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Inphysics, particularly inquantum field theory, configurations of a physical system that satisfy classicalequations of motion are calledon the mass shell (on shell); while those that do not are calledoff the mass shell (off shell).

In quantum field theory,virtual particles are termed off shell because they do not satisfy theenergy–momentum relation; real exchange particles do satisfy this relation and are termed on (mass) shell.^[1][2][3] Inclassical mechanics for instance, in theaction formulation, extremal solutions to thevariational principle are on shell and theEuler–Lagrange equations give the on-shell equations.Noether's theorem regarding differentiable symmetries of physical action andconservation laws is another on-shell theorem.

Mass shell is a synonym formass hyperboloid, meaning thehyperboloid inenergy–momentum space describing the solutions to the equation:

${\displaystyle E^2-|\overrightarrow{p}{|}^2{c}^2=m_0^2{c}^4}$

Themass–energy equivalence formula which gives the energy${\displaystyle E}$ in terms of the momentum${\displaystyle \overrightarrow{p}}$ and therest mass${\displaystyle m_0}$ of a particle. The equation for the mass shell is also often written in terms of thefour-momentum; inEinstein notation withmetric signature (+,−,−,−) and units where thespeed of light${\displaystyle c=1}$, as${\displaystyle p^{\mu }{p}_{\mu }\equiv p^2=m_0^2}$. In the literature, one may also encounter${\displaystyle p^{\mu }{p}_{\mu }=-m_0^2}$ if the metric signature used is (−,+,+,+).

The four-momentum of an exchanged virtual particle${\displaystyle X}$ is${\displaystyle q_{\mu }}$, with mass${\displaystyle q^2=m_X^2}$. The four-momentum${\displaystyle q_{\mu }}$ of the virtual particle is the difference between the four-momenta of the incoming and outgoing particles.

Virtual particles corresponding to internalpropagators in aFeynman diagram are in general allowed to be off shell, but the amplitude for the process will diminish depending on how far off shell they are.^[4] This is because the${\displaystyle q^2}$-dependence of the propagator is determined by the four-momenta of the incoming and outgoing particles. The propagator typically hassingularities on the mass shell.^[5]

When speaking of the propagator, negative values for${\displaystyle E}$ that satisfy the equation are thought of as being on shell, though the classical theory does not allow negative values for the energy of a particle. This is because the propagator incorporates into one expression the cases in which the particle carries energy in one direction, and in which itsantiparticle carries energy in the other direction; negative and positive on-shell${\displaystyle E}$ then simply represent opposing flows of positive energy.

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An example comes from considering ascalar field inD-dimensionalMinkowski space. Consider aLagrangian density given by${\displaystyle \mathcal{L}(\varphi ,{\mathrm{\partial }}_{\mu }\varphi )}$. Theaction is:

${\displaystyle S=\int d^{D}x\mathcal{L}(\varphi ,{\mathrm{\partial }}_{\mu }\varphi )}$

The Euler–Lagrange equation for this action can be found byvarying the field and its derivative and setting the variation to zero, and is:

${\displaystyle {\mathrm{\partial }}_{\mu }\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }\varphi )}=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }\varphi }}$

Now, consider an infinitesimal spacetimetranslation${\displaystyle x^{\mu }\to x^{\mu }+{\alpha }^{\mu }}$. The Lagrangian density${\displaystyle \mathcal{L}}$ is a scalar, and so will infinitesimally transform as${\displaystyle \delta \mathcal{L}={\alpha }^{\mu }{\mathrm{\partial }}_{\mu }\mathcal{L}}$ under the infinitesimal transformation. On the other hand, byTaylor expansion, we have in general:

${\displaystyle \delta \mathcal{L}=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }\varphi }\delta \varphi +\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }\varphi )}\delta ({\mathrm{\partial }}_{\mu }\varphi )}$

Substituting for${\displaystyle \delta \mathcal{L}}$ and noting that${\displaystyle \delta ({\mathrm{\partial }}_{\mu }\varphi )={\mathrm{\partial }}_{\mu }(\delta \varphi )}$ (since the variations are independent at each point in spacetime):

${\displaystyle {\alpha }^{\mu }{\mathrm{\partial }}_{\mu }\mathcal{L}=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }\varphi }{\alpha }^{\mu }{\mathrm{\partial }}_{\mu }\varphi +\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\nu }\varphi )}{\alpha }^{\mu }{\mathrm{\partial }}_{\mu }{\mathrm{\partial }}_{\nu }\varphi }$

Since this has to hold for independent translations${\displaystyle {\alpha }^{\mu }=(ϵ,0,...,0),(0,ϵ,...,0),...}$, we may "divide" by${\displaystyle {\alpha }^{\mu }}$ and write:

${\displaystyle {\mathrm{\partial }}_{\mu }\mathcal{L}=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }\varphi }{\mathrm{\partial }}_{\mu }\varphi +\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\nu }\varphi )}{\mathrm{\partial }}_{\mu }{\mathrm{\partial }}_{\nu }\varphi }$

This is an example of an equation that holdsoff shell, since it is true for any fields configuration regardless of whether it respects the equations of motion (in this case, the Euler–Lagrange equation given above). However, we can derive anon shell equation by simply substituting the Euler–Lagrange equation:

${\displaystyle {\mathrm{\partial }}_{\mu }\mathcal{L}={\mathrm{\partial }}_{\nu }\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\nu }\varphi )}{\mathrm{\partial }}_{\mu }\varphi +\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\nu }\varphi )}{\mathrm{\partial }}_{\mu }{\mathrm{\partial }}_{\nu }\varphi }$

We can write this as:

${\displaystyle {\mathrm{\partial }}_{\nu }\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\nu }\varphi )}{\mathrm{\partial }}_{\mu }\varphi -{\delta }_{\mu }^{\nu }\mathcal{L}\right)=0}$

And if we define the quantity in parentheses as${\displaystyle T^{\nu }{}_{\mu }}$, we have:

${\displaystyle {\mathrm{\partial }}_{\nu }{T}^{\nu }{}_{\mu }=0}$

This is an instance of Noether's theorem. Here, the conserved quantity is thestress–energy tensor, which is only conserved on shell, that is, if the equations of motion are satisfied.

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