1. Not finding help here? Sign up for a free 30min 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. Aug 28, 2012 #1
    Dimensional analysis....

    I am having a real hard time wrapping my head around dimensional analysis. Can someone please explain it in english? Thanks.
     
  2. jcsd
  3. Aug 28, 2012 #2
    Re: Dimensional analysis....

    Every physical quantity f has dimensions ([f]) w.r.t. the seven base physical quantities:

    • time T
    • length L
    • mass M
    • electric current I
    • thermodynamic temperature [itex]\mathrm{\Theta}[/itex]
    • amount of substance N
    • luminous intensity J

    Luminous intensity is a photometric quantity. Its unit is the candela (cd). It is a subjective measure of how the human eye perceives light. If one is concerned with objective measures of radiation fluxes, then this unit is not used.

    Amount of substance is directly proportional to the number of elementary entities in the ensemble. The unit is called a mole. The actual number of particles present in a mole of substance is considered a physical constant, although I would think that it makes more sense to call it a conversion factor. It is the Avogadro's constant:
    [tex]
    N_A = 6.022 \times 10^{23} \, \mathrm{mol}^{-1}
    [/tex]

    Similarly, due to developments in Statistical Physics, it had been shown that temperature is directly proportional to some energy ("thermal energy"). The unit of thermodynamic temperature is kelvin (K), whereas the unit of energy is joule (J). A conversion factor that converts any temperature (T) into energy units ([itex] \theta = k_B \, T[/itex]) is the Boltzmann's constant:
    [tex]
    k_B = 1.3806 \times 10^{-23} \, \frac{\mathrm{J}}{\mathrm{K}}
    [/tex]

    The above two constants combine to give another famous constant - the Universal gas constant:
    [tex]
    R = k_B \, N_A = 8.314 \, \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}
    [/tex]

    The unit of electric current is the ampere (A). If you look at its definition, it is defined through the force (a mechanical quantity) between two long straight conductors with given length, and a certain distance apart (both mechanical quantitites). Thus, there must be a conversion factor between mechanical quantities (time, length, and mass), and the electric quantitiy. Indeed, there is such a conversion factor, called the vacuum permeability:
    [tex]
    \mu_0 = 4\pi \times 10^{-7} \, \frac{\mathrm{N}}{\mathrm{A}^2}
    [/tex]
    Thus, in principle, all quantities have dimension only w.r.t. to the three mechanical base units:
    [tex]
    [f] = \mathrm{T}^{\tau} \, \mathrm{L}^{\lambda} \, \mathrm{M}^{\mu}
    [/tex]
    The exponents (the τ, λ, and μ, which are pure rational numbers) are the corresponding dimensions w.r.t. to each base quantity.

    Some rules for dimensions are:
    • A dimensionless quantity has zero dimension w.r.t. each base physical quantity [itex][f] = \mathrm{T}^0 \, \mathrm{L}^0 \, \mathrm{M}^0 = 1[/itex].
    • Both sides of an equality have the same dimension, i.e. [itex]f = g \Rightarrow [f] = [g][/itex]
    • You may only add (or subtract) quantities with the same dimension, i.e. [itex]h = f \pm g \Leftrightarrow [f] = [g] = [h][/itex].
    • The dimension of a product (quotient) of two physical quantites is he product of their respective dimensions, i.e. [itex][f \cdot g] = [f] \cdot [g][/itex] (the exponents for each base quantity add (subtract).
    • A mathematical function (exp, log, sin, cos, etc.) accepts a dimensionless argument and returns a dimensionless result, i.e. [itex]y = f(x), \Rightarrow [x] = [y] = 1[/itex].
     
    Last edited: Aug 28, 2012
  4. Aug 28, 2012 #3
    A short and simple explanation? Break down the units of some quantities, cancel some units if possible, and make some kind of comparison of the units. Don't let the word dimensional scare you. Just think "do the units make sense?"
     
  5. Aug 28, 2012 #4
    Re: Dimensional analysis....

    Example:
    Let us look at the following formula:
    [tex]
    v = v_0 + a \, t
    [/tex]
    according to the summation rule, we know that we must have:
    [tex]
    [v] = [v_0] = [a \, t]
    [/tex]
    Further, according to the product rule, we also have:
    [tex]
    [a \, t] = [a] \cdot [t]
    [/tex]
    Thus, if we want to express the dimension of a, we would have:
    [tex]
    [a] = \frac{[v]}{[t]}
    [/tex]
     
  6. Aug 29, 2012 #5
    Re: Dimensional analysis....

    its just weird because for kinematics I am having a way easier time then the majority of the homework in chapter one which involves alot more than just changing units.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dimensional analysis
  1. Dimensional Analysis (Replies: 0)

  2. Dimensional Analysis (Replies: 5)

  3. Dimensional Analysis (Replies: 1)

  4. Dimensional analysis (Replies: 3)

Loading...