Why is there no constant factor in the equation $E=mc^2$?

Question

As I understand, the metric system starts with an arbitrary weight, and calls it $1\, \mathrm{kg}$, then the volume of this weight in water gives the meter. And then the energy is defined based on the meter, second and kilogram.

My question is: given that the relation between meter, kg and joule relies on the density of water (or did so originally), how can we possibly tie them again with $E=mc^2$ by adding the speed of light? The density of water seems completely uncorrelated to the speed of light.

Intuitively I would have added an extra constant to Einstein's equation.

Answer 1

The assumptions on the SI units are not correct.

The kilogram isn't 'defined by an arbitrary weight', it's defined using Planck's constant since 2018:

The kilogram is defined by setting the Planck constant h exactly to 6.62607015×10−34 J⋅s (J = kg⋅m2⋅s−2), given the definitions of the metre and the second

Hence, the kilogram is now essentially defined in terms of the second and the metre. Though it did used to be defined by a platinum-iridium cylinder (and before that a pure platinum cylinder), whose mass was (as close as possible) to the absolute weight of a volume of pure water equal to the cube of the hundredth part of the metre, at 4 Celsius.

The metre is not a derived unit in SI, it's one of the base units, defined in 1983 as:

The metre is the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second.

I'm not too sure where the water based definition you mention comes from, but if you can elaborate, I can see if I can see where it fits into the story for you.

You are correct that the Joule is based on the kilogram, metre and second (since $J = kg \cdot m^2/s^2$). And if you did a dimensional analysis on the Joule, you would find it based on the above combination of the units you mention, it doesn't necessarily mean whatever energy you are measuring (in this case some particle, i presume) is correlated to that cylinder of platinum or any other thing used to make a definition.

Put simply, a unit is just shorthand for saying "is equal to this many of a quantity". For example we can define our own unit, the Dave, as equal to 63 times the average mass of a male platypus. I weigh 1 Dave (yeh, I need to lose some mass...). It doesn't mean that me and the average male platypus are in anyway correlated.

To answer, the 'extra constant' bit, what I think you're asking is:

Why is there no constant of proportionality used in $E=mc^2$ to relate the Joule to kilogram, metre, and second, since their definitions seem unrelated?

Feel free to correct me if I misunderstood. Constants of proportionality (and other constants) are used, as a way to get a certain unit out of an equation. $E=mc^2$ is just an equation, it doesn't require any constant until you choose a system of units to work in. Let's try adding the constant you mentioned, we will call it, k, so now the equation is $E=kmc^2$. For getting an answer in Joules, from kilograms, metres and seconds, k would be 1 (it is how the Joule is defined). If you wanted an answer in milliJoules ($1/1000$ Joules) from kilograms, metres and seconds, k would now have to be 1000, to balance the units. If you wanted an answer in milliJoules, from grams ($1/1000$ Kilograms), metres and seconds, k would be back to being 1, since the alterations made to both sides balance.

A good real world example of this, is natural units, where c=1 (whereas, in SI it's 299792458 $m/s^2$, but here the value and units are adjusted to output an energy in electronVolts, instead of Joules), and Energy and mass are measured in eV, but the equation is still $E=mc^2$. All the equation says, regardless of any constants, is it takes this much energy to create this much mass from the vacuum.

Hope that helps, feel free to ask more if I was unclear or didn't cover some of your questions - either units or platypus related.

Sources:

SI Unit Definitions

Platypus Mass

Answer 2

Short answer: there are no additional factors in this formula because unit of measure for energy is consistent with units of measure for mass, time and distance.

Suppose you decided to create your own system of units and let's forget about relativity theory so far.

You can choose arbitrary unit for mass. You can choose arbitrary unit for time. You can choose arbitrary unit for distance.

But after that you can't arbitrarily choose unit of measure for speed. You'd better use (unit of distance)/(unit of time). Otherwise formulas in your system of units will have some weird constant factors in it.

And you can't use arbitrary choose unit of measure for energy. You'd better use (unit of mass) * (unit of speed)^2. For the same reason.

And in this designed from scratch system of units the classic formula for energy $E = m * v^2/2$ will hold. No additional factors appear in this formula, because you have chosen units of measure for speed and energy corresponding to (arbitrarily selected) units for mass, time and distance.

Similarly, numeric values of mass and the speed of light will be different in your system of units, but the equation $E = m * c^2$ will hold.

UPDATE (answer to a question in comment).

So, if formulas $E = m * v^2/2$ and $E = m * c^2$ are correct with one system of units of measure, these formulas would be correct with all other systems of units of measure if the unit of measure of energy "corresponds" to units of measure of time, distance and mass. This is just a pure math. From your question I guess that this is clear. Let me know if it is not. Please note, that these two formulas have very different physical meaning, but when switching from one units of measure to another the math is the same.

The classic formula $E = m * v^2/2$ is correct with all units of measure. It does not matter if the unit of distance was defined as some fraction of distance from pole to equator of some planet or as a distance between ears of some species living on it.

