Question

In the theory of relativity, the mass of a particle with velocity v is

\(m = \frac{{{m_{\bf{0}}}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^{\bf{2}}}}}} }}\)

Where \({m_{\bf{0}}}\) is the mass of particle at rest and c is the speed of light. What happens as \(v \to {c^ - }\)?

Short answer

As \(v \to {c^ - }\), the value of mapproaches to infinity.

Step-by-step solution

Write the equation of m when \(v \to {c^ - }\)

The limit expressionfor m can be written as shown below:

\(\mathop {\lim }\limits_{v \to {c^ - }} m = \mathop {\lim }\limits_{v \to {c^ - }} \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)

Solve the limit in step 1

For the limit,\(\mathop {\lim }\limits_{v \to {c^ - }} \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\) as \(v \to {c^ - }\), the value of \(m\) will approach to infinity.

\(\mathop {\lim }\limits_{v \to {c^ - }} \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} = \infty \)

So, \(\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} \to {0^ + }\), then \(m \to \infty \).