PX154 - A2 - dimensions

we use dimensions to mean quantities from the SI system

quantities

luminous intensity is "visual effect of radiation"

• from these seven, we can derive all others
  • eg:
    • area - $m^2$ - $L^2$
    • density - $kg{m}^{-3}$ - $M{L}^{-3}$
    • kinetic energy - $kg{m}^2{s}^{-2}/joules$ - $M(L{T}^{-2})$
      • potential energy must be the same
    • charge - $C$ - $IT$

• homogeneity

  • total energy = kinetic energy + potential energy
  • $M(L{T}^{-2})=M(L{T}^{-2})+M(L{T}^{-2})$
  • all additive terms must have the same dimensions

• units of other quantities

angles

• using units of radians
  • $\varphi =\frac{l}{r}=$ no unit : use radians
  • "dimension of..." $=[\varphi ]$
  • angles have no dimensions

trigonometry

• $sin\varphi =\frac{opp}{hyp}$
  $\frac{[opp]}{[hyp]}=\frac{L}{L}=1⟹$ no dimensions

exponentials

• $e^x,exp(x)$
• $x$ has no dimensions
• eg: radioactive decay:
  • intensity of radiation emitted by a radioisotope
  $$I(t)=I_0{e}^{-\lambda t}$$$t=time(s),\lambda =decayconstant(s^{-1})$
  So, $[t\lambda ]=T{T}^{-1}=1⟹$ no dimensions
  -> argument for exponentials and logs has no dimensions
• eg: natural log:

might write this as:

• ideal gas law: $PV=nRT$

  • to find : dimensions of R
    $[P]=M.L^{-1}{T}^{-2}$
    $[V]=L^3$
    $[n]=mol$
    $[T]=\theta$
    so, $M.L^2{T}^{-2}=mol.\theta .[R]$
    $⟹[R]=M.L^2{T}^{-2}mo{l}^{-1}{\theta }^{-1}$
    $\therefore [R]=M{L}^2{T}^{-2}mo{l}^{-1}{\theta }^{-1}$

• use homogeneity to show that $sin(x)$ or $exp(x)$ has no dimensions

  • hint: taking series expansions
  $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$
  • in the RHS, various degrees of x are added.
  • if x had a unit, this would not be possible
  • $\therefore \sin x$ has no units
  $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
  • in the RHS, various degrees of x are added.
  • if x had a unit, this would not be possible
  • $\therefore e^x$ has no units

dimensional analysis

• let's be a little more rigorous:
• for our pendulum, we had 4 variables:

• so we should have set of variables $Q_{1...N}$
  • we also said: $T=f(l,g,m)$
    • we can write: $Q_1=f(Q_2,Q_3,Q_4)$
      • if we were interested in g: $Q_3=f(Q_1,Q_2,Q_4)$
• we wrote our answer for the pendulum:

• now we can write:

- *$\mathrm{\Pi }\ne 3.1415...$ but a convention to denote constants
- dimensions:
$$[\Pi] = [Q_1]^\alpha [Q_2]^\beta [Q_3]^\gamma [Q_4]^\delta$$
- for the pendulum:
$$[\Pi] = [T]^\alpha [l]^\beta [g]^\gamma [m]^\delta$$
$$[\Pi] = T^\alpha L^\beta {(LT^{-2})}^\gamma M^\delta$$
- LHS is dimensionless $\therefore$ RHS is dimensionless too
$$= T^\alpha L^\beta L^{\gamma}T^{-2\gamma} M^\delta$$
- $T$: $0=\alpha -2\gamma$
- $L$: $0=\beta +\gamma$
- $M$: $0=\delta$
- we have simultaneous equations for the exponents
- we can choose a value for one of the unknowns:
let $\alpha =1$
$⟹\gamma =\frac{+1}{2}$
$⟹\beta =\frac{-1}{2}$
$⟹[\pi ]=T^1{L}^{\frac{-1}{2}}{(L{T}^{-2})}^{\frac{1}{2}}{M}^0$
$\therefore \pi =T{g}^{\frac{1}{2}}{l}^{\frac{-1}{2}}=T\sqrt{\frac{g}{l}}$
$$[\Pi] = T^\alpha L^\beta {(LT^{-2})}^\gamma M^\delta$$
- $T$: $0=\alpha -2\gamma$
- $L$: $0=\beta +\gamma$
- $M$: $0=\delta$
- let $\alpha =2$
$⟹\gamma =1$
$⟹\beta =-1$
$⟹[\pi ]=T^2{L}^{-1}{(L{T}^{-2})}^1{M}^0$
$⟹[\pi {]}^{\frac{1}{2}}=T^1{L}^{-\frac{1}{2}}{(L{T}^{-2})}^{\frac{1}{2}}{M}^0$
$\therefore \sqrt{\pi }=T{g}^{\frac{1}{2}}{l}^{\frac{-1}{2}}=T\sqrt{\frac{g}{l}}$