Do These Equations Hold True Beyond Special Relativity?

[olgerm]

are these equstions true(not just in SR)
$g^{\nu m_2}g^{\mu m_1}g^{j_1 m_3} \frac{\partial \Gamma_{m_3 m_2 m_1}}{\partial x^{j_1}}=\frac{\partial{\Gamma}^{j1\nu\mu}}{\partial x^{j_1}}$

$g^{i_2 m_2}g^{i_1 m_1}g^{j_1 m_3}{\Gamma_{m_3 j_1 j_2}}{\Gamma_{m_4 m_2 m_1}}={\Gamma^{j_1}}_{j_1 j_2}\Gamma^{j_2 i_2 i_1 }$
?
$\Gamma$ is christoffel symbol.
$g$ is metric tensor.

contrvariance and covariance of chistoffel symbol indices have same meaning like in ricci tensor${\displaystyle R_{\alpha \beta }=2{\Gamma ^{\rho }}_{\alpha [\beta ,\rho ]}+2{\Gamma ^{\rho }}_{\lambda [\rho }{\Gamma ^{\lambda }}_{\beta ]\alpha }.}={R^{\rho }}_{\alpha \rho \beta }=\partial _{\rho }{\Gamma ^{\rho }}_{\beta \alpha }-\partial _{\beta }{\Gamma ^{\rho }}_{\rho \alpha }+{\Gamma ^{\rho }}_{\rho \lambda }{\Gamma ^{\lambda }}_{\beta \alpha }-{\Gamma ^{\rho }}_{\beta \lambda }{\Gamma ^{\lambda }}_{\rho \alpha }$.

 

[Ibix]

Your passion for reducing expressions to their component parts seems to have deserted you at a point when it would actually have helped you. Look at the definition of the Christoffel symbols $\Gamma^i_{jk}$. They include terms with $\partial_j$ and $\partial_k$, so raising the lower indices implies raising an index on a partial derivative. As discussed on one of your other threads, this only works in flat spacetime (where the connection coefficients are zero anyway).

 

[olgerm]

Look at the definition of the Christoffel symbols $\Gamma^i_{jk}$. They include terms with $\partial_j$ and $\partial_k$, so raising the lower indices implies raising an index on a partial derivative. As discussed on one of your other threads, this only works in flat spacetime (where the connection coefficients are zero anyway).

I used the relation that wipedia gave about Christoffel symbol(that first index of Christoffel symbol tranforms like Christoffel symbol were tensor). Just want to get confirmation wheter all is correct or made some mistakes.

upper and lower indices also appeat in GR equation so there must be some way of transorming these (co- to contravariant) in GR.

 

[Ibix]

Again, look at the definition of the Christoffel symbols. The first index does not appear on a partial derivative, only on the metric. You can raise or lower this index using the metric tensor, and the mixed form of the Christoffel symbols does indeed appear in the mixed Riemann tensor. But I've never seen an "all upper index" Christoffel symbol and as far as I'm aware such a thing doesn't really make sense.

 

[olgerm]

first index does not appear on a partial derivative, only on the metric.

it does $\Gamma _{cab}={\frac {1}{2}}\left({\frac {\partial g_{ca}}{\partial x^{b}}}+{\frac {\partial g_{cb}}{\partial x^{a}}}-{\frac {\partial g_{ab}}{\partial x^{c}}}\right)$

look the general deinition not definition for euclidean space.

 

[Ibix]

it does $\Gamma _{cab}={\frac {1}{2}}\left({\frac {\partial g_{ca}}{\partial x^{b}}}+{\frac {\partial g_{cb}}{\partial x^{a}}}-{\frac {\partial g_{ab}}{\partial x^{c}}}\right)$

look the general deinition not definition for euclidean space.

It doesn't in the mixed version, which I was looking at. So you can manifestly lower an index, but presumably there's some work to be done to show that you can raise it.

I'm bowing out now. I've never seen an "all upper" Christoffel symbol and as far as I know it doesn't make sense. We'll see if anyone else chips in.

 

[olgerm]

Are the equation in my 1. post true or not?

 

[Orodruin]

Are the equation in my 1. post true or not?

It is nonsensical. As you have already been told, we typically do not raise the two last indices of the Christoffel symbols because it makes very little sense to do so. Even if we did, the metric components do not commute with the partial derivative.

 

[olgerm]

contravairant ricci tensor includes these terms.
$R^{\alpha \beta }=R(\alpha,\beta)=\frac{\partial{\color{red}\Gamma}^{j1 \beta \alpha }}{\partial x^{j_1}}-\frac{{\partial{\Gamma^{j_1}}_{j_1}}^{\alpha}}{\partial x_{\beta}}+\sum_{j_2=0}^D(\color{red}{\Gamma^{j_1}}_{j_1 j_2}\Gamma^{j_2 \beta \alpha }\color{black}-{\Gamma ^{j_1 \beta}}_{j_2 }{{\Gamma^{j_2}}_{j_1}}^{\alpha }))$

 

[Pencilvester]

contravairant ricci tensor includes these terms.
$R^{\alpha \beta }=R(\alpha,\beta)=\frac{\partial{\color{red}\Gamma}^{j1 \beta \alpha }}{\partial x^{j_1}} \dots$

Many people have already told you that $\Gamma^{ijk}$ is a nonsensical term in the context of Riemannian and pseudo-Riemannian geometry. If you want to create your own symbols for your own personal use, that’s your prerogative. But don’t be surprised when we can’t tell you anything about the correctness of equations using those symbols, especially when you don’t define them, or if you do, you define them using incorrect presuppositions.

