# Either my professor made a boo-boo or I don't get it(regarding E=mc^2)

1. Aug 29, 2005

### schattenjaeger

either way, you can help! so w=change in KE, and W=the line integral from 0 to s of FdotdS(I never learned how to use the latex thingy, so bear with me, the question isn't that hard...)

So then he wrote KE=lineintegral from 0-S, d/dt(gamma*mv)ds

and then from there you end up with gamma*mc^2-mc^2 after a whole bunch of math, which you take to mean the total energy - the "rest" energy? So...shouldn't up there you write change in KE=lineintegral etc...? If it IS just KE, I dunno why, since the line integral definition is for work anyways...well that may be a little confusing in just words, sorry

2. Aug 29, 2005

### lightgrav

At the time when that he replaced Work with Delta KE,
there didn't seem to be any Potential Energies involved.
And with no "non-conservative" processes, KE_0 = 0.

If you want to interpret mc^2 as a PE, go ahead -
the important thing is the limit v = 0 .
I've never heard mc^2 called
a "KE of the object at speed v=0" .

If you need something to ponder here, how about
whether "non-conservative" is well-defined anymore.

3. Aug 30, 2005

### rbj

schattenjaeger, first i would suggest looking at some posts with tex in it (press the quote button to look at the syntax, but don't save) to get this math formatting down. and there are also some sticky posts that describe it.

second, forget about vectors and three dimensions for a moment. in fact, forget about relativity for the time being. given a single dimension (the "x" dimension), can you take a force field, with constant force $$F$$ and known length $$l$$, let a particle of mass $$m_0$$ enter this force field at the, say, left side with velocity zero, and then determine what the velocity is when it comes out the right side? do it first for the classical case where the mass is constant $$m = m_0$$:

$$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m_0 \frac{dv}{dt} = m_0 a$$

then do it for the relativistic case:

$$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m \frac{dv}{dt} + v \frac{dm}{dt}$$

where

$$m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$$

use all the theorems you learned in calculus to do derivatives. i think then you'll figure it out.

rots o' ruk.

4. Aug 30, 2005

### ZapperZ

Staff Emeritus
When you edit your post, there's an option to delete it.

Zz.

5. Aug 30, 2005

### rbj

thanks, Zz. maybe someday i'll look at these things. (i never read directions either.)

6. Aug 30, 2005

### pervect

Staff Emeritus
I'm not sure where this is leading is, or what the question is. Doing what you suggest (with a change back to the original posters notation of m=$m_0$), I get:

$$p = \frac{m v}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
$$F = \frac{dp}{dt} = \frac{dp}{dv} \, \frac{dv}{dt} = \frac{m }{\left( 1-\left( \frac{v}{c} \right)^2 \right)^{\frac{3}{2}}} \, \frac{dv}{dt}$$

7. Aug 30, 2005

### rbj

i think the OP wants to know how to get the relativistic kinetic energy to come out as:

$$T = m c^2 - m_0 c^2 = E - E_0$$

where

$$m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$$

(and then you interpret that first term to be the total energy $E$ and the second term to be the rest energy $E_0$.)

in the classical case where the mass is constant $m = m_0$, then you can show that with a force field with constant force $F$ and length $l$, a object of mass $m_0$ enters it at velocity $v_1$ and exits at velocity

$$v_2 = \sqrt{\frac{2 F l}{m_0} + v_1^2}$$

or

$$\frac{1}{2} m_0 v_2^2 = \frac{1}{2} m_0 v_1^2 + F l$$

which is where we get the concept of kinetic energy from. if the initial velocity is zero, the final velocity depends only on the product of $F$ and $l$ or the work performed on the object and not on the any other function of $F$ and $l$.

$$T_2 = T_1 + F l = T_1 + W$$

$T_1$ and $T_2$ are the initial and final kinetic energies.

to do that in the classical mechanics case ain't too hard. to do it in the reletivistic case, where the mass is not constant is harder, of course. if no one else picks this up, i'll work on it and if i get it, i'll post it later.