# There may be some problems with dimensional analysis within the stress energy momentum tensor

1. Oct 2, 2014

### space-time

Recently, I used the metric for the traversable wormhole (the one in this link):
http://www.spacetimetravel.org/wurmlochflug/wurmlochflug.html

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

I derived the metric tensor from this space-time interval and then from there, I derived the Christoffel symbols, the Ricci tensor, the curvature scalar, and ultimately the Einstein tensor. Here was my Einstein tensor:

G00=(-b2c2)/(b2 + l2)2
G11=(-b2)/(b2 + l2)2
G22=(b2)/(b2 + l2)
G33=(b2sin2(Θ))/(b2 + l2)

Every other element was 0.

I then multiplied this Einstein tensor by (c4)/(8πG) in order to derive my stress energy momentum tensor. I did this because the equations (without the cosmological constant) are:

Rμν - (1/2) gμνR =[ (8πG)/(c4)]Tμν

Here was the stress energy momentum tensor that I derived:

T00 = (-b2c6)/(8πG)(b2 + l2)2
T11 = (-b2c4)/(8πG)(b2 + l2)2
T22 = (b2c4)/(8πG)(b2 + l2)
T33 = (b2c4sin2(Θ))/(8πG)(b2 + l2)

Every other element was 0.

Now here is where the problem comes in. I used SI units and did some dimensional analysis on the terms of the stress energy momentum tensor to see if their ultimate dimensions would come out to be the dimensions of whatever quantity they represent (such as energy density in the case of T00). It turned out that most of the dimensions did not come out to represent the right quantities. For example:

T00 = (-b2c6)/(8πG)(b2 + l2)2

Now considering that b is the radius of the throat of the wormhole, its unit would be meters.
l is the radial coordinate (or so I am told). Therefore, its unit is meters.
c is m/s of course.
G= m3/ (kg * s2 ) You can do dimensional analysis on Newton's gravitational force equation to derive the units of G.

Now, the units for T00 would turn into:

(m8/s6) / (m7/(kg * s2)) = (m * kg) / s4

Those are not the units for energy density. The units for energy density are:

(kg* m2)/s2

These units can be derived by doing dimensional analysis on common energy formulas such as kinetic energy, gravitational potential energy, or Einstein's famous E= mc2.

The case of incorrect units is the case with most of the other elements as well.

This leads me to the following possible alternative conclusions:

1. When you plug a metric into the Einstein field equations and derive the Einstein tensor, you don't actually multiply it by (c4)/(8πG) to get the stress energy momentum tensor. Instead, you just set the stress energy momentum tensor to any one of your choosing (whether it be the electromagnetic stress energy momentum tensor, dust, vaccum, etc...) and then just set each element of your Einstein tensor equal to (8πG)/(c4) multiplied by the corresponding element of the stress energy momentum tensor of your choosing. You then just solve for whatever variables need to be solved for.

In other words, if I wanted to use the electromagnetic stress energy momentum tensor, then I would have just set my:

T00 = (-b2c6)/(8πG)(b2 + l2)2

equal to:
(8πG)/(c4) times the T00 of the electromagnetic stress energy momentum tensor which equals: (1/2)ε0E2 + (1/2) (B20)

Conclusion #2: l is not the radial coordinate after all. It is something else.

Conclusion # 3: I made some calculation errors (which I don't think I did and I hope that I didn't because these calculations took a long time).

Conclusion #4: There should be a cosmological constant involved in this scenario.

Can somebody please tell me which of these conclusions (if any) is correct? If conclusion # 2 is the correct one, then can somebody please tell me what l is? If it is #4, then can you please tell me how to determine what the cosmological constant is?

2. Oct 2, 2014

### michael879

I can tell you your problem is most likely here. In general, elements like $T_{\phi\phi}$ do NOT have the same units as $T_{00}$. Its clear from your metric you are using a spherical basis, so how did all of your G/T elements end up having the same units???

