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Background

Here's something I was wondering about. Let the line element between $2$ infinitesimally close points be given by:

$$ ds^2 = (a_s)^2 g^{\mu \nu} dx_\mu dx_\nu$$

Where $g^{\mu \nu}$ is the metric, $dx_i$ is an infinitesimal and $a_s$ is an arbitrary number at each point $s$. Due to the factor of $a_s$ this is obviously discontinuous.

Using this (for absolutely convergent $a_s > 0$) the distance between $2$ points is given by:

$$ d(x_1,x_2) = \lim_{\lambda \to 1} \frac{1}{\zeta(\lambda)} \sum_{s=1}^\infty \frac{a_s}{s^{\lambda}} \times \int_{x_1}^{x_2} \sqrt{g^{\mu \nu} dx_\mu dx_\nu}$$

Question

Is this a valid definition of distance? Is there an analog of the covariant derivative with this metric?

More Anonymous
  • 1,183
  • It's not clear to me what your infinitesimal description means; conventionally the line element denotes the map $v\mapsto\sqrt{g(v,v)}$. You can, of course, define the distance function to have the form you describe, in which case the corresponding metric tensor would be $C^2g$ where $C$ is the prefactor in the expression for $d(x_1,x_2)$. – Kajelad Jul 18 '20 at 05:42
  • @kajelad but the usual prefactor is just a coordinate transformation ... This one is discontinuous at every point – More Anonymous Jul 18 '20 at 05:46
  • Is $a_s$ a function on $\mathbb{N}$ or on the underlying space? The second expression suggests it is a fixed sequence, as in the linked question. – Kajelad Jul 18 '20 at 06:50
  • @kajelad it's on the underlying space – More Anonymous Jul 18 '20 at 07:47
  • Then what does the sum over $s$ in the second expression mean? – Kajelad Jul 18 '20 at 09:04
  • @Kajelad The $a_s$ in the second expression are the values of $a_s$ on the points transversed by the path. – More Anonymous Jul 19 '20 at 12:14
  • The path contains uncountably many points. There's no way of labeling the points with elements of $\mathbb{N}$. – Kajelad Jul 19 '20 at 19:42
  • @kajelad so if I have regions of space with different $a_s$ would it then work? – More Anonymous Jul 20 '20 at 11:18
  • Possibly. Geodesic distance is defined using length integrals, so some conditions on $a$ would be needed to ensure that the resulting integrals are well defined. – Kajelad Jul 20 '20 at 12:00

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