What does G mean in general relativity?

[Dale]

What, we can have a consistent system of units where G is a force?

In geometrized units both G and force are dimensionless, so yes.

Or c is a charge?

I don't know, I haven't tried to do that. Maybe it is not possible. Just because something is a convention doesn't mean that everything is possible. It just means that there are at least two options that are physically the same.

 

[ohwilleke]

Why not? Dimensions are part of the definition of a system of units, not a part of nature. So if your units say that a quantity is dimensionless then it is dimensionless.

No. Dimensionless means not corresponding to any physically defined unit, and one isn't free to define it otherwise. The fact that one declares that numbers without an expressly stated unit are lengths in your system, doesn't make a number without a unit shown dimensionless.

Also, please use the quote functionality. @vanhees71 may not see that you are quoting him since you didn't use the standard functionality.

My bad. I'll try to use the standard functionality in the future.

 

[Dale]

Dimensionless means not corresponding to any physically defined unit, and one isn't free to define it otherwise.

But units aren't defined by physics, they are defined by convention. Quantities that are dimensionless in one unit system are not necessarily dimensionless in others. For example $G$ is dimensionless in geometrized units but not in SI units.

 

[ohwilleke]

But units aren't defined by physics, they are defined by convention. Quantities that are dimensionless in one unit system are not necessarily dimensionless in others. For example $G$ is dimensionless in geometrized units but not in SI units.

I agree that units are defined by physics, but the dimensions to which units may be affixed are defined by physics. Quantities that are dimensionless in one unit system are necessarily dimensionless in others.

You can use a different physical constants than Newton's constant defined in a dimensionless way (as quantum theories of gravity often do) and shift the dimensions that are captured by the dimensions in Newton's constant to some other part of the equation, but you can't simply change units and eliminate their dimensionality.

Sometimes the dimensions aren't expressly stated, but the dimensions themselves are physical and can't be defined away.

 

[Vanadium 50]

I think there is a difference between a unit that is truly dimensionless, like alpha, and one that is only dimensionless in a particular choice of units, like c.

I do not believe that the strength of the electromagnetic force (a simpler analogy to the OP's question) can be defined by a dimensionless constant classically. I am not convinced by the suggestion we can add an hbar and keep things classical. The correspondence principle says we can take the limit as h goes to zero and recover classical physics. Requiring a non-zero h or hbar violates that.

 

[Dale]

the dimensions to which units may be affixed are defined by physics. Quantities that are dimensionless in one unit system are necessarily dimensionless in others

I have already given a counterexample. $G$ is dimensionless in geometrized units, and it is not dimensionless in SI units.

There are quantities that are dimensionless in all unit systems, like $\alpha$, but not all dimensionless quantities are like that.

You can use a different physical constants than Newton's constant defined in a dimensionless way (as quantum theories of gravity often do) and shift the dimensions that are captured by the dimensions in Newton's constant to some other part of the equation, but you can't simply change units and eliminate their dimensionality.

This doesn’t work because the equations are different in different unit systems. In those other systems there aren’t any other parts of the equation to shift them to.

For example in SI units Newton’s 2nd law is $\Sigma \vec f =m\vec a$ but you could make a system of units (Dale units) where force is its own base dimension. In those units Newton’s 2nd law is $\Sigma \vec f = k m \vec a$ where $k$ is a dimensionful universal constant with dimensions of $F^{1}\ M^{-1}\ L^{-1}\ T^{2}$.

When we convert from Dale units to SI units we set $k$ to be a dimensionless 1. In the SI system force is not a base unit, it is a derived unit. That $k$ factor is not hidden in any of the remaining terms. It is simply gone. It doesn’t exist at all in SI units. Natural units do the same thing for $c$ and $G$. They are not hidden somewhere in the equations, they are simply gone in those units, and the equations are simpler with nowhere to hide them anyway.

 

[PeroK]

I have already given a counterexample. $G$ is dimensionless in geometrized units, and it is not dimensionless in SI units.

The first example that most students meet is where velocities are expressed dimension-lessly as a proportion of the speed of light. As in$$\gamma = \frac 1 {\sqrt{1- v^2}}$$

 

[Ibix]

Quantities that are dimensionless in one unit system are necessarily dimensionless in others.

I have already given a counterexample. $G$ is dimensionless in geometrized units, and it is not dimensionless in SI units.

You seem to me to be using "dimensionless" in two different senses. Ohwilleke seems to regard only quantities that are invariant under all possible unit systems as dimensionless (e.g. $\alpha$). Dale additionally seems to regard quantities that are just numbers in some unit systems as dimensionless in those unit systems.

I guess your view depends on, for example, in units where $c=1$ whether you consider that there is a distinction between one unit of time and one unit of distance, other than what is encoded in the metric.

 

[pbuk]

you could make a system of units (Dale units) where force is its own base dimension

Yes of course you could...

