Is the physical law unit-free?

[evinda]

Hello! (Wave)

A small sphere with radius $1$ and density $p$ moves downwards with constant velocity $v$, under the influence of the gravity $g$, at a liquid of density $p_l$ and viscosity coefficient $\mu$. (The units of $\mu$ are mass per unit of length per unit of time). From results of experiments, we get the relation:

$v=\frac{2}{9} r^2 p g \mu^{-1} \left( 1-\frac{p_l}{p} \right)$

I want to check if the physical law $v=\frac{2}{9} r^2 p g \mu^{-1} \left( 1-\frac{p_l}{p} \right)$ is unit-free.

I have tried the following:

The physical law is: $f(v,r,p,g,p_l, \mu)=v-\frac{2}{9} r^2 p g \mu^{-1} \left( 1-\frac{p_l}{p} \right)$.

The quantities are:

Length: $L$
Time: $T$
Mass: $M$

So:

$[v]=LT^{-1}$
$[r]=L$
$[p]=ML^{-3}$
$[g]=M$
$[p_l]=ML^{-3}$
$[\mu]=LT^{-1}$

Let $\lambda_1, \lambda_2, \lambda_3>0$ and we consider the transformation of system of units:$$\overline{L}=\lambda_1 L, \ \overline{T}=\lambda_2 T, \ \overline{M}=\lambda_3 M $$

Then:

$[v]=\lambda_1L(\lambda_2T)^{-1}$
$[r]=\lambda_1L$
$[p]=\lambda_3M(\lambda_1L)^{-3}$
$[g]=\lambda_3M$
$[p_l]=\lambda_3M(\lambda_1L)^{-3}$
$[\mu]=\lambda_1L(\lambda_2T)^{-1}$

$$f(\overline{v}, \overline{r}, \overline{p}, \overline{g}, \overline{p_l}, \overline{\mu})=\overline{u}-\frac{2}{9} \overline{r}^2 \overline{p} \overline{g} \overline{\mu}^{-1} \left(1-\frac{p_l}{\overline{p}} \right)=\lambda_1 (\lambda_2)^{-1}u-\frac{2}{9} \lambda_1^2 r^2 \lambda_3 \lambda_1^{-3} \lambda_3 \lambda_1^{-1} \lambda_2 p g \mu^{-1} \left( 1-\frac{\lambda_3 \lambda_1^{-3} p_l}{\lambda_3 \lambda_1^{-3}p}\right)=\lambda_1 (\lambda_2)^{-1}u-\frac{2}{9} \lambda_1^{-1} \lambda_3^3 \lambda_2 r^2 p g \mu^{-1} \left( 1-\frac{p_l}{p}\right)$$

Is it right? If so, do we deduce that the physical law isn't unit-free?
Or have I done something wrong? (Thinking)

 

[jvicens]

Hello! Great job on trying to determine if the physical law is unit-free. Your approach looks correct, but there are a few things to consider.

First, the transformation of units that you have done is not a general transformation. It only works for length, time, and mass, but there may be other quantities involved in the physical law that have different units. So, we cannot assume that the transformation you have done is applicable to all quantities.

Second, the physical law itself contains a mixture of quantities with different units, such as length, density, and viscosity coefficient. This means that the physical law cannot be unit-free, as there is no way to eliminate all units from the equation.

However, we can check if the physical law is dimensionally consistent, meaning that the units on both sides of the equation are the same. In this case, the units on the left side are $LT^{-1}$, while the units on the right side are $\frac{1}{L^2}M^2$. This means that the physical law is not dimensionally consistent and there may be a mistake in the equation or the units used.

I would recommend double-checking the equation and the units to make sure they are correct. If they are, then the physical law may not be valid. I hope this helps!