Why is $E=mc^2$ true when the units are arbitrarily defined?

Question

I am a little confused on how $E=mc^2$ holds true, when a kilogram is defined arbitrarily as:

$$1 \ \mathrm{kg} = \frac{(299\ 792\ 458)^2}{(6.626\ 070\ 15\times10^{-34})(9\ 192\ 631\ 770)}\frac{h\Delta \mathscr{v}_{\mathrm{Cs}}}{c^2} \ ;$$

and a joule is defined as "the work done when the point of application of 1 MKS unit of force [newton], moves a distance of 1 meter in the direction of the force." The meter is defined in terms of the size of the earth.

Shouldn't the ratio between the two be some random constant? How can you get $c^2$ as a ratio, when both units are defined arbitrarily?

Answer 1

The speed of light, $c$, is also always expressed in units that are fundamentally "arbitrary", that is, they're a matter of convention. So there's no problem, if you choose other "arbitrary" units to express your masses and your energies, the ratio $E/m$ will still have units of speed squared!

A way to see this, is to express the quantities via the more general technique of "dimensional analysis", which doesn't commit to any specific choice units. We know that: \begin{align*} \text{Energy} &= \text{Force}\times\text{Length} \\&= \text{Mass}\times\text{Acceleration}\times\text{Length} \\&= \text{Mass}\times\text{Speed}\times\text{Time}^{-1}\times\text{Length} \\&= \text{Mass}\times\text{Length}^2\times\text{Time}^{-2} \\&= \text{Mass}\times\left(\frac{\text{Length}}{\text{Time}}\right)^2 \\&= \text{Mass}\times\left(\text{Speed}\right)^2 \end{align*}

which I have written in this jarring notation just for the sake of those less familiar with the shorthands for the basic dimensions (MLT - Mass, Length, Time).

Anyway, the above shows you that mass is related to energy by a factor of speed squared, and that is true regardless of the choice of units.

A separate thing is, that in the rest frame of a mass, we have exactly $E=mc^2$ with $m$ the rest mass. There are many, many posts here about why that equation isn't always true if $m$ is not at rest, and various historical aspects leading to that equation still being so famous. See for example this post.

Answer 2

Assume that the price of an apple is \$3. You buy 120 apples. Therefore you owe $360, which you calculate as follows:

(Quantity) $\times$ (Price) = (Money owed)

(120 apples) $\times$ (\$3 per apple) = $360

I prefer to measure money in cents, not dollars. I also prefer to measure apples in dozens. So I think of the price as 3600 cents per dozen. And I calculate the money I owe as follows:

(Quantity of apples) $\times$ (Price) = (Money owed)

(10 dozen apples) $\times$ (3600 cents per dozen) = 36,000 cents

Now $360 is exactly the same as 36,000 cents, so both equations give the same answer. My arbitary choice to measure apples in single-apples or in dozens, and my arbitary choice to measure prices in dollars or in cents, do not affect what my equation is telling me.

Answer 3

You seem to be confusing the quantity with its numerical value.

The variables $E$, $m$ and $c$ are dimensionful quantities and not just numbers. Therefore, they are independent of the units used. For example, $1\,\text{km} = 1000\,\text{m}$ as lengths even though their numerical values, $1$ and $1000$, are different.

The equation $E=mc^2$ is a physical statement relating dimensionful quantities. It holds regardless of the choice of units. It's just that when the equation is used in practice, such as entering it into a typical calculator, the numerical values are often taken, and in SI units, the numerical value of $E$ is $299792458^2$ times the numerical value of $m$. But the units are always there in the underlying equation. So the ratio will be $c^2$, but the ratio of the numerical values need not be $299792458^2$ if other units are used.

In addition, the meter is no longer defined in terms of the size of the Earth but rather in terms of the speed of light.

Answer 4

Remove all the units, so that $c=h=G=1$. You can do this in the same way that you can switch from metric to Imperial/American units and back.

The unit of energy is $[E]=[M][L]^2[T]^{-2}$, for the units of mass, length, and time $[M]$, $[L]$, and $[T]$ This is equivalent to force times distance, $[F][L]$, where $[F]=[M][L][T]^{-2}$.

Then, if you take a mass and want to get to the corresponding energy, you multiply by the “fundamental” velocity $c$:

$$[M]\times [c]^2=[M]\times\big([L][T]^{-1}\big)^2=[M][L]^2[T]^{-2}=[E].$$

Switch into any coordinate system you like, and the implicit $c^2$ in there becomes apparent:

$$Mc^2=E.$$

The speed of light is fundamental for a number of reasons. You can verify this experimentally in nuclear physics by measuring how the energy released in a reaction corresponds to the mass defect between the reactants and products (with a constant of proportionality, of course, of $c^2$).

Answer 5

Two important points that answer your questions:

1. The standard system of units is made to be consistent. Energy is units of $J = N \cdot m = \frac{kg \cdot m}{s^2} \cdot m$. Thus $E = mc^2$ can be expressed in units of $\frac{kg \cdot m^2}{s^2} = kg \cdot (\frac{m}{s})^2$.

2. The underlying principle is that Energy is proportional to mass times the speed of light squared. If you wanted, you could make up your own system of inconsistent units (or use "watt-hours" for energy) and the equation would still be true up to a coefficient of proportionality.