*edit* Of course, I'm neglecting factors of c. You can make the case that it is wrong that people like to set many of the dimensionful constants to 1. However, in a relativistic theory there is absolutely no reason not to set c to one (i.e. distance = time). An analogy I once heard was:
Lets say you pick some coordinate basis and decide to measure distances along the x-axis in mm, but along the y and z axes you measure distance in meters. The conversion factor of 1000 you need to apply to the x component of everything is ABSOLUTELY meaningless. That is what c really is (in a relativistic framework at least). We discovered the nature of spacetime well after we chose a set of units, so the units for time and distance don't match up. However, in relativity they are the same thing and its easy to show that every time "t" appears in an equation it is accompanied by a factor of c

**edit** ok I see your problem now. Removing factors of c your argument becomes:
$$T_{00}\sim\dfrac{E}{\ell^3}$$
from the metric, and
$$T_{00}\sim E$$
from the EM stress-energy tensor.

I think its clear where your confusion is now. The "energy density" you got from EM was actually an energy (integrated energy density). I'm not sure how you got this conclusion, but my guess is that you were looking at the integrated EM stress-energy tensor. The stress-energy tensor you derived for the metric DOES have the correct units.

In fact, if you look at the term E2, the units ARE an energy density. $\epsilon_0E^2\sim\epsilon_0F^2\sim \ell/M^{-1}(M\ell/t^2)^2\sim M/\ell^3$ So the answer is that you miscalculated the units of the EM stress energy density.

Last edited: Oct 2, 2014
3. Oct 2, 2014

### space-time

I'm not saying that $T_{\phi\phi}$ had the same units as T00. T00 ended up having the units as
(m* kg)/s4 while T11 had units of kg/(m * s2 ). T22 and T33 had
(m* kg)/s2.

I am saying that these units are wrong in representing the quantities that those particular elements should represent. I don't know what quantity that T00 would represent with those units, but it is definitely not energy density. Ironically, the units for T11 are the units for energy density (which by the way I made a slight mistake in the OP when stating what the units for energy density should be. Instead, I stated the units for energy itself).

As for T22 and T33 having (m* kg)/s2 as the units, you will notice that these units simply equal the Newton (the unit of force). That is not the same as momentum flux, which is what those terms are supposed to be.

4. Oct 2, 2014

### Staff: Mentor

But if this is the case, how do you calculate things like covariant divergences and traces? To do these calculations you have to add different SET elements together. How can you do that if they don't have the same units?

5. Oct 2, 2014

### Staff: Mentor

As I noted in my previous post, all of the SET components should have the same units, because the SET appears in equations where different components have to be added together. However, "the same units" does not necessarily mean the units are expressed the same way in English or have the same physical meaning in layman's terms: for example, energy density and pressure are different things in layman's terms, but they both have the same units when you expand them out into mass, length, and time: mass / (length * time^2).

From the metric you wrote down, it's evident that $b$ and $l$ have units of length, $t$ has units of time (so $ct$ has units of length), and $\theta$ and $\phi$ have no units (they are angular coordinates). This makes each term in the metric has units of length^2, which is what we want because the point of the metric is to give the squared length of intervals.

The Einstein tensor has units of curvature, or inverse length^2. Your $G_{11}$ has those units and looks OK. Your $G_{00}$ has an extra factor of $c^2$ in the numerator, but I think that's because you took derivatives with respect to $t$ instead of with respect to $ct$. To compute the Einstein tensor correctly, all the derivatives have to be taken with respect to length, not time. (Another way of putting this is that since we are computing a feature of spacetime geometry, all the dimensions in the geometry have to have the same units, so the real unit of time is $ct$, not $t$. This is a major reason why relativists like to use units where $c = 1$.) So instead of $d / dt$ you have to use $d / d (ct)$. This will remove the factor of $c^2$ and make the units of $G_{00}$ the same as the units of $G_{11}$.

Your $G_{22}$ and $G_{33}$ also have the wrong units, but again, I think this is because you didn't take derivatives with respect to length. Since $\theta$ and $\phi$ are angular coordinates, you can't take derivatives with respect to them alone; you have to take derivatives with respect to something like $(b + l) \theta$ and $(b + l) \phi$. I think that will put an extra factor of $(b^2 + l^2)$ in the denominator of $G_{22}$ and $G_{33}$, which will make their units the same as the other components above. However, I haven't checked that by an actual computation. (If I have a chance later today, I'll plug the metric into maxima to see what it gives.)

