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I admit this sounds like a dumb question. I know very little about physics, but I've always wondered that given the fact that:

  • A unit of measurement is a very specific constant value.

Then how can a formula like: $$E=mc^2$$

Be so precise and so clean when it has to use already pre-defined unit of measurements as values. I assume (I'm googling this and I know almost no physics, just what I remember from high-school) That this formula claims that $1 \text{joule} = 1 \text{kg} * (\text{speed of light}) ^ 2 \frac{m}{s}$

The meter, the kilogram, the second, the joule (composed of the base units) and so on are very very specific values defined by the International System of Units. How can one find a law, that just so happens to find a relationship between these exact values, without needing a custom made constant value or set of values to multiply or add by?

Edit: So I understand that my understanding of the equation itself was off. It does not claim that $1 \text{joule} = 1 \text{kg} * (\text{speed of light}) ^ 2 \frac{m}{s}$ but instead that a mass of $1 \text{kg} * (\text{speed of light}) ^ 2$ = A certain amount of energy expressed in joules.

My understanding of the equation itself improved, but my original question still stands. How is such a clean relationship found between the kg and the speed of light without multiplying or summing by a custom constant or set of constants?

If you were to change to a different system, wouldn't you be unable to neatly say that exactly 1 unit multiplied by the speed of light = a certain amount of energy? You would need to translate your metric for mass to 1 kg. In pounds for example I think you would have to say:

E = $2.20462pounds * c^2 m/s$

Is this off? And so on.. If you want to keep it clean with 1 unit you would have to change the constant from the speed of light to a different value (a smaller value in the case of pounds at least).

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    $\begingroup$ I don't understand what role you think the units play here - the equation $E=mc^2$ holds regardless of whether you measure distance in meters, inches or whatever else. $\endgroup$
    – ACuriousMind
    Mar 10, 2021 at 16:33
  • $\begingroup$ @ACuriousMind I assume that whatever you measure your values in, you would need to translate them to the unit that the formula is described with no? If I measured 10g of mass I would assume I would plug in 0.01kg into the formula which would definitely change your result. Do correct me if I'm spewing nonsense, I don't know any physics, these are just my assumptions at this point. $\endgroup$
    – Horace
    Mar 10, 2021 at 16:42
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    $\begingroup$ @Horace, what ACuriousMind is saying is this: the formula $E=mc^2$ is valid on earth with SI units, English units, or any other units. That formula is also valid for an outer-space alien visiting the earth, who has never heard of ANY earth units. The formula holds true independent of ALL units of measurement, but if one is interested in calculating a specific value, units must be carefully defined for each variable in the formula, and those units must be consistent. $\endgroup$ Mar 10, 2021 at 16:58
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    $\begingroup$ @Horace, given just a unit for length, I could use meters, centimeters, inches, fathoms, angstroms, light years, etc., and all of those units have different lengths. ALL of those units can be used in any given length measurement, and most people typically choose the most convenient unit for the measurement that they are taking. The fact that a particular unit has a well defined length ensures reproducibility in experiments, but it is important to remember that a unit is not a variable in an equation. $\endgroup$ Mar 10, 2021 at 17:28
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    $\begingroup$ Related, possible duplicate: physics.stackexchange.com/q/112959/44126 $\endgroup$
    – rob
    Mar 11, 2021 at 20:14

9 Answers 9

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The elegance of the equation $E = mc^2$ doesn't have much to do with the unit system. It's a great equation because the speed $c$ that shows up on the right-hand side is exactly the same as the speed of light, but this follows from the postulates of relativity, not the unit system used.

But you might wonder: the unit system does matter when we're computing with numbers, right? That's right: unit systems are not completely arbitrary, and if you set them up differently then you can get worse results. Fortunately, we never do.

For example, consider the formula for the area of a circle, $$A = \pi r^2.$$ In physics, we view this formula as a relation between physical quantities. The quantity $A$ is an area, and $r$ is a length. The purpose of a unit system is to convert these physical quantities to raw numbers, which we do by dividing both sides by a suitable unit, $$\frac{A}{1 \text{ square meter}} = \pi \, \left( \frac{r}{1 \text{ meter}} \right)^2.$$ This is now an equation relating two numbers. It says that the area of the circle (in square meters) is equal to $\pi$ times the radius (in meters) squared.

