you'll never get your particle up to the speed of light.