Gravity only illusion due to expansion of universe?

[Gerenuk]

Hi,

I read some earlier threads, but didn't quite understand them. But they gave me an idea:

Maybe there is not gravitational force, but just the space expands in a way as to create the illusion of an attractive force, i.e. things accelerating towards each other?

Are there such ideas?
I once knew the very basics of GR, but that's a while ago :)

Gerenuk

 

[Vanadium 50]

This is not possible, as it predicts attraction to be independent of mass.

 

[Mentz114]

I just fell down the stairs and my broken leg is not an illusion ! ( Oh no, sorry, I merely had an illusion that I fell down the stairs. I'm fine now).

There are models of gravity (on the fringes of accepted physics) that say this.

This is not possible, as it predicts attraction to be independent of mass.

I thought gravitational acceleration is independent of mass as in the equivalence principle.

 

[Ich]

Maybe there is not gravitational force, but just the space expands in a way as to create the illusion of an attractive force, i.e. things accelerating towards each other?

Are there such ideas?

There are such ideas, usually inconsistent crackpottery. There is one exception, GR, which states that a stationary observer near a mass actually is accelerating outwards. This idea is formulated consistently in 4 dimensions. Acceleration is defined as a local concept and does not necessarily imply changing distances.

 

[ManDay]

I thought gravitational acceleration is independent of mass as in the equivalence principle.

He said attraction, not acceleration (I admit both starts with an A and ends in an N, though).

 

[Mentz114]

He said attraction, not acceleration (I admit both starts with an A and ends in an N, though).

I don't see the difference in this context. Attraction=force which causes acceleration, and they are independent of mass in for the gravitational field.

 

[MeJennifer]

Acceleration is obviously not independent of mass.

Two heavy masses will come together quicker than two light masses.

 

[Mentz114]

MJ,
the strength of the field depends on the mass of the source. The acceleration imparted by the field is independent of mass, and depends only on field strength.

If Vanadium 50 is talking about gravitational collapse, let him say so.

M

 

[Vanadium 50]

Acceleration is obviously not independent of mass.

Two heavy masses will come together quicker than two light masses.

Exactly - and this idea does not predict that.

 

[MeJennifer]

MJ,
the strength of the field depends on the mass of the source. The acceleration imparted by the field is independent of mass, and depends only on field strength.

Experiment 1

Take two equal masses of mass M in empty space one meter apart and time how long it takes before they come together.

Experiment 2

Take two equal masses of mass M/2 in empty space one meter apart and time how long it takes before they come together.

Are you suggesting that both times will be the same? If so you are completely wrong.

 

[Mentz114]

OK, Vanadium, you didn't make that clear in your first post.

MeJennifer, I repeat

"the strength of the field depends on the mass of the source. The acceleration imparted by the field is independent of mass, and depends only on field strength."

In your example, the influence of body A on body B depends only on A's mass, and vice-versa.

Your example in no way contradicts that.

M

 

[yogi]

you can arrive at a plausible value for the gravitational constant based upon expansion of the Hubble sphere. For simplicity, take the Hubble sphere as dilating at a constant radial rate c so dV/dt = 4c(pi)R^2 and therefore the volumetic acceleration d^2V/dt^2 for constant radial dilation is 8(pi)(c^2)R ...now apply Gausses' theorem to make a volume to surface transformation ...this leads to division by the effective area which for a sphere is 4(pi)R^2 and therefore the isotropic acceleration is 2(c^2)/R

Multiply by the inertia to get the gravitational force

 

[granpa]

while what has been said about gravity isn't exactly wrong nevertheless there is no reason to see gravity as being any different from other forces. replace mass with change and reverse the direction of the resulting force and you simply have the electric field. (excluding relativistic effects like time dilation of course.

 

[Mentz114]

... there is no reason to see gravity as being any different from other forces. replace mass with change and reverse the direction of the resulting force and you simply have the electric field. (excluding relativistic effects like time dilation of course.

I have to disagree with that. Attempts to make gravity theories analogous with EM founder because gravitational 'forces' are independent of mass, unlike EM, where charge and mass are not the same thing. The acceleration of a charged particle in an electric field is F/m, but the acceleration from the gravitational field does not depend on m, only field stringth.

