Dust in special relativity - conservation of particle number

Pentaquark5
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Homework Statement


My textbook states:
Since the number of particles of dust is conserved we also have the conservation equation

$$\nabla_\mu (\rho u^\mu)=0$$

Where ##\rho=nm=N/(\mathrm{d}x \cdot \mathrm{d}y \cdot \mathrm{d}z) m## is the mass per infinitesimal volume and ## (u^\mu) ## is the four velocity of the dust particles.

Homework Equations



$$ \nabla_\mu A^\nu=\partial_\mu A^\nu+\Gamma^\nu_{\;\; \mu \gamma} A^\gamma $$

The Attempt at a Solution


$$\nabla_\mu (\rho u^\mu)= \underbrace{m \partial_\mu n u^\mu}_{=0} + m n \underbrace{\partial_\mu u^\mu}_{=0}+\Gamma^\mu_{\;\;\mu \gamma} mnu^\gamma$$

Where the first underbrace is zero since the divergence of the particle number is zero, and the second underbrace is zero due to the partial derivative of the velocity.

I don't understand why the last term should be zero, however?
 
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Just an ignorant guess: Isn't ## \nabla_\mu u^\mu=0## rather than ## \partial_\mu u^\mu##?
 
DrDu said:
Just an ignorant guess: Isn't ## \nabla_\mu u^\mu=0## rather than ## \partial_\mu u^\mu##?

The Christoffel symbols vanish in Minkowski space, so this would hold for flat spacetime. Unfortunately, I need the more general form where the Christoffel symbols are non-zero.
Thus, I do not believe the covariant derivative of the four velocity is generally zero, no?
 
Even in flat spacetime, you can have non-vanishing Christoffel symbols.
 
DrDu said:
Even in flat spacetime, you can have non-vanishing Christoffel symbols.
Right. My bad.

But do you see any argument as to why the identity above should be generally zero, then?
 
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