The coefficient is still exactly $\pi$, but we can screw it up if we use a different unit for area. For example, let's define the $\textit{Horace}$ to be a unit of area precisely equal to $1.39$ square meters. Then multiplying the left-hand side by $1 = (1.39 \text{ square meters}) / (1 \text{ Horace})$ gives $$\frac{A}{1 \text{ Horace}} = \frac{\pi}{1.39} \, \left( \frac{r}{1 \text{ meter}} \right)^2.$$ Now an ugly numeric constant appears, because the units for area are not compatible with those for length. The relation between $A$ and $r$ when those quantities are expressed in this unit system is no longer elegant, even though it's still true that $A = \pi r^2$.

Of course, in physics, we avoid this by defining all units to be products of powers of a set of base units, like how a square meter is the square of a meter.


There's a subtlety that I swept under the rug, which doesn't apply to $E = mc^2$ but does to other equations. When we talk about a "unit system", we don't just mean a set of base units. We often also include an entire set of conventions about how to define physical quantities in the first place. Different conventions can lead to differences in the appearances of the equations themselves.

For example, consider Faraday's law, $$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}.$$ I could define an alternative quantity $\tilde{\mathbf{B}} = c \mathbf{B}$. The advantage of this new definition is that $\tilde{\mathbf{B}}$ has the same dimensions as $\mathbf{E}$, which leads to a pleasing symmetry between Faraday's law and Ampere's law in vacuum, $$\nabla \times \mathbf{E} = - \frac{1}{c} \frac{\partial \tilde{\mathbf{B}}}{\partial t}, \quad \nabla \times \tilde{\mathbf{B}} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}.$$ But it also makes factors of $c$ pop up elsewhere. For example, the magnetic force becomes $$\mathbf{F} = q \mathbf{v} \times \mathbf{B} = \frac{q}{c} \mathbf{v} \times \tilde{\mathbf{B}}.$$ In the so-called Gaussian units in electromagnetism, the base units are the centimeter, gram, and second. That alone wouldn't change the forms of the equations, just the numbers you get when you express them in terms of the base units. However, in Gaussian units you additionally define the "magnetic field" to be $\tilde{\mathbf{B}}$, not $\mathbf{B}$ (along with a similar redefinition for electric charge and electric field). It's these conventions that change the forms of the equations.

This isn't relevant to $E = mc^2$ because usually we only define energy, mass, and speed in a single way, even if we can use different units to quantify them. For example, the speed $c$ is defined as the distance traveled per time. You could define an alternative speed $\tilde{c}$ as, say, half the distance traveled per time, but this would be incredibly confusing, and require changing results as basic as the kinematics equations. However, it's not as confusing to redefine electric charge and electromagnetic fields, because we never measure them directly, only the forces induced by them. (And there's always an exception -- in natural units we define energy as $\tilde{E} = E/c^2$, getting the simpler relation $\tilde{E} = m$.)

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  • $\begingroup$ Can an equation hold regardless of the unit system? I don't quite understand that. Could you give me an example of an unit system this particular equation could be translated to and it would still hold? Or at least a more in-depth description of the underlying idea here. $\endgroup$
    – Horace
    Mar 10, 2021 at 17:12
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    $\begingroup$ @Horace If an object is square, its width is equal to its height, $w = h$. This is true no matter what the units are. One meter equals one meter. 27 inches equals 27 inches. 395 nanometers equals 395 nanometers. $\endgroup$
    – knzhou
    Mar 10, 2021 at 17:25
  • $\begingroup$ I understand. I think I realized my confusion goes back to a much more fundamental understanding of what the equation initially stated, or what an equation is allowed to state. I thought we were equating 2 concrete values. Where in reality there is a concrete value on the right: $mc^2$ and , almost like a ... ? function output on the left? To me it literally seems like you have f(m) = $mc^2$ , where the value of the function is the amount of energy you can translate that amount of mass to. $\endgroup$
    – Horace
    Mar 10, 2021 at 17:51
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    $\begingroup$ @Horace Indeed, that's a question that requires special relativity to answer! Search on the site, I bet it's been discussed before. $\endgroup$
    – knzhou
    Mar 11, 2021 at 7:56
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    $\begingroup$ Or you can define a Horameter as sqrt(1.39) meters, and then the original equation works out when you measure the radius in Horameters. $\endgroup$
    – user253751
    Mar 11, 2021 at 9:42
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The equation

