Why isn't it $E \approx 27.642 \times mc^2$?

2,112

Solution 1

It's a side effect of the unreasonable effectiveness of mathematics. You are in good company thinking it is a little strange.

Many quantities in physics can be related to each other by a few lines of algebra. These tend to be the models that we think of as "pretty." Terms manipulated by pure algebra tend to pick up integer factors, or factors that are integers raised to integer powers; if only a few algebraic manipulations are involved, the integers and their powers tend to be small ones.

Other quantities may be related by a few lines of calculus. From calculus you get the transcendental numbers, which can't be related to the integers by solving an algebraic equation. But there are lots of algebraic transformations you can do to relate one integral to another, and so many of these transcendental numbers can be related to each other by factors of small integers raised to small integer powers. This is why we spend a lot of time talking about $\pi$, $e$, and sometimes Bernoulli's $\gamma$, but don't really have a whole library of irrational constants for people to memorize.

Most of constants with many significant digits come from unit conversions, and are essentially historical accidents. Carl Witthoft gives the example of $E=mc^2$ having a numerical factor if you want the energy in BTUs. The BTU is the heat that's needed to raise the temperature of a pound of water by one degree Fahrenheit, so in addition to the entirely historical difference between kilograms and pounds and Rankine and Kelvin it's tied up with the heat capacity of water. It's a great unit if you're designing a boiler! But it doesn't have any place in the Einstein equation, because $E\propto mc^2$ is a fact of nature that is much simpler and more fundamental than the rotational and vibrational spectrum of the water molecule.

There are several places where there are real, dimensionless constants of nature that, so far as anyone knows, are not small integers and familiar transcendental numbers raised to small integer powers. The most famous is probably the electromagnetic fine structure constant $\alpha \approx 1/137.06$, defined by the relationship $\alpha \hbar c = e^2/4\pi\epsilon_0$, where this $e$ is the electric charge on a proton. The fine structure constant is the "strength" of electromagnetism, and the fact that $\alpha\ll1$ is a big part of why we can claim to "understand" quantum electrodynamics. "Simple" interactions between two charges, like exchanging one photon, contribute to the energy with a factor of $\alpha$ out front, perhaps multiplied by some ratio of small integers raised to small powers. The interaction of exchanging two photons "at once," which makes a "loop" in the Feynman diagram, contributes to the energy with a factor of $\alpha^2$, as do all the other "one-loop" interactions. Interactions with two "loops" (three photons at once, two photons and a particle-antiparticle fluctuation, etc.) contribute at the scale of $\alpha^3$. Since $\alpha\approx0.01$, each "order" of interactions contributes roughly two more significant digits to whatever quantity you're calculating. It's not until sixth- or seventh-order that there begin to be thousands of topologically-allowed Feynman diagrams, contributing so many hundreds of contributions at level of $\alpha^{n}$ that it starts to clobber the calculation at $\alpha^{n-1}$. An entry point to the literature.

The microscopic theory of the strong force, quantum chromodynamics, is essentially identical to the microscopic theory of electromagnetism, except with eight charged gluons instead of one neutral photon and a different coupling constant $\alpha_s$. Unfortunately for us, $\alpha_s \approx 1$, so for systems with only light quarks, computing a few "simple" quark-gluon interactions and stopping gives results that are completely unrelated to the strong force that we see. If there is a heavy quark involved, QCD is again perturbative, but not nearly so successfully as electromagnetism.

There is no theory which explains why $\alpha$ is small (though there have been efforts), and no theory that explains why $\alpha_s$ is large. It is a mystery. And it will continue to feel like a mystery until some model is developed where $\alpha$ or $\alpha_s$ can be computed in terms of other constants multiplied by transcendental numbers and small integers raised to small powers, at which point it will again be a mystery why mathematics is so effective.


A commenter asks

Isn't α already expressible in terms of physical constants or did you mean to say mathematical constants like π or e?

