My Zero Gravity Challenge

[AV1611VET]

I voted "Don't Know" because I don't know.

Thank you for voting.

It's your civic duty.

 

[SelfSim]

Thank you for voting.

It's your civic duty.

.. (whereas our votes, and hence our civic duties, have been suppressed by the lack of context).

 

[Nithavela]

Shouldn't be too hard.

Thirteen posts now, and not one vote yet.

What? I voted before posting.

 

[AV1611VET]

What? I voted before posting.

Ya ... sorry.

Since I didn't vote myself, no votes show until I click the button.

 

[durangodawood]

I vote no.

Gravity exists, at least as an effect, everywhere - as I understand it. The effect never diminishes to zero over distance.

Would be interesting to know if Im wrong about this.

 

[sjastro]

I vote no.

Gravity exists, at least as an effect, everywhere - as I understand it. The effect never diminishes to zero over distance.

Would be interesting to know if I'm wrong about this.

If you are into Newtonian gravity the answer is no.
Newtonian gravity follows an inverse square law in our 3D spatial universe.

r → ∞, F → 0

Since the observable universe is finite in size, gravity can never be zero.

On the other hand the theory of gravity in general relativity is derived from the Einstein-Hilbert action S, the answer is yes.

Since gravity is not a "real" force in GR and does not follow the inverse square law, the bending of space-time due to gravity when mass is present has a finite range leading to asymptotically flat space-time.
When space-time becomes flat at a certain distance from the source such as a star or planet the gravitational field is for all intents and purposes zero.

 

[durangodawood]

If you are into Newtonian gravity the answer is no.
Newtonian gravity follows an inverse square law in our 3D spatial universe.

View attachment 322259

r → ∞, F → 0

Since the observable universe is finite in size, gravity can never be zero.

On the other hand the theory of gravity in general relativity is derived from the Einstein-Hilbert action S, the answer is yes.

View attachment 322262

Since gravity is not a "real" force in GR and does not follow the inverse square law, the bending of space-time due to gravity when mass is present has a finite range leading to asymptotically flat space-time.
When space-time becomes flat at a certain distance from the source such as a star or planet the gravitational field is for all intents and purposes zero.

But "negligible" is not the same as zero, right?

 

[sjastro]

But "negligible" is not the same as zero, right?

Negligible usually means the effects are so small they can be neglected as they are beyond measurement.
An example of a metric based on asymptotically flat space-time is the Schwarzschild metric.
This metric predicted space-time curvature caused by the Sun's gravity resulted in a planet's perihelion or closest approach to the Sun to advance by an angle Δ with each orbit.
Newtonian gravity cannot account for this effect.

Actual measurements of the perihelion advance Δϕ in units of arcseconds per century for each planet.

Mercury 43.1000 ± 0.5000
Venus 12.00000 ± 3.00000
Earth 5.0000 ± 1.0000
Mars 1.3624 ± 0.0005
Jupiter 0.0700 ± 0.0040
Saturn 0.0140 ± 0.0020
Uranus ?
Neptune ?

Mercury has the largest perihelion advance because it is closest to the Sun where the effects of space-time curvature is at its greatest.
As one moves further out into the solar system, space-time becomes flatter and flatter and the value of Δϕ decreases.
Measurements for Uranus are so small as to be unreliable while Neptune cannot even be measured even if it was possible as it has not yet completed a full orbit when Einstein made the prediction in 1915.

What the data shows is space-time rapidly converges to a flat geometry even at relatively short distances from the gravity source of our Sun.
While Neptune is only 4.545 billion km from the Sun, one the furthest objects in a Newtonian orbit around the Sun, Sedna is 143.73 billion km distant.
The theoretical value for the perihelion advance at such distances is essentially zero indicating while Newtonian gravity still exerts an influence, space-time curvature caused by Sun's gravity does not extend out this far.

 

[AV1611VET]

Negligible usually means the effects are so small they can be neglected as they are beyond measurement.

As I understand it, an electron on one side the the universe exerts a pull on an electron on the other side.

