What is energy density in the SR equation for energy-momentum tensor?

[George Keeling]

TL;DR Summary

What is energy density in the SR equation for energy-momentum tensor for a perfect fluid?

In Special Relativity I'm given the energy-momentum tensor for a perfect fluid:$$
T^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p\eta^{\mu\nu}
$$where $\rho$ is the energy density, $p$ is the pressure, $U^\mu=\partial x^\mu/\partial\tau$ is the four-velocity of the fluid. In the non-relativistic limit with $$
U^\mu=\left(1,v^i\right)\ \ ,\ \ v^i\ll1\ \ ,\ \ p\ll\rho
$$it is easy to get to a continuity equation for energy flow which is fine and dandy.

But I understand the first two non-relativistic limits but not really the third, $p\ll\rho$. It is 'because pressure comes from random motions of the individual particles, and in this limit these are taken to be small.' I know what pressure is and it's about ${10}^5\ kg\ m^{-1}s^{-2}$ at sea level. But what is the energy density? Should I take the density of the kinetic energy of all the molecules of air? I calculate that at about ${10}^3\ kg\ m^{-1}\ s^{-2}$. So $p>\rho$. Must I include energy according to $E=mc^2$? Then I get $\rho={10}^{17}\ kg\ m^{-1}\ s^{-2}$ which fits the bill nicely but it's not very non-relativistic . I realize that I am not really using the correct units (because $c=1$ here), nevertheless the comparison should work in any units.

 

[Paul Colby]

Summary:: What is energy density in the SR equation for energy-momentum tensor for a perfect fluid?

Must I include energy according to $E=mc^2$?

I believe yes is the correct answer. Why wouldn't the rest energy be included?

 

[pervect]

Summary:: What is energy density in the SR equation for energy-momentum tensor for a perfect fluid?

In Special Relativity I'm given the energy-momentum tensor for a perfect fluid:$$
T^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p\eta^{\mu\nu}
$$where $\rho$ is the energy density, $p$ is the pressure, $U^\mu=\partial x^\mu/\partial\tau$ is the four-velocity of the fluid. In the non-relativistic limit with $$
U^\mu=\left(1,v^i\right)\ \ ,\ \ v^i\ll1\ \ ,\ \ p\ll\rho
$$it is easy to get to a continuity equation for energy flow which is fine and dandy.

But I understand the first two non-relativistic limits but not really the third, $p\ll\rho$. It is 'because pressure comes from random motions of the individual particles, and in this limit these are taken to be small.' I know what pressure is and it's about ${10}^5\ kg\ m^{-1}s^{-2}$ at sea level. But what is the energy density? Should I take the density of the kinetic energy of all the molecules of air? I calculate that at about ${10}^3\ kg\ m^{-1}\ s^{-2}$. So $p>\rho$. Must I include energy according to $E=mc^2$? Then I get $\rho={10}^{17}\ kg\ m^{-1}\ s^{-2}$ which fits the bill nicely but it's not very non-relativistic . I realize that I am not really using the correct units (because $c=1$ here), nevertheless the comparison should work in any units.

In geometric units your expression for T is correct, but you are mixing in SI units, i.e. kg and meters, where the expressions you wrote are not correct. You need to settle on one system, and stick with it.

I don't have time to work through more than this at the moment, I'd have to focus not to make a silly conversion error as well :(.

 

[PeterDonis]

Must I include energy according to $E=mc^2$?

Yes; the relativistic energy density $\rho$ includes rest energy density.

 

[pervect]

I believe yes is the correct answer. Why wouldn't the rest energy be included?

Since geometric units are simpler, and all of the equations you wrote are in geometric units, I'll give the conversion factors to and from geometric units, per <wiki link>.

Density has a conversion factor of $G/c^2$. This converts the density in $kg/m^3$ into $1/m^2$, the geometric unit of density.

Pressure has a conversion factor of $G/c^4$. This converts force/area = $kg \, m / s^2 m^3$ into $1/m^2$

You can also use the non-geometric unit formulas, but then you'll have to revise a lot of your equations.

The short version is that pressure / c^2 is comparable to kg / m^3. You're omitting the factor of c^2, making the pressure appear much more significant than it is.

 

[PeterDonis]

Density has a conversion factor of G/c2G/c^2.

To be clear, this is mass density. Energy density has the same conversion factor as pressure, $G / c^4$.

 

[Paul Colby]

Since geometric units are simpler, and all of the equations you wrote are in geometric units, I'll give the conversion factors to and from geometric units, per <wiki link>.

To be clear, I didn't write any equations. I assume this is a misfire.

 

[George Keeling]

As we haven't done any calculations for the whole tensor yet all the talk about units is unnecessary. I could have compared pressure and energy density in imperial units. I have my answer: Energy density includes the $mc^2$ part, even in pounds shillings and pence (£sd). Thanks!

