Continuity Equation for fluid in a curved spacetime

[aliveone]

TL;DR Summary

Suggested the Continuity Equation derivation.

The current of fluid is the vector J^{\nu}. In free-falling laboratory due to Equivalence principle holds the know Continuity Equation
J^{\nu}_{, \nu}=0, where the ordinary 4-divergence is used. Latter equation was derived in Minkowski spacetime, thus the Christoffel Symbols are all zero for that equation to hold true. Let us denote in curvature coordinates the covariant divergence as scalar function K=J^{\nu}_{; \nu}. Then transforming latter expression into laboratory coordinates (where the Christoffel Symbols are zeroes), one gets K=J^{\nu}_{, \nu}=0, where above equations were used. Thus, the answer is J^{\nu}_{; \nu}=0.

 

[vanhees71]

Looks good!

 

[aliveone]

Looks good!

However, if you calculate the current for so called "perfect fluid", you will find out, that $$J^{\nu}_{;\nu}=p\,u^{\nu}_{;\nu}.$$ Thus, if I am correct and always $$J^{\nu}_{;\nu}=0,$$ then the fluid is not fluid, but dust: $$p=0.$$ Good thing, that it solves negatively the Millennium Prize problem for Fluid.

 

[pervect]

However, if you calculate the current for so called "perfect fluid", you will find out, that $$J^{\nu}_{;\nu}=p\,u^{\nu}_{;\nu}.$$ Thus, if I am correct and always $$J^{\nu}_{;\nu}=0,$$ then the fluid is not fluid, but dust: $$p=0.$$ Good thing, that it solves negatively the Millennium Prize problem for Fluid.

I see you've figured out the "tex" flags. You can also use a pair of $\$$#\$# to do in-line LaTex, and a pair of $\$$$\$$ to do non-inline LaTex.

As far as fluids goes, I believe one can get away with using $J^{\mu}$ as you did above (I could be mistaken on that, but I think it's OK). However, if one has a fluid with pressure, one needs to use the stress-energy tensor $T^{\mu\nu}$. Specifying the 4-velocity or current of the fluid does not specify the pressure in the fluid, one needs a different approach.

For a presureless fluid, $T^{\mu\nu} = \rho u^{\mu} u^{\nu}$, where u is the 4-velocity of the fluid. If $J^\mu = \rho u^\mu$, that would be $T^{\mu\nu} = \frac{1}{\rho} J^\mu J^\nu$

For a perfect and isotropic fluid, the standard expression for the stress-energy tensor is https://en.wikipedia.org/wiki/Perfect_fluid

$$T^{\mu\nu} = \left( \rho + P \right) u^\mu u^\nu + P g^{\mu \nu}$$

where again u is the 4-velocity, and where g^{ab} is the (inverse) metric tensor, written with the -+++ sign convention.

Note that one can get the first expression for $T^{\mu\nu}$ by subst6iting P=0 into the second expression.

The expression for conservation of momentum using the stress energy tensor is
$$T^{\mu\nu}{}_{;\mu} =0 $$

See for instance https://en.wikipedia.org/wiki/Stress–energy_tensor#Conservation_law

 

[vanhees71]

One-component relativistic ideal fluid equations involves three equations (working in natural units with $c=1$ and HEP conventions of the metric, i.e., for Minkowski space $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$):

The equation for the conservation of (net) charges (e.g., baryon number or electric charge),
$$\nabla_{\mu} J^{\mu}=0, \quad J^{\mu}=n u^{\mu},$$
where $n$ is the proper net-charge density (i.e., the charge density in the local rest frame of the fluid) and $u^{\mu}$ is the four-velocity of the fluid cell (obeying the constraints $u^{\mu} u_{\mu}=1$).

Conservation of energy and momentum:

$$\nabla_{\mu} T^{\mu \nu}=0,$$
with the energy-momentum tensor
$$T^{\mu \nu}=(u+P)u^{\mu} u^{\nu}-P g^{\mu \nu},$$
where $u$ is the internal-energy density (i.e., $h=u+P$ the enthalpy density) and $P$ the pressure as measured in the local restframe of the fluid cell.

Finally you need an equation of state, relating $n$ and $P$. Note that for an ideal fluid by assumption the entropy is conserved (adiabatic process).

 

[aliveone]

if one has a fluid with pressure, one needs to use the stress-energy tensor $T^{\mu\nu}$. Specifying the 4-velocity or current of the fluid does not specify the pressure in the fluid, one needs a different approach.

Dear pervect, according to the https://en.wikipedia.org/wiki/Navier–Stokes_equations
the fluid with the non-zero pressure uses the "mass continuity equation", which is $J^{\nu}_{,\nu}=0\,,J^{\nu}=\rho\,u^{\nu}$. Latter does not contain the pressure P, but describes the fluid with non-zero pressure.

