Register to reply 
Why does E=mc^2by ZeroPivot
Tags: emc2 
Share this thread: 
#1
Sep913, 04:55 PM

P: 54

explain it like im 12 years old.



#3
Sep913, 05:02 PM

P: 148

Good luck with that.



#4
Sep913, 06:21 PM

Mentor
P: 16,171

Why does E=mc^2



#5
Sep1013, 09:15 AM

P: 54




#6
Sep1013, 10:01 AM

Sci Advisor
Thanks
P: 3,458

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. 


#7
Sep1013, 10:34 AM

Mentor
P: 11,619

In both nonrelativistic 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 workenergy theorem. In the nonrelativistic 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 oldfashioned interpretation of ##m##). Warning: This derivation uses calculus. It also assumes that we already know the nonrelativistic 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 nonrelativistic and relativistic mechanics. To proceed further, we need an equation for ##p## in terms of ##v##. Here the relativistic and nonrelativistic cases diverge. Nonrelativistically, ##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 workenergy 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, $$p = \frac{mv}{\sqrt{1v^2/c^2}}$$ where ##m## is what is often called the "rest mass." I leave it as a calculus exercise for the reader to evaluate the derivative: $$\frac{dp}{dv} = \frac{m}{(1v^2/c^2)^{3/2}}$$ so that $$W = \int^{v_2}_{v_1} {\frac {mvdv}{(1v^2/c^2)^{3/2}}}$$ I leave it as another calculus exercise for the reader to evaluate the integral above to get: $$W = \frac{mc^2}{\sqrt{1v_2^2/c^2}}  \frac{mc^2}{\sqrt{1v_1^2/c^2}}$$ We define the energy of the object as $$E = \frac{mc^2}{\sqrt{1v^2/c^2}}$$ so that ##W = E_2  E_1## (the workenergy theorem), just like in the nonrelativistic 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: $$K = \frac{mc^2}{\sqrt{1v^2/c^2}}  mc^2$$ so that $$E = E_0 + K = \frac{mc^2}{\sqrt{1v^2/c^2}}$$ 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 $$m = \frac{m_0}{\sqrt{1v^2/c^2}}$$ so that ##E = mc^2## and ##E_0 = m_0 c^2##. Nowadays, most physicists don't do this, but popularlevel books and even some introductory textbooks still do. 


Register to reply 