High School Proof of Specific Covariant Divergence

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SUMMARY

The discussion centers on the proof of specific covariant divergence involving vector fields, specifically the expression ##(A_\mu A_\nu^{,\nu} - A_\mu^{,\nu} A_\nu)^\mu##. It is established that if the comma denotes an ordinary derivative, the equation simplifies to ##A_\mu^{,\mu}A_\nu^{,\nu} - A_\mu^{,\nu}A_\nu^{,\mu}##. However, when considering covariant derivatives, the product rule's applicability is questioned, leading to the conclusion that the relation does not hold due to the non-commutativity of covariant derivatives in non-flat spaces.

PREREQUISITES
  • Understanding of vector fields in differential geometry
  • Knowledge of ordinary and covariant derivatives
  • Familiarity with the product rule in calculus
  • Concept of flat vs. non-flat spaces in Riemannian geometry
NEXT STEPS
  • Study the properties of covariant derivatives in Riemannian geometry
  • Learn about the implications of the product rule for covariant derivatives
  • Explore examples of vector fields in curved spaces
  • Investigate the significance of index notation in tensor calculus
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Mathematicians, physicists, and students studying differential geometry, particularly those focusing on the behavior of vector fields and covariant derivatives in curved spaces.

gerald V
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If the comma means ordinary derivative, then ##(A_\mu A_\nu^{,\nu} - A_\mu^{,\nu} A_\nu)^\mu = A_\mu^{,\mu}A_\nu^{,\nu} + A_\mu A_\nu^{,\nu,\mu} - A_\mu^{,\nu,\mu} A_\nu - A_\mu^{,\nu}A_\nu^{,\mu} = A_\mu^{,\mu}A_\nu^{,\nu} - A_\mu^{,\nu}A_\nu^{,\mu} ##, where ##A## is some vector field. Does the same hold if the comma means covariant derivative (then usually denoted as semicolon)?
Please forgive me for this simple question. Due to my considerations the answer is yes, but I don't trust in this completely.
 
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Yes, because the product rule holds, and you've nowhere interchanged the order of differentiation, which holds for partial but not for covariant derivatives (if the space is not flat).
 
Thank you very much. But I did change the order of differentiation in the third term to make it cancel against the second. First, I just renamed the indices ##A_\mu^{,\nu,\mu}A_\nu \equiv A_\nu^{,\mu,\nu}A_\mu##, but then I replaced ##A_\nu^{,\mu,\nu} \rightarrow A_\nu^{,\nu,\mu}##. So I now assume the relation does not hold if the derivatives are covariant.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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