# Coordinate values of G can vary from standard value

1. Dec 2, 2014

### Jonathan Scott

I was recently checking out a paper about gravitational energy and found a unexpected minor inconsistency when I did the calculation entirely in isotropic coordinates. I later tracked down what caused it, and it surprised me a bit, so I'm wondering if others were aware of this. Basically, I'd forgotten to allow for the fact that when all other values are expressed as coordinate values, $G$ (or more accurately $G\hbar/c^4$) has to vary slightly as well.

The problem is with terms of the form $Gm/rc^2$ which must have a dimensionless numerical value.

This expression can be split up into the following three parts, consisting of an adjusted form of the gravitational constant with additional constants to get the dimensions right, a frequency (which is purely affected by the time part of the metric) and an inverse distance (which is purely affected by the space part of the metric): $$\frac{G \hbar}{c^4} \frac{m c^2}{\hbar} \frac{1}{r}$$ When studying equivalence between GR and Newtonian terms, it is often useful to look at quantities expressed entirely in isotropic coordinates, including replacing the standard $c$ with the coordinate speed of light (which is only meaningful for isotropic coordinates as otherwise it can be different in different directions at a given point).

In the weak field approximation for the Schwarzschild solution, the scale factor between coordinate and local frequency values cancels with the scale factor between coordinate and local rulers, so the overall term $G\hbar/c^4$ has approximately the same effective value both for local calculations and for calculations in isotropic coordinates.

However, the space and time factors in the metric are not actually exactly the same, so overall the product of the second and third terms varies very slightly with potential. The time factor in the Schwarzschild metric in isotropic coordinates is $$\frac{1-Gm/2rc^2}{1+Gm/2rc^2}$$ and the space factor is $$(1+Gm/2rc^2)^2$$ To get the view from some other potential, we have to divide the frequency and distance values by the metric factors for that location, as follows:
$$\frac{1+Gm/2rc^2}{1-Gm/2rc^2} \frac{1}{(1+Gm/2rc^2)^2} = \frac{1}{1-(Gm/2rc^2)^2}$$
In order for this overall to remain a dimensionless value which is unaffected by the potential at the relevant location, this means that the $G\hbar/c^4$ term must vary with observer potential by the inverse of this factor.

This also highlights a side-effect of using geometric units for gravity, where $G=1$ and $c=1$. I have always disliked this convention because these are quantities with dimensions which can change from their standard values in different frames of reference, but if they are omitted, it is not clear which dimensions apply in a given case. With geometric units, a term of the form $m/r$ has to be dimensionless, which means that the value of $m$ must behave like a length, not like a frequency as would normally be expected for an energy value, so a mass term expressed in these units would have a local value determined not by the time factor of the metric but rather by the space factor. This seems very confusing to me!

2. Dec 2, 2014

### Staff: Mentor

This depends on your choice of units. If you can certainly choose a system such that G=c=1 are dimensionless. Then in the above you have a dimensionless r as well. If r is dimensionless then there is no need to consider that G varies.

My preference is to use dimensionless coordinates as well as geometrized units.

3. Dec 2, 2014

### Jonathan Scott

I'm not entirely clear on what you think a dimensionless coordinate speed of light would mean. If you start trying to measure time in geometrized space units or vice versa then it seems to me that in the presence of a gravitational potential you will either find that "time" will no longer match what clocks measure or that "space" will no longer match what rulers measure. That doesn't seem a very sensible definition.

4. Dec 2, 2014

### Staff: Mentor

That is already true in most coordinate systems anyway, so you haven't lost anything.

Regarding the meaning of a dimensionless c. It would be like a conversion between different units with the same dimension, like 12 inches per 1 foot. Usually that is considered silly to do so you use units where the factor is 1 to make things easier.

5. Dec 2, 2014

### Jonathan Scott

I don't have a problem with space and time being measured in the same units. I do however have a problem with the way in which geometric units mean that "energy" is assumed to vary with the scale factor from the space part of the metric, instead of the inverse of the scale factor from the time part of the metric. This may only be a small difference, negligible in the weak field approximation, but it seems conceptually wrong. In contrast, I see "energy" $mc^2$ as varying strictly with the time part of the metric but multiplying by $G/c^4$ then matches it up to the space part.

Of course, if I work entirely in isotropic coordinates, the $c$ in $G/c^4$ also varies, so the coordinate value of $G$ on its own varies approximately as the 8th power of the time dilation factor!

6. Dec 2, 2014

### Staff: Mentor

I am afraid that I cannot help you here. If time and space use the same dimensions and units then it seems clear to me that there is nothing conceptually wrong with it at all. I don't understand the conceptual objection you have.

Let me explain how I think of these things. Please note, that this is my personal thought process, there is nothing which mandates this approach so several other approaches are equally valid, however I have found this approach clear and useful in organizing my thoughts.