Once we have the correct Einstein tensor, with all units inverse length^2, then we multiply by $c^4 / 8 \pi G$, as you say, to get the stress-energy tensor components. In terms of units, we take 1 / length^2 and multiply it by (length^4 / time^4) / [length^3 / (mass * time^2)], which gives mass / (length * time^2). For $T_{00}$, we interpret this as energy density; for $T_{11}$, $T_{22}$, and $T_{33}$, we interpret it as pressure (force per unit area); for other components, we interpret it as energy flux, momentum density, or shear stress, as appropriate. All of these quantities have the same units when you expand them out into mass, length, and time.

6. Oct 2, 2014

### michael879

If you had kept reading, I set c=1, m=s in which case those two elements do have the same units, and they shouldn't unless you transformed the metric into a cartesian-like coordinate system where each dimension has units of length
Sorry! I reread, you are correct the phi/theta components did have the correct dimensions. Please read below though, because that is not the problem anyway:

@PeterDonis: Think about how you take traces and derivatives in GR:
$$T = g_{\mu\nu}T^{\mu\nu} \\ \partial_\mu T^{\mu\nu} = \partial_tT^{t\nu} - \partial_rT^{r\nu} - \partial_\theta T^{\theta\nu} - \partial_\phi T^{\phi\nu}$$
Since theta and phi don't have units of length and the metric has the same phenomenon ($g_{\phi\phi}$ doesnt have the same units as $g_{tt}$), any time you need to "add" the components you get a cancellation because you are summing over the product of objects with the inverse unit discrepancy.

Anyway, all of this is irrelevant if you read my edit. The OP got the dimensions of the EM stress-energy tensor wrong, and found them to just be energies. The dimensions of the stress-energy tensor for the metric he showed WERE energy densities

Last edited: Oct 2, 2014
7. Oct 2, 2014

### michael879

This is your mistake. Those are the units for energy not energy density. Energy density has units energy/length^3

8. Oct 2, 2014

### Staff: Mentor

Ah, you're right, the mismatch in units with components gets cancelled out by the mismatch in units with derivatives.

But they weren't. And by your argument quoted above, they shouldn't be. At the very least, the angular components, $T_{\theta \theta}$ and $T_{\phi \phi}$, should have units of "force per solid angle" instead of "force per area", i.e., they should have units of pressure * length^2, or mass * length / time^2 (which is just force--to get the actual pressure measured in a tangential direction, you would have to divide the SET component by the actual physical area per unit solid angle, which would convert force into pressure, as desired.

Looking at the OP's $T_{22}$ and $T_{33}$, they have the same units as $c^4 / 8 \pi G$ (since the units of the factors in $b$ and $l$ cancel out), which are indeed mass * length / time^2, i.e., force. So these units are correct for the angular components. Looking at his $T_{11}$, it has units of force per area, or pressure, as it should since the $r$ coordinate has units of length. This is equivalent to energy density, so that's ok too.

The one that still confuses me is $T_{00}$. By your divergence argument, since a time derivative would be taken of this term rather than a space derivative, it should indeed have the extra factor of $c^2$ (one factor of $c$ for each index that corresponds to time instead of distance). But this makes the units energy density * velocity^2, or mass * length^3 / time^4, which doesn't make sense to me. In units where $c = 1$, it still works out ok, but not in ordinary units like SI. I'm not sure how to resolve this.

9. Oct 2, 2014

### Staff: Mentor

Ok, having plugged the metric into maxima, I am now unsure again. See below.

Plugging the metric into maxima unconfused me on one aspect of this, but only at the expense of re-confusing me about something else. Here's what maxima gives for the Einstein tensor of the line element given in the OP:

