Simplifying Equation (14) from Zee's Einstein Gravity for PhD Students

[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 without 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.

 

[WannabeNewton]

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

 

[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.

 

[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 = g_μν dX^μ/dλ dX^ν/dλ,

δA = (δg_μν) dX^μ/dλ dX^ν/dλ + g_μν δdX^μ/dλ dX^ν/dλ + g_μν dX^μ/dλ δdX^ν/dλ

= g_μν,σ dX^σ dX^μ/dλ dX^ν/dλ + g_μν dδX^μ/dλ dX^ν/dλ + g_μν dX^μ/dλ dδX^ν/dλ

 

[sardar]

Equation (14) in Zee's book can be simplified using the definition of the derivative from basic calculus. The first step is to use the definition of the derivative to write the variation in the action as a difference between the action at a point and the action at a nearby point:

ΔS = S(x + δx) - S(x)

Next, we can expand the action at the nearby point using a Taylor series:

S(x + δx) = S(x) + δx∂S/∂x + O(δx²)

Substituting this into the first equation, we get:

ΔS = δx∂S/∂x + O(δx²)

Now, we can use this expression for ΔS in the equation (14) from Zee's book:

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

Substituting ΔS for δX(λ), we get:

∂gκμδX = gκμ(X(λ) + ΔS) - gκμ(X(λ))

Expanding the first term using the Taylor series, we get:

∂gκμδX = gκμ(X(λ)) + gκμ∂S/∂X(λ) + O(ΔS²) - gκμ(X(λ))

The second term, gκμ∂S/∂X(λ), is the one that eliminates the square root. This is because the derivative of the action, ∂S/∂X(λ), is a linear function of ΔS, and the square root in the equation (14) arises from the quadratic term in ΔS². Therefore, when we substitute the linear term in place of ΔS, the square root disappears.

Finally, we can simplify the equation further by recognizing that the dummy variables κ and μ are just indices and can be freely chosen. This allows us to use different symbols for them, making the equation easier to write out and understand.

In summary, the square root in equation (14) from Zee's book disappears because of the linear term in the Taylor series expansion of the action, and the dummy variables can be chosen freely to simplify the equation.