Deriving Nonrelativistic Euler Equations from Stress Energy Tensor

[Ondrej Certik]

Hi,

I would like to start from the stress energy tensor for the perfect fluid:

$$T^{\mu\nu}=\begin{pmatrix} \rho c^2 & 0 & 0 & 0\cr 0 & p & 0 & 0\cr 0 & 0 & p & 0\cr 0 & 0 & 0 & p\cr\end{pmatrix}$$

where $$\rho$$ is the mass density and $$p$$ is the pressure, and I would like to derive the nonrelativistic Euler equations: continuity equation, momentum equation and the energy equation:

$$\partial_t\rho + \partial_i(\rho v^i) = 0$$
$$\partial_t (\rho v^i) + \partial_j (\rho v^i v^j + p\delta^{ij}) = 0$$
$$\partial_t E + \partial_j\left(v^j\left(E + p \right)\right) = 0$$
in the above, $$v^i$$ is the nonrelativistic velocity, $$E={1\over2}\rho v^2 + \rho e$$ the kinetic plus internal energy density, $$\partial_t\equiv {\partial\over\partial t}$$ is the time derivative and $$\partial_i\equiv {\partial\over \partial x^i}$$ are the spatial derivatives and we sum over $$i$$.
Those are the equations found for example on wikipedia:

http://en.wikipedia.org/wiki/Euler_equations_(fluid_dynamics)

Here are my attempts to derive it:

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids
http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#relativistic-derivation-of-the-energy-equation

essentially, I start from

$$\partial_\nu T^{\mu\nu} = 0$$

and for $$\mu=i$$ we get the momentum equation in the limit $$c\to\infty$$, and for
$$\mu=0$$ we get:

$$\partial_t \left(\left(\rho c^2 + p{v^2\over c^2}\right)\gamma^2\right) + \partial_i\left(\left(\rho c^2 + p\right)v^i \gamma^2\right) =0$$

where $$\gamma=(1-{v^2\over c^2})^{-1/2}$$. If we let $$E=\rho c^2$$, and neglect the term $$p {v^2\over c^2}$$, we almost get the right energy equation, the only problem is that this energy $$\rho c^2$$ also contains the rest mass energy, and thus the pressure $$p$$ is negligible, and the nonrelativistic limit of this equation just gives the equation of the continuity.

How do I derive the energy equation? Somehow, I guess I need to separate the rest mass, use the equation of continuity for the rest mass (that follows from the conservation of the baryon number), and then I should be just left with the kinetic+internal energy and the right equation for nonrelativistic energy. Does anyone know how to do that in all details?

I tried to find this question already answered, but didn't find exactly what I want (I am new to the forums, so I might have missed that --- please point me to the right direction).

Thanks,
Ondrej Certik

 

[George Jones]

The relativistic Euler equations are derived in

https://www.physicsforums.com/showthread.php?t=367369.

It should be possible to make a non-relativistic approximation.

 

[Ondrej Certik]

Thanks for the link and the hint.

So I think the energy equation must follow from this:

$$\dot{\varrho} + (\varrho+p)\nabla_au^a = 0$$

but that's where I got stuck, because if I do the nonrelativistic approximation, I have to introduce units, so I get:

$$\dot{\varrho c^2} + (\varrho c^2+p)\nabla_au^a = 0$$

then I divide by $$c^2$$ and neglect $${p\over c^2$$ compared to $$\rho$$ and I obtain the equation of continuity, but not the equation for energy, see my original post.

I also believe it must be possible somehow to do this right, but so far I wasn't successful.

Ondrej

 

[George Jones]

I obtain the equation of continuity, but not the equation for energy, see my original post.

This looks right. What happens when you work with

$$(\varrho+p)\dot{u}^a = \nabla^ap - u^a\dot{p}?$$

 

[Ondrej Certik]

This looks right. What happens when you work with

$$(\varrho+p)\dot{u}^a = \nabla^ap - u^a\dot{p}?$$

Then I get the following equation:

$$\partial_t (\rho u^i) + \partial_j (\rho u^i u^j + p\delta^{ij}) = 0$$

which is just the Euler momentum equation:

$${\partial (\rho{\bf u})\over\partial t} + \nabla \cdot (\rho {\bf u}{\bf u}^T) + \nabla p = 0$$

and that is fine too. However, I also need to obtain the energy equation somehow and that's what I don't know how to do it.

