11
$\begingroup$

In equations that have quantities with physical dimension.

Example: $\mathrm{Force} = (\mathrm{mass})(\mathrm{acceleration})$ or $F=ma$

I know that we use that (mass, force...) to help what we should use in the equation. But, why don't we just use $N = \mathrm{kg}\cdot \mathrm{m}/\mathrm{s}^2$ or $\mathrm{Newton} = (\mathrm{kilograms})(\mathrm{meter})/ \mathrm{seconds}^2$ as the equation?

Can any equation be wrong by doing that? I know that some equations are used in specific places/ranges/situations. But even with words instead of units in equations, we need to know the application boundaries of that equation.

$\endgroup$
  • 2
    $\begingroup$ I've deleted some comments. Please remember that comments are not to be used for answering the question. $\endgroup$ – David Z Jul 8 '16 at 7:00
  • 2
    $\begingroup$ How will you write the area of a rectangle? $m \cdot m$? Which meter? And what about triangle? Also $m \cdot m$... So are they equal? $\endgroup$ – Superbest Jul 8 '16 at 9:08
  • 1
    $\begingroup$ It's similar to not relying on parameter names in software programming. Relying on type checking in the call is handy but not always practical. $\endgroup$ – JDługosz Jul 8 '16 at 13:18
  • $\begingroup$ Back in the day when I was tutoring physics, variants of this question were probably the MOST frequently asked question I received. Must be something about human nature. +1 for the question just due to its popularity. $\endgroup$ – Ghotir Jul 8 '16 at 17:57
  • $\begingroup$ This question does not deserve to be put on hold. It's perfectly clear what's being asked, hence the multitude of answers below. $\endgroup$ – Mark H Jul 9 '16 at 12:11

10 Answers 10

47
$\begingroup$

A specific parameter might correspond to a specific (SI) unit, but not all units correspond to a specific parameter!

Kinetic energy is

$$\begin{align} K&=\frac{1}{2} mv^2 \\ [\text{Joules}]&=\frac{1}{2}[ \text{kilograms}\times\text{meters}^2/\text{seconds}^2] \end{align}$$

We also have gravitational potential energy:

$$\begin{align}U&=mgh \\ [\text{Joules}] &= [\text{kilograms} \times(\text{meters} / \text{seconds}^2) \times \text{meters}]\\ &= [\text{kilograms} \times \text{meters}^2 / \text{seconds}^2] \end{align}$$

So, is Joules both $\frac{1}{2} \text{kilograms}\times\text{meters}^2/\text{seconds}^2$ and $ \text{kilograms}\times\text{meters}^2/\text{seconds}^2$ at the same time? If you have a value in Joules and you need to find the number of kilograms, then how would you go backwards? How would you do the algebra?

You could start from any of these unit-formulations, and you would get difference answers for the number of kilograms. The answer is not unique seen from the units since the original formula could have contained unit-less parameters.

The problem is that there are many kinds of energy with the same unit. In general, parameters have unique units, but units don't belong to unique parameters. You cannot go "backwards" from the unit formulation of a formula.

| cite | improve this answer | |
$\endgroup$
  • 10
    $\begingroup$ In the case of energy one might add that torque would also share this unit (Nm), but actually is a vastly different concept. $\endgroup$ – Chieron Jul 8 '16 at 7:48
  • $\begingroup$ It seems like a bit of cheating to leave $1/2$ untouched in your example instead of stripping it down to its (lack of) unit. $\endgroup$ – user2357112 supports Monica Jul 9 '16 at 0:53
  • $\begingroup$ @user2357112 Will you elaborate, please? $\endgroup$ – Steeven Jul 9 '16 at 8:34
  • 1
    $\begingroup$ @Steeven It is refreshing occasionally to read from someone who actually read the OP's question. Note, by the way, that, other than that of user1717828 below, there does not seem to be here an actual answer to the question as to what, other than single symbols, can be used as a variable. $\endgroup$ – schremmer Jul 12 '16 at 18:36
  • 2
    $\begingroup$ @Steeven Not a compliment. Gratitude. $\endgroup$ – schremmer Jul 13 '16 at 1:21
7
$\begingroup$

For simple equations, the two might be equivalent. Certainly, dimensionally an equation must always be correct. But there are plenty of situations where the units may not obviously reflect a particular quantity; and clarity of communication improves understanding.

Take electrostatics. If I say "1 Volt" you know what I mean; an electric field is "Volts per meter" - still OK. But what if I used a quantity with units $\rm{kg~m^2~s^{-3}A^{-1}}$? Would you know if that was a voltage, or an electric field?

Conventions develop because when everybody "speaks the same language" you spend less time decoding, and more time thinking about the underlying physics.

