# Understanding the stress-energy tensor

• I
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
Mentor
2020 Award
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.

vanhees71
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?

stevendaryl
Staff Emeritus
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).

Ibix and vanhees71
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.