 

[granpa]

replace mass with charge and reverse the direction of the resulting force

independent of mass? that's like saying that the electric force is independent of charge.

the gravitational force is proportional to m1*m2 in exactly the same way that the electric force is proportional to q1*q2

and the gravitational field of an object is proportional to its mass just as an electric field is proportional to its charge.the biggest difference is that masses attract while like changes repel.

the fact that you have an additional mass term in your electric interactions no doubt makes the math very different but fundamentally the forces themselves are similar.

 

[Mentz114]

granpa,
I don't agree with that. Gravity is nothing like the electric field. Consider this - two different masses with the same charge move in an electric field. They will have different accelerations. Two different masses moving in a gravitational field will have identical accelerations. That is why I say that the effect of the field is independent of the mass of the body being influenced.

Also, there is only one sign on the 'mass charge'.

This is sufficient to indicate that the two fields are very different in nature ( and I haven't even gone in the quantum side of things).

M

 

[granpa]

nothing that you just said contradicts anything I said an post 15. yet you come to the opposite conclusion. i think we will just have to disagree. have a nice day.

 

[malawi_glenn]

Look at how different the field equations for gravity and EM are granpa, there are more things that they don't have in common then what they do have in common.

 

[Mentz114]

granpa:

but fundamentally the forces themselves are similar.

I suppose if you persist in thinking of gravity as the Newtonian theory only, it makes no difference. Seriously, you're onto a loser if you think the electric field is fundementally like the gravitational field.

Here's another one - how do you define free-fall in an electric field ? You can't. There is no transformation that takes away the force, unlike gravity, which is not a force.

Nice day to you too.

M

 

[MeJennifer]

In your example, the influence of body A on body B depends only on A's mass, and vice-versa.

General relativity, unlike Newtonian gravity, is a non-linear theory.

You cannot isolate the influence of A on B and vice versa. Both A and B influence spacetime. Only the combined influence determines the dynamics of A with respect to B and vice versa.

 

[Jonathan Scott]

Another difference between gravity and electromagnetism is that as seen in a typical background coordinate system a static gravitational field changes the momentum but does not change the total energy (so it also changes the rest energy relative to those coordinates), but a static electric field changes the kinetic energy and total energy but doesn't change the rest mass.

However, there's still quite a lot in common between gravity and electromagnetism in the weak field non-relativistic case.

 

[granpa]

does an object in a gravitational field experience a force proportional to its gravitational mass (just like a charge in an electric field experiences a force proportional to its charge) or an acceleration proportional to the gravitational field (as the posters below suggest)? obviously it makes no difference unless gravitational mass is not equal to inertial mass.

if gravity does produce acceleration instead of force then it would be extremely difficult to tell if gravitational mass is different from inertial mass.

one also has to take self inductance into account.

 

[MackBlanch]

granpa,
two different masses with the same charge move in an electric field. They will have different accelerations. Two different masses moving in a gravitational field will have identical accelerations. That is why I say that the effect of the field is independent of the mass of the body being influenced.

Also, there is only one sign on the 'mass charge'.

This is sufficient to indicate that the two fields are very different in nature ( and I haven't even gone in the quantum side of things).

M

Hmmm. Interesting, in an electric field, the smaller mass will have a greater acceleration, and emit EM waves of a higher frequency. In a gravitational field, they have the same acceleration and will emit the same frequency.

I think a good argument for them being incomparable is that charge is invariant and mass is not. You can increase the mass of an object by increasing its potential energy. If you squash a spring and tie it down, it will have a (quite negligible) mass increase. There is no analogous scenario for charge.

 

[Mentz114]

In a gravitational field, they have the same acceleration and will emit the same frequency.

A freely falling body does not radiate gravitational waves, as far as I know.

This is off-topic in this thread. Would you like to start a new thread in the 'Classical' forum ?

 

[Vanadium 50]

A freely falling body does not radiate gravitational waves, as far as I know.

Why not? The object and the object it's falling towards have a changing mass quadrupole moment. Why wouldn't it radiate?

 

[MackBlanch]

A freely falling body does not radiate gravitational waves, as far as I know.

This is off-topic in this thread. Would you like to start a new thread in the 'Classical' forum ?

I wasn't even touching on gravitational waves. Sorry for the confusion. I was hoping the "EM" would be implied on the second set of waves.

 

[DaveC426913]

the effect of the field is independent of the mass of the body being influenced.