First, I have a strong preference to consider all coordinates to be dimensionless. They are like street addresses or zip codes: simply a means to label locations with no intrinsic meaning.

Second, as a result of the first, this means that all components of any tensor have the same dimensions. So you can look at the units and dimensionality of a given expression at the tensor level rather than at the component level. This simplifies tracking units tremendously. For example since $ds^2$ has dimensions of length^2 and since $dx$ is dimensionless then $ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}$ immediately implies that $g$ has dimensions of length^2 also.

Third, because all components of a tensor have the same units it is easier to check each component for dimensional consistency. I don't have to consider what the final units are on each specific component, I can do that once for the whole tensor. Otherwise, you need to determine the correct units for each component and only then check the component for dimensional consistency.

Fourth, I have a strong preference for natural units and a slight preference for geometric units. This requires that I be careful in my algebra because I won't be able to catch many mistakes by looking at the units. However, it does simplify the algebra and makes it easier to do correctly. If I need to go to SI units then I can always add the factors back in to achieve dimensional consistency.

7. Dec 2, 2014

### Jonathan Scott

I don't understand what you mean by "dimensionless" coordinates.

It would never have occurred to me to try to form a tensor from components with mixed dimensions! I've always assumed factors of $c$ or whatever as needed to make things consistent.

Anyway, most of the time I prefer to include all the factors of $c$ and $G$ explicitly, even if the standard value is 1, because when values are expressed in some other coordinate system, the relevant term is no longer equal to the standard value. It seems more natural to me to refer to the "coordinate value" of the relevant constant (with slightly modified notation if necessary, e.g. using primed symbols) rather than to have to introduce a separate new factor for the ratio to the standard constant, especially since these new factors go in exactly the same places as $G$ and $c$ would go.

I must admit that when it came to $G$, I normally measure mass or energy in frequency units and replace $G$ with $G\hbar/c^4$ which is essentially constant and equal to the standard value even when using isotropic coordinates, at least in the weak field approximation. When I recently tried extending a calculation to the stronger field case I forgot to use the coordinate value, and ended up being puzzled about a missing factor of $1-(Gm/2rc^2)^2$.

8. Dec 2, 2014

### Staff: Mentor

I mean that, for example, ${t,r,\theta,\phi}$ are all pure numbers, without units. Specifically, $t$ does not have units of time, $r$ does not have units of length, and ${\theta, \phi}$ do not have any angular units. If you are giving units to your coordinates then your tensors will necessarily have components with mixed dimensions.

Consider the Schwarzschild coordinates $dx=(dt,dr,d\theta,d\phi)$ and $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$. All of the off-diagonal metric components are 0 so this expands to $ds^2=g_{tt}dt^2 + g_{rr}dr^2 + g_{\theta \theta}d\theta^2 + g_{\phi \phi}d\phi^2$. Now, since $ds^2$ has units of length^2 if the coordinates are dimensionless then all of the metric components have units of length^2. However, if $r$ has dimensions of length then $g_{rr}$ must be dimensionless, and if $t$ has dimensions of time then $g_{tt}$ must have dimensions of length^2/time^2, and so forth.

Unless the coordinates all have the same dimensionality, the components of the other tensors will have mixed dimensions. It is a necessary by-product of assigning different dimensions to the coordinates.

9. Dec 2, 2014

### Jonathan Scott

OK,
OK, thanks, now I see what you mean. I was thinking of the space coordinates described by the tensors, not the tensors themselves, but of course I use spherical coordinates. However, I still think there's a problem with geometric units as I said before.

10. Dec 2, 2014

### Staff: Mentor

Which problem is that?

11. Dec 2, 2014

### Jonathan Scott

I just don't like the confusing way that geometric units quietly turn energy into something which transforms as a length rather than as a frequency. And when I'm looking at actual calculations in isotropic coordinates, using explicit factors of $c$ and $G$ makes it clear where all the metric factors need to be applied to get a consistent result.

12. Dec 2, 2014

### Staff: Mentor

Energy is the timelike component of the four momentum, so it transforms as a time. It does not transform as a length, regardless of what units you use.

The units you use have nothing to do with how it transforms. A time measured in meters still transforms as a time.

I agree with that. There is nothing actually wrong with geometrized units, but (in my experience) they are certainly far less "fault tolerant". You can make a mistake and find it very difficult to clearly see where you made the mistake.

Last edited: Dec 2, 2014
13. Dec 3, 2014

### TrickyDicky

Aren't you simply absorbing the spatial curvature effects into the constants by using the isotropic radial coordinate in the dimensionless expression?
For example if one tries to get a dimensionless quantity from an expression like $\pi r \lambda$ in a context where there is curvature like it is the case in the Schwarzschild metric, and using a Euclidean distance as $r$ one could reach the
conclusión that $\pi$ is no longer a constant when all it happens is that it has absorbed the spatial change from the different geometric setting.

