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

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?

• When an object has mass, it has energy. And the mass depends on the frame of reference, so the energy due to the mass term depends on the path (velocity). – Kalpak Gupta Apr 11 '17 at 7:27
• but in many book about special relativity, $m c^2$ term is rest energy, even the object did not moving, it still has that term as their energy component. Sorry I do not intend to make a debat about the thing that obviously, I just do not understand, what Einstein need with the derivation was to calculate how much kinetic energy of an object from special relativitu point of view, and he make radical assumption about additional term emerged. – Mohammad Fajar Apr 11 '17 at 7:34
• Maybe you can just answer the question – Mohammad Fajar Apr 11 '17 at 7:36
• @KalpakGupta Mass does not depend on frame of reference or velocity. – garyp Apr 13 '17 at 12:05
• @Kalpak Gupta The concept of relativistic mass has long since been discarded, and it should always be referenced as “relativistic mass”. The unqualified term “mass” should always refer to the invariant mass. – Dale Mar 22 '19 at 21:01

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.

• you make statement $E_0 = m_0 c^2$ as a postulat, whereas my question is how you can derive this equation from basic principle. – Mohammad Fajar Apr 11 '17 at 8:50
• In fact with that approach, you cannot derive relativistics form of energy. Did you say that energy for relavistics particle just their relativistics mass multiplied by $c^2$, then where $c^2$ term come from? – Mohammad Fajar Apr 11 '17 at 8:52
• Secondly, despite it's continuing prevalence in pop-sci and secondary-school textbook, the "rest mass" and "relativistic mass" formulation of relativity has been largely abandoned by working physicist. In the modern formulation there is just one mass (which is what was previously called the rest mass, but is now called the invariant mass if you need to be specific) and the quantity $\gamma m$ is not given a name at all. – dmckee --- ex-moderator kitten Apr 11 '17 at 15:30
• @MohammadFajar The relativistic momentum is $\gamma m \vec{v}$, and it forms a Lorentz (four-)vector with the total energy $\gamma mc^2$. None of the mathematics change, only the terminology and (critically) the mode of thinking. There are statements that are obviously false if you look closely, but which sound OK with the old terminology; the new terminology keeps them from be stated in a way that sound like it might be true. – dmckee --- ex-moderator kitten Apr 12 '17 at 21:16
• Here are some "for instance" questions from the site: physics.stackexchange.com/questions/3436/… physics.stackexchange.com/questions/180698/… The answer to both question is "no", but both sound reasonable in the old terminology, while the it is much easier to reason correctly about them in the new terminology. This one too: physics.stackexchange.com/questions/222809/…. – dmckee --- ex-moderator kitten Apr 12 '17 at 21:21

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.