Question

I was watching the video and i still dont get the dimensional analysis? Can someone help

Answer 1

What's up Vrex717? to answer your question

Physical quantities, as the name implies are measurable properties that describe a system's physical state; a system being a portion of the universe chosen for analysis.

All physical quantities are derivations of the three fundamental quantities: Length (L), time (T), and mass (M). We use physical quantities in calculations involving the physical realm.

For example, if we choose a block of wood to be our system, a few physical quantities of this block of wood and their dimensions would include the surface area of the block (which has dimensions L^2) and its volume (which has dimensions L^3).

We can determine the dimensions of these physical quantities by looking at their formulas:

The Volume of a cube, for example, we calculate by multiplying the value of its length by its width and then by its height; because volume is the amount of three dimensional space that an object encloses (how far this object extends in the x, y, and z directions of a 3d coordinate system, and the subsequent space it encloses)

Therefore the Volume of a cube=length*width*height or V= side*side*side (because all sides are the same length)

The measurement of each side (for example 20cm) would have the dimension of length (intuitive right?)

So we can substitute into the volume formula the dimension of each measurement
V=l*w*h which then becomes Volume= L*L*L
and we can see that physical quantity Volume would have the dimensions of Length cubed or L^3.

Now, this ability to evaluate the physical quantities of systems in terms of their dimensions is very valuable. Because if we had wanted to find out whether or not one physical quantity affects the outcome of another, we would be able to set up a relation connecting all viable factors to the expected result. For example, (citing professor Lewin's example) if we had wanted to find out whether or not an object's mass determines how long it will take to fall toward the Earth (under the influence of the Earth's gravitational pull), we could just set up a simple relation between factors that we think would influence the time it will take for an object to fall.
so we begin by rationalizing that the time an object takes to fall is proportional to the height of that object, or h, to some power lambda (which we are unsure of)
\[t \alpha h ^{\lambda}\]

we then assume that the time it takes for an object to fall under the influence of gravity is also proportional to the mass of that object, or m, to some power beta (which we are unsure of), so our relation becomes
\[t \alpha h ^{\lambda}m ^{\beta} \]

lastly, we assume that the time it takes for an object to fall under the influence of gravity is also proportional to the value of the gravitational acceleration present, or g, to some power omega (which we are unsure of), so our relation now becomes
\[t \alpha h ^{\lambda}m ^{\beta}g ^{\omega} \]

we then express this relation of physical quantities in terms of their dimensions so that we may evaluate the corresponding powers of lambda, beta, and omega.
Time has dimensions T, height has dimensions L, mass has dimensions M, and g (which is an acceleration) has dimensions \[L/T ^{2}\]

we substitute these dimension into our previously established relation:
\[t \alpha h ^{\lambda}m ^{\beta}g ^{\omega} \]

and obtain
\[T \alpha L ^{\lambda}M ^{\beta}(L/T ^{2})^{\omega} \]

*note dimensions work exactly like units, if we are multiplying two lengths 20cm and 10cm, we obtain a final answer of 200 cm squared, the units are equal on both sides of the equation. Like units, dimensions must be consistent on both sides of an equality or a relation. Just as we can't have only cm on one side and end up with an answer in kilograms on the other, we can't have only Time dimensions on one side and end up with a Length dimension on the other.

with that being said , we see that there is no Mass dimension on the other side of the relation, so therefore, we can take lambda to be 0 and mass is eliminated from the relation. As we can already see, the time it takes for an object to fall under the force of gravity is not proportional to the mass of the object.
our equation becomes:
\[T \alpha L ^{\lambda}(L/T ^{2})^{\omega} \]
we now distribute omega with the acceleration dimensions and obtain
\[T \alpha L ^{\lambda}(L ^{\omega}/T ^{2\omega})\]
We know that in order to have a balance of dimensions on both sides of the relation (T=T) the following conditions need to be met:

\[\lambda + \omega = 0\] (so that L dimensions cancel) and

\[-2\omega=1 \]
(so that there is one dimension of T on each side)

*the value of -2omega happens to be negative due to the fact that we brought the power of T upstairs in the equation( from the denominator) for analysis.

if \[-2\omega=1 \]
then\[\omega=-1/2\]

and if
\[\omega=-1/2\]
then \[\lambda=1/2\]
because \[\lambda + \omega = 0\]

therefore, the time it takes for an object to fall is proportional to some constant C (to allow for any factors we cannot determine at the moment), times its height to the power of one half, times the gravitational acceleration to the power of negative one half

\[t \alpha Ch ^{1/2}g ^{-1/2}\]
which can also be written as
\[t \alpha C \sqrt{h/g}\]
we were able to determine this because earlier we established the relation between the dimensions of time, height, mass, and gravitational acceleration of a falling object, and we had also determined the value of their powers lambda, beta, and omega. The relation above is just a combination of all that we have determined thus far.

This process of checking relations between physical quantities through use of their dimensions is known as dimensional analysis, and as you’ve seen it is a very useful and powerful tool when working with physical quantities and phenomena.

Answer 2

Sorry that was so long, I just wanted to make sure it was thorough enough.
I hope this helps :D

Answer 3

and as a correction, earlier in the response I said we take lambda to be zero so that mass will be eliminated, I meant to type beta, sorry. Oh, and I really didn't mean to spell things out, I just wanted to stress important parts so that this question can be used as a reference for others who might be weaker in the subject.

Answer 4

there are seven basic dimentional quantities.....you are jst goin to play wid that important being kg(M),meters(L),seconds(T),kelvin(K),ampere(A)....i wuould jst like to explain you wid 1 example....suppose they ask you the dimention of force....you know that F=ma where m is mass and a is acceleration.......
now mass is kg(M)
acceleration=velocity/time=distance/time/time=lenght/second squared(L/T*T)
this is nothin but newtons......
work done(W)=F*d=newtom metre.
so now you tell me its dimentions........