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Don't the Pythagorean Theorem and calculus depend upon a right angle? Or should I say rely on...
Riemannian metrics of the form ##adx^2+bdy^2+cdz^2 ## where ##a,b,c## not all ##1## refer to, or depict non-Euclidean geometries, or geometries whereDon't the Pythagorean Theorem and calculus depend upon a right angle? Or should I say rely on...
https://en.wikipedia.org/wiki/Relativistic_Doppler_effectOK, how do you measure 90° angles in GR?
That looks like a generalized inner product using a quadratic form:Riemannian metrics of the form ##adx^2+bdy^2+cdz^2 ## where ##a,b,c## not all ##1## refer to, or depict non-Euclidean geometries, or geometries where
##a^2+b^2=c^2## does not quite hold. But, yes, in the Calculus done in Euclidean space, the generalized Pythagorean theorem does hold. But the Pythagorean theorem does not hold in general when your Calculus is done on a general manifold.
Yes, you can see it that way; this is the Riemannian metric tensor, a quadratic form , often represented as a matrix ##g(X_i, X_j)## where the {## X_i ##} is a basis for the tangent space. The Riemannian metric tensor associated with the standard Euclidean metric is, like you said, the identity. The Riemannian metric is an inner product, defined on tangent vectors at each point in a manifold.That looks like a generalized inner product using a quadratic form:
d×(x,y,z)⋅((a,0,0),(0,b,0),(0,0,c))⋅(x,y,z)^{T}, where (...) indicates a row matrix and ((...),(...)) is a matrix, in this case a diagonal matrix. Is this the right way to think of the metrics you refer to?
I occurred to me after I wrote that post that by dx^{2}, you meant the the square of the differential of x, not multiplication by the scalar, d. Is that correct? And so, as you imply, the vector of differentials, (dx,dy,dz) is a basis for the tangent space at (x,y,z)? Thanks for your help. I only know enough differential geometry to be dangerous, as they say...Yes, you can see it that way; this is the Riemannian metric tensor, a quadratic form , often represented as a matrix ##g(X_i, X_j)## where the {## X_i ##} is a basis for the tangent space. The Riemannian metric tensor associated with the standard Euclidean metric is, like you said, the identity. The Riemannian metric is an inner product, defined on tangent vectors at each point in a manifold.
No problem, feel free to ask, I will do my best, and ask for clarification if needed.I occurred to me after I wrote that post that by dx^{2}, you meant the the square of the differential of x, not multiplication by the scalar, d. Is that correct? And so, as you imply, the vector of differentials, (dx,dy,dz) is a basis for the tangent space at (x,y,z)? Thanks for your help. I only know enough differential geometry to be dangerous, as they say...