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Feynstein100

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In summary, the G constant in general relativity is the same as in Newtonian gravity. It is 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?

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Feynstein100

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jedishrfu

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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.

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PeterDonis

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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 convertFeynstein100 said:how is G defined in GR?

In GR, no, not really, since our choice of units is a human convention.Feynstein100 said:Is it a property of spacetime

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.

Unit conversions would probably be better described asFeynstein100 said:or just some useless mathematical artefact?

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Ibix

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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.

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Dale

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

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Feynstein100

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

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

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

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jedishrfu

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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.Feynstein100 said: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

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

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Feynstein100

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

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ohwilleke

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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.

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Dale

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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.Feynstein100 said:I wasn't referring to the value, but rather what the thing actually signifies

In both cases it is only there to make the SI units work. You get rid of it by going to natural units.Feynstein100 said: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

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Feynstein100

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That's kind of what I don't understand G is theohwilleke said:"There's a physical significance to it."

It is the same physical significance and necessarily so.

G would now need to be redefined according to the things postulated by GR, for eg spacetime curvature or something like that.

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But it appears that ##G## sticks for a lot of time with us, so simply use natural units in GR, too.

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pbuk

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No it isn't, it's a constant with dimensions of MFeynstein100 said:That's kind of what I don't understand G is theattractive force.

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

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PeterDonis

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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.Feynstein100 said: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?

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.Feynstein100 said:couldn't we extend that to G as a property of spacetime, since it plays the same mathematical role?

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Vanadium 50

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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 [itex]\alpha = \frac{1}{4\pi\epsilon_0} \frac{e^2}{\hbar c}[/itex], Ignoring the fact there is no [itex]\hbar[/itex] 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?PeterDonis said:the property of the electromagnetic field is the fine structure constant

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

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PeterDonis

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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.Vanadium 50 said:We keep saying this. I'm not sure it is more enlightening than confusing.

It appears in the appropriate classical equations, such as Coulomb's law, once artifacts of the choice of units are factored out.Vanadium 50 said:I'm also not sure it's true classically.

What do you mean by "the electron charge doubles or halves"? How are you going to make this happen?Vanadium 50 said:what happens classically if the electron charge doubles or halves?

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.

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Vanadium 50

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PeterDonis

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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.Vanadium 50 said:

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Dale

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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.Vanadium 50 said:what happens classically if the electron charge doubles or halves? Nothing, right?

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

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Vanadium 50

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Great - that will help. What is it? (Without any h's, since it's classical)Dale said:here is one dimensionless quantity

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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.

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Dale

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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.Vanadium 50 said:Great - that will help. What is it? (Without any h's, since it's classical)

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ohwilleke

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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.

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Dale

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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.ohwilleke said: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.

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

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pbuk

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

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Dale

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

*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

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pbuk

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What, we can have a consistent system of units where G is a force? Or ## c ## is a charge?Dale said:Yes, we can.

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

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PeterDonis

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In geometrized units mass and distance have the same dimension, so ##G## does have the same dimension as force.pbuk said:we can have a consistent system of units where G is a force?

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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.ohwilleke said:

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.

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##.

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Dale

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In geometrized units both G and force are dimensionless, so yes.pbuk said:What, we can have a consistent system of units where G is a force?

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.pbuk said:Or c is a charge?

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ohwilleke

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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.Dale said: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.

My bad. I'll try to use the standard functionality in the future.Dale said:Also, please use the quote functionality. @vanhees71 may not see that you are quoting him since you didn't use the standard functionality.

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Dale

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

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ohwilleke

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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.Dale said: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.

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.

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Vanadium 50

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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.

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