quangtu123
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First I want to consider an example of 1D motion. Lagrange equation:
$$ \frac{d}{dt} \frac{\partial L}{\partial \dot x} - \frac{\partial L}{\partial x} = 0 $$
If we transform $$L \rightarrow L+a$$ with a is constant, the equation of motion remains unchanged. This is global symmetry.
To obtain local symmetry we want when transforming $$L \rightarrow L+a(x) $$ we still have the same equation. To obtain that we introduce the "total derivative":
$$\frac{Df}{dt} = \frac{df}{dt} + \frac{\partial a}{\partial x}$$
Then the equantion of motion would be unchanged under any local transformation:
$$\frac{D}{dt} \frac{\partial L}{\partial \dot x} - \frac{\partial L}{\partial x} = 0$$
The quantity $$\frac{\partial a}{\partial x}$$ is similar to the Christoffel symbols in general relativity.
Is there anyway to construct General Relativity by demanding local symmetry like this?
$$ \frac{d}{dt} \frac{\partial L}{\partial \dot x} - \frac{\partial L}{\partial x} = 0 $$
If we transform $$L \rightarrow L+a$$ with a is constant, the equation of motion remains unchanged. This is global symmetry.
To obtain local symmetry we want when transforming $$L \rightarrow L+a(x) $$ we still have the same equation. To obtain that we introduce the "total derivative":
$$\frac{Df}{dt} = \frac{df}{dt} + \frac{\partial a}{\partial x}$$
Then the equantion of motion would be unchanged under any local transformation:
$$\frac{D}{dt} \frac{\partial L}{\partial \dot x} - \frac{\partial L}{\partial x} = 0$$
The quantity $$\frac{\partial a}{\partial x}$$ is similar to the Christoffel symbols in general relativity.
Is there anyway to construct General Relativity by demanding local symmetry like this?