What is the nature of gravity constant?

[freesom]

what is the nature of gravity constant?

i just ask that?

and if we can know the nature of it

can we understand the meaning of gravity?

 

[russ_watters]

Huh? The gravitational constant? It is number. A physical constant.

Physicists look for order, they don't look for "meaning". "Meaning" doesn't really mean anything anyway.

 

[rbj]

what is the nature of gravity constant?

i just ask that?

and if we can know the nature of it

can we understand the meaning of gravity?

since it is unit-dependent, there is no meaning in the numerical value that you normally see (except, i s'pose, knowing that using the anthropometric units we usually use, gravity can be said to be a "weak force" because G appears to be very small in these human-oriented units). but that meaning really has to do with scale that seems meaningful to humans, which is sort of arbitrary.

Huh? The gravitational constant? It is number. A physical constant.

it's not just a number, it's a dimensionful number. one that normally has units attached to it. so the value of the number attached to those units change if the units change. we can define a set of units to make the numerical value of G anything we want. Planck units (and some other definitions of "natural units") are chosen so that G = 1.

Physicists look for order, they don't look for "meaning". "Meaning" doesn't really mean anything anyway.

oh, Russ, certainly sometimes they look for meaning. no? isn't that what physicists do when they try to simplify models (like Maxwell's eqs.)? or come up with a completely different paradigm (like "gravity = curvature of space-time"). that stuff is about "meaning", isn't it?

 

[yogi]

The units of Big G, the so called "Gravitational Constant" are most interesting: cubic meters per second squared per kilogram. Translated, this means Volumetric Acceleration per unit mass. The first thing that might occur to one looking at this is why would the volumetric accelertion be the same for a much smaller universe, say one only the size of a basketball as compared to one having a Hubble radius of 10^26 meters. The other thing that should be suggested is how do you get the number 6.67 x 10^-11 out of what we think we know about the size and rate of expansion of the cosmos. Specifically, how does the volumetric acceleration come into being if the Hubble expansion rate is uniform?

 

[gamburch]

The gravitational constant is equal to (3Pi/4)H(squared)/density of matter in the universe. I hope this helps.

 

[yogi]

Hi gamburch

How did you arrive at this - I assume you used Friedmann's relationship that relates G and density. If the universe has critical density mass - then the factor would be 3pi/8 rather than 3pi/4.

Yogi

 

[rbj]

The gravitational constant is equal to (3Pi/4)H(squared)/density of matter in the universe. I hope this helps.

How did you arrive at this - I assume you used Friedmann's relationship that relates G and density. If the universe has critical density mass - then the factor would be 3pi/8 rather than 3pi/4.

in addition, gamburch, can you use the LaTeX markup so we what you are talking about? and can you identify all of your symbols? is it this:

$$G = \frac{3 \pi}{4} \frac{H^2}{\rho}$$ ??

where $\rho$ is the "density of matter in the universe". if so, can you identify H?

 

[pervect]

This appears to be an incorrect derivation from The Friedmann equations.

The correct expression would be

$$G = \frac{3 H^2}{8 \pi \rho}$$

In any event, this equation is not always true - it requires that the cosmological constant $\Lambda$ be zero, that the spatial curvature of the universe K=0, and of course the assumptions that GR is correct and the cosmological principle holds so that the universe is homogenoeus and isotropic.

H here would be Hubble's constant.

 

[pervect]

I've moved a post by rbj to the cosmology forum as a new thread https://www.physicsforums.com/showthread.php?t=198513

 

[gamburch]

Yogi: Thanks for your interest. Gravitational potential has the units of velocity squared. If one uses the value of the density of matter of the universe published by Peeblers some time ago, one can compute the gravitational potential locally arising from all matter in the universe by merely integrating over a sphere of size equal to the speed of light divided by the Hubble constant. This computation yields a number close to the speed of light squared. I like this way of doing things because it rather nicely explains the relationship of mass to energy; the potential times the matter being the cost of its assembly.

If in the computation one substitutes c/H for the radius one has c squared on both sides yielding a relationship between the gravitational constant, the density of the universe and Hubble's constant. The constant in the equation comes from the volume of the sphere and has no magic. A more careful worker than I would have accounted for the Doppler effect on the local field so the criticisms of my simple computation are probably very apt. I did not get the equation from elsewhere, however. What you see is what you get.