 

[PeterDonis]

according to the https://en.wikipedia.org/wiki/Navier–Stokes_equations

The Navier-Stokes equations are not relativistic, so they are not applicable in this discussion.

 

[Orodruin]

The Navier-Stokes equations are not relativistic, so they are not applicable in this discussion.

However, it should be pointed out that the mass continuity equation and the Cauchy momentum equations together comprise the non-relativistic limit of the relativistic conservation of the energy-momentum tensor, i.e., of $\nabla_\mu T^{\mu\nu} = 0$.

 

[pervect]

As others have mentioned, the Navier-Stokes equation are not relativistic.

The simplest procedure would be to start with the 4-vector covariant version of the conservation of energy and momentum, namely $T^{ab}{}_{;a}=0$ and derive the non-covariant Cauchy momentum equations as a non-relativistic approximation. Then the Navier Stokes equation can be further derived from the Cauchy momentum equations in flat space-time as is done in Wikipedia by assuming the Cauchy stress tensor is given as a the sum of a viscosity term and a pressure term.

It's unclear to me what adjustments might need to be made to the Navier-Stokes equations to account for curved space-time, which is the title of the thread. Starting with the covariant formulation in flat space-time to confirm that one gets the Cauchy momentum equation, then extending the results to see what happens in curved space-time, seems to me to me to be the most likely way to proceed. I am not sure if the assumption that the Cauchy stress tensor is the sum of a viscosity term and a pressure term will still hold. I am concerned about what one might call "tidal forces" adding additional terms to the Cauchy stress tensor.

If one imagines a blob of fluid floating in space near a large mass in the non-relativistic limit, the shape of the blob of fluid will be distorted by tidal gravitatioanl forces. I don't believe these tidal forces would fit into the restrictive mold of the Navier-Stokes equations. Unless one includes self-gravitational forces of said blob of fluid the blob of fluid would be torn apart by said tidal forces. It seems to me the Navier-Stokes equations would need some extension just to deal with a static blob of fluid in the presence of curved space-time. One might or might not want to include the self-gravitation of the fluid - if one does, this would be another addition to the Navier-Stokes equations, I think.

There might be a reference somewhere that does the work of going from the covariant formulation to the non-covariant Navier-Stokes formulation, somewhere, but I wouldn't know where.

Note that the covariant formulation $T^{ab}{}_{;a}=0$ also includes the conservation of energy and well as the conservation of momentum, another reason to start with the relativistic equation rather than the non-relativistic ones, as one only needs only to drop the energy equation.

Note that the stress-energy tensor formulation would from the standpoint of the usual relativsitic onventions, most naturally generate what Wiki calls the "conservation form" of the Cauchy momentum equations. To quote Wiki again, from their article on the stress-energy tensor

In solid state physics and fluid mechanics, the stress tensor is defined to be the spatial components of the stress–energy tensor in the proper frame of reference. In other words, the stress energy tensor in engineering differs from the relativistic stress–energy tensor by a momentum convective term.

Some confusion may arise from the differing conventions here, the convective vs non-convective forms of the equations, depending on what one is used to.

Trying to go the other way round - getting the correct relativistically covariant formulation from the non-covariant formulation - seems to me to be much harder and error prone.

 

[pervect]

I was looking at carrying out my proposal above. The first comment I have is there is some confusion in notation with the same name being used for 3-vectors / tensors and 4-vectors.

There are a few things that still puzzle me about the notation for the 3-vector form of the Cauchy momentum equation, though. I'll work through what I have, it may be helpful.

If we stick with cartesian coordinates in an inertial frame, we can write the energy equation (not neeed ) from $\partial_a T^{ab} = 0$ in terms of the 4 dimensional rank 2 stress energy tensor T as:

$$\partial_0 T^{00} + \partial_1 T^{10} + \partial_2 T^{20} + \partial_3 T^{30} = 0$$

Then the three momentum equations are:

$$\partial_0 T^{01} + \partial_1 T^{11} + \partial_2 T^{21} + \partial_3 T^{31} = 0$$
$$\partial_0 T^{02} + \partial_1 T^{12} + \partial_2 T^{22} + \partial_3 T^{33} = 0$$
$$\partial_0 T^{03} + \partial_1 T^{13} + \partial_2 T^{23} + \partial_3 T^{33} = 0$$

We can break down the 4-tensor T into the following:
$\rho$, a number, equal to $T^{00}$. Because we are doing a non-relativsitic analysis, $\rho$ doesn't depend on the frame of reference, though in a true relativistic analysis it does.

j, a three-vector defined as
$$j = \begin{bmatrix} T^{01} \\ T^{02} \\ T^{03} \end{bmatrix}$$

In the non-relativistic version, j is the non-relativistic momentum $\rho u$, where u is also a 3-vector.

