Antimatter collision - Energy Released

Question

I think it's about time for me to ask this question, as I've been contemplating this for a while.

$$E = mc^2.$$

This is Einstein's most famous equation. But what does it mean?

On my own, I had to think a lot about it, since no one in my close surroundings can give me good answers.

To me it means that the energy contained inside an object is proportional to the mass. Thus mass is energy.

And if that object gets annihilated then its mass will be converted back into pure energy, such as light.

Isn't that what happens when Antimatter collides with our ordinary matter? The masses cancel out and they turn into light (photons)?

Here is a little fun I had last year in my Grade 9 Science class.

100g of H.
100g of anti-H.

They collide, and their masses, convert into energy. E = mc^2.
E = (100)*((299792458)^2);
E = 8.987551788E18 joules.

So that's the amount of energy produced.

Is this correct? Did I use the wrong equation? What else can E=mc^2 be used for. If this is correct, boy that's a lot of energy contained in a hundred grams of Hydrogen.

Answer 1

You're on the right track; there are just a few things I would add.

The first is that you've made a slight mistake in your calculation. Remember that a joule is 1kg m^2 / s^2. 100g of H is 0.1 kg; and since you have H and anti-H there are 0.2kg total. The total mass corresponds to an amount of energy

$$ E = (0.2 kg) \times (3 \times 10^8 m/s)^2 = 1.8 \times 10^{16} J $$

Now about $E = mc^2$ itself. Before Einstein, we knew that objects could have energy due to motion (kinetic energy), due to interaction with external fields (gravitational and electrical potential energy), due to energy stored in internal forces (e.g. harmonic potential energy in a spring), due to the kinetic energy of its microscopic constituents (i.e. internal energy, proportional to temperature, etc.). Einstein added something to this list - any object with mass has a rest energy proportional to its mass given by $m c^2$. In any process where mass is conserved (as of the year ~ 1900, all known physical processes), this energy is locked away and undetectable. Einstein's work showed that it must exist, and equivalently that processes violating conservation of mass could be possible as long as energy was still conserved.

Another important note: Before Einstein, the energy of some object (ignoring all the potential energies and internal energy) is

$$ E = \frac{1}{2} mv^2 $$

After Einstein, you don't just add $m c^2$, i.e.,

$$ E = mc^2 + \frac{1}{2} mv^2 $$

but instead

$$ E = \frac{1}{\sqrt{1 - v^2/c^2}} m c^2 $$

At first this formula looks extremely different from $mc^2 + 1/2 mv^2$, but it is equivalent at small velocities. For a small number $\epsilon$,

$$ \frac{1}{\sqrt{1 - \epsilon}} \approx 1 + \frac{\epsilon}{2} $$

which you can check with a calculator.

In fact, there is a much better expression for the energy than $E = \frac{1}{\sqrt{1 - v^2/c^2}}m c^2$. This is

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

Here $p$ is the momentum of the object. For a particle with mass $m$, this is

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

The reason this expression works better is that there are many particles, in particular the photons that make up light, which are massless but still carry momentum. A photon with wavelength $\lambda$ has momentum $2 \pi \hbar / \lambda$. Using the above, its energy is

$$ E = \sqrt{m^2 c^4 + p^2 c^2} = p c = 2 \pi \hbar \nu = \hbar \omega $$