I A sufficient condition for integrability of equation ##\nabla g=0##

[Jianbing_Shao]

My citation in post #22 explicitly states that the equation:$$\partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$$is nothing more than the definition of $\omega$ in terms of the arbitrary chosen fields $e,\Gamma$. (Equivalently, it can serve to define $\Gamma$ in terms of arbitrary $e,\omega$.)

In your definition, the fields $e,\Gamma$ is chosen independently. But if we demand that the connection $\Gamma$ to be the Levi-Civita connection compatible with the metric field defined using tetrad field $e$. Then what is the property of spin connection $\omega$.

 

[PeterDonis]

I really think there is contradiction here

You missed my point. My statement in post #29, which is a completely uncontroversial fact, contradicts your claim that I quoted. That means your claim is wrong. It doesn't mean there's a contradiction in the literature. There isn't.

 

[renormalize]

But if we demand that the connection $\Gamma$ to be the Levi-Civita connection compatible with the metric field defined using tetrad field $e$. Then what is the property of spin connection $\omega$.

Just look at Wikiedia: https://en.wikipedia.org/wiki/Spin_connection!

So for the Levi-Civita connection $\left\{ _{\mu\nu}^{\alpha}\right\}$, the spin-connection $\omega$ is expressed uniquely and unambiguously in terms of $e$ and its first derivatives. For more general connections $\Gamma_{\mu\nu}^{\alpha}\,$, $\omega$ will involve the torsion $T$ and nonmetricity $Q$ tensors as well.

 

[Jianbing_Shao]

You missed my point. My statement in post #29, which is a completely uncontroversial fact, contradicts your claim that I quoted. That means your claim is wrong. It doesn't mean there's a contradiction in the literature. There isn't.

Then return to the question we have discussed. From a non-zero curvature Levi-Civita connection and metric compatible equation, then how to get a metric field. and why the path dependent property disspeared?
A right theory can answer any question we ask.

 

[renormalize]

From a non-zero curvature Levi-Civita connection and metric compatible equation, then how to get a metric field. and why the path dependent property disspeared?

You avoided my request above for a reference, so I request again: please supply a credible published reference that explicitly discusses the possibility, existence and consistency of a path-dependent metric/tetrad in 4D spacetime. Otherwise you waste everyone's time trying to debate an ill-conceived personal theory.

 

[Jianbing_Shao]

You missed my point. My statement in post #29, which is a completely uncontroversial fact, contradicts your claim that I quoted. That means your claim is wrong. It doesn't mean there's a contradiction in the literature. There isn't.

Another problem ,Do you think a non trivial metric field can also be compatible with a zero curvature connection. they call it teleparallel connection. If it is true. then why we must believe we can use it to tell if a metric space is curved.

 

[Jianbing_Shao]

Just look at Wikiedia: https://en.wikipedia.org/wiki/Spin_connection!
View attachment 365158
So for the Levi-Civita connection $\left\{ _{\mu\nu}^{\alpha}\right\}$, the spin-connection $\omega$ is expressed uniquely and unambiguously in terms of $e$ and its first derivatives. For more general connections $\Gamma_{\mu\nu}^{\alpha}\,$, $\omega$ will involve the torsion $T$ and nonmetricity $Q$ tensors as well.

Simply from a tetrad field , then we can calculate the difference between teleparallel connection and Levi-Civita connection:
If we define:
$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a} + c_{abc} + c_{acb} - c_{bca})$
and
$g^{ab} = {e^a}_I {e^b}_J \eta^{IJ}$,

${c_{ab}}^c e_c = [e_a, e_b]$
We can get:
$\Gamma_{abc} = \frac{1}{2} ( {{e_a}^I}_{,c} e_{bI} + {{e_b}^I}_{,c} e_{aI} + {{e_a}^I}_{,b} e_{cI} + {{e_c}^I}_{,b} e_{aI} - {{e_b}^I}_{,a} e_{cI} - {{e_c}^I}_{,a} e_{bI} + {{e_b}^I}_{,a} e_{cI} - {{e_a}^I}_{,b} e_{cI} + {{e_c}^I}_{,a} e_{bI} - {{e_a}^I}_{,c} e_{bI} - {{e_c}^I}_{,b} e_{aI} + {{e_b}^I}_{,c} e_{aI} )$
At last we can find that:
$\Gamma_{abc} = e_{aI} {{e_b}^I}_{,c}$
So the difference is very clear:
$c_{abc} + c_{acb} - c_{bca})$.
Then is the tern equal the spin connection? and if the tetrad $e_a$ field is a coordinate basis, then spin connection is zero?