Isn't this more to do with the fact that the greater attraction is perfectly canceled by the greater inertia?

A satellite of mass m will fall to Earth as a = F/m.
A satellite of mass 2m will fall to Earth as a = 2F/2m.

i.e. it's not that the effect is independent of the object's mass, its that the object's mass cancels out of the result.

I realize that this is a completely semantic argument, but it seems to clarify the topic at-hand.

 

[D H]

I don't see the difference in this context. Attraction=force which causes acceleration, and they are independent of mass in for the gravitational field.

Isn't this more to do with the fact that the greater attraction is perfectly canceled by the greater inertia?

A satellite of mass m will fall to Earth as a = F/m.
A satellite of mass 2m will fall to Earth as a = 2F/2m.

Both of you are ignoring that gravitation works both ways: object A attracts object B gravitationally, and object B attracts object A gravitationally. Suppose you measure the time for two masses each of mass M to collide after starting at some distance with a relative velocity of zero. If you perform the same experiment but with one of the masses now having mass m<<<M you will get a different (longer) time to collision.

 

[TalonD]

Acceleration is obviously not independent of mass.

Two heavy masses will come together quicker than two light masses.

I dissagree. Drop a cannon ball and a feather off the top of the Eifel tower (in a vacum) and see which hits the ground first.

 

[D H]

The cannon ball will (assuming we are talking about normal-sized cannon balls and normal-sized feathers). The difference in this case is immeasurably small.

 

[Mentz114]

Why not? The object and the object it's falling towards have a changing mass quadrupole moment. Why wouldn't it radiate?

True. I wasn't including the attracting body. If it's a test particle. though, it won't radiate.

 

[TalonD]

The cannon ball will (assuming we are talking about normal-sized cannon balls and normal-sized feathers). The difference in this case is immeasurably small.

You may be correct, but that's not the way I leaned it in school. See Galileo and the leaning tower of pizza.

Yeah, I know I spelled Pisa wrong, I'm hungry.

 

[D H]

The Earth's acceleration toward a 100 kg cannon ball is 100,000 times that of the Earth's acceleration toward a 1 gram feather. However, either acceleration is very, very, very very tiny compared to the ~9.8 m/s² acceleration of the cannon ball (or feather) toward the Earth: 10^-18 m/s² in the case of the cannon ball, 10^-23 m/s² in the case of the feather.

 

[DaveC426913]

You may be correct, but that's not the way I leaned it in school. See Galileo and the leaning tower of pizza.

Yeah, I know I spelled Pisa wrong, I'm hungry.

What did you learn in school? That, even in vacuum, the cannon ball would hit first?

 

[TalonD]

The Earth's acceleration toward a 100 kg cannon ball is 100,000 times that of the Earth's acceleration toward a 1 gram feather. However, either acceleration is very, very, very very tiny compared to the ~9.8 m/s² acceleration of the cannon ball (or feather) toward the Earth: 10^-18 m/s² in the case of the cannon ball, 10^-23 m/s² in the case of the feather.

What did you learn in school? That, even in vacuum, the cannon ball would hit first?

What I learned in grade school is that they would both hit the ground at the same time without any air resistance. Remember Galileo's famous experiment? Even the astronauts demonstrated it on the moon. But now that I'm all grown up with a higher education, it suddenly occurs to me after reading the previous post that the cannonball would hit first. And I am assuming it is because the cannonball would have it's own gravitational field which is stronger than the gravitational field of the feather, is that right?

 

[TalonD]

The OP's title of this thread asks if gravity is due to the expansion of the universe. That's something I would be interested in responding to, but I think my post would be removed as being too speculative.
:(

 

[DaveC426913]

What I learned in grade school is that they would both hit the ground at the same time without any air resistance. Remember Galileo's famous experiment? Even the astronauts demonstrated it on the moon. But now that I'm all grown up with a higher education, it suddenly occurs to me after reading the previous post that the cannonball would hit first. And I am assuming it is because the cannonball would have it's own gravitational field which is stronger than the gravitational field of the feather, is that right?

Oh I see what you're getting at. The cannonball would pull the Earth toward it faster than the feather would, making for a slightly shorter delay before contact. (You'd have to do the two tests sequentially, rather than simultaneously.)

Yes. This might be easier to visualize of you substituted a moon for a cannonball.

