Recent content by Jianbing_Shao

Sorry, I have repeated several times, ##e_a## can also be a non-coordinate basis, and ##e_a^I## is arbitrary, no matter how you would like to call it. and no constrain on it.

Sorry, I just don't how to rewrite it, I seems that I have no right to do make some correction.

From the discussion above, I just follow the notation in Carroll's book, He call ##e_a^I## a tetrad, and he didn't say it only can take values in ##SO(1,3)## group, So I think it can be chosen arbitrary. In fact in the calculation there is no need to put such a constrain on it.

Ok, I just suggest if we start from a arbitrary basis ##e_a##, so ##e_a^I## is arbitrary, from the definition of coordinate basis. it just means commutation coefficients ##{c_{ab}}^c## is not always zero. then we calculate the cuvature, it is not strange the curvature involve the term...

I just start from a arbitrary ##e_a^I##, there is no constrain on it, but according to your opnion, ##e_a^I## only can be ##\delta_a^I##, In fact, if you demand ##g_{ab}=\eta_{ab}##, ##e_a^I## also can be arbitrary element of group ##SO(1.3)## , but in my calculation the choose of ##e_a^I## is...

Yes if we define such a basis, in most caese it can not be a coordinate basis if: $${c_{ab}}^c e_c = [e_a, e_b]$$ $$ {c_{ab}}^c {e_c}^K \frac{\partial}{\partial x^K} = {e_a}^I {{e_b}^J}_{,I} \frac{\partial}{\partial x^J} + {e_a}^I {e_b}^J \frac{\partial^2}{\partial x^I \partial x^J}...

Yes ,you are right, I didn't distinguish the difference between the two。 ##\hat e_{(a)}## and ##e^a_\mu## are just some vectors and the components of these vectors.

It is just the definition of a non-coordinate basis. Thanks a lot, I have found this book, my problem have a close relation with exercise I.9.1. and using 'tetrad' ##e_a^I## we will arrive at a strange result. My definition of connection: $$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} -...

From the eqn J.3 in Carroll, we can know that ##{e_I}^a## form an inverse with ##{e^I}_a##, So the two equations in eqn J.3 are equivalent with each other. isn't it? Then what is the difference ?

The notations I used just comes from the Sean Carroll's book 'SPACETIME AND GEOMETRY An introduction to General Relativity'. In the appendix J(p483) in which noncoordinate bases is discussed. In the book above, there he said: The point of havvng a basis is that any vector can be expressed as a...

It is just the components of a vector, the components combine with the corresponding basis then give a vector, we can't call the components of vector vector.

So ##\frac{\partial}{\partial x^I}## is vector, ## {e_a}^I \frac{\partial}{\partial x^I}## the sum of some vectors is also vector so ##e_a## is also vector, but ##e_a^I## is not.

Then a vector can be described with a basis ##V=v^ie_i##, then how can you describe ##e_a^i## with basis ##e_i##?

Vector ##\frac{\partial}{\partial x^I}## itsself is a derivative. From vector ##e_a## we can define derivative ##\partial_a={e_a}^I \frac{\partial}{\partial x^I}##, Then from a tetrad ##{e_a}^I##, How to define derivative corresponding to it?

Conbine the two formula: $$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$ $${c_{ab}}^c e_c = [e_a, e_b]$$ I don't think it imply a tetrad is vector, If I say so it is not right. ##e_a## is vector, ##\frac{\partial}{\partial x^I}## is vector, but ##{e_a}^I## is not.