 

[weirdoguy]

A right theory can answer any question we ask.

Can you answer questions you've been asked?

 

[PeterDonis]

From a non-zero curvature Levi-Civita connection and metric compatible equation, then how to get a metric field

Um, solve the differential equation? If you know the connection is the Levi-Civita connection for some metric, then you have a differential equation for the metric in terms of the connection.

 

[PeterDonis]

Do you think a non trivial metric field can also be compatible with a zero curvature connection. they call it teleparallel connection.

Mathematically, sure. But math isn't the same as physics.

If it is true. then why we must believe we can use it to tell if a metric space is curved.

In physics, we pick a specific connection for physical reasons, regardless of the fact that it's only one of many connections we could use mathematically. In General Relativity, we use the Levi-Civita connection for the metric, and the curvature of that connection, to tell whether spacetime is physically curved, and we derive the physical properties we associate with gravity from the curvature. The fact that there are other connections we could use is irrelevant, because we don't use them.

In teleparallel gravity, which is a different theory from GR, the properties we associate with gravity are derived from the torsion, not the curvature. So the fact that the connection used in teleparallel gravity has zero curvature doesn't matter, because we aren't deriving the physical properties of gravity from the curvature. We're deriving them from the torsion. As @ergospherical told you way back in post #2 of this thread.

 

[Jianbing_Shao]

Um, solve the differential equation?

Yes, I think we should treat the metric compatible equation as differential equations. To do so we at least can understand two problems:
1, Why a connection can be compatible with many metric fields, Because the different initial conditions we choose.
2, Why a metric field can be compatible with many connections, Because the inner-symmetric structure of the metric field. From the symmetric group of metric we can know what is the difference between the different connections.

If you know the connection is the Levi-Civita connection for some metric, then you have a differential equation for the metric in terms of the connection.

If we know how to use the connection to describe the mtric field. then we can know the physical meaning of Levi-Civita connection and metric field better.

 

[PeterDonis]

Why a connection can be compatible with many metric fields, Because the different initial conditions we choose.

No, because "compatible" isn't just one thing. "Compatible" if the connection is the Levi-Civita connection gives you one equation that relates the metric and the connection. But any other connection compatible with the same metric will have a different equation that relates the metric and the connection. It has nothing to do with different initial conditions; it has to do with different equations relating the metric and the connection.

Why a metric field can be compatible with many connections, Because the inner-symmetric structure of the metric field.

No, because, as above, "compatible" isn't just one thing. The answer here is the same as the answer above.

From the symmetric group of metric we can know what is the difference between the different connections.

How? Do you have a reference?

 

[Jianbing_Shao]

No, because "compatible" isn't just one thing. "Compatible" if the connection is the Levi-Civita connection gives you one equation that relates the metric and the connection. But any other connection compatible with the same metric will have a different equation that relates the metric and the connection. It has nothing to do with different initial conditions; it has to do with different equations relating the metric and the connection.

Because compatible is defined with metric compatible equation, and this equation is different with the normal differential equation in $IR^n$. But we can analyze the equation in such a way. At first we can write the metric compatible equation in the combination of two equations:
$\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$ or $e_a^J\partial_J {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$.
and
$g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$.

Then I am confused by such a problem: Does the operator $\partial_a$ associate with a coordinate system. or at least is defined with a non-coordinate basis field. What is your opinion on this problem?

 

[PeterDonis]

compatible is defined with metric compatible equation

Which "metric compatible equation:"? There isn't just one. There are as many different metric compatible equations as there are connections. But you don't even seem to recognize this.

Does the operator $\partial_a$ associate with a coordinate system

Of course. That's what it means. It's a partial derivative with respect to the coordinates.

at least is defined with a non-coordinate basis field.

What "non-coordinate basis field"? There isn't one as far as the operator $\partial_a$ is concerned.

What is your opinion on this problem?

My opinion is that there is no problem at all, just your continued misunderstandings.

 

[Jianbing_Shao]

Which "metric compatible equation:"? There isn't just one. There are as many different metric compatible equations as there are connections. But you don't even seem to recognize this.