Finally, we have F, a rank 2 3-tensor, which I am defining as
$$F = \begin{bmatrix} T^{11} & T^{12} & T^{13} \\ T^{21} & T^{22} & T^{23} \\ T^{31} & T^{32} & T^{33} \end{bmatrix}$$see for instance the wikiepdia image <link> for motivations for this breakdown of T into \rho, j and F

we see most of the conservatio form of the Cauchy momentum equation from the wiki <link>

$$\frac{\partial}{\partial t} j + \nabla \cdot F = s$$

What's a bit of a mystery is why $F = \rho u \otimes u - \sigma$, in particular the minus sign of $\sigma$. The $\rho u \otimes u$ makes sense to me, it's just the dynamic pressure in the conservation form of the stress-energy tenso in the frame where $\rho$ is moving, though I'm not sure if that description will make sense to others. What I don't quite understand is the minus sign in $\sigma$. I thought $\sigma$ was just a 3x3 submatrix in the 4x4 stress energy tensor, but perhaps there is some sign difference? Or maybe I'm missing something else that accounts for the sign.

Additionally, the approach I used doesn't really incorporate the body force s, due to $\rho g$. By using an inertial frame and no gravity (we can't do gravity with flat space-time), the divergence of the tensor is zero, not s. I assume g here is "gravity" (I usually use g for the metric tensor, not gravity). However, it's not really a surprise to find $\rho g$ on the right hand side rather than zero. Use of a non-inertial accelerating frame to simlulate "gravity" and replacing the ordinary derivative with the covariant derivative might be one way to resolve the issue, but I suspect it'd be not worth the effort, as my treatment of accelerated frames would be based on 4-vectors, and the goal is to express the expression in a 3-vector form.

 

[Orodruin]

What I don't quite understand is the minus sign in σσ\sigma. I thought σσ\sigma was just a 3x3 submatrix in the 4x4 stress energy tensor, but perhaps there is some sign difference?

Yes, there is a sign in the definition of the stress tensor. To get the correct stress tensor the way it is usually defined in non-relativistic physics what appears in the energy-momentum tensor is $-\sigma$. For example, the pressure part in the stress tensor is -p on the diagonal because the pressure force on a body is directed in the opposite direction of its surface normal with the normal pointing away from the body. At the same time, what appears in the ideal fluid energy-momentum tensor is +p on the spatial diagonal.

 

[vanhees71]

The Navier-Stokes equations are not relativistic, so they are not applicable in this discussion.

There are relativistic Navier-Stokes equations, but they violate causality. That's why you need to go at least to 2nd order in the gradient expansion. The most simple example for such a theory are the Israel-Stewart equations.

https://iopscience.iop.org/article/10.1088/0143-0807/29/2/010/meta

There are also newer developments in relativistic fluid dynamics in the context of relativistic heavy-ion collisions based on systematic moment expansions of the relativistic Boltzmann equation. See e.g.,

http://de.arxiv.org/abs/1004.5013
https://arxiv.org/abs/1202.4551

 

[aliveone]

As others have mentioned, the Navier-Stokes equation are not relativistic.

Trying to go the other way round - getting the correct relativistically covariant formulation from the non-covariant formulation - seems to me to be much harder and error prone.

The formula $J^{\nu}_{,\nu}=0,\,J^{\nu}=\rho\,u^{\nu}$ for a fluid ("perfect" or else) is known to be relativistic as well: https://en.wikipedia.org/wiki/Relativistic_Euler_equations
As well section "Relativistic version" in
https://en.wikipedia.org/wiki/Continuity_equation

 

[Orodruin]

The formula $J^{\nu}_{,\nu}=0,\,J^{\nu}=\rho\,u^{\nu}$ for a fluid ("perfect" or else) is known to be relativistic as well: https://en.wikipedia.org/wiki/Relativistic_Euler_equations
As well section "Relativistic version" in
https://en.wikipedia.org/wiki/Continuity_equation

You are misunderstanding or misrepresenting (it is one or the other) what is written on those Wikipedia pages. Wikipedia is not a textbook.

 

[pervect]

The formula $J^{\nu}_{,\nu}=0,\,J^{\nu}=\rho\,u^{\nu}$ for a fluid ("perfect" or else) is known to be relativistic as well: https://en.wikipedia.org/wiki/Relativistic_Euler_equations
As well section "Relativistic version" in
https://en.wikipedia.org/wiki/Continuity_equation

Are we looking at the same version? I'm looking at

https://en.wikipedia.org/w/index.php?title=Relativistic_Euler_equations&oldid=857905150

and what I see is:

In fluid mechanics and astrophysics, the relativistic Euler equations are a generalization of the Euler equations that account for the effects of special relativity.