In those units Newton’s 2nd law is $\Sigma \vec f = k m \vec a$ where $k$ is a dimensionful universal constant with dimensions of $F^{1}\ M^{-1}\ L^{-1}\ T^{2}$.

...but you could not also have M, L, and T as base dimensions because F, M, L and T are not linearly independent.

 

[pbuk]

I guess your view depends on, for example, in units where $c=1$ whether you consider that there is a distinction between one unit of time and one unit of distance, other than what is encoded in the metric.

I do not see how you can not have that distinction, otherwise how can you have a distinction between one unit of area and one unit of "squared time"?

 

[Dale]

but you could not also have M, L, and T as base dimensions because F, M, L and T are not linearly independent.

They are independent in Dale units.

I do not see how you can not have that distinction, otherwise how can you have a distinction between one unit of area and one unit of "squared time"?

Just because two quantities have the same dimensions doesn’t mean they are indistinguishable:
Torque and energy
Counts and angles
Frequencies and time constants
(The above in SI units)

 

[PeroK]

I do not see how you can not have that distinction, otherwise how can you have a distinction between one unit of area and one unit of "squared time"?

The basis of SR and GR is that spacetime is a 4D manifold. And then it is natural to measure time and space in the same units, as you have single geometry.

 

[pbuk]

They are independent in Dale units.

No, you can't get away from $F = MLT^{-2}$.

Just because two quantities have the same dimensions doesn’t mean they are indistinguishable:
Torque and energy...

Hmmm, this is true, I may be wavering...

Edit: but I don't think that gets you any closer to a consistent system of units where G and force have the same dimensionality.

 

[Dale]

No, you can't get away from $F = MLT^{-2}$.

Why not? SI units come from the BIPM, not from God.

If your assertion were correct then why is the Gaussian unit of charge a derived unit but the SI unit of charge is not derived? If charge can be derived in one and a base unit in another then why not force?

 

[Vanadium 50]

SI units come from the BIPM, not from God.

The difference is that God doesn't think he's the BIPM.

I think the point @pbuk is making is that while one is free to define units any way they like, the relationship between the units is fixed. V = L/T. That is true whether we define L and T (as we used to) or V and T (as we do today).

 

[pbuk]

The difference is that God doesn't think he's the BIPM.

I think the point @pbuk is making is taht while one is free to define units any way they like, the relationship between the units is fixed. V = L/T. That is true whether we define L and T (as we used to) or V and T (as we do today).

Yes this makes my point more clearly - I think it is past my bedtime.

 

[Dale]

the relationship between the units is fixed. V = L/T. That is true whether we define L and T (as we used to) or V and T (as we do today)

That simply is not true for dimensions. Those can and do change with different unit systems. In some unit systems $v$ is dimensionless, in some it has dimensions of $L^{1}\ T^{-1}$, and in some it has its own base dimension of $V^{1}$.

In most systems of units you would only have two of those as base dimensions, and the other would be derived. But it is possible to have a system of units where $V$, $L$, and $T$ are all independent base dimensions. That would be weird, but in principle it is no different than how SI and Gaussian units treat charge.

 

[PeterDonis]

That simply is not true for dimensions. Those can and do change with different unit systems.

Dimensions can change, but relationships between dimensions still have to be consistent with actual physical observations.

In some unit systems $v$ is dimensionless, in some it has dimensions of $L^{1}\ T^{-1}$, and in some it has its own base dimension of $V^{1}$.

And in each of these unit systems, the relationship between the dimension of $v$ and the dimension of a measured length and the dimension of a measured time is the same:

In units where $v$ is dimensionless (such as the "natural" units used in relativity), length and time both have the same dimension, so the dimension of $v$ is indeed the dimension of length divided by the dimension of time--which gives the result "dimensionless".

In units where $v$ has its own base dimension, the dimension of length will be that base dimension times the dimension of time.

There is one other alternative, though, which may be what you are implicitly thinking of here:

it is possible to have a system of units where $V$, $L$, and $T$ are all independent base dimensions.

In such a unit system, how would we express the physical fact that, for an object moving at a constant speed, the length it covers is its speed times the time it travels?

If the dimensions of the units of these three quantities do not have the relationship described earlier in this post, then the mathematical equation describing the above relationship between physical measurements would have to have an extra constant in it with dimension $L\ V^{-1}\ T^{-1}$.

 

[Dale]

the relationship between the units is fixed. V = L/T. That is true whether we define L and T (as we used to) or V and T (as we do today)

This is really not true in general. Let's start with Gaussian units vs SI units and go to your example here by analogy.

In Gaussian units Coulomb's law is $$f=\frac{q_1 q_2}{r^2}$$ where $f$ is force, $q$ is charge, and $r$ is distance. In Gaussian units $[f]=M^{1}\ L^{1}\ T^{-2}$ and $[r]=L^1$ so $[q]=L^{3/2}\ M^{1/2}\ T^{-1}$. Importantly, charge does not have an independent base dimension. Electric charge is a derived unit.