In an "Earth-feather system", your total mass (and thus your total gravitational attraction) is equal to 1 Earth + 1 feather.
In an "Earth-moon system", your total mass (and thus your total gravitational attraction) is equal to 1 Earth + 1 moon.

It becomes intuitively obvious now that the Earth-Moon system should make contact in less time.

 

[neopolitan]

Oh I see what you're getting at. The cannonball would pull the Earth toward it faster than the feather would, making for a slightly shorter delay before contact. (You'd have to do the two tests sequentially, rather than simultaneously.)

Yes. This might be easier to visualize of you substituted a moon for a cannonball.

In an "Earth-feather system", your total mass (and thus your total gravitational attraction) is equal to 1 Earth + 1 feather.
In an "Earth-moon system", your total mass (and thus your total gravitational attraction) is equal to 1 Earth + 1 moon.

It becomes intuitively obvious now that the Earth-Moon system should make contact in less time.

I wanted to reply to Metz' https://www.physicsforums.com/showpost.php?p=1897967&postcount=11":

In your example, the influence of body A on body B depends only on A's mass, and vice-versa.

but you seem to have gone most of the way towards addressing it.

The influence is, of course, related to the mass of the total system. We usually are talking about systems like "Earth" and "feather" where M>>m and we can approximate and simplify by saying (M+m)=M.

The question then arises, is the OP's original question related to crackpottery if we consider the mass of the total system, ie the whole universe, rather than just considering local (and open) subsystems. Naturally, we would need to carefully consider what we would treat a mass and I would suggest that we look at a mass as a concentration of energy (or concentration of mass-energy, if you prefer) rather than a point mass notionally located at a body's centre. A body's concentration of mass-energy falls away as you move away from that body (the rate at which the concentration falls away is easily calculated).

If you can visualise the effects of two concentrations of mass-energy in an expanding universe, accept that concentrations of mass-energy would resist expansion in proportion to the concentration of mass-energy (if they didn't the gravitational "illusion" would never eventuate) and do the sums, you will find that the effects would be the same as gravity (and G would be related to a coefficient of "expansion resistance" which would in turn be inversely related to the speed of light squared. If Planck units are used, the coefficient of "expansion resistance" resolves back to unity).

The equations are available, but as it is not considered to be mainstream physics this is not the correct forum for providing them or even posting links to them.

Even if it were mainstream physics, you would still be left with the questions: what causes the expansion? (an answer is available, but may not be mainstream) and what causes the resistance to expansion of concentrations of mass-energy? (Although, to be fair, without gravity or some similar phenomena which leads to the "illusion" of gravity, the universe would be smooth and there would be no lumpy bits like ourselves to ponder the question.)

cheers,

neopolitan

 

[Akgcube]

you can arrive at a plausible value for the gravitational constant based upon expansion of the Hubble sphere. For simplicity, take the Hubble sphere as dilating at a constant radial rate c so dV/dt = 4c(pi)R^2 and therefore the volumetic acceleration d^2V/dt^2 for constant radial dilation is 8(pi)(c^2)R ...now apply Gausses' theorem to make a volume to surface transformation ...this leads to division by the effective area which for a sphere is 4(pi)R^2 and therefore the isotropic acceleration is 2(c^2)/R

Multiply by the inertia to get the gravitational force

Hi Yogi:
I did not undestand the last line ' multiply by inertia...
Can you kindly give me some link or reference where I can learn more about this point of view... Thanks

 

[atyy]

Maybe there is not gravitational force, but just the space expands in a way as to create the illusion of an attractive force, i.e. things accelerating towards each other?

This is not possible, as it predicts attraction to be independent of mass.

How about construing "expansion of space" more broadly as the "metric"? Then given a metric, the field equations (and equations of state) give the stress-energy-momentum distribution.

 

[Phrak]

Maybe there is not gravitational force, but just the space expands in a way as to create the illusion of an attractive force, i.e. things accelerating towards each other?

This idea may not be as dismissable as it seems. In some form it might be found to be equivalent to general relativity, or similar to general relativity.

 

[Vanadium 50]

How about construing "expansion of space" more broadly as the "metric"? Then given a metric, the field equations (and equations of state) give the stress-energy-momentum distribution.

That's a pretty broad way of looking at it, as under this definition space isn't necessarily expanding at all, and indeed can be contracting, and you'd still have gravity.

