A Visualizing Arbitrary Coordinate System - Example Needed

BLevine1985
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Hi I'm wondering if someone can illustrate with an example what I bracketed in blue? I'm having a hard time visualizing how it is that the accelerations of the components are NOT necessarily equal to the components of the acceleration...Much appreciated!
relative acceleration of geodesic in 3+ dimensions.png
 
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In plane polars, the position vector is r\mathbf{e}_r(\theta) = r(\cos \theta, \sin \theta). The accelerations of the components are therefore \ddot r and zero. However, the acceleration is <br /> \ddot{\mathbf{r}} = \ddot r\mathbf{e}_r + 2\dot r \dot \theta \mathbf{e}_\theta + r(-\dot \theta^2 \mathbf{e}_r + \ddot \theta \mathbf{e}_\theta) = (\ddot r - r\dot \theta^2)\mathbf{e}_r + (2\dot r \dot \theta + r\ddot \theta)\mathbf{e}_\theta.
 
How did you get the acceleration of the components as r-double-dot and zero?

I understand how you got the general acceleration of r-double dot but not the first part. Sorry it's been like 10 years since I took classical mechanics...
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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