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how to find tetrad of this metric
the tetrad given
As Peter says, there are infinitely many possible tetrads for any given metric. A tetrad at a point is just four mutually orthogonal vectors at that point, and you can always rotate or Lorentz boost any set of four to make another.how to find tetrad of this metric
As Peter says, there are infinitely many possible tetrads for any given metric. A tetrad at a point is just four mutually orthogonal vectors at that point, and you can always rotate or Lorentz boost any set of four to make another.
You have a specific tetrad field (i.e., a recipe for writing down a tetrad at any event such that nearby events have similarly oriented tetrads). Somebody must have made a choice on how to go about that, picking one field from the infinite choices, and it's that context that's needed.
Where? What reference are you getting this from?
Which one? We need a link or other reference.Actually I found it in a research paper.
Somebody decided what tetrads they wanted. Then they wrote down an expression for those four tetrad vectors in terms of the coordinate basis vectors. I don't think I can be more specific than that without knowing what the reasoning behind that selection of tetrad was.But Will you please help me out how to find these four tetrads.
Provide a link to the paper and we might be able to help.Actually I found it in a research paper.
This oneWhich one? We need a link or other reference.
Yup this oneJust a link is fine: https://arxiv.org/abs/1710.09880
how to find tetrad of this metric
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the tetrad given is this one
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I m a newly born in General Relativity
please help me out how this tetrad is derived
Your tetrad is just an orthornormal basis of one forms.
In index-free notation, the four one-forms are ##e^0, e^1, e^2, e^3##. Note that here the superscripts are not tensor indices, they are selectors, so that ##e^0## is the first memeber of the set, ##e^1## is the second member of the set, etc. In index-free notation we don't need the subscript.
The components of ##e^0## are written by the author of the paper you reference as ##e^{(0)}_a##, similarly for the other members of the tetread.
Note that dt, ##d\theta##, ##d\phi##, and dr as I write them also one forms. The author of the paper writes the components, so where I write dt he writes ##dt_a##, and similarly for the other one-forms.
The point then is the metric is just
$$-e^0\otimes e^0+ e^1 \otimes e^1 + e^2 \otimes e^2 + e^3 \otimes e^3$$
(I actually only verified the component for dt^2, but I think the others work out as well, but you should double check this).
Deriving this is basically just a matter of factoring the line element with algebra. It may help to re-write the original Kerr metric in the same index free notation that I use. If you go this route, don't forget that the metric must be symmetric, ##dt \otimes d\phi## is not the same as ##d\phi \otimes dt##. So the line element written as (...)dt##d\phi## becomes ##\frac{1}{2}## (...) (dt ##\otimes d\phi + d\phi \otimes## dt)
I humbly request you to show me how you verified it?(I actually only verified the component for dt^2, but I think the others work out as well, but you should double check this).
I'll so a short, symbolic version
e0 and e3 are one forms, as are dt and d##\phi##. I'm using index free notation, you can imagine that everything has a subscript if you prefer abstract index notation. So my e0 is your paper's ##e^{(0)}_a##, my dt is your paper's ##dt_a##
I'm going to omit e1 and e2 from what I write for brevity, they can be handled in an even simpler manner
Then
$$
e0 = A\,dt + B\,d\phi \quad e3 = C\,dt + D\,d\phi
$$
I haven't written out the values of A,B,C,D, I've left them symbolic. You can get them from the paper. But I'll write out the value of A so you can compare.
$$A = \sqrt{\frac{\Delta}{\Sigma}}$$
Then
$$-e0 \otimes e0 + e3 \otimes e3 = -(A\,dt + B\,d\phi ) \otimes (A\,dt + B\,d\phi ) + (C\,dt + D\,d\phi) \otimes (C\,dt + D\,d\phi)
$$
The tensor product ##\otimes## is distributive, though it doesn't commute. Of course, A,B,C,and D are just numeric expressions, so their multiplication does commulte. Carrying this out, and collecting terms, we get..
$$
(C^2-A^2)\, dt \otimes dt + (C\,D - A\,B) ( dt \otimes d\phi + d\phi \otimes dt) + (D^2- B^2) d\phi \otimes d\phi
$$
In matrix notation, this just means that
$$
g_{tt} = C^2 - A^2 \quad g_{t\phi} = g_{\phi t} = C\,D - A\,B \quad g_{\phi\phi} = D^2 - B^2
$$
I've taken the liberty of using the symbolic subscripts for the metric coefficients rather than the numeric, hopefully it's clear what I meant.
And you can see that the resulting matrix is symmetric because ##g_{t\phi} = g_{\phi t}##
e1 and e2 , which I omitted, just give you ## g_{rr} ## and ##g_{\theta\theta}## in a very straightforwards manner, you just have to square the appropriate expressions in the paper.
I know how to solve e0⊗e3 but don't know how g⊗e0⊗e3=0 is solved .g⊗e0⊗e3=0g⊗e0⊗e3=0g \otimes e0 \otimes e3=0
I know how to solve e0⊗e3 but don't know how g⊗e0⊗e3=0 is solved .
will you please tell me how this is solved
##g \otimes e0 \otimes e3## is just index free notation for ##g^{ab}e0_a\,e3_b## in abstract index notation
##g^{ab}## is the inverse metric. The operatior ##\otimes## is the tensor product, not the dot product. I suspect you might be confusing the two.
Got it.You should get a single number - there are no free indices in this expression. It's an inner
Definition.A square matrix A is orthogonally diagonalizableif if there exists an orthogonal matrix Q such that ##Q^TAQ=D## is a diagonal matrix.
Remarks.Since ##Q^T=Q^{−1}## for orthogonal Q, the equality ##Q^TAQ=D## is the same as ##Q^{−1}AQ=D##, so A∼D, so this a special case of diagonaliza-tion: the diagonal entries of D are eigenvalues of A, and the columns of Qare corresponding eigenvectors. The only difference is the additional requirement that Q be orthogonal, which is equivalent to the fact that those eigenvectors – columns of Q – form an orthonormal basis of ##R^n##