This is a continuation of this question.

http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-1/ skip this lecture to around 25:50.

After doing dimensional analysis on $t\propto h^\alpha m^\beta g^\gamma$ Lewin concludes that:

$$\alpha = \frac{1}{2}, \gamma = -1/2, \beta = 0$$

This is all fully understood, but he then goes to conclude from this that:

$$t = C \sqrt{\frac{h}{g}}$$

How did he get to this? And why is he allowed to just assume that there is a constant, C, there when he doesn't even know its value or what it is?

Keep things as simple as possible please, I'm 16.

  • $\begingroup$ or alternatively watch from 24minutes if you want easier understanding of my question. $\endgroup$
    – ODP
    Mar 13, 2012 at 23:12
  • $\begingroup$ and can someone PLEASE tell me how to type the god forsaken mathematical symbols like alpha and square route sign ahhahaa $\endgroup$
    – ODP
    Mar 13, 2012 at 23:14
  • $\begingroup$ Regarding the symbols, use TeX. Enclose your symbols in $.....$ and $$....$$ (this one makes it larger and on its own line). You may want to click "edit" on your question to see what I did. $\endgroup$ Mar 14, 2012 at 0:35
  • 1
    $\begingroup$ The randomly capitalized words, and demands for people to watch a video rather than simply summarizing the problem do not help attract people to answer your question. $\endgroup$
    – Colin K
    Mar 14, 2012 at 2:19
  • $\begingroup$ I've +1'd this to remove the downvote because I think it's a fair question, and because I asked which bit of the video to watch for the last question, so it's partly my fault Olly has asked us to watch the video :-) @Olly: is Manishearth's reply understandable? I worry his rant (which I basically agree with) may have clouded his answer a it. I'm happy to reply if you want more detail, but it would be plagiarising Manishearth! $\endgroup$ Mar 14, 2012 at 12:02

4 Answers 4


As requested, but please don't mark this as the answer because it's a rehash of Manishearth's post:

When physicists say for example "time is proportional to $\sqrt{h}$" what they mean is that:

$$t = C \times \sqrt{h}$$

where C is just a number, called the constant of proportionality, and it doesn't change. The only way to find out the value of the constant $C$ is to measure $t$ for a variety of values of $\sqrt{h}$. Typically you would draw a graph of $t$ against $\sqrt{h}$ and you should get a straight line. The gradient of the line gives you the value of $C$. If you don't get a straight line then you've made a mistake somewhere and $C$ isn't constant.

So the lecturer in the video starts by guessing that $t$ is proportional to $h^\alpha m^\beta g^\gamma$, and what he means by this is:

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

where $C$ is the constant of proportionality. Dimensional analysis can't tell you what the constant of proportionality is. The only way to find the value of $C$ is to do the experiment.

  • $\begingroup$ Don't worry about plagiarizing my answer. If you can write a clearer version, then just do so :). I don't care much about rep, so the best answer should get the green tick, not the confusing, rant-y one. $\endgroup$ Mar 15, 2012 at 15:50
  • $\begingroup$ I just realised that I end up saying 'this is in addition to JohnRennie's excellent answer' often, and the same goes for you. It's happened atleast twice both ways. Pretty strange =P $\endgroup$ Mar 15, 2012 at 17:20

The next answer is wrong and should not be considered, please.

Why a constant like the one $C$ in: $t = C \sqrt{\frac{h}{g}}$

You can express time in hours, or minutes or years, ...
You can express length in meters, miles or light-years, or ...

Clearly the constant we have to use is dependent on the choice of the units. We can also choose a system of units (existent or created ad-hoc because it is a convention) in such a way that $C=1$.

Edit add, after the correct remark of Mr Mark.
Indeed C is dimensionless. I beleive that it is tied to the geometry of the problem and, in this case, I think that its value is $2\pi$, normally associated to a planar problem (2D) involving angles. The value of $2\pi$ is not only a number but the expression of the measure of an angle (360°) (the length of the circle with radius=1 u is 2$\pi\cdot u$, it have length units!) .
I will show with an example why the above answer is wrong.
The formula in question is the pendulum equation for small apertures and gives its period in function of the lenght $h$ and the $g$ gravitational acceleration.
For instance using $100\:\rm{m}$ and $9.8 \:\rm{m/s^2}$, it is $ t=2\pi\sqrt{\frac{100}{9.8}}, t=20.07s$ the period of such pendulum (in the usual units meter and second).
Now if we adopt a unit of measure of time ($ut$) with the double of the extension of one second, say $2s=ut$, and keep measuring in meters. The period in this units has the value $10.035\:\rm{ut}$ (half of the $20.07$ but it has the same extension of time) . Clearly we can not use the value $9.8\:\rm{m/s^2}$ for the acceleration $g$. We have to express the acceleration in the units $\:\rm{m/ut^2} $ which is $g = 39.20 \:\rm{m/ut^2}$.

