Widdekind Posted November 23, 2012 Share Posted November 23, 2012 The equations for Fluid Dynamics, on static grids, are well known, e.g. mass continuity: [math]\frac{\partial \rho}{\partial t} + \vec{\nabla} \circ \left( \rho \vec{v} \right) = 0[/math] Those equations must (?) remain the same, on expanding grids, for comoving quantities. For example, the amount of mass within a comoving cell (derivative of comoving density) can only change, by (net) mass flow into / out of the same cell, from surrounding cells, which (net) flow is measured by the divergence. If you measured all distances with an expanding ruler, then you would never notice that your fluids were evolving on an expanding grid. The only "clue" you might get, would be that the fluid would exhibit a seemingly strange sort of dissipation, such that comoving speeds decayed away -- for a steady absolute space speed, the expansion of space [math]\left( a(t) \rightarrow \infty \right)[/math] would translate the same absolute space speed [math]\left( v \right)[/math], to fewer and fewer (larger & larger) comoving distance units [math]\left( \tilde{v} = v / a(t) \rightarrow 0 \right)[/math]. Please ponder, that the scale factor a(t) possesses important physical units, namely "actual meters of distance per comoving meters of distance" (say). Thus, the scale factor is not dimensionless. So, to translate the equations of FD, from comoving quantities (on a comoving grid) to actual quantities, translate the comoving quantities to actual quantities, within the comoving equations: [math]\frac{\partial \tilde{\rho}}{\partial t} + \tilde{\nabla} \circ \left( \tilde{\rho} \tilde{v} \right) = 0[/math] [math]\tilde{\rho} = \rho \times a(t)^3[/math] [math]\tilde{v} = v \div a(t)[/math] [math]\tilde{\nabla} = \nabla \times a(t)[/math] [math]\therefore \frac{\partial ( \rho a^3 )}{\partial t} + a \vec{\nabla} \circ \left( a^3 \frac{v}{a} \right) = 0[/math] [math]\boxed{\frac{\partial \rho}{\partial t} + 3 H \rho + \vec{\nabla} \circ \left( \rho v \right) = 0}[/math] where the last line derives from the spatial invariance of the scale factor (which depends only upon time), and then by dividing the entire previous equation, by a couple of powers of the scale factor. Please pay particular attention, to the role of the scale factor ("actual length / comoving length"), in translating from comoving to actual quantities. Similarly, for momentum, [math]\frac{\partial \tilde{v}}{\partial t} + \left( \tilde{v} \circ \vec{\nabla} \right) \tilde{v} = -\frac{\vec{\nabla} \tilde{P}}{\tilde{\rho}}[/math] For the LHS, we proceed as per previous: [math]LHS = \frac{\partial (v a^{-1})}{\partial t} + \left( (v a^{-1}) \circ (a \nabla) \right) (v a^{-1})[/math] [math]= a^{-1} \times \left( \frac{\partial v}{\partial t} - Hv + \left( v \circ \nabla \right) v \right)[/math] For the RHS, we first notice the units -- dividing out the density from denominator & numerator, the relative pressure amounts to the comoving gradient, of comoving temperature. Thermodynamically, temperature, definable as the square of a sound speed, measures the square of the characteristic (microscopic) space speed of the particles. As alluded to above, if the actual speed is steady, the comoving speed, measured with expanding comoving rulers, seems to decay away: [math]\tilde{T} \approx \frac{<\tilde{u}^2>}{m} = a(t)^{-2} \times \frac{<u^2>}{m} = a^{-2} T[/math] So, [math]RHS = -\frac{a \nabla (a^{-2} P)}{\rho}[/math] [math]= a^{-1} \times \left( -\frac{\nabla P}{\rho} \right)[/math] And so (?) [math]\boxed{\frac{\partial v}{\partial t} - Hv + \left( v \circ \nabla \right) v = -\frac{\nabla P}{\rho}}[/math] i want to question the equations stated on the following webpage: http://ned.ipac.caltech.edu/level5/March03/Bertschinger/Bert2_2.html In particular, please pay particular attention to the units embodied in the scale factor ("actual distance / comoving distance"). So, i perceive, that the correct equations cannot rescale some quantities, and not others -- to me, such equations seem "confused", "muddling" comoving lengths, in some terms, with actual lengths, in other terms. Because the units balanced in the original equations (comoving quantities, in comoving coordinates), so all length scales must all be equally translated, from comoving distances, to actual distances, by the same number of powers of the scale factor. The correct equations cannot (?) mix and match, with some terms involving, and others omitting, the scale factor. Would somebody please explain, if (and if so, then why) my boxed equations are (or are not) correct, as compared to those in the aforecited website? i feel i have made no errors in physical analysis, or mathematical derivation. Link to comment Share on other sites More sharing options...