 

[vanhees71]

Summary:: What is energy density in the SR equation for energy-momentum tensor for a perfect fluid?

In Special Relativity I'm given the energy-momentum tensor for a perfect fluid:$$
T^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p\eta^{\mu\nu}
$$where $\rho$ is the energy density, $p$ is the pressure, $U^\mu=\partial x^\mu/\partial\tau$ is the four-velocity of the fluid. In the non-relativistic limit with $$
U^\mu=\left(1,v^i\right)\ \ ,\ \ v^i\ll1\ \ ,\ \ p\ll\rho
$$it is easy to get to a continuity equation for energy flow which is fine and dandy.

But I understand the first two non-relativistic limits but not really the third, $p\ll\rho$. It is 'because pressure comes from random motions of the individual particles, and in this limit these are taken to be small.' I know what pressure is and it's about ${10}^5\ kg\ m^{-1}s^{-2}$ at sea level. But what is the energy density? Should I take the density of the kinetic energy of all the molecules of air? I calculate that at about ${10}^3\ kg\ m^{-1}\ s^{-2}$. So $p>\rho$. Must I include energy according to $E=mc^2$? Then I get $\rho={10}^{17}\ kg\ m^{-1}\ s^{-2}$ which fits the bill nicely but it's not very non-relativistic . I realize that I am not really using the correct units (because $c=1$ here), nevertheless the comparison should work in any units.

Obviously you are using the east-coast convention (i.e., $\eta_{\mu \nu}=\mathrm{diag}(-1,1,1,1)$). To understand the meaning of the variables in the energy-momentum tensor of a perfect fluid, note that in the local fluid rest frame the four-velocity $U^{\mu}=(1,0,0,0)$ and thus in this reference frame $T^{00}=\rho$ and $T^{11}=T^{22}=T^{33}=p$, $T^{\mu \nu}=0$ for all index pairs with $\mu \neq \nu$. So $\rho$ is the internal energy density of fluid in its rest frame, including the rest energy but also the thermal energy from the thermal motion.

So to make contact with the corresponding non-relativistic quantities one has to note that $\rho=\mu c^2 +u$, where $\mu$ is the rest-mass density as measured in the local fluid rest frame and $u$ the usual internal-energy density as defined in non-relativistic physics. Since the thermal velocities for the non-relativistic limit $k_{\text{B}} T \ll m_{\text{molecules}}$ are much smaller than $c$ and $p$ is of the same order of magnitude as $u$, you have $u \ll \rho$ and thus also $u \ll p$.

To see this take as an example an ideal gas in the non-relativistic limit. The usual internal energy density is $u=\frac{3}{2} n k_{\text{B}} T$, where $n$ is the number density of the fluid, and from the ideal-gas law you have $p=n k_{\text{B}} T$, i.e., $p$ is of the same order of magnitude as $u$. On the other hand the rest-mass density is $m c^2 n$. For a non-relativistic ideal gas you must have $\langle v^2 \rangle \ll c^2$, where $v$ is the thermal velocity of the atoms, but that implies $k_{\text{B}} T \sim m \langle v^2 \rangle \ll m c^2$, and thus in the non-relatistic limit $\rho \simeq m c^2 n$.

The units (with all constants like $c$ and $k_{\text{B}}$ kept in and using the SI of units) are of course $[\rho]=[p]=\text{J}/\text{m}^3$, $[n]=1/\text{m}^3$ (the same is also the unit of $u$, but one cannot write for some reason a $u$ in square brackets within the forum software ;-)), $[T]=\text{K}$, and $[k_{\text{B}}]=\text{J}/\text{K}$.

 

[pervect]

Yes; the relativistic energy density $\rho$ includes rest energy density.

Yes, you're right - and that's where I went wrong.

Energy density and pressure, force/unit area have the same units, $kg \, m^{-1} s^{-2}$ so the unit mismatch I was concerned about is a non-issue. I was thinking of the wrong density, in spite of the fact the OP clearl spelled out they were interested in the energy density.

 

[SiennaTheGr8]

... one cannot write for some reason a $u$ in square brackets within the forum software ;-)), $[T]=\text{K}$, and $[k_{\text{B}}]=\text{J}/\text{K}$.

$[ u ]$

 

[vanhees71]

How did you do it? If I try, I get

####

 

[Ibix]

How did you do it? If I try, I get

####

u in square brackets is BBCode for underline. Anything that does nothing visually but breaks up the string will work. @SiennaTheGr8's approach is to write spaces between the brackets and the u (i.e. [ u ]), which works because LaTeX ignores spaces but BBCode doesn't. Outside LaTeX, changing text colour to black is the usual trick - [[COLOR=black]u[/COLOR]] renders as [u].