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If $c$ is just an arbitrary constant, why don't we say $E=mc$ and define the value of $c$ to be $\sqrt{299 792 458} \approx 17314$ meters per second? Or, why not use $E=mc^3$?

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    $\begingroup$ I am not sure whether OP is joking or under influence $\endgroup$ Commented Apr 4, 2015 at 6:11
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    $\begingroup$ $E=mc$ and $E=mc^3$ not even dimensionally correct. $\endgroup$ Commented Apr 4, 2015 at 6:49
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    $\begingroup$ possible duplicate of Einstein's mass-energy equivalence $\endgroup$ Commented Apr 4, 2015 at 9:20
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    $\begingroup$ can we close that since OP doesn't state with which physical phenomena he has problem/misunderstanding. To me he takes square root of speed and still gets m/s, not square root of (m/s)... $\endgroup$ Commented Apr 4, 2015 at 10:40
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    $\begingroup$ possible duplicate of How do physicists justify using $c^2$ in equations? $\endgroup$ Commented Apr 4, 2015 at 13:37

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All these kinds of questions boil down to a pure question of convention: there is no in-principle reason why one couldn't think of your proposed $c^2$ as the "fundamental" constant: it's simply that generations of people have found it less awkward to use $c$ instead because one needs a parameter in special relativity that makes a time interval into an equivalent length to unite space and time into Minkowski spacetime. Accordingly, it is this co-ordinate transformation parameter, which must have the units length/time, that is focussed on as the fundamental entity, not its square.

Ultimately, special relativity can be derived from Galilean relativity if one relaxes the assumption of absolute time. In other words, one derives the most general possible transformation between inertial co-ordinate frames given the assumption of the first relativity postulate, (which is essentially that there is no experiment that an inertial observer can do making measurements completely confined to their own frame that would detect inertial motion) together with basic symmetry assumptions. From these axioms Galilean relativity uniquely follows if one assumes absolute time (i.e. that all inertial observers will agree on the time elapsed between two events). If, however, you drop the last assumption, then a whole family of transformation laws follows, parameterised by different values of a universal speed limit $c$: Galilean relativity is the family member with $c\to\infty$. In any member of this family, all inertial observers agree on the "length" of the vector $(\Delta x,\,\Delta y,\,\Delta z,\,c\,\Delta t)$ between two events in spacetime, where the parameter $c$ "makes the time interval $\Delta t$ into a length". It is this fundamental agreement - or conservation of the pseudonorm $\Delta x^2+\Delta y^2+\Delta z^2 - c^2\,\Delta t^2$ (note the minus sign - but it doesn't really come into this answer) by any Lorentz transformation that is the most striking, pithy and straighforward description of the most general transformation in keeping with the above simple, generalised Galilean assumptions. I discuss this more if you're interested in my answer to the Physics SE question "What's so special about the speed of light?" I also give references to Einstein's derivation of his $E=m\,c^2$ formula, which is a special case of the invariant 4-momentum pseudonorm $E^2/c^2 - p^2 = m^2 \,c^2$

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Where $E_0 = mc^2$ comes from:

From elementary physics, the work $W$, done on an object by a constant force of magnitude $F$ that acts through distance $s$, is given by $W = Fs$. By work-kinetic energy theorem, $$KE = \int_{0}^{s} F ds$$. In non-relativistic physics, the KE of an object of mass $m$ & speed $v$ is $$KE = \dfrac{mv^2}{2}$$ To find the correct relativistic formula, we start from the relativistic form of the second law which gives $$KE = \int_{0}^{s} \dfrac{{\gamma}{mv}}{dt} ds \implies KE = \int_{0}^{mv} v d(\gamma \cdot mv) \implies KE = \int_{0}^{v} v \cdot d{\left( \dfrac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}} \right)}$$.

