FieldEqIntro

# FieldEqIntro - Physics 131 UCSB Spring 2008 A Zee...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 131 UCSB Spring 2008 A. Zee Department of Physics, University of California, Santa Barbara, CA 93106 Kavli Institute for Theoretical Physics, University of California, Santa Barbara, CA 93106 5/24/08 To Einstein’s field equation, as quickly as possible! I. TO EINSTEIN’S FIELD EQUATION, AS QUICKLY AS POSSIBLE! ∞ Let me take you to Einstein’s field equation as quickly as possible starting with what you already know. My pedagogical philosophy is to keep things as simple as possible. I could elaborate and expand on each point, but we could always come back and do that later. We start with the covariant derivative of a vector with a lower index: D ν S ρ = ∂ ν S ρ- Γ σ νρ S σ (1) For our calculation below, we will need to know the covariant derivative of a tensor T νρ with two lower indices. To figure this out, we recognize from the discussion in which we obtain the covariant derivative that what is essential is merely the transformation property of T νρ under a coordinate change. Since T νρ transforms like the product of two vectors U ν W ρ we define the covariant derivative of T νρ as if it is made of the product of two vectors (even though that is certainly not the case in general.) On the other hand, we want the covariant derivative to be distributive, just like the ordinary derivative; in other words, we impose the distributive rule D μ ( U ν W ρ ) = ( D μ U ν ) W ρ + U ν ( D μ W ρ ) = ∂ μ ( U ν W ρ )- Γ σ μν U σ W ρ- Γ σ μρ U ν W σ (2) Thus, we are led to define D μ T νρ = ∂ μ T νρ- Γ σ μν T σρ- Γ σ μρ T νσ (3) II. BREAKING THE NEWTON-LEIBNIZ RULE To derive the Riemann curvature tensor, let us, once again, start from elementary calculus. Given the value of a function f ( x ) and its derivative at some point x , then the value of the function at the point x + δx a short distance away is of course given by (1 + δx d dx ) f ( x ) ∼ f ( x + δx ). Think of the operation (1 + δx d dx ) as a translation operator: it moves or translates the function from the point x to the point x + δx . Generalize to two dimensional flat space. Consider a small rectangle with opposite corners at ( x,y ) and ( x + δx, y + δy ). To find the value of a function at ( x + δx, y + δy ) we could translate first in the x direction and then in the y direction, thus (1 + δy ∂ ∂y )(1 + δx ∂ ∂x ) f ( x ) = (1 + δx ∂ ∂x + δy ∂ ∂y + δxδy ∂ ∂y ∂ ∂x ) f ( x,y ) (4) We apply the translation operator in the x direction, followed by the translation operator in the y direction. Now let us ask a seemingly pointless question: suppose we translate first in the y direction and then in the x direction. We could travel from the corner ( x,y ) to the other corner ( x + δx, y + δy ) along the edges of the rectangle in two different ways. You say, we would get the same answer of course. Indeed, the difference between the two ways of getting f ( x + δx, y + δy ) is equal to...
View Full Document

## This note was uploaded on 07/15/2008 for the course PHYS 105 taught by Professor Martinis during the Spring '08 term at UCSB.

### Page1 / 5

FieldEqIntro - Physics 131 UCSB Spring 2008 A Zee...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online