$$ E=mc^2$$

holds regardless of the unit system you choose. More precisely, it does not tell you

$$ 1 \,\mathrm{joule}= 1 \,\mathrm{kg} \cdot \textrm{(speed of light)}^2$$

but rather: an object at rest with a mass $m$ has an energy equal to $mc^2$, so an object with mass $1$ kilogram has an energy of $299 792 458^2\approx 9\cdot 10^{16} $ joules, because $c=299 792 458 \textrm{ m }\mathrm{s}^{-2}$. What has to hold is an equality between units, i.e.

$$ 1 \,\mathrm{joule}= 1 \,\mathrm{kg} \cdot (\textrm{1 m/s})^2$$

this tells you that $E$, something in units of energy, is equal to $mc^2$, something that is also in units of energy. $\mathrm{kg\,m^2 s^{-2}}$ is exactly the definition of joule. In fact, "one joule" is nothing but an abbreviation of "one kilogram times one meter squared over a second squared", that one holds exactly by definition.

If you choose furlongs to measure distance, fortnights to measure time and pounds to measure mass, you have yourself that 1 pound of mass has an energy of approximately $3.24 \cdot 10^{24}$ Horaces, where I used that the speed of light is $c\approx 1.8 \cdot 10^{12}$ furlongs per fortnight and I named a new unit of energy in your honor

$$ \textrm{1 Horace}= 1 \,\mathrm{pound} \cdot (\textrm{1 furlong/fortnight})^2.$$

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    $\begingroup$ I understand. This does make sense. I misunderstood the meaning of E in $E = mc^2$. This might sound dumb but that looks more like a function to me rather than an equation, since E itself is not a defined quantity that we're equating to another quantity. I chose a poor equation to prove my point. Point taken. $\endgroup$
    – Horace
    Mar 10, 2021 at 17:32
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    $\begingroup$ Well, it is a function. The energy of a body at rest is a function of its mass, $E(m)=mc^2$. Of course you can rearrange it as $m(E)=E/c^2$ if you like, but all these equations are stating is that some measurable quantities are related to one another by functions, i.e. that given $m$ you can plug it in some function and obtain $E$, or vice versa. $\endgroup$ Mar 10, 2021 at 18:02
  • $\begingroup$ But it is an equation. It can be used in the other direction if the energy was known, but not the mass directly. And this is actually how things are typically weighed; how we determine the mass of produced particles in an accelerator. We measure the energy the impart on a detector and then can use E = mc^2 along with kinematic equations to solve for m. $\endgroup$ Mar 11, 2021 at 6:27
  • $\begingroup$ @Shufflepants Yea I realized that for some reason I had a fixation on treating the unit of measurement almost like an implicitly inserted value into the equation for some reason. And I was fixating on the right side of the equation for some reason too. It's an equation. You can re-write it like an equivalent function if you so desire. At least that's my current understanding $\endgroup$
    – Horace
    Mar 11, 2021 at 7:34
  • $\begingroup$ Your very first sentence is wrong: If the unit of energy were not defined the way it is ($1J = 1kg\frac{m^2}{s^2}$), you would need to add a unit conversion factor to the equation. $\endgroup$ Mar 11, 2021 at 16:45
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In addition to the answers already given, I would point out to, in response to this specific part of your question,

How can one find a law, that just so happens to find a relationship between these exact values, without needing a custom made constant value or set of values to multiply or add by?

