March 3, 2011


Hurray, an educational post.

Now, many of us might have problems in understanding this simple topic in the STPM physics syllabus (myself formerly included).

[Note: You can skip over the ugly maths parts, don't worry]

1.2 Dimensions of Physical Quantities

One might not understand it due to missed classes, but it's actually very, very simple. The definition in the book is complex and will confuse newcomers, therefore I will rely on one simple sentence to assist us in our understanding of this topic.


Dimension, in physics, an expression of the character of a derived quantity in relation to fundamental quantities, without regard for its numerical value.

Now, in the world where the metric system is applied, certain quantities are considered fundamental, and other quantities maybe derived from them.

We now introduce the fundamental physical quantities-
Mass kg
Length m
Time s
Electric Current A
Temperature (°C/K/°F)
where the quantities in bold are the ones we will be focusing on.

I will now give the quantities symbols-
Mass kg                   M
Length m                  L
Time s                       T

All other quantities in physics can be derived from these fundamental quantities. In example:

[speed] <==This means the dimensions of speed. Therefore,

[speed] = [L]/[T]

Whether it's metres per second, or inches per year, that's not what matters. It has the dimension length per time. Similarly,

[Volume] = [L]^3
[Density] = [M] / [L]^3 (mass per unit volume)
[Acceleration] = [L] / [T]^2 = [L][T]^-1
[Pressure] = ( [M][L][T]^-2 ) / [L]^2 = [M][L]^-1[T]^-2

There, not so hard, is it?
Note: Dimensions are NOT units.
Note2: Some quantities do not have units and are therefore, dimensionless. (numbers, ratios)

Congratulations, you have just completed a topic in STPM Physics.

Next, I want to show you the Uses of Dimensions.

-Dimensions can be used to determine the units of a physical quantity. However, the formula for the quantity must first be known.

-Dimensions can also be used to check the homogeneity of equations(each term in the equation must have the same dimensions)

(To elaborate on both of these would be to stick the contents of the textbook on this page, and that's a little counter-productive. Believe me, I tried.)

Using dimensions, we can derive an expression to show how a physical quantity is related to other different physical quantities.

[t] = T
where the period t of a loaded spring depends on the load m and the force constant k of the spring.

By letting t be cm^(x)k^(y) where c is a dimensionless constant and x and y are powers that are yet to be determined, we have-
[cm^(x)k^(y)] = M^(x)(MT^(-2))^(y)

Confused? Let me show you.

c = numerical constant = no dimensions
[cm^(x)k^(y)] = M^(x)(MT^(-2))^(y) 

where [m] = M and [k] = MT^(-2)

Got it? Let's proceed.

T = M^(x)(MT^(-2))^(y) 
expand, and we get
T = M^(x+y)T^(-2y

Equating the indices of each dimension will give
T: 1 = -2y => y = -(1/2)
M: 0 = x+y => x = (1/2)                            (anything to the power of 0 will equal to 1)

Hence, T = cm^(1/2)k(-1/2) 

T = c sqrt(m/k)  

And all others can be done in similar fashion.

That's it for dimensions~! Not so hard, isn't it? 

1 comment:

  1. Holy jesus that's a lot of math.
    A. Lot. Of. Math.


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