Keeping units in mind

One of the most powerful tools in all of physics is dimensional analysis. Not to be confused with unit conversion, which is where you take one thing (meters, for example) and convert it to another unit (feet, for example). Dimensional analysis is more powerful than that.

Using dimensional analysis, we can look at the units of what we are given in a problem and arrive at the exact result multiplied by a number.

Let’s look at an example: 

Say you want to find the time it takes to drop a ball from height h, and you know the acceleration due to gravity, g. h has units [meters] and g has units [meters/second²]. Maybe you don’t know how to solve this question. But we can see how we can put together h and g to get units of time: (h/g)^(½). So, we have a decent approximation to our actual answer. And it’s actually pretty close: the exact answer is (2h/g)^(½).  

This technique is useful in that if we are solving a question, we have an idea of what the answer should look like. I particularly like this in E&M when you are solving for electric field, because in the end we know we are going to get something like Q/(ε₀R²) where Q is charge, R is distance, and ε₀ is vacuum permittivity. This is extremely helpful when you are getting in the nitty gritty of a problem and want to check that you’re on the right track! 

However, in some examples we might not be able to see the result as quickly as we did here. So what should you do in that case? We can make a system of equations.

Using the same example as above, we can write [seconds] = [meters]^x * [meters / seconds²]^y where x and y are unknown.

(Note the left-hand side is the units for t what we want, and the right-hand side is the product of h and g raised to some powers we will solve for.)

Let's rewrite our equation: [seconds]¹ * [meters]⁰ = [meters]^(x+y) * [seconds]^(-2y). Here, we have used the fact on the LHS that 1 = [meters]⁰, and that [1/seconds²]^y = [seconds]^(-2y) on the RHS.

Now we can just solve these equations! We have 1 = -2y and 0 = x+y. This tells us y = -1/2 and x = 1/2. Which is exactly our result from two paragraphs ago! 

So, what are the steps to dimensional analysis? 

1. Write down your known quantities and their dimensions
2. Write down what quantity you want and its dimension
3. Write down your system of equations for the units, where the left hand side is your desired quantity and the right hand side is the quantities you know and raise to some unknown power
4. Solve the system of equations!