There are indeed laws where things did not look so clean when first derived. Both Newton's law of gravitation and Coulomb's Law are examples. For Newton's law we have: $$F=\frac{Gm_1m_2}{r^2}$$where in MKS units G is $6.674\times10^{-11} \mathrm{m^3kg^{-1}s^{-2}}$, or $1.068846\times10^{-9} \mathrm{ft^3lb^{-1}s^{-2}}$ in English units. This unwieldy number was given its own letter, G, and accepted as a fundamental constant, like c. So sometimes things don't work out as cleanly as $E=mc^2$!

{Thanks to the comments below! I have cleaned up my answer to make it better.}

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    $\begingroup$ @Horace I would also point out that just like the formula $F = G m_1 m_2 / r^2$ needs the custom constant $G$, the formula $E = mc^2$ also needs a custom constant $c^2$. The fact that the custom constant $c^2$ happens to be the square of the vacuum speed of light is miraculous! $\endgroup$
    – d_b
    Mar 10, 2021 at 17:49
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    $\begingroup$ @Horace note that most of the time in SR we assume c = 1 to make formulae look less "messy". What that means is that we have defined c as 1 light year per year, or approximately one foot per nanosecond, or another equivalent combination of time and distance units. The point is that custom constants are not fundamental, and we have some latitude to choose them. $\endgroup$
    – m4r35n357
    Mar 10, 2021 at 18:13
  • $\begingroup$ I thought I had a comment on this particular answer. Maybe it got deleted. Regardless, thank you @CGS, very good answer! Gave me a lot of clarity. $\endgroup$
    – Horace
    Mar 11, 2021 at 10:08
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    $\begingroup$ Indeed: Had the unit of energy not been defined to equal the multiplied units of mass and speed squared ($1J = 1kg\frac{m^2}{s^2}$), but rather some unrelated unit $u_E$, Einsteins equation would need to be written with a conversion factor $[\alpha_{con}] = \frac{u_E s^2}{kg m^2}$ and look like this: $E = \alpha_{con} m c^2$. $\endgroup$ Mar 11, 2021 at 16:39
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    $\begingroup$ @cmaster-reinstatemonica This is precisely what happens with temperature, resulting kTs and RTs in all the equations. $\endgroup$ Mar 11, 2021 at 16:56
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If we measure $m$ in kilograms and $c$ in meters/second, then when we multiply $mc^2$ we get some number for Energy that has units $\frac{kg\ m^2}{s^2}$. That number is the energy in those units.

If we instead want the energy in Joules, we need to multiply by some conversion factor relating J to $\frac{kg\ m^2}{s^2}$. As it happens, $$1\ \frac{kg\ m^2}{s^2} = 1\ J$$ because that is how we defined Joules.

However, if we wanted the energy in some other units, like Kilowatt-hours, we'd need to use a different conversion factor.

$$1\ \frac{kg\ m^2}{s^2} = \frac{1}{3600000}\ kWh$$

And if we used different units to initially measure $m$ and $c$ (say, pounds and miles/hour), we'd get a different initial number with units $\frac{lbs\ mi^2}{hr^2}$. We'd need another conversion factor to convert that to Joules or kilowatt-hours.

The point is that every one of these answers is correct because they all represent the same amount of energy, just written in different units.

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  • $\begingroup$ All these answers are correct, but I assume only one of them is going to look clean. My initial confusion is how does $E = mc^2$ look so clean only on a single measurement system (S.I.) Or maybe my assumption that only in S.I. it looks clean is incorrect? If I were to use a different measurement system and I wanted to convert say: m pounds to energy expressed in kWh, I assume I wouldn't just : $m pounds * c^2 (m/s)^2 = E kWh$ I would assume you would need to change the $c^2$ constant to some other value? Cleanliness relative to a specific measurement system is my confusion. $\endgroup$
    – Horace
    Mar 11, 2021 at 10:23
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The equation

$$E = mc^2$$

says that energy is mass multiplied by speed of light squared.

This equation does not mention any particular unit of measurement, and holds irrespective of how we choose to measure energies, masses, distances, and time.

