What does G mean in general relativity?

[Feynstein100]

In Newtonian mechanics, G is simply a proportionality constant or the force with which two bodies of unit mass attract each other. However, GR doesn't treat gravity as a force. So how is G defined in GR? Is it a property of spacetime or just some useless mathematical artefact? What does G actually mean in general relativity? Or is it even needed?

 

[jedishrfu]

The G constant has the same value as in Newtonian gravity.

https://en.wikipedia.org/wiki/General_relativity

Einstein's equations
Main articles: Einstein field equations and Mathematics of general relativity

Having formulated the relativistic, geometric version of the effects of gravity, the question of gravity's source remains. In Newtonian gravity, the source is mass. In special relativity, mass turns out to be part of a more general quantity called the energy–momentum tensor, which includes both energy and momentum densities as well as stress: pressure and shear.[38] Using the equivalence principle, this tensor is readily generalized to curved spacetime. Drawing further upon the analogy with geometric Newtonian gravity, it is natural to assume that the field equation for gravity relates this tensor and the Ricci tensor, which describes a particular class of tidal effects: the change in volume for a small cloud of test particles that are initially at rest, and then fall freely. In special relativity, conservation of energy–momentum corresponds to the statement that the energy–momentum tensor is divergence-free. This formula, too, is readily generalized to curved spacetime by replacing partial derivatives with their curved-manifold counterparts, covariant derivatives studied in differential geometry. With this additional condition—the covariant divergence of the energy–momentum tensor, and hence of whatever is on the other side of the equation, is zero—the simplest set of equations are what are called Einstein's (field) equations:
Einstein's field equations

G μ ν ≡ R μ ν − 1 2 R g μ ν = 8 π G c 4 T μ ν {\displaystyle G_{\mu \nu }\equiv R_{\mu \nu }-{\textstyle 1 \over 2}R\,g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }\,} G_{\mu \nu }\equiv R_{\mu \nu }-{\textstyle 1 \over 2}R\,g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }\,

On the left-hand side is the Einstein tensor, G μ ν {\displaystyle G_{\mu \nu }} G_{\mu \nu }, which is symmetric and a specific divergence-free combination of the Ricci tensor R μ ν {\displaystyle R_{\mu \nu }} R_{\mu \nu } and the metric. In particular,

R = g μ ν R μ ν {\displaystyle R=g^{\mu \nu }R_{\mu \nu }\,} R=g^{\mu \nu }R_{\mu \nu }\,

is the curvature scalar. The Ricci tensor itself is related to the more general Riemann curvature tensor as

R μ ν = R α μ α ν . {\displaystyle R_{\mu \nu }={R^{\alpha }}_{\mu \alpha \nu }.\,} R_{\mu \nu }={R^{\alpha }}_{\mu \alpha \nu }.\,

On the right-hand side, T μ ν {\displaystyle T_{\mu \nu }} T_{\mu \nu } is the energy–momentum tensor. All tensors are written in abstract index notation.[39] Matching the theory's prediction to observational results for planetary orbits or, equivalently, assuring that the weak-gravity, low-speed limit is Newtonian mechanics, the proportionality constant is found to be 8 π G c 4 {\textstyle {\frac {8\pi G}{c^{4}}}} {\textstyle {\frac {8\pi G}{c^{4}}}}, where G {\displaystyle G} G is the gravitational constant and c {\displaystyle c} c the speed of light in vacuum.[40] When there is no matter present, so that the energy–momentum tensor vanishes, the results are the vacuum Einstein equations,

R μ ν = 0. {\displaystyle R_{\mu \nu }=0.\,} R_{\mu \nu }=0.\,

In general relativity, the world line of a particle free from all external, non-gravitational force is a particular type of geodesic in curved spacetime. In other words, a freely moving or falling particle always moves along a geodesic.

