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### Can you get $g_{\mu\nu}$ from $R^\lambda_{\alpha\beta\gamma}$?

Although you can definitely express the Riemann tensor in first and second derivatives of the metric tensor, the metric tensor is not uniquely determined by it. The Riemann tensor is given by the ...

### Can you get $g_{\mu\nu}$ from $R^\lambda_{\alpha\beta\gamma}$?

No, this is not possible. Finding the metric tensor requires finding a solution to the EFEs. It's important to understand that the Riemann tensor tells us how the metric tensor varies across some ...
Accepted

### Why was $U=\frac{1}{2}QV$ not listed in my book as the potential energy of a capacitor?

It's in your book somewhere for sure. Just maybe not in the section about capacitors because it's much more general than that. In Griffith's "Introduction to Electrodynamics" it's equation 2....

### Difference and meaning of index the derivative operator

You can eventually (if you need to) learn a more rigorous treatment later, so let me instead provide a cookbook approach: An object with an open index means that its value changes when the observer ...
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### Why can the dot product of two vectors be expressed as a differential?

I would imagine that the simplest way to show this is to note that the position vector $\mathbf x$ can be expressed in either basis: $$x'^j \hat e_j' = \mathbf x = x^i \hat e_i$$ A given set of ...

### Can you get $g_{\mu\nu}$ from $R^\lambda_{\alpha\beta\gamma}$?

Although, the question posed by OP is an overdetermined problem, it is interesting to note that the number of independent components of Weyl Curvature (only the trace free part of $R_{abcd}$) in $d$ ...
1 vote

### Can you get $g_{\mu\nu}$ from $R^\lambda_{\alpha\beta\gamma}$?

A very simple counterexample: let $g_{\mu\nu}$ be a flat metric (i.e. $R\equiv 0$). Then for $\alpha\in\mathbb R_{>0}$, if $\alpha\ne1$, $\alpha g_{\mu\nu}$ is a different metric, but still flat.
1 vote

### Why can the dot product of two vectors be expressed as a differential?

First observe, that for any matrix, we can pick up its $i,j$ entry by applying it first to a column vector that is zero apart from $1$ at its $j$'th position and then dotting it into a similar vector ...
1 vote

Here is a brief summary: In (relativistic) physics, it is standard to adorn a local coordinate $x^{\mu}$ with a superindex. We define a shorthand notation for the partial derivative $\partial_{\mu}:=... 1 vote ### General Relativity manipulating tensors, tensor indices meaning About the first part, you nearly answered your own question: it is indeed exactly because the definition of tensors arises naturally out of multilinear maps such as$t(e^a,e^b)$that the order of ... 1 vote ### Relation between rank of tensor and electromechanical property of a material (Disclaimer: Applications of material symmetry and point group analysis lie outside my research focus, and so I'm relying on class-gained background knowledge here along with a textbook survey to at ... 1 vote Accepted ### Is there a way to split a Lorentz index into spinor indices? According to your notation (which is different from the standard Gauge Theory Gravity notation, and assuming$\mu$is the tangent flat spacetime index), the$\overline{h}$field is translated into a ... 1 vote ### Does Lie Bracket take basis changes into account? The Lie bracket of two vector fields is a vector field whose components depend on the choice of basis vectors. So if you change coordinate systems and therefore change the basis vectors for every ... 1 vote Accepted ### Difference between$R^{a}_{bcd}$and$R_{abcd}$Riemann tensor types There is no deep intuitive geometrical meaning behind a Riemann tensor with some indices moved up/down. You could say that the two variants are "dual" to each other, loosely speaking. The ... 1 vote ### Equating 2 sides of EFE First, note that for scalar functions, the covariant derivative reduces to the partial derivative. So for scalar functions, it is true that if the covariant derivative is zero at a point, then the ... 1 vote Accepted ### Raising and lowering indices to second order The objects$h_{\mu\nu}^{(1)}$and$h_{\mu\nu}^{(2)}$(and higher order counterparts) are tensors on the background spacetime (i.e. the spacetime you are expanding about). Consequently, their indices ... 1 vote ### Raising and lowering indices to second order The full GR theory is defined by the metric$g_{\mu\nu}\$ and contains a set of the tensors (wrt. general coordinate transformations), such as e.g. the Riemann curvature tensor. Tensors indices are ...

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