Is there a "force of gravity"?

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Offline Geezer

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Is there a "force of gravity"?
« on: 29/08/2011 07:44:46 »
It seems to me that there is no "force of gravity", although gravity clearly accelerates mass. What's wrong with this reasoning?

We know that;

1) F = m.a

2) We can measure the acceleration (a) of a free falling object in a gravitational field, and the acceleration is independent of the mass (m) of the object.

3) We can consistently measure the mass (m) of an object independently of any gravitational field.

We can conclude that a free falling object in a gravitational field is experiencing an accelerative force which is the product of its mass (m) and the acceleration (a) that we observe.

Except that;

We also know by observation that the acceleration of the object is constant, regardless of its mass, so the accelerative force (F) must be proportional to the mass of the object.

We might express this as;

F[prop]m

Therefore, F = K.m (K is some constant)

However, we also know that F = m.a

So we have just proved that a is a constant! (Which was obvious in the first place.)

What it does not seem to prove is that there is any "force of gravity". All it seems to tell us is that gravity accelerates matter. Either that, or F does not equal m.a

Where am I going wrong?

 
« Last Edit: 04/09/2011 15:37:41 by JP »
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Offline Soul Surfer

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Re: Is there a "force of gravity"?
« Reply #1 on: 29/08/2011 08:23:05 »
You are forgetting that the acceleration due to gravity is not constant and varies as you approach the body and the equations are only true for one instant the moment the body falls a little towards the gravitating mass the acceleration and therefore the force increases.
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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #2 on: 29/08/2011 08:37:42 »
You are forgetting that the acceleration due to gravity is not constant and varies as you approach the body and the equations are only true for one instant the moment the body falls a little towards the gravitating mass the acceleration and therefore the force increases.

But I'm talking about a particular distance and a specific instant in time. Why does your argument prove that there is a force of gravity? My reasoning and math is not very complicated. If it's wrong, it should not be too difficult to shoot it down in flames (which, BTW, is fine with me - I'd just like to be able to understand this because it has been bugging me for a long time.)
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Robro

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Re: Is there a "force of gravity"?
« Reply #3 on: 29/08/2011 12:50:02 »
I do not think that there is a "Force" of gravity. I tend to think of gravity as a consequential artifact of electromagnetism. There is plenty of mathematics to describe what happens in a gravitational field, but it leaves the definition up to "curved space-time", which in itself is not really a definition but just another description of what happens. I do not think that mainstream offers a "definition" of gravity or space-time. Same goes with "magnetic force" and "electric charge". I do not think anybody really knows "what" these things are, but their effect have many good descriptions.
"Consciousness is the Universe viewing itself through a microscope."

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Offline JP

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Re: Is there a "force of gravity"?
« Reply #4 on: 29/08/2011 13:39:09 »
Geezer, I think you're basically reasoning along the same lines as one of the postulates of general relativity: the equivalence principle.  In GR, gravity isn't a force, either.  Your proof above is similar to Einstein's famous elevator experiment.  If you're in an elevator, the "force" you experience as it accelerates upward at 9.8 m/s2 is indistinguishable from gravitational acceleration at the earth's surface (if you look only at a point, clearly, since you can tell that the earth's gravity is decreasing as you move away from it).  If you follow through to the logical conclusion, our "force" of gravity is just an acceleration, so there's some reference frame (a freely falling one) in which this acceleration vanishes.  Then if you add in lots of fancy math, you end up at general relativity.
« Last Edit: 29/08/2011 17:19:16 by JP »

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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #5 on: 29/08/2011 17:18:49 »
Geezer, I think you're basically reasoning along the same lines as general relativity, and in GR, gravity isn't a force, either.  Your proof above is similar to Einstein's famous elevator experiment.  If you're in an elevator, the "force" you experience as it accelerates upward at 9.8 m/s2 is indistinguishable from gravitational acceleration at the earth's surface (if you look only at a point, clearly, since you can tell that the earth's gravity is decreasing as you move away from it).  If you follow through to the logical conclusion, our "force" of gravity is just an acceleration, so there's some reference frame (a freely falling one) in which this acceleration vanishes.  Then if you add in lots of fancy math, you end up at general relativity.