Units are only involved when we want to plug in actual numbers. For instance, we might know that in our experiment, $m = 1 \ \textrm{kg}$ and $c = 299792458 \ \textrm{m/s}$. Plugging that into our equation, we get:

$$ E = 1 \ \textrm{kg} \ (299792458 \ \textrm{m/s})^2$$

which simplifies to

$$ E = 89875517873681764 \ \frac{\textrm{kg} \ \textrm{m}^2}{\textrm{s}^2} $$

This gives us the energy in a rather esoteric looking unit of measurement. What if we wanted to express the energy in Joule instead? That is, we want to know which $x$ satisfies

$$ x \ \textrm{J} = E = 89875517873681764 \ \frac{\textrm{kg} \ \textrm{m}^2}{\textrm{s}^2}$$

This turns out a lot simpler than we might expect, because $J$ is defined as the energy imparted by a force of $1 \textrm{N}$ over a distance of $1 \textrm{m}$, and $1 \textrm{N}$ is defined as the force that accelerates an object of mass $1 \ \textrm{kg}$ by $1 \ \frac{\textrm{m}}{\textrm{s}}$ in $1 \ \textrm{s}$. Putting all that together, we have:

$$\textrm{J} = \textrm{N} \ \textrm{m} = \textrm{kg} \frac{\textrm{m}}{\textrm{s}^2} \ \textrm{m} = \frac{\textrm{kg} \ \textrm{m}^2}{\textrm{s}^2}$$

Plugging this into our equation we get

$$ x \ \frac{\textrm{kg} \ \textrm{m}^2}{\textrm{s}^2} = E = 89875517873681764 \ \frac{\textrm{kg} \ \textrm{m}^2}{\textrm{s}^2}$$

which simplifies to

$$ x = 89875517873681764 $$

That is, it doesn't matter how long a meter is, or how long a second takes, or how much a kilogram of matter is, because a Joule is defined in terms of that same meter, that same second, and that same kilogram.

Put differently, the reason we don't need to do any unit conversion here is that our choice of units was not arbitrary, because we chose to measure energy in a unit that is derived from our chosen units for mass, distance and time.

This is, of course, intentional. The goal of the international system of units is to simplify calculations about the natural world, and eliminating unnecessary unit conversions by choosing coherent units of measurement definitely helps with that.

The flip side of striving for coherent systems of measurement is that systems of measurement must evolve alongside our understanding of the natural world. For instance, distance and time were thought to be unrelated until the speed of light was discovered to be constant. In a fully coherent system of measurement, we'd set this constant to 1 by measuring distances in light seconds rather than meters, which would simplify Einstein's famous formula to

$$ E = m $$

However, since light seconds make for an unwieldy large unit of measurement for our everyday lives and most phenomena of interest to us, the SI system chose to keep the meter, but redefined it as:

The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

thereby ensuring that the conversion factor is at least an integer :-)

In summary, while natural laws hold irrespective of units of measurement, calculations involving actual physical quantities can be simplified by using coherent systems of measurement, such as the international system of units.

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I'll target your edited/clarified question:

How is such a clean relationship found between the kg and the speed of light without multiplying or summing by a custom constant or set of constants?

This is because both the numbers and concepts of the International System of Units have specifically been designed and tuned by physicists to turn out such that they work well (read: easy) together.

There are two different aspects here:

  • The quantity (i.e., time, length, mass)
  • The unit (i.e., second, metre, kilogram) of each quantity

The whole system consists of seven base quantities/units (meters, seconds and such) and many derived units (e.g., radian, hertz and so on). There are many possible ways to chose both the quantities and the units making up such a system. Both aspects have been chosen for the ISU so that as many aspects as possible "cancel out".

For example, you could have removed time from the set of base quantities, and instead have used frequency. Then, instead of deriving frequency as the inverse of time, you would have derived time as the inverse of frequency. This would lead to a mild inconvenience, because now everywhere we multiply by time, we would then instead divide by frequency; and it would be mightily unintuitive. Instead of saying "I will be there in 5 minutes", I would need to say "I will be there in however long it takes a 0.2 Hz signal to repeat".

