Understanding Linear Momentum Dimensions with New Fundamentals

Dimensional analysis is a powerful tool in physics that helps us understand the relationship between different physical quantities. Normally, mass (M), length (L), and time (T) are considered the fundamental dimensions. However, we can choose other quantities as fundamental, such as Force (F), Velocity (V), and Time (T), as requested in this problem to find the dimensions of linear momentum.

Standard Dimensions of Relevant Quantities

Before we find the linear momentum dimensions in the new system, let's recall their standard dimensions in terms of M, L, and T:

• Mass (M): $\mathtt{[M]}$
• Length (L): $\mathtt{[L]}$
• Time (T): $\mathtt{[T]}$
• Force (F): $\mathtt{[MLT^{-2}]}$ (from Newton's second law, F=ma)
• Velocity (V): $\mathtt{[LT^{-1}]}$ (distance/time)
• Linear Momentum (P): $\mathtt{[MLT^{-1}]}$ (mass $\times$ velocity)

Deriving Linear Momentum Dimensions in F, V, T System

We want to express the dimensions of linear momentum (P) in terms of the new fundamental dimensions: Force (F), Velocity (V), and Time (T). Let's assume the dimensions of linear momentum in this new system are $\mathtt{[F^a V^b T^c]}$.

We can write this relationship using standard dimensions:

$\mathtt{[P] = [F^a V^b T^c]}$

Substitute the standard dimensions for P, F, V, and T:

$\mathtt{[MLT^{-1}] = [MLT^{-2}]^a [LT^{-1}]^b [T]^c}$

Now, distribute the powers to each fundamental dimension (M, L, T) on the right side:

$\mathtt{[MLT^{-1}] = [M^a L^a T^{-2a}] [L^b T^{-b}] [T^c]}$

Combine the powers of the same fundamental dimensions on the right side:

$\mathtt{[MLT^{-1}] = [M^a L^{a+b} T^{-2a-b+c}]}$

Solving for the Exponents (a, b, c)

By the principle of dimensional homogeneity, the powers of each fundamental dimension must be equal on both sides of the equation. This gives us a system of linear equations:

For M: $\mathtt{a = 1}$

For L: $\mathtt{a + b = 1}$

For T: $\mathtt{-2a - b + c = -1}$

Now, we solve this system:

1. From the first equation, we directly get $\mathtt{a = 1}$.
2. Substitute the value of $\mathtt{a}$ into the second equation: $\mathtt{1 + b = 1}$. Subtracting 1 from both sides gives $\mathtt{b = 0}$.
3. Substitute the values of $\mathtt{a}$ and $\mathtt{b}$ into the third equation: $\mathtt{-2(1) - 0 + c = -1}$. This simplifies to $\mathtt{-2 + c = -1}$. Adding 2 to both sides gives $\mathtt{c = -1 + 2 = 1}$.

So, the exponents are $\mathtt{a = 1}$, $\mathtt{b = 0}$, and $\mathtt{c = 1}$.

Final Dimensions of Linear Momentum

Substituting these values back into our assumed dimensional form $\mathtt{[F^a V^b T^c}]$, we get:

$\mathtt{[P] = [F^1 V^0 T^1]}$

Since any quantity raised to the power of 0 is 1, $\mathtt{V^0 = 1}$.

Therefore, the dimensions of linear momentum in the system where Force (F), Velocity (V), and Time (T) are fundamental are $\mathtt{[FT]}$.

This result confirms the linear momentum dimensions in the new framework.