This thread is supposed to be a continuation of the discussion of this thread: (1) https://www.physicsforums.com/showthread.php?t=88570.(adsbygoogle = window.adsbygoogle || []).push({});

The previous thread was closed but there was a lot of things I did not understand.

This is also somewhat related to a recent thread I created: (2) https://www.physicsforums.com/showthread.php?t=729135.

So, starting with the discussion in thread (1):

Physics Monkey wrote,

This obviously begs the question, "How do I know which of the raised or lowered components are the ones independent of the metric?" The answer has to come from context. Suppose I define some vector as tangent to a curve. Well this object naturally comes with a raised index and isn't defined using the metric, so the raised components must be metric independent. However, when you use the metric to find the equivalent 1 form, the 1 form you obtain obviously depends to the metric you use. Thus the lowered components do depend on the metric. This subtlety means that you have to be very careful about how you define various tensors. For instance, is the electromagnetic potential most naturally a 1 form or a vector? Elementary treatments often start with it as a vector, but it gauge theory it appears as a connection 1 form. Consistency must be you guide.Then, from what I understood, if the electromagnetic potential is "most naturally" a 1-form and the most natural way to define a derivative is also through a 1-form, it is the [itex] F_{\mu\nu} [/itex] part of the lagrangian that is constant with respect to the metric, therefore:

[tex] t^{\alpha\beta} \equiv \frac{\partial \left( F_{\mu\nu}F^{\mu\nu} \right)}{\partial g_{\alpha\beta}} =

\frac{\partial}{g_{\alpha\beta}} \left( g^{\mu\lambda} g^{\nu\kappa} F_{\mu\nu} F_{\lambda\kappa} \right) =

F_{\mu\nu} F_{\lambda\kappa} \left[ \frac{\partial g^{\mu\lambda}}{\partial g_{\alpha\beta}} g^{\nu\kappa} + g^{\mu\lambda} \frac{\partial g^{\nu\kappa}}{\partial g_{\alpha\beta}} \right] =

[/tex]

[tex]

=-F_{\mu\nu} F_{\lambda\kappa} \left[ g^{\mu\alpha} g^{\lambda\beta} g^{\nu\kappa} + g^{\mu\lambda} g^{\nu\alpha} g^{\kappa\beta} \right] =

-2 g_{\mu\nu} F^{\alpha\mu}F^{\beta\nu}

[/tex]

And:

[tex]

t_{\alpha\beta} \equiv \frac{\partial \left( F_{\mu\nu}F^{\mu\nu} \right)}{\partial g^{\alpha\beta}} =

\frac{\partial}{g^{\alpha\beta}} \left( g^{\mu\lambda} g^{\nu\kappa} F_{\mu\nu} F_{\lambda\kappa} \right) =

F_{\mu\nu} F_{\lambda\kappa} \left[ \frac{\partial g^{\mu\lambda}}{\partial g^{\alpha\beta}} g^{\nu\kappa} + g^{\mu\lambda} \frac{\partial g^{\nu\kappa}}{\partial g^{\alpha\beta}} \right] =

[/tex]

[tex]

=F_{\mu\nu} F_{\lambda\kappa} \left[ \delta^{\mu}_{\alpha} \delta^{\lambda}_{\beta} g^{\nu\kappa} + g^{\mu\lambda} \delta^{\nu}_{\alpha} \delta^{\kappa}_{\beta} \right] =

2 g^{\mu\nu} F_{\alpha\mu}F_{\beta\nu}

[/tex]

But we also have:

[tex]

t_{\alpha\beta} = g_{\alpha\mu} g_{\beta\nu} t^{\mu\nu} = -2 g^{\mu\nu} F_{\alpha\mu}F_{\beta\nu} \neq 2 g^{\mu\nu} F_{\alpha\mu}F_{\beta\nu}

[/tex]

Which is clearly a contradiction. I don't know if I missed or misunderstood something in his explanation, but as I see it, that is the consequence of what he explained.

Also, from thread (1):

lonelyphysicist asks about terms like [itex] V^{\alpha} V_{\alpha} [/itex] in the lagrangian:

what is

[tex] \frac{\partial}{\partial g_{\mu \nu}} \left( g_{\alpha \beta} V^{\alpha} V^{\beta} \right) [/tex]

Is it

[tex] \frac{\partial}{\partial g_{\mu \nu}} \left( g_{\alpha \beta} V^{\alpha} V^{\beta} \right) = \delta^{\mu}_{\phantom{\mu}\alpha} \delta^{\nu}_{\phantom{\mu}\beta} V^{\alpha} V^{\beta} = V^{\mu} V^{\nu} [/tex]

or is it

[tex] \frac{\partial}{\partial g_{\mu \nu}} \left( g^{\alpha \beta} V_{\alpha} V_{\beta} \right) = - g^{\mu \alpha} g^{\nu \beta} V_{\alpha} V_{\beta} = -V^{\mu} V^{\nu} [/tex]

?Physics Monkey promptly replies him, but what if instead of terms like [itex] V^{\alpha} V_{\alpha} [/itex] we had something like [itex] A^{\alpha} B_{\alpha} [/itex], where A and B are arbitrary vector fields.

This question is related to my other thread (2) and to Barbarabados post in thread (1) (last post).

The point of all of this being:

I'm not sure what are the technical procedures for deriving the stress-energy tensor for an arbitrary lagrangian using the usual definition given in GR books. It seems to me that the usual method for differentiating with respect to the the metric lacks consistency sometimes. Although, I think it's much more probable that I'm just not getting something very simple.

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# Stress-Energy Tensor from Lagrangian: Technical Question II

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