Kewel! Would it be a bit late to expect a jaunt to Norway to pick up the prize?
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Offline JP

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Re: Is there a "force of gravity"?
« Reply #6 on: 29/08/2011 17:27:37 »
Probably.  By the way, I was editing something into my post as you quoted it, so you probably missed it.  You might want to check this out:
http://en.wikipedia.org/wiki/Equivalence_principle

You're assuming the equivalence of inertial mass and gravitational mass in your equation.  There's no fundamental reason we know of yet why they should be the same, but they appear to be. 

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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #7 on: 29/08/2011 18:29:51 »
Probably.  By the way, I was editing something into my post as you quoted it, so you probably missed it.  You might want to check this out:
http://en.wikipedia.org/wiki/Equivalence_principle

You're assuming the equivalence of inertial mass and gravitational mass in your equation.  There's no fundamental reason we know of yet why they should be the same, but they appear to be. 

Thanks!

(I wonder if he thinks I'm really going to read that. There's an awful lot of words and math and stuff  [::)])
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Offline Soul Surfer

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Re: Is there a "force of gravity"?
« Reply #8 on: 29/08/2011 23:41:13 »
What I was pointing out was that this force is not a fixed value but bears a relationship with distance from the gravitating object.  This is an inverse square law which corresponds with the normal dimensionality of things because the surface area of a sphere increases as the square of its radius.

All you are pointing out with your mathematical arguments is that there is a common constant of proportionality in this relationship this is the quantity which we call mass.  That is, mass is not something fundamental but just comes out of the way things behave.
If you are familiar with dimensional analysis where we express all measurements and observations in terms of mass, length and time,  if one includes gravity into all the relationships it is possible to eliminate mass entirely and express everything in terms of length and time.  Nowadays even these measurements are being looked on as not the fundamental properties of the universe and it is suggested that everything should be really expressed in terms of momentum and energy.  note this does reintroduce the concept of mass but this can be easily removed by replacing it with its energy equivalent via the famous e=mc^2.  that is mass is really just a form of localised energy.
« Last Edit: 29/08/2011 23:43:12 by Soul Surfer »
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Offline Phractality

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Re: Is there a "force of gravity"?
« Reply #9 on: 30/08/2011 04:41:02 »
F = ma only works at non-relativistic speeds. As you near the speed of light, m increases, so for a constant acceleration, the force must also increase. The more general definition of force is f = δp/δt; i.e., force is the time rate of change of momentum.

At non-relativistic speeds, δm = 0, so vδm/δt = 0. The change of momentum at non-relativistic speed is simply δp = mδv; at relativistic speed, it is δp = mδv + vδm. So δp/δt = mδv/δt + vδm/δt. Acceleration is a = δv/δt, so mδv/δt + vδm/δt = ma + vδm/δt.

F = δp/δt works not only for objects with rest mass, but also for photons. Although a photon is not considered to have mass and force in general relativity, it certainly could by this definition.
« Last Edit: 30/08/2011 04:51:58 by Phractality »
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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #10 on: 30/08/2011 07:21:06 »
F = ma only works at non-relativistic speeds. As you near the speed of light, m increases, so for a constant acceleration, the force must also increase. The more general definition of force is f = δp/δt; i.e., force is the time rate of change of momentum.

At non-relativistic speeds, δm = 0, so vδm/δt = 0. The change of momentum at non-relativistic speed is simply δp = mδv; at relativistic speed, it is δp = mδv + vδm. So δp/δt = mδv/δt + vδm/δt. Acceleration is a = δv/δt, so mδv/δt + vδm/δt = ma + vδm/δt.

F = δp/δt works not only for objects with rest mass, but also for photons. Although a photon is not considered to have mass and force in general relativity, it certainly could by this definition.

Thanks! So does this mean you believe there is a "force of gravity"?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Phractality

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Re: Is there a "force of gravity"?
« Reply #11 on: 30/08/2011 09:18:36 »
So does this mean you believe there is a "force of gravity"?
I believe masses accelerate toward one another, resulting in a rate of change of momentum which can be calculated from Newton's law of gravitation or from general relativity. You may explain that acceleration in terms of a gravitational field, but what is a field but a mathematical description of an effect?

I prefer to explain the cause in terms of interacting ćthereal waves, but I'm not permitted to describe that model except in the New Theories section.
Imagination is more important than knowledge. Einstein

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Offline MikeS

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Re: Is there a "force of gravity"?
« Reply #12 on: 01/09/2011 08:46:50 »
So does this mean you believe there is a "force of gravity"?
I believe masses accelerate toward one another, resulting in a rate of change of momentum which can be calculated from Newton's law of gravitation or from general relativity. You may explain that acceleration in terms of a gravitational field, but what is a field but a mathematical description of an effect?