 

[Altabeh]

This looks right. What happens when you work with

$$(\varrho+p)\dot{u}^a = \nabla^ap - u^a\dot{p}?$$

I have a little problem here with the derivation of energy tensor from Lagrange equations. How'd one be possibly able to convert the equation
http://certik.github.com/theoretical-physics/book/images/math/cba9e5e539e0fd54c7fa70def2082965767e1bbe.png
into
http://certik.github.com/theoretical-physics/book/images/math/56db3e95c06721e962f220763e3c8240cf581d9b.png ?[/URL]
This is obtained here: http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids

Thanks
AB

 

[Ondrej Certik]

I have a little problem here with the derivation of energy tensor from Lagrange equations. How'd one be possibly able to convert the equation
http://certik.github.com/theoretical-physics/book/images/math/cba9e5e539e0fd54c7fa70def2082965767e1bbe.png
into
http://certik.github.com/theoretical-physics/book/images/math/56db3e95c06721e962f220763e3c8240cf581d9b.png ?[/URL]
This is obtained here: http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids

Thanks
AB

Just look at the first equation, take the d L / dx^\mu from the left hand side, put it to the right hand side, and then insert it into the brackets, where you differentiate with respect to \nu, so you also have to insert the \delta_\mu^\nu. then you get the second equation. It's a standard derivation of the Noether conserved current.

This is unrelated to this discussion though, I just put it into my notes, because I thought I could derive the tensor somehow from the Lagrange equations, but it turns out it is not easy.

 

[Altabeh]

Just look at the first equation, take the d L / dx^\mu from the left hand side, put it to the right hand side, and then insert it into the brackets, where you differentiate with respect to \nu, so you also have to insert the \delta_\mu^\nu. then you get the second equation. It's a standard derivation of the Noether conserved current.

This is unrelated to this discussion though, I just put it into my notes, because I thought I could derive the tensor somehow from the Lagrange equations, but it turns out it is not easy.

My point was ahead to motivate you to start from a Lagrangian to get that equation. But as you have noticed, it does not sound so easy. I don't know what else can be considered an alternative here to derive energy equation other than Lagrange. Maybe George knows...

 

[Ondrej Certik]

My point was ahead to motivate you to start from a Lagrangian to get that equation. But as you have noticed, it does not sound so easy. I don't know what else can be considered an alternative here to derive energy equation other than Lagrange. Maybe George knows...

No, I appreciate the suggestion. :) I run out of all ideas myself, so any fresh insight from you and others is appreciated.

The way I understand it is that there are 3 nonrelativistic equations (continuity, momentum and energy) and they follow from the conservation law of the stress energy tensor $$\partial_\nu T^{\mu\nu}$$ and the conservation of the baryon number $$\partial_\mu (n U^\mu)$$. The spatial compents of the stress energy tensor give the momentum equation. So we are left with the temporal component of the stress energy tensor and the baryon number law. Somehow we need to get the continuity + energy equations from them, I guess by somehow combining them.

I don't think anything else is needed. Of course, you can also derive the energy equation just purely from nonrelativistic considerations, that is shown here:

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#energy-equation

and I am sure there are other ways to get it too. But I just want to see how it follows from the relativistic conservation laws.

 

[Altabeh]

No, I appreciate the suggestion. :) I run out of all ideas myself, so any fresh insight from you and others is appreciated.

The way I understand it is that there are 3 nonrelativistic equations (continuity, momentum and energy) and they follow from the conservation law of the stress energy tensor $$\partial_\nu T^{\mu\nu}$$ and the conservation of the baryon number $$\partial_\mu (n U^\mu)$$. The spatial compents of the stress energy tensor give the momentum equation. So we are left with the temporal component of the stress energy tensor and the baryon number law. Somehow we need to get the continuity + energy equations from them, I guess by somehow combining them.

I don't think anything else is needed. Of course, you can also derive the energy equation just purely from nonrelativistic considerations, that is shown here:

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#energy-equation

and I am sure there are other ways to get it too. But I just want to see how it follows from the relativistic conservation laws.

Look at the pages 49 and 50 of Weinberg's "Gravitaion and Cosmology Principles" book...
There you can find a relativistic proof of energy equation.

AB

 

[Ondrej Certik]

Look at the pages 49 and 50 of Weinberg's "Gravitaion and Cosmology Principles" book...
There you can find a relativistic proof of energy equation.