The geodesic equation is:

d 2 x μ d s 2 + Γ μ α β d x α d s d x β d s = 0 , {\displaystyle {d^{2}x^{\mu } \over ds^{2}}+\Gamma ^{\mu }{}_{\alpha \beta }{dx^{\alpha } \over ds}{dx^{\beta } \over ds}=0,} {\displaystyle {d^{2}x^{\mu } \over ds^{2}}+\Gamma ^{\mu }{}_{\alpha \beta }{dx^{\alpha } \over ds}{dx^{\beta } \over ds}=0,}

where s {\displaystyle s} s is a scalar parameter of motion (e.g. the proper time), and Γ μ α β {\displaystyle \Gamma ^{\mu }{}_{\alpha \beta }} {\displaystyle \Gamma ^{\mu }{}_{\alpha \beta }} are Christoffel symbols (sometimes called the affine connection coefficients or Levi-Civita connection coefficients) which is symmetric in the two lower indices. Greek indices may take the values: 0, 1, 2, 3 and the summation convention is used for repeated indices α {\displaystyle \alpha } \alpha and β {\displaystyle \beta } \beta . The quantity on the left-hand-side of this equation is the acceleration of a particle, and so this equation is analogous to Newton's laws of motion which likewise provide formulae for the acceleration of a particle. This equation of motion employs the Einstein notation, meaning that repeated indices are summed (i.e. from zero to three). The Christoffel symbols are functions of the four spacetime coordinates, and so are independent of the velocity or acceleration or other characteristics of a test particle whose motion is described by the geodesic equation.

 

[PeterDonis]

how is G defined in GR?

It's part of a unit conversion factor between conventional units for mass/energy and geometric units, where mass/energy have units of length. The full conversion factor, to convert to units of length, is $G / c^2$ for mass or $G / c^4$ for energy.

Is it a property of spacetime

In GR, no, not really, since our choice of units is a human convention.

It should be noted that in quantum gravity, at least as far as we can tell with our current knowledge, $G$ functions similarly to coupling constants for non-gravitational interactions (such as the fine structure constant for electromagnetism). But that's beyond the scope of classical GR.

or just some useless mathematical artefact?

Unit conversions would probably be better described as useful mathematical artifacts. They're not strictly necessary (since we can always write our equations in "natural" units so that the conversion factors disappear--in the case of GR those would be "geometric units"), but they can be very helpful when we humans need to match up the equations with actual measurements that use conventional units.

 

[Ibix]

G does appear in GR. The Einstein field equations are $G^{ab}=\frac{8\pi G}{c^4}T^{ab}$. The $G^{ab}$ is the Einstein tensor, but the $G$ on the right is just Newton's gravitational constant.

Some authors set it to 1, arguing that it's just a constant of proportionality between units of mass and units of distance and time, and you can just pick a "geometric" unit of mass so that G=1. Some don't like to do that and keep G in the maths explicitly. Setting $c$ to 1 is almost universal in relativity, and I'm not sure why setting $G$ to 1 is not so popular.

 

[Dale]

So how is G defined in GR? Is it a property of spacetime or just some useless mathematical artefact?

G has nothing to do with GR vs Newtonian gravity. It has to do with SI units vs natural units. If you are using SI units you need G regardless of what theory you are using. If you use natural units you don’t need G regardless of what theory you are using.

 

[Feynstein100]

The G constant has the same value as in Newtonian gravity.

https://en.wikipedia.org/wiki/General_relativity

I wasn't referring to the value, but rather what the thing actually signifies Perhaps I should've made this clearer. For instance, if we say that the value of g near Earth's surface is 9.81 m/s^2, it's not just a random number. It means that if you release an object near Earth's surface, it will accelerate at a rate of 9.81 m/s^2.
In case of G, in Newtonian mechanics, it means that two objects of 1 kg mass at a distance of 1 m will each feel a gravitational force of 6.67 * 10^-11 N. There's a physical significance to it.
I was wondering if something similar exists in GR too. Like idk maybe G is the unit curvature created in spacetime by a unit mass or something like that

 

[jedishrfu]

I wasn't referring to the value, but rather what the thing actually signifies Perhaps I should've made this clearer. For instance, if we say that the value of g near Earth's surface is 9.81 m/s^2, it's not just a random number. It means that if you release an object near Earth's surface, it will accelerate at a rate of 9.81 m/s^2.
In case of G, in Newtonian mechanics, it means that two objects of 1 kg mass at a distance of 1 m will each feel a gravitational force of 6.67 * 10^-11 N. There's a physical significance to it.
I was wondering if something similar exists in GR too. Like idk maybe G is the unit curvature created in spacetime by a unit mass or something like that

Sadly, it is the constant that is needed to complete Einsteins equations. It has the same meaning as Newtons gravitational constant and is measured the same way.

https://en.wikipedia.org/wiki/Gravitational_constant?wprov=sfti1

 

[Feynstein100]

In GR, no, not really, since our choice of units is a human convention.

