DIMENSIONAL ANALYSIS
What it is:
• In science, dimensional analysis is a tool to
find or check relations among physical
quantities by using their dimensions.
What it is:
• In science, dimensional analysis is a tool to
find or check relations among physical
quantities by using their dimensions.
• The dimension of a physical quantity is the
combination of (up to) 7 fundamental
dimensions which describe it.
– Ex: speed has the dimension length per time, and
volume has dimensions of length cubed.
What it is:
• In science, dimensional analysis is a tool to find
or check relations among physical quantities by
using their dimensions.
• The dimension of a physical quantity is the
combination of the 7 fundamental dimensions
which describe it.
– Ex: speed has the dimension length per time, and
volume has dimensions of length cubed.
• Dimensional analysis is based on the fact that a
physical law must be independent of the units
used to measure the physical variables.
What it is:
• In science, dimensional analysis is a tool to find or check
relations among physical quantities by using their
dimensions.
• The dimension of a physical quantity is the combination of
the 7 fundamental dimensions which describe it.
– Ex: speed has the dimension length per time, and volume has
dimensions of length cubed.
• Dimensional analysis is based on the fact that a physical law
must be independent of the units used to measure the
physical variables.
• It follows that any meaningful equation must
have the same dimensions in the left and right
sides. Checking this is the basic way of
performing dimensional analysis.
EXAMPLE I
• Let’s check if our instincts about speed lead to
a sensible equation. We calculate distance by
multiplying the speed of travel by the time of
travel. Is this equation dimensionally correct?
𝑑 =𝑣∙𝑡
EXAMPLE I
• To check, let’s first replace distance by its
dimension, [L] (length):
𝑑 =𝑣∙𝑡 → 𝐿 =𝑣∙𝑡
EXAMPLE I
• To check, let’s first replace distance by its
dimension, [L] (length):
𝑑 =𝑣∙𝑡 → 𝐿 =𝑣∙𝑡
• Next, let’s replace speed by its dimension,
[L]/[T] (length/time):
[𝐿]
𝐿 =
∙𝑡
[𝑇]
EXAMPLE I
• Next, let’s replace speed by its dimension, [L]/[T]
(length/time):
[𝐿]
𝐿 =
∙𝑡
[𝑇]
• Finally, we replace time by its dimension (of time,
of course!):
[𝐿]
𝐿 =
∙ 𝑇
[𝑇]
EXAMPLE I
• Finally, we replace time by its dimension (of
time, of course!):
[𝐿]
𝐿 =
∙ 𝑇
[𝑇]
• We can now simplify the right side, to obtain:
𝐿 =
[𝐿]
[𝑇]
∙ 𝑇 = 𝐿
EXAMPLE I
• We can now simplify the right side, to obtain:
𝐿 =
[𝐿]
[𝑇]
∙ 𝑇 = 𝐿
• Both sides of the equation have the same
dimension; the equation is, in principle,
correct.
EXAMPLE II
• Let’s look at a chemistry equation: the ideal
gas law. It states that the pressure, volume
and temperature of n moles of gas is given by:
𝑃∙𝑉 =𝑛∙𝑅∙𝑇
• To make it easier to see, we’ll replace
quantities by their SI units (instead of
dimensions). Furthermore, we’ll do it one at a
time.
IDEAL GAS LAW (cont’d)
• Pressure is by definition measured in N/m2:
[𝑁]
∙V
2
[𝑚]
= n∙R∙T
IDEAL GAS LAW (cont’d)
• Pressure is by definition measured in N/m2:
[𝑁]
∙V
2
[𝑚]
= n∙R∙T
• Volume in measured in m3:
[𝑁]
3
∙[
m
]
[𝑚]2
= n∙R∙T
IDEAL GAS LAW (cont’d)
• Volume in measured in m3:
[𝑁]
3
∙[
m
]
[𝑚]2
= n∙R∙T
• n is the number of moles, and therefore
measured in mol:
[𝑁]
3
∙[
m
]
[𝑚]2
= [mol]∙R∙T
IDEAL GAS LAW (cont’d)
• n is the number of moles, and therefore
measured in mol:
[𝑁]
3
∙[
m
]
[𝑚]2
= [mol]∙R∙T
• Temperature is measured in K(elvin):
[𝑁]
3
∙[
m
]
[𝑚]2
= [mol]∙R∙[K]
IDEAL GAS LAW (cont’d)
• Temperature is measured in K(elvin):
[𝑁]
3
∙[
m
]
[𝑚]2
= [mol]∙R∙[K]
• Finally, R is the universal gas constant,
measured in [J]/([mol]∙[K]):
[𝑁]
3
∙[
m
]
[𝑚]2
= [mol]∙
[𝐽]
∙[K]
𝑚𝑜𝑙 ∙[𝐾]
IDEAL GAS LAW (cont’d)
• Now we simplify both sides of the equation:
[𝑁]
3
∙[
m
]
[𝑚]2
= [mol]∙
[𝐽]
∙[K]
𝑚𝑜𝑙 ∙[𝐾]
• We obtain:
[N]∙[m] = [J]
IDEAL GAS LAW (cont’d)
• We obtain:
[N]∙[m] = [J]
• Now, it turns out that 1J = 1 N∙m (one Joule
equals one Newton times meter). So:
[J ] = [J ]
IDEAL GAS LAW (cont’d)
• Now, it turns out that 1J = 1 N∙m (one Joule
equals one Newton times meter). So:
[J ] = [J ]
• This means that our equation is dimensionally
correct!