$$G_{tt} = \frac{b^2}{\left( b^2 + l^2 \right)^2}$$

$$G_{ll} = - \frac{b^2}{\left( b^2 + l^2 \right)^2}$$

$$G_{\theta \theta} = \frac{b^2}{\left( b^2 + l^2 \right)^2}$$

$$G_{\phi \phi} = \frac{b^2}{\left( b^2 + l^2 \right)^2}$$

Multiplying each of these by $c^4 / 8 \pi G$ gives the corresponding SET components. In other words, if we define an "energy density" $\rho = \left[ b^2 / \left( b^2 + l^2 \right)^2 \right] \left( c^4 / 8 \pi G \right)$, then the SET is diagonal with elements $\left( \rho, - \rho, \rho, \rho \right)$. This describes a weird sort of "fluid" with negative energy density (yes, I know the Einstein tensor component is positive, but with both indexes as lower indexes that turns out to mean negative energy density for reasons that are too long to fit in the margin of this post ;) ), negative radial pressure (i.e., radial tension), and positive tangential pressure, all equal in magnitude; in other words, this is a sort of "exotic matter", which is to be expected since something like that is required to have a traversable wormhole. (It is also considered to be highly unrealistic physically, but that's a whole separate issue.)

The key point for this discussion is that the units of all the Einstein tensor components above are the same: they all have units of inverse length^2. Looking back through the calculations of the Christoffel symbols, Riemann tensor, and Ricci tensor, it looks like the unit differences disappear in going from the Riemann tensor to the Ricci tensor. The only nonzero Ricci tensor component is $R_{ll}$, which we can write out as:

$$R_{ll} = R^{\mu}{}_{l \mu l} = R^{t}{}_{l t l} + R^{\theta}{}_{l \theta l} + R^{\phi}{}_{l \phi l}$$

The first of the three terms in the sum, $R^{t}{}_{l t l}$, is zero, which is why nothing related to the $c^2$ in the $g_{tt}$ metric coefficient shows up anywhere (in fact it doesn't even show up in the Christoffel symbols since none of the ones with a $t$ index are nonzero). The second two of the three terms in the sum have the first index raised, and it appears like that's where the units get corrected: the Riemann tensor components computed by maxima have all lower indexes, i.e., they are $R_{\theta l \theta l}$ and $R_{\phi l \phi l}$. These components are unitless (they are ratios like $b^2 / \left( b^2 + l^2 \right)$), so when the first index is raised, that involves multiplying by the inverse metric coefficient, which adds a factor $\left( b^2 + l^2 \right)$ in the denominator, correcting the units to inverse length^2. (The raising of the index also cancels out a factor of $sin^2 \theta$ in the components where $\phi$ appears, which is why that factor doesn't appear anywhere in the above.)

All this seems to make sense, but now I'm re-confused about how the covariant divergence works, since michael879's point, that the derivative operators for different coordinates can have different units, so the units of the components whose derivatives are being taken should vary as well, still appears to me to be valid. I need to think about this some more.

Last edited: Oct 2, 2014
10. Oct 2, 2014

### space-time

Hey, did maxima give you the metric tensor in matrix form? If so, then what were the individual metric tensor elements? I ask because that might be my problem with what I have been hearing from you and Michael thus far. From the space-time interval, I had deduced that the metric tensor would be:

g00= - c2
g11= 1
g22 = (b2 + l2)
g33= (b2 + l2)sin2(Θ)

every other element was 0.

The coordinate axes were:
x0= t
x1= l
x2 = Θ
x3 = Φ

With this metric tensor, I may have gotten slightly different Christoffel symbols and Ricci tensor elements than maxima.

My Christoffel Symbols were as follows:

All Christoffel symbols with 0 as the upper index Γ0ij were equal to 0.

Γ122 = - l
Γ133 = - lsin2(Θ)
Γ212 , Γ221, Γ313, and Γ331 = l/(b2 + l2)
Γ233= - sin(Θ)cos(Θ)
Γ323 and Γ332 = cot(Θ)

every other Christoffel symbol was 0.

My Ricci tensor only had one nonzero element and that was:

R11 = (-2b2)/ (b2 + l2)2

Of course the curvature scalar under this circumstance would be the same as my sole Ricci tensor component.

That was my work. I really don't understand what I did wrong. I checked and didn't see any basic calculation errors, but feel free to check behind me.

11. Oct 2, 2014

### Staff: Mentor

Yes, that's the metric tensor I input into maxima, and it gave me the appropriate matrix back in the output.

Yes, that's what maxima gave me.

Yes, that's what maxima gave me.

In general this wouldn't be true, because you have to raise an index in order to take the trace. But in this case, since $g_{11} = g^{11} = 1$, it works out that way, yes.

