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In many physics textbooks dimensional analysis is introduced as a valid method for deducing physical equations. For instance, it is usually claimed that the period of a pendulum cannot possibly depend on its mass because if it did the units would not match.

However, I think that this kind of argument is not correct.

Let's imagine we were trying to deduce Coulomb's law. We would make an educated guess by stating that the force between two charges depends on its charge and the distance. Nevertheless it is obvious that it is not possible to obtain the unit Newton from Coulombs and Meters. If we repeated the argument we used with the pendulum, we would end up with a different law for the force between two charges.

Am I wrong? If not, why and when is dimensional analysis used?

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    $\begingroup$ A newton isn't a base unit in SI, so yes, you're wrong. Dimensional analysis is another tool to analyze a problem. If the units for the "answer" are wrong, then the answer can't possibly be right. So if you divide "4 apples" by "two pears" you don't get "two bananas." $\endgroup$ – MaxW Mar 1 '18 at 22:18
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    $\begingroup$ In the case of Coulomb's law, it is possible to deduce its form in Gaussian electromagnetic units based purely on dimensional analysis. This is because, in Gaussian units, charge has units of g$^{1/2}$cm$^{3/2}$/s. You know that the Coulomb force depends on the two charges and their separation; the only way to get the units of force from those quantities is by taking $Q_1Q_2/r^2$. Even better, in Gaussian units this dimensional-analysis result is exact; the proportionality constant is 1. $\endgroup$ – probably_someone Mar 1 '18 at 22:30
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    $\begingroup$ Dimensional analysis is not all that useful for deriving equations, but it is quite useful for "sanity checking" $\endgroup$ – Hot Licks Mar 1 '18 at 22:47
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    $\begingroup$ Similar question here, but for $G$ instead of $1/4\pi\epsilon_0$. $\endgroup$ – knzhou Mar 2 '18 at 16:45
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Dimensional analysis is used when you're trying to figure out how a certain set of parameters (your "inputs") can be combined to yield a quantity with a particular set of units (your "output"). It only works under the following assumptions:

  • You know what all of your inputs are, and the list of inputs is finite;

  • There are no redundancies in your list of inputs (i.e. each input has different units), or, if there are redundancies, there must be additional information that constrains the behavior of the redundant inputs (e.g. "the Coulomb force must involve both $Q_1$ and $Q_2$");

  • Each of these quantities are expressed in units that are compatible with each other and with your output (in practice, this means that each of them should be expressible in only SI base units);

  • Any constant factors are assumed to be pure numbers, and

  • The number of additive terms is constrained to be finite.

If these assumptions are satisfied, dimensional analysis will yield a set of possible combinations of inputs ("formulas") that will have the same units as your output. If you're lucky, there will only be one; if you're not, there will be a few, which serve as arguments for an arbitrary function (this follows from the Buckingham Pi theorem).

Let's try this procedure on your two examples. In the first one, we have a pendulum. Our output is the period, which has units of time ($T$). The inputs are:

  • The length of the pendulum rod, which has units of length ($L$);

  • The local gravitational acceleration, which has units $\frac{L}{T^2}$, and

  • The mass of the pendulum, which has units of mass ($M$) (since we don't know a priori that it can't be an input, we include it as a possible input).

Using dimensional analysis, we can constrain the possible forms for the formula using the following equation and solving for the powers $a$, $b$, and $c$:

$$T=L^a\left(\frac{L}{T^2}\right)^bM^c$$

where $k$ is some unitless proportionality constant. Equating the powers of $L$, $T$, and $M$ on the left-hand side with their powers on the right-hand side:

$$0=a+b\quad\quad\quad 1=-2b\quad\quad\quad 0=c$$

Solving this system gives you $a=\frac{1}{2}$, $b=-\frac{1}{2}$, $c=0$. So, even though we didn't know that the period didn't depend on mass before, we have just proven that this is the case, assuming that our list of inputs was complete. Substituting the powers back into the original expression, we get an expression for the period $\tau$:

$$\tau=k \ell^{1/2}g^{-1/2}m^0=k\sqrt{\frac{\ell}{g}}$$

Since the solution to the linear system above is unique, there is only one term.

