By S. Waner

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**Additional resources for Intro to Differential Geometry and General Relativity**

**Example text**

Assuming it persists for a period of time, we can give it spatial coordinates (x1, x2, x3) at every instant of time (x4). Since the first three coordinates are then functions of the fourth, it follows that the particle determines a path in M given by x1 x2 x3 x4 = = = = x1(x4) x2(x4) x3(x4) x4, so that x4 is the parameter. This path is called the world line of the particle. Mathematically, there is no need to use x4 as the parameter, and so we can describe the world line as a path of the form xi = xi(t), where t is some parameter.

Thus, we can think of covariant tangent fields as nothing more than 1forms. Proof Here is the one-to-one correspondence. Let F be the family of 1-forms on M (or U) and let C be the family of covariant vector fields on M (or U). Define ∞: C’F by ∞(Ci)(Vj) = Ck Vk . In the homework, we see that Ck Vk is indeed a scalar by checking the transformation rule: C—k V—k = ClVl. The linearity property of ∞ now follows from the distributive laws of arithmetic. We now define the inverse 30 §: F’C by (§(F))i = F(∂/∂xi).

Its norm-squared is (1 - ∫2), and we want this to be 1, so we replace the vector by “ 1 1-∫2 , 0, 0, - ∫/c 1-∫2 ‘. This is the first column of D. To keep things simple, let us take the next two columns to be the corresponding basis vectors e2, e3. Now we might be tempted to take the forth vector to be e4, but that would not be orthogonal to the above first vector. By symmetry (to get a zero inner product) we are forced to take the last vector to be “- ∫c 1-∫2 , 0, 0, 1 1-∫2 ‘ This gives the transformation matrix as 1 1-∫ 0 0 D= ∫/c - 1-∫ 2 2 0 0 - 1 0 0 1 0 0 ∫c 1-∫ 0 0 1 2 1-∫2 .

### Intro to Differential Geometry and General Relativity by S. Waner

by David

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