AB

I obtained the book. Weinberg essentially derives the same equation as Schutz:

$$u^\alpha \partial_\alpha S = 0$$

which says, that the flow conserves specific entropy. This equation is obtained from a thermodynamic relation, e.g. eq. 2.10.18 in Weinberg:

$$k T d \sigma = pd \left(1\over n\right) + d \left(\rho\over n\right)$$

where the right hand side follows from the 0th component of

$$\partial_\nu T^{0\nu}$$

However, do you know how to obtain

$$\partial_t E + \partial_j\left(v^j\left(E + p \right)\right) = 0$$

from that? One might need the equation of state for that, so let's use the ideal gas.

 

[Altabeh]

I obtained the book. Weinberg essentially derives the same equation as Schutz:

$$u^\alpha \partial_\alpha S = 0$$

which says, that the flow conserves specific entropy. This equation is obtained from a thermodynamic relation, e.g. eq. 2.10.18 in Weinberg:

$$k T d \sigma = pd \left(1\over n\right) + d \left(\rho\over n\right)$$

where the right hand side follows from the 0th component of

$$\partial_\nu T^{0\nu}$$

However, do you know how to obtain

$$\partial_t E + \partial_j\left(v^j\left(E + p \right)\right) = 0$$

from that? One might need the equation of state for that, so let's use the ideal gas.

The below picture shows the page the page 98 of Warner's "Into to Differential Geometry and GR".

http://img508.imageshack.us/img508/3976/scanjo.jpg

Without having the internal energy involved, this 'energy equation' has been obtained through a geometric definition of flow of energy in three directions plus the direction of time, and this can be combined with the components of stress-energy tensor to give us the 'energy equation' as in the required form. But I am really out of any idea about how internal energy can be considered here...

I've spent much of today on this thing, and checked more than 40 books regarding relativity, but none of them obtain a good and vivid derivation and most of them address the reader to have a look at another book for a proof (e.g. Weinberg's proof) which makes it even worse because you learn at the end that it is just your time getting wasted...

 

[Altabeh]

I obtained the book. Weinberg essentially derives the same equation as Schutz:

$$u^\alpha \partial_\alpha S = 0$$

which says, that the flow conserves specific entropy. This equation is obtained from a thermodynamic relation, e.g. eq. 2.10.18 in Weinberg:

$$k T d \sigma = pd \left(1\over n\right) + d \left(\rho\over n\right)$$

where the right hand side follows from the 0th component of

$$\partial_\nu T^{0\nu}$$

However, do you know how to obtain

$$\partial_t E + \partial_j\left(v^j\left(E + p \right)\right) = 0$$

from that? One might need the equation of state for that, so let's use the ideal gas.

Take a look, Certik:

http://img509.imageshack.us/img509/5189/97652996.jpg

AB

 

[George Jones]

Take a look, Certik

Great!

A few details: $0 = \partial_\beta T^{\alpha \beta}$ is true in all coordinate systems, and, in particular, is true in the global inertial coordinate system of an inertial observer who watches the fluid stream by. The result follows when this equation is contracted with the 4-velocity of the observer instead of the 4-velocity of the fluid. In the observer's inertial coordinate system with $x_0 = ct$, $u^0 = c \gamma$, and $u^i = \gamma v^i$. This gives

$$T^{00} = \rho' \left( u^0 \right)^2 = \rho' \gamma^2 c^2 = \rho c^2$$

and

$$T^{0i} = \left( \rho' + p'/c^2 \right) u^0 u^i = \left( \rho' + p'/c^2 \right) \gamma^2 c v^i = \left( \rho + p/c^2 \right) c v^i$$

I have taken the unprimed frame to be the observer's frame and the primed frame to be a frame (momentarily) comoving with the fluid.

 

[Ondrej Certik]

Take a look, Certik:

http://img509.imageshack.us/img509/5189/97652996.jpg

AB

Excellent, I really appreciate your help. You derived the equation that I have here, eq. (3):

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#relativistic-derivation-of-the-energy-equation

But as I write there (and also above in this thread), the problem with it is that $$\rho c^2$$ is the total energy, including the rest mass, and thus the "p" next to it is completely negligible. So what you got is just the continuity equation for rho.

