The expression of dimensional representation is also called “dimensional formula” of the given physical quantity. For brevity, we do not include basic dimensions, which are not part of derived quantity, in the dimensional formula.
Force
,
[
F
]
=
[
M
L
T

2
]
Force
,
[
F
]
=
[
M
L
T

2
]
Velocity
,
[
v
]
=
[
L
T

1
]
Velocity
,
[
v
]
=
[
L
T

1
]
Charge
,
[
q
]
=
[
A
T
]
Charge
,
[
q
]
=
[
A
T
]
Specific heat
,
[
s
]
=
[
L
2
T
2
K

1
]
Specific heat
,
[
s
]
=
[
L
2
T
2
K

1
]
Gas constant
,
[
R
]
=
[
M
L
2
T

2
K

1
m
o
l

1
]
Gas constant
,
[
R
]
=
[
M
L
2
T

2
K

1
m
o
l

1
]
It is conventional to omit power of “1”. Further, in mechanics, we can optionally use all three symbols corresponding to three basic quantities, which may or may not be involved. Even if one or two of them are not present, it is considered conventional to report absence of the dimension in a particular basic quantity. It is done by raising the symbol to zero as :
[
v
]
=
[
M
0
L
T

1
]
[
v
]
=
[
M
0
L
T

1
]
It should be noted here that dimensional representation of a physical quantity does not include “magnitude (n)” of the physical quantity. Further, dimensional formula does not distinguish nature of quantity. For example, the nature of force (or types of force) has no bearing on its dimensional representation. It, then, also follows that variants of a physical quantity bears the same dimensional representation. For example, velocity – whether instantaneous, average, relative – has the same dimensional representation
[
L
T

1
]
[
L
T

1
]
. Moreover, dimensional representation of a vector and its scalar counterpart is same. For example, the dimensions of velocity and speed are same. Further, dimensions of the difference of a physical quantity are same as that of physical quantity itself. For this reason, dimensions of velocity and difference of two velocities are same.
A physical quantity need not have dimensions in any of the basic quantities. Such is the case, where physical quantities are equal to the ratio of quantities having same dimensions. Take the case of an angle, which is a ratio of arc(length) and radius(length). The dimensions of such physical quantity are zero in each of the basic quantities.
Angles
,
[
θ
]
=
[
M
0
L
0
T
0
]
Angles
,
[
θ
]
=
[
M
0
L
0
T
0
]
Similar is the case with Reynold’s number, used in fluid mechanics. It also does not have dimensions in basic quantities. Thus, a physical quantity can be either “dimensional” or “dimensionless”.
On the other hand, there are numerical constants like trigonometric function, pi etc, which are not dependent on the basic physical quantities. The dimensions of such a constant are zero in each of the basic quantities. They are, therefore, called “dimensionless” constants.
pi
,
[
π
]
=
[
M
0
L
0
T
0
]
pi
,
[
π
]
=
[
M
0
L
0
T
0
]
However, there are physical constants, which appear as the constant of proportionality in physical formula. These constants have dimensions in basic quantities. Such constant like Gravitational constant, Boltzmann constant, Planck's constant etc. are, therefore, “dimensional” constant.
In the nutshell, we categorize variables and constants in following categories :
 Dimensional variable : speed, force, current etc.

Dimensionless variable : angle, reynold’s number, trigonometric, logarithmic functions etc.
 Dimensional constant : universal gas constant, permittivity, permeability
 Dimensionless constant : numerical constants, mathematical constants, trigonometric, logarithmic functions etc.
Dimensional equation is obtained by equating dimension of a given physical quantity with its dimensional formula. Hence, followings constitute a dimensional equation :
[
F
]
=
[
M
L
T

2
]
[
F
]
=
[
M
L
T

2
]
[
v
]
=
[
M
0
L
T

1
]
[
v
]
=
[
M
0
L
T

1
]
Dimensions of a derived quantities can always be expressed in accordance with their dimensional constitutions. Consider the example of force. Its dimensional formula is
M
L
T
2
M
L
T
2
. Its SI unit, then, can be “kgm/s2”
kgm
/
s
2
kgm
/
s
2
. Instead of using dimensional unit, we give a single name to it like Newton to honour his contribution in understanding force. Such examples are prevalent in physics.
A dimensionless physical quantity, on the other hand, has no dimensional unit as no dimension is involved. Consider the case of reynold’s number. It is a number without unit. However, there are exceptions. Consider measure of angle. It is a dimensionless quantity, but has radian as unit.
Each of the basic dimensions form a group in dimensional formula. Consider a physical quantity work. The work is product of force and displacement. The basic dimension “length” is present in both force and displacement. Dimension in length, therefore, forms a group – one from force and one from displacement. The dimensions in the group is operated by multiplication following rule of indices
x
n
x
m
=
x
m
+
n
x
n
x
m
=
x
m
+
n
.
The basic approach to determine dimensional formula of a physical quantity is to use its defining equation. A defining equation like that of force in terms of mass and acceleration allows us to determine dimensional formula by further decomposing dependent quantities involved. This is the standard way. However, this approach is cumbersome in certain cases when dependencies are complicated. In such cases, we can take advantage of the fact that a physical quantity may appear in simpler relation in some other context. Let us consider the case of magnetic field strength. We can determine its dimensional expression, using BiotSavart’s law. But, we know that Magnetic field strength appears in the force calculation on a wire carrying current in a magnetic field. The force per unit length :
F
l
=
B
I
t
F
l
=
B
I
t
⇒
[
B
]
=
[
F
]
[
A
L
T
]
=
[
M
L
T