 

[durangodawood]

Negligible usually means the effects are so small they can be neglected as they are beyond measurement.
An example of a metric based on asymptotically flat space-time is the Schwarzschild metric.
This metric predicted space-time curvature caused by the Sun's gravity resulted in a planet's perihelion or closest approach to the Sun to advance by an angle Δ with each orbit.
Newtonian gravity cannot account for this effect.

Actual measurements of the perihelion advance Δϕ in units of arcseconds per century for each planet.

Mercury 43.1000 ± 0.5000
Venus 12.00000 ± 3.00000
Earth 5.0000 ± 1.0000
Mars 1.3624 ± 0.0005
Jupiter 0.0700 ± 0.0040
Saturn 0.0140 ± 0.0020
Uranus ?
Neptune ?

Mercury has the largest perihelion advance because it is closest to the Sun where the effects of space-time curvature is at its greatest.
As one moves further out into the solar system, space-time becomes flatter and flatter and the value of Δϕ decreases.
Measurements for Uranus are so small as to be unreliable while Neptune cannot even be measured even if it was possible as it has not yet completed a full orbit when Einstein made the prediction in 1915.

What the data shows is space-time rapidly converges to a flat geometry even at relatively short distances from the gravity source of our Sun.
While Neptune is only 4.545 billion km from the Sun, one the furthest objects in a Newtonian orbit around the Sun, Sedna is 143.73 billion km distant.
The theoretical value for the perihelion advance at such distances is essentially zero indicating while Newtonian gravity still exerts an influence, space-time curvature caused by Sun's gravity does not extend out this far.

Interesting. Thanks!

But for this poll it depends on how seriously you take the option "zero". I mean "effectively zero" or "zero for intents and purposes" is not strictly zero necessarily.

 

[driewerf]

As I understand it, an electron on one side the the universe exerts a pull on an electron on the other side.

Yes. But the repellent force due to the same charge would be much bigger.

 

[AV1611VET]

Yes. But the repellent force due to the same charge would be much bigger.

Of the four forces, gravity is the weakest, but stretches the farthest.

 

[sjastro]

Interesting. Thanks!

But for this poll it depends on how seriously you take the option "zero". I mean "effectively zero" or "zero for intents and purposes" is not strictly zero necessarily.

Science is not about semantics.
We can use AV’s example of two electrons separated by 92 billion light years and calculate the gravitational force between them.
Using the inverse square law we find it is a non-zero value which is so ridiculously small it is impossible to measure.
In a scientific model infinitesimally small values are treated as zero as scientific models by definition are approximations.

The situation is different if we consider the matter density parameter of the universe Ωₘ.
Before dark energy was discovered Ωₘ ≈ 0.7 and it was thought Newtonian gravity was sufficiently strong to affect expansion by slowing down its rate to a point where in the infinite future it would come to a halt.
After the discovery Ωₘ ≈ 0.3, the expansion of the universe was found to be accelerating which could not be countered by Newtonian gravity.

While Newtonian gravity exists even at cosmological scales in an accelerating universe, evidence points to the universe being flat where gravity in the general relativity sense as bending space-time at cosmological scales is zero.
The measured curvature of the observable universe Ωₖ = 0.0007 ± 0.0019 is obtained by measuring the angular separation of structures in the CMB.
A flat universe has a value of Ωₖ = 0, the measured values are not only extremely small but a value of Ωₖ = 0 also falls within the uncertainty limits of the measurements.
A non zero value of Ωₖ more likely indicates uncertainty in the measurements and Ωₖ = 0 reflects the geometry of the observable universe as being flat where gravity does not play a role at cosmological scales.

 

[durangodawood]

Science is not about semantics.
We can use AV’s example of two electrons separated by 92 billion light years and calculate the gravitational force between them.
Using the inverse square law we find it is a non-zero value which is so ridiculously small it is impossible to measure.
In a scientific model infinitesimally small values are treated as zero as scientific models by definition are approximations.

