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?

EXAMPLE I

• To check, let’s first replace distance by its dimension, [L] (length):

EXAMPLE I

• To check, let’s first replace distance by its dimension, [L] (length):

• Next, let’s replace speed by its dimension, [L]/[T] (length/time):

EXAMPLE I

• Next, let’s replace speed by its dimension, [L]/[T] (length/time):

• Finally, we replace time by its dimension (of time, of course!):

EXAMPLE I

• Finally, we replace time by its dimension (of time, of course!):

• We can now simplify the right side, to obtain: =

EXAMPLE I

• We can now simplify the right side, to obtain: =

• 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:

• 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:∙V = n R T∙ ∙

IDEAL GAS LAW (cont’d)

• Pressure is by definition measured in N/m2:∙V = n R T∙ ∙

• Volume in measured in m3:∙[m]3 = n R T∙ ∙

IDEAL GAS LAW (cont’d)

• Volume in measured in m3:∙[m]3 = n R T∙ ∙

• n is the number of moles, and therefore measured in mol:

∙[m]3 = [mol] R T∙ ∙

IDEAL GAS LAW (cont’d)

• n is the number of moles, and therefore measured in mol:

∙[m]3 = [mol] R T∙ ∙

• Temperature is measured in K(elvin):∙[m]3 = [mol] R∙ ∙[K]

IDEAL GAS LAW (cont’d)

• Temperature is measured in K(elvin):∙[m]3 = [mol] R∙ ∙[K]

• Finally, R is the universal gas constant, measured in [J]/([mol] [K]):∙

∙[m]3 = [mol]∙∙[K]

IDEAL GAS LAW (cont’d)

• Now we simplify both sides of the equation:∙[m]3 = [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:

,where a and b are the lengths of the two sides.

EXAMPLE III (cont’d)

• Let’s replace a and b by their dimensions:

.

EXAMPLE III (cont’d)

• Let’s replace a and b by their dimensions:

.

• Now, 2 times a length is still a length, so:

.

EXAMPLE III (cont’d)

• Now, 2 times a length is still a length, so:

.

• Finally, two lengths added still have dimensions of length, so the right side is:

.

EXAMPLE III (cont’d)

• Finally, two lengths added still have dimensions of length, so the right side is:

.

• By definition, the perimeter of any geometric figure is the length along its periphery, so:

.

EXAMPLE III (cont’d)

• By definition, the perimeter of any geometric figure is the length along its periphery, so:

.

• The equation is dimensionally correct, because both its sides have the same dimension:

.

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:

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:

EXAMPLE IV (cont’d)

• Replace each of the three sides by their dimensions of length:

• All lengths squared add up to, well, length squared:

EXAMPLE IV (cont’d)

• All lengths squared add up to, well, length squared:

• The right side becomes, after simplification:

EXAMPLE IV (cont’d)

• The right side becomes, after simplification:

• The left side has dimensions of cubic length, so…

EXAMPLE IV (cont’d)

• The left side has dimensions of cubic length, so…

• Simplifying the left side:

EXAMPLE IV (cont’d)

• Simplifying the left side:

• 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

and

THE END

Lilian Wehner ©