I prefer to explain the cause in terms of interacting ćthereal waves, but I'm not permitted to describe that model except in the New Theories section.

What acceleration?
"An object in free-fall is in actuality inertial, but as it approaches the planetary object the time scale stretches at an accelerated rate, giving the appearance that it is accelerating towards the planetary object when, in fact, the falling body really isn't accelerating at all. This is why an accelerometer in free-fall doesn't register any acceleration; there isn't any."
http://en.wikipedia.org/wiki/Equivalence_principle
 [;D]

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Offline imatfaal

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Re: Is there a "force of gravity"?
« Reply #13 on: 01/09/2011 11:42:19 »
Mike - go to the top of the newly reopened Blackpool Tower Eye (crazy place crazy name) - step off.    So why are you gonna be unable to post the results of your experiment?  We are conflating GR and Newtonian mechanics - to describe an object falling off a desk as not accelerating is futile, whilst quite clearly also correct.  GR is a model that works very well - especially in cosmological distances and masses, Newtonian mechanics is another model (or in fact a limit of GR) that works very well in human distances and masses; but let's be clear - neither of them are correct in absolute terms.  you use the model that is most appropriate for your situation
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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #14 on: 01/09/2011 18:51:35 »
"This is why an accelerometer in free-fall doesn't register any acceleration; there isn't any."

Which only goes to prove you should not believe everything in Wikipedia. That statement is complete bollocks nonsense.

The reason an accelerometer does not measure acceleration in free fall is because there is no difference between the "sensor" and the "mass" of the accelerometer. They are both experiencing the same acceleration. That could be any value, including zero, but it sure as heck does not prove there is no acceleration.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline MikeS

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Re: Is there a "force of gravity"?
« Reply #15 on: 02/09/2011 11:09:39 »
"This is why an accelerometer in free-fall doesn't register any acceleration; there isn't any."

Which only goes to prove you should not believe everything in Wikipedia. That statement is complete bollocks nonsense.

The reason an accelerometer does not measure acceleration in free fall is because there is no difference between the "sensor" and the "mass" of the accelerometer. They are both experiencing the same acceleration. That could be any value, including zero, but it sure as heck does not prove there is no acceleration.

If you correct for time dilation within the fall, there will be no acceleration, only a constant speed.

This should be easily testable.  We now have the capability to measure time dilation over distances as small as a meter.  It should therefore be possible to drop an optical atomic clock through a few meters and measure whether it has accelerated or not.

We, as a species do not have the ability to differentiate small differences in the gravitational field.  Neither, we do not have the ability to differentiate small differences in the passage of time.  Instead we 'smear' the passage of time to an average.  This was obviously beneficial to our species.
« Last Edit: 02/09/2011 11:56:02 by MikeS »

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Offline imatfaal

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Re: Is there a "force of gravity"?
« Reply #16 on: 02/09/2011 11:39:09 »
Drop an atomic clock?  They have to custom-build hovercarts to move them around to avoid the bumps of going over door lintels.  The most recent and accurate atomic clocks rely on fountains of matter falling under gravity and constrained influenced by lasers and microwaves - not sure that would take well to being dropped, my casio f91-w would be more accurate and might not need millions to rebuild afterwards.

The best accuracy over a long period of an atomic clock (USNO rubidium fountain) is around 1*10^-16 of a second.  Even if the experiment introduced no error what so ever this would not allow the time differential to be measured. 

More importantly you do not just need to calculate the gravitational time dilation you need to calculate the correct four vector in a solution of the einstein equations/tensors - which to all intents is impossible, the assumptions you would have to make would drown any actual figures.  do the sums for just the time dilation and you get nonsense.  Drop something from 20m high, in the first second you will travel 5m, the second second you will travel 15m yet the variation in time dilation is unmeasurable.
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Offline MikeS

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Re: Is there a "force of gravity"?
« Reply #17 on: 02/09/2011 12:03:03 »
Drop an atomic clock?  They have to custom-build hovercarts to move them around to avoid the bumps of going over door lintels.  The most recent and accurate atomic clocks rely on fountains of matter falling under gravity and constrained influenced by lasers and microwaves - not sure that would take well to being dropped, my casio f91-w would be more accurate and might not need millions to rebuild afterwards.