Hmm but can't we? I mean, G plays the same role as permeability and permittivity in the expressions for electric and magnetic forces. And those are properties of the electromagnetic field, right? So couldn't we extend that to G as a property of spacetime, since it plays the same mathematical role?

 

[ohwilleke]

"There's a physical significance to it."

It is the same physical significance and necessarily so.

Newtonian gravity and GR are equivalent in the non-relativistic limit, which basically means when the stress-energy tensor components other than mass are negligible, and mass-energy density isn't unduly compact.

So, to recover Newtonian gravity from GR the physical constant G (called "Newton's Constant"), with the same units and value, has to be in there somewhere.

Of course, if Einstein has been perverse, he could have used instead, a constant gamma, for instance, defined to be something weird like G times the reduced Planck's constant divided by Euler's constant e, times the fine structure constant, just to be annoying. But, thankfully, he didn't.

 

[Dale]

I wasn't referring to the value, but rather what the thing actually signifies

The thing that it signifies has nothing to do with which theory of gravity you are using. It is only about the units you are using. So it has the same meaning.

In case of G, in Newtonian mechanics, it means that two objects of 1 kg mass at a distance of 1 m will each feel a gravitational force of 6.67 * 10^-11 N. There's a physical significance to it.
I was wondering if something similar exists in GR too. Like idk maybe G is the unit curvature created in spacetime by a unit mass or something like that

In both cases it is only there to make the SI units work. You get rid of it by going to natural units.

 

[Feynstein100]

"There's a physical significance to it."

It is the same physical significance and necessarily so.

That's kind of what I don't understand G is the attractive force when all quantities are 1 in Newtonian mechanics. However, GR doesn't have attractive forces, just spacetime curvature. Doesn't this invalidate the Newtonian definition of G? You can't say "G is the attractive force felt by two unit bodies" when said attractive force doesn't exist in the GR formulation. That would make no sense and would kind of be like defining G as the strength of the Higgs field in Newtonian mechanics. You can't define things in your theory in terms of other things that don't exist in that theory. That's the problem I've been struggling with.
G would now need to be redefined according to the things postulated by GR, for eg spacetime curvature or something like that.

 

[dextercioby]

Just a side remark. Writing G specifically in the "shorthand version of Einstein's equations" may be misleading, as G is also the trace of the Einstein tensor, just as R is the trace of the Ricci tensor. So one is advised to use $\gamma_G$ or simply $\gamma$ to account for Cavendish constant. Also one can directly use $\kappa = \frac{8 \pi G}{c^4}$.
But it appears that $G$ sticks for a lot of time with us, so simply use natural units in GR, too.

 

[pbuk]

That's kind of what I don't understand G is the attractive force.

No it isn't, it's a constant with dimensions of M^-1L³T^-2. Forces have dimensions M¹L¹T^-2.

I suggest you read about G on Wikipedia: https://en.wikipedia.org/wiki/Gravitational_constant#Definition

 

[PeterDonis]

G plays the same role as permeability and permittivity in the expressions for electric and magnetic forces. And those are properties of the electromagnetic field, right?

No, the property of the electromagnetic field is the fine structure constant $\alpha$. Things like "permeability" and "permittivity" only arise when you make particular choices of units that split up $\alpha$ into multiple pieces.

couldn't we extend that to G as a property of spacetime, since it plays the same mathematical role?

It doesn't play the same mathematical role in GR that $\alpha$ plays in electromagnetism. The mathematical role you are thinking of (appearing as a constant in the equation for static force between two objects) is an artifact of Newtonian gravity.

 

[Vanadium 50]

the property of the electromagnetic field is the fine structure constant

We keep saying this. I'm not sure it is more enlightening than confusing. I'm also not sure it's true classically. In MKSA \alpha = \frac{1}{4\pi\epsilon_0} \frac{e^2}{\hbar c}, Ignoring the fact there is no \hbar classically, and ignoring the fact that classically atoms don't even have energy levels, much less fine structure, what happens classically if the electron charge doubles or halves? Nothing, right?

Now, if you want to argue there is no fully consistent classical theory of EM, fair enough. But GR has all the same problems.

 

[PeterDonis]

We keep saying this. I'm not sure it is more enlightening than confusing.

The reason the fine structure constant is the appropriate thing to focus on is that it is the constant associated with electromagnetism that is the same regardless of one's choice of units.

I'm also not sure it's true classically.

It appears in the appropriate classical equations, such as Coulomb's law, once artifacts of the choice of units are factored out.

what happens classically if the electron charge doubles or halves?

