# Baez's vizualisation of Ricci tensor

1. May 17, 2015

### exponent137

I am reading Baez's article http://arxiv.org/pdf/gr-qc/0103044v5.pdf and I do not understand paragraph before equation (10), page 18.

Equation (9) will be true if any one component holds in all local inertial coordinate systems. This is a bit like the observation that all of Maxwell’s equations are contained in Gauss’s law and ∇B= 0. Of course, this is only true if we know how the fields transform under change of coordinates. Here we assume that the transformation laws are known. Given this, Einstein’s equation is equivalent to the fact that ...

1. If any one local component holds in all inertial systems, does this means that we look only $R_{tt}$, as eq. 10 shows?
2. Why all of Maxwell's equation's are contained in Gauss law and in $\nabla \bf{B}$? Are both rotor equations unnecessary? Is $\nabla \bf{B}$ not a Gauss law?
3 Is equation 11 mentioned as local inertial systems, not as approximation?
4. Are transformation laws mentioned as pages 88 to 94 in http://arxiv.org/pdf/gr-qc/9712019v1.pdf?

2. May 17, 2015

### Staff: Mentor

Not really. Suppose we know that the $R_{tt}$ equation holds in one local inertial frame. If we transform that equation to another local inertial frame, we will obtain an equation in the new (primed) frame that contains other components of the Ricci tensor besides $R_{t' t'}$. Conversely, if we know that the equation for $R_{t' t'}$ holds in the primed frame, and we transform back to the unprimed frame, we will obtain an equation in the unprimed frame that contains other components of the Ricci tensor besides $R_{tt}$. So knowing that the $R_{tt}$ equation holds in every local inertial frame is equivalent to knowing that all of the components of the Einstein Field Equation hold in any chosen local inertial frame.

They're implied by the two divergence equations, in the same manner I described above. The equation $\nabla \cdot \vec{E} = 4 \pi \rho$ in one inertial frame, with charge density on the RHS, when transformed into another inertial frame, produces, in addition to the equation $\nabla' \cdot \vec{E}' = 4 \pi \rho'$ in the new (primed) frame, the equation $\nabla' \times \vec{B}' - \partial \vec{E}' / \partial t' = 4 \pi \vec{J}'$, with current density in the new frame on the RHS. Similarly, $\nabla \cdot \vec{B} = 0$ in the original frame, when transformed into the new frame, gives, in additional to $\nabla' \cdot \vec{B}' = 0$, the equation $\nabla' \times \vec{E}' + \partial \vec{B}' / \partial t' = 0$. Try it! (Note that I'm using units in which $c = 1$.)

Technically, it is, since it has the same form, but the source on the RHS is always zero, so it's not usually referred to that way (the term "Gauss's Law" is usually used for equations where the source on the RHS can be anything).

Yes.

No. They're local Lorentz transformations between two local inertial frames that are both centered on the same event.

Last edited: May 17, 2015
3. May 17, 2015

### exponent137

On page 17, http://arxiv.org/pdf/gr-qc/0103044v5.pdf, is formula (6) and the third formula is similar as (6), but without $u^\gamma$.
Above it, it is written
"with $u$ being a vector of lenght $\epsilon$ in ..."
It seems to me, that $u=1$, not $\epsilon$, otherwise part $-R^j_{\beta j\delta} v^\beta v^\delta$ would not be such as it is.

Do you think that this is written wrongly, or awkwardly?

4. May 17, 2015

### Staff: Mentor

No. Equation (6) is valid in any coordinate chart. The equations below it, which are written in terms of $t$, are in local inertial coordinates in which the center of the ball is at rest, and the vector $u$ is pointing in the $j$ spatial direction; so the index $j$ in the equations serves the same function as the vector $u$ does in equation (6). The vector $u$ is still of length $\epsilon$, because that is the initial radius of the ball. Read the paragraph right after equation (6) carefully.

5. May 18, 2015

### exponent137

According to (6), (http://arxiv.org/pdf/gr-qc/0103044v5.pdf) and three equations below it is evident that $R^\alpha_{\beta\gamma\delta} u^\gamma= R^j_{\beta\ j \delta}$, if $v$'s are not written. But, if $u=\epsilon$, I expect $R^\alpha_{\beta\gamma\delta} u^\gamma= \epsilon R^j_{\beta j \delta}$. Besides, according to figures in page 16, the distance between $p$ and $q$ is $\epsilon u$, and top equation in page 17 includes $\epsilon^2$, so why to introduce another $\epsilon$?

You should respect that my main knowledge about tensors is only from Carroll's book, so my knowledge is more intuitive than exact.

Thanks for explanation above.

Last edited: May 18, 2015
6. May 29, 2015

7. Jun 2, 2015

### victorneto

Hello,
After best thinking, concludes that ε², ε being the geodesic tiny parallelogram side, the right of which is made parallel displacement of vector w, is the area of that parallelogram infinisitesimal so that w'can be compared at the same point where w giving w'-w as a vector. Do ε tend to zero, will make infinisitesimal area be as small as it can. The Riemann tensor can, from this variation in vector w, be clearly defined, as is done in the article, excellent from Baez.