For Coulomb's law, we must use a system in which the units of each quantity are compatible; as such, we use Gaussian electromagnetic units, in which charge has units of $M^{1/2}L^{3/2}T^{-1}$ (i.e. units of g$^{1/2}$ cm$^{3/2}$/s). The units of the separation are, of course, $L$. In this case, we have three inputs: the two charges $Q_1$ and $Q_2$, which both have the above units, and the separation between the charges. Our output is force, which has units of $MLT^{-2}$ (i.e. g cm/s$^2$). Setting up dimensional analysis again:

$$MLT^{-2}=(M^{1/2}L^{3/2}T^{-1})^a(M^{1/2}L^{3/2}T^{-1})^bL^c$$

Solving for the powers:

$$1=\frac{a}{2}+\frac{b}{2}\quad\quad\quad 1=\frac{3a}{2}+\frac{3b}{2}+c\quad\quad\quad -2=-a-b$$

This system is degenerate, and gives you one free parameter, so $(a,b,c)=(a,2-a,-2)$. As such, the abstract general form of the force that you get with dimensional analysis is

$$F=f\left(\left\{k_a\frac{Q_1^aQ_2^{2-a}}{r^2}\right\}_{a\in\mathbb{R}}\right)$$

for some arbitrary function $f$. Note that we immediately get the $1/r^2$ nature of the force just through dimensional analysis. It is also very easy to experimentally eliminate all but one of these possible arguments, by invoking one piece of additional information, that can easily be gleaned from experiment: exchange symmetry. The force on two charges does not change if you swap the two charges with each other. This means that the powers on $Q_1$ and $Q_2$ must be equal. As such, the only possible powers are $(a,b,c)=(1,1,-2)$. This additional empirical information eliminates the redundancy in this dimensional analysis, and so we arrive at the correct formula:

$$F=k\frac{Q_1Q_2}{r^2}$$

This all comes with one major caveat: if there's an input that you don't know about and don't include, then you could get an entirely different formula. For example, let's look at the pendulum again. Let's assume that in this case, the rod is very slightly elastic, with an effective spring constant $K$ that has the usual units of N/m, or abstractly $MT^{-2}$. Now, redoing the dimensional analysis yields:

$$T=L^a\left(\frac{L}{T^2}\right)^b M^c \left(\frac{M}{T^2}\right)^d$$

Which yields the following system:

$$0=a+b\quad\quad\quad 1=-2b-2d\quad\quad\quad 0=c+d$$

Note that there are 4 variables and 3 equations, so there is 1 free parameter, which I will take to be $c$. As such, we solve the system to obtain $(a,b,c,d)=(-c+1/2,c-1/2,c,-c)$, which gives us an abstract general formula:

$$\tau=f\left(\left\{ k_c\left(\frac{\ell}{g}\right)^{1/2-c}\left(\frac{m}{K}\right)^c\right\}_{c\in\mathbb{R}}\right)$$

again for some arbitrary function $f$. Now, if $c$ is nonzero, then the period of a slightly elastic pendulum does depend on the mass of the pendulum, and the particular way that it does must be measured.

But now let's consider the limit of very slight elasticity (i.e. the limit of large $K$). Equivalently, suppose $\tau$ varies slowly enough with $m/K$ that $\log\tau$ vs. $\log(m/K)$ is well-approximated by a line. This means that there is only one nonzero term in the above set of arguments, since a linear log-log plot corresponds to power-law behavior. This simplified our expression considerably:

$$\log\tau=\log k+\left(\frac{1}{2}-c\right)\log\left(\frac{\ell}{g}\right)+c\log\left(\frac{m}{K}\right)$$

Therefore, we have reduced a complicated physical task (finding the period of a pendulum with a slightly elastic rod) into the much easier task of empirically finding the two constants $k $ and $c $.

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  • $\begingroup$ "there will be a few, which will all be added together" - that's completely wrong. The core applicable result if there are redundant combinations is the Buckingham Pi theorem, which allows for arbitrary functional dependence (on suitable dimensionless combinations). $\endgroup$ – Emilio Pisanty Mar 24 '18 at 23:38
  • $\begingroup$ Sorry about that. Edited to fix. $\endgroup$ – probably_someone Mar 24 '18 at 23:57
  • $\begingroup$ Happens to all of us - that's why we have edits ;-). Great answer. $\endgroup$ – Emilio Pisanty Mar 25 '18 at 1:46
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In engineering and science, dimensional analysis is the analysis of the relationships between different physical quantities by identifying their base quantities (such as length, mass, time, and electric charge) and units of measure (such as miles vs. kilometers, or pounds vs. kilograms vs. grams) and tracking these dimensions as calculations or comparisons are performed. Converting from one dimensional unit to another is often somewhat complex. Dimensional analysis, or more specifically the factor-label method, also known as the unit-factor method, is a widely used technique for such conversions using the rules of algebra.

