Astrophysics/Physical dimensions & analysis
Hello. I was fascinated by the exchange of Prof. James Gort with someone named Brianna to do with how certain units and such just don't work out, though they look like they do. I am concerned about falling into a similar trap, say by venturing into a website that might not have all its marbles and wonder if you could illustrate techniques using actual physics examples - say for dimensional analysis. Thanks!
Thanks for your question. You may also have seen that this same individual had an exchange with me, e.g.
which also didn't end too auspiciously for her. I noted similar problems to those in the exchange you cited, as well as the fact the cited website she used was bare balderdash and only fit for cranks. But these are the problems we now confront, with spurious websites that profess "real" science but actually only a sham. Sadly, they lure too many to expend their mental talents on nonsense instead of directing those energies to legitimate science.
As I also noted, if she had any rudimentary skills at units or dimensional analysis none of those problems would have materialized. Hence, you are right to be concerned yourself and to want to learn of the methods of unit and dimensional analysis which one can say are the groundwork for physics, since above all physics is a science of measurement. And one must know what units, dimensions one is measuring for the end quantity to have any meaning.
Let me begin by giving a few basic examples on how one can check physical units for consistency first noting the basic S.I. units (I Table I) as well as the derived units in Table II, and the related dimensional quantities.
You can consult either or both tables as we look at the examples which follow:
1) Current as function of drift velocity:
We're told: 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.
If then all the above factors are multiplied together they ought to give the current in amperes (A) or Coulombs (charge) per second. For this first example let us include numbers as well as units.
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 1^0 -19 C
Area A is: 2 x 10^-5 m^ 2
The number density is: n = 7.4 x 10^ 28 /m3
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 ^2 )
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:
(m^3) / (m^3) x (C/ s)
Since the m^3 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) Find the correct units for the thermal conductivity k, if we have the relation:
Q = k A T(t2 - t1)/L
We will leave out the precise numbers and just work with the units here. Q will be in joules (J), A the area in m^2, T the temperature in C degrees (Celsius), L in meters, m,and the time difference in seconds, s.
Then we can write for k:
k = QL/ A T(t2 - t1)
Writing in terms of the units only:
k(?) = (J) (m)/ (m^2) (s) (C)
k (?) -> (J/s)/ mC = W/m C
Since joules per second is the same as watts. SO the units for k will be watts per meter per degree Celsius (or also per degree Kelvin)
3) Another example for a number which should be dimensionless. This is ehe magnetic Reynolds number (used in solar physics), defined:
R_m = L V_A / n
Where L denotes a scale length (m), V_A is the Alfven velocity (m/s) and n is the magnetic diffusivity in m^2/ s
Check the end units, which ought to be NO units:
(m) (m/s)/ (m^2 /s)
So: (m^2 /s) / (m^2 /s) = 1
We see m^2 cancels with m^ 2 and s with s leaving a dimensionless result.
There is also more detailed dimensionless 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.
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, c , of the liquid
b) The radius, r, of the pipe
c) delta P/L the pressure gradient arising from the flow
Then we may write as a first approximation to the actual equation: V = k c^x r^y (delta P/L)^z
Where k denotes an unknown constant, and x, y and z are indices whose values are to be found.
We also know (check the tables):
V has dimensions L^3 T^- 1 (and is also per time in this case)
c has dimensions M L^-1 T^ -1
r has dimension L
delta P has dimensions M L^-1 T^- 2
So, delta P/L has dimensions: M L^-2 T^- 2
Then, equating dimensions:
L^3 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 c^-1 r^4 (delta P/L)
Or: V = k (delta P) r^4 /c L
We note here that k cannot be obtained by the method of dimensions, but a fuller analysis would reveal k = pi /8, so:
V = pi (delta )P r 4 /8 c L (Poiselle’s formula)
These methods are employed by real world scientists to check the validity of any type of equation proposed to account for some process, phenomenon, relationship. They are therefore extremely useful in separating 'wheat from chaff' as a first approximation. And they are certainly useful in exposing bogus equations such as present on the website "Bri" referenced and which she believes to be the "last word".