# Which units has the relation $E=mc^2$? [closed]

Fast everyone knows the relation $E = mc^2$ but I still ask me, which units the relation has.

$c$ is the velocity of light. It means, a constant, isn't it? We therefore just have the relation $E = \text{constant} \times m$

Ok, nice to know the mass and the energy are directly proportional. But the constant is only a constant; with another unit system, it would be another one...

What is wrong to my reflection?

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## closed as unclear what you're asking by Dilaton, Qmechanic♦Jul 6 '13 at 16:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Hi Olivier. Welcome to Physics.SE. I can't get a grip on your question. What's your question? Are you asking, "Why does $c$ have an unit as it's a constant?" ? – Waffle's Crazy Peanut Mar 27 '13 at 8:02
It's difficult to understand what you're asking. In SI units, $m$ is measured in $\mathrm{kg}$ and $E$ is measured in $\mathrm{kgm^{-2}s^{-2}}$, so whatever constant relates them, it must have units of $\mathrm{m^{-2}s^{-2}}$. Since $c$ is a speed ($\mathrm{ms^{-1}}$), $c^2$ has units $\mathrm{m^{-2}s^{-2}}$, so it's all good. (If this answers your question, let me know and I'll post it as an answer.) – Nathaniel Mar 27 '13 at 8:03
@CrazyBuddy d'oh, you're right, that's what they call a thinko. I meant $E$ has units $\mathrm{kgm^2s^{-2}}$, $c$ has units $\mathrm{ms^{-1}}$, $c^2$ has $\mathrm{m^2s^{-2}}$, so $mc^2$ has $\mathrm{kgm^2s^{-2}}$, the same as $E$. Sorry, Olivier, for any confusion I might have caused! – Nathaniel Mar 27 '13 at 8:25
Thanks Nathaniel, it was my question. Because the energy is given in joules and the joule depends of kg, meters and seconds, the constant cannot be another one at it is. I had a problem as I though that the energy unit doesn't depend of other ones. – Jean Louis Mar 27 '13 at 12:20

Yes, the proportionality constant $c^2$ is a dimensionful constant which means that its numerical value depends on the choice of units. In SI, it's about $10^{17}\,{\rm{ m^2/s^2}}$. However, theoretical physicists routinely use units in which simply $c=1$ which simplifies many equations.

If a quantity has different values in different unit systems, it doesn't mean that the numerical value is irrelevant. For example, one's height may be 5.9 feet or 180 cm – different numbers – but it's still different from a human whose height is 4 feet = 122 cm. The point is that the information about the units has to be attached to the number and the combination "number, unit" preserves all the physical information and it is independent of the units.

$E=mc^2$ says that one kilogram of matter carries the latent energy of $10^{17}$ joules. one may express the same sentence in many other units. For example, one pound of mass carries $5\times 10^{23}$ ergs of latent energy or so. One may use many conventions but this fact doesn't mean that there is no information in the sentence.

The only wrong thing about your reflexion is that you seem to think that something is wrong.

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I guess you mean $10^{17}\mathrm{m^2/s^2}$ – Andre Holzner Mar 27 '13 at 20:32

The point is that the constant $c$ doesn't only appear in this equations but also in a whole bunch of other formulas in the theory from which it's derived from. Specifically, $c$ appears in the concellation $\frac{x'(t)}{c}$, which has to be unitless, as you e.g. can see from

$$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}.$$

Hence in the theory, $c$ has to have units of length devided by units of time. You're right that it's possible to choose different units for the involved variables. E.g. there are unit systems where [length]=[time]=s and there $c$ is unitless, hence the unit of mass is the unit of energy.

"Fast everyone knows the relation E = mc² but I still ask me, ..."

Wunderschöne deutsche Grammatik hast du da ;)

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