The second formula ($E = m * c^2$) can be formulated in units-of-measure-independent form. It says: there is some universal speed (speed of light), and the total energy of any object is twice as much as you would get if you use the classic formula of kinetic energy to calculate the kinetic energy of that object moving with our universal speed.

Why "twice as much" - it's a different question. It's relativity theory. But as long as there are no weird constant factors depending on units of measure in the formula $E = m * v^2/2$ there would be no such factors in $E = m * c^2$.

Answer 3

Well, $E=mc^2$ doesn't mean that a Joule is equal to a kilogram times (the speed of light) squared. Actually, if a Joule is defined to be $$\operatorname{Joule} =\operatorname{Newton}\cdot \operatorname{meter} = (\operatorname{kg}\cdot \frac{\operatorname{meter}}{\operatorname{second}^2})\cdot \operatorname{meter}=\operatorname{kg}\cdot \left( \frac{\operatorname{meter}}{\operatorname{second}} \right) ^2$$ Then by plugging 1 kilogram to Einstein's equation, you get that rouphly

$$E = 1\operatorname{kg}\cdot (3\cdot 10^8 \frac{\operatorname{meter}}{\operatorname{second}})^2 = 9 \cdot 10^{16} \operatorname{kg}\cdot \left( \frac{\operatorname{meter}}{\operatorname{second}} \right) ^2=9\cdot10^{16}\operatorname{Joule}$$ Meaning the energy you need to create from nothing a kilogram is that much of energy.

Constants come in handy when we specifically wanted to create a relation that when you plug in some units you will get other units.

For example, if Newtonian gravity says that $$F\propto \frac{m_1 m_2}{r^2}$$ That means that on the left side we have the units $\operatorname{kg}\cdot \frac{\operatorname{meter}}{\operatorname{second}^2}$ and on the right we have $\frac{\operatorname{kg}^2}{\operatorname{meter}^2}$. If we want the units to match we need a constant with the units of $\frac{\operatorname{meter}^3}{\operatorname{kg}\cdot \operatorname{second}^2}$ and give it a value that really gives the force between two masses of a kg one meter apart (in Newtons). In $E=mc^2$ you don't need such a constant to balance the units, this formula is derived no matter which units you use.

This concept if constants in physical relations as a way to work with a specific unit system is actually what makes the equations of electromagnetism differ depending on which units you use. In the SI system (meters, kilograms, seconds), one likes to work with the "Coulomb" unit for electric charge, but then needs a constant in Coulomb's law. In the cgs system of units (centimeter, grams, seconds), one defines the electric charge to be such that you don't need a constant. I hope I didn't confuse you with the last paragraph. You are welcome to ask more questions :)

Answer 4

This is completely wrong. The international system of units SI

Defines :

Time: $1$ second equals:

The duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.

Length: $1$ meter equals:

The distance travelled by light in vacuum in $(1/299792458) \times$ second.

Mass: $1$ kilogram equals:

The kilogram is defined by setting the Planck constant h exactly to $6.62607015×10^{−34}\,\mathrm{J\cdot s}\;(\mathrm J = \,\mathrm{kg⋅m^2⋅s^{−2}}$), given the definitions of the metre and the second.

So we are in the twenty-first century, and the definition of the kilogram depends on the Planck constant, it ain't simple, and it takes care of densities and such.

In addition, the m in $E=mc^2$ is not constant anyway, because it is velocity dependent. It is called the relativistic mass and is no longer used in physics, as it is useful to calculate the inertial mass for velocity nearing the speed of light, not the mass of the particle and the weight we find in our scales. One uses the invariant mass given by the length of the four vector:

$$\sqrt{P\cdot P}=\sqrt{E^2-(pc)^2}=m_0c^2.$$

The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.

Answer 5

Historically, the kilogram was defined in terms of the metre and not the other way around. Originally, the metre was defined as "one ten-millionth of the shortest distance from the North Pole to the equator passing through Paris." Based on this definition of the metre, the kilogram was defined as the mass of one litre of water. So the kilogram was not just some arbitrary mass.

Of course, these definitions are historically where they came from. Today, these unit are defined in more sophisticated ways, as explained in some of the other answers provided here.

About the lack of constants in the famous equation $E=m c^2$. Remember, this equation is not proposed as a definition of units and should not be interpreted that way. Instead, it gives the relationship between mass and energy. It shows that one is proportional to the other. The role of the $c^2$ is to match the dimensions. In that sense, a numerical dimensionless constant factor would be meaningless. In fact, arbitrary numerical factors would appear when one assigns various different units of measure to the different quantities in the equation.

That does not mean that one cannot use the equation for the definition of units, by relating a new unit of measure to existing units. In terms of the SI units, the Joule is related to the kilogram and one metre-per-second for velocity. One could in principle use this equation to define a new unit of energy associated with one kilogram.