It's certainly true that $$ \alpha \equiv \frac{e^2}{4\pi\epsilon_0} \frac1{\hbar c} $$ defines $\alpha$ in terms of other experimentally measured quantities. However, one of those quantities is not like the other. To my mind, the dimensionless $\alpha$ is the fundamental constant of electromagnetism; the size of the unit of charge and the polarization of the vacuum are related derived quantities. Consider the Coulomb force between two unit charges: $$ F = \frac{e^2}{4\pi\epsilon_0}\frac1{r^2} = \alpha\frac{\hbar c}{r^2} $$ This is exactly the sort of formulation that badroit was asking about: the force depends on the minimum lump of angular momentum $\hbar$, the characteristic constant of relativity $c$, the distance $r$, and a dimensionless constant for which we have no good explanation.

Solution 2

It's all in how you define the units. $E = mc^2$ in nice MKSA units; but then change energy into BTUs and you'll need the ever-lovable "fudge factor" in there.

People spent a lot (well, some) of time developing self-consistent sets of units largely to keep equations simple, tho' as Rijul pointed out, assigning ugly numbers to known constants hides a lot as well.

Solution 3

I would not say I know the answer, but in my belief we tend to hide the ugly numbers.

See Rydberg's equation $$\frac {1}{\lambda} = R (\frac {1}{n_1^2}-\frac {1}{n_2^2}) $$ where $ R $ hides the ugly number $R = 1.0973731568539×10^{-7}\ \mathrm{m}^{-1}$

Similarly, observe Bohr's second postulate $$L = \frac{nh}{2\pi}$$ where $h$ hides the ugly number $h=6.62606957×10^{-34}\ \mathrm{kg}\cdot\mathrm{m}^2\cdot\mathrm{s}^{-1} $

There can be many more examples, but I guess this is enough to make my point!

Note: the numbers you might call ugly others might consider extremely beautiful, as some people might consider Planck's constant as divine beauty!

Addendum: the number of equations with and without such numbers should be considered, I do not think there would be more beautiful relations than the ugly ones!

Also, you need to start including all numbers even simple natural numbers like 1 and 2 in the list of ugly numbers! Then by that definition even the equation of time dilation has a "1" hiding there!

Added with respect to comment: the numbers you referred to as ugly in your question were uncommon and complicated, I have pretty much found beauty in symmetry both complete and partial, I read somewhere that plants and all are aesthetically pleasing because of partial symmetry! So maybe the simplicity of rational numbers and familiarity with constants makes them less ugly than others.

Solution 4

It seems to me that there's two ways of looking at this question, depending on what you view as fundamental. At the end of the day it's all tied up with the surprising power of dimensional analysis.

In classical dynamics there are 3 independent dimensionful quantities. These are simply mass ($M$), length ($L$) and time ($T$). Once you've chosen a standard unit for each of these quantities then all other dimensionful quantities are uniquely represented by a number and appropriate units.

For example energy has units $ML^2T^{-2}$. That means that once you've set standard quantities of mass, length and time then you can talk about energy unambiguously. We usually use SI units where energy comes with the unit Joule ($\mathrm J$) and

$$1\ \mathrm J = 1\ \mathrm{kg\ m^2\ s^{-2}}$$

From this perspective then, it's very puzzling that $E = mc^2$ works out exactly right. Put another way, the quantity $E/mc^2$ is dimensionless - so altering your standard definitions of mass, length and time doesn't affect it! So why on earth should we find that

$$E/mc^2 = 1$$

rather than the less elegant

$$E/mc^2 = 27.1252$$

The answer lies in a more detailed understanding of Einstein's special theory of relativity. Basically what Einstein achieved was reconciling the following ideas

  1. Physics looks the same no matter what speed you are travelling
  2. The speed of light is a universal constant

Einstein's solution exactly requires that $E/mc^2=1$. In fact you can derive this equation by assuming Lorentz symmetry and Einstein's notion of proper time. There's a couple of good accounts available online - here and here.

But you could possibly have guessed $E=mc^2$ was true, just from your knowledge of dimensions. Think about a decaying nucleus. This loses mass and emits energy in the form of electromagnetic waves. So the three quantities involved are (naively) $E$,$m$ and $c$.

