Tensor Field Notation: Einstein Gravity Explained

[Noobnoob]

TL;DR

need help for basic calculus

Hi there,
I'm just starting Zee's Einstein Gravity in a Nutshell, and I'm stuck on a seemingly very easy assumption that I can't figure out. On the Tensor Field section (p.53) he develops for vectors x' and x, and tensor R (with all indices being upper indices) : x'=Rx => x=RT x' (because R-1=RT in that case, that's ok) => ∂xk/∂x'h = (RT)kh = Rhk
On page 45 in the Index&Notations section he develops, for vectors u=Mv (M matrix) : ui=Mij vj => dui = Rij dvj (R rotation matrix) => dx'i = Rij dxj
So should not it be ∂xk/∂x'h = Rkh ?
Thanks for enlighting me

 

[PeterDonis]

If you're going to ask a question about tensor notation, which depends critically on correctly putting indexes "up" or "down", you really, really need to use the PF LaTeX feature. You can find help on it here:

https://www.physicsforums.com/help/latexhelp/
As it stands, I can't even tell for sure what indexes in your question are supposed to be up vs. down.

 

[Noobnoob]

Thanks for the tips, and so sorry about the poor notation. I'll try to be more explicit.
In his paragraph he demonstrates that the derivative of a vector field
\begin{equation}
V'^{i}(\vec x') = R^{ij}V^{j}(\vec x)
\end{equation}
transforms like
\begin{equation}
\frac {\partial V'^{i}(\vec x')} {\partial x'^{h}} = R^{hk}R^{ij}\frac {\partial V^{j}(\vec x)} {\partial x^{k}}
\end{equation}
so that it transforms like a tensor.

To get the R^hk tensor he writes (with R element of the simple orthogonal group) :
\begin{equation}
\vec x' = R\vec x => \vec x = R^{-1}\vec x' = R^{T}\vec x' => \frac {\partial x^{k}} {\partial x'^{h}} = (R^{T})^{kh} = R^{hk}
\end{equation}

What puzzles me is that earlier he defines vectors as entities that transform under rotations like (M transformation matrix, R rotation matrix) :
\begin{equation}
\vec u = M\vec v => u^{i} = M^{ij}v^{j} => dx'^{i} = R^{ij}dx^{j}
\end{equation}

so to me it looks like I should rather get
\begin{equation}
\frac {\partial x^{k}} {\partial x'^{h}} = (R^{T})^{hk} = R^{kh}
\end{equation}

I'm definitely missing something since the final demonstration is right...

 

[Pencilvester]

\begin{equation}
\vec u = M\vec v => u^{i} = M^{ij}v^{j} => dx'^{i} = R^{ij}dx^{j}
\end{equation}

so to me it looks like I should rather get
\begin{equation}
\frac {\partial x^{k}} {\partial x'^{h}} = (R^{T})^{hk} = R^{kh}
\end{equation}

$$dx’^i = \frac{\partial x’^i}{\partial x^j} dx^j \neq \frac{\partial x^i}{\partial x’^j} dx^j$$ So your thinking is incorrect.

 

[Noobnoob]

So I'm just trying to figure it out, from :
\begin{equation}
dx’^i = \frac{\partial x’^i}{\partial x^j} dx^j = R^{ij} dx^j => \frac{\partial x’^i}{\partial x^j} = R^{ij} ,
\end{equation}
I can't write :
\begin{equation}
dx^j = \frac{\partial x^j}{\partial x'^i} dx'^i = (R^T)^{ij} = R^{ji} => \frac{\partial x^j}{\partial x'^i} = R^{ji} ,
\end{equation}
that's where I'm wrong ?

 

[Orodruin]

The partial derivatives of the components of a vector field most certainly are not the components of a rank 2 tensor. At least not unless you restrict yourself to coordinate systems where all Christoffel symbols are identically zero (this requires a flat space and affine coordinates).

Also note that $R$ is not a tensor.

 

[Noobnoob]

Yes I assume this is the case at this stage of the book

 

[Noobnoob]

And I still don't get it...

 

[Pencilvester]

And I still don't get it...

Do you still not understand why $\frac {\partial x^{k}} {\partial x'^{h}} \neq R^{kh}$?

