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I'm reading Friedman and Susskind's Special Relativity and Classical Field Theory.

They define the Lagrangian of a free particle

$$\mathcal L = -mc^2\sqrt{1-{v^2\over c^2}}$$

and then derive the corresponding Hamiltonian to be

$$H = \frac{mc^2}{\sqrt{1-{v^2\over c^2}}} .$$

Then they note that in the non-relativistic limit

$$v \ll c \\H \to mc^2+{1\over 2}mv^2.$$

Also, for $v=0$, $H = mc^2$. They then identify this $mc^2$ as "energy of assembly" of the particle.

Now this SE post's answers suggest that the process of conversion of mass into energy follows from particle physics. But particle physics can't be just explained by the two postulates of special relativity! How can then this "rest energy" be derived from just postulating the free particle Lagrangian? What is so non-trivial about it?

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  • $\begingroup$ arxiv.org/abs/1204.6576 $\endgroup$ Commented Jul 12, 2020 at 12:42
  • $\begingroup$ If you define rest mass as $m_0$, which is just m in your expression, and say the entire term without $c^2$ as mass, you could just say E=mc² $\endgroup$
    – SK Dash
    Commented Jul 12, 2020 at 13:01
  • $\begingroup$ Related: physics.stackexchange.com/q/43813/2451 , physics.stackexchange.com/q/178960/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jul 12, 2020 at 13:19
  • $\begingroup$ The particle physics comment probably meant that the rest system equality $E=m$ doesn't represent a conversion of mass into energy. Any difference in energy between two nuclei just depends on the difference in static Coulomb energy between the two nuclei. In this sense, nuclear fission is no different than burning log where the energy released also equals the difference in atomic masses, but in that case is too small to be measured. A true conversion of mass into energy is in something like $n\rightarrow p+e+\nu$. $\endgroup$ Commented Jul 14, 2020 at 12:31

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You are fundamentally correct, but are mixing up what's necessary for each. Special relativity is both necessary and sufficient to construct the idea of rest energy, but it is merely necessary for constructing a model of conversion of rest energy to other forms of energy. In order to do that, you need particle physics, and the only way we know how to construct a good model of particle physics is quantum field theory.

To elaborate a little: in special relativity, you can derive the rest energy of a particle as you have shown above! Special relativity is the source of the idea of rest energy. However, in special relativity, there is no concept of particle creation and destruction; there is no concept of converting this energy into anything else. You need quantum field theory, the marriage of special relativity and quantum mechanics, to do that.

Check out this question for more information.

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  • $\begingroup$ Is special relativity really necessary for the idea of rest energy? I was expecting particle physics to be able to show that on its own... $\endgroup$
    – Atom
    Commented Jul 12, 2020 at 14:36
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    $\begingroup$ @Atom It very well be able to, but only because particle physics is based on special relativity! The only way we know how to make a consistent model of particle physics is by using quantum field theory, which combines quantum mechanics and special relativity. The only reason rest energy shows up in particle physics is because of special relativity! Without special relativity, you have neither particle physics nor rest energy itself. $\endgroup$ Commented Jul 12, 2020 at 17:36
  • $\begingroup$ Thanks tons! That answers my question! Answer accepted. $\endgroup$
    – Atom
    Commented Jul 12, 2020 at 17:46
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Written as $ΔE = (Δm)c^2$ the $ΔE$ is the increase in the kinetic energy of a mass being accelerated, and the $Δm$ is the corresponding increase in the inertial mass (as predicted by special relativity).

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  • $\begingroup$ That ist false, if you increase the kinetic energy by accelerating the object. Your equation is only correct if you increase the internal energy of the object without changing the total momentum. $\endgroup$
    – Azzinoth
    Commented Jul 13, 2020 at 11:00
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One of the first steps in developing the kinematics and dynamics of special relativity is to extend three-dimensional momentum to a four dimensional four-vector. This is done by multiplying the four-velocity $(\gamma,\gamma{\bf v})$ by a constant, $m$, with the dimensions of energy (in natural units with c equal to 1). The invariant length squared of the momentum four-vector is $E^2-{\bf p}^2=m^2$. When this equation is reduced in the limit $p<<E$, all the non-relativistic formulas up show. The relation $E^2-{\bf p}^2=m^2$ shows that $m$ is the invariant length of the four-vector, and I hope most people agree should be called the 'invariant mass'. It also shows that $m$ and $E$ are two quite different things. $E$ is the fourth component of the four-vector and $m$ is its invariant length. $m$=$E$ in the rest system, but that doesn't make them the same thing.

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