Metric tensor, infitesimal transformation

[gouranja]

Hi,

I don't think this belongs in the homework section since this is a graduate course.
My question is regarding making a field theory generally covariant by including a metric tensor $$g_{\mu\nu}(x)$$in the Lagransian density, and it's transformation under infinitesimal coordinate change.

If I transform the coordinates according to:
$$x^{\mu}\rightarrow x^{\mu}'=x^{\mu}+\epsilon^{\mu}(x)$$

The metric must be transformed according to:
$$g_{\mu\nu}(x)\rightarrow g'_{\mu'\nu'}(x')=\frac{\partial x^{\alpha}}{\partial x^{\mu}'}\frac{\partial x^{\beta}}{\partial x^{\nu}'}g_{\alpha\beta}(x)$$

which I understand well. But according to the textbook I'm reading the infitesimal result is:
$$\delta g_{\mu\nu}(x)=\epsilon_{\mu;\nu}+\epsilon_{\nu;\mu}$$
(with covariant derivatives).

I have tried using:
$$\frac{\partial x^{\alpha}}{\partial x^{\mu}'}\simeq\delta_{\mu}^{\alpha}-\frac{\partial\epsilon^{\alpha}}{\partial x^{\mu}}+O(\epsilon^{2})$$

to obtain:
$$\delta g_{\mu\nu}(x)=-g_{\mu\alpha}\partial_{\nu}\epsilon^{\alpha}-g_{\nu\alpha}\partial_{\mu}\epsilon^{\alpha}$$

but I don't know how to obtain the textbook result.
Can someone clue me in on how to do it?

Thanks

 

[gouranja]

Never mind, I solved it...

 

[denian]

for your question. The metric tensor is an essential tool in the study of general relativity and field theories. It describes the local geometry of spacetime and allows us to calculate distances and angles in curved spacetime. In order for a field theory to be generally covariant, meaning that it is invariant under arbitrary coordinate transformations, the metric tensor must be included in the Lagrangian density. This ensures that the theory will give the same predictions regardless of the choice of coordinates.

When we make an infinitesimal transformation of the coordinates, we must also transform the metric tensor accordingly. This is because the metric tensor is a function of the coordinates and any change in the coordinates will result in a change in the metric. The formula you have provided for the transformation of the metric tensor is correct and comes from the chain rule for partial derivatives.

The infinitesimal result you have shown, \delta g_{\mu\nu}(x)=\epsilon_{\mu;\nu}+\epsilon_{\nu;\mu}, comes from the definition of the covariant derivative. The covariant derivative of a tensor is defined as the partial derivative plus a term involving the Christoffel symbols, which are related to the metric tensor. In this case, the covariant derivative of the metric tensor is given by \nabla_{\mu}g_{\nu\alpha}=\partial_{\mu}g_{\nu\alpha}-\Gamma^{\beta}_{\mu\nu}g_{\beta\alpha}-\Gamma^{\beta}_{\mu\alpha}g_{\nu\beta}. When we make an infinitesimal transformation, the Christoffel symbols do not change, but the metric tensor does. This results in the formula you have shown above.

I hope this helps clarify the derivation of the textbook result. If you have any further questions, please feel free to ask. Good luck with your studies!