Dimensional Analysis with $\alpha$, $\beta$, and $\gamma$ Powers

Question

In Prof. Walter Lewin's Dimensional Analysis lecture, he stated that:

$$t ~\propto~ h^α m^β g^γ$$ ($\alpha$, $\beta$ and $\gamma$ all to some power of their unit).

Why does he put $h$, $m$ and $g$ to some power of $\alpha$, $\beta$ and $\gamma$?

Answer 1

The argument is that the time it takes the apple to fall must depend on the height, because that's common sense. But we don't know how $t$ varies with $h$. For example we could have:

$$t \propto h$$

or

$$t \propto h^2$$

or

$$t \propto h^{\frac{1}{2}}$$

and so on. So the lecturer is saying, because we don't know how t depends on h let's just put:

$$t \propto h^{\alpha}$$

and then we'll use dimension analysis to calculate $\alpha$. Likewise the lecturer is assuming that mass and gravity are also important, and again we don't know how the time depends on them, so he's putting them in to the power $\beta$ and $\gamma$ and then he'll calculate what $\beta$ and $\gamma$ are. That's why you get:

$$t = h^\alpha \times m^\beta \times g^\gamma$$

The cunning trick is that $h$ has units of length ($L$), $m$ has units of mass ($M$) and $g$ (an acceleration) has units of length per second$^2$ ($LT^{-2}$), and of course time has units of time ($T$). If we put these into our equation we get:

$$T = L^\alpha \times M^\beta \times (LT^{-2})^\gamma$$

The only way this can be true is if $\beta = 0$, $\gamma = -\frac{1}{2}$ and $\alpha = \frac{1}{2}$ i.e.

$$t \propto \sqrt{\frac {h}{g}}$$

However there is a sneaky trick here, and I remember this annoyed me when I was first learning about dimensional analysis. The analysis gives you three simultaneous equations for $M$, $T$ and $L$, so when you have three variables like $\alpha$, $\beta$ and $\gamma$ you can solve the simultaneous equations and find $\alpha$, $\beta$ and $\gamma$. But suppose you put in a fourth parameter to the power $\delta$. Then you have four unknowns and only three equations and you can't solve them.