Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
What is the modern way of proving E=MC^2 using SR?
Replies:
3
Last Post:
Feb 6, 2013 4:50 PM




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, start with the geometry of Minkowski spacetime.
** 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 EulerLagrange 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



