# Something is wrong

1. Jul 18, 2013

### russelljbarry15

I am looking for help on page 128 equation (14) from zee's book Einstein Gravity.
A lot of you may not have the book. I have a phd and cannot see it, feel really really stupid.

How did he get rid of the square roots. I know that he used the definition of the derivative from basic calculus with out the delta x on the bottom. In the first square root he varied the action so that L moves to the bottom. You need the second g so the taylor series starts off with a derivative.

How does the square root go away when the second part is not varied.

I will use different symbols from the book, but they are only dummy variables so I can change them.
Plus it makes it easier to write out the equation.

∂gκμδX = gκμ(X(λ) + δX(λ)) - gκμ(X(λ))

so you need the second term. Please remember κ and μ are just dummy variables so I am free to choose them, as long as I carry them through. The κ and μ are indices.

2. Jul 18, 2013

### WannabeNewton

Yeah I'm not seeing what he did there either...

3. Jul 18, 2013

### martinbn

You expand what's under the first radical, ignoring higher order variations, then you rationalize the expression and you get in the numerator what he has in the parentheses in the last line. The denominator is different not just L, but I suppose it is of the same order.

4. Jul 18, 2013

### Bill_K

Isn't it just straightforward differentiation? With L = √A he used δ(√A) = (1/√A)(½δA). That puts the L in the denominator. Then with A = a product, A = gμν dXμ/dλ dXν/dλ,

δA = (δgμν) dXμ/dλ dXν/dλ + gμν δdXμ/dλ dXν/dλ + gμν dXμ/dλ δdXν/dλ

= gμν,σ dXσ dXμ/dλ dXν/dλ + gμν dδXμ/dλ dXν/dλ + gμν dXμ/dλ dδXν/dλ

Last edited: Jul 18, 2013