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: $M$


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

Let $\lambda_1, \lambda_2, \lambda_3>0$ and we consider the transformation of system of units:

$$\overline{L}=\lambda_1 L, \ \overline{T}=\lambda_2 T, \ \overline{M}=\lambda_3 M $$


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

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

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


You have to take the dynamic velocity (and not the kinematic velocity) :

$$[v]=L.T^{-1}$$ $$[r]=L$$ $$[\rho]=M.L^{-3}$$ $$[g]=L.T^{-2}$$ $$[\mu]=M.L^{-1}.T^{-1}$$ $$[P]=[P_i]=M.L^{-3}$$

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 $$0=\lambda_1\lambda_2^{-1}(v-\frac{2}9r^2\rho g\mu^{-1}(1-\frac{P_i}P))$$

Hence your relation is unit-free.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.