 

[Naty1]

Here's what Peter Bergmann (a student of Einsteins) had to say about gravity and electromagnetism in THE RIDDLE OF GRAVITATION (1992) :

...large velocities affect masses differently from electric charges. Whereas a bodies electric charge has the same value for all observers, it's mass depends on its speed relative to the observer...Because the magnitudes of the sources of gravitation depend so much on the frame of reference in which they are measured, the resulting field is bound to be more complex than the electromagnetic field...Einstein concluded the gravitational field was probably a...tensor field...

 

[Dale]

Out of curiosity, since the universe is expanding isotropically how could that be the explanation for gravity which points down rather than being isotropic?

 

[Phrak]

...large velocities affect masses differently from electric charges. Whereas a bodies electric charge has the same value for all observers,...

Though charge density is not Lorentz invariant.

 

[Phrak]

The Friedmann-Lemaître-Robertson-Walker (FLRW) metric

\Large{c^2 d\tau^2 = c^2dt^2 - A(t)^2 d \Sigma^2}

This looks suspiciously as if it gives preferencial treatment to particular inertial frames, nominally at rest with the cosmic background radiation, perhaps. Does anyone know?

 

[Dale]

The Friedmann-Lemaître-Robertson-Walker (FLRW) metric

\Large{c^2 d\tau^2 = c^2dt^2 - A(t)^2 d \Sigma^2}

This looks suspiciously as if it gives preferencial treatment to particular inertial frames, nominally at rest with the cosmic background radiation, perhaps. Does anyone know?

Any metric is written in a specific coordinate system and represents the underlying spacetime geometry as described by that specific coordinate system. You can transform the metric to another coordinate system and get a different "reference frame" that describes the same spacetime. Neither is preferred or in any way describes a different spacetime.

 

[neopolitan]

Out of curiosity, since the universe is expanding isotropically how could that be the explanation for gravity which points down rather than being isotropic?

When you say gravity "points down" you are thinking of it as a force, I thought the whole idea of the OP was that it is an illusion. Expansion of the universe may be isotropic (specifically in all directions), but it is not smooth. It it were smooth, we'd not notice it because we are largely empty space ourselves. What we see expanding are the gaps between masses, or concentrations of mass-energy.

What then gets difficult to explain is why we see expansion between big lumps (galaxies) and not so much between smaller lumps (planets in the solar system). It could be that the universe would expand both isotropically and smoothly, if it weren't for mass (or concentrations of mass-energy). Concentrations of mass-energy seem to do something, it could be that they bend space or it could be that they resist expansion, the overall effect would be the same.

If concentrations of mass-energy did resist expansion, then gravity would then "point" along a line of increasing concentration, ie towards the (other) mass. Put two masses close enough to each other and you will see a line of maximum resistance to expansion linking them, because the centre of their combined mass (combined mass-energy) lies on that line.

How would this differ from "normal" gravity? Not much, it just wouldn't be a force, it would be a phenomenon - and you wouldn't, therefore, have a gravity field or gravitons. Gravity lensing would still happen if a photon passed through a region of resistance to expansion. Space would still seem to be bent.

Oh, and some physicists would be upset since it would play havoc with some pet theories.

cheers,

neopolitan

 

[Phrak]

Any metric is written in a specific coordinate system and represents the underlying spacetime geometry as described by that specific coordinate system. You can transform the metric to another coordinate system and get a different "reference frame" that describes the same spacetime. Neither is preferred or in any way describes a different spacetime.

It's the appearance of the A(t) term that looks questionable, as if space has a preferred coordinate system in which it expands isotropically. I suppose it depends upon what motivates its inclusion. Is it ad hoc, to explain the expansion of the Universe; does it require a cosmological constant added to the Einstein tensor?

 

[Austin0]

Isn't this more to do with the fact that the greater attraction is perfectly canceled by the greater inertia?

A satellite of mass m will fall to Earth as a = F/m.
A satellite of mass 2m will fall to Earth as a = 2F/2m.

i.e. it's not that the effect is independent of the object's mass, its that the object's mass cancels out of the result.

.

Hi---What you have said here seems, to me, to be completely logical and valid.
That the acceleration is always going to be a result of reciprocal interaction but if inertial mass is actually equivalent to gravitational mass, this shouldn't make any difference.
But subsequent posts seemed to question this so now I am curious as to which is correct.