Understanding the stress-energy tensor

[Higgsono]

I have trouble understanding some terms in the stress-energy-tensor. For instance T^(12) stands for the flux of the x-component of momentum in the y-direction. But what does it means for the x-component of momentum to flow in the y direction? Since momentum is a vector should't the x-component always point in the x-direction?

 

[Dale]

But what does it means for the x-component of momentum to flow in the y direction? Since momentum is a vector should't the x-component always point in the x-direction?

This is a shear stress. In a shear stress you have a force in the x direction on a face which is normal to the y direction. The x component of the force does always point in the x direction, but it can cross a face which is normal to the y direction.

 

[Higgsono]

This is a shear stress. In a shear stress you have a force in the x direction on a face which is normal to the y direction. The x component of the force does always point in the x direction, but it can cross a face which is normal to the y direction.

So what does this mean for say an electromagnetic wave propagating through vacuum or a medium?

 

[Dale]

So what does this mean for say an electromagnetic wave propagating through vacuum or a medium?

Same thing. It is still a stress. The stress due to an electromagnetic field is given by the Maxwell stress tensor

https://en.m.wikipedia.org/wiki/Maxwell_stress_tensor

The Maxwell stress tensor is the space space part of the electromagnetic stress energy tensor

https://en.m.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

 

[stevendaryl]

I have trouble understanding some terms in the stress-energy-tensor. For instance T^(12) stands for the flux of the x-component of momentum in the y-direction. But what does it means for the x-component of momentum to flow in the y direction? Since momentum is a vector should't the x-component always point in the x-direction?

In 3 dimensions, imagine that you have lots of particles moving around. You want to compute the flux of y-momentum in the x-direction. The way you do it is to consider a membrane that is set up parallel to the y-z plane, as shown in the figure. Let $\Delta \overrightarrow{P}$ be the momentum that passes through the membrane from left to right during a small time interval $\Delta t$. The momentum flux in the x-direction will then be: $\frac{\Delta \overrightarrow{P}}{A \Delta t}$ (change in momentum per unit area per unit time).

So how do we calculate $\Delta \overrightarrow{P}$? Well, in the simplest case, the particles are non-interacting. So the amount of momentum passing through the membrane is just the average momentum of each particle that passes through times the number of particles.

So we have: $\Delta \overrightarrow{P} = \overrightarrow{P}_{av} \Delta N$

The number of particles that pass through the membrane during time $\Delta t$ is just:

$\Delta N = \rho V^x_{av} A \Delta t$

It's proportional to the area of the membrane, $A$, the average velocity in the direction perpendicular to the membrane, $(V^x)_{av}$ and the number of particles per unit volume, $\rho$, and the time $\Delta t$. So putting it all together:

$\frac{\Delta \overrightarrow{P}}{A \Delta t} = \frac{\overrightarrow{P}_{av} \rho V^x_{av} A \Delta t}{A \delta t} = \overrightarrow{P}_{av} \rho V^x_{av}$

The flux of y-momentum in the x-direction would then be:

$P^y_{av} V^x_{av} \rho$

Nonrelativistically, $P^y = m V^y$, so you have:

$T^{yx} = V^y_{av} V^x_{av} \rho_{m}$ (where $\rho_m$ is $m \rho$, the mass density, rather than the number density).

Relativistically, $P^y = E/c^2 V^y$ (where $E$ is the relativistic energy), so you have:

$T^{yx} = V^y_{av} V^x_{av} \rho_{E}/c^2$ (where $\rho_E$ is $E \rho$, the energy density, rather than the number density).

 

[Higgsono]

In 3 dimensions, imagine that you have lots of particles moving around. You want to compute the flux of y-momentum in the x-direction. The way you do it is to consider a membrane that is set up parallel to the y-z plane, as shown in the figure. Let $\Delta \overrightarrow{P}$ be the momentum that passes through the membrane from left to right during a small time interval $\Delta t$. The momentum flux in the x-direction will then be: $\frac{\Delta \overrightarrow{P}}{A \Delta t}$ (change in momentum per unit area per unit time).

So how do we calculate $\Delta \overrightarrow{P}$? Well, in the simplest case, the particles are non-interacting. So the amount of momentum passing through the membrane is just the average momentum of each particle that passes through times the number of particles.

So we have: $\Delta \overrightarrow{P} = \overrightarrow{P}_{av} \Delta N$

The number of particles that pass through the membrane during time $\Delta t$ is just:

$\Delta N = \rho V^x_{av} A \Delta t$

It's proportional to the area of the membrane, $A$, the average velocity in the direction perpendicular to the membrane, $(V^x)_{av}$ and the number of particles per unit volume, $\rho$, and the time $\Delta t$. So putting it all together:

$\frac{\Delta \overrightarrow{P}}{A \Delta t} = \frac{\overrightarrow{P}_{av} \rho V^x_{av} A \Delta t}{A \delta t} = \overrightarrow{P}_{av} \rho V^x_{av}$

The flux of y-momentum in the x-direction would then be:

$P^y_{av} V^x_{av} \rho$

Nonrelativistically, $P^y = m V^y$, so you have:

$T^{yx} = V^y_{av} V^x_{av} \rho_{m}$ (where $\rho_m$ is $m \rho$, the mass density, rather than the number density).

Relativistically, $P^y = E/c^2 V^y$ (where $E$ is the relativistic energy), so you have:

$T^{yx} = V^y_{av} V^x_{av} \rho_{E}/c^2$ (where $\rho_E$ is $E \rho$, the energy density, rather than the number density).

Thanks, I think it made it clearer.

 

[vanhees71]

Wikipedia has it quite nicely:

https://en.wikipedia.org/wiki/Cauchy_stress_tensor