Problem 58 - Calculus Single Variable

Question

The basic "dimensions" of mechanics are length, mass, and time. Trigonometric functions are unitless because they represent the ratio of two lengths-the units cancel. Arguments of trigonometric functions are angles and so unitless. Likewise, exponential functions are unitless and take unitless arguments. According to the Theory of Relativity, the mass \(m\) of an object at speed \(v\) is given by $$ m(v)=\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}} $$ where \(c\) is the speed of light, and \(m_{0}\) is its rest mass. There is a function \(\varphi\) such that $$ m^{\prime}(v)=m_{0} \varphi(v)\left(1-\frac{v^{2}}{c^{2}}\right)^{-3 / 2} $$ What are the dimensions of \(m^{\prime}(v), \varphi(v),\) and \(\left(1-v^{2} / c^{2}\right)^{-3 / 2} ?\) Calculate \(m^{\prime}(v),\) identifying each application of the Chain Rule. What is \(\varphi(v)\) ? Verify that \(\varphi(v)\) has the expected dimensions.

Step-by-Step Solution

Answer

Dimensions: \(m'(v) = [M], \varphi(v) = [T]^{-1}, (1-v^2/c^2)^{-3/2} = 1\). \(\varphi(v) = \frac{v}{c^2}\) with dimension \([T]^{-1]\). Verified.

Step 1: Dimension Analysis of Functions

Start by analyzing the function and terms provided. Since both trigonometric and exponential functions are unitless, any function argument must also be unitless. The expression for mass \(m(v)\) is unitless in its square root term because it involves the ratio of \(v^2\) and \(c^2\), both of which are speeds and thus cancel out. Mass \(m_0\) has unit of mass \(M\). Thus, \(m(v)\) also has the dimension of mass, \([M]\).

Step 2: Chain Rule Application on Mass Function

To find \(m'(v)\), differentiate \(m(v) = \frac{m_0}{\sqrt{1 - v^2/c^2}}\). Apply the chain rule:\[- \frac{d}{dv}\left( \sqrt{1 - \frac{v^2}{c^2}} \right) = \frac{1}{2} \cdot (1 - \frac{v^2}{c^2})^{-3/2} \cdot \left(-\frac{2v}{c^2}\right) \text{.}\] This gives \[m'(v) = m_0 \cdot \left(1 - \frac{v^2}{c^2}\right)^{-3/2} \cdot \frac{v}{c^2}\text{.}\]

Step 3: Expression of m'(v) Function Components

The expression for \(m'(v)\) can be simplified and rewritten as given by the problem statement: \[m'(v) = m_0 \cdot \varphi(v) \cdot \left(1-\frac{v^2}{c^2}\right)^{-3/2}\text{.}\] This allows us to identify \(\varphi(v) = \frac{v}{c^2}\).

Step 4: Dimension Analysis of m'(v) and \(\varphi(v)\)

For \(m'(v)\), the dimensions come from the product \(m_0\cdot\varphi(v)\), which must match \([M]\) (since it must be the same dimension as \(m(v)\)). Thus, the dimensions of \(m'(v)\) are \([M]\). \(\varphi(v) = \frac{v}{c^2}\) has dimensions \([T]^{-1}\), as velocity has dimensions \([LT^{-1}]\) and \(c^2\) has dimensions \([L^2T^{-2}]\). The term \(\left(1-\frac{v^2}{c^2}\right)^{-3/2}\) is unitless as it only involves the unitless ratio of two speeds squared.

Step 5: Verification of Expected Dimensions

Verify that \(\varphi(v) = \frac{v}{c^2}\) has expected dimensions \([T]^{-1}\). Given that \(v\) has dimensions \([LT^{-1}]\) and \(c^2\) has dimensions \([L^2T^{-2}]\), dividing these gives \([T]^{-1]\), which aligns with our earlier analysis.

Key Concepts

Theory of Relativity

Einstein's Theory of Relativity fundamentally changed our understanding of space and time. In this theory, the mass of an object increases with its speed. The provided equation \( m(v) = \frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}} \) tells us how the effective mass \( m(v) \) of an object changes as it moves at speed \( v \). Here, \( m_0 \) represents the rest mass, or the mass when the object is stationary. The denominator \( \sqrt{1 - v^2/c^2} \) shows us the relativistic factor influencing mass.

chain rule

The chain rule in calculus is a method for understanding how changes in a composite function affect its derivative. In this case, we used the chain rule to find the derivative of the mass function \( m'(v) \):

• Differentiate the outer function first \((f(x) = \frac{1}{x})\), which became \(-\frac{1}{2} (1 - \frac{v^2}{c^2})^{-3/2}\).
• Then differentiate the inner function \((g(v) = 1 - \frac{v^2}{c^2})\), resulting in \(-\frac{2v}{c^2}\).

The product of these differentiations gives us the rate of change of the mass \( m'((v) \).

mass-energy equivalence

Mass-energy equivalence is a profound principle from the Theory of Relativity asserting that mass and energy are interchangeable. This idea is encapsulated in the famous equation \( E = mc^2 \).

This concept is crucial for understanding why mass increases with speed. As an object moves faster, its energy due to motion increases, and thus its effective mass \( m(v) \) changes. The relationship between mass and energy means that even though mass seems intrinsic, it varies with kinetic energy.

unit analysis

Unit analysis is a handy method to ensure calculations make sense, by analyzing units on both sides of an equation.\( m(v) \) and \( m_0 \), having units of mass, proves correct after unit analysis.

For \( \varphi(v) \), we determined it as \(\frac{v}{c^2}\). Here:

• The numerator \(v\) has dimensions \( [LT^{-1}]\).
• The denominator \(c^2\) leads to \([L^2T^{-2}]\).

Upon simplifying \( \frac{[LT^{-1}]}{[L^2T^{-2}]} \), we are left with \([T^{-1}]\), confirming \(\varphi(v)\) behaves as anticipated.