You could have done that with every derived unit that also contains a base unit. For example, the equivalent dose (sievert $m^2s^{−2}$) could easily replace either the length or the time in any equation containing length or time, but make it stupendously complicated to write the formulas correctly and to explain them.

Also, the constants going with it have finely been chosen so that they cancel out as much as possible, which is what gives us very easy conversions for the most important units.

As a comparison, in some regimens of physics they chose their units differently. When people are talking about relativity all the time, they usually scale values so that c=1. Nothing changes, but c (the constant, not the phenomenon) disappears from all formulas... as long as everybody knows what they're doing this is fine and makes calculations much easier.

The topic is historically very interesting, and they are tuning that stuff all the time. Give that Wikipedia page I linked a good read, it's quite fascinating.

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The speed of light $c$ is the "custom made constant value". Also, the equation does not tell you 1 Joule equals that. (It does say that many Joules equal 1kg*$c^2$

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    $\begingroup$ So then the equation is claiming that by multiplying 1 kg by $c^2$ you get an amount of energy that the 1kg corresponds to expressed in Joules? $\endgroup$
    – Horace
    Mar 10, 2021 at 16:56
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Nice question!
Most symbols used in physical formulae refer to physical quantities that can be measured. Hence the name quantity. They are measured in units. If the symbols in the formulae stand in a certain relation to each other then so should the measured values. In your example, if the mass is one kilogram (one can measure this), and if this mass (so the value, not the symbol) is multiplied by the speed of light squared (you can measure this speed) then you can calculate the value of the mass' energy. To see if the relation between the symbols conjectured by the formula is correct one can do a measurement of the energy (though making a measurement of the value of the rest energy of some mass is quite difficult). On this basis, you can accept a formula or reject it.
In mathematical physics (where symbols are manipulated all the time), most symbols do not refer to measurable quantities. For example in quantum field theory. Of course, the final result of all this manipulating must refer to measurable quantities (in quantum field theory these quantities are mostly cross-sections of particle reactions and decay rates of particles) to determine if all the manipulating was worthwhile unless you care about totally imaginary situations.
I think it is clear now why physics theorems (formulae) don't have to be accurate always. Only when the relationship between the symbols is confirmed by measurements then this is true.
The formulae are clean, the corresponding relationship between the measured values to which the symbols in the formulae refer will be not. Well, the cleaner the last the more precise the formulae are confirmed.
We can see also that the formulae of physics hold regardless of the units we use. The formulae are objective manipulations of symbols (of course we do this manipulating), while the measuring units are invented by us. You can say that the unit of distance is a parsec or a Planck length. This doesn't change the validity of $E=mc^2$. If we change the unit (measure) of one of the quantities on one side of the mathematical formula (in this case the measure of$c$), the measure of the unit on the other side will change accordingly ($E$ in this case).

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Think of it this way:

A formula is a relationship between parameters. This relationship holds true regardless of the actual values.

Newton's 2nd law, for instance, tells that total force is proportional to acceleration:

$$\sum F=m a$$

If you have a mass $m$ and its acceration $a$, the value you'll find for the force depends on the units you used for mass and acceleration. But the relationship is the same regardless.

  • If you used $\mathrm{kg }$ for the mass and $\mathrm{ m/s^2}$ for the acceleration, then the force gets a value on $\mathrm{ kg\cdot m/s^2}$. (This chunk of units has been named the Newton $\mathrm{N }$, but that's just this specific case.)
  • If you instead used, say, miles-per-hour-squared $\mathrm{ mi/hr^2}$ for the acceleration, then the force gets a value on $\mathrm{ kg\cdot mi/hr^2}$.

This other value belongs to another value for the force. But the forces are still the same - they are just measured differently (they refer to different units).

How we choose to measure the mass and the acceleration (which units we choose), does not change the relationship of total force being mass times acceleration - the result we get just has to be interpreted according to those same units, according to how we measured the parameters that we plugged in.

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