# Do all elements of the stress energy tensor really have the same units?

1. Oct 17, 2014

### space-time

Some people may remember awhile back when I made a thread showing how when I derived the Einstein tensor and the stress energy momentum tensor for a certain traversable wormhole metric, that the units of the energy momentum tensor were not the same for each element and how a couple of the elements had the wrong units.

Now here was the metric:

ds2= -c2dt2 + dl2 + (b2 + l2)(dΘ2+ sin2(Θ)dΦ2)

Now in my first attempt, here was the coordinate basis and the metric tensor that I used:

x0= ct
x1= l (the radial coordinate)
x2= Θ
x3= Φ

b is just the radius of the throat of the wormhole.

g00= -1
g11= 1
g22= b2 + l2
g33= (b2 + l2)sin2(Θ)

Every other element was 0.

When I derived the Einstein tensor using this metric and then multiplied it by c4/(8πG) to get the stress energy momentum tensor, I got:

T00= (-b2c4)/[(8πG)(b2 + l2)2]

T11 just happened to equal the same thing as T00.

T22= (b2c4)/[(8πG)(b2 + l2)]

T33= (b2c4sin2(Θ))/[(8πG)(b2 + l2)]

Every other element was 0.

Now if you do dimensional analysis on these elements (I used SI units), you will notice that only T00 and T11 have the units of energy density/ momentum flux. Those units are:

Kg/(m * s2)

T22 and T33 however, had units of force instead:

(kg* m)/ s2 which is equal to the newton.

I was soon informed within the thread that I mentioned earlier, that the reason that this occurred was because Θ and Φ were angles whereas ct and l were lengths. The person then told me that I should rearrange my coordinate basis so that I was differentiating with respect to lengths rather than angles.

This led to my 2nd attempt with a slightly modified coordinate basis and metric tensor:

x0= ct
x1= l
x2= Θ(b2 + l2)¼
x3= Φ[(b2 + l2)sin2(Θ)]¼

g00= -1
g11= 1
g22= (b2 + l2)½
g33= sin(Θ)[(b2 + l2)½]

Now when I derived the stress energy momentum tensor in the same manner as before with this particular metric, I got this:

T00= [(-8b2 + 4l2)c4]/[(64πG)(b2 + l2)2]

T11= 0 (It just came out this way.)

T22= (4b2c4(b2 + l2)½)/[(64πG)(b2 + l2)2]

T33= [(4b2 - 4l2)c4sin(Θ)]/[(64πG)(b2 + l2)(b2 + l2)½]

Now when you do dimensional analysis on this stress energy momentum tensor, you will see that T00 once again has the proper units of energy density:

Kg/(m * s2)

T11 was 0, so that one is fine as well.

However, once again it is T22 and T33 that have different units. This time, the units for both of these two elements were:

kg / s2

Now these units do not equal those of force like in the first case, but I did notice something interesting here.

In both cases, it was always the T22 and T33 elements that had the wrong units. Additionally, the wrong units for these two elements would always be the same as each other in both cases. Furthermore, T00 and T11 in both cases were always the acceptable ones with the right units.

Having noticed this recurrence in both attempts, I deduced that this could not be coincidence and that perhaps the two angular elements were meant to have slightly different units after all.

What do you guys think of this? Do you have any explanation for the strange occurrence that I have noticed in both attempts? Am I doing something wrong?

P.S. Some people have recommended that I try using an orthonormal basis. I have tried to learn how to do that, but the farthest I've gotten is learning how to derive the 4 basis vectors. I don't know how to use that to get me to the Einstein tensor or even the Christoffel symbol for that matter. In fact, I've noticed that if I dot product all of those basis vectors with each other, it just gives me the inverse metric tensor that I already had in the first place. Also, how does this get me around the unit problem?

2. Oct 18, 2014

### aleazk

The asymmetry you get between the temporal/radial and the angular components of the SET has to do with the original asymmetry between the temporal/radial and the angular coordinates. If you use an orthonormal basis, you eliminate this asymmetry (this orthonormal basis is a non-coordinate basis, but keep in mind the usual inertial coordinates in flat spacetime, t, x, y, z, note the symmetry there). Nevertheless, what we know about tensors is that if we write the linear combination of the basis vectors with the components of the tensor, we get the real, invariant abstract geometric object.