The best accuracy over a long period of an atomic clock (USNO rubidium fountain) is around 1*10^-16 of a second.  Even if the experiment introduced no error what so ever this would not allow the time differential to be measured. 

More importantly you do not just need to calculate the gravitational time dilation you need to calculate the correct four vector in a solution of the einstein equations/tensors - which to all intents is impossible, the assumptions you would have to make would drown any actual figures.  do the sums for just the time dilation and you get nonsense.  Drop something from 20m high, in the first second you will travel 5m, the second second you will travel 15m yet the variation in time dilation is unmeasurable.

I don't see why there would be a problem as time differential has already been measured over distances as small as one meter as previously mentioned.

posted on August 29th, 2011
“(…) if one clock is placed one centimeter higher than another clock, the higher clock is affected by less gravity, so it goes faster. That difference could be read out in the 18th decimal place of the clocks in one second averaging time. Until now, clocks have been thought of as tools for sharing a common time. But with clocks like this, conversely, we can understand that time passes at different speeds, depending on the time and place a clock is at.”
http://techcrunch.com/2011/08/29/video-new-atomic-clock-reaches-a-100-quadrillionth-of-a-second-accuracy/
« Last Edit: 02/09/2011 12:10:57 by MikeS »

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Offline JP

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Re: Is there a "force of gravity"?
« Reply #18 on: 02/09/2011 14:14:04 »
Here's another interesting point, Geezer.  Say we decide to use Newton's laws in an accelerating reference frame.  For the sake of making your grumpy, let's assume it's a rotating reference frame.  Suddenly we find that everything is experiencing a centrifugal force that is proportional to mass. 

Now let's say we spent our entire lives in this rotating reference frame.  We might not even realize it's rotating and we might assume that centrifugal force is some fundamental force in the universe.  Eventually someone bright would come along and figure out that the magic centrifugal force can be explained by the fact that all our experiments and observations have been done in a rotating reference frame, and that centrifugal force is just an artifact of us having formulated all our theories in rotating coordinates.

Now imagine instead we grew up in a reference frame that was accelerating with respect to the natural, inertial, free-falling frame.  We didn't realize  we were in an accelerating reference frame, so when we formulated all our physical laws, we got this fictitious force that was always proportional to mass, and called it gravity.  Then one day, Einstein came along and figured out that the "force" of gravity appears because we're working in reference frames that can be thought of as accelerating with respect to the inertial one in which gravity disappears as a force.

The analogy is quite good, actually.  Fictitious forces are proportional to mass, and pop up whenever you try to formulate Newton's laws in an accelerating reference frame.  Einstein's genius was to realize that gravity is also a fictitious force if you regard free-fall as the appropriate "non-accelerating" reference frame. 

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Offline imatfaal

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Re: Is there a "force of gravity"?
« Reply #19 on: 02/09/2011 16:47:01 »
I don't see why there would be a problem as time differential has already been measured over distances as small as one meter as previously mentioned.
That's the problem Mike - you don't see any problems that contradict your preconceptions.  Do the maths! 
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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #20 on: 02/09/2011 17:25:09 »
I'm not grumpy!

I was only pointing out that the twit who wrote that in Wiki obviously has not the faintest idea how an accelerometer actually works.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline imatfaal

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Re: Is there a "force of gravity"?
« Reply #21 on: 02/09/2011 17:46:20 »
I'm not grumpy!

I was only pointing out that the twit who wrote that in Wiki obviously has not the faintest idea how an accelerometer actually works.

well no more than usual anyway
There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #22 on: 02/09/2011 19:45:10 »
Here's another interesting point, Geezer.  Say we decide to use Newton's laws in an accelerating reference frame.  For the sake of making your grumpy, let's assume it's a rotating reference frame.  Suddenly we find that everything is experiencing a centrifugal force that is proportional to mass. 

Now let's say we spent our entire lives in this rotating reference frame.  We might not even realize it's rotating and we might assume that centrifugal force is some fundamental force in the universe.  Eventually someone bright would come along and figure out that the magic centrifugal force can be explained by the fact that all our experiments and observations have been done in a rotating reference frame, and that centrifugal force is just an artifact of us having formulated all our theories in rotating coordinates.

