# Do we deduce that the physical law isn't unit-free?

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?

$$[v]=L.T^{-1}$$ $$[r]=L$$ $$[\rho]=M.L^{-3}$$ $$[g]=L.T^{-2}$$ $$[\mu]=M.L^{-1}.T^{-1}$$ $$[P]=[P_i]=M.L^{-3}$$
Thus our relation becomes $$0=\lambda_1\lambda_2^{-1}v=\frac{2}9\lambda_1^1r^2\lambda_1^{-3}\lambda_3\rho\lambda_1\lambda_2^{-2}g\lambda_1\lambda_2\lambda_3{-1}\mu^{-1}(1-\frac{\lambda_1^{-3}\lambda_3P_i}{\lambda_1^{-3}\lambda_3P})$$
Which is $$0=\lambda_1\lambda_2^{-1}(v-\frac{2}9r^2\rho g\mu^{-1}(1-\frac{P_i}P))$$