 

[yogi]

Interesting Gamburch. I have seen the same formula derived using Hubble's law v = Hr, the procudure the poster used was to differentiate and plug in the value v = c at the Hubble limit getting dv/dt = Hc and then re substituting c = Hr giving an effective acceleration = (H^2)R If I remember, the fellow actually got the paper published in an electronics magazine - probably because he extrapolated the result in terms of electric fields, which I could never figure out. Maybe I can dig up the old thread, you might be interested.

regards

yogi

 

[yogi]

This is a follow-up to post 11. I don't think the old thread is still available - at least I can't find it - but here is the rest of what the poster did - he sets the force of gravity GMm'/R^2 equal to the F = m'a force. For "a" he uses the (H^2)/R that he derived from from Hubbles law - so he sets (GM/R^2) = (H^2)R where M is the mass of the universe
and since M = V(rho) and V = 4/3(pi)R^3 ...vola, G = [3/4(pi)](H^2)

I am sure the math types will have nothing but condemnation for this sort of procedure but I think these simple relationships are telling us something about the universe that is being overlooked

 

[gamburch]

Thank you Yogi. I truly appreciate this recollection and will be mulling it over; however, today is the day for starting my Christmas home brew. I'll be back soon. Thanks again.

 

[yogi]

Home Brew - sounds like something with profound cosmological intrigue. When we talk later I will show you my own derivation of G - and some consequences thereof. Seems there are many ways to skin the same cat.

Happy Holidays

Yogi

 

[exponent137]

I have almost the same question as this one of Freesom. It is known what means G in Newtonian gravity. (for instance: acceleration 1 m from 1 kg heavy (point) ball) Maybe there is something still more clear. But, did G get any clearer meaning in Einsteinian curved space?

"Physicists look for order, they don't look for "meaning". "Meaning" doesn't really mean anything anyway."
I never agreed with this style of thinking. This style make unclear, what can be easily clear.

 

[Vanadium 50]

I have almost the same question as this one of Freesom. It is known what means G in Newtonian gravity. (for instance: acceleration 1 m from 1 kg heavy (point) ball) Maybe there is something still more clear. But, did G get any clearer meaning in Einsteinian curved space?

I don't see how it can be made much clearer than "if you make this measurement, you'll get this result".

 

[Dale]

In both theories G has the same role. GR made two principle improvements over Newtonian gravity. First, it is more general, applying in situations where Newtonian gravity was wrong (Mercury) or unclear (bending light). Second, it makes the relationship between inertia and gravitation more clear (equivalence of gravitational and inertial mass).

G itself, as a dimensionful constant, can be seen as simply relating different units. You cannot really attach any physical significance to any dimensionful constant as anything other than a unit conversion. This independent of which theory you use.

 

[exponent137]

I don't see how it can be made much clearer than "if you make this measurement, you'll get this result".

Still more elementary than acceleration, is impact on light. So maybe G which causes horizon of a black hole?

Every new interpretation of G can maybe "open eyes" for new physics.

And, how to connect G with equivalence principle. In the most simple example (elevator) G can be ignored.

 

[ZachN]

The true nature of the constant is that it is there to make up for a lack of understanding for the true mechanism for gravity. Constants are used to express values we can't explain through other means. That is the dirty secret of science.

e.g. - the resistance(drag) of an object falling through the atmosphere is proportional to the velocity. we get an equation to model the system which looks like this:

mdx/dt = mg - kv

we could use this generic equation to model the falling of most objects with decent results. However, much work has been done to study the drag on specifically shaped objects and so k can be replaced with more accurate models.

the kv term can be replaced in the 2-d case with d=q_inf*C_D*A_frontal. But now look at this we have a new constant C_D (drag coefficient) which is essentially another dimensionless quantity to make up for more complex behavior which would be too difficult to discern at this time.

So that is the story of coefficients and constants. Whenever you see one remember that it represents something we don't yet understand.

 

[A.T.]

e.g. - the resistance(drag) of an object falling through the atmosphere is proportional to the velocity.

Drag is proportional to velocity squared.

So that is the story of coefficients and constants. Whenever you see one remember that it represents something we don't yet understand.

We do understand how drag works. But the computations for specific shapes are very complex, even for computers. The drag coefficient is a practical engineering approach.

 

[ZachN]

Proportional to the velocity was just an example.

We do understand how drag works. But the computations for specific shapes are very complex, even for computers. The drag coefficient is a practical engineering approach.[quote/]

Exactly the point, in order to calculate drag exactly we would need to no the position and properties of every molecule. Because that is difficult for us to do at this point we use constants such as the drag coefficient.

Because we don't actually understand the true nature and cause of gravity we have a gravitational constant which works for what we need it for. But the point is that constants are a way for us to make up for what we can't completely explain or calculate.

 

[Dale]

The true nature of the constant is that it is there to make up for a lack of understanding for the true mechanism for gravity. Constants are used to express values we can't explain through other means. That is the dirty secret of science. ...

So that is the story of coefficients and constants. Whenever you see one remember that it represents something we don't yet understand.

What you say is correct in some instances, but you are vastly over-generalizing.

