<p>See also <a href="https://physics.stackexchange.com/questions/11614/do-the-laws-of-physics-work-everywhere-in-the-universe">Do the laws of physics work everywhere in the universe?</a> on the Physics SE.</p>
<p>If the gravitational constant were not constant, then the laws of physics work differently elsewhere in the universe. If that were the case, we would not measure conservation of linear momentum (by Noether's theorem). Since we measure conservation of linear momentum, we have reason to believe <span class="math-container">$G$</span> is constant across space.</p>
<p>Besides, if <span class="math-container">$G$</span> were not constant over our universe, we ought to be able to observe it directly - for example, galaxies that are far away would behave in ways that General Relativity cannot explain.</p>

CC BY-SA 4.0
2024-04-04T22:54:25.273Z
Thanks, I am thinking the main body of universe contained inside is almost all a steady state and if variations occur it is under the extreme conditions of the boundaries, and perhaps in a black hole etc.
Bryan Major
CC BY-SA 4.0
2024-04-05T03:23:13.400Z
@ScienceSnake I'm not aware of any such test, and I'm sure you are right, but I just meant to say that we have experimental reason to believe that the laws of physics do not change much over time & space.
Allure
CC BY-SA 4.0
2024-04-05T03:24:42.200Z
@BryanMajor that's too vague - you need more details like what is a "steady state", how you define "boundary" (are you postulating that the universe has a boundary?), etc.
Allure
CC BY-SA 4.0
2024-04-05T05:37:49.483Z
"*galaxies that are far away would behave in ways that General Relativity cannot explain*" We do observe this. We can make it fit with GR by hypothesizing something that has mass but doesn't interact with light: Dark Matter. Others are trying to explain it with changes to the laws of gravity. I don't know if anyone is proposing a change to G.
Schwern
CC BY-SA 4.0
2024-04-04T03:16:42.810Z
2024-04-04T03:16:42.810Z
Allure
<p>Within <a href="https://en.wikipedia.org/wiki/General_relativity" rel="nofollow noreferrer">General Relativity (GR)</a>, the gravitational constant <span class="math-container">$G$</span> is postulated to be a fundamental constant having the same value over our Universe.</p>
<p>Within the <a href="https://en.wikipedia.org/wiki/Brans%E2%80%93Dicke_theory" rel="nofollow noreferrer">Brans-Dicke theory</a>, which is a competitor to GR, the gravitational constant is not actually a constant. Instead, its inverse, <span class="math-container">$ 1/G $</span> is set to be equal to a scalar field <span class="math-container">$\varphi$</span> whose values can vary over the space-time.  A similar approach is taken in a more general <a href="https://en.wikipedia.org/wiki/Scalar%E2%80%93tensor_theory" rel="nofollow noreferrer">scalar-tensor theory</a>, of which the Brans-Dicke theory is a particular case.</p>

CC BY-SA 4.0
2024-04-04T03:04:49.703Z
Thanks, my reasoning is that if it is not constant,  we cannot be sure that seeing an expanding universe really means it is expanding, and energy balance calculations become very tricky.
Bryan Major
CC BY-SA 4.0
2024-04-04T02:38:23.473Z
2024-04-04T02:45:27.717Z
2024-04-04T02:45:27.717Z
Michael_1812
<p>One of the big problems with the gravitational constant is that its units are not dimensionless. And so one expects it to represent some physical quantity. What is that?</p>
<p>Another problem is that observations of gravity outside our solar system fit <a href="https://wikipedia.org/wiki/Modified_Newtonian_dynamics" rel="nofollow noreferrer">MOND theory</a> much better than Newton's theory.</p>
<p>A third problem is that <span class="math-container">$\dfrac{g}{G}=R_\odot$</span>. And this coheres with the previous statement.</p>
<p>A fourth problem is that despite a huge effort to unify gravity and the <a href="https://en.wikipedia.org/wiki/Standard_Model" rel="nofollow noreferrer">Standard Model</a>, no unification has been found.</p>
<p>Harmony may be found by explaining the gravitational force within the framework of <a href="https://en.wikipedia.org/wiki/Standard_Model" rel="nofollow noreferrer">quantum mechanics</a>, not outside it.</p>
<p>The growing <a href="https://en.wikipedia.org/wiki/Hubble%27s_law#Hubble_tension" rel="nofollow noreferrer">crisis in cosmology</a> demands a paradigm shift in physical theories and the observation that G isn't universal is part of that.</p>

