Dimensional analysis - When can you introduce constants that make dimensions compatible?

I have just read this question: What justifies dimensional analysis. One statement was:

Maybe the speed of a comet is given by its period multiplied by its mass. Why not?

As a formula this is $v=mT$. How do we know that this is wrong? I am not asking for the standard answer concerning the incompatibility of the dimensions. Suppose $v\propto m$ and $v \propto T$, then one could argue that $v\propto mT$ and so $v=CmT$ where $C$ is a constant that fixes the dimensions.

I will give another example: $F_G=Gm_1 m_2/r^2$ - Newton's law of gravity. As far as I know, Newton knew the following: $F \propto m_1m_2/r^2$ and he didn't know the value of $G$ so he simply stated $F=Gm_1m_2/r^2$ with $G$ fixing the dimensions.

Now comes my real question: My intuition tells me that $v\propto mT$ is not right but can you exclude it with dimensional analysis? Then you would also have to deny Newton's law of gravity. How do you know that $v=CmT$ is incorrect but $F=Gm_1m_2/r^2$ is not? Especially: When can you "invent" a constant in dimensional analysis which fixes the dimensions? If you could do it all the time, dimensional analysis would not be helpful...

• The speed of a comet is not its period multiplied by its mass because a little rock falls at the same speed as a big rock. Commented Aug 25, 2015 at 18:47

2 Answers

Dimensional analysis should never take the place of common sense - it should be used to inform it.

Let's take Newton's example. He knew that objects of different mass fell at the same speed - which had to mean they experienced a gravitational force proportional to their mass. Hence

$$F \propto m$$

If gravitational attraction between two bodies is symmetrical, the equation must be symmetrical in their mass. So if it's proportional to $m$ it must be proportional to $M$ (we can switch subscripts of the bodies without changing the physics) so

$$F \propto Mm$$

Looking at the motion of celestial bodies, you can deduce from the elliptical orbits that the force must go as the inverse of the distance between them, so

$$F \propto \frac{Mm}{r^2}$$

But dimensionally, that does not make sense. Since there are no other quantities around (properties of the bodies involved) that we can recruit to make the expression work, we need "something else" to come to the rescue - a factor that gives us the scaling (since we are working with different units) and that captures "everything else we don't know about gravity". Enter G, with units of $\frac{m^3}{kg m^2}$. Not satisfying, but necessary.

So I propose

you can introduce a constant with dimensionality when you have exhausted all the factors that physics tell you should be included, and you were not able to solve for the dimensionality of the problem

Note - this is not a "text book" answer AFAIK; it is based on my own experience only.

• Thanks for your answer! I have got another question: Suppose you do dimensional analysis on a problem and you find an answer with correct dimensions on both sides. Now, 1) is there any reason for this answer to be incorrect (are there examples where this is incorrect)? 2) Can you be sure that there is no need to introduce a constant that is dimensional and therefore change your previous answer (do you maybe have an example)? Commented Aug 26, 2015 at 8:32
• @user50224 1) I can't think of any but that doesn't mean there couldn't be one, and 2) no. I can't be sure but I have no example. Commented Aug 26, 2015 at 11:12

Since proportionality is NOT the same as equality, you cannot say that v proportional to mT is incorrect or false solely based on dimensional analysis. But when we deal with equalities, the situation is quite different, as both members of the equation, to be equal, must be of the same dimensions.

Consider the equation 3x = 9x where x is length (a dimension). This equation is dimensionally correct, but it is false or wrong (since 3x is obviously not equal to 9x).

What dimensional analysis tells us is that if the dimensions of each member of an equation are NOT equal, then what we have is an inequality. Thus something is wrong with our "equation".

We pass from a proportionality to an equality by introducing an appropriate constant in the above expression, but that does not mean that v = CmT, although made dimensionally correct by introducing the constant C expresses any true physics.