Why c2 (speed of light squared)?

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Discussion Overview

The discussion centers around the equation E=mc², specifically exploring why the speed of light is squared in this equation. Participants seek to understand the implications of this formulation, its derivation, and its significance in the context of physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that dimensional analysis shows the necessity of squaring the speed of light to match units of energy and mass.
  • Others propose that the derivation of the equation inherently leads to the conclusion that energy and mass are equivalent, with c² being a fundamental constant in this relationship.
  • One participant mentions that the equation can be derived from the Work-Energy Theorem using four-momentum and proper time.
  • Another participant discusses the concept of relativistic mass and how it relates to kinetic energy, suggesting that the increase in mass at high speeds contributes to the equation.
  • Some contributions question the deeper meaning of c² and its role as a universal constant, contrasting it with other potential constants.
  • There is a suggestion that the speed of light's significance might be tied to its role as a speed limit for energy and information transmission.
  • One participant expresses uncertainty about whether the derivations can be understood by a layman and emphasizes that the equation's validity is ultimately based on observation.

Areas of Agreement / Disagreement

Participants express a range of views on the significance and implications of c² in the equation. While there is some agreement on the necessity of c² for dimensional consistency, the deeper philosophical and physical implications remain contested and unresolved.

Contextual Notes

Limitations include varying levels of understanding among participants regarding the mathematical derivations and physical concepts involved. Some assumptions about the nature of light and its relationship to the equation are not fully explored.

Who May Find This Useful

This discussion may be of interest to those exploring the foundations of physics, particularly in relation to energy-mass equivalence, special relativity, and the philosophical implications of physical constants.

  • #61
Vorde said:
In relativistic physics momentum isn't defined as ##p=mv## but rather as ##p= \gamma m v## where ##\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}##.

You are right ! Maybe i have to search some more about this derivation .
 
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  • #62
You can't do something like that because In E=mc^2 relation 'm' stands for the mass of light !

That's about as wrong as it gets in special relativity. E=mc^2 is actually the abbreviated form of the full equation which is E=mc^2+pc. The abbreviated version is used in cases that deal with energies that specifically EXCLUDE light photons, that is, rest mass. Situations dealing with light quanta will typically exclude the mc^2 term and be in the form E=pc.

And in this equation v=E/Mc , 'M' stands for the mass of the cylinder ! Even if you put 'c' in the place of 'v' it won't become E=mc^2

Of course it will, this is simple algebra.
 
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  • #63
DiracPool said:
Of course it will, this is simple algebra.

You actually mistook the whole derivation . v=E/Mc not E/mc ! 'M' and 'm' are different ... Go through the derivation once more . This derivation is not circular .
 
  • #64
The Lagrangian L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}, p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}, E = p\cdot v - L = \frac{mv^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} + mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} so in the rest frame we have that E = mc^{2}. The speed of light squared quantity is just a unit conversion. If you work in natural units then you won't even see it explicitly.
 
  • #65
WannabeNewton said:
The Lagrangian L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}, p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}, E = p\cdot v - L = \frac{mv^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} + mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} so in the rest frame we have that E = mc^{2}. The speed of light squared quantity is just a unit conversion. If you work in natural units then you won't even see it explicitly.

Agreed . But c^2 is not just an unit conversion . It is also known "the specific energy" .

And I don't know what is "ρ" in the equation . Could you explain?
 
  • #66
P is momentum: which is standard notation.
 
  • #67
Vorde said:
P is momentum: which is standard notation.

Sorry bro ! I mistook it as "rho(ρ)"
 

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