Integrating by parts

$$\int x dy = xy - \int y dx \quad, KE =\dfrac{mv^2}{\sqrt{1 - \dfrac{v^2}{c^2}}} - m\int_{0}^{v} \dfrac{v \cdot dv}{\sqrt{1 - \dfrac{v^2}{c^2}}} \implies KE = \dfrac{mc^2}{\sqrt{1 - \dfrac{v^2}{c^2}}} - mc^2$$

Therefore $$KE = (\gamma - 1)mc^2$$ The result states that the kinetic energy of an object is equal to the difference between $\gamma mc^2$ & $mc^2$. If we write the former as total energy of the object, we see that when it is at rest, the object, nevertheless, possesses energy $mc^2$ which is called the rest energy . Therefore, rest energy is given by $$ E_0 = mc^2$$.

Response to @aandreev:

By Newton's Second law, the force is measured as $$F = m\dfrac{dv}{dt}$$. Now, we multiply the force by the distance over which it acts. In this case we obtain:

$$F\cdot \mathit{x} = ma \mathit{x} \implies F\cdot \mathit{x} = ma\dfrac{v_1 + v_2}{2} t = \frac{1}{2} m(at)(v_1 + v_2) \implies F\cdot \mathit{x} = \frac{1}{2} m(v_2 - v_1)(v_1 + v_2)$$. Therefore, $$F\cdot \mathit{x} = \dfrac{m{v_2}^2}{2} - \dfrac{m{v_1}^2}{2}$$ which is equal to the work done.

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  • $\begingroup$ I'm not understanding why the latex form isn't working. I urge anyone to please help. $\endgroup$ Commented Apr 4, 2015 at 9:16
  • $\begingroup$ You forget some }s. $\endgroup$ Commented Apr 4, 2015 at 9:24
  • $\begingroup$ but why is work not W=Fs^2 or something? That seems to be among OP's questions as well $\endgroup$ Commented Apr 4, 2015 at 10:39
  • $\begingroup$ @aandreev: Now, ok, sir? $\endgroup$ Commented Apr 4, 2015 at 11:10
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Since unit of $c$ is $\text{ms^{-1}}$ and unit of $m$ is $kg$. hence to get energy in $kg \frac{m^2}{s^2}$ we assume $c$ to be as it is, $\approx 290000~\text{km/s}$

$c$, $E$ and $m$ are not unitless values. By redefining mass unit (say, you start to measure everything tomorrow in $\sqrt {kg}$'s) you can change given formula to $E = kg^2c^2$

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  • $\begingroup$ This seems somewhat circular to me. The only reason that M and E have commensurable units is because they're defined that way through E=mc^2. Before Einstein, they were unrelated, right? $\endgroup$ Commented Apr 4, 2015 at 5:48
  • $\begingroup$ mass, speed, and energy were related since Newton. More like impulse and energy $\endgroup$ Commented Apr 4, 2015 at 6:10
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It's what we get through the derivation(s). If you look it up, we derive the equations and they don't give you $mc^3 \quad \text{or} \quad mc$.

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  • $\begingroup$ No need to downvote,it is a good ans. $\endgroup$ Commented Apr 4, 2015 at 6:53
  • $\begingroup$ I was wondering why I got down voted lol! $\endgroup$ Commented Apr 4, 2015 at 6:54
  • $\begingroup$ +1: This is the accurate answer. I've deduced it in my answer; just quantified yours:) $\endgroup$ Commented Apr 4, 2015 at 9:56
  • $\begingroup$ πŸ˜„πŸ˜„πŸ˜„πŸ˜„πŸ˜„πŸ˜„πŸ˜„πŸ˜„ I honestly felt it was a little too smart elic-y, lol! $\endgroup$ Commented Apr 4, 2015 at 10:04
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Firstly we can discuss the definition of the kinetic energy. Because, we have v^2 in the kinetic energy so a new definition for the speed of light leads to new definition for whole unit system. I mean if you write c instead of c^2, you get energy in a different unit system so you should redefine velocity, energy, etc otherwise they are not compatible....

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