So a equation which is dimensionally correct can be wrong. But equation which is dimensionally wrong is definitely wrong.

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Obviously is not the dimensional analysis to be wrong. It is wrong if one tries to rely only on it to find physical laws, despite it clearly represents a very good help to us.

In the example of the pendulum, the argument is: since the system is described by only three parameters which are the length of the pendulum, its mass and the gravity, then the expression of the period (which at this point can depend only on them) must be something proportional to $\sqrt{L/g}$. But that is a very particular case and you can't use the same reasoning in all situations. With the Coulomb law, you just know that it may depend on the distance and that involves charges, but these two quantities alone are not enough to determine a formula for a force. For this case to be similar to that of the pendulum, you should have the information of what are all the parameters entering the physics of the system. And indeed, if someone gives us the information that the other (only) parameter you need is the Coulomb constant $k$, then using the dimensional analysis we finally end up with the expression of the Coulomb force.

However, this use of the dimensional analysis cannot work in all cases. If in the phenomenon of study there are a lot of quantities, then probably there are a lot of combinations of them giving a plausible result for the expression we are looking for. At this point I would say that this kind of role of the dimensional analysis is perfectly licit a posteriori and must be handled with care. And let me add, in all cases there must be a physical intuition behind.

As for your last question, behind the problem above, dimensional analysis is a useful tool to check if a result is correct or not, or to get some hints about the possible dependences of a quantity and so on.

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Either your book is misleading you, or you are misreading the book. Either way, dimensional analysis is not a tool for deducing physical equations. It can be used as a validity check on your use of the equations, to make sure that you used the right ones, but if you did not know the law, you would not be able to infer it from units.

Consider Hook's law. It's the easiest. It is a relationship between the force on a spring and its displacement. We typically learn it as $F=kX$, where $k$ is the spring constant for this particular spring. By the logic you describe, you should be able to infer that the force does not depend on the mass of the spring because the units on $k$ are $\frac{N}{m}$. But as you point out, this is backwards. If you already knew that Hooks law was valid for springs and you knew that the dimensionality of $k$ is $\frac{force}{length}$ then you might be able to come to some conclusion. But, as you notice, that's kind of hazy.

If we were deducing laws from observations, what we would observe is actually $F\propto X$: the force is proportional to the deflection. This could be observed by deflecting a spring to various positions and observing the forces.

When we have such a proportionality, we can always rewrite it as an equality by adding a constant. Thus $F\propto X \Rightarrow F=kX$ We then define the units of k to be the correct units to make the dimensional analysis work.

So when coming up with the equations, dimensional analysis does not help. However, when using them, it's an incredibly good way to make sure you did not miss anything. For example, its very easy to confuse weight and mass. But weight is a force and mass is mass, so if you confuse one with the other and do the unit analysis, you'll quickly realize that you made a mistake somewhere.

Likewise, if you make a simple algebra error, dimensional analysis can help. If you do a bunch of math and were supposed to reach the equation $E=\frac{1}{2}mv^2$, but due to your mistake you got $E=\frac{1}{2}mv$, a dimensional analysis will quickly show that something is amiss.

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    $\begingroup$ You can deduce some equations with dimensional analysis. For instance, since in QFT the Lagrangian has to have mass dimension 4, and renormalizable theories don't have coupling constants with negative mass dimension, with a bunch of fields you can write the most general renormalizable Lagrangian just by dimensional analysis. $\endgroup$ – Chris Mar 1 '18 at 23:08
  • $\begingroup$ Dimensional analysis is sometimes a tool for deducing physical equations; see my example of a pendulum with an elastic rod below, in which I determine the log of its period up to proportionality constants. You just have to make sure the assumptions I listed are valid in your particular case. $\endgroup$ – probably_someone Mar 1 '18 at 23:40
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Dimensional analysis is just a tool for checking correctness of a equation not for deducing one and that too when you have got all the quantities in place with proper units.The example you gave about simple pendulum is irrelevant because going by your words I can simply introduce a constant say the a constant so that the equation becomes dimensionally correct .

Moreover as I said it is just a tool for checking correctness of a equation so if an equation is dimensionally correct doesn't simply mean it is correct as there can be dimensionalless constants and excess terms which cancel out dimensionally.

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protected by Qmechanic Mar 2 '18 at 5:52

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