Extracting PDE's from the energy-momentum tensor

Question

I'm trying to understand part of the question below, I even have the solution for this question, but the solution doesn't matter for now since I cannot understand the question.

The energy-momentum tensor of a perfect fluid is given by $$T^{\mu\,\nu}=\left(\rho +\frac{p}{c^2}\right)u^\mu u^\nu- p\eta^{\mu\,\nu}\tag{1}$$ Where $\rho$ is the proper density, $p$ is the pressure and $u^{\mu}$ is the four-veloctity of the fluid.

Assuming that $u^iu^i\leqslant c^2$, and $p\leqslant \rho c^2$, the non-relativistic limit, then,

$$c^{-2}T^{0\,0}=\rho\tag{a}$$ $$c^{-1}T^{i\,0}=c^{-1}T^{0\,i}=\rho u^i\tag{b}$$ $$T^{i\,j}=\rho u^iu^j + p\delta^{i\,j}\tag{c}$$ Where $i=1,2,3$ only has spatial components, $(\hat e_x,\hat e_y,\hat e_z)$ in Cartesians, the $\mu$ and $\nu$ indices still carry four-components.

Extract the 4 PDEs contained in the tensor equation $\partial_{\nu}T^{\mu\, \nu} = 0$ (convert derivatives with respect to $x^0$ into derivatives with respect to time $t$).

To try to understand this question properly I want to check I can obtain eqns $(\mathrm{a})$, $(\mathrm{b})$, and $(\mathrm{c})$. I can see that if $\mu=\nu=0$ is substituted into $(1)$ then

$$T^{\mu=0,\,\nu=0}=\left(\rho +\frac{p}{c^2}\right)u^0 u^0- p\eta^{0,\,0}$$ Now, using the $p\leqslant \rho c^2$ approximation and the metric, $\eta^{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$, for which $\eta^{00}=1$, I find that $$c^2 T^{00}=\left(\rho c^2 + p\right)(u^0)^2-pc^2\ne (\mathrm{a})$$

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Proceeding anyway to try to reach $(\mathrm{b})$,

$$T^{\mu=i,\,\nu=0}=\left(\rho +\frac{p}{c^2}\right)u^i u^0- p\eta^{i,\,0}$$

Using $p\leqslant \rho c^2$ and $\eta^{i,\,0}=0$. Then multiplying both sides by $c^2$ as before, I find that $$c^2 T^{i0}=\left(\rho c^2 + p\right)u^iu^0\ne (\mathrm{b})$$

[The same story goes for $T^{0,i}$]

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As for $(\mathrm{c})$, I have no idea where to even start.

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To summarize, why can't I verify (or reproduce) eqns $(\mathrm{a})$ and $(\mathrm{b})$?

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Edit

In response to the comment below, I think I am starting to understand what it meant now, as the four-velocity, $u$ is,

$$u=\left(u^0, u^1, u^2, u^3\right)=\left(c, \frac{dx^1}{t}, \frac{dx^2}{t}, \frac{dx^3}{t}\right)$$

But the only way to get the norm, $(u^0)^2 - (u^1)^2 - (u^2)^2 - (u^3)^2$ equal to $c^2$ is if $u^1=u^2=u^3=0$ as $u^0 = c$.

The question now is, why should $u^1=u^2=u^3=0$?

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Edit 2

Now I can derive $(\mathrm{a})$ and $(\mathrm{b})$.

So to reach $(\mathrm{a})$ starting from $(1)$

$$T^{00}=\left(\rho + \frac{p}{c^2}\right)\left(u^0 \right)^2-p$$ as $\eta^{00}=1$, and using the fact that $u^0=c$ (first element of four-velocity). Now using the $\rho \gg \frac{p}{c^2}$ approximation $$T^{00}\approx\rho c^2-p$$ and then using $\rho c^2 \gg p$ means that $$T^{00}\approx \rho c^2$$ which leads straight to $$\color{blue}{c^{-2}T^{0\,0}=\rho}\tag{a}$$

To obtain $(\mathrm{b})$, starting from $(1)$,

$$T^{i\,0}=\left(\rho +\frac{p}{c^2}\right)u^i u^0- p\eta^{i\,0}$$ and using $\rho \gg \frac{p}{c^2}$ with the condition that $\eta^{i,0}=0$ leads to $$T^{i\,0}\approx\rho u^i u^0$$ insertion of $u^0 = c$ gives $$T^{i\,0}=\rho u^i c$$ for which the result $$\color{darkgreen}{c^{-1}T^{i\,0}=\rho u^i}\tag{b}$$ immediately follows. The precise same steps can be applied to show that $c^{-1}T^{0,i}=\rho u^i$ also.