12. Oct 2, 2014

### space-time

Good. So then if I didn't make any computational mistakes, what did I do wrong exactly to get the wrong units?

13. Oct 2, 2014

### Staff: Mentor

Looking at that Ricci tensor again made me wonder why the metric coefficients didn't appear in the einstein tensor I posted earlier (which is what I got from maxima). Looking in the maxima manual, I see why: the maxima function einstein() gives the mixed Einstein tensor, with one upper and one lower index. In other words, the components I posted were really $G^t{}_t$, $G^l{}_l$, $G^{\theta}{}_{\theta}$, and $G^{\phi}{}_{\phi}$.

In order to get the covariant Einstein tensor, with two lower indexes, you have to lower an index with the metric tensor; when you do that, you get the components that the OP posted, with the factor of $c^2$ in $G_{00}$, and the appropriate factors in $G_{22}$ and $G_{33}$. So it looks like the OP's components were correct; my apologies for the confusion.

This still leaves a question about how the covariant divergence works, however; for the mixed Einstein tensor, that looks like $\nabla_{\mu} G^{\mu}{}_{\nu}$, and since all the $G$ components have the same units, all the $\nabla$ components should too; but they don't. I'm still pondering that.

14. Oct 2, 2014

### Staff: Mentor

I'm not sure you did anything wrong. See my previous post.

15. Oct 2, 2014

### Staff: Mentor

I generally like to think of the coordinates as being dimensionless numbers, so that all of the entries in any given tensor have the same units.

16. Oct 2, 2014

### pervect

Staff Emeritus
Comment. The web reference you cite also derived the Einstein tensor, in an orthonormal basis. (They did omit the factors of c, though). Many of the unit issues and the issues of physical interpretation you are struggling with can be greatly eased by working in an orthonormal basis rather than a coordinate basis. Is there some reason you are avoiding orthonormal bases? They'd help you a lot.

If you use an orthonormal basis, then you have a set of orthonormal basis vectors $\hat{t}$, $\hat{l}$, $\hat{\theta}$, $\hat{\phi}$ which are all of unit length and orthogonal. In your local tangent space, you can write an arbitrary vector as a sum of your orthonormal basis vectors as $T \hat{t} + X \hat{l} + Y \hat{\theta} +Z \hat{\phi}$, and you can regard T,X,Y,Z as being Cartesian displacements in the flat tangent space.

Note that this is rather similar to defining unit vectors $\hat{r}$ and $\hat{\theta}$ in polar coordinates, $d\theta$ doesn't have a unit length, but $r d\theta$ does. Furthermore, and more relevant to the point, $\hat{r}$ has units of length and $\hat{\theta}$ ALSO has units of length, while $\theta$ would be expressed in radians. So by going to an orthonormal basis, we can make all the position coordinates be measured in meters, simplifying our interpretation.

Now, T has units of time, so you need to do the same thing you do in special relativity if you want SI units. This is a pain, so I rarely do it.

See for instance http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html for a review of 4-vectors with SI units in SR though

(T,X,Y,Z) has mixed units, so instead you use (ct, [X,Y,Z]) for your 4-vector, and all components of the 4-vector are expressed as distances (meters in SI units).

For the energy-momentum 4-vector, E and P have different units, so you use (E, $c \,\vec{P})$ for the 4-vector.

Your main question, what the SI units of the stress energy tensor are, I don't have a reference for at this time. I can say that you'll make all of the components have the same unit , though, just as we did above with the other 4-vectors. You can figure out what unit that is from the Einstein field equation, if you find version that has all the factors of c included.

The Riemann and Einsten tensors, R and G, both have units of 1/m^2 in SI units

Recall you have to use an orthonormal basis for any of this to be true.

17. Oct 2, 2014

### Staff: Mentor

Yes, but most commonly used charts don't do this. Even in SR this isn't true: the standard Minkowski chart has coordinate with length and time units.

Yes, but again, most commonly used charts don't have an orthonormal basis as their coordinate basis. Certainly the chart used in the OP does not.

18. Oct 2, 2014

### Staff: Mentor

You can choose to do it with any chart, including commonly used ones like the Minkowski chart. It certainly isn't the usual approach with those well-known charts, but it is perfectly justified by Riemannian geometry even for them.

