1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dimensional Analysis

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data

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

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

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

    If we know the dimensions: [R]=[L] [[itex]\mu[/itex]]=[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:


    or the system


    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!
  2. jcsd
  3. May 30, 2009 #2


    User Avatar
    Gold Member

    Ignore length real quick. [tex][MT^{ - 1} ]^a [MT^{ - 2} ]^b [/tex] can NEVER be dimensionless (except for a=b=0)
  4. May 30, 2009 #3
    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.
  5. May 30, 2009 #4


    User Avatar
    Gold Member

    Yes, that method works.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Dimensional Analysis
  1. Dimensional analysis (Replies: 4)

  2. Dimensional Analysis (Replies: 5)

  3. Dimensional Analysis (Replies: 1)

  4. Dimensional analysis (Replies: 3)