2
]
[
A
L
T
]
=
[
M
T

2
A

1
]
⇒
[
B
]
=
[
F
]
[
A
L
T
]
=
[
M
L
T

2
]
[
A
L
T
]
=
[
M
T

2
A

1
]
Problem : Find the dimensional formula of thermal conductivity.
Solution : The thermal heat of conduction (Q) is given as :
Q
=
k
A
θ
2
−
θ
1
t
d
Q
=
k
A
θ
2
−
θ
1
t
d
where “k” is thermal conductivity, “Q” is heat, “A” is area of crosssection, “θ” is temperature and “t” is time.
⇒
[
k
]
=
[
Q
d
]
[
A
θ
2
−
θ
1
t
]
⇒
[
k
]
=
[
Q
d
]
[
A
θ
2
−
θ
1
t
]
We make use of two facts (i) heat is energy and (ii) dimension of difference of temperature is same as that of temperature.
⇒
[
k
]
=
[
M
L
2
T

2
]
[
L
]
[
L
2
K
T
]
⇒
[
k
]
=
[
M
L
2
T

2
]
[
L
]
[
L
2
K
T
]
[
k
]
=
[
M
L
T

3
K

1
]
[
k
]
=
[
M
L
T

3
K

1
]
Problem : Find the dimension of expression,
C
V
ε
0
ρ
C
V
ε
0
ρ
where “C” is capacitance, “V” is potential difference, “
ε
0
ε
0
” is permittivity of vacuum and “ρ” is specific resistance.
Solution :
By inspection of expression, we observe that capacitance of parallel plate capacitor is given by the expression containing “
ε
0
ε
0
”. This may enable us to cancel “
ε
0
ε
0
” from the ratio. The capacitance is given as :
C
=
ε
0
A
d
C
=
ε
0
A
d
On the other hand, specific resistance,
ρ
=
R
A
L
ρ
=
R
A
L
Hence,
⇒
[
C
V
ε
0
ρ
]
=
[
ε
0
A
L
V
]
[
ε
0
d
R
A
]
⇒
[
C
V
ε
0
ρ
]
=
[
ε
0
A
L
V
]
[
ε
0
d
R
A
]
Applying Ohm’s law,
⇒
[
C
V
ε
0
ρ
]
=
[
ε
0
A
L
V
]
[
ε
0
L
R
A
]
=
[
V
]
[
R
]
⇒
[
C
V
ε
0
ρ
]
=
[
ε
0
A
L
V
]
[
ε
0
L
R
A
]
=
[
V
]
[
R
]
Applying Ohm's law,
⇒
[
C
V
ε
0
ρ
]
=
[
V
]
[
R
]
=
[
A
]
⇒
[
C
V
ε
0
ρ
]
=
[
V
]
[
R
]
=
[
A
]
Problem : Find the dimension of term given as :
E
2
μ
0
E
2
μ
0
where “E” is electric field and “
μ
0
μ
0
” is absolute permeability of vacuum.
Solution : We can approach the problem by evaluating each of the quantities in the expression terms of basic quantities. Alternatively, however, we can evaluate the dimension of the given ratio by expressing the given expression in terms of physical quantities, whose dimensions can be easily found out. For illustration purpose, we shall use the second approach.
Here we can write given expression in the following manner :
E
2
μ
0
=
ε
0
E
2
ε
0
μ
0
E
2
μ
0
=
ε
0
E
2
ε
0
μ
0
where “
ε
0
ε
0
” is absolute permittivity of vacuum.
However, we know that electric energy density of a medium is
1
2
ε
0
E
2
1
2
ε
0
E
2
. It means that numerator has the same dimension as “energy/volume”. On the other hand, speed of light in vacuum is given as :
c
=
1
ε
0
μ
0
c
=
1
ε
0
μ
0
Thus, dimension of denominator is same as that of
1
c
2
1
c
2
.
Therefore,
⇒
E
2
μ
0
=
[
energy/volume
]
[
1
c
2
]
⇒
E
2
μ
0
=
[
energy/volume
]
[
1
c
2
]
⇒
E
2
μ
0
=
[
M
L
2
T

2
L
3
]
[
L
T

1
]
2
=
[
M
L
T

4
]
⇒
E
2
μ
0
=
[
M
L
2
T

2
L
3
]
[
L
T

1
]
2
=
[
M
L
T

4
]