# Extracting PDE's from the energy-momentum tensor

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 $$c^2 T^{00}=\left(\rho c^2 + p\right)(u^0)^2-pc^2\ne (\mathrm{a})$$

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 $$T^{0,i}$$]

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

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

## 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 $$u^1=u^2=u^3=0$$?

## 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.

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

## 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 $$(\mathrm{B)}$$.
• Notice $u^\mu u_\mu = (u^0)^2 - (u^1)^2-(u^2)^2 - (u^3)^2 = c^2$. In non-relativistic limit, $u^0 \sim c$. So in $(a)$, the leading piece of $p$ cancel out. Other equations should be similar. Commented May 30, 2022 at 9:37
• @achillehui Hi thanks for your reply, however, the indices are both upper indices in $u^\mu u^\nu$, so there is no summation as far as I can see. It also doesn't explain where the other $\left((\mathrm{b})\, \text{&}\, (\mathrm{c}) \right)$ equations come from. Commented May 30, 2022 at 10:10
• Question 1 in Edit 3 is answered by realizing that the switch from Greek letters $\mu, \nu$ to Latin letters $i, j$ indicates a switch from space-time coordinates $\mu, \nu\in \{0,1,2,3\}$ to just spacial coordinates $i,j\in\{1,2,3\}$. So, $-\eta^{ij}\ = \delta^{ij}$ when $i,j\in \{1,2,3\}$. This is a common convention to use. Commented Jun 4, 2022 at 23:15
• You have to use the two facts just before (a) to get to the final result (continuity equation + Euler equation for the perfect fluid). See PhysicsSE physics.stackexchange.com/q/351727/226902 physics.stackexchange.com/q/185833/226902 Commented Jun 8, 2022 at 14:47

## 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.

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.

• Thank you very much for your answer, I'm sorry it took a while for me to respond. Just out of curiosity, when you write "Please first note that you should write $\color{red}{\nabla}_\mu T^{\mu\nu}=0$ and not $\partial_\mu T^{\mu\nu}=0$" Be that as it may, I can only type (word-for-word), an exact quote to the information I have been provided with, here is a screenshot image of the solution I typeset in my answer by the author. Are you suggesting that this screenshot (and therefore the quote at the end of my post are wrong)? Commented Jun 8, 2022 at 23:20
• @FutureCop Not wrong per se, but a bit sloppy in my opinion. Commented Jun 8, 2022 at 23:38

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).

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.

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.

*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!

• Hi there, thanks for your answer and sorry it took me a while to respond. I understand most of your answer apart from the last part. I have found an alternative way to derive $(\mathrm{a})$ and $(\mathrm{b})$, but I cannot get equation $(\mathrm{c})$. Commented May 31, 2022 at 12:59
• What about the last part? Commented May 31, 2022 at 18:26
• In the last line you write "continue evaluating according to which of these has $x^0$ dependence....", but I don't see any dependence on $x^0$, I'm assuming in space-time coordinates $x^0=ct$, but we don't have that in the equations in the question. Instead there is $u^0$, $u^1$, etc. I see only velocities not distances, that last sentence of the question I don't understand either, as it also mentions derivatives with respect to $x^0$. I'm just trying to understand the beginning of the question before getting to the end of it. Commented May 31, 2022 at 19:15
• So it seems the constants are $c$ and the metric expression (on manifolds with curvature the metric is spatially dependent), but aside from those we are left with $\rho$, $p$, and $u$. Each of these I would imagine have spacetime dependence. So for example if say $\rho=\rho(x_0,...,x_3) = x_1^2+3x_3-x_0$ or something then $\partial_0\rho = -1$ etc. So to continue you have to distribute the derivative and maybe apply product rule etc. At the time of writing, I figured that was best left for later! Commented May 31, 2022 at 23:59
• Okay thanks, when I get some free-time I will typeset the authors solution in my post. Commented Jun 1, 2022 at 0:07