19. Oct 2, 2014

### Staff: Mentor

This is really a question of terminology, but to me the specific definition of the coordinates, including their units, is part of the definition of a chart. So a chart which had a line element that looked like $ds^2 = - dt^2 + dx^2 + dy^2 + dz^2$, but where we define $t$, $x$, $y$, $z$ as being dimensionless, is not, in my view, the same as the standard Minkowski chart. (Such a line element would also raise the question of how $ds^2$ can have units of length squared, as it must, physically; presumably each term on the RHS would have to have a coefficient with units of length squared, whose numerical value just happens to be 1.) Of course this is perfectly consistent mathematically; I'm just not sure it would properly be called "the Minkowski chart".

20. Oct 3, 2014

### pervect

Staff Emeritus
I am not saying that most commonly used charts have an orthonormal basis. I would suggest simply not using units for coordinates in the general case.. Defining just the units is insufficient to completely define the coordinates in any event,, and once you provide enough information to completely specify the coordinates - (for instance by giving a metric, which is the most concise way to completely specify coordinates), - then units are redundant anyway.

For what it's worth, the IAU seems to agree with me, though there appears to be a surprising amount of debate on the issue from people who apparently don't like the IAU recommendations. The opponents of the IAU apparently want to create a whole bunch of different units, one for each coordinate system that is in use (TCB seconds and meters, TCG seconds, and meters, etc). I'm not clear why, I think this is simply a result of being taught that "all measurements need to have units". The issue as I see it is that the coordinate themselves do not represent anything physical (like distances or Lorentz intervals) until you process them through the metric. Therefore it makes little sense to me to assign units to these coordinates that imply otherwise.. It's better not to give the coordinates units, and to save the units for cases where the quantity in question DOES represent a distance (or more generally some proper interval) because the coordinate changes have been processed via a metric to give the physically meaningful and invariant interval.

The other point I am trying to make is that while it doesn't IMO make sense to use units for generalized coordinates, it DOES make sense to use units to describe the components of tensors expressed in an orthonormal basis. Actually, one might be able to make a case for dropping the "orthogonality" condition if one is pressed, the "normal" condition is the most important part of giving things units, because normality insures that a 1 unit multiple of a basis vector represents a physical interval of 1 unit in some direction.

However, when one can create a set of basis vectors that is both orthogonal and normal, I would say that that is by far the easiest way to get physical insight of the significance of the tensor components.

The remaining issue involves the mechanics of how to do this, for instance with Maxima. I could go into the details if needed, but at the moment I am trying to motivate the use of orthonormal basis vectors, and to ascertain why the OP was apparently avoiding them (as the reference already gave him the orthonormal components, but he went and calculated the coordinate components). I would guess that the issue was unfamiliarity, but I don't want to speculate too much about it, rather my goal is to highlight the use and importance of this technique.

[add]I've taken the liberty of appending a Maxima batch file that calculates the Einstein tensor for the wormhole metric in an orthonormal frame field basis, just to illustrate the process. Note that Maxima uses rather strange orderings, and calculates $G^{\mu}{}_{\nu}$ for the Einstein tensor rather than the more usual $G_{\mu\nu}$.

#### Attached Files:

• ###### Maxima_Worm.txt
File size:
2.4 KB
Views:
51
Last edited: Oct 3, 2014
21. Oct 3, 2014

### Ben Niehoff

I may be repeating some things you have already figured out in this thread, but here are my responses:

Not really. An expression like $T_{11} + T_{23} + T_{01}$ is not covariant, so it shouldn't be turning up anywhere. Instead you have quantities like $g^{\mu\nu} T_{\mu\nu}$ and $T_{\mu\nu} u^\mu u^\nu$ which are contractions, and the discrepancies in units in $T_{\mu\nu}$ should be exactly cancelled by inverse discrepancies in the objects with upper indices.

Remember also that $g_{\mu\nu}$ does not have units of length^2, but rather the line element $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$ does. If you start assigning your coordinates units, then you are introducing mixed units into $g_{\mu\nu}$.

As I just said, careful doing things like that.

