# Dimension Analysis

## Homework Statement

In the following expression x is a position and t represents time. What are the physical dimensions of each of the constants α and β? Also, for each what are the corresponding SI units?

## Homework Equations

x=α+(2/3)βt^(3/2)

## The Attempt at a Solution

I know that since x is a position, it must have the dimension of L and the unit of m. That makes the right have a dimension of L as well. That said [α]=L, but the β is throwing me off because of the t^(3/2). I can't for the life of me figure out how to manipulate that exponent to end of with a dimension of L or L/T or L/T^2.

Basically, the strategy for finding the units of β is to make sure all other units cancel besides one power of [x].

Since the factor of 2/3 doesn't matter (it is unitless or "dimensionless"), and you already have figured out α, we can write:

[x]=[βt3/2] = [β] [t]3/2
where the square brackets mean "the units of" whatever is enclosed. One way to solve for the units of [β] is to use algebra (really some sort of pseudo algebra, but everything works out just fine) like this:

[x]/[t]3/2 = [β]
or
[β]=[x][t]-3/2

So this means the units of β are units of length times time to the power -3/2, or meters per root second cubed.

If you plug this in you can verify the result:
[β t3/2] = [β] [t]3/2 = ([x][t]-3/2) [t]3/2 = [x]
as desired.

Try to figure out what must be the dimensions of β so that the dimensions of βt$^{3/2}$ ends up as L.

I tried this approach and it made sense algebraically. However, I don't know of an SI unit associated with [x][t]^-3/2. Am I missing something?

pasmith
Homework Helper
I tried this approach and it made sense algebraically. However, I don't know of an SI unit associated with [x][t]^-3/2. Am I missing something?

Yes; the SI unit is $\mathrm{m}\,\mathrm{s}^{-3/2}$.

The formula $y = x^a t^b$ makes dimensional sense for any real $a$ and $b$, and if $x$ and $t$ are measured in metres and seconds respectively then $y$ will be measured in units of $\mathrm{m}^a\,\mathrm{s}^b$.

SteamKing
Staff Emeritus