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18. [-/1 Points] DETAILS MY NOTES SCALCET9 2.2.051. In the theory of relativity, the mass \( m \) of a particle with velocity \( v \) is \( m=\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}} \), where \( m_{0} \) is the mass of the particle at rest and \( c \) is the speed of light. What happens as \( v \rightarrow c^{-} \)? \( m \rightarrow-\infty \) \( m \rightarrow m_{0} \) \( m \rightarrow 0 \) \( m \rightarrow \infty \)

Ask by Paul Riley. in the United States
Feb 21,2025

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As \( v \) approaches \( c \), \( m \) approaches infinity.

Solution

According to the formula, as the velocity v approaches c from below, the denominator √(1 - v²/c²) approaches zero. This causes the mass m to grow without bound. Therefore, m → ∞ as v → c⁻.

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As \( v \) approaches the speed of light \( c \), the denominator of the fraction \( \sqrt{1 - v^{2}/c^{2}} \) approaches zero. This means that the mass \( m \) becomes increasingly larger, tending towards infinity. Essentially, as something moves closer to the speed of light, it requires more and more energy to continue accelerating, implying that \( m \rightarrow \infty \). This concept completely revolutionized our understanding of physics! When you think about the implications, it means that nothing with mass can actually reach the speed of light because it would require infinite energy. This hints at why particles like photons, which are massless, can travel at light speed without any of these constraints.

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