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I was wondering if there were any ways of representing the metric tensor with a vector or scalar field and started calculating some potential ways. I recently stumbled across the equation $$\widetilde{g}=\left(\nabla\vec{v}\right)\left(\nabla\vec{v}\right)^T\gets\ Matrix\ Multiplication$$ on Wikipedia’s page for the metric tensor(it used Jacobians instead of vector gradients). I saw the example $$\vec{v}=\left[\begin{matrix}y\cos{x}\\y\sin{x}\\\end{matrix}\right]$$ for polar coordinates. What is the proper term for this kind of vector field which describes the metric tensor and where can I learn more about it? I also graphed this field. It has divergence and isn’t conservative.

-ln(|cos(x)|) I also calculated the field for the Minkowski metric. $$\vec{v}=\left[\begin{matrix}ict\\x\\y\\z\\\end{matrix}\right]$$ Before I came across this method though, I was analyzing the viability of the equation $$\widetilde{g}=\nabla\vec{v}+\left(\nabla\vec{v}\right)^T $$ I found that if it could represent the metric tensor, it would also need to have divergence and not necessarily be conservative. I haven’t yet disproved the viability of this equation and I was wondering if it could also work for all possible metric tensor configurations(smoothly varying nxn real symmetric matrix fields). If this equation is viable, it would be rather similar to the formula for the Faraday tensor. $$\widetilde{F}=\nabla\vec{A}-\left(\nabla\vec{A}\right)^T $$ Any information and help are appreciated, thank you for your time.

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  • $\begingroup$ Be careful with notation, $(\nabla\vec{v})(\nabla\vec{v})^T$ looks like an tensor product to some people (like me!), i.e. a rank-4 tensor, so perhaps consider writing it as a dot product (i.e. a contraction) - I just deleted a comment where I said it was a rank-4 tensor before realising you probably meant it as matrix multiplication. $\endgroup$
    – FizzKicks
    Apr 2, 2022 at 23:54
  • $\begingroup$ Why would I describe simple matrix multiplication with a dot product? I don't see how that would actually clarify things to most people. $\endgroup$
    – Laff70
    Apr 3, 2022 at 0:30
  • $\begingroup$ When you are working with tensors, if $\mathbf{A}$ is a rank-$m$ tensor and $\mathbf{B}$ is a rank-$n$ tensor, the quantity $\mathbf{A}\otimes\mathbf{B}$, which is sometimes written as $\mathbf{AB}$, where $\otimes$ is a the tensor product, is not the standard multiplication of matrices, but instead is a rank-$m+n$ tensor. In the language of tensor algebra$(\mathbf{AB})_{ij}\neq A_{ik}B_{kj}$; in fact the quantity $(\mathbf{AB})_{ij}$ will only make sense if $\mathbf{A}$ and $\mathbf{B}$ are rank-1 tensors, i.e. vectors. $\endgroup$
    – FizzKicks
    Apr 3, 2022 at 12:14
  • $\begingroup$ I would strongly suggest reading an introductory book to general relativity, it will clear up much of your confusion. $\endgroup$
    – FizzKicks
    Apr 3, 2022 at 12:26
  • $\begingroup$ I edited it to make it clear that matrix multiplication is being used. $\endgroup$
    – Laff70
    Apr 3, 2022 at 17:28

