What is the actual equation of e=mc^2?

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    E=mc^2
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Discussion Overview

The discussion revolves around the equation E=mc², with participants exploring its actual form and implications. The conversation includes theoretical considerations, definitions of terms, and differing perspectives on mass and energy in the context of physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the simplified nature of E=mc² and seeks the actual equation.
  • Another participant proposes the equation E² = p²c² + m²c⁴ as a more complete form.
  • Clarification is provided that 'p' stands for momentum, and a question about the homogeneity of the equation is raised but not addressed.
  • A distinction is made between "massists," who attribute mass values in all states of motion, and "energists," who attribute energy values, leading to different interpretations of the equations.
  • For massists, E² = p²c² + m₀²c⁴ holds true in any inertial frame, while for energists, the relationship is framed differently, emphasizing the role of energy in defining mass.
  • One participant corrects a previous statement regarding the conditions under which E = mc² applies, specifying it relates to particles with non-zero mass.
  • A later contribution references the relationship between mass, energy, and velocity, providing a formula for tardyons (particles traveling slower than light).

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of mass and energy, with no consensus reached on a singular "actual" equation or the definitions of terms like mass and momentum.

Contextual Notes

There are unresolved assumptions regarding the definitions of mass and energy, as well as the conditions under which the proposed equations apply. The discussion reflects varying interpretations within the physics community.

alchemist
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what is the actual equation of e=mc^2? this is only the simplified equation, and i have forgotten the actual one already...
 
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You probably mean

E2 = p2c2 + m2c4

- Warren
 
what does the P stand for? and is that equation homogenous?
 
It's a lowercase p, and it stands for (linear) momentum. I don't know what you mean by "homogenous."

- Warren
 
It all depends on whether you are a massist or an energist.

------

A massist is willing to attribute mass values to anything, in any state of motion. For a massist, this m is really m0, a mass attributed to something in its rest frame of reference. For a massist

E2 = p2c2 + m02c4
p = mv

are always true in any inertial frame. For light quanta,

E = pc
p = mc

, because m0 = 0 for light quanta. But m = p/c = E/c2, a mass value dependent upon total energy of a quantum.
So E = mc2 is true for a light quantum as well as a particle with a non-zero rest mass.

------

An energist is willing to attribute energy values to anything, in any state of motion. For an energist, m can only be attributed to something in its rest frame, so the subscript 0 is never needed. For an energist,

p2 = E2/c2 - m2c2

is always true in any inertial frame. The energy E must come from other physics. For light quanta, p = E/c is a given, so

p2 = p2 - m2c2

, so

m2c2 = 0

. Since c > 0,

m = 0 for a light quantum.
So, E = mc2/(1 - v2/c2)1/2 only in the case of a particle with non-zero rest mass.

------

Most modern day physicists, especially high-energy physicists, tend to be energists rather than massists.
 
I said:

So, E = mc2/(1 - v2/c2)1/2 only in the case of a particle with non-zero rest mass.

I should have said:

So, E = mc2/(1 - v2/c2)1/2 only in the case of a particle with non-zero mass.
 


Originally posted by alchemist
what is the actual equation of e=mc^2? this is only the simplified equation, and i have forgotten the actual one already...

The equation E = mc2 is the mass-energy equation relating the mass m of a particle to the free-particle energy E. The proof can be found here

www.geocities.com/physics_world/sr/mass_energy_equiv.htm

If the particle is a tardyon (i.e. a particle which travels at speeds less than light) then

m = m0/sqrt[1-(v/c)2]

Multiply both sides by c2

mc2 = m0c2/sqrt[1-(v/c)2]

Substitute in E = mc2 to get

E = m0c2/sqrt[1-(v/c)2]

This equation can be rewritten as

E2 - (pc)2 = (m0c2)2

Pete
 

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