Question

Review: Limits

In the theory of relativity, the mass of a particle with velocity v is
\(\sf m = \large \frac{m_0}{\sqrt{1 − v^2/c^2}}\)
where \(\sf m_0\) is the mass of the particle at rest and c is the speed of light. What happens as \(\sf v → c^−\)?

Answer 1

Let me rephrase the question: What happens to m as
\(\sf v → c^−\)?

Answer 2

so m will approach 0 or +infinity? not sure. i think positive infinity right?

Answer 3

as \(v\to c\), the denominator is approaching \(0\), so the "relativistic" mass approaches +infinity

Answer 4

You guys may probably have heard that small particles having small mass are more likely to travel close to the speed of light. But that small mass is their rest-mass.

But if we somehow keep \(m_0\) constant and large enough during the entire process, then yes, the relativistic mass does approach infinity!

Answer 5

Let \(\dfrac{v}{c}=x\),

then the limit is same as \(\lim\limits_{x\to 1} \dfrac{m_0}{\sqrt{1-x^2}}\)
which may be slightly easy to think of..

Answer 6

\[m=mo \left( 1-\frac{ v^2 }{ c^2 } \right)^{\frac{ -1 }{ 2 }}=mo \left( 1+\left( \frac{ -1 }{ 2 } \right)\left( -\frac{ v^2 }{ c^2 } \right) +... \right)\]
so \[\rightarrow \infty ~as~ v \rightarrow~c\]

Answer 7

Okay.. I think I understand now. Thanks!

I was just confused before because I know \(\sf v^2/c^2\) must be less than one, but it should be positive.. oh nvm lol I got it. Thank you so much!