The E in the energy equation in the euler equations on the other hand is equal to the kinetic energy $${1\over2}\rho v^2$$ and the internal energy.

We would get what we want, if we could substract the continuity equation for rho, from the equation that you just derived. If you see what I mean. But my question is then --- how do we derive the continuity equation for rho? Surely we cannot use the same 0th component of the stress energy tensor to derive *both* the continuity equation and the energy equation, can we?

Somehow we need to use the conservation of particles (baryon) law, but in there there is the number of particles $$n$$ and so we need to somehow connect $$n$$ with $$\rho$$, so we need to use the equation of state somehow. It's really complex. :(


Let me know what you think. I think we should start from the equations in Weinberg or Schutz, and write down the thermodynamic relations for the ideal gas, connecting the number of particles "n", density "rho", entropy "s", etc. And somehow, we might get it.

 

[George Jones]

Here is something sloppy and simple-minded, probably too sloppy and simple-minded.

Represent the energy density of the internal and kinetic energy by (changing the meaning of $E$) by $E = \rho c^2 - \rho' c^2$ and use $\rho c^2 = E + \rho' c^2$ in what we have. This gives

$$0 = \frac{\partial E}{\partial t} + \nabla \cdot \left[ \left( E + p \right) \vec{v} \right] + \frac{\partial \rho' c^2}{\partial t} + \nabla \cdot \left[ \left( \rho' c^2 \right) \vec{v} \right]$$

and by the a non-relativistic approx. to the continuity equation?

I haven't had time to think if this is legit., as I'm scrambling to prepare to put food on my table (i.e., do real work.)

Back in a few hours

 

[Ondrej Certik]

Here is something sloppy and simple-minded, probably too sloppy and simple-minded.

Represent the energy density of the internal and kinetic energy by (changing the meaning of $E$) by $E = \rho c^2 - \rho' c^2$ and use $\rho c^2 = E + \rho' c^2$ in what we have. This gives

$$0 = \frac{\partial E}{\partial t} + \nabla \cdot \left[ \left( E + p \right) \vec{v} \right] + \frac{\partial \rho' c^2}{\partial t} + \nabla \cdot \left[ \left( \rho' c^2 \right) \vec{v} \right]$$

and by the a non-relativistic approx. to the continuity equation?

I haven't had time to think if this is legit., as I'm scrambling to prepare to put food on my table (i.e., do real work.)

Back in a few hours

Yes, that's my idea too --- but now we need to get rid of the two terms on the right --- the continuity equation for rho --- who do we derive that those are 0?

 

[Altabeh]

Great!

A few details: $0 = \partial_\beta T^{\alpha \beta}$ is true in all coordinate systems, and, in particular, is true in the global inertial coordinate system of an inertial observer who watches the fluid stream by. The result follows when this equation is contracted with the 4-velocity of the observer instead of the 4-velocity of the fluid. In the observer's inertial coordinate system with $x_0 = ct$, $u^0 = c \gamma$, and $u^i = \gamma v^i$. This gives

$$T^{00} = \rho' \left( u^0 \right)^2 = \rho' \gamma^2 c^2 = \rho c^2$$

and

$$T^{0i} = \left( \rho' + p'/c^2 \right) u^0 u^i = \left( \rho' + p'/c^2 \right) \gamma^2 c v^i = \left( \rho + p/c^2 \right) c v^i$$

I have taken the unprimed frame to be the observer's frame and the primed frame to be a frame (momentarily) comoving with the fluid.

Nice attempt. Your derivation is even true in GR but locally not globally. As you know, any metric tensor at a given point would be cast into the form of a Minkowski metric (not by a coordinates transformation but rather by manipulation of the Jaccobi matrices or, to better put it into words, by making an involutive matrix out of the metric tensor as all eigenvalues of this new matrix consist of-1 and +1).

 

[Altabeh]

Yes, that's my idea too --- but now we need to get rid of the two terms on the right --- the continuity equation for rho --- who do we derive that those are 0?

And what about a transformation?! Gotta think of it...

AB

 

[Ondrej Certik]

And what about a transformation?! Gotta think of it...

AB

Also a related question -- the nonrelativistic energy $$E$$ is composed of the kinetic and an internal one.