Dimensional analysis says that they must be related by an equation

$$E/mc^2 = K$$

where $K$ is some dimensionless number. A strong philosophical principle called naturalness says that $K$ must be roughly $1$. Over the years physicists have found that naturalness is an incredibly good guide to our understanding. If formulas are natural, like $E=mc^2$ it's usually a sign that the underlying theory is sound.

This links in nicely with Rob's answer. He mentions a few dimensionless quantities which aren't close to $1$. Some physicists feel that this shows our models are incomplete. There are many proposed solutions that explain these unnatural quantities. Quite a few might be ruled out if the LHC doesn't see any new particles when it turns on again next year. So perhaps by 2016 we'll be abandoning naturalness as a guide!

I mentioned that there was another way to look at the question. Suppose we take our fundamental units to be mass ($M$), energy ($E$) and speed ($V$). Now of course, $E/mc^2$ is no longer a dimensionless quantity. It has units of $EM^{-1}V^{-2}$ and we can adjust our standard measures so that we get exactly

$$E = mc^2$$

This is precisely what Carl and Rijul were talking about. In a world where your units are fundamentally $M$,$E$ and $V$ there's no mystery - the formula is just a useful mnemonic for an experimental fact.

Let me know if you want any more details!

Solution 5

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).

Share:
2,112

Related videos on Youtube

badroit
Author by

badroit

Updated on March 12, 2021

Comments

  • badroit
    badroit over 2 years

    Why is it that many of the most important physical equations don't have ugly numbers (i.e., "arbitrary" irrational factors) to line up both sides?

    Why can so many equations be expressed so neatly with small natural numbers while recycling a relatively small set of physical and mathematical constants?

    For example, why is mass–energy equivalence describable by the equation $E = mc^2$ and not something like $E \approx 27.642 \times mc^2$?

    Why is time dilation describable by something as neat as $t' = \sqrt{\frac{t}{1 - \frac{v^2}{c^2}}}$ and not something ugly like $t' \approx 672.097 \times 10^{-4} \times \sqrt{\frac{t}{1 - \frac{v^2}{c^2}}}$.

    ... and so forth.


    I'm not well educated on matters of physics and so I feel a bit sheepish asking this.

    Likewise I'm not sure if this is a more philosophical question or one that permits a concrete answer ... or perhaps even the premise of the question itself is flawed ... so I would gratefully consider anything that sheds light on the nature of the question itself as an answer.


    EDIT: I just wished to give a little more context as to where I was coming from with this question based on some of the responses:

    @Jerry Schirmer comments:

    You do have an ugly factor of $2.997458 \times 10^8 m/s$ in front of everything. You just hide the ugliness by calling this number c.

    These are not the types of "ugly constants" I'm talking about in that this number is the speed of light. It is not just some constant needed to balance two side of an equation.

    @Carl Witthoft answers:

    It's all in how you define the units ...

    Of course this is true, we could in theory hide all sorts of ugly constants by using different units on the right and the left. But as in the case of $E=mc^2$, I am talking about equations where the units on the left are consistent with the units on the right, irrespective of the units used. As I mentioned on a comment there:

    $E=mc^2$ could be defined using units like $m$ in imperial stones ($\textsf{S}$), $c$ in cubits/fortnight ($\textsf{CF}^{−1}$) and $E$ in ... umm ... $\textsf{SC}^2\textsf{F}^{−2}$ ... so long as the units are in the same system, we still don't need a fudge factor.

    When the units are consistent in this manner, there's no room for hiding fudge constants.

