Homogeneous Quantities
Two quantities are considered homogeneous if they share the same unit of measurement or have different but convertible units, allowing them to be directly compared or combined.
For example, two lengths (5 meters and 3 meters) are homogeneous because they can be compared:
$$5 m > 3m $$
and can also be added together:
$$ 5m + 3m = 8m $$
On the other hand, it doesn’t make sense to add two non-homogeneous quantities, such as a length (5 meters) with a temperature (30°C) or a mass (10 kg).
Note: Generally, lengths are considered homogeneous even if they have different units (e.g., 5 meters and 3 kilometers, or 3 inches) because they can always be converted to the same unit of measurement.
Why is it important to understand homogeneous quantities?
Understanding homogeneous quantities is crucial because it allows us to check the consistency of physical equations through dimensional analysis.
Dimensional analysis is a method used to examine the relationships between different physical quantities by analyzing their units of measurement.
In a physical equation, every term must be homogeneous with the others.
This ensures that the equation is logically consistent regardless of the specific units used and helps verify the equation’s correctness.
If non-homogeneous terms are present in an equation, it’s a clear indication that something is wrong.
Note: Of course, dimensional analysis alone isn’t enough to completely eliminate the risk of errors. There could still be simple calculation mistakes, even if the dimensional analysis is correct. However, checking the dimensions of the quantities is a useful initial step to ensure everything is in order. Once this check is passed, it’s still necessary to verify that the units are compatible and that the calculations are accurate.
A Practical Example
Example 1
Let’s consider this physical equation:
$$ 50 \ km/h = \frac{50 \ km}{1 \ h} $$
On the left side, we have a speed of 50 km/h, which is determined by the average speed formula $ v = \frac{s}{t} $, as the ratio between the distance traveled (s=50 km) and the time taken (t=1 h).
We can replace 50 km/h with the dimension [L]/[T], where [L] represents a generic length (distance traveled, s=50 km) and [T] represents a generic time (time taken, t=1 h).
$$ \frac{[L]}{[T]} = \frac{50 \ km}{1 \ h} $$
On the right side, we have the ratio of a length (50 km) to a time (1 h).
We substitute these two terms with their respective dimensions in the equation: [L] instead of 50 km and [T] instead of 1 h.
$$ \frac{[L]}{[T]} =\frac{[L]}{[T]} $$
Both sides of the equation have the same dimensions, indicating that they involve homogeneous quantities.
This is because speed is the result of dividing a length (in meters) by a time (in seconds).
So, in both sides of the equation, there is a ratio of two homogeneous quantities: distance and time, which together produce a new quantity, speed.
From this, we can conclude that the equation is physically correct.
Of course, this doesn’t rule out the possibility of calculation errors or incompatible units of measurement, but that’s another matter.
Note: For simplicity, we can rewrite the ratio [L]/[T] as the product [L][T]-1. It’s the same thing. $$ [L][T]^{-1} = [L][T]^{-1} $$ In general, in physics, it’s preferable to write the ratio in this latter form because it’s more compact and easier to read.
Example 2
Now, let’s consider Newton’s second law of motion:
$$ F = m \cdot a $$
Where \( F \) is the force, \( m \) the mass, and \( a \) the acceleration.
The unit of force \( F \) in the International System of Units (SI) is the newton (N).
A newton is defined as the force required to accelerate a mass of one kilogram (kg) at a rate of one meter per second squared (m/s²).
So, dimensionally:
$$ [F] = [M][L][T]^{-2} $$
Where \( [M] \) represents mass, \( [L] \) represents length, and \( [T] \) represents time.
Mass (m) is measured in kilograms (kg) in the SI. Dimensionally:
$$ [m] = [M] $$
Acceleration (a) is defined as the rate of change of velocity over time. Its unit is meters per second squared (m/s²).
Thus, dimensionally:
$$ [a] = [L][T]^{-2} $$
Putting this all together in the equation \( F = m \cdot a \), we get:
$$ [F] = [m][a] = [M][L][T]^{-2} $$
This shows that the equation is dimensionally consistent. Each term has the same dimensions as force, confirming the equation’s validity from a dimensional perspective.
And so forth.