The situation is different if we consider the matter density parameter of the universe Ωₘ.
Before dark energy was discovered Ωₘ ≈ 0.7 and it was thought Newtonian gravity was sufficiently strong to affect expansion by slowing down its rate to a point where in the infinite future it would come to a halt.
After the discovery Ωₘ ≈ 0.3, the expansion of the universe was found to be accelerating which could not be countered by Newtonian gravity.

While Newtonian gravity exists even at cosmological scales in an accelerating universe, evidence points to the universe being flat where gravity in the general relativity sense as bending space-time at cosmological scales is zero.
The measured curvature of the observable universe Ωₖ = 0.0007 ± 0.0019 is obtained by measuring the angular separation of structures in the CMB.
A flat universe has a value of Ωₖ = 0, the measured values are not only extremely small but a value of Ωₖ = 0 also falls within the uncertainty limits of the measurements.
A non zero value of Ωₖ more likely indicates uncertainty in the measurements and Ωₖ = 0 reflects the geometry of the observable universe as being flat where gravity does not play a role at cosmological scales.

Again really interesting.

You say science isnt about semantics.... but then you provide me with a functional scientific definition for term zero. We do have to agree on a definition for the term in the poll question.

I will defer to the definition you provided, but I do have to be dragged kicking and screaming because I instinctually consider zero to mean not "negligible", but... "none"

 

[SelfSim]

Again really interesting.

You say science isnt about semantics.... but then you provide me with a functional scientific definition for term zero. We do have to agree on a definition for the term in the poll question.

I will defer to the definition you provided, but I do have to be dragged kicking and screaming because I instinctually consider zero to mean not "negligible", but... "none"

Its an example of where math concepts, appearing in math models, don't always make the transition into physical models like the universe.
'Zero' may appear in math models but 'zero' of anything cannot be tested for existence beyond the precision limits of measurement.

 

[sjastro]

Again really interesting.

You say science isnt about semantics.... but then you provide me with a functional scientific definition for term zero. We do have to agree on a definition for the term in the poll question.

I will defer to the definition you provided, but I do have to be dragged kicking and screaming because I instinctually consider zero to mean not "negligible", but... "none"

Zero is a numerical value which can be based on a measurement.
The temperature can be 0⁰ C but it doesn’t make sense to refer to the temperature as "none" as it implies temperature is a countable physical object.
Mathematicians and scientists use the term "negligible" to exclude values which are too small or insignificant in their models in which case they take on the value of zero.

 

[Bradskii]

Zero is a numerical value which can be based on a measurement.
The temperature can be 0⁰ C but it doesn’t make sense to refer to the temperature as "none" as it implies temperature is a countable physical object.
Mathematicians and scientists use the term "negligible" to exclude values which are too small or insignificant in their models in which case they take on the value of zero.

I'm with @durangodawood on this. If the question was 'Are there situations where the effects of gravity are so small that they can't be measured?' then the answer would be 'Yes, the effects are there but can't be measured in any meaningfull way'.

 

[SelfSim]

If the question was 'Are there situations where the effects of gravity are so small that they can't be measured?' then the answer would be 'Yes, the effects are there but can't be measured in any meaningfull way'.

.. and how do you know this?

 

[sjastro]

I'm with @durangodawood on this. If the question was 'Are there situations where the effects of gravity are so small that they can't be measured?' then the answer would be 'Yes, the effects are there but can't be measured in any meaningfull way'.

Like trying to test the gravitational force between two electrons 92 billion light years apart?

If the effects of gravity can't be measured in a meaningful way then you can't infer the effects are there.
If you had the capability of making measurements, you don't know if the measurements made are statistically significant or nothing but noise composed of random and non random errors.
Even after eliminating all the non random errors in your measurements the danger is the random or statistical error might resemble a real signal.
This is why scientists analyze very large numbers of measurement data to minimize this pitfall.
What does 5-sigma mean in science?.

 

[Bradskii]

Like trying to test the gravitational force between two electrons 92 billion light years apart?

If the effects of gravity can't be measured in a meaningful way then you can't infer the effects are there.

And you can't deny that they might be there. So the very best that you can say is that we can't measure them. But they theoretically exist. End of story.