$E=mc^2$ is the famous mass-energy equation of Albert Einstein. I know that it tells that mass can be converted to energy and vice versa. I know that $E$ is energy, $m$ is mass of a matter and $c$ is speed of light in vacuum.

What I didn't understood is how we will introduce speed of light?

Atom bomb is made using this principle which converts mass into energy; in that the mass is provided by uranium but where did speed of light comes into play? How can speed of light can be introduced in atom bomb?

  • $\begingroup$ Possible duplicate of this question; you may want to take a look at that. $\endgroup$ – Mark Beadles Jan 22 '12 at 5:23
  • $\begingroup$ Not really a dupe - that question is how can you have c^2 if c is the fastest speed. $\endgroup$ – Martin Beckett Jan 22 '12 at 5:47
  • $\begingroup$ N.B.: the relation is a specific case of the relativistic dispersion relation, $E^2 = m^2c^4 + p^2 c^2$. $\endgroup$ – JamalS Mar 28 '14 at 14:19
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    $\begingroup$ c is just a conversion constant or you can say it proportionality constant. Energy and mass are the same. When c=1, E=m $\endgroup$ – Self-Made Man Mar 28 '14 at 14:21

c is a priori not the speed of light. It is the speed of massless particles. The way it comes about is as follows: You construct the Lorentz-transformations as the symmetry transformations of Minkowski space. The group has one parameter, that's c. You have to fix it by physical means. You can look at the dynamics of massive particles and massless particles and find that massive particles will approach c asymptotically only at infinite energy, and massless particles always move with c.

Since to our best knowledge photons are massless, c is also the speed of light. Also, that was historically Einstein's motivation, which is why it's usually motivated in textbooks this way. However, should it turn out one day that photons do have very tiny masses, then c will still be there, it will just no longer be called the speed of light.

  • $\begingroup$ Bravo, I was also thinking how can c, speed of light come in play. But sir i have a question : if tachyons really exist, then what will be said about their mass - something beyond infinity??, and what about the interesting notion "result coming before the cause".. Pardon me if I am asking like this (in comment section) but want to get to know your views $\endgroup$ – Pranjal Rana Oct 14 '17 at 7:11

The reason $c$ is important is not because it is the speed of light. It's important because it is a universal conversion factor between time and distance. If you have a certain amount of time $t$, you can calculate the corresponding amount of distance by multiplying it by $c$.

Note: I'm not talking about the distance any particular object travels in the time $t$. If you have a car traveling at speed $v$, you can find out how far the car moves in time $t$ by multiplying by $v$, but that's not converting time into distance. The conversion is something more fundamental.

The fact that time and distance can be converted into each other like this is one of the ways relativity changed our view of the world. One of its consequences is that speeds can now be measured as pure, unitless numbers. How so? Well, normally, we might measure distance in meters and time in seconds. So when you calculate a velocity as distance divided by time, you get an answer in meters per second. But because time can be converted into distance, now you can measure time in meters as well. So if you divide distance (in meters) by time (in meters), the meters cancel out and you get a pure number.

As a pure number, $c = 1$. There are a few things that travel at speed 1, including light and gravity. Light was simply the first one that we discovered, which is the only reason $c$ is called the "speed of light."

Once you see that $c$ is important for reasons having nothing to do with the fact that light travels at that speed, hopefully it seems less strange that it enters into the formula $E = mc^2$. Just as you can convert time into distance, you can also convert mass into energy. You just have to multiply by $c$ twice, not just once, to make the units work out.


The constant $c$ is the speed of light, namely the constant $$ c= 299,792,458\,\,{\rm m/s} $$ exactly (using the current definition of a meter and a second). As David mentions, it is called the speed of light only because light was the first entity that was known to move by that speed. But it is also the speed of gravitons or anything else that moves by the maximum speed. It is the most important value of a speed in the Universe.

It appears in the relationship between mass and energy – which is valid for nuclear bombs, thermonuclear bombs, particle accelerators, stars, as well as everything else in this Universe – because $c$ is a fundamental universal constant that is important everywhere in the basic laws of this Universe. That's a conclusion (and partly also assumption) of the special theory of relativity by Albert Einstein. For example, the speed of light in the vacuum is always measured to be the same, regardless of the speed of the source and the observer, and it is the maximum speed that can be achieved by information (or "almost" achieved by moving matter).

Mature physicists understand that the conserved mass and conserved energy are really the same thing and only one independent quantity of this kind is conserved; mass may be converted via $E=mc^2$ and vice versa. After all, adult physicists (in particle physics and other fields that depend on relativity) often use units in which $c=1$. One light second and one second is fundamentally the same thing. The fact that $c$ has an awkward numerical value in SI units is just because the SI units (and other units) carry an awkward cultural baggage. Fundamentally, $c$ is very clean and crisp and it should be $c=1$. In these units, $E=m$ simply holds. Mass and energy is the same thing, when converted by the natural conversion factor.