1 Answer 1

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Found the answer to my question as well as some new stuff. For starters, whilst $\widetilde{g}=\left(\nabla\vec{v}\right)\left(\nabla\vec{v}\right)^T$ is a correct equation, it isn’t what I’m looking for. The correct usage of this would be to describe the surface of a sphere in 3D with the coordinates u and v. The vector gradient works properly with this, and the resulting metric tensor is 2D and has the curvature of the sphere. You can see what’s actually happening here. We take a surface and merely change the coordinate system. But this cannot affect the surface’s curvature though. By default, we were remapping a flat 4D space to 4D space. So, this method cannot create a curved 4D metric with a 4D vector field. We could create any metric we wanted using a higher dimensional vector field due to the Nash embedding theorem, where the dimension required for an nD metric tensor is at most $\frac{n\left(3n+11\right)}{2}$, so 46D for a 4D metric tensor. Obviously, this is the antithesis of what I wanted. The formula $\widetilde{g}=\nabla\vec{v}+\left(\nabla\vec{v}\right)^T$ has also been ruled out. The reasons why are as follows. The formula is linear, so $$\widetilde{g}=\nabla\vec{v}+\left(\nabla\vec{v}\right)^T=\nabla\left({\vec{v}}_1+{\vec{v}}_2\right)+\left(\nabla\left({\vec{v}}_1+{\vec{v}}_2\right)\right)^T=\nabla{\vec{v}}_1+\nabla{\vec{v}}_2+\left(\nabla{\vec{v}}_1\right)^T+\left(\nabla{\vec{v}}_2\right)^T=\left(\nabla{\vec{v}}_1+\left({\vec{v}}_1\right)^T\right)+\left(\nabla{\vec{v}}_2+\left(\nabla{\vec{v}}_2\right)^T\right)={\widetilde{g}}_1+{\widetilde{g}}_2$$ All metric tensor fields should be possible, and their configurations determine the form of the stress energy tensor. As such, taking this and the linearity, we add various metric tensor fields together to get a metric tensor with only 1 non-zero unique term. We could create 10 of these fields and add them together to create any 4D metric tensor field. Now we need to check if this formula has the capacity to create such a field. We will start, and end, with the following case. $$g_{11}=f\left(\vec{x}\right)=\sum_{m_1,m_2,m_3,m_4=0}^{\infty}{c_{m_1,m_2,m_3,m_4}x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}}$$ $$v_1=\frac{1}{2}\int{f\left(\vec{x}\right)dx_1}$$ $$\left(\frac{\partial v_1}{\partial x_2}\right)=\frac{1}{2}\left(\frac{\partial}{\partial x_2}\int{f\left(\vec{x}\right)dx_1}\right)=\frac{1}{2}\int{\left(\frac{\partial f\left(\vec{x}\right)}{\partial x_2}\right)dx_1}=-\left(\frac{\partial v_2}{\partial x_1}\right)$$ $$v_2=-\frac{1}{2}\iint{\left(\frac{\partial f\left(\vec{x}\right)}{\partial x_2}\right){dx_1}^2}$$ $$0=\left(\frac{\partial v_2}{\partial x_2}\right)=-\frac{1}{2}\iint{\left(\frac{\partial^2f\left(\vec{x}\right)}{\partial x_2^2}\right){dx_1}^2}=-\frac{1}{2}\sum_{m_1,m_2,m_3,n_4=0}^{\infty}{\frac{m_2\left(m_2-1\right)}{\left(m_1+1\right)\left(m_1+2\right)}c_{m_1,m_2,m_3,m_4}x_1^{m_1+2}x_2^{m_2-2}x_3^{m_3}x_4^{m_4}}$$ Obviously, this formula only equals 0 if no values of $m_2$ other than 0 or 1 are permitted. The same logically applies for the other coordinates. As such, $f\left(\vec{x}\right)$ must have the form $$f\left(\vec{x}\right)=f_{000}\left(x_1\right)+x_2\left(f_{100}\left(x_1\right)+x_3\left(f_{110}\left(x_1\right)+x_4f_{111}\left(x_1\right)\right)+x_4f_{101}\left(x_1\right)\right)+x_3\left(f_{010}\left(x_1\right)+x_4f_{011}\left(x_1\right)\right)+x_4f_{001}\left(x_1\right)$$ which cannot describe all metric tensor fields. Therefore, $\widetilde{g}=\nabla\vec{v}+\left(\nabla\vec{v}\right)^T$ doesn’t describe all metric tensor fields. These finding are relevant to all linear formulas. If they cannot produce a field for one unique component, they aren’t general. The search continues! (or ends here)

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    $\begingroup$ In two dimensions, a vector field has only two independent components at each point. However, the metric tensor in two dimensions has three independent components at each point. In n-dimensions, the metric has n(n+1)/2 independent components. Thus, a vector field is not sufficient to describe the general metric tensor. $\endgroup$
    – robphy
    Apr 6, 2022 at 20:52
  • $\begingroup$ @robphy That rules out using only the 0th order derivatives of the vector field. That doesn't rule out higher order derivatives. $\endgroup$
    – Laff70
    Apr 6, 2022 at 21:46

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