What do you mean by "the electron charge doubles or halves"? How are you going to make this happen?

If you just mean redefining units without changing anything physical (for example, defining a new set of "SI" units that change around everything that would need to be changed to make the defined value of $e$ twice or half what it is in our current SI unit system, but still giving correct answers for all actual physical observables in our current universe), then of course nothing observable would change, including $\alpha$, because you can't change observables by redefining units. (This means, of course, that you would have to change more than just $e$ in this unit redefinition; at a minimum, you would have to change either $\hbar$ or $c$. But those redefinitions would also have other cascading effects.)

If you mean imagining a hypothetical universe in which the actual physical phenomena were what they would be if the electron charge were twice what it is in our universe but all other fundamental constants were the same, then $\alpha$ would be $4$ times what it is in our universe and all of the changes in actual physical phenomena in that universe as compared with ours would properly be attributed to the change in $\alpha$, since that is the unit-independent physical constant.

 

[Vanadium 50]

Classically, there is no α. Further, α is essentially the electron's charge, and classically there is no electron - just a continuous fluid of charge. So how can one say that α has any significance classically? How would you even measure it?

 

[PeterDonis]

Classically, there is no α. Further, α is essentially the electron's charge, and classically there is no electron - just a continuous fluid of charge. So how can one say that α has any significance classically? How would you even measure it?

Classically, $\alpha$ is the coupling constant that appears in, for example, Coulomb's law, or at least the part of it that is independent of one's choice of units. Laws like Coulomb's law are perfectly valid classically, and the concept of a coupling constant in them makes perfect sense, even though classically we treat charge as a continuous fluid and we can't make any connection between the coupling constant in that law and, for example, the fine structure of atomic spectra.

 

[Dale]

what happens classically if the electron charge doubles or halves? Nothing, right?

I did these calculations once a long time ago. The thing is that it is not possible for only the electron charge to double. There is one dimensionless quantity and four dimensionful quantities. If you change the electron charge and keep the dimensionless quantity the same then at least one of the other dimensionful quantities must change. If you keep all of the other dimensionful quantities the same then the dimensionless quantity must change.

The changes that are physically detectable are exactly the ones where the dimensionless quantity changes.

 

[Vanadium 50]

here is one dimensionless quantity

Great - that will help. What is it? (Without any h's, since it's classical)

 

[vanhees71]

Classically you can't define a dimensionless quantity specifying the electromagnetic coupling constant, and the same is true for Newton's gravitational constant, $G$.

Also in classical electrodynamics there's no hint at a discreteness of the electric charge. Even within the Standard Model it's not particularly evident that you need an "elementary electric charge". One argument is that in the Standard model the electroweak interaction is described by a local gauge symmetry with a chiral gauge group, $\mathrm{SU}(2)_{\text{L}} \times \mathrm{U}(1)_{Y}$, and this group must not be anomalously (explicitly) broken. As it turns out with the observed charge pattern for the quarks (coming in pairs iwth charges $-e/3$ and $2e/3$ and 3 colors in each family) and leptons (coming in pairs of charge -e and 0) the gauge symmetry is safe against being anomalously broken, and together with quark confinement you have only integer multiples of the elementary charge $e$ (charge of a proton). Written down in the usual natural units of HEP physics ($\hbar=c=1$) the electric charge is dimensionless, and $e^2/4 \pi=\alpha$ is the Sommerfeld fine-structure constant, taking the approximate value of about 1/137. There's nothing in the fundamental laws that fixes it to this value and must be empirically determined.

 

[Dale]

Great - that will help. What is it? (Without any h's, since it's classical)

It is $\alpha$, but I reject your restriction of "without any $h$" in principle, even classically. We are using SI units so we cannot help but bring in $h$ with our kilograms, and hyperfine transitions with our seconds. We may use a classical theory, but since we are using SI units we cannot avoid $h$. Nor should we.

 

[ohwilleke]

"Written down in the usual natural units of HEP physics (ℏ=c=1) the electric charge is dimensionless"

Calling a physical constant dimensionless because the units are implied by a notation convention rather than explicitly stated in writing doesn't really pass the smell test.

 

[Dale]

Calling a physical constant dimensionless because the units are implied by a notation convention rather than explicitly stated in writing doesn't really pass the smell test.

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.