14. Dec 3, 2014

### Jonathan Scott

I'm not sure what you mean here, but I don't think so!

The specific example I showed above was to show the conversion factor between evaluating the same $Gm/rc^2$ term in isotropic coordinates or local coordinates at the same point (which are also isotropic but with a different scale factor). It is clear how local time and space factors need to be adjusted in this case, which determines the coordinate values of $m$, $r$ and $c$, and that therefore determines how the effective coordinate value of $G$ (or more accurately $G\hbar/c^4$) must vary with the change of observer potential. That is, all I am doing is taking the same expression (including the same definition of the isotropic radial coordinate) and rescaling it to a different observer potential.

Obviously if I try to convert between isotropic coordinates and local coordinates when describing anything other than the vicinity of a single point, I have to take into account curvature of space, but I don't see how that would apply here.

15. Dec 3, 2014

### TrickyDicky

Wouldn't curvature show up in a potential rescaling, like it happens with say grav. redshift?

16. Dec 3, 2014

### Jonathan Scott

I think you'd need to be much more specific before I could answer that.

The context is that I'm working with the Schwarzschild solution in isotropic coordinates. If everything is expressed in isotropic coordinates (including using the coordinate speed of light instead of the standard one) in the weak field approximation then various formulas become very simple to understand in Newtonian terms, such as the equation of motion: $$\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )$$ where $\mathbf{g} = -Gm/r^2$. That's sufficiently accurate to reproduce the standard solar system GR predictions including the perihelion precession of Mercury.

I recently saw a paper that used isotropic coordinates to describe an effect which matches my Newtonian result in the weak field case but which is also valid for the strong field case, so I converted that to express everything in terms of coordinate values, but I ended up with a small difference until I realised that I also needed to use a modified coordinate value of $G$. In Schwarzschild coordinates this doesn't appear to apply (in that dt and dr are scaled relative to local coordinates by exactly matching factors) and it also doesn't apply in the weak field approximation for isotropic coordinates. However, as the dt and dr factors in isotropic coordinates do not exactly match, a correction is needed in the strong field case. One could of course split this "coordinate value of G" into "standard G" and a correction factor, very close to 1, for the specific coordinate system, but one way or another the correction is required.

17. Dec 3, 2014

### Staff: Mentor

Frankly, I don't think that this is physically meaningful in any way. G is entirely an artifact of your units. If you use Planck units then it has a dimensionful value of 1. If you use geometrized units then it is a dimensionless 1. If you use SI units then it is the standard SI value.

The value and the dimensionality of G is entirely an artifact of your chosen units. Thus, the only way that a spatially varying G would make sense is if you chose a system of units that had a spatial variation. While you certainly could do so, it would be a very strange thing to do and not at all required by GR.

Last edited: Dec 3, 2014
18. Dec 4, 2014

### Jonathan Scott

That's exactly what happens when you use coordinate measurements for time and space. For example, the coordinate speed of light at a given point is no longer the standard value $c$ but depends on the potential, yet in isotropic coordinates, quantities of the form $v^2/c^2$ work just as well as in local coordinates if $c$ is allowed to denote the coordinate speed of light (which can vary and hence contribute additional terms to the derivative of any expression involving $c$). As shown above the GR equation of motion can be expressed in a surprisingly similar way to the Newtonian one, showing for example that the rate of change of coordinate momentum is not affected by the direction of travel, and is always exactly towards the source, preserving angular momentum, and there are various other related semi-Newtonian formulae which are very easy to use in the same way.

I admit that simply using the same names $c$ and $G$ when switching to using purely coordinate values is a bit controversial, and needs to be made very clear, but the alternative of something like putting primes on every variable seems unnecessary, or similarly using some sort of associated scale factor combined with the standard value.

19. Dec 4, 2014

### Staff: Mentor

Yes, but again, that is not physically meaningful in any way. A varying c or G simply means that you have a funny set of units, specifically a set of units which varies. When you are calculating your "coordinate c" or "coordinate G" the reason that it varies is because you are basing your units on your coordinates.

You certainly can do so, but it doesn't mean anything physically. It is purely an artifact of your conventions. By choosing different conventions you can make it be any value you want, without changing the outcome of any physical measurement.

20. Dec 4, 2014

### Jonathan Scott

I don't know what you mean by "doesn't mean anything". I am taking physical equations and expressing them systematically in terms of different coordinates in order to make the correspondence with Newtonian theory much easier to understand. Each term transforms relative to local terms according to the metric expressed in those coordinates. When one does that in isotropic coordinates, one gets a "coordinate speed of light" and a "coordinate gravitational constant" which are slightly different from the standard values. I like to keep the same form for the equations but say that they are expressed entirely in coordinate values. Others might like to include various terms extracted from the metric, or at least to mark all coordinate values (that is, just about everything) with primes or similar.