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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.

BioPhysicist
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Manu de Hanoi
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  • Also what is the constant you propose adding? That might help make the question clearer. – BioPhysicist Oct 21 '19 at 06:04
  • Dear all, I feel all the answers arent addressing my issue. I'm not asking wether units are consistent in the sense that they boil down to the same sub units but rather that we have units that were defined initially around the density of water (and then redefined but it still matches with water) and that in E=mc2 we are matching mass & energy around the unrelated speed of light. – Manu de Hanoi Oct 22 '19 at 01:14
  • @ManudeHanoi I don't quite understand your confusion. The equation e=MC2 is no different in principle to the equation for KE =1/2MV2- in both equations the physical quantity of energy is the product of a mass and a velocity squared. The choice of units is irrelevant. – Marco Ocram Oct 22 '19 at 08:50

5 Answers5

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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

Epideme
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  • if you define 1 Joule as equal to platypus times mass. and then comes e=mc2 that can be used to define 1 joule as the energy contained in a mass of 1kg/c2 you end up with having 2 redundant definitions of energy implying that 1 platypus =1/c2 . My question is how could platypus(=density of water) and inverse square of the speed of light be related – Manu de Hanoi Oct 22 '19 at 01:32
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    @ManudeHanoi The point that everybody is telling you is that the numeric value of $c^2$, in the units you are using, depends on the units you are using. If you change what "platypus" means, that indirectly changes what number $c^2$ is equal to, in platypus units. – knzhou Oct 22 '19 at 05:02
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    @ManudeHanoi This is like being amazed that the height and width of a wooden square are always the same in any units, $h = w$, even though you could have measured them in, say, platypus lengths. What do platypuses have to do with the sides of a wooden square? Nothing, but still $h = w$. – knzhou Oct 22 '19 at 05:04
  • @knzhou, if so, can you prove it ? Can you prove that 1 joule =1kg(1m/s)^2 is equivalent to 1 joule = energy of (1/c^2) kg ? – Manu de Hanoi Oct 22 '19 at 08:14
  • Hey Manu -- You can have multiple compatible equivalencies of a unit, and you are right, this would make them equivalent to each other. But that doesn't make the phenomena they are based off of related to one another. The numeric values in the equations (in this case, c) change to fit the new units. Whilst c=299792458m/s^2 for when you work in Joules, Metres, Seconds... it might equal an entirely different figure for when working in platypus units. If it helps, and I'm a bit reluctant to say it like this, see the c as including the constant of proportionality you mentioned. c is only... (1/2) – Epideme Oct 22 '19 at 09:36
  • ...c is only 299792458 when working in m/s^2. It is a different numeric value if you want to work in say parsec/hour^2, or natural units etc. This is what keeps those various unit definitions equivalent.

    The h=w thing is a bit different, thats more due to the properties of a square than anything related to the units.

    As for the proof, Ofek Gillon did the proof above for the number of joules required to make a 1kg mass from the vacuum. (2/2)

     – Epideme Oct 22 '19 at 09:41
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    @ManudeHanoi For the proof, see any special relativity textbook. You are correct that it is a nontrivial fact that there is no numeric coefficient in $E = mc^2$. But this has nothing to do with the unit system: if there were a coefficient, it would be the same in every reasonable unit system. – knzhou Oct 22 '19 at 16:13
  • As a side point, I am so happy that Platypus units have taken off so well – Epideme Mar 06 '20 at 14:23
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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 $E = m * c^2$.