Widdekind Posted November 26, 2012 Author Share Posted November 26, 2012 On second thought, i think the equations from the cited website, represent the comoving quantities, on the comoving grid. For the comoving quantities, on the comoving grid, i suspect that the continuity equation holds, in its regular form. For, the product of comoving velocity x comoving density represents the flux of mass, per comoving area; and the comoving divergence of the same, seemingly "must" represent the amount of mass accumulated into, or dispersed from, a given comoving grid cell. As for the "force" equation (which ultimately derives from Newton's law, ma=F), i suspect the proper procedure, is to proceed from the "real" equation, and substitute in for the "real" quantities, their corresponding equivalents, in comoving quantities x appropriate powers of the scale factor. For, the force equation models real actual physical forces applied, to fluid parcels (RHS), which generate real actual physical acceleration of said parcels (LHS). So, the equation is valid in "real" actual physical quantities & units; the proper procedure "must" then be to substitute in for the real quantities [math]\rho \left( \frac{\partial}{\partial t} + v \circ \nabla \right) v = - \nabla P[/math] [math]\frac{\tilde{\rho}}{a^3} \left( \frac{\partial}{\partial t} + a \tilde{v} \circ \frac{\tilde{\nabla}}{a} \right) a \tilde{v} = - \frac{\tilde{\nabla}}{a} P[/math] [math]a^{-2} \times \tilde{\rho} \left( \frac{\partial}{\partial t} + H + \tilde{v} \circ \tilde{\nabla} \right) \tilde{v} = - \frac{\tilde{\nabla}}{a} P[/math] The above equation more closely resembles those from said cited website, e.g. adding the term involving "H", instead of subtracting. However, on the LHS, i still do not understand why some terms would involve the scale factor, whilst others would not. And, on the RHS, considerable care must be applied, to what is meant by "Pressure", and (possibly) converting the same (Newtons per square real meter) to comoving coordinates ([comoving?] Newtons per square comoving meter). Separately, in the gravity equation: [math]- \nabla \circ g = 4 \pi G \rho[/math] the Gravity constant (G) has physical units (real meters cubed / mass / time squared). So, on a comoving grid, perhaps the real physical G must be converted, to a comoving [math]\tilde{G} = G a^{-3}[/math] ? Link to comment Share on other sites More sharing options...

juanrga Posted November 28, 2012 Share Posted November 28, 2012 The equations for Fluid Dynamics, on static grids, are well known, e.g. mass continuity: [math]\frac{\partial \rho}{\partial t} + \vec{\nabla} \circ \left( \rho \vec{v} \right) = 0[/math] Those equations must (?) remain the same, on expanding grids, for comoving quantities. For example, the amount of mass within a comoving cell (derivative of comoving density) can only change, by (net) mass flow into / out of the same cell, from surrounding cells, which (net) flow is measured by the divergence. Why would remain the same? In the first place the spatial coordinates are not Galilean, therefore one would wait a change in the form of the nabla operator. In the second place, there is not conservation of energy in an Friedmann universe. Concretely the variation of energy in each cell cannot be accounted by flows to/from adjacent cells, because all the cells are losing the same amount of energy at each instant due to cosmological expansion. Using the mass-rest-energy relationship we find the same conclusion for m inside each cell. I think that all of this is accounted in the link that you give. For instance pay attention to "Terms such as cosmological expansion [...] may be included as source terms". Link to comment Share on other sites More sharing options...

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