  • $\begingroup$ This is completely wrong. $C$ doesn't have any units in the equation and is independent of what units you use to measure. $\endgroup$ Apr 4, 2012 at 3:52
  • $\begingroup$ @Mark Eichenlaub - You have reason. I posted a wrong answer. Sorry to all. $\endgroup$ Apr 5, 2012 at 0:49

Remember there's a proportionality sign $\propto$ $$t\propto h^\alpha m^\beta g^\gamma$$ Whenever we have $A\propto B$, then $A=CB$, where C is some constant.For example, gravitational force on the surface of the earth $\propto m$, but it's actually $mg$, where $g$ is (sort of ) constant.

For a pendulum, in SI units, the $c$ is $2\pi$.

The point is, while using dimensional analysis, you know which quantities your variable depends on, but not how. Generally, a dependency is of the form $kA^\alpha$, where A is the variable, and the other two are constants that you need to determine. You can determine $\alpha$ via dimensional analysis, but $k$ requires experimental data. The $c$ comes directly from the product of the $k$s.

  • 3
    $\begingroup$ Your whole "what I hate about this" section is wrong, or at least very confused. If you have constants of proportionality with unknown units, there are some serious problems. Nobody would ever do that; maybe you just never had that pointed out, because a lot of people consider that obvious (no offense). $\endgroup$
    – Colin K
    Mar 14, 2012 at 2:26
  • $\begingroup$ @ColinK: what I'm trying to say is, how do we know that the constant of proportionality for time period is dimensionless? There hypothetically could be some 'universal pendulum constant' with dimensions.. I know there isn't, but I feel that is only obvious in retrospect. Basically I don't like the idea that dimensional analysis is made to look like it can be used to derive formulae.. None taken :D $\endgroup$ Mar 14, 2012 at 3:02
  • $\begingroup$ @ColinK: Could you specify the unclear/confused parts and why exactly "nobody would do that"? Also the "serious problems" (aside from the one I pointed out in my answer.. $\endgroup$ Mar 14, 2012 at 3:06
  • $\begingroup$ If the result must be in units of time, then there is no uncertainty about the units of the constant. How could there be? If there were, it would imply that you don't know the units of one side of your equation, of of your other variables. That doesn't happen. $\endgroup$
    – Colin K
    Mar 14, 2012 at 3:10
  • $\begingroup$ @ColinK "it would imply that you don't know the units of one side of your equation". Could you explain? For example, why can't the pendulum equation be $k\sqrt{\frac{l^3}{g}}$, where $k$ is some universal wavenumber? Yes, the example is a bit absurd, and we don't see this happening too much in physics. My point is that dimensional analysis is portrayed as a way of deriving equations, as well as a sanity check. I've even seen it used to "derive" Stokes' law (the viscosity one). I feel that this is an erronous way of looking at it.. $\endgroup$ Mar 14, 2012 at 3:20

Have not watched your lesson but your question was:

$$t= C\sqrt{\frac{h}g}$$

How did he get to this?

$h^{1/2} = \sqrt{h}$

$m^0 = 1$ (not used)

$g^{-1/2} = \sqrt\frac1g$

$t$ could therefore be said, to be proportional to $\sqrt{h}$, since, $\sqrt{h}$ is the only variable besides $t$.

$\therefore t= C\sqrt{h}$ where $C = \sqrt\frac1g$ (a constant)

Not sure why g was left in the equation, rather than including it in with constant C, but it was, so

$t= C\sqrt\frac{h}{g} $

This is the standard formula for,

Time $t$,taken for an object to fall distance h.

In this formula $C = \sqrt2$

$g$ is a constant, if distance $h$, is not excessive.

Hope that makes sense to you.


Left this as an answer, to your other similar question.

Yes $g$ is a constant, so can be dropped from the equation.

So $t$ is proportional to $\sqrt h$

or $h$ is proportional to $t^2$

I.E. Distance $h$,is proportional to the square of the elapsed fall time, $t$.

  • $\begingroup$ For all near Earth tests, small g, is a constant, which is known as “standard gravity” (Google this) and its value is: g=35.30394 (km/h)/s (≈32.174 ft/s2). Because you are 16, I assumed you would be using standard gravity. I did say in my answer above, that I was surprise your teacher had left g within the square root. It could have been included in the constant C, and would then, not have caused you confusion. Once you get further away, from Earth, variations in g become greater. A big G is use for this new variable. This can also cause confusion if g and G are mixed up. $\endgroup$ Apr 4, 2012 at 14:47

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