How to derieve newtons lawE=m/c
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Answers
Answer:
There are quite a few ways to derive Einstein’s famous equation E=mc^{2}. I am going to show you what I consider to be the simplest way. Feel free to comment if you think you know of an easier way.
We will start off with the relationship between energy, force and distance. We can write
dE = F dx \text{ (1) }
Where dE is the change in energy, F is the force and dx is the distance through which the object moves under that force. But, force can also be written as the rate of change of momentum,
F = \frac{dp}{dt}
Allowing us to re-write Equation (1) as
dE = \frac{dp}{dt}dx \rightarrow dE = dp \frac{dx}{dt} = vdp \text{ (2) }
Remember that momentum p is defined as
p =mv
In classical physics, mass is constant. But this is not the case in Special Relativity, where mass is a function of velocity (so-called relativistic mass).
m = \frac{ m_{0} }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } \text{ (3) }
where m_{0} is defined as the rest mass (the mass of an object as measured in a reference frame where it is stationary).
Assuming that both m \text{ and } v can change, we can therefore write
dp =mdv + vdm
This allows us to write Equ. (2) as
dE = vdp = v(mdv + vdm) = mvdv + v^{2}dm \text{ (4) }
Differentiating Equ. (3) with respect to velocity we get
\frac{dm}{dv} = \frac{d}{dv} \left( \frac{ m_{0} }{ \sqrt{ (1 - v^{2}/c^{2}) } } \right) = m_{0} \frac{d}{dv} (1 - v^{2}/c^{2})^{-1/2}
Using the chain rule to differentiate this, we have
\frac{dm}{dv} = m_{0} \cdot - \frac{1}{2} (1 - v^{2}/c^{2})^{-3/2} \cdot (-2v/c^{2}) = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-3/2} \text{ (5) }
But, we can write
(1 - v^{2}/c^{2})^{-3/2} as (1-v^{2}/c^{2})^{-1/2} \cdot (1-v^{2}/c^{2})^{-1}
This allows us to write Equ. (5) as
\frac{dm}{dv} = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-1} \cdot (1 - v^{2}/c^{2})^{-1/2}
From the definition of the relativistic mass in Equ. (3), we can rewrite this as
\frac{dm}{dv} = \frac{ m v }{ c^{2} }(1-v^{2}/c^{2})^{-1}
Which is
\frac{dm}{dv} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}} - \frac{ v^{2}}{c^{2} } \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}-v^{2}}{c^{2}} \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}-v^{2}} \right)
\frac{dm}{dv} = \frac{ m v }{ (c^{2}-v^{2}) } \text{ (6) }
So we can write
c^{2}dm - v^{2}dm = mvdv
Substituting this expression for mvdv into Equ. (4) we have
dE = vdp = vd(mv) = mvdv + v^{2}dm = c^{2}dm - v^{2}dm + v^{2}dm
So
dE = c^{2} dm
Integrating this we get
\int_{E_{0}}^{E} dE = c^{2} \int_{m_{0}}^{m} dm
So
E - E_{0} = c^{2} ( m - m_{0} ) = mc^{2} - m_{0}c^{2}
E - E_{0} = mc^{2} - m_{0}c^{2}
This tells us that an object has rest mass energy E_{0} = m_{0}c^{2} and that its total energy is given by
\boxed{ E = mc^{2} }
Answer:
First of all this was not given by Newton
It was given by Einstein
E = mc2 derive. When Einstein first proposed his Special Theory of Relativity in 1905 few people understood it and even fewer believed it. It wasn't until 1919 that the Special Theory was "proved by inference" from an experiment carried out on his General Theory of Relativity.