lesnik
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  • Can you reconcile the fact that : 1) 1 joule is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre AND that: 1)1 joule is the energy in 1/c2 kg ? Yes the units match but I see no correlation between the speed of light and the acceleration of 1 newton over 1 meter – Manu de Hanoi Oct 22 '19 at 02:00
  • @ManudeHanoi I tried to answer your question and updated my original post - the answer did not fit into comment. – lesnik Oct 22 '19 at 07:48
  • I have no problem with and never mentionned the kinetic energy definition, I dont know why your edit mentions it. My problem is that we have defined the joules using 2 definitions, one that involves the density of water and one that involves the speed of light. And these 2 constants are unrelted – Manu de Hanoi Oct 22 '19 at 07:57
  • @ManudeHanoi Note, that definition of Newton is that this is the force which is required to accelerate 1 kg with 1 $(meter/sec)^2$. It also involves kilogram, so the first of your definitions of Joule (1 Newton * 1 meter) also involves density of water! – lesnik Oct 22 '19 at 08:08
  • both definitions involve water and one involves the speed of light on top of it – Manu de Hanoi Oct 22 '19 at 08:12
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    @ManudeHanoi So, density of water is involved in both definitions, so there are no problems with it? Ok. And only one of these formulas mentions speed of light. This second definition ($E=mc^2$) brings some new physical meaning, it establishes some natural correspondence between mass and energy. It tells that there is a natural unit-of-measure-independent way to measure energy in kilograms or mass in Joules or electron-vaults or whatever. Why the factor between mass and energy is exactly $c^2$ - well, this is subject of relativity theory. Absence of UOM-dependent constants in it - pure math. – lesnik Oct 22 '19 at 08:29
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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 :)

Ofek Gillon
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  • Can you reconcile the fact that :
    1. 1 joule is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre

    AND that: 1)1 joule is the energy in 1/c2 kg ? Yes the units match but I see no correlation between the speed of light and the acceleration of 1 newton over 1 meter

     – Manu de Hanoi Oct 22 '19 at 01:59
  • I think what the OP is really asking is why is it $E=mc^2$ and not, say, $E=\frac{1}{2}mc^2$, which would be similar to Newtonian kinetic energy which has the same kinds of variables on both sides of the equation. – Mark H Oct 22 '19 at 03:13
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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.

Ruslan
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anna v
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    dear Anna, the constants used to define the units were choosen so the side of a cube of 1000 kg of water still measures 1meter. Using arbitrary constants like "The duration of 9192631770 periods of the radiation" instead of the historical arbitrary choice of water doesnt change my question – Manu de Hanoi Oct 22 '19 at 01:37
  • @ManudeHanoi Of course it changes it. The new definitions do not depend on density, this was the correction for the desnity,getting a stable to external influences definition. Worth reading the history of SIhttps://usma.org/si-unit-definitions – anna v Oct 22 '19 at 03:49
  • it doesnt matter Anna, you've replaced the arbitrary constant of density of water with the arbitrary constant of "9192631770" and "1/299792458" in either case, you end up with 2 definitions of the joule, one based directly on the above constants (1joule=1kg x (1m/s)^2) and one based on the speed of light (1joule = energy of (1/c2) kg). So in order to reconcile the 2 definitions you'd have to somehow prove that they are the same mathematically. And I can't do that, can you ? – Manu de Hanoi Oct 22 '19 at 08:08
  • @ManudeHanoi No, these definitions are not mathematically the same! You can not establish natural unit-of-measure-independent correspondence between mass and energy using the first definition. Second definition is not actually a mathematical definition, it says "energy of $1/9 * 10^{-16}$ kg" - but how much is it? Would it be enough to lift some object, or heat some water? It's more like not a definition of Joule, but a statement, that if object of mass $m$ decays into massless objects, an energy $m*c^2$ would be released. – lesnik Oct 22 '19 at 09:10
  • Think a bit: there used to be an independent unit, the foot. Obviously it is connected with the human foot. Whose foot? my foot, your foot? An averaged foot?https://en.wikipedia.org/wiki/Foot_(unit)#Historical_origin . It is obvous that the older defintions have an error, dependent on the width of the average value of a male foot : thats why it was different from city to city. To request that the old foot measure can be connected with the new SI meter unit through a fomula is obviously not possible. The new SI definitions try to use theory constants whose measurement errors are very small. – anna v Oct 22 '19 at 12:01
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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.

flippiefanus
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  • we do define the joule based on water and paris position (with the proxy of wavelengths) but we could define the same joule as the energy in 1/c2 kg (E=mc2). How come the 2 definitions amount to the same joule ? – Manu de Hanoi Oct 22 '19 at 08:20
  • The current system of units has been defined to be consistent with previous definitions. Remember that we do not use the water and the Paris position anymore, because they don't provide the same accuracy and precision that we can get with the current definitions. – flippiefanus Oct 22 '19 at 09:15
  • I understand that, but the constants used to define the units are just as arbitrary as paris and density of water as explained in OP. Be it paris position or 184154131101 the wavelength of khsdfgkjbasdfuh – Manu de Hanoi Oct 22 '19 at 09:58