Not necessarily. The units of a curvature tensor depend on the basis you've chosen! In an orthonormal basis, each component should have the same units, and you can choose units such that they are inverse length^2. However, if you've assigned your coordinates units, then in a coordinate basis, curvature might have different units depending on the units of the coordinates! In particular, angular coordinates such as $\theta, \phi$ will have different units from length coordinates, which will have different units still from time coordinates.

Have to disagree there. My above point aside, the real reason for setting $c = \hbar = k_B = 1$ is because there is always a unique way to restore them at the end of a calculation, and omitting them in calculations not only saves loads of time, but makes the mathematical structure of those calculations more transparent. I also work in units where $G = 1/4\pi$, which removes a lot of extra symbols from the Einstein equations.

It should not be necessary to do such things. The math should work out the same in the end anyway. In fact, one of the nice things about units is that you should be able to ignore them, do a calculation, and at the end you should get something that is dimensionally consistent anyway. It's a great way to check for errors.

22. Oct 3, 2014

### Staff: Mentor

I agree it is a question of terminology. It is interesting that I have a very different impression of the definition of a chart. My impression of the definition of a chart is simply that it is a smooth invertible mapping from events in spacetime to points in R4. From Carroll I certainly never got the impression that units were part of the chart, although he didn't explicitly say so one way or the other. Do you know which author gave you the impression that the units are part of the definition of a chart?

You can write the line element $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$, where it is clear that it doesn't matter if the units are in g or in dx. When you write the line element as $ds^2 = - dt^2 + dx^2 + dy^2 + dz^2$ you are actually writing $ds^2 = g_{tt} dt^2 +g_{xx} dx^2 + g_{yy} dy^2 +g_{zz} dz^2$, and just suppressing the 1's, but there is no reason that the 1's in the metric cannot have units.

To me, neither way is wrong, but I think that the unitless-coordinates way is more consistent and straightforward once you get over the initial "woah, that is wierd" reaction.

23. Oct 3, 2014

### Staff: Mentor

Can you give a link? I'm not familiar with the IAU's take on this, but it sounds interesting.

Agreed.

Agreed. In fact I would say something a bit stronger: the "components" of tensors in an orthonormal basis are actually the invariants that express physical observables, because they are really contractions of tensors with 4-vectors representing an observer and their measuring apparatus. For example, $T^0{}_0$ in an orthonormal basis is the contraction of $T^{\mu}{}_{\nu}$, the stress-energy tensor, with $u^{\mu}$, the observer's 4-velocity, which in turn is just the energy density measured by the observer.

24. Oct 3, 2014

### Staff: Mentor

In your opinion, does it make sense to use units to describe the tensor itself, not the components? I.e. in a dimensionless coordinates approach, would you assign units of length^2 to $g$, $g_{\mu\nu}$, or both, or something else?

25. Oct 3, 2014

### Staff: Mentor

This makes sense; what you're saying is that, ultimately, any expression for a physical observable has to be a scalar, with no free indices, so any unit discrepancies in objects with indices, vectors, tensors, etc., has to ultimately get resolved when we contract with some other object with indices to get a scalar.

However, this still leaves me wondering about something. If I write down the covariant divergence of the mixed Einstein tensor I computed before, it is $\nabla_{\mu} G^{\mu}{}_{\nu}$. This expression still has one free index, so I have to contract it with something like $u^{\nu}$ to get a scalar, and the units of each component of the divergence will combine with the units of each component of $u$ to give an observable with the correct units. All that is fine, but what about the contraction of the $\mu$ index in the expression above? If I expand that out, it is $\nabla_0 G^0{}_{\nu} + \nabla_1 G^1{}_{\nu} + \nabla_2 G^2{}_{\nu} + \nabla_3 G^3{}_{\nu}$, and the problem is that all the Gs have the same units but the different $\nabla$ operators do not, so how can I add them together in the contraction?

In the particular case under discussion here, this issue doesn't arise, because $G$ is diagonal, so only one term of the contraction above survives anyway. But that won't always happen. I guess I need to work this through for a case where $G$ is not diagonal so I can see how it goes.

Agreed.

Agreed; I should have said that the units are inverse length^2 in the particular basis being used for this problem.