# If the mass of an object or system is a measure of its energy content, multiplied by speed of light squared…?

If the mass of an object or system is a measure of its energy content, multiplied by speed of light squared, does it mean that the total potential energy of the rest mass of object or entire system can only be realized when velocity is similar to speed of light?

What I am looking for is a clear and simple explanation on the exact purpose 'c square' serves in this equation. I need to have a basic explanation suitable for a 10 year old without too many complicated Special Relativity equations.

What he understands is that $E = mc^2$ oblivious of the units of measurement.

Is this the way the equation was derived?

Work (W) is a measure of the energy (E) expended in applying a force (F) to move an object.

And, Work (W) is defined as force (F) times distance (D).

So $E = W = D * F$

And Force (F) is Mass (M) times Acceleration (A).

So, $E = D * M * A$

And Acceleration (A) is the rate at which the velocity (V) of the object changes over time.

So, A = Change in velocity / Time taken (T).

And V = Distance (D) / Time (T).

So, $Energy = Distance \times Mass \times ((Distance/Time)/Time)$, viz.,

$E = (M x D^2) / T^2$ or $Energy = Mass \times (Distance/Time)$ squared.

Since $D/T$ = Velocity or Speed

The potential Energy (E) of an object is made up of its mass times its velocity squared.

And since the object is at rest, and we are trying to obtain 'Potential Energy' (any type of stored energy), I guess we can mention here the 'Potential Velocity' which cannot exceed the universal constant 'c' representing 'speed of light'.

So, $E = mc^2$.

Is this understanding correct?

• possible duplicate of Does $E$ really equal $mc^2$? – Brandon Enright Aug 17 '14 at 17:16
• @BrandonEnright I don't think so - this question deals more with derivation than actual validity. – Dave Coffman Aug 17 '14 at 17:38

I'll make a few general comments first and then try to answer "What I am looking for is a clear and simple explanation on the exact purpose 'c square' serves in this equation."

First off, while $E=mc^2$ is still widely used in the media, it is not widely used amongst physicists. That equation uses the concept of relativistic mass, which is more than a bit outdated. A much better expression is $$E^2 = (mc^2)^2 + (pc)^2$$ where $m$ is the intrinsic mass (also called rest mass, but nowadays often just called "mass") and $p$ is the relativistic momentum. I'll be using the above expression as the basis for my answer.

The second general comment is that the second part of your question uses concepts from Newtonian mechanics. You can't do what you did. You cannot mix and match Newtonian mechanics and relativity.

Finally, you wanted an answer that is amenable to a ten year old. That means I can't show the derivations, I can't use anything more complex than the expression I wrote above. (And perhaps even that is too much for a ten year old.)

Suppose you observe an object at rest. In this case, the above equation reduces to $E_0=mc^2$. This is called the rest energy of that object. In a very real way, this says that mass is a form of energy. There are many different forms of energy; mass is but one of them. We can convert energy from one form to another. For example, if you turn on a gas stove, the chemical potential energy in the gas is converted into heat and light through combustion.

So can we change mass into another form of energy? The answer is yes. The mass of the natural gas and oxygen is a tiny, tiny bit more than is the water and carbon dioxide that result from the combustion. The change is very small in this case, about one part in ten billion. The change in mass is more significant in nuclear explosions and nuclear power plants. It's still a rather small fraction of the mass that is converted to heat and light. The conversion is complete when matter and anti-matter annihilate one another.

The reverse can happen, to. The big bang, very large stars, and very powerful particle colliders can create matter from energy.