anonymous
  • anonymous
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 \(\sf v → c^−\)?
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Let me rephrase the question: What happens to m as \(\sf v → c^−\)?
anonymous
  • anonymous
so m will approach 0 or +infinity? not sure. i think positive infinity right?
ganeshie8
  • ganeshie8
as \(v\to c\), the denominator is approaching \(0\), so the "relativistic" mass approaches +infinity

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ParthKohli
  • ParthKohli
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!
ganeshie8
  • ganeshie8
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..
anonymous
  • anonymous
\[m=mo \left( 1-\frac{ v^2 }{ c^2 } \right)^{\frac{ -1 }{ 2 }}=mo \left( 1+\left( \frac{ -1 }{ 2 } \right)\left( -\frac{ v^2 }{ c^2 } \right) +... \right)\] so \[\rightarrow \infty ~as~ v \rightarrow~c\]
anonymous
  • anonymous
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!

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