Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the first lecture of MIT's Classical Mechanics Professor Lewin talks about Dimensional Analysis.He talks about an apple being dropped from a certain height can be quantitatively expressed as the following.

$$t\propto h^{\alpha }m^{\beta }g^{\gamma }$$ (t is proportional to height mass and gravity)

In the above equation t = time,h = height raised to some power alpha,m = mass raised to some power beta and g = acceleration due to gravity raised to some power gamma.

I understand that if you increase the height then it will take a longer time for the apple to hit the ground and that's what h raised to the power alpha indicates and same with mass too. How about g? g is suppose to be a constant right? How can professor Lewin think of raising g to some power gamma?

One other thing.Professor Lewin then evaluates $$t\propto h^{\alpha }m^{\beta }g^{\gamma }$$ by substituting $$[T]^1$$ for time,$$[L]^\alpha$$ for height(h) and $$[M]^\beta$$ for mass(m) and $$\frac{[L]^\gamma }{[T]^\gamma2}$$ for acceleration(g).

how did he end up getting $$t = c\sqrt{\frac{h}{g}}\propto \sqrt{h}$$ ? can someone explain clearly?

share|cite|improve this question
Sorry.I didn't get it. – alok Oct 26 '11 at 10:02
A different $g$ on e.g. a different planet affects $t$ in the same way. Also exponents $\alpha$, $\beta$, $\gamma$ could be negative, zero... – liberias Oct 26 '11 at 10:21
yeah i thought so.I was thinking about our moon and the jupiter! – alok Oct 26 '11 at 10:29
up vote 4 down vote accepted

In this technique you can put all variables which you think will affect time. Here, Prof. has assumed that the time taken by apple to reach the ground may depend on $h$, $m$ and $g$. Let's see on what it depends, we already have as assumed,

$t \propto h^\alpha m^\beta g^\gamma $

$ \therefore t = c \; h^\alpha m^\beta g^\gamma \qquad \text{where} \; c \;\; \text{is} \; \text{constant} \qquad \dots \dots (1)$

Putting the dimensions of $ t, h, m, g$ in above equation,

$ [T^1] = [L^1]^\alpha [M^1]^\beta [L^1 T^{-2}]^\gamma \quad c \; \text{is dimensionless}$

$ \therefore [T^1] = [L^{\alpha+\gamma}][M^\beta][T^{-2\gamma}]$

comparing the powers, we get,

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

$ \Rightarrow \alpha = \frac{1}{2} $

Putting the value of $ \alpha,\; \beta\; \text{and} \; \gamma$ in equation $(1)$, we obtain

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

Here, you can observe, how mass is out of the equation. Similarly, dimensional analysis provides you the power to assume that a quantity $x$ depends on $y$ $z$ $\dots$ At the end you will be left out with only those quantities, on which $x$ depends.

share|cite|improve this answer
Excellent.This is what i was looking for.You have explained it clearly.Thank you very much and appologies for the delayed response. – alok Oct 28 '11 at 15:25
You are welcome. – orion Oct 28 '11 at 16:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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