# Can special relativity be deduced from $E=mc^2$?

So instead of assuming that the velocity $c$ is a maximal velocity, proving that while assuming $E=mc^2$.

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Intuitively, I would say no. Mass-energy equivalence can be deduced from special relativity and probably from other (incorrect) theories too. –  Noldorin Nov 24 '10 at 15:48
assuming that the velocity c is a maximal velocity is not needed to obtain Lorenz transformations and SRT. Einstein deduces that from Clerk-Maxwell equations and "assumption that c=const is the basis of STR" is kind of useful folklore with mainly education purposes. –  kakaz Mar 10 '11 at 13:46

Usually, the cornerstone of Special Relativity is the constancy of the speed-of-light: form this fact alone, much of Special Relativity can be "deduced". (For a very interesting article about this point, see Relativity without light, by David Mermin.)

Further, this formula for energy is just an approximation of the fully relativistic expression, cf Mass-Energy Equivalence.

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$E = mc^2$ is not even an approximation, it's just a special case that applies only when $p = 0$. –  David Z Nov 25 '10 at 5:05
@David: I had the low speed expansion in mind when i made the comment above — of course, the case where $p = 0$ is contained in this expansion. ;-) –  Daniel Nov 25 '10 at 5:16
@Daniel: OK, I suppose I could have phrased that better. The point I wanted to make is that $E = mc^2$ is completely inaccurate at relativistic speeds. Yes, it is a valid approximation when $v \ll c$, but if you're going to call it an approximation at all, you really should include the domain of applicability. Otherwise you might wind up confusing people into thinking that $E = mc^2$ applies in every case. –  David Z Nov 26 '10 at 2:30
Um, I never understood why people talk about approximations when considering $E = mc^2$. The equation holds in full generality but the $m$ in Einstein's equation is not an invariant mass but an apparent mass. It's very confusing when people try to interpret that $m$ as invariant mass and the whole equation as either some approximation or else just to be valid in the particle's own reference frame (where both concepts of mass coincide). –  Marek Nov 26 '10 at 17:37
@Marek: in modern physics parlance, nobody (to leading order :-P) uses relativistic (or "apparent") mass anymore. $m$ always means rest mass. Instead of relativistic mass, you talk about energy, or $E / c^2$ if you are concerned about units. (And of course, laypeople will always consider $m$ to be rest mass) –  David Z Nov 27 '10 at 4:28

These two concepts are not related at all, so it's not possible to deduce anything either way.

Special relativity in its bare form talks only about space-time, i.e. geometry. You don't have any energy or mass there without first somehow postulating what it is.

For the reverse direction: you can't really deduce anything about SR from $E = mc^2$ because $E$ and $m$ could be anything (in particular you don't know whether $E$ is a component of a four-momentum $p$ or something completely different). And even if you knew what $E$ and $m$ was that would still give you no clue about what $c$ is (and that it should be constant in every inertial frame).

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Isn't special relativity all in a "flat" space, i.e. without geometry. And E=mc² appears in special relativity, which can be deduced from the invariance of c. In other words, I think your answer is wrong. –  Frédéric Grosshans Nov 24 '10 at 18:29
"without geometry" isn't quite right. While Minkowski space has no curvature, it DOES have a novel geometric feature--its inner product is not positive definite. In fact, you can work out all of the properties of special relativity from knowing that the inner product of the time time basis vector is negative, the inner product of all other basis vectors is positive, and the spacetime is flat. Though it should be noted that this structure is just a restatement of the constancy of the speed of light. –  Jerry Schirmer Nov 24 '10 at 20:09
@Frédéric: flat space does have a geometry. You have rotations and translations there (in Euclidean space). In Lorentz space you have boost (which is an imaginary rotation). So there is lot of geometry even in flat space! Arguably, most of the geometry people think about comes from flat spaces (just think about all the problems with triangles, parallel lines, etc., etc.). –  Marek Nov 25 '10 at 0:12
@Frédéric: as to the mass: yes, it can be deduced if you tell me what $E$ and $m$ is. You can also deduce lot of properties of quantum field theory just by using Lorentz (and Poicare) invariance, if you tell me, what quantum mechanics is. But special relativity at its core is just about geometry. No mass, no energy, no quanta... That is special stuff built on top of it. –  Marek Nov 25 '10 at 0:14

Here is a quick result that might interest you:

Let's begin from the equation of relativistic energy in terms of the rest mass, $m_0$, $E=\gamma m_oc^2$, then by inserting $\gamma = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$, we find:

$\hspace{6cm}E^2-E^2\frac{v^2}{c^2}=(m_0)^2c^4$

Hence:

$\hspace{6cm}E^2=(E_0)^2+p^2c^2$

which is the equation that relates relativistic energy and momentum. So maybe the best we can say is if you start from the equations $E^2=(E_0)^2+p^2c^2$, and $E=\gamma m_oc^2$, we can derive the Lorentz contraction.

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+1 This answer might not appear right or the reasoning might be circular, but at least it is an honest attempt to answer the question rather than evade it. –  user346 Dec 13 '10 at 4:27