In SI units Coulomb's law is $$f=k\frac{q_1 q_2}{r^2}$$ where $k$ is a universal constant that is present only to make the units match. As before we have $[f]=M^{1}\ L^{1}\ T^{-2}$ and $[r]=L^1$, but now $[q]=I^{1}\ T^{1}$ so $[k]=M^{1}\ L^{3}\ T^{-4}\ I^{-2}$

Note that both systems use $M$, $L$, and $T$ dimensions, but SI introduces an extra dimension, $I$, and because of the additional dimension Coulomb's law in SI contains an extra constant that is not found in Coulomb's law in Gaussian units. This constant is a "universal" dimensionful constant that seems like it is telling you something about the universe or physics, but is actually only telling you that you are using SI units instead of Gaussian units. The constant serves only to bring in the SI base dimension $I$ for electrical current, which is absent in Gaussian units.

So, you say that in all units $V=L/T$, but that is not necessarily true. We could use Dale2 units where we measure $v$ in knots with $[v]=V^{1}$, $l$ in feet with $[l]=L^{1}$, and $t$ in seconds with $[t]=T^{1}$. Then in these units we would have $$v=k \frac{l}{t}$$ where $k$ is a dimensionful constant, with dimensions $[k]=V^{1}\ L^{-1}\ T^{1}$, that converts between the Dale2 base units for time, distance, and speed.

These three base units would be independent in in precisely the same way that $I$, $T$, $M$, and $L$ are in SI units. The fact that SI units do not have an independent dimension for speed does not make the Dale2 units invalid for the same reason that Gaussian units which do not have an independent dimension for current do not make SI units invalid.

If you just feel like the Dale2 unit convention must be wrong for having $V$ as an independent base dimension then craft your argument. Whatever argument you craft can also be used to show that the SI unit convention must be wrong for having $I$ as an independent base dimension.

If the dimensions of the units of these three quantities do not have the relationship described earlier in this post, then the mathematical equation describing the above relationship between physical measurements would have to have an extra constant in it with dimension L V−1 T−1.

Yes, exactly.

 

[Ibix]

I do not see how you can not have that distinction, otherwise how can you have a distinction between one unit of area and one unit of "squared time"?

As others have noted, you do it the same way you distinguish between a unit of energy and one of torque. If the distinction matters the information is carried elsewhere, usually in surrounding text or choice of symbol.

 

[vanhees71]

That simply is not true for dimensions. Those can and do change with different unit systems. In some unit systems $v$ is dimensionless, in some it has dimensions of $L^{1}\ T^{-1}$, and in some it has its own base dimension of $V^{1}$.

In most systems of units you would only have two of those as base dimensions, and the other would be derived. But it is possible to have a system of units where $V$, $L$, and $T$ are all independent base dimensions. That would be weird, but in principle it is no different than how SI and Gaussian units treat charge.

In all unit systems $[V]=[L]/[T]$. In some unit systems (HEP natural units as well as Planck units) [L]=[T] and thus [V]=1.

 

[Dale]

In all unit systems $[V]=[L]/[T]$.

See post 49 for a counterexample

 

[Vanadium 50]

See post 49 for a counterexample

That counter-example works by takling an external constant and running it into (or, I suppose, out of) the unit definition. This is actually fine and not a counter-example at all, because everything measurable still has the relationship between measurable quantities.

By the way, I see a definite confusion in this thread about units and dimensions. They are not the same thing: torque and energy have the same dimensions (in MKSA) but are different things with different units.

 

[vanhees71]

In the SI torque and energy have the same dimensions. Of course this doesn't imply that they are the same as a physical quantity. In natural units everything is measured in dimensionless numbers but that doesn't imply at all that all quantities have the same physical meaning.

 

[Dale]

This is actually fine and not a counter-example at all, because everything measurable still has the relationship between measurable quantities.

It is a counter-example for the specific claim that "in all unit systems [V]=[L]/[T]". In the Dale2 units $$v=k\frac{l}{t}$$$$[V]=[V^{1}\ L^{-1}\ T^{1}]\frac{[L]}{[T]}$$$$[V]=[V]$$ so $V$ is an independent base dimension.

Yes, of course it is fine and everything measurable still has the relationship between measurable quantities. The point is exactly that the dimensionality of a quantity is NOT something measurable because the dimensionality can be revised by convention without affecting anything measurable.

When you measure a charge, $q$, in SI vs in Gaussian units you can use the same experiments and the same measurements even though the formulas you use are different and the dimensionality of the resulting $q$ is different.

 

[ohwilleke]

For example in SI units Newton’s 2nd law is $\Sigma \vec f =m\vec a$ but you could make a system of units (Dale units) where force is its own base dimension. In those units Newton’s 2nd law is $\Sigma \vec f = k m \vec a$ where $k$ is a dimensionful universal constant with dimensions of $F^{1}\ M^{-1}\ L^{-1}\ T^{2}$.

Dimensionless doesn't distinguish between a "base dimension" and a "derived dimension."