Editing On shell and off shell (section)

==Scalar field ==
{{unreferenced section|date=December 2019}}
An example comes from considering a [[scalar field theory|scalar field]] in ''D''-dimensional [[Minkowski space]]. Consider a [[Lagrangian density]] given by <math>\mathcal{L}(\phi,\partial_\mu \phi)</math>. The [[Action (physics)#Action (functional)|action]] is

:<math>S = \int d^D x \mathcal{L}(\phi,\partial_\mu \phi)</math>.

The Euler–Lagrange equation for this action can be found by [[Calculus of variations#Euler.E2.80.93Lagrange equation|varying the field and its derivative and setting the variation to zero]], and is:

:<math>\partial_\mu \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} = \frac{\partial \mathcal{L}}{\partial \phi}</math>

Now, consider an infinitesimal  spacetime [[Translation (geometry)|translation]] <math>x^\mu \rightarrow x^\mu +\alpha^\mu</math>. The Lagrangian density <math>\mathcal{L}</math> is a scalar, and so will infinitesimally transform as <math>\delta \mathcal{L} = \alpha^\mu \partial_\mu \mathcal{L}</math> under the infinitesimal transformation. On the other hand, by [[Taylor expansion]], we have in general

:<math>\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta( \partial_\mu \phi) </math>

Substituting for <math>\delta \mathcal{L}</math> and noting that <math>\delta( \partial_\mu \phi) = \partial_\mu ( \delta \phi)</math> (since the variations are independent at each point in spacetime):

:<math>\alpha^\mu \partial_\mu \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi} \alpha^\mu \partial_\mu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)} \alpha^\mu \partial_\mu \partial_\nu \phi </math>

Since this has to hold for independent translations <math>\alpha^\mu = (\epsilon, 0,...,0) , (0,\epsilon, ...,0), ...</math>, we may "divide" by <math>\alpha^\mu</math> and write:

:<math> \partial_\mu \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi} \partial_\mu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)} \partial_\mu \partial_\nu \phi </math>

This is an example of an equation that holds ''off 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 an ''on shell'' equation by simply substituting the Euler–Lagrange equation:

:<math> \partial_\mu \mathcal{L} = \partial_\nu \frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)}  \partial_\mu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)} \partial_\mu \partial_\nu \phi </math>

We can write this as:

:<math> \partial_\nu \left (\frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)} \partial_\mu \phi -\delta^\nu_\mu \mathcal{L} \right) = 0 </math>

And if we define the quantity in parentheses as <math>T^\nu{}_\mu</math>, we have:

:<math>\partial_\nu T^\nu{}_\mu = 0</math>

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