# Physical interpretation of the stress energy momentum tensor

1. Nov 13, 2014

### space-time

First, I'd like to thank everyone that has helped me thus far in deriving the general relativistic tensors for the Morris-Thorne wormhole metric in an orthonormal basis. I have finally done it and grasped that concept. Now that I have done that, my new stress energy momentum tensor for this metric is as follows:

T00 and T11 = -b2c4 / (8πG)(b2 + l2)2

T22 and T33 are the same thing but they are positive.

Now that this has been derived, I would like to know how to physically interpret this tensor. Does this mean that if I obtain a ball of a ball of negative energy that is of the density -b2c4 / (8πG)(b2 + l2)2 , that I could create a wormhole of radius b?

For example, if I want to create a wormhole of radius 10 (not worrying about units right now), then I'd need an energy ball that has a density of:

(-100c4) / (8πG)(100 + l2)2

Also, if I got a ball of such energy density, would that guarantee that the ball would automatically the angular momentum fluxes of the same magnitude that is depicted in the SET?

Finally, I know that l is the radial coordinate, but the radial coordinate with respect to what? Example: In normal spherical coordinates, the radial coordinate is the distance from the center of the sphere.

In this wormhole, what would be the "center"? Is the radial coordinate the distance away from the center of the wormhole itself? Is the actual wormhole assumed to be spherical since this metric is in a spherical basis?
Or perhaps is l the length of the wormhole (ie. If I want to take a shortcut through space that would normally be a distance of 100 meters that l = 100) ?

2. Nov 14, 2014

### space-time

Since it has been well over 24 hours and nobody has replied, perhaps I should try to rephrase my questions in a clearer fashion.

1. We know that l is the radial coordinate. We know that the radial distance is the distance away from the center. What exactly would be the center in the case of this wormhole metric? Are the ends of the wormhole assumed to be spherical, and the radial distance is the distance away from the center of these ends of the wormhole? Please tell me what exactly is the center in this case.

2. Alternatively, l could be the length of the wormhole tunnel. Is this the case as opposed to option 1?

3. Every element on the diagonal of my stress energy momentum tensor (orthonormal basis) is:

b2c4 / (8πG)(b2 + l2)2

Note: The angular elements are positive and the temporal/radial elements are negative, but they all look exactly like this aside from sign differences.

Now, if I want a wormhole with a throat that has a radius b = 10 and and a radial distance l = 10, would I need an energy ball with an energy density of:

-100c4 / (320000πG) (I am ignoring units for now) ? Note: These are the numbers you get if you plug 10 into b and l.

4. If I were to obtain said ball of negative energy, would I have to make extra manual effort to make sure that the momentum fluxes in the radial and angular directions were equal to the elements in the stress energy momentum tensor, or would obtaining this negative energy ball with the specifications listed in number 3 already make it so that the ball already inherently had the correct momentum fluxes?

3. Nov 14, 2014

### pervect

Staff Emeritus
I don't have anything more to add than what I tried to say last time. I have to conclude that it wasn't understood as you keep asking :(.

4. Nov 14, 2014

### Staff: Mentor

space-time, some of your questions are really questions about the coordinates and the metric, not about the stress-energy tensor. Rather than try to address them one by one, let me run through how one would investigate the meaning of the coordinates, given an expression for the metric. (Note that I'm using units where $G = c = 1$ to keep the formulas from being cluttered.)

The wormhole metric in question is

$$ds^2 = - dt^2 + dl^2 + \left( b^2 + l^2 \right) \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$$

Now we consider intervals with various coordinates held constant. First, consider an interval with all coordinates except $t$ held constant, i.e., only $dt$ is nonzero. Then we have $ds^2 = - dt^2$, indicating that the coordinate time $t$ is equivalent to proper time for an observer at rest in this coordinate chart (i.e., whose spatial coordinates are all constant).

Next, consider an interval with only $dl$ nonzero. We'll defer for a moment the question of whether this is a "radial" spacelike curve, and just note that for such an interval we have $ds^2 = dl^2$, indiating that the coordinate $l$ directly represents spatial distance in the $l$ direction.

Next, consider a 2-surface at constant $t$ and $l$. This gives an interval $ds^2 = \left( b^2 + l^2 \right) \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$, and if we integrate this over the entire range of $\theta$ and $\phi$, we get the area of the 2-surface: $A = 4 \pi \left( b^2 + l^2 \right)$. This is clearly a 2-sphere with radius $R = \sqrt{ b^2 + l^2 }$. The entire spacetime is simply a 2-parameter group of such 2-spheres, with each 2-sphere in the group being labeled by the pair of coordinates $t, l$. Furthermore, since the metric is independent of $t$, we can consider each hypersurface of constant $t$ as an identical one-parameter group of 2-spheres, labeled by $l$.

Now, what range of areas does this one-parameter group of 2-spheres cover? Obviously the area increases without bound as $l \rightarrow \infty$. But what happens when $l = 0$? As you can see from the above, at $l = 0$ we have, not a point, but a 2-sphere with radius $R = b$ and area $A = 4 \pi b^2$. The 2-spheres can never get any smaller than that, because $l^2$ can never be negative. So there is no "center" in the usual sense in this metric. Instead, as you proceed from $l = \infty$ to $l = 0$, you are passing through 2-spheres with smaller and smaller area; then, at $l = 0$, you are at a 2-sphere with minimum area; then, as you pass from $l = 0$ to $l = - \infty$, you pass through 2-spheres of larger and larger area again, but in a different "universe" (or at least a different part of the universe) than the one you started out in.

I'll address the stress-energy tensor in a separate post.

Last edited: Nov 14, 2014
5. Nov 15, 2014

### Staff: Mentor

For the physical interpretation of any stress-energy tensor, the best components to use are the mixed components, with one upper and one lower index. (Again, I'm using units where $G = c = 1$.)

The mixed components all have the same magnitude, as you noted; we'll call this magnitude $\rho = b^2 / \left( b^2 + l^2 \right)^2$. Then the mixed components are

$$T^0{}_0 = \rho$$

$$T^1{}_1 = - \rho$$

$$T^2{}_2 = \rho$$

$$T^2{}_2 = \rho$$

(Note the sign difference in the 0-0 component, compared to your OP.)

The physical interpretation of this is simple, since it is diagonal, so there is no momentum or energy flow and no shear stress. This is simply a "fluid" at rest in these coordinates, with positive energy density $\rho$, negative "radial" pressure (i.e, tension) $\rho$, and positive tangential pressure $\rho$. ("Radial" is in quotes because, as shown in my previous post, there is no "center" in this spacetime in the ordinary sense, so $l$ is not quite a standard radial coordinate.)

In other words, if you are moving through the wormhole, starting at $l = \infty$ and moving inward towards $l = 0$, you will measure an increasing energy density around you, and increasing pressure in the tangential direction (i.e., perpendicular to your motion), but increasing tension in the direction of your motion (i.e., you will feel the "fluid" pulling you along instead of pushing against you, as an ordinary fluid's pressure would). When you reach $l = 0$, the energy density, pressure, and tension are all at a maximum value of $1 / b^2$ (or $c^4 / 8 \pi G b^2$ in conventional units). So the constant $b$, in addition to being the radius of the smallest 2-sphere (the one at $l = 0$), is also a measure of the maximum energy density, pressure, and tension associated with the wormhole.