The kinetic energy can be obtained by:

$$\rho c^2 \gamma={\rho c^2\over \sqrt{1-{v^2\over c^2}}} = \rho c^2 \left(1 + {v^2\over 2 c^2}+\cdots\right) = \rho c^2 + {1\over 2} \rho v^2 + \cdots$$

now I am not sure what exactly the internal energy is, but I think it is composed of the random motion of the molecules in the liquid (e.g. a temperature), i.e. everything besides the rest mass and the kinetic energy, which is obtained by boosting into some other frame by $$\gamma$$. How do we obtain the rest mass energy? Well, I would guess it's:

$$\rho' = n m \gamma ^2$$

where $$n$$ is the particle number, $$m$$ is the mass of one particle and the two $$\gamma$$ factors are necessary (see e.g. Schutz). We then substract $$\rho'$$ as George did. What remains is to prove, the continuity equation for $$\rho'$$. From the conservation of the Baryon number, it follows the continuity equation for $$n$$:

$$0 = \frac{\partial (n\gamma)}{\partial t} + \nabla \cdot \left[ n\gamma \vec{v} \right]$$

so by multiplying by $$m$$, you get the continuity equation for $$\rho_0$$, well, up to one $$\gamma$$ factor, but I think it might be ok, since we want a nonrelativistic limit at the end anyway.

So we are almost there. Do you think the kinetic+internal energy is equal to $$\rho c^2 - nm\gamma^2 c^2 = \rho c^2 - \rho' c^2$$?

 

[Altabeh]

Also a related question -- the nonrelativistic energy $$E$$ is composed of the kinetic and an internal one.

The kinetic energy can be obtained by:

$$\rho c^2 \gamma={\rho c^2\over \sqrt{1-{v^2\over c^2}}} = \rho c^2 \left(1 + {v^2\over 2 c^2}+\cdots\right) = \rho c^2 + {1\over 2} \rho v^2 + \cdots$$

now I am not sure what exactly the internal energy is, but I think it is composed of the random motion of the molecules in the liquid (e.g. a temperature), i.e. everything besides the rest mass and the kinetic energy, which is obtained by boosting into some other frame by $$\gamma$$. How do we obtain the rest mass energy? Well, I would guess it's:

$$\rho' = n m \gamma ^2$$

where $$n$$ is the particle number, $$m$$ is the mass of one particle and the two $$\gamma$$ factors are necessary (see e.g. Schutz). We then substract $$\rho'$$ as George did. What remains is to prove, the continuity equation for $$\rho'$$. From the conservation of the Baryon number, it follows the continuity equation for $$n$$:

$$0 = \frac{\partial (n\gamma)}{\partial t} + \nabla \cdot \left[ n\gamma \vec{v} \right]$$

so by multiplying by $$m$$, you get the continuity equation for $$\rho_0$$, well, up to one $$\gamma$$ factor, but I think it might be ok, since we want a nonrelativistic limit at the end anyway.

So we are almost there. Do you think the kinetic+internal energy is equal to $$\rho c^2 - nm\gamma^2 c^2 = \rho c^2 - \rho' c^2$$?

We are there! This idea came to my mind too, but I was afraid that Lorentz factor is not that much small so setting $$\rho' = n m \gamma ^2$$ (I just thought of one factor of Lorentz, though Schutz claims two which makes it work) would not neccessarily be non-relativistic... but it could be so by Schutz's treatment as $$\gamma ^2<< gamma$$, so we are more closer to Newton than before. Excellent idea that exactly maches mine...

So yes, it is kinetic+internal energy because the total energy $$E=\rho c^2$$ is just composed of these two and a rest mass energy but I think we better say $$\rho c^2\gamma - nm\gamma^2 c^2 = \rho c^2\gamma - \rho' c^2$$ to have a more correct analysis just because as George showed, now we are looking through an observer's eyes who is not co-moving with the fluid. This does not make a difference in non-relativistic limit!

AB

 

[Ondrej Certik]

We are there! This idea came to my mind too, but I was afraid that Lorentz factor is not that much small so setting $$\rho' = n m \gamma ^2$$ (I just thought of one factor of Lorentz, though Schutz claims two which makes it work) would not neccessarily be non-relativistic... but it could be so by Schutz's treatment as $$\gamma ^2<< gamma$$, so we are more closer to Newton than before. Excellent idea that exactly maches mine...

