Starter Physics - SI Units and Dimensions

SI units and Dimensions

SI units are the basis of physics.

Physical properties have standard measurement units.
In physics we use a system known as SI units.
The basic SI units are kilograms (kg), metres (m), seconds (s), and Coulomb (C).
Physics equations all work correctly (give the right answers) if we use these units.
This makes things much easier, as there is no need to convert to other units.

We know that both sides of an equation must have exactly the same units.
This is useful because we can use this to check that the equations are correct and that the units we are using will give the right answers.
We can also use these ideas to create or guess how an equation might look.
If there are two or more terms on one side, e.g. s = vt + ½at², then vt and ½at² (and s) should all have the same units (in this case, metres).

Dimensions simplify this further - we don’t even need to specify whether we’re measuring kg or g or lbs or oz - if we see mass on one side of an equation we know there must be mass on the other side.
We just need to look at each side of the equation and check how many M, L, T and Q there are (where M is Mass, L is Length, T is Time, and Q is Charge).
For example: Energy is ML²/T² , and mc² is M(L/T)² = ML²/T² , so E = mc² is a valid equation.
The equation is balanced with the correct numbers and positions of M’s, L’s, and T's on both sides.
An equation is only valid if both sides have the same M's, L's, and T's.
Balancing the units and/or dimensions is a good check to do when we see an unfamiliar equation.
Dimensions are usually taught at advanced level, but it's useful to know that this system is behind all physics relationships.

Lets take the example of E = mc²
The terms on the right hand side are mass x speed x speed.
When terms are multipled we can add up all their MLTQ codes.
The total MLTQ codes on both sides match.
So we prove that E = Amv² is a valid equation, where A is a constant, such as 1 or ½.
Examples are: E = ½mv²and E = mc²

Or if we want to find what could be the missing term X in E = mX, we can work this out by re-organising the equation, and writing X = E/m.
When terms are divided, we subtract the MLTQ code of terms on the bottom from the MLTQ code of those on the top.
So, we substract the MLTQ code of m from that of E.
This tells us that the MLTQ code of X is the same as speed².
So we know that E = Amv², where A is a constant, is a possible answer.

The terms on the right hand side are multiplied so they can be added together:

In the case below, the terms on the right hand side are E/m so the MLTQ code of m is subtracted from E: