Dimensional Formulas and Equations

TL;DR

Dimensional analysis is a powerful tool to check the consistency of equations and derive relationships between physical quantities by ensuring units on both sides match. A dimensional formula represents a physical quantity in terms of fundamental dimensions like mass (M), length (L), and time (T). Dimensional equations set these formulas equal to each other to ensure dimensional homogeneity.

1. The Mental Model

Think of dimensional analysis as a unit checker for physics. Just like you can't add apples and oranges, you can't add a length to a time, or have an equation where one side is an area and the other is a volume. Every valid physics equation must have the same "types" of stuff on both sides.

2. The Core Material

When we talk about dimensional formulas, we're breaking down a physical quantity into its fundamental, most basic building blocks:
* Mass (M): How much "stuff" something has.
* Length (L): How long or wide something is.
* Time (T): How long an event lasts.

Sometimes, we also include:
* Electric Current (A)
* Temperature (K)
* Amount of Substance (mol)
* Luminous Intensity (Cd)

For most mechanics problems, M, L, and T are enough. The dimensional formula for a quantity tells you how these fundamental dimensions combine to make it. For example, velocity is length per unit time, so its dimensional formula is [L][T]$^{-1}$. Area is length times length, so it's [L]$^2$.

A dimensional equation is simply an equation where we replace all the physical quantities with their dimensional formulas. The crucial principle here is dimensional homogeneity: for any physically correct equation, the dimensions of the terms on both sides of the equation must be identical.

How to Find a Dimensional Formula

1. Start with the definition or standard formula for the quantity.
2. Substitute the dimensional formulas for any known quantities.
3. Simplify using basic algebra for exponents.

Let's look at a common example: Force.
We know from Newton's second law that Force (F) = mass (m) × acceleration (a).
* Mass (m) has dimension [M].
* Acceleration (a) is change in velocity over time. Velocity is length/time [L][T]$^{-1}$. So acceleration is ([L][T]$^{-1}$)/[T] = [L][T]$^{-2}$.

Therefore, the dimensional formula for Force is [M] × [L][T]$^{-2}$ = [M][L][T]$^{-2}$.

graph TD
    A["Physical Quantity (e.g., Force)"] --> B["Definition/Formula (e.g., F = ma)"];
    B --> C["Identify Component Quantities (m, a)"];
    C --> D["Substitute Known Dimensions"];
    D -- "Mass (m) --> [M]" --> E("Intermediate: [M]");
    D -- "Acceleration (a) --> [L][T]^-2" --> F("Intermediate: [L][T]^-2");
    E & F --> G["Combine & Simplify"];
    G --> H["Dimensional Formula ([M][L][T]^-2)"];
    H --> I["Use in Dimensional Equations"];

Checking Equations for Dimensional Consistency

This is where dimensional equations shine. If you're unsure about a formula, or if you've derived one, you can run a quick dimensional check.

Rule: For an equation to be dimensionally correct, the dimensions of every term additively (on both sides) must be the same.
* If you have A = B + C, then [A] must equal [B] and [C].
* If you have D = E × F, then [D] must equal [E] × [F].

Example: Is the formula for kinetic energy, KE = 1/2 mv$^2$, dimensionally consistent?
* Left Side (KE): Kinetic energy is a form of energy. Energy is typically defined as Force × Distance.
* [Force] = [M][L][T]$^{-2}$
* [Distance] = [L]
* So, [KE] = [M][L][T]$^{-2}$ × [L] = [M][L]$^2$[T]$^{-2}$.

• Right Side (1/2 mv$^\text{2}$):
  • The constant '1/2' is dimensionless (it has no units).
  • [m] = [M]
  • [v]$^2$ = ([L][T]$^{-1}$)$^2$ = [L]$^2$[T]$^{-2}$
  • So, [1/2 mv$^\text{2}$] = [M][L]$^2$[T]$^{-2}$.

Since [KE] = [M][L]$^2$[T]$^{-2}$ and [1/2 mv$^\text{2}$] = [M][L]$^2$[T]$^{-2}$, the equation is dimensionally consistent.

Important Note: Dimensional consistency only tells you if an equation could be correct. It doesn't guarantee it is correct (e.g., it can't check dimensionless constants like the 1/2 in KE = 1/2 mv$^\text{2}$, or whether a plus should be a minus).

3. Worked Example

Let's test the dimensional correctness of the equation for the period of a simple pendulum:
T = 2π√(L/g)
Where:
* T = Period (time)
* L = Length of the pendulum string
* g = Acceleration due to gravity

First, identify the dimensions of each variable:
* [T] = [T] (since Period is a measure of time)
* [L] = [L] (since Length is a measure of length)
* [g] = [L][T]$^{-2}$ (since acceleration is length per time squared)
* The constant 2π is dimensionless.

Now, let's substitute these into the equation for both sides:

Left Side:
[T] = [T]

Right Side:
Dimensions of √(L/g)
= √([L] / ([L][T]$^{-2}$))
= √(1 / [T]$^{-2}$)
= √([T]$^2$)
= [T]

Since the dimensions of the Left Side ([T]) are equal to the dimensions of the Right Side ([T]), the equation T = 2π√(L/g) is dimensionally consistent.

4. Key Takeaways

• A dimensional formula expresses a physical quantity in terms of fundamental dimensions like M, L, T.
• Dimensional homogeneity means all additive terms in a valid equation must have the exact same dimensions.
• You can use dimensional analysis to check if an equation is potentially correct, but not definitively.
• Dimensionless constants (like 1/2, π, angles) are ignored in dimensional analysis.
• Quantities can only be added or subtracted if they have the same dimensions.

Common Mistakes:
- Forgetting that exponents apply to all parts of a dimensional formula (e.g., [v]$^2$ is [L]$^2$[T]$^{-2}$, not [L]$^2$[T]$^{-1}$).
- Treating dimensionless constants as having dimensions.
- Confusing units with dimensions (units are specific measures, dimensions are categories).
- Assuming a dimensionally correct equation is automatically physically correct.
- Not checking every term in an equation (especially those connected by + or -).

5. Now Try It

Derive the dimensional formula for Power. Then, check the dimensional consistency of the equation P = Fv, where P is power, F is force, and v is velocity.

What to do:
1. Start with the definition of Power: Power = Work / Time.
2. Recall the dimensional formulas for Work (which is Force × Distance) and Time.
3. Substitute and simplify to get the dimensional formula for Power.
4. Then, take the given equation P = Fv.
5. Substitute the dimensional formulas for Force and Velocity into the right side (Fv).
6. Compare the dimensions of the left side (your derived [P]) with the right side ([F][v]).

What success looks like:
You should be able to state the dimensional formula for Power and confirm whether the equation P = Fv is dimensionally consistent.