So yes, it is kinetic+internal energy because the total energy $$E=\rho c^2$$ is just composed of these two and a rest mass energy but I think we better say $$\rho c^2\gamma - nm\gamma^2 c^2 = \rho c^2\gamma - \rho' c^2$$ to have a more correct analysis just because as George showed, now we are looking through an observer's eyes who is not co-moving with the fluid. This does not make a difference in non-relativistic limit!

AB

I am still not comfortable with those $$\gamma$$ factors -- because the kinetic energy is generated by one such gamma factor. So if we are comoving with the fluid, there is no kinetic energy (that makes sense), but I think we should do the analysis in the laboratory frame (as you said), and thus we expand one gamma factor, this gives the kinetic energy, then we substract the rest mass, and we are left with kinetic+internal energy. Roughly.

I will try to write this down in details in the evening.

I still have the confusion about the energies.So the rest mass (i.e. when comoving with the fluid) does, or does not contain the internal energy (=random motion of the molecules and their interaction)? The rho in the stress energy tensor is the total energy? Let's think: the total energy (e.g. all energies) that one measures is:
$$\rho = T_{\alpha\beta} u^\alpha u^\beta$$
this follows from the definition of the stress energy tensor. So when comoving with the fluid, the spatial components of u^\alpha are zero and thus what we measure is the \rho in the stress energy tensor. So I think that rho contains all energies. We can convert between energy and the mass by E = mc^2. Right? So the mass, that we measure in the comoving fluid contains all the contribution (including internal). However, in the comoving fluid, the mass coming from $$n m$$ only contains the real mass of the particles and it doesn't contain the internal energy. And I think this is what we need to substract. And then when we boost to a different frame, we can also see the kinetic energy, and that follows from the gamma factor.

However, in this equation:
$$\rho c^2 \gamma={\rho c^2\over \sqrt{1-{v^2\over c^2}}} = \rho c^2 \left(1 + {v^2\over 2 c^2}+\cdots\right) = \rho c^2 + {1\over 2} \rho v^2 + \cdots$$
the "rho" contains both the "n*m" energy and also the internal energy. But the mass contribution coming from the internal energy is very small, and in the nonrelativistic limit, we neglect that.

So I think things start to make sense, more or less. We still need to write this down very carefully with all details. I'll try to do that soon and we'll see if we have this right or not. As I understand it, one has to be very careful how the nonrelativistic limit is done -- the internal energy and pressure are comparable, while the mass energy mc^2 and the mass "m*n" are also comparable but both are the order of magnitude (c^2) larger (their difference is the internal energy). This mass energy is needed to get the kinetic energy.

Well, I hope it will work.

 

[Altabeh]

I am still not comfortable with those $$\gamma$$ factors -- because the kinetic energy is generated by one such gamma factor. So if we are comoving with the fluid, there is no kinetic energy (that makes sense), but I think we should do the analysis in the laboratory frame (as you said), and thus we expand one gamma factor, this gives the kinetic energy, then we substract the rest mass, and we are left with kinetic+internal energy. Roughly.

I will try to write this down in details in the evening.

I still have the confusion about the energies.So the rest mass (i.e. when comoving with the fluid) does, or does not contain the internal energy (=random motion of the molecules and their interaction)? The rho in the stress energy tensor is the total energy? Let's think: the total energy (e.g. all energies) that one measures is:
$$\rho = T_{\alpha\beta} u^\alpha u^\beta$$
this follows from the definition of the stress energy tensor. So when comoving with the fluid, the spatial components of u^\alpha are zero and thus what we measure is the \rho in the stress energy tensor. So I think that rho contains all energies. We can convert between energy and the mass by E = mc^2. Right? So the mass, that we measure in the comoving fluid contains all the contribution (including internal). However, in the comoving fluid, the mass coming from $$n m$$ only contains the real mass of the particles and it doesn't contain the internal energy. And I think this is what we need to substract. And then when we boost to a different frame, we can also see the kinetic energy, and that follows from the gamma factor.

However, in this equation:
$$\rho c^2 \gamma={\rho c^2\over \sqrt{1-{v^2\over c^2}}} = \rho c^2 \left(1 + {v^2\over 2 c^2}+\cdots\right) = \rho c^2 + {1\over 2} \rho v^2 + \cdots$$
the "rho" contains both the "n*m" energy and also the internal energy. But the mass contribution coming from the internal energy is very small, and in the nonrelativistic limit, we neglect that.