CC BY-SA 4.0
2024-04-06T19:12:46.723Z
What is $\dfrac{g}{G}=R_\odot$ ? It doesn't make sense, given the usual meanings of those symbols.
PM 2Ring
CC BY-SA 4.0
2024-04-08T06:03:10.530Z
9.81/6.674e-11 ≈ 1.47e11 which is the distance to the sun at perihelion.
David
CC BY-SA 4.0
2024-04-08T06:23:06.310Z
Ok, but that's just a cute numerical coincidence. And it's dimensionally inconsistent, so the coincidence relies on measuring with SI units. $g/G$ has units of $\rm{kg/m^2}$, but the perihelion distance has units of $\rm{m}$.
PM 2Ring
CC BY-SA 4.0
2024-04-08T07:39:05.450Z
It's the ULG that's dimensionally inconsistent. The units of G are there to compensate for this inconsistency. Which suggests there are hidden factors of the equation that are hidden behind G.
David
CC BY-SA 4.0
2024-04-08T07:49:27.590Z
It's just a coincidence that ULG works within the solar system but not outside it. Those who believe in a pattern and order to the universe therefore ask why, seeking true science, and not leaving chaos and coincidence as an explanation. One fruit of this enquiry is MOND theory which fits observational data without special appeal to dark matter. (At least according to some papers.)
David
CC BY-SA 4.0
2024-04-04T05:42:34.303Z
2024-04-05T08:28:13.433Z
2024-04-05T08:28:13.433Z
David
<p>For instance, if objects are travelling at relatively high speed around 'c', is G the same value for them as for us on earth?</p>
<p>I believe a lot of matter is travelling fast away from other matter, so is G the same for that matter? Space-time may be involved in any answer.</p>

CC BY-SA 4.0
2024-04-05T05:34:06.230Z
"*I believe a lot of matter is travelling fast away from other matter*" The appearance of distant matter tending to be moving away from us is due to the [expansion of space itself](https://en.wikipedia.org/wiki/Expansion_of_the_universe). Like dots on an expanding balloon, the dots don't move, the space between them gets bigger. G is the same. Velocity also does not affect G, light is affected by gravity the same as everything else.
Schwern
CC BY-SA 4.0
2024-04-06T15:29:35.830Z
@Schwern expansion of space is not a physical phenomenon distinct from objects moving apart (see [this answer](https://astronomy.stackexchange.com/a/53836/47607) or your link)
Sten
CC BY-SA 4.0
2024-04-06T16:18:17.223Z
Nothing depends on speed, because there is no absolute speed.
ProfRob
CC BY-SA 4.0
2024-04-06T18:20:51.103Z
The gravitational force within an isotropic mass distribution must be zero, so the value of $G$ should not really matter globally. Gravity manifests itself only in case of anisotropies i.e. usually only on a small scale.
Thomas
CC BY-SA 4.0
2024-04-07T19:29:01.623Z
@Thomas Correct that gravitational force at the center of an isotropic distribution is zero, but incorrect that this means gravity does not manifest. For example, $G$ enters into the Friedmann equations. The resolution is that different observers do not agree on acceleration measurements because their own reference frames are accelerating differently. Each observer says that they are unaccelerated but the others are accelerated.
Sten
CC BY-SA 4.0
2024-04-08T07:17:17.150Z
@Sten The Friedman equations purportedly assume global isotropy and homogeneity ( https://en.wikipedia.org/wiki/Friedmann_equations#Assumptions )
Thomas
<astrophysics><gravity><universe><space-time>
Is the gravitational constant really constant over our Universe?
Bryan Major
# Is the gravitational constant really constant over our Universe?

(Asked by: username_0 on 2024-04-03)

For instance, if objects are travelling at relatively high speed around 'c', is G the same value for them as for us on earth?

I believe a lot of matter is travelling fast away from other matter, so is G the same for that matter? Space-time may be involved in any answer.

**Tags:** astrophysics, gravity, universe, space-time
> **Comments:**
> **username_1**: "*I believe a lot of matter is travelling fast away from other matter*" The appearance of distant matter tending to be moving away from us is due to the [expansion of space itself](https://en.wikipedia.org/wiki/Expansion_of_the_universe). Like dots on an expanding balloon, the dots don't move, the space between them gets bigger. G is the same. Velocity also does not affect G, light is affected by gravity the same as everything else.  
> **username_2**: @Schwern expansion of space is not a physical phenomenon distinct from objects moving apart (see [this answer](https://astronomy.stackexchange.com/a/53836/47607) or your link)  
> **username_3**: Nothing depends on speed, because there is no absolute speed.  
> **username_4**: The gravitational force within an isotropic mass distribution must be zero, so the value of $G$ should not really matter globally. Gravity manifests itself only in case of anisotropies i.e. usually only on a small scale.  
> **username_2**: @Thomas Correct that gravitational force at the center of an isotropic distribution is zero, but incorrect that this means gravity does not manifest. For example, $G$ enters into the Friedmann equations. The resolution is that different observers do not agree on acceleration measurements because their own reference frames are accelerating differently. Each observer says that they are unaccelerated but the others are accelerated.  
> **username_4**: @Sten The Friedman equations purportedly assume global isotropy and homogeneity ( https://en.wikipedia.org/wiki/Friedmann_equations#Assumptions )  

---

## 3 answers

### Answer by username_5 on 2024-04-04. Score: 12

See also [Do the laws of physics work everywhere in the universe?](https://physics.stackexchange.com/questions/11614/do-the-laws-of-physics-work-everywhere-in-the-universe) on the Physics SE.

