4
votes
2answers
120 views

Geometric meaning of parallel transport

The definition of parallel transport of a vector $v^b$ along a curve $C$ with tangent field $\it{t}^a$ is given by Wald's GR as $$t^a \nabla_a v^b = 0$$ Is it correct to think of $\nabla_a v^b$ as ...
2
votes
1answer
87 views

Can these two terms cancel out?

In trying to prove that $$\Gamma_{\mu\nu\lambda} = \eta_{ab}J^a_bJ^b_{\nu\lambda}.$$ The author canceled out while expanding the first equation $$J^a_{\mu\lambda}J^b_\nu$$ with ...
1
vote
2answers
221 views

Differentiation in general relativity

If we have: $$ \frac{d\phi^a}{d\tau}= \frac{\partial \phi^a}{\partial x^\mu} \frac{dx^\mu}{d\tau} \tag{1}$$ Differentiating it, we get: $$ \frac{\partial \phi^a}{\partial ...
3
votes
1answer
85 views

Why do we do partial and not covariant differentiation with $x^{\nu}$?

Why when taking the velocity vector we make $$u^{\nu}=\frac{d}{d\tau}x^{\nu}$$ and not $$u^{\nu}=\frac{\nabla}{d\tau}x^{\nu}$$ where in the last equation I meant the covariant derivative. Why?
2
votes
2answers
101 views

Covariant derivative of a covariant tensor wrt superscript

Is it true that when you take the covariant derivative of a covariant tensor, do you always have to do with a subscript? What if you do it wrt a superscript?Does the first term (with the partial ...
2
votes
1answer
94 views

Are covariant derivatives of Killing vector fields symmetric?

I'm reading the Lecture Notes on General Relativity by Matthias Blau, and in section 9.1 (point 1) he writes: Let $K^\mu$ be a Killing vector field, and ${x^\mu(\tau)}$ be a geodesic. Then the ...
0
votes
1answer
72 views

How to derive the schwarzchild metric?

I'm having trouble differentiating the following when making a change of co-ordinates to determine the Schwarzchild metric. $$r'^{2}=r^{2}C(r)$$ Then taking the total derivative of both sides, the ...
0
votes
1answer
67 views

Covariant derivative of a vanishing tensor component [closed]

Is the covariant derivative of a vanishing tensor component necessarily zero?
1
vote
1answer
84 views

Covariant derivative as a tensor

$$\nabla_{j} v^{i}~=~g^{ik}\nabla_{j}v_{k}.$$ Does this equality involve an intermediate step, where I take the metric inside the derivative, and then use the fact that covariant derivative of the ...
1
vote
2answers
156 views

What is path of light in the accelerating elevator?

Mathematically, (by mathematically I means by equations) what is path of light in the accelerating elevator? What is the difference between an ordinary derivative and covariant derivative (which is ...
0
votes
1answer
287 views

How to find the intrinsic covariant derivative component?

How to find the intrinsic covariant derivative component? In general relativity the elements of the acceleration four-vector are related to the elements of the four-velocity through a covariant ...