The instance that you are correct is the fundamental dimensionless constants. Those do represent features of the universe that fundamentally cannot be explained by any modern theory. This is no secret, scientists are quite open about it and the search for a "Theory of Everything" is largely motivated by this problem.

However, this is not the case with the dimensionful "fundamental" constants (e.g. c or G), those are completely understood as they are simply artifacts of our choice of units and essentially just allow us to convert between different units.

A second category of constant that does not represent a fundamental lack of understanding is what A.T. mentioned above (e.g. your drag coefficient). They represent a situation where the fundamental rules that govern the system are understood, but the boundary conditions or the system are too complicated to apply the fundamental rules given current computational limitations. Such constants abound in biology and engineering.

A final type of constant that is not due to a lack of knowledge is when a quantity is defined as a proportionality. For example, resistance is defined as the constant R in V=IR. Such constants would remain even with a tractable system and a perfect Theory of Everything.

 

[exponent137]

What you say is correct in some instances, but you are vastly over-generalizing.

The instance that you are correct is the fundamental dimensionless constants. Those do represent features of the universe that fundamentally cannot be explained by any modern theory. This is no secret, scientists are quite open about it and the search for a "Theory of Everything" is largely motivated by this problem.

However, this is not the case with the dimensionful "fundamental" constants (e.g. c or G), those are completely understood as they are simply artifacts of our choice of units and essentially just allow us to convert between different units.

A second category of constant that does not represent a fundamental lack of understanding is what A.T. mentioned above (e.g. your drag coefficient). They represent a situation where the fundamental rules that govern the system are understood, but the boundary conditions or the system are too complicated to apply the fundamental rules given current computational limitations. Such constants abound in biology and engineering.

A final type of constant that is not due to a lack of knowledge is when a quantity is defined as a proportionality. For example, resistance is defined as the constant R in V=IR. Such constants would remain even with a tractable system and a perfect Theory of Everything.

Yes, Theory of Everything will tell more precisely, what is the value for G and what is its best intepretation.
For instance m^2 G/(hbar c), where m are masses of particles, are the best interpretation of gravitational constant.

But what is the best interpretation in macro world!?

And what is the best possible precise theoretical measurement of G? I think that it is measurement of photons of Hawking radiation.

 

[Vanadium 50]

Exponent137, I don't see any reason to replace the existing description of G, which describes what happens when you hold two known masses a fixed distance apart, something which can easily be done experimentally, by something which relies on the gravitational attraction of light (which is known to 10x worse precision), or on unmeasured properties of black holes (like the exact location of the event horizon) or on unobserved phenomena (like Hawking radiation).

It looks to me that your "simplification" drags in everything except the kitchen sink into the picture, and so is not any simpler after all.

 

[exponent137]

Exponent137, I don't see any reason to replace the existing description of G, which describes what happens when you hold two known masses a fixed distance apart, something which can easily be done experimentally, by something which relies on the gravitational attraction of light (which is known to 10x worse precision), or on unmeasured properties of black holes (like the exact location of the event horizon) or on unobserved phenomena (like Hawking radiation).

It looks to me that your "simplification" drags in everything except the kitchen sink into the picture, and so is not any simpler after all.

I have two reasons.
1. Better understand General relativity GR theory. GR is a little different than Newton gravity, so it need different intepretation. (I think it need impact of gravity on light ray).
2. Better understand theory of everything, where G is important at understanding of it.

I am not agains classical interpretation. I only wish still other interpretations.
Regards

 

[Vanadium 50]

Exponent, the people who study GR for a living use the same old G as everyone else: determined by measuring the force on a pair of masses. Why do you think the professionals are wrong and you are right?

As far as a Theory of Everything is concerned, it may well be that it leads us to a better way of defining G. But we don't have one yet, so it seems premature to a) declare that this will be the case, and b) to decide what that better way is.

 

[yogi]

From post 26

"Exponent, the people who study GR for a living use the same old G as everyone else: determined by measuring the force on a pair of masses. Why do you think the professionals are wrong and you are right?"


The experts don't have a theory of why the gravitatinal constant has the value we observe - so the measured value is inserted to get to the answer - in this regard, the General Theory is incomplete - and it will probably be that way for some time - if you wanted to pursue a theoretical value, look to the units and see what they say: "volumetric acceleration per unit mass" - if we are talking about the Hubble sphere expanding, then the volumetric acceleration is is calculatable if it is assume the rate of dilation is c.
This leads a model that predicts
isotropic acceleration at all points normal to the Hubble
surface - so if we had a perfect model of the universe and an accurate measurement of the Hubble parameter H we would be able to compute G
analytically - as it turns out using this dimensional analysis, G looks to be in the vicinity of (c^2)H on the low side and 2(c^2)H on the upper limit...depending upon what value one assigns to the q factor that determines the degree of comic slowing or acceleration..