    • diffeomorphism
      diffeomorphism over 9 years
      "Why is time dilation describable by something as neat as..." well, because $t' = t$ if $v=0$, so the expression in a Taylor series of v must begin with a factor of one near the $t$. This is true for all transformations that are described by small movements around the identity (Lie groups)
    • Robert Mastragostino
      Robert Mastragostino over 9 years
      These formulas are derived from more basic assumptions; they're not just random numerical coincidences we came up with. And normally when we have only a few variables and a few simple operations there aren't really any opportunities for weird numbers to appear. You could ask the same thing about the quadratic formula or Pythagorean theorem; I don't think this phenomenon is peculiar to physics.
    • zzz
      zzz over 9 years
      There's a cheap answer to the first question and the title: suppose we ended up with a numerical coefficient, we can just take its square root and absorb it into c.
    • badroit
      badroit over 9 years
      @bechira, yep but as you say that's a bit cheap though since c is the speed of light.
    • evil999man
      evil999man over 9 years
      $E=27.642mc^2$ where $E$ is in $\text{Joules}$, $m$ is in $\text{cool unit}$ and $c$ in $ms^{-1}$. $1 \text{cool unit}=\frac{1}{27.642} kg$
    • Jerry Schirmer
      Jerry Schirmer over 9 years
      You do have an ugly factor of $2.997458\times 10^{8}$ m/s in front of everything. You just hide the ugliness by calling this number $c$
    • Lagerbaer
      Lagerbaer over 9 years
      But isn't it then a surprise that this ugly number that we call $c$ is also the speed of light?
    • John Alexiou
      John Alexiou over 9 years
      No ugly factors are needed when using consistent units. So it depends on the definition of meter and Joule and second etc.
    • badroit
      badroit over 9 years
      @ja72, but for some equations, it doesn't matter what the units are since they cancel out. As I mentioned in a comment below, doesn't $E = mc^2$ work even if you consider units like imperial stones, cubits/fortnight and (imperial stones)$\cdot$(cubits/fortnight)$^2$? In other words, all you have to do is use consistent units (not SI units) for many such equations to work, right?
    • John Alexiou
      John Alexiou over 9 years
      Yes as I said the key is to use consistent units.
    • badroit
      badroit over 9 years
      @ja72, ah yes, sorry I misread your comment! :) For some reason I read it as "No. Ugly factors ...".
    • Jerry Schirmer
      Jerry Schirmer over 9 years
      @Lagerbaer: $c$ is the conversion factor between distance and time. That light goes at that speed is a derived result. It's not coincidence.
    • Jerry Schirmer
      Jerry Schirmer over 9 years
      @badroit: it doesn't work if you don't use consistent units -- say, if you measure $E$ in calories, but everything else in standard SI units. One can imagine a world where we discovered natural atomic reactors before we figured out the equivalence of thermodynamic and mechanical energy, for example.
    • Admin
      Admin over 9 years
      The answer has nothing to do with unit choices or "naturalness," as everyone seems to think. And in fact the lack of a constant in the formula for time dilation directly leads to the lack of a constant in $E=mc^2$. A good place to start is Einstein's (unfortunately rather terse) 1905 letter. If I find more time later (and if no one else has done it yet), I can write up Einstein's ideas in more modern parlance.
    • Edward Hughes
      Edward Hughes over 9 years
      @ChrisWhite - you are correct that the underlying reason for $E=mc^2$ is just the postulates of special relativity. I've detailed a couple of less terse references in my answer. But I wouldn't say this has nothing to do with naturalness. In fact $E=mc^2$ is a great example of where the principle of naturalness works. Of course, naturalness cannot explain why $E=mc^2$ at any deep level, but it does serve as an intuitive guide towards the formula.
    • Déjà vu
      Déjà vu almost 3 years
      @JerrySchirmer Nice remark (+1), but I think $c \approx 2.99792458\times 10^8ms^{-1}$ :-)
    • user76284
      user76284 almost 3 years
      Using rationalized Planck units might make things clearer.
  • Lagerbaer
    Lagerbaer over 9 years
    But MKSA wasn't explicitly invented to make $E = mc^2$ a prefactorless equation. It's not like the atomic units theorists use where $\hbar$ and $m_e$ and a bunch of other constants come out to $1$.
  • Rijul Gupta
    Rijul Gupta over 9 years
    Will the downvoter be kind enough to explain what problem he/she had with the answer?
  • badroit
    badroit over 9 years