For beginners or historically, there are various ways to derive $E=mc^2$. Einstein was considering an accelerating mass object. He did work ${\rm d}E$ on this object of rest mass $M_0$ which increased its velocity $v$. One may prove that the inertial mass of the object $M$ also had to increase by ${\rm d}M={\rm d}E/c^2$ for certain things to work – i.e. to prevent the mass object from surpassing the speed of light by extra acceleration.

The keyword you should look for if you want to understand $E=mc^2$ is the "special theory of relativity", the broader theory implying $E=mc^2$ and other things and discovered by Einstein's 1905 paper.

Let me add a caricature of the explanation which contains all the relevant things. Special relativity shows that various things – time, distance – get inflated or contracted by the Lorentz factor of $$ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} >1 $$ That's also true for the mass. If you accept that the total mass is $$ M = M_0\gamma $$ where $M_0$ is the rest mass, then you may Taylor expand the total $Mc^2$ for small $v$ $$ Mc^2 = M_0 c^2 + \frac{1}{2} M_0 v^2 + \dots $$ So the first subleading term is exactly the usual kinetic energy from Newton's world, $M_0v^2/2$, but there's also an even greater term, the latent energy stored in the rest mass, $M_0 c^2$. This latent energy is constant for "non-nuclear" processes that don't change the internal character of the matter. Because it's constant, this term in the energy is physically inconsequential. Then the most important term is the kinetic energy $mv^2/2$, and it has the right coefficient assuming we have $E=mc^2$ to start with. However, $M_0c^2$ is there and it may get changed to other forms of energy if we do change the fundamental character of matter, e.g. if we split uranium nuclei to other nuclei. Relativity guarantees that we release $E=\delta m\cdot c^2$ of energy where $\delta m$ is the change of the rest mass of the radioactive matter.


c: “Light Velocity” (c: celeritas, Latin: “swift”),
is the speed of photons in a straight, flat vacuum:
299, 792, 458 meters per second ( 3e^8 mps).
This reference is a constant that never varies.
(Does NOT mean mass is moving at light speed!)

c^2 = 89, 875, 517, 873, 681, 764 ( 9e^16 ), where
“9e^16” becomes the “mass to energy ratio”
(nucleons to photons) but NOT a velocity.

Each kg of matter is composed of
6e^26 nucleons (N) and every nucleon contains
1.5e^−10 joules (J) of energy.
The total energy (E) in a mass (m) of 1 kg
is equal to the number of nucleons (N) per kg
multiplied by the energy (J) per nucleon (J/N):
E: 9e^16 J = 6e^26 N/kg x 1.5e^−10 J/N
Compare this to:
E=mc^2: E: 9e16 J = m: 1 kg x c^2: 9e^16
m=E/ c^2: m: 1 kg = E: 9e^16 J / c^2: 9e^16

This energy is released by atomic annihilation
(using anti-matter). The energy released
would be Gamma Rays.

If this energy is released all at once:
9e^16 J would equal approximately
21.5 Mt (Megatons of TNT).
This is 1,000 times greater than either
the Trinity or Nagasaki devices.

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    $\begingroup$ I do not see how this answers the OP, or what this adds to existing answers. $\endgroup$ – ZeroTheHero Feb 1 '20 at 2:21

A previous answer has provided a beautiful explanation of what $c$ represents, and that is not necessarily related to light. This answer just adds a bit from the nuclear reaction perspective.

You start with a large nucleus, say uranium. Once it splits, it forms two smaller nuclei. The masses of these smaller nuclei put together are less than the mass of the original nucleus. The 'missing mass' is what we use as an energy source, because as previously mentioned, mass not conserved means energy also not conserved.

Einstein's equation is simply a tool for us to calculate how much energy we are getting through this process. In that sense, $c$ can be thought of as a conversion factor without physical significance.

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    $\begingroup$ I wouldn't say that $c$ has no physical significance. As DavidZ & Lubos said, it's the conversion factor that you use to compare the sizes of space distances & time durations, i.e., that in a very real sense 1 light-second of space has the same magnitude as 1 second of time. But it's certainly an "accident of history" that we first encountered $c$ as the speed of light. $\endgroup$ – PM 2Ring Feb 1 '20 at 3:21
  • $\begingroup$ @PM 2Ring- I agree completely. My point is, in this specific nuclear reaction context, one can understand what is happening without invoking the physical significance. Hence "can be thought of as..." instead of "has no physical significance". $\endgroup$ – AppliedAcademic Feb 1 '20 at 3:36

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