So I think things start to make sense, more or less. We still need to write this down very carefully with all details. I'll try to do that soon and we'll see if we have this right or not. As I understand it, one has to be very careful how the nonrelativistic limit is done -- the internal energy and pressure are comparable, while the mass energy mc^2 and the mass "m*n" are also comparable but both are the order of magnitude (c^2) larger (their difference is the internal energy). This mass energy is needed to get the kinetic energy.

Well, I hope it will work.

To my knowledge, internal energies always exist in a fluid, though so small, but definitely they are present all the time because interactions cannot be stopped by any means. The rho is the total energy if and only if it is measured by an observer co-moving with the fluid. All other observers must see those gamma factors in their measurements... as you noticed, this can easily be seen by \rho = T_{\alpha\beta} u^\alpha u^\beta. The last thing to say is that you understand things so much faster than I do. Don't worry about these confusions... you are not the only one. My eyes are getting closed gradually because it is 2 AM here. I'll write things carefully early today and then I think we can could make a conclusion.

AB

 

[Ondrej Certik]

To my knowledge, internal energies always exist in a fluid, though so small, but definitely they are present all the time because interactions cannot be stopped by any means. The rho is the total energy if and only if it is measured by an observer co-moving with the fluid. All other observers must see those gamma factors in their measurements... as you noticed, this can easily be seen by \rho = T_{\alpha\beta} u^\alpha u^\beta. The last thing to say is that you understand things so much faster than I do. Don't worry about these confusions... you are not the only one. My eyes are getting closed gradually because it is 2 AM here. I'll write things carefully early today and then I think we can could make a conclusion.

AB

You are right. Also with your definition of E in your previous post. I wrote everything in detail here:

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids

and it seems everything nicely fits together. Even all gammas play well. Questions/comments that remain:

1) what exactly is "m" in the "n * m"? Is it the mass of the isolated particle? E.g. let's say I take one atom of Hydrogen with mass m, then the H2 molecule has n*m equal to the 2*m. The energy of the H2 molecule however is less than 2*m. So in some cases I guess n*m can actually be more than "rho", that measures the total energy (or mass). However, for liquids, there is always some internal motion etc., so I guess rho is always more than n*m, is that right?

2) in the Newtonian limit, "rho" becomes the density of the fluid, and n*m also becomes the density of the fluid? E.g the "rho - n*m" is equal to E/c^2, so it's virtually zero. However, if we are interested in E, as in the energy equation, then we have to treat rho-n*m carefully.

3) the 0th component of the stress energy tensor is in fact the energy equation, if the difference rho-n*m is treated precisely (with the help of the conservation of baryons), but apparently, if we just do a crude Newtonian limit, we get the continuity equation from it. So indeed, the 0th component gives both continuity and energy equations.

4) If we do the Newtonian approximation to the stress energy tensor as in the MTW, we never get the right energy equation, because they neglect the pressure to early. One has to keep the terms with pressure there, even though at first sight it's way smaller than the other terms. But the pressure gets relevant, once we substract n*m*c^2.

Essentially, the mistake that I was doing in my very first post to this thread is that I wrongly defined the nonrelativistic energy.

I think we made a huge progress. Thanks a lot guys! I am going to bed now (11pm here), and in the coming days I'll think about this some more and try to polish it. Let me know if you have some comments to this.

 

[Altabeh]

You are right. Also with your definition of E in your previous post. I wrote everything in detail here:

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids

and it seems everything nicely fits together. Even all gammas play well. Questions/comments that remain:

1) what exactly is "m" in the "n * m"? Is it the mass of the isolated particle? E.g. let's say I take one atom of Hydrogen with mass m, then the H2 molecule has n*m equal to the 2*m. The energy of the H2 molecule however is less than 2*m. So in some cases I guess n*m can actually be more than "rho", that measures the total energy (or mass). However, for liquids, there is always some internal motion etc., so I guess rho is always more than n*m, is that right?

2) in the Newtonian limit, "rho" becomes the density of the fluid, and n*m also becomes the density of the fluid? E.g the "rho - n*m" is equal to E/c^2, so it's virtually zero. However, if we are interested in E, as in the energy equation, then we have to treat rho-n*m carefully.

3) the 0th component of the stress energy tensor is in fact the energy equation, if the difference rho-n*m is treated precisely (with the help of the conservation of baryons), but apparently, if we just do a crude Newtonian limit, we get the continuity equation from it. So indeed, the 0th component gives both continuity and energy equations.