 

[Noobnoob]

Yes, I surely don't derive it correctly, but I don't see how I can't get (9) from (8)

 

[Pencilvester]

$$dx’^i = \frac{\partial x’^i}{\partial x^j} dx^j$$ is a foundational calculus theorem, so that plus eq. (4) should be enough to show you that your eq. (5) is generally incorrect since in general, $\frac{\partial x’^i}{\partial x^j} \neq \frac{\partial x^i}{\partial x’^j}$. And in post 5 you said “I can’t write [eq. (9)]”, but in fact, (9) does follow directly from (8) since R represents a rotational transformation.

 

[Noobnoob]

So if in that particular case (9) follows (8), by changing indices it's equivalent to \begin{equation}\frac {\partial x^{k}} {\partial x'^{h}} = R^{kh} ,\end{equation} so I don't see what's wrong
Besides I understand that \begin{equation}\frac{\partial x’^i}{\partial x^j} \neq \frac{\partial x^i}{\partial x’^j}\end{equation} ,
and there is no really use of the right hand side of this inequality in the previous equations. In fact I imagine that if
\begin{equation}\frac{\partial x’^i}{\partial x^j} = R^{ij} then \frac{\partial x^j}{\partial x’^i} = (R^{ij})^{-1} = (R^{T})^{ij} = R^{ji} \end{equation}

If you could detail what happens in eq (3) I may see it more clearly

 

[Pencilvester]

Okay, I see the issue now. Yes, you’re right, Zee is not consistent with the order of indices between those two sections to which you’re referring. I don’t think he intends for the reader to refer directly to the notation from pg. 45 to get context for pg. 53. He probably does not feel the need to be consistent with the order of indices across different sections since he has yet to introduce the distinction between contravariant indices (superscripts) and covariant indices (subscripts). If you keep reading, you’ll soon find that coordinate transformations like $R$ should be written with one superscript and one subscript. The “order of indices” does not really make sense between upper and lower indices.

 

[Noobnoob]

Okey, that's a relief I started going crazy.
Since I'm not familiar with those notations I'm trying to get everything step by step and all was clear till this point.
I'll take a more relaxed approach :)
Thanks !

 

[vanhees71]

So if in that particular case (9) follows (8), by changing indices it's equivalent to \begin{equation}\frac {\partial x^{k}} {\partial x'^{h}} = R^{kh} ,\end{equation} so I don't see what's wrong
Besides I understand that \begin{equation}\frac{\partial x’^i}{\partial x^j} \neq \frac{\partial x^i}{\partial x’^j}\end{equation} ,
and there is no really use of the right hand side of this inequality in the previous equations. In fact I imagine that if
\begin{equation}\frac{\partial x’^i}{\partial x^j} = R^{ij} then \frac{\partial x^j}{\partial x’^i} = (R^{ij})^{-1} = (R^{T})^{ij} = R^{ji} \end{equation}

If you could detail what happens in eq (3) I may see it more clearly

It is utmost (!) important to be very pedantic with placing indices of tensor components. It's important to keep both the horizontal and the vertical position under control.

Concerning the transformation properties, everything can be explained with vector fields, and I assume a flat spacetime and that $x^{\mu}$ ($\mu \in \{0,1,3,\}$) are pseudocartesian components.

Then any object transforming under Lorentz transformations as the $x^{\mu}$ or rather the "infinitesimal displacements" denote contravariant components:
$$\mathrm{d} x^{\prime \mu}=\frac{\partial x^{\prime \mu}}{\partial x^{\nu}} \mathrm{d} x^{\nu}.$$
Here
$$\frac{\partial x^{\prime \mu}}{\partial x^{\nu}}={\Lambda^{\mu}}_{\nu}$$
is a Lorentz-transformation matrix.

Now consider a scalar field $\Phi(x)$, which transforms as
$$\Phi'(x')=\Phi(x).$$
Now let's see how the partial derivatives transform:
$$\frac{\partial}{\partial x^{\mu}} \Phi \rightarrow \frac{\partial}{\partial x^{\prime \mu}} \Phi'(x')= \frac{\partial x^{\nu}}{\partial x^{\prime \mu}} \frac{\partial}{\partial x^{\nu}} \Phi(x),$$
This means the partial derivatives transform with the inverse matrix, and such an object gets lower indices. Objects with lower indices are called covariant objects. Thus we write
$$\frac{\partial}{\partial x^{\mu}}=\partial_{\mu}$$
with a lower index!

The transformation property for the covariant components is thus like
$$\partial_{\mu} ' \Phi'=\frac{\partial x^{\nu}}{\partial x^{\prime \mu}} \partial_{\nu} \Phi = {(\Lambda^{-1})^{\nu}}_{\mu} \partial_{\nu} \Phi.$$