If the gravitational constant were not constant, then the laws of physics work differently elsewhere in the universe. If that were the case, we would not measure conservation of linear momentum (by Noether's theorem). Since we measure conservation of linear momentum, we have reason to believe $G$ is constant across space.

Besides, if $G$ were not constant over our universe, we ought to be able to observe it directly - for example, galaxies that are far away would behave in ways that General Relativity cannot explain.

> **Comments:**
> **username_0**: Thanks, I am thinking the main body of universe contained inside is almost all a steady state and if variations occur it is under the extreme conditions of the boundaries, and perhaps in a black hole etc.  
> **username_5**: @ScienceSnake I'm not aware of any such test, and I'm sure you are right, but I just meant to say that we have experimental reason to believe that the laws of physics do not change much over time & space.  
> **username_5**: @BryanMajor that's too vague - you need more details like what is a "steady state", how you define "boundary" (are you postulating that the universe has a boundary?), etc.  
> **username_1**: "*galaxies that are far away would behave in ways that General Relativity cannot explain*" We do observe this. We can make it fit with GR by hypothesizing something that has mass but doesn't interact with light: Dark Matter. Others are trying to explain it with changes to the laws of gravity. I don't know if anyone is proposing a change to G.  

---

### Answer by username_6 on 2024-04-04. Score: 4

Within [General Relativity (GR)](https://en.wikipedia.org/wiki/General_relativity), the gravitational constant $G$ is postulated to be a fundamental constant having the same value over our Universe.

Within the [Brans-Dicke theory](https://en.wikipedia.org/wiki/Brans%E2%80%93Dicke_theory), which is a competitor to GR, the gravitational constant is not actually a constant. Instead, its inverse, $ 1/G $ is set to be equal to a scalar field $\varphi$ whose values can vary over the space-time. A similar approach is taken in a more general [scalar-tensor theory](https://en.wikipedia.org/wiki/Scalar%E2%80%93tensor_theory), of which the Brans-Dicke theory is a particular case.

> **Comments:**
> **username_0**: Thanks, my reasoning is that if it is not constant,  we cannot be sure that seeing an expanding universe really means it is expanding, and energy balance calculations become very tricky.  

---

### Answer by username_7 on 2024-04-04. Score: -7

One of the big problems with the gravitational constant is that its units are not dimensionless. And so one expects it to represent some physical quantity. What is that?

Another problem is that observations of gravity outside our solar system fit [MOND theory](https://wikipedia.org/wiki/Modified_Newtonian_dynamics) much better than Newton's theory.

A third problem is that $\dfrac{g}{G}=R\_\odot$. And this coheres with the previous statement.

A fourth problem is that despite a huge effort to unify gravity and the [Standard Model](https://en.wikipedia.org/wiki/Standard_Model), no unification has been found.

Harmony may be found by explaining the gravitational force within the framework of [quantum mechanics](https://en.wikipedia.org/wiki/Standard_Model), not outside it.

The growing [crisis in cosmology](https://en.wikipedia.org/wiki/Hubble%27s_law#Hubble_tension) demands a paradigm shift in physical theories and the observation that G isn't universal is part of that.

> **Comments:**
> **username_8**: What is $\dfrac{g}{G}=R_\odot$ ? It doesn't make sense, given the usual meanings of those symbols.  
> **username_7**: 9.81/6.674e-11 ≈ 1.47e11 which is the distance to the sun at perihelion.  
> **username_8**: Ok, but that's just a cute numerical coincidence. And it's dimensionally inconsistent, so the coincidence relies on measuring with SI units. $g/G$ has units of $\rm{kg/m^2}$, but the perihelion distance has units of $\rm{m}$.  
> **username_7**: It's the ULG that's dimensionally inconsistent. The units of G are there to compensate for this inconsistency. Which suggests there are hidden factors of the equation that are hidden behind G.  
> **username_7**: It's just a coincidence that ULG works within the solar system but not outside it. Those who believe in a pattern and order to the universe therefore ask why, seeking true science, and not leaving chaos and coincidence as an explanation. One fruit of this enquiry is MOND theory which fits observational data without special appeal to dark matter. (At least according to some papers.)  

---

