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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 \(\sf v → c^−\)?

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  1. anonymous
    • one year ago
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    Let me rephrase the question: What happens to m as \(\sf v → c^−\)?

  2. anonymous
    • one year ago
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    so m will approach 0 or +infinity? not sure. i think positive infinity right?

  3. ganeshie8
    • one year ago
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    as \(v\to c\), the denominator is approaching \(0\), so the "relativistic" mass approaches +infinity

  4. ParthKohli
    • one year ago
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    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!

  5. ganeshie8
    • one year ago
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    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..

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

  7. anonymous
    • one year ago
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    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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