    Thanks! Indeed there are formulas with weird physical constants but the fact that there's so many without is really interesting to me ... and it' an interesting point about what is the difference between beautiful numbers (1, 2, $\pi$) and ugly ones ... perhaps it's a question of "beauty is in the eye of the beholder".
  • badroit
    badroit over 9 years
    Thanks! I had considered this before ... maybe it's some trick with the units ... but I came to the same conclusion as @Lagerbaer. $E= mc^2$ could be defined using units like $m$ in imperial stones ($\mathsf{S}$), $c$ in cubits/fortnight ($\mathsf{CF}^{-1}$) and $E$ in ... umm ... $\mathsf{S}\mathsf{C}^2\mathsf{F}^{-2}$ ... so long as the units are in the same system, it's still $E = mc^2$, right?
  • John Alexiou
    John Alexiou over 9 years
    -1 because the example R has units, and thus not an ugly number but a meaningful quantity. The poster is asking about coefficients without units like $F = 2.39872 m a$.
  • badroit
    badroit over 9 years
    This is a very nice answer! I struggle to understand some of the technical aspects in paragraph 5 but I get the gist that even seemingly arbitrary constants of nature have a way to be computed from more elemental structures. This brings me to a related (and very naive) question: are such constants always computable on paper or are some experimentally determined? If all such constants are computable on paper, for me that's a good answer to my overall question: it means they must be formed from integers or well-known transcendental numbers/mathematical constants.
  • rob
    rob over 9 years
    There is (as yet) no known theory that predicts the size of the fine structure constant, or the ratios of the masses of the quarks and leptons. Those numbers are, for now, simply experimental facts.
  • rob
    rob over 9 years
    I downvoted because I thought the answer missed the point of the question. Both of your examples are dimensionful quantities which have lots of digits for historical reasons. While the Rydberg factor was initially an empirical constant, modern theory gives us $R = m_e c \alpha^2/2h$, a product of several meaningful experimental quantities and a rational number. No hard feelings, I hope.
  • Rijul Gupta
    Rijul Gupta over 9 years
    I would criticise your downvotes because you have simply downvoted because I did not give dimensionless constants! When carl talks about the fudge factor, it may be dimensionless but it is no less meaningful than the ones I gave! Moreover the question is clearly about encountering more constants than numbers in equation as one could have written the numerical value for "c" and easily lose the beauty of the eqn, even though the number would have meaning and dimension both! If you find my reasoning justified, I would ask you to retrieve your downvotes.
  • rob
    rob over 9 years
    @rijulgupta I have elaborated in another answer. I highly recommend you follow the first link in that answer, to Wigner's famous essay. badroit's question has some real philosophical meat to it.
  • Rijul Gupta
    Rijul Gupta over 9 years
    @rob : does that mean you disagree with my justification, or were you the one who retrieved the down vote?
  • AJS
    AJS over 9 years
    In the last paragraph, what kind of constants did you mean in "$\alpha$ or $\alpha_s$ can be computed in terms of other constants"? Isn't $\alpha$ already expressible in terms of physical constants or did you mean to say mathematical constants like $\pi$ or $e$?
  • rob
    rob over 9 years
    @A.J.Shajib Edited.
  • rob
    rob over 9 years
    Very nice answer!
  • Edward Hughes
    Edward Hughes over 9 years
    @badroit - essentially you are right about $E=mc^2$ being surprising if you've chosen units of mass, length and time consistently. I've elaborated on how exactly this works in my answer below.
  • Bob Bee
    Bob Bee over 7 years
    You got it exactly right. It no mathematical coincidence, Einstein derived it from the equality of the speed of light in all inertial frames. c came into the relatively simple calculation.
  • Bob Bee
    Bob Bee over 7 years
    In fact, it also had nothing to do with choosing units. M and E already had their units, and c was measured already. It came from Lorentz symmetry. Like others said, it is not always that nice, like for alpha.
  • Bob Bee
    Bob Bee over 7 years
    Another lack of coincidence was how Maxwell's equation have a c squared in them. In fact the derivations have the epsilon 0 and mu 0 in them, vacuum permitivity and (I forget what the epsilon is). It was quite a victory that those came out to be c squared. It showed that light is electromagnetic.
  • badroit
    badroit over 2 years
    Though I accepted Rob's answer, I think this answer complements Rob's answer very nicely. Many years later, I'm slightly puzzled that this answer remains very much underrated.