PS - it's voltage.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, you can open a derived unit into other units that multiplies and divides and etc... but, that's adding complexity where isn't needed. What I mean is changing kg to mass. Because a lot of equations in Portuguese ( my native language ) have different letters ( based in Portuguese ). When I have to read other language books/pages I have to learn that word and adapt. It's kind hard to change to equations in that way today, some unit will get complicated, others simplified. $\endgroup$ – Kaique Gomes Jul 7 '16 at 19:03
  • $\begingroup$ And the question is more about if there is any "real" problem. Because I know that change it's made to make equation more easy to understand, but, there is any mathematical or physical problem by using units instead of names/words. $\endgroup$ – Kaique Gomes Jul 7 '16 at 19:09
7
$\begingroup$

For one, the laws of physics are the same whether you're working in SI or Imperial. $F = ma$, regardless if your m is in kg, pounds mass, or solar masses. Actually, in college we had a bonus challenge to give the answer to a problem in craziest units of energy that we could come up with that actually worked. "Slug lightyears" was a pretty good one, but the winner was "Lizard foot slaps per workweek" (making use of the impulse of a basilisk's foot slap when running across the water, which was given in the textbook.)

For another, you're setting up an equivalence relation that, while true, isn't useful. Consider energy. It is absolutely true that $J = N \cdot m$. But this requires us to now perform the calculus every time we want to calculate energy, since $K = \frac{1}{2}mv^2$. The $\frac{1}2$ that results from the integration of $K = \int F \cdot dx$ is problematic now because $J = \frac{1}{2} kg \cdot \left(\frac{m}{s}\right)^2$ is absolutely not true.

To make matters worse, certain formulae now appear to be tautological when they are anything but, such as $\tau = r \times F$, which will now appear to be $N \cdot m = N \cdot m$, making no mention of the angle between $N$ and $m$ or whether the product should be scalar or vector.

| cite | improve this answer | |
$\endgroup$
7
$\begingroup$

Writing equations using only units would not work at all for dimensionless equations. For example the Snell's law $$n_1 \sin\theta_1=n_2 \sin\theta_2.$$ You would also lose many of the dimensionless (but usefull) parameters in physics such as the Lorentz factor $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}.$$

Also consider equations whose all variables have the same units. For instance the (truly fundamental) first law of thermodynamics, $$\Delta U=Q-W.$$ It would be meaningless if we write it only in terms of units.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I get it, dimensionless quantities (ratio, factor, pi )have no unit. So my question should specify that is about quantities that have physical dimension. $\endgroup$ – Kaique Gomes Jul 7 '16 at 19:42
  • 3
    $\begingroup$ I think your last point is the most convincing. To put it in simpler terms, suppose I drive 10 km north, then 3 km south and I want to find out how far I am from my starting point. If I just write km = km - km, I explain nothing. If I want to write this algebraicly, I need to make a different symbol for each variable, like $s_{tot}=s_1 - s_2$. $\endgroup$ – The Photon Jul 7 '16 at 21:21
  • $\begingroup$ @Steeven He added that part after my answer. $\endgroup$ – Kaique Gomes Jul 7 '16 at 21:47
  • $\begingroup$ @ThePhoton Yes, but you are adding subscript letter to the words, could add the "tot", "1" and "2" to m(meter). Like: mtot = m1 - m2 $\endgroup$ – Kaique Gomes Jul 7 '16 at 21:47
  • 3
    $\begingroup$ @KaiqueGomes I see, I apologize and removed the comment. To your other comment: Here you are exactly reaching the point! You need to assign names to the values - units are simply not enough. You can choose to call them $m_{tot} =m_1-m_2$ or you can choose to call them $\Delta U =Q-W$ or anything else. And this is what is called parameters / variables / properties etc. - the name it is given. $\endgroup$ – Steeven Jul 7 '16 at 21:57
6
$\begingroup$

Disclaimer: I have no idea what question everyone else is answering; none of the other answers seem to address the question as I understand it.


We don't use units in formula because not everything we want a variable to represent (like $a$ for acceleration) has its own nice unit. Acceleration is a perfect example: it's just too long to say "meters per second squared".

Additionally, what about unitless variables? For example,

$$F_f=\mu\cdot F_N$$

where $\mu$ is the friction coefficient. How would you write that equation with units instead of names?

