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## Main Question or Discussion Point

explain it like im 12 years old.

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explain it like im 12 years old.

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jtbell

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So you don't know any calculus?

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Good luck with that.

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Vanadium 50

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https://www.physicsforums.com/blog.php?bt=6374 [Broken]

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i do know upto multivariable.So you don't know any calculus?

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Nugatory

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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.i do know upto multivariable.

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

jtbell

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OK, try this. It's a draft of an FAQ that I've been working on.i do know upto multivariable.

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##).

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.

$$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

$$E = K = \frac{1}{2} mv^2$$

$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$

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

$$E_0 = mc^2$$

which doesn't depend on the motion, and the

$$K = \frac{mc^2}{\sqrt{1-v^2/c^2}} - mc^2$$

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

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

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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It seems to me that the work-energy theorem

is a relation between the net-work done by the forces on the object

and the change in

in spite of this common name for the theorem.).

In intro physics, we obtain the total-Mechanical-Energy when we identify the

"portion of the net-work that is done by conservative forces"

as "minus the change in Potential Energy". When we move this portion to the other side,

we then have

the work-done-by-nonconservative-forces is equal to the change-in-total-Mechanical-Energy.

Let's not make that split now.

If you consider the case when it

Following the same scheme for the relativistic case,

you get (using your integration) [itex]\frac{mc^2}{\sqrt{1-(V/c)^2}}-mc^2=(\frac{1}{\sqrt{1-(V/c)^2}}-1)mc^2[/itex], which is

(From that you can build an expression for the relativistic-energy by grouping K and mc^2.).

Then, more generally,

you can express the work done in changing the velocity from [itex]v_1[/itex] to [itex]v_2[/itex]

as the difference of their relativistic-kinetic-energies.

It's true that each of those relativistic-kinetic-energies has a constant [itex]mc^2[/itex], which

cancel when the difference is calculated, ending up with something that resembles the

In addition, the work-energy theorem suggests that we can apply forces to change the kinetic energy of a particle of constant-mass m, possibly bringing the kinetic energy to zero.

However, we can't apply forces to bring the relativistic-energy of a particle of constant-mass m (here, $\gamma mc^2) to zero.

So, it seems to me that

the work-energy-theorem is really about changes in

- #9

jtbell

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So is it just a lucky coincidence that the constant term turns out to be the correct rest-energy?

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Perhaps this derivation indicates that there is an intrinsic amount of "kinetic energy" in an object with some mass. I.e. ##K=mc^2##, and this "kinetic energy" increases as velocity is added to the object. Maybe the "total rest energy" of an object is it's "rest kinetic energy" plus any internal potential energy or other possible forms of energy present.

(furthermore...)

Furthermore, I would intuitively think that the "total energy in some volume" would be dependent of what is immediately outside that volume. Obviously, in a more pressurized environment, it takes more energy to open up some volume of space. I find it puzzling that the entirety of energy in some object at rest could be dependent only on the mass of the object, and not any other properties of the object and its environment.

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