# Dimensional Analysis: Combining 3 Variables

In summary, the conversation discusses choosing three variables out of four (Q, R, \mu, dp/dx) that cannot be combined into a dimensionless product. The method of using the dimensions and setting their power product equal to zero is a valid approach. However, inspection must also be used to ensure that the chosen variables make sense physically.

## Homework Statement

Hi. I have a function that contains 4 variables: Q, R, $\mu$, dp/dx

I wish to choose 3 of them, such that they cannot be combined into a dimensionless product.

I have chosen (correctly) R, $\mu$, dp/dx and I would like to know if my method sounds correct:

If we know the dimensions: [R]=[L] [$\mu$]=[ML-2T-2] and [dp/dx]=[ML-1T-1]

and I know that in order for them to be dimensionless, their power product must equal zero:

[L]a[ML-1T-1]b[ML-2T-2]c=[MLT]0

or the system

a-b-2c=0
b+c=0
-b-2c=0

By inspection, this system can only be satisfied if a=0 but that does not make any sense since R is a physical quantity.

Hence I have reasoned that these 3 variable cannot form a non-dimensional parameter by themselves.

Does this work? Thanks!

Ignore length real quick. $$[MT^{ - 1} ]^a [MT^{ - 2} ]^b$$ can NEVER be dimensionless (except for a=b=0)

Pengwuino said:
Ignore length real quick. $$[MT^{ - 1} ]^a [MT^{ - 2} ]^b$$ can NEVER be dimensionless (except for a=b=0)

Okay cool!

Also, though longer and more tiresome, my method above works right? For the same reason.

I just want a general method just in case inspection is not that obvious.

Yes, that method works.