I Covariant divergence of vector; physical meaning with contracted Tuv

  • I
  • Thread starter Thread starter Tertius
  • Start date Start date
Tertius
Messages
57
Reaction score
10
TL;DR Summary
The covariant divergence of a vector has a simplified form. I am discussing this in relation to a contraction of the SEM tensor and its meaning.
I'm studying Carroll's section on covariant derivatives, which shows that the covariant divergence of a vector ##V^\mu## is given by $$\nabla_\mu V^\mu = \partial_\mu V^\mu + \Gamma^\mu_{\mu\lambda}V^\lambda$$. Because ##\Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{g}}\partial_\lambda \sqrt{g}## we can write $$\nabla_\mu V^\mu = \frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g}V^\mu)$$. If we say ##V^\mu = U_\nu T^{\mu\nu}##, then the covariant divergence looks like $$\frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g} U_\nu T^{\mu\nu})$$. The rank-1 tensor ##U_\nu T^{\mu\nu}## should represent the energy and momentum densities in each of the 4 coordinate directions. The covariant divergence of this rank-1 tensor should then be the sum of the changes of the energy and momentum densities along each coordinate. The simplified form $$\nabla_\mu V^\mu = \frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g}V^\mu)$$ looks to be volume independent in that the partial derivative is taken of the volume element multiplied by the vector, and then divided again by ##\sqrt{g}## after the change is computed. From this perspective, it seems like ##\frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g}V^\mu)## should represent changes to the components of ##V^\mu## independent of volumetric changes. Is this a correct interpretation?
For a flat FRW universe of only dust and a timelike observer, it would look like $$\frac{1}{a^3}\partial_t(-\rho a^3)$$
 
Last edited:
Physics news on Phys.org
I realized the problem was quite simple. The covariant derivative is "correcting" for changes to the metric. The changes in the FRW metric look volumetric because ##\sqrt{g}=a^3##.
 
It's, of course, ##\sqrt{-g}## everywhere, since ##g=\mathrm{det} g<0##. Otherwise it's correct.
 
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right. Couls someone point me in the right direction? "What have you tried?" Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason. I thought it would be a bit of a challenge so I made a derivation or...
Back
Top