Now a problem arises when I try to apply the same logic to reach equation $(\mathrm{c})$, so as usual, starting from $(1)$:

$$T^{ij}=\left(\rho + \frac{p}{c^2}\right)u^i u^j - p\eta^{ij}$$ using the $\rho \gg \frac{p}{c^2}$ approximation: $$\color{red}{T^{ij}\approx\rho u^i u^j - p\eta^{ij}}\ne \mathrm{(c)}$$ There are now two problems in trying to reach eqn $(\mathrm{c})$ from the red equation.

1. I don't understand how or why the $\eta$ became $\delta$ as their matrix representations are totally different: $$\eta=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix},\quad \delta=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$
2. Eqn $(\mathrm{c})$ has a positive sign but the equation in red has a negative sign.

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Once I have understood how to obtain eqn $(\mathrm{c})$ I will attempt to extract the differential eqns.

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Edit 3

For the sake of completeness, I have typeset the authors full solution:

$$\partial_ \nu T^{\mu\nu}=0$$ For $\mu=0,$ $$\partial_{\mu}T^{0 \nu}=c^{-1}\partial_t T^{00}+\partial_i T^{0 i}=c\left(\partial_t \rho+\partial_i\left[\rho u^i\right]\right)=0\tag{A}$$ whereas for $\mu=i,$ $$\partial_{\nu}T^{i \nu}=c^{-1}\partial_t T^{i0}+\partial_j T^{ij}=\partial_t \left[\rho \mu^i\right] +\partial_j\left(pu^iu^j+p\delta^{ij}\right)\tag{B}$$ where $\delta^{ij}=1$ for $i=j$ and is zero otherwise. Accordingly, $$\partial_{\nu}T^{i \nu}=\rho\partial_t u^i+\rho u^j \partial_j u^i+\partial_i p+\left[\partial_t\rho+\partial_j\left(\rho u^j\right)\right]u^i=0\tag{2}$$ by substituting $(\mathrm{A})$ into $(\mathrm{B})$, and $(\mathrm{B})$ simplifies to $$\partial_{\nu}T^{i \nu}=\rho\partial_t u^i+\rho u^j \partial_j u^i+\partial_i p=0$$ These are simply the continuity and Euler equations of fluid mechanics. In vector calculus notation: $$\frac{\partial\rho}{\partial {t}}+\nabla\cdot\left(\rho {\bf {u}}\right)=0,$$ $$\frac{\partial{\bf{u}}}{\partial t}+\left({\bf{u}}\cdot \nabla\right){\bf {u}}+\frac{\nabla \rho}{\rho}=0$$

Now I have just 2 concerns regarding question and solution in this post.

1. How did the $\eta^{\mu\nu}\,$in eqn $(1)$ becomes $\delta^{ij}$ in $(\mathrm{c})$?
2. Why can equation $(\mathrm{A)}$ be set to zero but not eqn $(\mathrm{B})$? I ask this because in order for $\partial_ \nu T^{\mu\nu}=0$ to be satisfied we must sum over $\nu$ for $\mu = 0$ and sum over $\nu$ also for $\mu=i = 1,2,3$ (this ensures everything is in the sum). So I think the eqn $(2)$ above should be the sum of $(\mathrm{A)}$ and $(\mathrm{B)}$.

Answer 1

Addressing $\boldsymbol 1$:

There are two conventions for the Minkowski metric. Because it avoids as many minus signs as possible, the convention used in applied mathematics (and as such, the one I am more comfortable with) is $$\boldsymbol \eta=\operatorname{diag}(-1,1,1,1)$$ However, in physics, the more common convention is $$\boldsymbol \eta=\operatorname{diag}(1,-1,-1,-1)$$ The reason that this choice is used is so that the line element can be written as $$\mathrm ds^2=\eta_{\mu\nu}\mathrm dx^\mu\mathrm dx^\nu$$ And not with the somewhat awkward minus sign: $$-\mathrm ds^2=\eta_{\mu\nu}\mathrm dx^\mu\mathrm dx^\nu$$ So, starting with $$T^{\alpha\beta}=\left(\rho+\frac{p}{c^2}\right)u^\alpha u^\beta-p\eta^{\alpha\beta}$$ Considering the spacial components $i,j\in\{1,2,3\}$, and using the physics convention for the metric signature, you get $$T^{ij}=\left(\rho+\frac{p}{c^2}\right)u^i u^j\color{red}{+}p\delta^{ij}$$ Since $\eta^{11}=\eta^{22}=\eta^{33}=-1$, under this convention.

