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 $\gamma$?

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    $\begingroup$ Can you provide a link to the lecture and what time in the lecture he gives this equation. $\endgroup$ Commented Mar 10, 2012 at 15:38
  • $\begingroup$ ocw.mit.edu/courses/physics/… skip to around 22.40 $\endgroup$
    – ODP
    Commented Mar 10, 2012 at 15:44
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    $\begingroup$ Turn the question around: "Why would you assume that simple factors (like length, mass, ...) enter linearly?" $\endgroup$ Commented Mar 10, 2012 at 16:53
  • $\begingroup$ Thank you very much John Rennie (and I now understand what you meant in your comment dmckee). Thanks $\endgroup$
    – ODP
    Commented Mar 11, 2012 at 9:43

1 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:

$$t \propto h$$


$$t \propto h^2$$


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

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.

  • $\begingroup$ More often than not, when you have four quantities, they are interrelated. Tell me one common formula utilising M,L,T that has 4 or more independant quantities. Or some EM formula with 5 or more independant quantities. $\endgroup$ Commented Mar 11, 2012 at 4:47
  • $\begingroup$ I take your point: my point was that when you learn dimensional analysis the examples you're taught are carefully chosen! How about trying to calculate the period of the pendulum when you need to consider g, the mass of the bob, the length of the string and the amplitude of the swing. $\endgroup$ Commented Mar 11, 2012 at 7:16
  • $\begingroup$ Ooh I didn't think of that. Indeed, a non-physicist using dimensional analysis would consider those extra variables (which have power 0). There's also the 'sneaky trick' of forgetting that constants of proportionality can have dimensions. Which makes dimensional analysis pretty useless. I always hated the proof of stokes law via dimensional analysis for this reason.. $\endgroup$ Commented Mar 11, 2012 at 10:46
  • $\begingroup$ With the pendulum-four-variables example, one could argue that since T does not depend on m,A (experimentally found), we need not consider the two. But I'm no fan of dimensional analysis, so I won't defend it too much :p. Would make for a good chat discussion, though. Will bring it up sometime. $\endgroup$ Commented Mar 11, 2012 at 10:50
  • $\begingroup$ Ah, but the period does depend on the amplitude of the swing. Real pendulums (penduli?) are anharmonic oscillators so the period is amplitude dependant. You can exclude the mass, but that still leaves you two simultaneous equations in three variables. Despite all this, I still consider dimensional analysis to be very important if only as a sanity check. $\endgroup$ Commented Mar 11, 2012 at 11:05

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