# Question about the stress-energy tensor

• I
• GR191511

#### GR191511

I came across a statement in《A First Course in General Relativity》on page 97 which confused me.It read:"if the forces are perpendicular to the interfaces,then##T^i{^j}##will be zero unless ##i=j##".
Where the ##T## is stress-energy tensor,##T^i{^j}##is the flux of i momentum across the j surface.

I came across a statement in《A First Course in General Relativity》on page 97 which confused me.It read:"if the forces are perpendicular to the interfaces,then##T^i{^j}##will be zero unless ##i=j##".
Where the ##T## is stress-energy tensor,##T^i{^j}##is the flux of i momentum across the j surface.
What confuses you about the statement?

What confuses you about the statement?
If the forces are perpendicular to the interfaces,then##T^i{^j}##will be zero?Why?

If the forces are perpendicular to the interfaces,then##T^i{^j}##will be zero?Why?
The spatial components ##T^{ij}## are in essence the stress tensor, which means that the components are the ##i## component of the force across an interface in the ##j## direction.

Ibix
The spatial components ##T^{ij}## are in essence the stress tensor, which means that the components are the ##i## component of the force across an interface in the ##j## direction.
Oh! I see...Thank you！

if the forces are perpendicular to the interfaces
"Perpendicular to the interface" is potentially a misleading phrase.

Does that mean perpendicular to the plane of the interface or perpendicular to the normal to the plane of the interface? Clearly the author means the former. But if I am trying to associate a direction with an interface so that I can judge perpendicularity, the direction I would would immediately choose is the normal. Perpendicular to that direction has a meaning opposite to the author's intent.

I struggled for a minute or two trying to figure out how the author's statement could possibly be correct until I caught on to the intended interpretation.

GR191511
The direction of a surface by definition is along the normal (with the sign choice arbitrary of course).

GR191511
with the sign choice arbitrary of course
Have you told the divergence theorem this?

vanhees71
In the divergence theorem we have a sign convention followed in all except of one textbook source I know, i.e., with the surface-normal vectors of the boundary ##\partial V## pointing out of the volume, and (in Cartesian coordinates) ##\mathrm{div} \vec{A}=+\vec{\nabla} \cdot \vec{A} = \partial_j A^j##.

The only exception is Max von Laue's textbook on relativity. There he uses the convention to let the surface-normal vectors pointing inwards. It drives you nuts, when used for years to the "standard convention", but after all, it's just a convention!

malawi_glenn and dextercioby