I Get Relation from Stress-Energy Tensor Def.

[fab13]

Starting from the following definition of stress-energy tensor for a perfect fluid in special relativity :

$${\displaystyle T^{\mu \nu }=\left(\rho+{\frac {p}{c^{2}}}\right)\,v^{\mu }v^{\nu }-p\,\eta ^{\mu \nu }\,}\quad(1)$$

with $v^{\nu}=\dfrac{\text{d}x^{\nu}}{\text{d}\tau}$ and

$V^{\nu}=\dfrac{\text{d}x^{\nu}}{\text{d}t}$ (we have $v^{\nu}=\gamma\,V^{\nu}$)

So, finally, I have to get the following relation :

$$\dfrac{\partial \vec{V}}{\partial t} + (\vec{V}.\vec{grad})\vec{V} = -\dfrac{1}{\gamma^2(\rho+\dfrac{p}{c^2})} \bigg(\vec{grad}\,p+\dfrac{\vec{V}}{c^2}\dfrac{\partial \rho}{\partial t}\bigg)\quad(2)$$

To get this relation, I must use the conservation of energy : $\partial_{\mu}T^{\mu\nu}=0\quad(3)$

If someone could help me to find the equation $(2)$ from $(1)$ and $(3)$, this would be nice to indicate the tricks to apply.

Regards

 

[Orodruin]

Can you show us what you have done so far? Also, I would suggest writing $\nabla$ instead of $\vec{grad}$ (or $\vec\nabla$ if you must).

 

[fab13]

I recognize in the left member of wanted relation $\quad(2)$ the Lagrangian derivative :

$$\dfrac{\text{D}\,\vec{V}}{\text{d}t}=\dfrac{\partial \vec{V}}{\partial t} + (\vec{V}.\vec{\nabla})\vec{V}\quad(4)$$

and I can rewrite $(1)$ with the $V^{\mu}$ components like :

$$T^{\mu\nu}=\left(\rho+\dfrac{p}{c^{2}}\right)\,\gamma^2\,V^{\mu}V^{\nu }-p\,\eta^{\mu\nu}\,\quad(5)$$

But from this point, I don't know how to make the link between $(4)$, $(5)$, $(3)$ (the divergence of stress-energy equal to zero), and $(1)$ ...

Any help is welcome