Thursday, March 19, 2015

An Introduction To Dimensional Analysis

In the post before last I cited a loony question received (at 'All Experts') from a clueless individual who sincerely believed her "pure light speed math" held the keys to the universe. She presented a link to a Canadian website 'Einstein now', e.g.

Which confirmed that all the units were wacky and the math actually improper. Moreover, none of the equations offered were complete which she defended in a subsequent comment. Of course, when a site presents incomplete equations we know we've ventured into crank territory, or should. And cranks often try to gain respect by pressing legit sites to bestow some gravitas - but we never do. What we do is call them "rubbish".

The young lady never would have been trapped in such bunkum had she possessed even rudimentary skills at dimensional analysis (the basis for all good physics) and also had an acquaintance with proper units as well as significant figures.  In this post we examine some examples of dimensional analysis which can show whether an equation that purports to show a force, velocity or current, really does. This ought to be in the "kit bag" of any person who claims to do physics, "pure light speed" (sic) or not.

Units and dimensional analysis lay the groundwork for physics, since above all physics is a science of measurement and obtaining correct values while recognizing the associated errors is essential. We begin with the base  and derived units for the S.-I. system given in the link below:

The units are all metric which is used because it has the greatest functionality, convenience as well as rationality, as opposed to English units.  Let's look at a few basic examples (the readers should confirm familiarity with the assorted unites used from the table first):

1) Current as function of drift velocity:

By definition  : I = nevA

where v is the drift velocity, n is the density of electrons per unit volume, A is the area of the conductor, and e, the electronic charge.

We know v will be in m/s (meters per second) and assume here: 8.4 x 10 -4  m/s

e = 1.6 x 10 -19 C

Area A is: 2 x 10  -5 m 2

The number density is: n =  7.4 x 10 28 /m3

If then all the above factors are multiplied together they ought to give the current in amperes (A) or Coulombs (charge) per second.

We write:

I  = n e v A =

( 7.4 x 10 28 /m3)(1.6 x 10 -19 C) ( 8.4 x 10 -4  m/s) (2 x 10  -5 m )

Note when multiplying the factors together one must also attend to units and how they combine on multiplication.  Paying attention to the units factors alone we will see:

(m3) / (m3)  C/ s

Since the cubic meters cancel out that leaves only C/s or coulombs per second which we note from the table is A (amperes).

Thus, I  is indeed confirmed as a current.

2) The wavelength of the photon emitted from an atom, going from higher energy E2 to lower E1

We know this should be in m.

The equation to test is:  l =   hc/ (E2 – E1) 

Where h = 6.626069 x 10- 34 J-s   (The Planck constant)

c = 3 x 10 8 m/s

E2 - E1 =   16.4 x 10-19 J 


l =    (6.626069 x 10- 34 J-s)(3 x 10 8 m/s)/ 16.4 x 10-19 J 

Check first the unit product - that it is really in m:

(J-s) (m/s) / J    =   (J/ J ) (s/s ) m  = m

so the end unit is m, making the dimensional character correct.

3) The magnetic Reynolds number is supposed to be dimensionless

 It is defined:   Âm  =  L VA / h

The units for  magnetic diffusivity,  h   are m2 /s

 V is in m/s

L is in m

Check the end units, which ought to be NO units:

(m)    (m/s)/  (m2 /s)

So:   (m2 /s) /   (m2 /s)  =   1

 We see m2 cancels with  m2  and s with s leaving a dimensionless result.

There is also more detailed dimensional analysis which focuses on treating the dimensions of the equation as opposed to units. Thus, this approach may be used to check the homogeneity of physical equations, or obtain a useful empirical equation from basic measurements.

Consider a volume V of a liquid flowing per second through a pipe under steady pressure as shown below:

It is reasonable to assume, in the first instance, to assume that V (volume flow per unit time) depends upon:

a)     The coefficient of viscosity, h, of  the liquid

b)     The radius, r, of the pipe

c)      DP/L the pressure gradient arising from the flow

Then we may write as a first approximation to the actual equation: V = k hx ry (DP/L)z

Where k denotes a constant, and x, y and z are indices whose values are to be found.

We know:

V has dimensions L3 T - 1

h has dimensions M L-1 T -1

r has dimension L

DP has dimensions M L-1 T - 2

So, DP/L has dimensions: M L-2 T - 2

Then, equating dimensions:

L3 T – 1  = [M L-1 T -1]x [L]y [M L-2 T - 2]z

We next equate indices for M, L, and T on both sides:

For M:  0 = x + z

For L: z = -x + y – 2z

For T: -1 = -x – 2z

Next, solve for each of the indices:
a)     x = -z
b)     -1 = -(-z)- 2z = -z or z = 1
c)      -1 = -x -2z
d)     3 = 1 + y -2 or y = 4

Finally:  V = k h- 1 r 4 (DP/L)1

Or: V =  k DP r 4 /h L

We note here that k cannot be obtained by the method of dimensions, but a fuller analysis would reveal k = p/8, so:

V = p DP r 4 /8 h L    (Poiselle’s formula)

1)     Use the method of dimensions to obtain a formula for the force experienced by a sphere of radius r moving at velocity v through a fluid of viscosity h.  (Hint: Set out the force as             F = k hx v y r z  )

2) Repeat the exercise above to obtain the equation  for the period (T) of a pendulum’s swing. (Hint: The pendulum's period T should depend on its length L and the acceleration of gravity, g.)

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