Of course. That's what it means. It's a partial derivative with respect to the coordinates.

So $e_a^J$ should be a tetrad field. then from the equation:
$\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$
If $e_a^J$ is a tetrad field, then the connection can be written as:
${\Gamma^c}_{ab}={e^c}_{I}\partial_a {e_b}^{I}$
It is a teleparallel connection, and the curvature of the connection is zero. is it right?

 

[Jianbing_Shao]

What "non-coordinate basis field"? There isn't one as far as the operator $\partial_a$ is concerned.

If the basis field $e_a={e_a}^I e_I$ is not a coordinate basis, then it means that:
${c_{ab}}^c e_c = [e_a, e_b]\neq 0$
Using definition of metric:
$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$
If we define the connection:
$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a} + c_{abc} + c_{acb} - c_{bca})$
Then we can get:
$e_{aI} {{e_b}^I}_{,c} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a} + c_{abc} + c_{acb} - c_{bca})$
So if the basis field is a coordinate basis, then:
$c_{abc} + c_{acb} - c_{bca}=0$.
So the Levi-Civita connection is a teleparallel connection, and the curvature is zero.
But if the basis field is a non-coordinate basis, then the Levi-Civita connection is only symmetric part of a teleparallel connection, Then if only given such a Levi-Civita connection, can we get the original basis field?

 

[PeterDonis]

the Levi-Civita connection is a teleparallel connection

No, it isn't. They're two different things.

and the curvature is zero

No, it isn't. The Levi-Civita connection has nonzero curvature in any metric that's not the flat Minkowski metric. Any GR textbook will tell you that.

I don't understand what you think you're trying to do, but since you appear to be working from false premises, whatever you're trying to do doesn't seem like it's going to work out well.

 

[Jianbing_Shao]

No, it isn't. They're two different things.

But does it means in all cases they can not be equal? Is there anything wrong in my calculations?

And you tell me the partial derivative in metric compatible equation is 'a partial derivative with respect to the coordinates.'. then does it means tetrad $e_a^J$ should be a tetrad field?
If this is true, of cours you can get a teleparallel connection. is there anything wrong here?

Just like we can not get a non-zero curl vector field purely from a scalar field. Logically we also can not get a non-zero curvature connection from a metric field, because change of metric field is path independent. If we can ,then where the infomation about the difference of different paths comes from?

since you appear to be working from false premises, whatever you're trying to do doesn't seem like it's going to work out well.

The premises is just the theorem I gave at first, Then someone told me that is just a teleparallel connection. If you think this is wrong. then perhaps I really was wrong.

 

[Jianbing_Shao]

No, it isn't. The Levi-Civita connection has nonzero curvature in any metric that's not the flat Minkowski metric. Any GR textbook will tell you that.

Is there any book tell us that a metric field can also be compatible with zero-curvature teleparallel connection? How to get the teleparallel connection from a metric?

I only believe in calculations, if you point out the falses in my calculations, I'll be gald to accept it.
In fact all the calculations I gave are not invented by me, I copied them from some papers.
Of course if you believe the metric compatible equation only can be delt with in ways mentioned in textbook. Perhaps I really don't think so.

 

[Jianbing_Shao]

No, it isn't. The Levi-Civita connection has nonzero curvature in any metric that's not the flat Minkowski metric. Any GR textbook will tell you that.

If you find that in my calculation I said if the basis field is a coordinate basis field, then it just means we only apply a coordinate transformation on metric space with $\eta_{IJ}$ , then isn't it natural that the cuavature of the Levi-Civita connection is zero?

 

[PeterDonis]

does it means in all cases they can not be equal?

They're not equal in any spacetime other than the flat Minkowski metric, since the whole point of the teleparallel connection is to put the effects of gravity into the torsion instead of the curvature.

if you point out the falses in my calculations, I'll be gald to accept it.

I can't even understand your calculations. The reason I know they're wrong is that they're giving you obviously wrong answers.

isn't it natural that the cuavature of the Levi-Civita connection is zero?

Not if the spacetime geometry is anything other than the flat Minkowski metric. If your math tells you that this is "natural", then your math is wrong. I can't understand what math you think you're doing, so I can't tell you exactly how your math is wrong, but since it's leading you to make this wrong claim, repeatedly, it's wrong somewhere.

