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Assuming that the speed of light was infinite, we would get back to Newtonian physics and special relativity wouldn't be as relevant anymore.

However, this would mean that the energy is infinite as well since E=mc^2+..., meaning that even the smallest amount of mass could even turn into infinite energy.

What is an explanation to this infinite energy?

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    $\begingroup$ The derivation for $E=mc^2$ builds on the assumption that $c$ is finite. $\endgroup$
    – lemon
    Apr 21, 2017 at 17:35
  • $\begingroup$ "assuming the speed of light is infinite" is a question of what-if type listed in the types of questions you shouldn't ask. If you want to know how to go back to Newtonian mechanics from special relativity, the proper limit is to assume that $v$ is much smaller than $c$, not that $c$ is infinite. There can be no physical explanation for the "infinite energy" you get for infinite speed of light because physically, the speed of light is finite, so it's not clear what you're looking for here. $\endgroup$
    – ACuriousMind
    Apr 22, 2017 at 13:15
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    $\begingroup$ I think it is a very legitimate question. There are always cases in physics in which yo can take known constants to zero or infinity in order to see what happens. Another example would be taking Planck's constant to zero in the path integral to receive classical mechanics from quantum mechanics. $\endgroup$
    – FlyGuy
    Apr 22, 2017 at 18:56

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Actually, the correct formula is

$$E^2-p^2c^2=m^2c^4$$

or, solved for $E$,

$$E=\sqrt{m^2c^4+p^2c^2}=mc^2\frac{1}{\sqrt{1-v^2/c^2}}=\gamma mc^2$$

If you Taylor expand $\gamma$ in powers of $v^2/c^2$, you will get

$$E=mc^2+\frac{1}{2}mv^2+\frac{3}{8}mv^2\frac{v^2}{c^2}+\frac{5}{16}mv^2\frac{v^4}{c^4}+\mathcal{O}(v^6/c^6)$$

You see that in the limit of large $c$ only the first two terms survive since $v^2/c^2$ becomes very small. The second term is the well known kinetic energy. The first term is a fixed constant which is actually indeed infinite. However, you might have already heard that energy, as well as potential, is only defined up to an additive constant. Therefore, it is possible to shift the energy scale such that the constant first term vanishes:

$$E'=E-mc^2=\frac{1}{2}mv^2+\mathcal{O}(v^2/c^2)$$

This is what we have in non-relativistic mechanics.

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  • $\begingroup$ Yes you're completely right about the energy. I should've written E=mc^2+... $\endgroup$
    – FlyGuy
    Apr 21, 2017 at 17:55
  • $\begingroup$ But is there a more physical explanation to infinite energy rather than it just being like reference energy? $\endgroup$
    – FlyGuy
    Apr 21, 2017 at 17:56
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    $\begingroup$ To be honest, I have no better explanation. But I need to note that this is not unusual in physics. Already in classical electromagnetism one cannot define the electric potential of a point charge such that it is finite (or even zero) at the origin: $\Phi(r)=\tfrac{q}{r} + C$ is such that no matter how you select C you will get an infinity at $r=0$. So you leave it and live with an infinity at $r=0$ defining the potential such that it is zero at $r=\infty$. Similar infinities frequently emerge in quantum field theories (see e.g. zero point energy). $\endgroup$
    – Photon
    Apr 21, 2017 at 18:02
  • $\begingroup$ It's not c that is infinite, the real problem has been that calculations of m in QFT led to infinite mass. That was fixed with renormalizarion $\endgroup$
    – Bob Bee
    Apr 21, 2017 at 18:04
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    $\begingroup$ @BobBee: Well, mass is not what we want to set to zero but a constant term in the energy. So, actually, one would need to look at the energy momentum tensor of a point particle and see how it behaves in this limit... A point particle has a delta-like density distribution already, so the $(00)$ component of the energy momentum tensor is ugly by construction, question is now what happens if we try to take the non-relativistic limit (which way ever we would want to do it)... $\endgroup$
    – Photon
    Apr 21, 2017 at 18:32

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