Metric tensor, infitesimal transformation

In summary, the conversation discusses the transformation of a metric tensor in a field theory to make it generally covariant. The textbook presents an infinitesimal result for this transformation using covariant derivatives, while the individual has attempted to derive it using partial derivatives and has asked for clarification. However, they have since solved the problem on their own.
  • #1
gouranja
11
0
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 [tex]g_{\mu\nu}(x)[/tex]in the Lagransian density, and it's transformation under infinitesimal coordinate change.

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

The metric must be transformed according to:
[tex]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)[/tex]

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

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

to obtain:
[tex]\delta g_{\mu\nu}(x)=-g_{\mu\alpha}\partial_{\nu}\epsilon^{\alpha}-g_{\nu\alpha}\partial_{\mu}\epsilon^{\alpha}[/tex]

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

Thanks
 
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  • #2
Never mind, I solved it...
 
  • #3
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!
 

Related to Metric tensor, infitesimal transformation

1. What is a metric tensor?

A metric tensor is a mathematical object used in the field of differential geometry to measure distances and angles between points in a curved space. It is represented by a symmetric matrix of functions that describe the local geometry of the space at each point.

2. How is the metric tensor related to the concept of curvature?

The metric tensor is closely related to the concept of curvature in that it determines how distances and angles change as you move along a curved space. In fact, the components of the metric tensor are used to calculate the curvature of a space.

3. What is an infinitesimal transformation?

An infinitesimal transformation is a small, continuous change in a system or object. In the context of metric tensors, it refers to the transformation of coordinates in a curved space, where small changes in one set of coordinates correspond to small changes in another set of coordinates.

4. How is the metric tensor used in general relativity?

In general relativity, the metric tensor is used to describe the curvature of spacetime and how it is affected by the presence of matter and energy. It is a crucial component of Einstein's field equations, which relate the curvature of spacetime to the distribution of matter and energy within it.

5. Can the metric tensor be used in flat spaces?

Yes, the metric tensor can also be used in flat spaces. In this case, the metric tensor reduces to the identity matrix, which indicates that the distances and angles between points are constant. This is known as the Euclidean metric tensor.

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