So, do the following: take the components you calculated at the beginning (I think the result is right) and just make the linear combination with the $$\left\{ \mathrm{d}t\otimes\mathrm{d}t,\mathrm{d}l\otimes\mathrm{d}l,\mathrm{d}\theta\otimes\mathrm{d}\theta,\mathrm{d}\varphi\otimes\mathrm{d}\varphi\right\}$$
and this will give you the actual SET T because the components you calculated are simply the components of this tensor in this basis.

Now, look at this expression and regroup the terms so that you get some coefficients but now in linear combination with the (covector) orthonormal basis of the old thread. The coefficients are the components of the SET in this new basis. Notice that they come with the right units. That's because the new basis is now 'symmetrical' in this sense.

Edit: I know this 'symmetry/asymmetry' explanation is a little sloppy, but try to make the calculation first so you can see it actually works. The origin of the 'symmetry/asymmetry' explanation surely has to do with some unit conventions that we are overlooking here. To be honest, I'm lazy to think about tricky unit conventions right now! :p

BTW, be sure to check the paper by Thorne and Morris, 'Wormholes in spacetime and their use for interstellar travel: a tool for teaching GR'. They first introduced this metric in that paper. It contains interesting exercises about how to interpret the metric physically, so it wil be useful for you if you are learning GR.

Last edited: Oct 18, 2014
3. Oct 18, 2014

### space-time

Just so I understand you, I need to verify what this symbol here is asking for: ⊗

For example: dt ⊗ dt What exactly do I do for this expression? Is this where I do g00dx0dx0 while having dx0 equal to some normalized vector like <1,0, 0, 0>? If not, then what exactly does that expression require me to do?

4. Oct 18, 2014

### aleazk

It's the tensor product. Those 'differentials' are actually dual vectors, dt ⊗ dt is a tensor of type (2,0), like the metric g, and these tensor products of the basis vectors form a basis of the space of tensors of type (2,0). The usual line element is just the expansion of the metric in this coordinate basis. That's the rigorous meaning of that expression. Don't be afraid, it's just linear algebra.

From what you say, I'm guessing you learned tensors using the old definition (components that change with some formula). You will need the definition from multilinear algebra in this case, since this orthonormal basis is not a coordinate basis. Check this: http://en.wikipedia.org/wiki/Multilinear_map

Last edited: Oct 18, 2014
5. Oct 18, 2014

### Staff: Mentor

Your coordinates do not have the same units so none of the GR tensors will have components with the same units either. Your results are correct as far as I can tell at a glance.

6. Oct 18, 2014

### pervect

Staff Emeritus
OK.

This isn't right, there should be no square root here. Because of the square root, you are analyzing a different metric :(.

In addition, $x^0, x^1, x^2, x^3$ are all coordinates. None of them is a basis vector or a cobasis vector.

A common way of writing cobasis vectors in index free notation is just dt, dx, dl, etc. There are apparently some conceptual issues here, I'm guessing your textbook isn't explaining them well or you are working on your own, and I'm not sure how to go about rectifying the issue

Meanwhile, I can say that all your 1/4 exponents should be 1/2.

I'll think about trying to explain the conceptual issues some, but you should already have noted that your results aren't the same as that of the webpage you cited for the source of this metric.

Basically, you should get plus or minus $\frac{b^2}{\left( b^2 + l^2 \right)^2}$ for all the components of the Einstein tensor. This means the curvature tensor has units of 1/lenght^2.

When you multiply the Einstein tensor by $\frac{c^4}{8 \pi G}$ you convert 1/m^2 to joules/m^3 which is equivalent in units to the units of the pascal unit of pressure.

I'll append a maxima batch file for this metric as an example. Maxima has some weird sign and ordering issues though. To run it, you'll need to rename the .txt file to a .mac file, then use the "batch file" option of maxima. I've tried to comment it.

#### Attached Files:

• ###### Maxima_Worm.txt
File size:
2.4 KB
Views:
63
Last edited: Oct 18, 2014
7. Oct 18, 2014

### pervect

Staff Emeritus
I'm going to use some non-tensor notation here and hope for the best.

Lets suppose you have polar coordinates, r and $\theta$. Both r and $\theta$ are scalars, not vectors. Unit wise, r has units of length, $\theta$ has units of radians.

Referring to the following image

$\hat{r}$ and $\hat{\theta}$ are vectors, drawn using the usual "arrow" notation. Both of them have units of length. You can see that they are not the same as the coordinates r and $\theta$, though they are related.

dr and d$\theta$ are a bit ambiguous. Usually they are just scalars. But if you interpret them as a map from a vector to a scalar, then you can regard them as dual vectors.

