Is true that moving object did not posses $mc^2$ as their component of energy?

Question

According to the special relativity theory, moving object have kinetic energy given by formula $KE = (\gamma - 1) m c^2 $. So this is equivalent with classical formulation $KE = (1/2) m v^2$ for the case $v^2 / c^2 \ll 1 $ (using binomial expansion). But in classical formulation, moving object just have kinetic energy and there are no form of energy other than that. For special relativity say it statement for KE is equivalent with classical formulation, then we can say there is no component of $m c^2$ for moving object too, because total energy for moving object in classical formulation is just their kinetik energy.

If not (there is component $m c^2$ in moving object), then what is equivalent statement for total energy $\gamma m c^2$ in classical formulation? I mean is there any procedur to exploit total energy other than making a bomb (complex nuclear reaction)?

Einstein derive his statement for energy just using integration to $F ds$, a classical defenition of work, but like a magic trick there is appear $m c^2 $ term, where $m c^2 \gg (1/2) m v^2 $ in resultant equation. So it is safe for me to say that when an object moving on $ds$ path it posses an intrinsic $m c^2$ property? Remember $F ds $ is statement for the work done by that object, so this defenition coupled with the path $ds $ and also movement of that object. If that object at rest, then $ds = 0 $, there is no work done by that object, and maybe there is nothing $m c^2$ too.

Precisely, is there any way to derive $m c^2$ term without using integration to $ds$ path or wihout assumption that the object is moving?

Answer 1

An object of a rest mass $m_o$ has always, at a frame of reference, an Energy if it moves or not, it always has a rest Energy $E_o=m_oc^2$. When it moves at a $v \ll c$,it also acquires a Kinetic Energy $(1/2)mv^2+ \text{negligible terms}$, but its total energy is $E=E_o+E_{kin}$. The term $(1/2)mv^2$ comes out from the approximation in the expansion of $1/\sqrt{1-(v/c)^2}$ when $v \ll c$. You may check the following,

An object of invariant mass $m_o$ has always, at a frame of reference, an Energy
$E_o=m_oc^2$ If the object moves at a velociy v, then its Energy becomes $E=γm_oc^2$ and by expanding $γ$ when $v \ll c$ we get the terms $E=m_oc^2$+$(1/2)m_ov^2$+... where the term $(1/2)m_ov^2$ represents the Kinetic Energy term.

Hope the above help.

Answer 2

A well-known pedagogic route to the idea that a body's total energy is given by $\gamma m c^2$ is this...

1. Newtonian momentum is $\Sigma m \vec u,$ in which $m$ is a constant for each body. It is easy to show that if Newtonian momentum is conserved in one inertial frame, it is not conserved in a frame moving at constant velocity with respect to the first frame if we use the Lorentz transformations (LTs). So the conservation of Newtonian momentum isn't consistent with the relativity principle if we use the LTs.

2. By considering a suitable collision between two bodies we can show that a body's momentum component in the y direction, that is at right angles to the relative motion of the frames, is unchanged between frames: a 'Lorentz invariant'. $m\frac{\Delta y}{\Delta \tau}=\gamma m u_y$ is a plausible candidate, making $\gamma m \vec u$ the relativistic formula for momentum.

3. Considering components parallel to the direction of relative motion of the frames (the x direction), we find that $$\Sigma_\text{before}\ \gamma m u_x=\Sigma_\text{after}\ \gamma m u_x$$is consistent with the Relativity Principle and the LTs if and only if $$\Sigma_\text{before}\ \gamma m =\Sigma_\text{after}\ \gamma m $$

4. We recognise $(\gamma m, \gamma m u_x, \gamma m u_y, \gamma m u_z)$ as the components of a 4-vector.

5. Bearing in mind that this 4-vector is conserved in all collisions, not just elastic ones, we must interpret the time-like component, $\gamma m$ (multiplied by the 'mere' constant, $c^2,$ so the units are right) as the total energy of a body. Therefore the energy of a stationary body ($\gamma=0$) is $mc^2.$ This can be made extremely plausible by looking at one or two simple elastic and inelastic collisions, but

6. This argument doesn't stand alone. Remember that Special Relativity, together with electromagnetism and dynamics is a coherent web of relationships, and that it has been tested by experiment many, many times.