1
$\begingroup$

I am slightly confused about the relationship between the stress energy tensor and the force on a boundary. From what I have read the stress energy tensor can be written as: $$ \Bbb{T}=\begin{pmatrix} u & \vec N/c\\ \vec N/c & \tilde \sigma\end{pmatrix} \tag{1}$$ where $\tilde \sigma$ is the stress tensor. Now consider a boundary of normal $\hat n$, from fluid mechanics I know that the force acting on this boundary is given by: $$\vec t=\tilde \sigma \hat n\tag{2}$$ And thus I would expect the force in relativity to be given by: $$ \vec t=(\Bbb{T})_{ss}\hat n \tag{3}$$ where the subscript $ss$ denotes the space-space part. Consider an ideal fluid in a box centered on the origin and with sides parallel to the coordinate planes. The stress-energy tensor for such an ideal fluid is (in its rest frame): $$\Bbb{T}_{IF}=\begin{pmatrix}\rho c^2 & 0 & 0 & 0\\ 0 & p & 0 & 0 \\ 0&0&p&0\\ 0&0&0&p\end{pmatrix}\tag{4}$$ Which would predict a force on the side of the box with normal (into the box) of $-\hat e_x$ (i.e. the side which lies at $+$ve $x$) of: $$\vec t=-p \hat e_x\tag{5}$$ I.e. the boundaries of the box feel a force into the box itself-which is clearly wrong. What is going on here? i.e. how do we actually find forces on boundaries in special relativity and why does (3) not hold?

$\endgroup$
2
$\begingroup$

In continuum mechanics the interpretation of $\sigma$ is the folliwing. Consider a finite portion $C$ of continuum and a point $p$ on the boundary $\partial C$ (a closed regular surface) of that finite portion. Suppose that $n$ is the outward unit vectors at $p$ normal to $\partial C$. Then $\sigma(n)$ is the surface density of force acting on $C$ at $p$ due to the remaining external part of the whole body. For a perfect fluid, this force enters the portion $C$ throught $\partial C$, if the pressure is positive. All that means that the correct correspondence between relativity and continuum mechanics is $$T_{ij} = -\sigma_{ij}$$ for $i,j =1,2,3$ (where I am using the metric $-+++$).

$\endgroup$
0
$\begingroup$

Here is my take on the situation (with thanks to Valter Moretti for his answer.) In this answer summation is not implied by repeated indices.

In continuum mechanics the stress tensor can be defined as follows:

Stress Tensor: The tensor, $\sigma$, for which $\sigma^{ij}$ is the force per unit area in the $i$ direction acting on a surface with normal in the $j$ direction. ($i, j \in \{1,2,3\}$)

Whilst in relativity the stress-energy tensor can be defined as follows:

Stress Energy Tensor: The tensor, $T$, such that $T^{\mu \nu}$ is the flux of the $\mu$th component of 4-momentum through a surface of constant $x^\nu$. ($\mu, \nu \in \{0,1,2,3\}$)

I will not address the meaning of this definition since a descent explanation is given here. I will however, demonstrate how we can relate the space-space component of $T^{\mu \nu}$ to $\sigma^{ij}$. We consider a surface of constant $x^i$ (assumed to increase to the right) as shown in the figure below and take the case of a stream of particles density $n$ all moving at the same constant velocity $u$.

enter image description here

The momentum flux through this surface is going to be: $$T^{ii}=mnu^2$$ However the the force per unit area acting on the right hand side surface (that with normal in the direction of increasing $x^i$) is given by: $$\sigma^{ii}=-mnu^2$$. Analogous arguments hold for other components giving us: $$(T^{\mu \nu})_{ss}=-\sigma^{ij}$$ This explains the sign error in my question above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.