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Is there some historical reasons or is there a specific reason behind it?

This question is connected to: Why on-shell vs. off-shell matters?

Community
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Gunn
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2 Answers2

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A particle is said to be on-shell if it satisfies the relativistic dispersion relation,

$$E^2 = p^2 +m^2$$

in units wherein $c=\hbar=1$. If you graph it, you obtain a hyperboloid surface (in 4-momentum space) for massive particles, and a cone for massless particles, like a photon. This is known as the mass shell, it is quite literally a shell or surface. The momentum of a real particle can be represented as a vector on the surface, hence the expression on shell. Virtual particles do not have these on the surface, hence they are off shell.

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Source: Perimeter Institute, A Deeper Dive: On Shell and Off Shell

Constantine D.
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JamalS
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  • Of course, the term "on-shell" has broadened significantly to refer field or particle trajectories that obey the EoMs. – AfterShave Mar 30 '24 at 03:47
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The term "shell" originally derives from the non-relativistic version of the answer by @JamalS. In a non-relativistic theory, a free particle satisfies the following dispersion relation $$ E = \frac{ {\bf p}^2 }{ 2m } $$ For a fixed energy a particle satisfies $$ {\bf p}_x^2 + {\bf p}_y^2 + {\bf p}_z^2 = 2 m E $$ In momentum space, this is precisely the equation of a spherical "shell" with radius $\sqrt{2mE}$. (See a plot here)

Now, in momentum space, the 3-momentum of a particle is described by a point. If $E$ is fixed, then the 3-momentum of the particle is only allowed to lie on the spherical shell described above, and is therefore said to be on-shell.

In quantum mechanics, a particle is allowed to not satisfy the dispersion relation. Thus, the momentum of a particle in quantum mechanics is allowed to not lie on the shell, and can be off-shell.

In a relativistic theory, the dispersion relation changes to $$ E^2 = {\bf p}^2 + m^2 \implies {\bf p}_x^2 + {\bf p}_y^2 + {\bf p}_z^2 = E^2 - m^2 $$ The corresponding constant $E$ surface is again a spherical shell, (this time of radius $\sqrt{ E^2 - m^2 }$)

Prahar
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