TransWikia.com

Is it mathematically wrong to use units instead of words/parameters/names in equations?

Physics Asked by Kaique Gomes on December 17, 2020

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.

10 Answers

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

Kinetic energy is

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

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 $frac{1}{2} text{kilograms}timestext{meters}^2/text{seconds}^2$ and $ text{kilograms}timestext{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.

Correct answer by Steeven on December 17, 2020

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.

Answered by Floris on December 17, 2020

Writing equations using only units would not work at all for dimensionless equations. For example the Snell's law $$n_1 sintheta_1=n_2 sintheta_2.$$ 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.

Answered by Diracology on December 17, 2020

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.

Answered by Mark H on December 17, 2020

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.

Answered by Devsman on December 17, 2020

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,

$$F_f=mucdot F_N$$

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

Answered by user1717828 on December 17, 2020

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.

Answered by schremmer on December 17, 2020

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:

$$N = frac{Ncdot m^2}{kg^2} frac{kgcdot kg}{m^2}$$

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

Where Force is the magnitude of the force experienced by both bodies, UniversalGravitationalConstant is $6.674cdot10^{−11}(frac{Ncdot m^2}{kg^2})$, $mass_A$ 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:

$$F=Gfrac{m_Acdot m_B}{r^2}$$

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 $kgcdot kg$, 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:

$$a_A=Gfrac{m_B}{r^2}$$

If we were to only include units, we would have

$$frac{m}{s^2} = frac{Ncdot m^2}{kg^2} frac{kg}{m^2}$$

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

Answered by Cort Ammon on December 17, 2020

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].

Answered by Abhinav on December 17, 2020

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.

Answered by luk32 on December 17, 2020

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP