Understanding Special Relativity: The Equation E = mc² Explained

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Discussion Overview

The discussion revolves around the equation E = mc² and its relativistic implications, specifically addressing the total energy of a moving body in the context of special relativity. Participants explore the correct formulation of the equation and its components, including rest mass and the role of velocity.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the equation E = \frac{m_{0}c^{2}}{1-v^{2}/c^{2}}, suggesting a specific form of the energy equation.
  • Another participant points out a potential omission of a square root in the denominator, proposing the expression E = \gamma mc² = \frac{mc²}{\sqrt{1-v²/c²}} as the correct relativistic energy equation.
  • A different participant comments on the terminology, suggesting that the term 'rest mass' (m₀) is less commonly used outside of high school, implying that mass is generally understood without that qualifier.
  • Another participant discusses the equation's appearance in the context of four-momentum, indicating that the formulation may depend on the conventions used for time and space coordinates.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of the energy equation and the terminology used, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

There are unresolved aspects regarding the definitions of mass and the implications of different coordinate conventions in the equations presented.

Lizwi
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What is E = \frac{m_{0}c^{2}}{1-v^{2}/c^{2}}
 
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Hi Lizwi,

I think you are missing a square root in the denominator? The expression:
E= \gamma m c^2 = \frac{mc^2}{\sqrt{1-v^2/c^2}}
Is the relativistic expression for the total energy of a moving body.
 
What Nabeshin said
You can also drop the 0 from the m_0, I've not seen the term 'rest mass' used since high school. Once you're out of high school it simply becomes mass as far as I know :p
 
Lizwi said:
What is E = \frac{m_{0}c^{2}}{1-v^{2}/c^{2}}

its a term that pops up in the four-momentum if you use the convention that x0 = ct rather that x0 = t

if you use x0 = t then you just get p0 = gamma*m
 

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