It is important to realise that it only makes sense to add the same sort of quantities, e.g. area may be added to area but area may not be added to temperature! These considerations lead to a powerful method to analyse scientific equations called dimensional analysis. One should note that while units are arbitrarily chosen (an alien civilisation will not use seconds or weeks), dimensions represent fundamental quantities such as time.
Basic dimensions are written as follows:
Dimension  Symbol 

Length  L 
Time  T 
Mass  M 
Temperature  K 
Electrical current  I 
See the module on Units for a review of SI units.
Example 1: An area can be expressed as a length times a length. Therefore the dimensions of area are L × L = L^{2}, which we can express as: [area] = L^{2}.
A given area could be expressed in the SI units of square metres, or indeed in any appropriate units.
In some equations symbols appear which do not have any associated dimension, e.g., in the formula for the area of a circle, πr^{2}, π is just a number and does not have a dimension.
Exercise 1: Calculate the dimensions of the following quantities:
Solution:
A volume is given by multiplying three lengths together:Dimension of volume  =  L × L × L 
L^{3} 
So [volume] = L^{3}
Solution:
Speed is the rate of change of distance with respect to time:Dimension of speed  =  LT 
LT^{−1} 
So [speed] = LT^{−1} (The SI units of speed are metres per second.)
Solution:
Acceleration is the rate of change of speed with respect to time:Dimension of acceleration  =  LT^{−1}T 
LT^{−2} 
So [acceleration] = LT^{−2} (The SI units of acceleration are metres per second per second)
Solution:
Density is the mass per unit volume, so using the dimension of volume we get:Dimension of density  =  
ML^{−3} 
So [density] = M^{−3} (The SI units of density are kg m^{−3})
Click on questions to reveal solutions
Quiz 1: Pick out the units that have a different dimension to the other three:
Explanation: kg^{2} m s^{−2} has dimensions = M^{2}LT^{−2}
It can be checked that all the other answers have dimension ML^{2}T^{−2}
Example 2: Consider the equation:
Since any terms which are added together or subtracted MUST have the same dimensions, in this case y, x and ½kx^{3} have to have the same dimensions. We say that such a scientific equation is dimensionally correct. However, if it is not true, then the equation must be wrong.
If in the above equation x and y are both lengths (dimension L) and ½ is a dimensionless number, then for the kx^{3} term to have the same dimension as the other two, we would need:
dimension of k × L^{3}  =  L 

∴ dimension of k = LL^{3}  =  L^{−2} 
So k would have dimensions of one over area, i.e., [k] = L^{−2}
Quiz 2: Hooke’s law states that the force, F, in a spring extended by a length x is given by F = − kx. From Newton’s second law F = ma, where m is the mass and a is the acceleration, choose the dimension of the spring constant k:
Explanation: From Hooke’s law, F = −kx, we see that we can write:
Now F = ma, so the dimensions of force are given by:
[F]  =  M × (LT^{−2}) 
=  MLT^{−2} 
Therefore the spring constant has dimensions:
[k]  =  MLT^{−2}L 
=  MT^{−2} 
Example 3: The expressions for kinetic energy E = ½mv^{2} (where m is the mass of the body and v is its speed) and potential energy E = mgh (where g is the acceleration due to gravity and h is the height of the body) look very different, although both describe energy.
One way to see this is to note that they have the same dimension:


Both expressions have the same dimensions, so they can therefore be added and subtracted from each other.
Exercise 2: Check that the dimensions of each side of the equations below agree:
Solution: A volume is given by multiplying three lengths together:
Dimension of volume  =  L × L × L 
=  L^{3} 
So [volume] = L^{3}
Solution: We want to check the dimensions of the equation v = u + at. Since v and u are both speeds, they have dimensions LT^{−1}. Therefore we only need to verify that at has this dimension. To see this consider:
So the equation is dimensionally correct because all the terms have dimensions of speed.
Solution: We want to check the dimensions of the equation E = mc^{2}. Since E is an energy it has dimensions ML^{2}T^{−2}. The right hand side of the equation can also be seen to have this dimension, if we recall that m is a mass and c is the speed of light with dimension LT^{−1}. Therefore:
[mc^{2}]  =  M(LT^{−1})^{2} 
=  ML^{2}T^{−2} 
So the equation is dimensionally correct as all terms have dimensions of energy.
Solution: We want to check the dimensions of the equation c = λν. Since c is the speed of light it has dimensions LT^{−1}. The right hand side of the equation involves wavelength [λ] = L and frequency [λ] = T^{−1}. We thus have:
 which indeed also has dimensions of speed, so the equation is dimensionally correct.
Click on questions to reveal solutions
Some quantities are said to be dimensionless. These are pure numbers which would be the same no matter what units were used (e.g., the mass of a proton is roughly 1850 times the mass of an electron, regardless of how you measure mass).
Example 4: The ratio of one mass m_{1} to another mass m_{2} is dimensionless:
The dimensions have canceled and the result is a number (which is independent of the units, i.e., it would be the same whether the masses were measured in kilograms or tonnes).
Note: angles are defined in terms of ratios of lengths. They are therefore dimensionless! Functions of dimensionless variables are themselves dimensionless (????).
Very many functions are dimensionless. The following quantities are important cases of dimensionless quantities:
Trigonometric Functions  Logarithms 

Exponentials  Numbers, e.g. π 
Note the following properties:
Quiz 3: If the number of radioactive atoms is found to be given as a function of time t by:
 where N_{0} is the number of atoms at time t = 0, what is the dimension of k?
Explanation: In N(t) = N_{0} exp (−kt) , the exponential and its argument must be dimensionless. Therefore kt has to be dimensionless. Thus:
So the dimension of k must be inverse time, i.e., [k] = T^{−1}.
Exercise 3: Determine the dimensions of the expressions below:
Solution: We want to check the dimensions of d sin(θ) = nλ. Both d and λ have dimensions of length. The angle θ and sin (θ), as well as the number n, must all be dimensionless. Therefore we have:
L × 1  =  1 × L 

∴ [d sin (θ)]  =  [nλ] 
Solution: The factor exp (−E ⁄ (kT)) is an exponential and so must be dimensionless. Therefore its argument −E ⁄ (kT) must also be dimensionless.
The minus sign simply corresponds to multiplying by minus one and is dimensionless. Energy, E has dimensions [E] = ML^{2}T^{−2} and temperature has dimensions K, so the dimensions of Boltzmann’s constant are:
The SI units of Boltzmann’s constant are therefore kg m^{2} s^{−2} K^{−1} or J K^{−1}.
Click on questions to reveal solutions
Quiz 4: Use dimensional analysis to see which of the following expressions is allowed if P is a pressure, t is a time, m is a mass, r is a distance, v is a velocity and T is a temperature:
Explanation: First note that the argument of a logarithm must be dimensionless. Now pressure is force over area, so it has dimensions:
[P]  =  MLT^{−2} L^{2} 
=  ML^{−1}T^{−2} 
Therefore the combination Prt^{2} ⁄ m is dimensionless since:
[Prt^{2} m]  =  (ML^{−1}T^{−2}) × L × T^{2} M 
=  M M  
=  1 
None of the other combinations are dimensionless and so it would be completely meaningless to take their logarithms.
Choose the solutions from the options given
Note: dimensional analysis is a way of checking that equations might be true: it does not prove that they are definitely correct. Dimensional analysis would suggest that both Einstein’s equation E = mc^{2} and the (incorrect) equation E = ½mc^{2} might be true. On the other hand, dimensional analysis shows that E = mc^{3} makes no sense.