# Question Concerning Dimensional Analysis

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?

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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

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.

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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