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Why does E=mc^2

by ZeroPivot
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ZeroPivot
#1
Sep9-13, 04:55 PM
P: 54
explain it like im 12 years old.
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jtbell
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Sep9-13, 05:00 PM
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So you don't know any calculus?
Turion
#3
Sep9-13, 05:02 PM
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Good luck with that.

Vanadium 50
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Sep9-13, 06:21 PM
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Why does E=mc^2

http://www.physicsforums.com/blog.php?bt=6374
ZeroPivot
#5
Sep10-13, 09:15 AM
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Quote Quote by jtbell View Post
So you don't know any calculus?
i do know upto multivariable.
Nugatory
#6
Sep10-13, 10:01 AM
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Quote Quote by ZeroPivot View Post
i do know upto multivariable.
OK, then you are prepared for something much better than a 12-year-old level explanation. There are a ton of good derivations online - find one, study it, come back with a more specific question.

One hint: You're really looking for the derivation of the equation ##E^2=(m_0{c}^2)^2+(pc)^2##, as ##E=mc^2## is just the ##p=0## special case of that more general relationship.
jtbell
#7
Sep10-13, 10:34 AM
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Quote Quote by ZeroPivot View Post
i do know upto multivariable.
OK, try this. It's a draft of an FAQ that I've been working on.

In both non-relativistic and relativistic physics, we can define an object's (kinetic) energy in terms of the work done on the object by a force, using the work-energy theorem. In the non-relativistic case, this leads to the familiar ##E = \frac{1}{2}mv^2##; in the relativistic case, this leads to the familiar ##E = mc^2## (using an old-fashioned interpretation of ##m##).

Warning: This derivation uses calculus. It also assumes that we already know the non-relativistic and relativistic formulas for an object's momentum.

Beginning physics students learn early on that when a constant force ##F## acts on an object, causing the object to move a distance ##\Delta x##, it does work ##W = F \Delta x##. If the force is not constant, we have to consider it as a function of position, and integrate to find the work:
$$W = \int^{x_2}_{x_1} {Fdx}$$
Inserting Newton's Second Law of Motion in the form ##F = dp/dt## (where ##p## is momentum) and playing some games with the differentials:
$$W = \int^{x_2}_{x_1} {\frac{dp}{dt} dx}
= \int^{x_2}_{x_1} {\frac{dp}{dv} \frac{dv}{dt} dx}
= \int^{v_2}_{v_1} {\frac{dp}{dv} \frac{dx}{dt} dv}
= \int^{v_2}_{v_1} {\frac{dp}{dv} v dv}$$
So far, this applies in both non-relativistic and relativistic mechanics. To proceed further, we need an equation for ##p## in terms of ##v##. Here the relativistic and non-relativistic cases diverge.

Non-relativistically, ##p = mv##, so ##\frac{dp}{dv} = m## and
$$W = \int^{v_2}_{v_1} {m v dv}
= \frac{1}{2} mv^2_2 - \frac{1}{2} mv^2_1$$
We define the energy of the object as
$$E = \frac{1}{2} mv^2$$
so that ##W = E_2 - E_1## (the work-energy theorem). When ##v = 0##, ##E = 0##, so the energy that we've defined here is due only to the object's motion, and we therefore call it kinetic energy.
$$E = K = \frac{1}{2} mv^2$$

Relativistically,
$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$
where ##m## is what is often called the "rest mass." I leave it as a calculus exercise for the reader to evaluate the derivative:
$$\frac{dp}{dv} = \frac{m}{(1-v^2/c^2)^{3/2}}$$
so that
$$W = \int^{v_2}_{v_1} {\frac {mvdv}{(1-v^2/c^2)^{3/2}}}$$
I leave it as another calculus exercise for the reader to evaluate the integral above to get:
$$W = \frac{mc^2}{\sqrt{1-v_2^2/c^2}} - \frac{mc^2}{\sqrt{1-v_1^2/c^2}}$$
We define the energy of the object as
$$E = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$
so that ##W = E_2 - E_1## (the work-energy theorem), just like in the non-relativistic case. When ##v = 0##, ##E = mc^2##, so the energy that we've defined here is not due only to the object's motion. We can separate it into two pieces: the rest energy
$$E_0 = mc^2$$
which doesn't depend on the motion, and the kinetic energy which does:
$$K = \frac{mc^2}{\sqrt{1-v^2/c^2}} - mc^2$$
so that
$$E = E_0 + K = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$
In the early days of relativity, physicists called the ##m## that I used above, the "rest mass" ##m_0##; and defined the "relativistic mass" as
$$m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$
so that ##E = mc^2## and ##E_0 = m_0 c^2##. Nowadays, most physicists don't do this, but popular-level books and even some introductory textbooks still do.


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