 

[olgerm]

$\Gamma^{ijk}$ is a nonsensical term in the context of Riemannian and pseudo-Riemannian geometry.

thats just what contravariant ricci tensor tensor includes.
from wiki:
$R_{\alpha \beta }=\sum_{j_1=0}^D({R^{j_1}}_{\alpha j_1 \beta })=\sum_{j_1=0}^D(\partial _{j_1 }{\Gamma ^{j_1 }}_{\beta \alpha }-\partial _{\beta }{\Gamma ^{j_1}}_{j_1 \alpha }+\sum_{j_2=0}^D({\Gamma ^{j_1 }}_{j_1 j_2 }{\Gamma ^{j_2}}_{\beta \alpha }-{\Gamma ^{j_1 }}_{\beta j_2 }{\Gamma ^{j_2 }}_{j_1 \alpha }))=\sum_{j_1=0}^D(\frac{\partial{\Gamma ^{j_1}}_{\beta \alpha }}{\partial x^{j_1}}-\sum_{j_2=0}^D(\frac{{\Gamma^{j_1}}_{j_1 \alpha}}{\partial x^{\beta}}+{\Gamma^{j_1}}_{j_1 j_2}{\Gamma ^{j_2}}_{\beta \alpha }-{\Gamma ^{j_1}}_{\beta j_2 }{\Gamma^{j_2}}_{j_1 \alpha }))$

now raising indices on both sides of equation:
$R^{\alpha \beta }=R(\alpha,\beta)=\sum_{j_1=0}^D(\frac{\partial{\Gamma}^{j1 \beta \alpha }}{\partial x^{j_1}}-\frac{{\partial{\Gamma^{j_1}}_{j_1}}^{\alpha}}{\partial x_{\beta}}+\sum_{j_2=0}^D({\Gamma^{j_1}}_{j_1 j_2}\Gamma^{j_2 \beta \alpha }-{\Gamma ^{j_1 \beta}}_{j_2 }{{\Gamma^{j_2}}_{j_1}}^{\alpha }))$

or are you saying that contravariant ricci tensor "doesnt make sense" ether?
anyway the term "doesnt make sense" ambigous.

 

[Orodruin]

now raising indices on both sides of equation:

For the hundredth time: You cannot just go around raising indices inside derivatives, even if you for some reason want to define Christoffel symbols with exclusively upper indices.

 

[Pencilvester]

...now raising indices on both sides of equation:
$R^{\alpha \beta }=R(\alpha,\beta)=\sum_{j_1=0}^D(\frac{\partial{\Gamma}^{j1 \beta \alpha }}{\partial x^{j_1}}-\frac{{\partial{\Gamma^{j_1}}_{j_1}}^{\alpha}}{\partial x_{\beta}}+\sum_{j_2=0}^D({\Gamma^{j_1}}_{j_1 j_2}\Gamma^{j_2 \beta \alpha }-{\Gamma ^{j_1 \beta}}_{j_2 }{{\Gamma^{j_2}}_{j_1}}^{\alpha }))$

or are you saying that contravariant ricci tensor "doesnt make sense" ether?
anyway the term "doesnt make sense" ambigous.

$$\Gamma^\alpha_{\mu \nu} \equiv \langle \mathbf{\omega}^\alpha ~ , ~ \nabla_{\mathbf{e}_\mu} \mathbf{e}_\nu \rangle$$

Or, if you like:
$$\partial_\nu \mathbf{e}_\mu \equiv \Gamma^\lambda_{\nu \mu} \mathbf{e}_\lambda + K_{\nu \mu} \mathbf{n}$$
where indices represent n coordinate directions on an n-dimensional manifold embedded in an (n+1)-dimensional Euclidean space, $\mathbf{e}_\mu$ are the basis vectors for those coordinates, and $\mathbf{n}$ is the normal vector to that manifold.

Can you, in a similar fashion, define $\Gamma^{ijk}$? (by the way, this question is rhetorical)

My Preview button isn't compiling my latex code, so if I made a mistake, I apologize.

 

[olgerm]

You cannot just go around raising indices inside derivatives

can you explain it more?

 

[olgerm]

indices represent n coordinate directions on an n-dimensional manifold embedded in an (n+1)-dimensional Euclidean space, $\mathbf{e}_\mu$ are the basis vectors for those coordinates, and $\mathbf{n}$ is the normal vector to that manifold.

I do not understand your post,but just mention, that you can not embed any maifold into more dimensional euclidean space. consider minkowsky spacetime:
for every point in minkowsky spacetime there exists infinitly different point to whom distance(interval) is 0. that can not be embeded into euclidean space, because i euclidean space for exery point $\vec{X}$ there is only one point to which distance from $\vec{X}$ is 0. It is $\vec{X}$ itself.
also interval may be negative, but distance in (whatever dimensional) euclidean space can not.

 

[Orodruin]

can you explain it more?

I already did:

the metric components do not commute with the partial derivative.

If you are not going to consider what you are being told, I see little meaning in continuing having this conversation in thread after thread.

 

[haushofer]

No, these equations are not right, unless you want to introduce highly confusing notation.

What you want to do, is similar to pouring out some beer, then grab a glass, and calling the resulting mess a glass of beer.

I'd be confused.

 

[PeterDonis]

This thread is going nowhere and is now closed.