# The Stress part of the Stress - Energy - Momentum Tensor

1. Dec 19, 2010

### WannabeNewton

Hey guys I'm having trouble understanding the full nature of the energy momentum tensor. I understand the whole "matter flow and energy flow" affects the curvature of the space time metric being used. What I don't understand is the role of stress and shear - stress. Is the stress being described the stress that the mass places on space - time? Thanks.

2. Dec 19, 2010

### zeromodz

Remember how 0,1,2,3 correspond to different components of the Stress energy momentum tensor. mu can be thought of the energy time component and nu can be though of the momentum, x,y,z components.

Also remember, that pressure is the external push and stress is internal and pull. So we can think of pressure as T11, T22, and T33. Ever component against itself that does not include energy density is stress. Energy density, pressure, and stress also have the same units.

Energy density = Energy / Area * Distance
Energy density = Force * Distance / Area * Distance
Energy density = Force / Area
Pressure = Force / Area
Stress = Force / Area

3. Dec 19, 2010

### WannabeNewton

Cool thanks. Just another quick (kind of unrelated) question:
If one wants to solve the field equations for a particular Energy Momentum Tensor then would one introduce that tensor into the equations solving for the metric? I'm asking because doesn't the metric also determine energy - momentum distribution? Or do you use a metric that would best suit the kind of mass distribution you wish to solve for and then solve to find the metric of the space - time with that specific mass distribution? (Like in the case of the schwarzschild metric where you use static spherical mass distribution and the metric for spherical coordinates) - sorry for sounding like a completely confused idiot.

4. Dec 19, 2010

### yuiop

Hi WN, this is my very casual superficial understanding that I am posting in the hopes that some real expert can correct/refine what I am saying here. The stress is mechanical stress acting on a physical object and its effect on gravity acts in the opposite sense to the pressure terms. For example if we had a spherical shell with a nuclear bomb inside it and if the shell was strong enough that it could contain the explosion, then the gravitational measurements outside the sphere would be the same before and after the explosion. The increase in pressure inside the sphere due to the explosion results in increase in the (shear?) stress in the skin of the shell as it resists the pressure. The increased stress has an "anti-gravitational" effect that offsets the effects due to increased pressure. I think it would be impossible to design an anti-gravity machine based on the negative stress effect because it would be impossible to design a system that had significant stress without it having more significant pressure/energy/mass. Like I said it is just a superficial understanding that may be way off the mark, so feedback would be welcome and might be helpful for both of us.

5. Dec 20, 2010

### Q-reeus

The stress in the shell will be bi-axial tension - which is not shear but negative pressure. Some links dealing with that sort of thing: http://arxiv.org/abs/gr-qc/0505040, http://arxiv.org/abs/gr-qc/0510041 The role of shear in the energy-momentum stress tensor was my first question on this forum. Apart from my own observation that static shear stress in a solid aught to make no contribution (shear can be resolved into orthogonal tensile and compressive stresses that cancel as sources of 'gravity'), no satisfactory answer was offered. There seems to be some link to viscosity (turbulence in fluids?), but it is hard to find anything clear and specific.

6. Dec 20, 2010

### yuiop

The wikipedia page on the stress-energy tensor http://en.wikipedia.org/wiki/Stress–energy_tensor sheds some light on this.

It states that "$T^{ii}$ (not summed) represents normal stress which is called pressure when it is independent of direction." where i ranges from 1 to 3 as indicated by the diagonal green "pressure" band in the diagram.

These may be a key observation. The "bi-axial" forces in the shell are classed as stress, but if the forces are equal in all directions then it is classed as pressure. Would I be right in thinking that if $T^{11}=T^{22}= T^{33}$ then that represents pressure, but if those elements are not equal, then it represents stress?

Wikipedia also states the orange triangular sections of the tensor diagram [itex]T^{ik}, \quad i \ne k[/tex] represents shear stress so the shear stress components are distinct from the normal stress components. The same components are sometimes described as viscosity in other references and presumably the distinction depends on whether we are talking about solids or liquids, even though that distinction is not always clear cut.

Wikipedia also hints that the "box" labelled "momentum flux" in the above diagram is basically the same as the stress tensor used in engineering.

Also, if we had a large planet sized sphere, would I be right that in thinking that a stress-energy tensor cannot represent the whole object but just some small finite (infinitesimal?) element of the large object? Is a metric basically the result of summing numerous stress-energy tensors that make up the composite whole?

Last edited: Dec 21, 2010
7. Dec 21, 2010

### Q-reeus

You are correct in that strictly speaking pressure is defined as uniform triaxial stress. What I had in mind was that for a thin shell incapable of supporting any radial stress, uniform bi-axial stress can be thought of as a 2D 'pressure'. For a thick shell circumferential stress differentials in the radial direction apply, though there is still no shear.
That's the point - it's all very vague as to how one specifically applies the off-diagonals as source terms in T. Haven't come across one single reference. If someone on this forum can point to one, please inform! As I have noted, uniform shear in solids is a null contribution. For fluids, what happens? There will be heat generation. But that represents energy which automatically is subsumed in the T00 term. There will be a velocity gradient - laminar flow or vortex motion (spin). But that is automatically subsumed in the momentum flow terms T10,T20,T30. Integrating a shear gradient in any direction leads to stress differentials, which are automatically subsumed in the diagonals. So what's left? Nothing as far as I can see. Can't help feeling these terms are there as formal 'padding', and have no fundamental significance in the way the other terms do. As an outsider though could be quite wrong.
Correct.
About the only objects I could think of where your observation would not be true would be say a rotating thin hoop under uniform circumferential tensile stress, or a balloon inflated under uniform pressure. Though even here one would still need to integrate over the entire object to get the contribution to the curvature terms exterior to the source.

Last edited: Dec 21, 2010