Covariant derivative for four velocity

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SUMMARY

The discussion focuses on proving that the covariant derivative of the four-velocity \( U^a \nabla_a U^b = 0 \) in Minkowski space. The four-velocity is defined as \( U^0 = \gamma \) and \( U^{1-3} = \gamma v^{1-3} \). The participant correctly identifies that the Christoffel symbols are zero in Minkowski space, simplifying the expression to \( U^a \partial_a U^b \). However, they express uncertainty regarding the interpretation of this result, particularly in relation to the four-acceleration.

PREREQUISITES
  • Understanding of four-velocity in special relativity
  • Familiarity with covariant derivatives and their notation
  • Knowledge of Minkowski space and its properties
  • Basic grasp of Christoffel symbols and their role in general relativity
NEXT STEPS
  • Study the properties of covariant derivatives in curved and flat spacetime
  • Learn about the implications of the four-acceleration being zero in special relativity
  • Explore the role of Christoffel symbols in different coordinate systems
  • Investigate the relationship between four-velocity and proper time
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Students of physics, particularly those studying general relativity and special relativity, as well as researchers interested in the mathematical foundations of spacetime and motion.

Kyrios
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Homework Statement


Show U^a \nabla_a U^b = 0

Homework Equations


U^a refers to 4-velocity so U^0 =\gamma and U^{1 - 3} = \gamma v^{1 - 3}

The Attempt at a Solution



I get as far as this:

U^a \nabla_a U^b = U^a ( \partial_a U^b + \Gamma^b_{c a} U^c)
And I think that the Christoffel symbol is 0 in minkowski space so that gives:
U^a \partial_a U^b

And I'm not certain what to do from this point
 
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Is this the problem statement exactly as given. At face value it just looks like the 4-acceleration which by no means needs to be zero.
 
Yes this is all that was given, unfortunately, so I'm having trouble explaining why exactly it is zero
 

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