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anonymous
 one year ago
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 \(\sf v → c^−\)?
anonymous
 one year ago
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 \(\sf v → c^−\)?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let me rephrase the question: What happens to m as \(\sf v → c^−\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so m will approach 0 or +infinity? not sure. i think positive infinity right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1as \(v\to c\), the denominator is approaching \(0\), so the "relativistic" mass approaches +infinity

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1You 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 restmass. But if we somehow keep \(m_0\) constant and large enough during the entire process, then yes, the relativistic mass does approach infinity!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Let \(\dfrac{v}{c}=x\), then the limit is same as \(\lim\limits_{x\to 1} \dfrac{m_0}{\sqrt{1x^2}}\) which may be slightly easy to think of..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[m=mo \left( 1\frac{ v^2 }{ c^2 } \right)^{\frac{ 1 }{ 2 }}=mo \left( 1+\left( \frac{ 1 }{ 2 } \right)\left( \frac{ v^2 }{ c^2 } \right) +... \right)\] so \[\rightarrow \infty ~as~ v \rightarrow~c\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay.. 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!
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