4) If we do the Newtonian approximation to the stress energy tensor as in the MTW, we never get the right energy equation, because they neglect the pressure to early. One has to keep the terms with pressure there, even though at first sight it's way smaller than the other terms. But the pressure gets relevant, once we substract n*m*c^2.

Essentially, the mistake that I was doing in my very first post to this thread is that I wrongly defined the nonrelativistic energy.

I think we made a huge progress. Thanks a lot guys! I am going to bed now (11pm here), and in the coming days I'll think about this some more and try to polish it. Let me know if you have some comments to this.

These are my ideas and answers to your questions:

1- This is more of a chemistry-related question, but let's see what molecular mass is in essence. Strictly speaking, the molecular mass is not precisely equal to the sum of the atomic masses or n*m, but slightly lower. This is because of the energy released upon the formation of chemical bonds, which is equivalent to "lost" mass according to SR. This difference is negligible however.

Even better, this is no big deal when discussed as a whole like a flowing fluid. For instance, the atomic mass of hydrogen is 1.00784 and that of oxygen is 15.9994; therefore, the molecular mass of water with formula H₂O is (2 × 1.00784) + 15.9994 = 18.01508. Therefore, one molecule of water weighs 18.01508 u, and one mole of water weighs 18.01508 grams. As you can see the rest mass of one mole of water is equal to the sum of masses of its total molecules, essentially, the Avogardo-number of 'em or 6.02 x 10²³. So n*m for one mole of water is 18.01508 grams and then m= 18.01508 u grams. Here u is the water molecular mass times the molar mass constant M_u per 6.02 x 10²³ of water molecules. For fluids with, permanent internal interactions between elements, the 'rho' is always greater than n*m because u*m does not contain internal energy but rho does.

2- In the Newtonian limit, IFF the internal energy is neglected, then rho - n*m vanishes and as you said it is zero virtually not in reality as some small contributions to energy was ruled out.

All calculations verify 3 and 4. So I don't go there.

AB

 

[Ondrej Certik]

These are my ideas and answers to your questions:

1- This is more of a chemistry-related question, but let's see what molecular mass is in essence. Strictly speaking, the molecular mass is not precisely equal to the sum of the atomic masses or n*m, but slightly lower. This is because of the energy released upon the formation of chemical bonds, which is equivalent to "lost" mass according to SR. This difference is negligible however.

Even better, this is no big deal when discussed as a whole like a flowing fluid. For instance, the atomic mass of hydrogen is 1.00784 and that of oxygen is 15.9994; therefore, the molecular mass of water with formula H₂O is (2 × 1.00784) + 15.9994 = 18.01508. Therefore, one molecule of water weighs 18.01508 u, and one mole of water weighs 18.01508 grams. As you can see the rest mass of one mole of water is equal to the sum of masses of its total molecules, essentially, the Avogardo-number of 'em or 6.02 x 10²³. So n*m for one mole of water is 18.01508 grams and then m= 18.01508 u grams. Here u is the water molecular mass times the molar mass constant M_u per 6.02 x 10²³ of water molecules. For fluids with, permanent internal interactions between elements, the 'rho' is always greater than n*m because u*m does not contain internal energy but rho does.

2- In the Newtonian limit, IFF the internal energy is neglected, then rho - n*m vanishes and as you said it is zero virtually not in reality as some small contributions to energy was ruled out.

All calculations verify 3 and 4. So I don't go there.

AB

Thanks for the comments. I agree with your conclusions. I made a few cosmetic changes to:

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids

it now derives the continuity equation with "rho", not "n*m".

I think we answered all my questions. I am now very happy with the results. This bothered me for couple months. And now I can finally sleep well. :)

Thanks for the help.

Ondrej

 

[Altabeh]

Thanks for the comments. I agree with your conclusions. I made a few cosmetic changes to:

http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids

it now derives the continuity equation with "rho", not "n*m".

I think we answered all my questions. I am now very happy with the results. This bothered me for couple months. And now I can finally sleep well. :)

Thanks for the help.

Ondrej

I'm, too, happy with all we did here in 3 days. Thanks to you and George, your question was resolved in such a short time.

AB

 

[George Jones]

I'm, too, happy with all we did here in 3 days. Thanks to you and George, your question was resolved in such a short time.

And to Altabeh