Dust in special relativity - conservation of particle number

AI Thread Summary
The discussion centers on the conservation of particle number in the context of special relativity, specifically addressing the equation $$\nabla_\mu (\rho u^\mu)=0$$ for dust particles. Participants analyze the terms involved, noting that the first two terms in the attempted solution are zero due to the conservation of particle number and the behavior of velocity. The main point of confusion arises around the last term, where the distinction between the covariant derivative and the partial derivative of the four-velocity is debated. It is clarified that while Christoffel symbols can vanish in Minkowski space, they can also be non-zero in more general contexts, influencing the covariant derivative. The conversation concludes with a request for further arguments regarding the general validity of the identity in question.
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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