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I disagree with this answer. Even in cases where the relevant physical quantities can be associated with one or more units we still wouldn't want to write the equation with any choice of units made in advance. $\endgroup$ – DanielSank Jul 7 '16 at 21:47
  • $\begingroup$ @user1717828 I added in the question that is about quantities with physical dimension. Unitless/Dimensionless numbers can't be done, but, I'm not asking to do it, you can leave a unitless value together with unit like Newton and kg in the equation. $\endgroup$ – Kaique Gomes Jul 8 '16 at 0:05
  • 1
    $\begingroup$ Parameter names are not just used to replace long and complicated units. Those two things do not replace each other and it is not simply a conveniency matter. If you removed parameter names and kept only units, you would be unable to distinguish parameters with same units from each other $\endgroup$ – Steeven Jul 8 '16 at 7:22
4
$\begingroup$

The purpose of units is to assign numbers to measurements. They are necessary but of secondary importance to the thing being measured. Scientists want to describe the real world with their equations, not just their measurement tools.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

If a physics equation is to be valid, it is necessary for the units to work out, but it is not sufficient.

The handling of dimensionality only provides part of the information. Units and dimensionality are good checks to make sure you did the equation right, but the mere fact that the units were right does not automatically mean the equation was right.

When we use words, we can describe what they mean later. If we use units, we've lost that opportunity. As an example, consider Newton's Universal Law of Gravitation:

$$N = \frac{N\cdot m^2}{kg^2} \frac{kg\cdot kg}{m^2}$$

Or should I say: $$(Force) = (Universal Gravitational Constant)\frac{(mass_A\cdot mass_B)}{(separation)^2}$$

Where Force is the magnitude of the force experienced by both bodies, UniversalGravitationalConstant is $6.674\cdot10^{−11}(\frac{N\cdot m^2}{kg^2})$, $mass_A$ and $mass_B$ are the masses of the two objects, and separation is the distance between those two objects.

We can also render this using variables for the same effect. They're just symbols after all:

$$F=G\frac{m_A\cdot m_B}{r^2}$$

Note that in both the case of the words and the variables we have more symbols to worth with, so we can be more specific. While the units just show $kg\cdot kg$, the words and variables specify which masses we are talking about. In this case, we happen to use an equation where you could be ambiguous and get away with it, but what if I used the equation for the acceleration experienced by mass A:

$$a_A=G\frac{m_B}{r^2}$$

If we were to only include units, we would have

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

In the latter case, it's not obvious that the final $kg$ term is actually the mass of object B.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

The basic fact is units are a demarcation of the quantity of something, so they can never be used as a fundamental equation. As you have mentioned F=ma can also be written as N=kg ms^-2 , then how can you say that the equation is of Newton's Law only? The equation can be used to define force dimensionally which is [M L T^-2].

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

It is not incorrect, it is incomplete.

A physical value consists of:

  1. Magnitude - this is "the number"
  2. Unit - "type" of the value

Physical values are not only magnitudes, they also have a unit. It is inseparable. Unit defines a physical meaning to a value. You could formally treat it as a pair (magnitude, unit). All calculations in physics are done this way, although people might not think about it. You can do the calculations separately on each part.

Your idea is doing only one part. So you could not do typical equations with actual magnitudes, e.g. how much of velocity a body has if it travels three meters in two seconds. You simply cannot skip magnitude.

Your question could be reduced a bit to "Why can we not drop this 'one' i.e. magnitude for unit definitions."

You are right that units are often defined like that, e.g. "One joule is equal to the energy transferred (or work done) to an object when a force of one newton acts on that object in the direction of its motion through a distance of one metre (1 newton metre or N·m)." Notice that they explicitly say "one".

Your idea would work but only if all the units were orthogonal to each other. You could normalize every unit then, so in every definition magnitude would always be 1. But this is not practical, and not true in real world.

E.g: Relation between angular velocity - $\omega$ and frequency -$f$ is: $\omega = 2 \pi f$. You could not express it using only units. Another, simpler, example would be conversion between different units having the same physical meaning e.g "one minute is 60 seconds", $1$cm = $0.01$m. Or converting between imperial units and metric system. Etc. etc.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

A "naked" number such as $3$ does not denote anything in the real world. In order to represent on paper a real world collection of items, we need a numerator and a denominator, e.g. $3$ Apples. But to represent on paper a real world amounts of substance we need in addition a unit to, so to speak, "discretize" the amount into a collection of units, e.g. $3$ liters of milk. (And this is where approximation comes in.)

In mathematics, a variable is usually all-including, for instance $x$ includes a sign as well as a size, e.g. $x$ may be replaced by $-3$, I suppose that in physics a variable includes a unit as well so that, e.g. $x$ may be replaced by $-3$ liters of milk.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Um, "numerator" and "denominator" are the top and bottom of a fraction. Also, "apples" is a perfectly fine unit to use. $\endgroup$ – Deusovi Jul 8 '16 at 8:48
  • $\begingroup$ @Deusovi Conflating units and denominators does not seem to help: A "naked" fraction such as 2/3 does not denote anything in the real world either. But 2/3 apple denotes "two of which three make up an apple" where 2 is the numerator and "of which 3 make up an apple" denominates what the 2 is numerating. $\endgroup$ – schremmer Jul 11 '16 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.