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Topic: What is the modern way of proving E=MC^2 using SR?
Replies: 3   Last Post: Feb 6, 2013 4:50 PM

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 Koobee Wublee Posts: 1,417 Registered: 2/21/06
Re: What is the modern way of proving E=MC^2 using SR?
Posted: Feb 4, 2013 12:55 PM

On Jan 30, 11:24 am, black head <larryhar...@softhome.net> wrote:

> How do physicists nowadays prove E=MC^2 using the model of SR?

To derive that, you must derive the relativistic momentum first. So,

** dS^2 = c^2 dt^2 ? ds^2

Where

** dS^2 = Spacetime geometry
** ds^2 = Spatial geometry

Claim all geodesic motions follow the path with the least accumulated
spacetime. In doing so, the Lagrangian of the geodesics can be
identified.

** dS = L dt

Where

** L = sqrt(1 ? (ds/dt)^2 / c^2), the Lagrangian

Derive the only Euler-Lagrange equation to the Lagrangian identified
above.

** d[- ds/dt / c^2 / sqrt(1 ? (ds/dt)^2)]/dt = 0

Or

** B / sqrt(1 ? B^2) = Constant

Where

** B = ds/dt / c

Interpret the constant identified related to the observed (or
relativistic) momentum.

** B / sqrt(1 ? B^2) = p / (m c)

Or

** p = m B c / sqrt(1 ? B^2)

Reason the total energy is the sum squared of the momentum times the
speed of light squared and the energy at null momentum.

** E^2(B) = p^2 c^2 + E^2(0)

Solve for E(B).

** E = m c^2 / sqrt(1 ? B^2)

Define the observed mass to be the following.

** M = m / sqrt(1 ? B^2)

Thus,

** E = M c^2

Date Subject Author
2/4/13 Koobee Wublee
2/4/13 Brian Q. Hutchings
2/6/13 Brian Q. Hutchings