 

[PeterDonis]

Is there any book tell us that a metric field can also be compatible with zero-curvature teleparallel connection?

You were given a link to a Wikipedia article that shows this way back in post #2 of this thread.

 

[Jianbing_Shao]

I can't even understand your calculations. The reason I know they're wrong is that they're giving you obviously wrong answers.

Are you serious to say so? If my answers is obviously wrong, then it is very easy to point out where is my wrong. In fact the mathematical technics I used is very easy.

ergospherical in post #10 derived the compatibility between teleparallel connection and a metric field.

1. The teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ is flat in the sense that it has vanishing Riemann, $R= 0$.
2. Metrics of the form $g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$ are compatible with the teleparallel connection. This is what follows from the definition of the teleparallel connection, $\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$, since
\begin{align*}
\partial_c g_{ab} &= \eta_{IJ}\left[ (\partial_c {e_a}^{I} ) {e_b}^{J} + {e_a}^{I} (\partial_c {e_b}^{J})\right] \\
&= \eta_{IJ} \left[ {\Gamma}^d_{ca} {e_d}^I {e_b}^J + {\Gamma}^d_{cb} {e_a}^I {e_d}^J\right] \\
&= {\Gamma^d}_{ca} g_{db}+ {\Gamma^d}_{cb} g_{ad} \\
\end{align*}
which is the metric compatibility equation $\nabla_c g_{ab} = 0$.

Can you find any problems in his calculation?

 

[PeterDonis]

If my answers is obviously wrong, then it is very easy to point out where is my wrong.

Not if your reasoning doesn't make sense to me. Your answer is obviously wrong because your answer is that the Levi-Civita connection has zero curvature, and that the Levi-Civita connection is the same as the teleparallel connection, which is wrong for any spacetime that isn't flat Minkowski spacetime. I don't have to know exactly where your math is wrong to know that those answers are wrong; those answers contradict everything in the literature about those two connections.

 

[PeterDonis]

ergospherical in post #10 derived the compatibility between teleparallel connection and a metric field.

@ergospherical is not making the two wrong claims you are making. He's not claiming that the Levi-Civita connection has zero curvature, and he's not claiming that the Levi-Civita connection is the same as the teleparallel connection. Nor does his calculation in post #10 show that either of those things are true (which is good, since they're not). So his calculation in post #10 is irrelevant to the wrong claims you are making.

 

[Jianbing_Shao]

@ergospherical is not making the two wrong claims you are making. He's not claiming that the Levi-Civita connection has zero curvature, and he's not claiming that the Levi-Civita connection is the same as the teleparallel connection. Nor does his calculation in post #10 show that either of those things are true (which is good, since they're not). So his calculation in post #10 is irrelevant to the wrong claims you are making.

Don't hurry,
if we start from the definition of Levi-Civita connection:
$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a})$
Then we can use the definition of metric field
$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$
and we demand that $e_a$ is a coordinate basis field;
$[e_a , e_b]=0$
Then you can find that:
$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a})= e_{aI} {{e_b}^I}_{,c}$

The calculation is just a reverse of ergospherical's.
You can check if I was wrong. But if my calculation is right. then how to explain the result?

 

[PeterDonis]

The calculation is just a reverse of ergospherical's.

Then it can't possibly support the two wrong claims you're making. I never said @ergospherical's calculation was wrong. I only said it doesn't support what you're claiming.

 

[PeterDonis]

and we demand that $e_a$ is a coordinate basis field

@ergospherical's calculation in post #10 doesn't make that assumption. Indeed, I don't think that assumption is even consistent with his calculation. So your apparent belief that you are just reversing his calculation appears to be wrong.

 

[Jianbing_Shao]

Then it can't possibly support the two wrong claims you're making. I never said @ergospherical's calculation was wrong. I only said it doesn't support what you're claiming.

So how to explain my result? if you think my explaination is wrong. then tell me the right one to explain the result I gave.

 

[PeterDonis]

So how to explain my result? if you think my explaination is wrong. then tell me the right one to explain the result I gave.

So far I haven't seen you give any valid "explanation" of anything. I just see you continuing to show math that leads you to results that are obviously wrong, and refusing to acknowledge that fact. Math that gives obviously wrong results can't be a valid explanation of anything.

If you're looking for a suggestion about one step in your latest "explanation" that looks wrong, see my post #58.