EXAMPLE III
• As our final example, let’s check a geometry
equation, for example, for the perimeter of a
rectangle:
𝑃 = 2𝑎 + 2𝑏,
where a and b are the lengths of the two sides.
EXAMPLE III (cont’d)
• Let’s replace a and b by their dimensions:
𝑃 = 2 𝐿 + 2[𝐿].
EXAMPLE III (cont’d)
• Let’s replace a and b by their dimensions:
𝑃 = 2 𝐿 + 2[𝐿].
• Now, 2 times a length is still a length, so:
𝑃 = 𝐿 + [𝐿].
EXAMPLE III (cont’d)
• Now, 2 times a length is still a length, so:
𝑃 = 𝐿 + [𝐿].
• Finally, two lengths added still have
dimensions of length, so the right side is:
𝑃= 𝐿.
EXAMPLE III (cont’d)
• Finally, two lengths added still have dimensions
of length, so the right side is:
𝑃= 𝐿.
• By definition, the perimeter of any geometric
figure is the length along its periphery, so:
𝐿 = 𝐿.
EXAMPLE III (cont’d)
• By definition, the perimeter of any geometric
figure is the length along its periphery, so:
𝐿 = 𝐿.
• The equation is dimensionally correct, because
both its sides have the same dimension:
𝐿 = 𝐿.
EXAMPLE IV
• As a last example, let’s say we need an
equation to figure out the volume V of a
geometric figure in terms of all three sides (for
example, a, b and c); we try the following one:
2
𝑉 =
3
𝑎2 + 𝑏2 + 𝑐 2
EXAMPLE IV (cont’d)
• Let’s check if it is dimensionally correct, that
is, if our “educated guess” has at least a
chance of being correct. Replace each of the
three sides by their dimensions of length:
2
𝑉 =
3
[𝐿]2 +[𝐿]2 +[𝐿]2
EXAMPLE IV (cont’d)
• Replace each of the three sides by their
dimensions of length:
2
𝑉 =
3
[𝐿]2 +[𝐿]2 +[𝐿]2
• All lengths squared add up to, well, length
squared:
2
𝑉 =
3
[𝐿]2
EXAMPLE IV (cont’d)
• All lengths squared add up to, well, length
squared:
2
𝑉 =
3
[𝐿]2
• The right side becomes, after simplification:
𝑉 2 = [𝐿]2/3
EXAMPLE IV (cont’d)
• The right side becomes, after simplification:
𝑉 2 = [𝐿]2/3
• The left side has dimensions of cubic length,
so…
([𝐿]3 )2 = [𝐿]2/3
EXAMPLE IV (cont’d)
• The left side has dimensions of cubic length,
so…
([𝐿]3 )2 = [𝐿]2/3
• Simplifying the left side:
[𝐿]6 = [𝐿]2/3
EXAMPLE IV (cont’d)
• Simplifying the left side:
[𝐿]6 = [𝐿]2/3
• The two sides are wildly different (in
dimension); therefore, our “guess” equation is
not dimensionally correct.
Procedure to check dimensionality
1. Eliminate all constants (numbers without
units), by “dropping” them out of the
equation. That is, re-write the equation,
leaving the constants out.
Procedure to check dimensionality
1. Eliminate all constants, by “dropping” them
out of the equation. That is, re-write the
equation, leaving the constants out.
2. Replace all quantities by their SI dimensions,
on both sides of the equation.
Procedure to check dimensionality
1. Eliminate all constants, by “dropping” them
out of the equation. That is, re-write the
equation, leaving the constants out.
2. Replace all quantities by their SI dimensions,
on both sides of the equation.
3. Simplify both sides.
Procedure to check dimensionality
1. Eliminate all constants, by “dropping” them
out of the equation. That is, re-write the
equation, leaving the constants out.
2. Replace all quantities by their dimensions, on
both sides of the equation.
3. Simplify both sides.
4. Compare both (simplified) sides. If they have
the same dimension, the equation is
dimensionally correct.
CAUTION!
• Because “constants” are dropped from
equations when testing their dimensionality,
the final answer will have the “correct form”,
but will not necessarily be the correct one.
• For example, both equations below have the
same dimensions, but only the one on the
right is correct when applied to a right angle
triangle:
CORRECT ONE
3𝑎2 + 2𝑏 2 = 4𝑐 2 and 𝑎2 + 𝑏 2 = 𝑐 2
THE END
Lilian Wehner ©