# Recent content by epovo

1. ### I Recovering Newton's energy conservation law for an Earth's lab

Thank you all. I think my problem is solved now.
2. ### I Recovering Newton's energy conservation law for an Earth's lab

ok, I am quoting from Schutz here: "For instance, suppose we have a stationary gravitational field. Then a coordinate field can be found in which the metric components are time independent, and in that system ##p^0## is conserved [...]. The system in which the metric components are stationary is...
3. ### I Recovering Newton's energy conservation law for an Earth's lab

My problem is the way the thing is presented. The lab on Earth is given as an example of $$m \frac {dp_\beta} {d\tau} = \frac 1 2 g_{\nu\alpha,\beta } p^\nu p^\alpha$$ Which in turn was obtained from ## p^\alpha p^\beta_{;\alpha}=0 ##, i.e. a geodesic. Maybe the result above is valid for any...
4. ### I Recovering Newton's energy conservation law for an Earth's lab

I'm looking at Schutz 7.4 where first he obtains the following expression for a geodesic: $$m \frac {dp_\beta} {d\tau} = \frac 1 2 g_{\nu\alpha,\beta } p^\nu p^\alpha$$ This means that if all the components of ##g_{\nu\alpha }## are constant for a given ##\beta##, then ##p_\beta## is also...
5. ### I Riemann curvature tensor on a unit radius 2D sphere

I see. If I had used unit basis instead of the usual coordinate basis I would not get this dependence on theta. Thanks!
6. ### I Riemann curvature tensor on a unit radius 2D sphere

I have worked out (and then verified against some sources) that ##R^\theta_{\phi\theta\phi} = sin^2(\theta)##. The rest of the components are either zero or the same as ##R^\theta_{\phi\theta\phi} ## some with the sign flipped. I was surprised at this, because it implies that the curvature...
7. ### I Flux of i-momentum

It does not mean that. The pressure is acted upon the (imaginary) pressure detector of unit magnitude. The particles do not have to collide. Pressure in a gas does not come from particles colliding with one another. Statistically, there will be collisions in a real situation, but that is not the...
8. ### I Flux of i-momentum

Why? Forget the cuboid. Just imagine a pressure sensor (which is placed perpendicular to the x axis at any point inside the gas) being hit by both particles. I don't think that the idealization works at the edges of the volume of gas. In fact, a gas of particles with random velocities would...
9. ### I Flux of i-momentum

I agree. So, going back to my original particle pair, a sensor would be hit by both particles, each contributing to the pressure in equal measure.
10. ### I Flux of i-momentum

It does seem however that you can have pressure with no collisions of particles.
11. ### I Flux of i-momentum

I found this exercise from Schutz: 4.22 Many physical systems may be idealized as collections of *noncolliding particles* (for example, black-body radiation, rarified plasmas, galaxies and globular clusters). By assuming that such a system has a random distribution of velocities at every point...
12. ### I Flux of i-momentum

If the net momentum is zero for my two particles, bouncing or not, how is there a momentum flux to account for the non-zero component of the tensor? Because it's obviously not zero. ##T^{11}## is by definition x-momentum flux. According to Pencilvester it should be zero.
13. ### I Flux of i-momentum

But if the two particles in my example collide and bounce back, isn't that situation indistinguishable from the one where each particle goes on its own way? Both represent the exact same transfer of momentum between the two boxes you alluded to.
14. ### I Flux of i-momentum

If the momentum cancels out, where does the pressure come from, if not from adding up the contribution from the particles traversing the x=constant surface?
15. ### I Flux of i-momentum

I am probably wrong (that's why I am asking!). But I thought ##T^{11} = T^{22} = T^{33} = p## where p is the pressure in a perfect fluid in the MCRF. The ##2pv/A## would correspond to ##T^{11}##