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Dimensions & Checking Equations
ECAT Interactive Study Tool
1. Core Theory
Principle of Homogeneity
The Principle of Homogeneity states that the dimensions of all the terms on both sides of a physical equation must be identical.
You can only add or subtract physical quantities if they have the exact same dimensions. For instance, you cannot add velocity to mass.
[LHS]
[RHS]
For an equation ( A = B + C ) to be dimensionally correct:
[ [A] = [B] = [C] ]
Checking the Correctness of Equations
Dimensional analysis is used to check if a derived physics formula is potentially correct. If the dimensions of the Left Hand Side (LHS) equal the Right Hand Side (RHS), the equation is dimensionally correct.
Note: Dimensional correctness does not guarantee physical correctness, because dimensional analysis ignores dimensionless constants like (pi) or (1/2).
Deriving Possible Formulae
We can use dimensions to derive relationships between physical quantities by setting up proportionalities and solving for exponents.
If we know a quantity (Y) depends on (A), (B), and (C), we can write:
( Y propto A^a B^b C^c )
( [Y] = [A]^a [B]^b [C]^c )
By substituting dimensions on both sides and equating the powers of ([M]), ([L]), and ([T]), we can find values for (a), (b), and (c).
2. Concept Check
1. True or False: Dimensional analysis can help us find the exact value of dimensionless constants like ( pi ) or ( 2 ).
2. True or False: If an equation is dimensionally correct, it is guaranteed to be a physically correct equation.
3. Solved Examples
Example 1: Checking Equation Homogeneity
Check the dimensional correctness of the equation ( v^2 = u^2 + 2as ).
Step 1: Write dimensions of LHS.
LHS is ( v^2 ). Since ( v = [L T^{-1}] ), we have ( [LHS] = [L T^{-1}]^2 = [L^2 T^{-2}] ).
Step 2: Write dimensions of RHS.
RHS has two terms: ( u^2 ) and ( 2as ).
( [u^2] = [L T^{-1}]^2 = [L^2 T^{-2}] )
( [2as] = [L T^{-2}] times [L] = [L^2 T^{-2}] ) (Note: 2 is dimensionless)
Step 3: Compare LHS and RHS.
Since ( [v^2] = [u^2] = [2as] = [L^2 T^{-2}] ), all terms have the same dimensions.
Conclusion: The equation is dimensionally correct.
Example 2: Finding Dimensions of Unknowns
The position of a particle is given by ( x = At + Bt^2 ), where ( x ) is in meters and ( t ) is in seconds. Find the dimensions of ( B ).
Step 1: Apply the Principle of Homogeneity.
Every term in the equation must have the same dimension as ( x ). So, ( [x] = [At] = [Bt^2] ).
Step 2: Isolate the term with the unknown.
We know ( [x] = [L] ) and we need to find ( B ). So we use ( [x] = [Bt^2] ).
Step 3: Solve for [B].
( [L] = [B] [T^2] implies [B] = frac{[L]}{[T^2]} = [L T^{-2}] )
Conclusion: The dimension of ( B ) is ( [L T^{-2}] ) (which is the dimension of acceleration).
4. MCQ Practice
1. The time period ( T ) of a simple pendulum depends on length ( l ) and gravity ( g ). Which equation is dimensionally correct?
2. Which of the following equations is dimensionally INCORRECT?
3. If force ( F ), velocity ( V ), and time ( T ) are taken as fundamental quantities, then the dimensions of mass are:
4. In the equation ( y = a sin(omega t – kx) ), what are the dimensions of ( k )? (where (x) is distance)
5. The dimensional formula for Universal Gravitational Constant ((G)) is:
5. Summary Table: Important Dimensional Formulas
| Physical Quantity | Formula | Dimensional Formula | SI Unit |
|---|---|---|---|
| Force | ( m times a ) | ( [M L T^{-2}] ) | Newton (N) |
| Work / Energy | ( F times d ) | ( [M L^2 T^{-2}] ) | Joule (J) |
| Power | ( W / t ) | ( [M L^2 T^{-3}] ) | Watt (W) |
| Momentum | ( m times v ) | ( [M L T^{-1}] ) | kg m/s |
| Pressure | ( F / A ) | ( [M L^{-1} T^{-2}] ) | Pascal (Pa) |
| Strain | ( Delta L / L ) | ( [M^0 L^0 T^0] ) | No Unit |