When regarded as maps from vectors (graphically represented as arrows with tails) to scalars, the cobasis vectors are dr and d$\theta$. They are frequently graphically represented via a stack-of-plates.

The intent of this image is to represent a geometrical structure that maps a vector (the thing with the arrow and tail) to a scalar (the number of plates pierced by the vector).

8. Oct 18, 2014

### aleazk

Ok, I will do the calculation then:

The SET is also a tensor of type (2,0); the components you calculated are the components of this tensor in the dt ⊗ dt, etc., basis, i.e.,

$$T=T_{tt}\mathrm{d}t\otimes\mathrm{d}t+T_{ll}\mathrm{d}l\otimes\mathrm{d}l+T_{\theta\theta}\mathrm{d}\theta\otimes\mathrm{d}\theta+T_{\varphi\varphi}\mathrm{d}\varphi\otimes\mathrm{d}\varphi$$

So, we simply make a change of basis (remember, the T is just a vector in some vector space of tensors of type 2,0):

$$T=T_{tt}\mathrm{d}t\otimes\mathrm{d}t+T_{ll}\mathrm{d}l\otimes\mathrm{d}l+T_{\theta\theta}\frac{(b^{2}+l^{2})}{(b^{2}+l^{2})}\mathrm{d}\theta\otimes\mathrm{d}\theta+T_{\varphi\varphi}\frac{\sin^{2}\theta(b^{2}+l^{2})}{\sin^{2}\theta(b^{2}+l^{2})}\mathrm{d}\varphi\otimes\mathrm{d}\varphi$$

$$=T_{tt}\mathbf{e}_{0}\otimes\mathbf{e}_{0}+T_{ll}\mathbf{e}_{1}\otimes\mathbf{e}_{1}+\frac{T_{\theta\theta}}{(b^{2}+l^{2})}\mathbf{e}_{2}\otimes\mathbf{e}_{2}+\frac{T_{\varphi\varphi}}{\sin^{2}\theta(b^{2}+l^{2})}\mathbf{e}_{3}\otimes\mathbf{e}_{3}$$

(check the formula for the orthonormal basis vectors in the other thread; without the c, I made a mistake there; also, the argument of the sin is just the angle, I should have written the bs and ls first)

This means that the components of the SET in this new basis are:

$$T_{00}=T_{tt}$$

$$T_{11}=T_{ll}$$

$$T_{22}=\frac{T_{\theta\theta}}{(b^{2}+l^{2})}$$

$$T_{33}=\frac{T_{\varphi\varphi}}{\sin^{2}\theta(b^{2}+l^{2})}$$

Which are the values in Thorne's paper, since he used the orthonormal basis, and have the right units (check @pervect's comments in the old thread about why one has to use an orthonormal basis for these units issues) . The other way is to calculate the curvature using this orthonormal basis with the tetrad method or with Maxima (which I guess uses the same method).

9. Oct 19, 2014

### space-time

I think this will help me understand what to do: First I should let you guys know that I am actually still a senior in high school whose most recent math was Calculus 2. Everything that I know about tensor analysis, linear algebra and most other university level maths are self taught by using this forum and video playlists from places like youtube and khan academy. Having said that, I will tell you where I am thus far in deriving tensors in an orthonormal basis.

First off: I am using my original metric tensor:

g00= -1

g11= 1

g22= b2 + l2

g33= (b2 + l2)sin2(Θ)

Step 2: Derive the basis vectors by doing gijdxidxj where
dxi and dxj are vectors that are orthogonal and normalized. Here goes:

ct = e0= <1, 0, 0, 0 >
L = e1= <0, 1, 0, 0 >
Θ= e2= < 0, 0, 1/squrt(b2 + l2) , 0 >
Φ= e3= <0, 0, 0, 1/ [sin(Θ) * squrt(b2 + l2)] >

Those are my basis vectors. If you ask how I got these, then here is how:

Take e2 for example:

g22* < 0, 0, 1/squrt(b2 + l2) , 0 > ⋅ < 0, 0, 1/squrt(b2 + l2) , 0 > = 1

Therefore e2 is normalized. Considering the fact that this particular metric tensor was only nonzero on the diagonal, I already know that these basis vectors are orthogonal.

Now that I have my basis vectors, where do I go from here? That is where I am having trouble. Also, if you look at my basis vectors, you'll notice that if you dot product them together, then they equal the elements of the inverse metric tensor.

10. Oct 19, 2014

### space-time

By the way, my reply above this particular post was for you mainly. I just didn't quote because anyone who has an answer will suffice.