The equations of motion are contained in the continuity equation of the stress–energy tensor
$$ \nabla _{\mu }T^{\mu \nu }=0$$
For a perfect fluid,
$$T^{\mu \nu }\,=(e+p)u^{\mu }u^{\nu }+p\eta ^{\mu \nu }$$

Which is what I've been saying all along. The second reference says similar things in the sections on special relativity

https://en.wikipedia.org/w/index.php?title=Continuity_equation&oldid=932025756#Special_relativity

Examples of continuity equations often written in this form include electric charge conservation
$$\partial _{\mu }J^{\mu }=0$$
where J is the electric 4-current; and energy-momentum conservation
$$\partial _{\nu }T^{\mu \nu }=0$$
where T is the stress-energy tensor.

Note that the example they give for $J^{\mu}{}_{;\mu}=0$ is the continuity equation for electric charge, for energy-momentum they invoke the stress-energy tensor.

I do agree with Orodruin that Wiki is not a textbook, and I think the discussion has reached the point where textbook references would be useful. Wiki is supposed to support it's expositions with sources in case of disagreements, so that is one place to start.

The first article in wiki has a tag where it does not cite references at all :(.

The second article in wiki cites a number of general sources, but the only one I am familiar with is MTW's "Gravitation", which I would suggest as a good place to look. Unfortunately my copy of MTW has wandered off, alas (I'm pretty sure it discusses the topic). I have some other sources, but they don't discuss this particular topic.

If you disagree with what we are trying to say, I encourage you to consult a textbook to check it out. (And if you don't already have a textbook on the topic, which seems likely , I would suggest MTW's Gravitation). I believe you'll find that we've been pointing you in the correct direction.

 

[aliveone]

If you disagree with what we are trying to say, I encourage you to consult a textbook to check it out. (And if you don't already have a textbook on the topic, which seems likely , I would suggest MTW's Gravitation). I believe you'll find that we've been pointing you in the correct direction.

My problem is solved in http://mathreview.uwaterloo.ca/archive/voli/2/olsthoorn.pdf
However, to be honest, I have made calculations with Navier-Stokes fluid energy-momentum tensor and found inconsistency. Do you want the file?

 

[vanhees71]

Just looking briefly over it, I think the linked pdf provides the standard treatment of the ideal (Euler) fluid correctly. It briefly mentions the relativistic Navier-Stokes (1st-order gradient correction) equations, which however, are well known to be acausal and have to be extended at least to the Israel-Stewart equation (taking into account a schematic version of the 2nd order gradient expansion). The newer developments in the field, I've already quoted above.

 

[pervect]

I don't follow everything in http://mathreview.uwaterloo.ca/archive/voli/2/olsthoorn.pdf , but it seems like a good source to me too. The part I do follow matches what I've been trying to say, with some important notational notational differences. In particular, section 3.2 seems key. This section makes a reference to 2.8, which is what I've been saying $\nabla_\mu T^{\mu\nu} = 0$.

The notational differences may be causing some of the confusion.

equation 3.1 matches wiki and my posts, with the exception that what I call $\rho$, they call e.

They define e (which I call $\rho$) as

$$ e = \rho \left(1 + \epsilon \right) $$

I haven't found any defintion of j in my brief reading., I believe they always use $\nabla_\mu \left(\rho u^\mu \right)$

This unfamiliar definition of $\rho$ allows them to talk about the conservation of mass, at least as long as the temperature of the fluid is low enough where there is no particle creation.

Without further work, It's a bit unclear to me at this point exactly how justified their claim is that with this set of definitons they have that can write the continuity equation for a perfect fluid as $\nabla_\mu \left(\rho u^\mu \right)$ is, they state it as a fact but they don't do the derivation. In particular it seems incomplete, as they don't have any conditions on $\epsilon$.

The have an equation below it which does have $\epsilon$, which they refer to as a conservation of energy equation. Perhaps they are conceptually splitting energy and momentum conservation into two equations, $\nabla_\mu \left(\rho u^\mu \right)$ being only the momentum part, and not valid for the "0" index? It's just not clear to me what their claim is, exactly, or how to justify it.

 

[vanhees71]

These are just conventions. In some texts they prefer to deal directly with $e$, which is the internal-energy density in the local restframe of the fluid, and $(T^{\mu \nu})=\mathrm{diag}(e,p,p,p)$ is in the local restframe of the fluid and not of the observer, by the way. Anything concerning intrinsic quantities of the matter is referenced in the local restframe of the matter and in this way are defined as scalar fields, particularly temperature, chemical potentials, pressure, densities of various thermodynamic potentials etc. Also some texts deal with the same quantities per unit mass.