- #1
wduff
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Hello, this should be an easy one to answer, hope it's in the right place.
I'm going through Sean M. Carroll's text on General Relativity, "Spacetime and Geometry." I'm working through calculating Christoffel connections (section 3.3, if you happen to have the book), which Carroll demonstrates generically by varying the proper time functional.
This yields an integral which he simplifies with "integration by parts," and he provides an example of the procedure for one of the integral's terms (equation 3.52): (sorry about the weird formatting, the integral's down there I promise)
\frac{}{}
\frac{1}{2}\int [g_{\mu \nu }\frac{dx^{\mu }}{d\tau }\frac{d(\delta x^{\nu })}{d\tau }]d\tau =-\frac{1}{2}\int [g_{\mu \nu }\frac{d^{2}x^{\mu }}{d\tau ^{2}}+\frac{dg_{\mu \nu }}{d\tau }\frac{dx^{\mu }}{d\tau }]\delta x^{\nu }d\tau
My problem is, this doesn't look like the "integration by parts" I'm familiar with it. How does sucking out that variational x^nu from the derivative yield the right side of the above equation?
Anyway, if you'd care to help it'd be much appreciated, thanks in advance.
I'm going through Sean M. Carroll's text on General Relativity, "Spacetime and Geometry." I'm working through calculating Christoffel connections (section 3.3, if you happen to have the book), which Carroll demonstrates generically by varying the proper time functional.
This yields an integral which he simplifies with "integration by parts," and he provides an example of the procedure for one of the integral's terms (equation 3.52): (sorry about the weird formatting, the integral's down there I promise)
\frac{}{}
\frac{1}{2}\int [g_{\mu \nu }\frac{dx^{\mu }}{d\tau }\frac{d(\delta x^{\nu })}{d\tau }]d\tau =-\frac{1}{2}\int [g_{\mu \nu }\frac{d^{2}x^{\mu }}{d\tau ^{2}}+\frac{dg_{\mu \nu }}{d\tau }\frac{dx^{\mu }}{d\tau }]\delta x^{\nu }d\tau
My problem is, this doesn't look like the "integration by parts" I'm familiar with it. How does sucking out that variational x^nu from the derivative yield the right side of the above equation?
Anyway, if you'd care to help it'd be much appreciated, thanks in advance.