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

 

[pbuk]

Dimensions are part of the definition of a system of units, not a part of nature

Whoa there, our choice of a particular set of dimensions as a basis is part of the definition of a system of units, but the relationship between physical quantities IS part of nature so we can't just choose a basis that, for instance, makes the gravitational constant G = 1 and call it a force, which is what the OP is trying to do.

 

[Dale]

the relationship between physical quantities IS part of nature so we can't just choose a basis that,

Yes, we can. In SI units a charge has dimensions of $Q$ which is considered a base or fundamental dimension* that cannot be expressed in terms of other dimensions. In cgs units a charge has dimensions of $L^{3/2}\ M^{1/2}\ T^{-1}$ so it is a derived quantity that is expressed in terms of other dimensions. In SI units the speed of light has dimensions of $L^{1}\ T^{-1}$ while in geometrized units it is dimensionless. The dimensions of a quantity are tied directly to the unit system you are using and are established by convention.

*EDIT: this is a mistake. In SI units current has dimensions of $I$ which is the base dimension. Charge has dimensions of $Q=I^1 \ T^1$ in SI units

 

[pbuk]

Yes, we can.

What, we can have a consistent system of units where G is a force? Or $c$ is a charge?

PS would that make acceleration a current or would there be no raisin to go that far?

 

[vanhees71]

We can have a system of units, where all quantities are measured in dimensionless numbers like Planck's natural system of units. It is defined by setting the following fundamental constants to 1: the speed of light $c=1$, the (modified) Planck constant, $\hbar=1$, and the Gravitational constant $G=1$. A more consistent choice is $G=1/(4 \pi)$ though.

 

[PeterDonis]

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

In geometrized units mass and distance have the same dimension, so $G$ does have the same dimension as force.

 

[vanhees71]

"Written down in the usual natural units of HEP physics (ℏ=c=1) the electric charge is dimensionless"

Calling a physical constant dimensionless because the units are implied by a notation convention rather than explicitly stated in writing doesn't really pass the smell test.

Why do you think natural units "don't pass the smell test"? They make a lot of things much simpler in theoretical physics, and that's why theoretical physicists use them all the time.

Admittedly it can be sometimes cumbersome to translate back to SI units which are what one should use when it comes to experimental physics, because they are a very carefully chosen system of units with clear definitions providing the utmost precision possible with current (and future!) technology without ever changing these definitions anymore, because since 2019 they are defined via fundamental constants, defining the values of the constants, $\nu_{\text{CS}}$, $c$, $\hbar$, $e$, $k_{\text{B}}$, and $N_{\text{A}}$ to define the base units Second, Meter, kilo gram, Ampere (or rather Coulomb=Ampere times Second), Kelvin, and Mol. All other physical units are derivable from those. In fact this is not so much different than natural units; it's only the numbers that are chosen such as to provide continuity in the definition of these units which partially reach back to the time of the French Revolution.

In the usual HEP natural units, one sets $\hbar=c=k_{\text{B}}=1$. Then everything is measured in powers of in principle only one unit, which is usually chosen as GeV. Quantities like energy, momentum, mass, and temperature have the dimension GeV then; times and lengths the dimension 1/GeV (since energy times time has the dimension of an action, which is dimensionless, because of $\hbar=1$, i.e., time must have the inverse dimension of an energy etc. etc.). Often one has more intuition when measuring times and lengths in Fermis (or femto-meters, i.e., $10^{-15} \; \text{m}$. Then you only need to use $\hbar c \simeq 0.197 \text{GeV} \; \text{fm}$ to convert the length given in 1/GeV into a length given in fm and vice versa. It's very convenient and you get used to it pretty quickly.

Now what about the dimension of charge. From the fundamental operational definition of charge via Coulomb's Law in these units you get for the Coulomb potential of a point charge $\Phi=Q/(4 \pi r)$. Since $\Phi$ is an energy and the distance from the charge, $r$, is measured in 1/GeV, the electric charge $Q$ must be a dimensionless quantity.

Now it turns out that the elementary charge is related to the fine structure constant by $\alpha=e^2/(4 \pi)$, and usually one quotes $\alpha$ (approximately 1/137).

To get rid of the last arbitrary unit (in this HEP example the GeV or fm) you can also set the gravitational constant to 1 (or better to 1/(4 \pi) to make the system "rational", i.e., no factors $4 \pi$ in the fundamental equations; in this case in Newton's Law for the gravitational potential $\Delta \Phi=4 \pi G \rho_m$, which in this natural units then simply becomes $\Delta \Phi=\rho_m$.

 

[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.