Addressing 2:

Please first note that you should write $\color{red}{\nabla}_\mu T^{\mu\nu}=0$ and not $\partial_\mu T^{\mu\nu}=0$. These are covariant derivatives, and should be written as such.

I think you are confusing yourself here. The equation $$\nabla_\mu T^{\mu\nu}=0$$ Means that $$\nabla_\mu T^{\mu 0}=\nabla_\mu T^{\mu 1}=\nabla_\mu T^{\mu 2}=\nabla_\mu T^{\mu 3}=0$$ So the two choices $\mu=0$ and $\mu=i\in\{1,2,3\}$ will produce two different equations for you. One is a scalar equation and one is a vector equation and they are $$\frac{\partial \rho}{\partial t}+\vec{\nabla}\cdot(\rho\vec{u})=0 \\ \frac{\partial \vec{u}}{\partial t}+\vec u\cdot \vec{\nabla}\vec{u}=-\frac{1}{\rho}\vec{\nabla}p$$ So in your question, $(\text{A})$ and $(\text{B})$ need to vanish independently, not just their sum.

Answer 2

Starting from where you left off for the first equation: $$c^2T^{00} = (\rho c^2+p)(u^0)^2 - pc^2,$$ we have the conditions: $$u^iu^i\leq c^2\text{ for }i\in\{1,2,3\}?\text{ }\text{ }\text{ and }\text{ }\text{ }p\leq \rho c^2.$$ If we assume that the first condition applies to $i=0$ as well, we get: $$c^2T^{00}\leq (pc^2 + \rho c^2)(c^2) - (\rho c^2)c^2 = 2\rho c^4 - \rho c^4 = \rho c^4.$$ Dividing both sides by $c^4$ gives $c^{-2}T^{00} \leq \rho$ as desired for (a).

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Then as you noted with the off diagonal components of the metric being zero, we get: $$c^2T^{i0} = (\rho c^2+p)u^iu^0\text{ }\text{ }\text{ and }\text{ }\text{ }c^2T^{0i} = (\rho c^2+p)u^0u^i.$$ The first condition $(u^i)^2 \leq c^2$ implies $|u^i|\leq |c|$ (taking square roots), but since we know both values are positive, this gives $u^i\leq c$. This (with $i=0$ again) together with the second inequality gives: $$c^2T^{i0} \leq (\rho c^2+\rho c^2)u^i(c)\text{ }\text{ }\text{ and }\text{ }\text{ }c^2T^{0i} \leq (\rho c^2+\rho c^2)(c)u^i$$ Which when dividing by $c^3$ gives: $$c^{-1}T^{i0} \leq 2\rho u^i\text{ }\text{ }\text{ and }\text{ }\text{ }c^{-1}T^{0i}\leq 2\rho u^i.$$

This unfortunately has a 2 in it.

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Lastly, for the third equation (having $i,j\neq 0$) the metric sub-matrix is just $\eta^{ij} = -\delta^{ij}$ so: $$T^{ij}\leq (\rho + \frac{\rho c^2}{c^2})u^iu^j - p\eta^{ij}$$ $$= 2\rho u^iu^j + p\delta^{ij},$$ which is what we want aside from the 2 again.

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*The only thing extra throughout the above that I assumed was that $(u^0)^2 \leq c^2$ as well. Physically this is a bound on time and is probably wrong but it makes the equations close to what you want!

As for the PDEs we are supposed to extract from $\partial_{\nu}T^{\mu \nu} = 0$, just calculate the partial for $\nu=0,i$ and add them together (as it is a summation). Since the result has variable $\mu$ still, there are four equations.

For $\nu = 0$ for example: $$\partial_0 T^{\mu 0} = \partial_0\bigg[(\rho+\frac{p}{c^2})u^{\mu}u^{0} - p\eta^{\mu 0}\bigg]$$ and continue evaluating according to which of these has $x^0$ dependence... good luck!