Now imagine instead we grew up in a reference frame that was accelerating with respect to the natural, inertial, free-falling frame.  We didn't realize  we were in an accelerating reference frame, so when we formulated all our physical laws, we got this fictitious force that was always proportional to mass, and called it gravity.  Then one day, Einstein came along and figured out that the "force" of gravity appears because we're working in reference frames that can be thought of as accelerating with respect to the inertial one in which gravity disappears as a force.

The analogy is quite good, actually.  Fictitious forces are proportional to mass, and pop up whenever you try to formulate Newton's laws in an accelerating reference frame.  Einstein's genius was to realize that gravity is also a fictitious force if you regard free-fall as the appropriate "non-accelerating" reference frame. 

Yes. That makes sense JP (except that I'm not sure there is such a thing as a rotating frame when you are looking at it from within a non-rotating frame  [:D])

I'm not sure that Newton's laws are necessarily incompatible with an accelerating frame. I think that's more a misapplication than anything else.

The only thing we know about gravity is that it produces acceleration. It's meaningless to apply F=ma to derive a "force" when the acceleration is obviously independent of mass. There is nothing wrong with using fictitious forces if they help in solving problems, but we should be aware that they are fictitious.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Pmb

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Re: Is there a "force of gravity"?
« Reply #23 on: 03/09/2011 22:45:12 »
It's quite all right to think of gravity as a force, so long as that force is not a 4-force. I explained all this at

http://home.comcast.net/~peter.m.brown/gr/grav_force.htm

Please forgive me for the misssing images. They are missing drawings and missing equations. I'll get it up to date sometime in the future.

The concept of force in GR is used in "Basics Relativity," by Richard M. Mould. Springer-Verlag Press

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Offline Geezer

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Re: Is there a "force of gravity"?
« Reply #24 on: 04/09/2011 00:44:32 »
It's quite all right to think of gravity as a force, so long as that force is not a 4-force. I explained all this at

http://home.comcast.net/~peter.m.brown/gr/grav_force.htm

Please forgive me for the misssing images. They are missing drawings and missing equations. I'll get it up to date sometime in the future.

The concept of force in GR is used in "Basics Relativity," by Richard M. Mould. Springer-Verlag Press

Well jolly good, but what exactly do you mean by 4-force? Einstein was pretty clear about gravity not being a force. Is this an alternative theory? 
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Pmb

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Is there a "force of gravity"?
« Reply #25 on: 05/09/2011 19:15:53 »
It's quite all right to think of gravity as a force, so long as that force is not a 4-force. I explained all this at

http://home.comcast.net/~peter.m.brown/gr/grav_force.htm

Please forgive me for the misssing images. They are missing drawings and missing equations. I'll get it up to date sometime in the future.

The concept of force in GR is used in "Basics Relativity," by Richard M. Mould. Springer-Verlag Press

Well jolly good, but what exactly do you mean by 4-force? Einstein was pretty clear about gravity not being a force. Is this an alternative theory? 
A 4-force is dwfined in the link above. I'll try to get it up today. In the mean time I'll just say that a 4-force is defined as F = DP/dt where D/dt is the absolute derivative defined here
http://home.comcast.net/~peter.m.brown/math_phy/absolute_derivative.htm

Also, Einstein never opposed the notion of gravity not being a force. In fact he used that notion in his text The Meaning of Relativity. Do you have a copy of this text? If so then turn to page 83 where Einstein answers your question. Einstein writes
Quote
The gravitational field transfers energy and momentum to the "matter" in that it exerts forces upon it and gives it energy:...
I wrote an article about how Einstein viewed gravity. It's online here
http://xxx.lanl.gov/abs/physics/0204044

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Offline Geezer

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Is there a "force of gravity"?
« Reply #26 on: 07/09/2011 07:24:07 »
Do you have a copy of this text? If so then turn to page 83 where Einstein answers your question. Einstein writes
The gravitational field transfers energy and momentum to the "matter" in that it exerts forces upon it and gives it energy:...

Found it, thanks! But now my brain hurts.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline JP

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Is there a "force of gravity"?
« Reply #27 on: 08/09/2011 00:11:39 »
Do you have a copy of this text? If so then turn to page 83 where Einstein answers your question. Einstein writes
The gravitational field transfers energy and momentum to the "matter" in that it exerts forces upon it and gives it energy:...

Found it, thanks! But now my brain hurts.

Like everything else, so much misunderstanding comes from not realizing that fundamental concepts in Newtonian mechanics don't generalize simply to modern physics, be it relativity or quantum mechanics.  The word "force" groups together a lot of things in Newtonian mechanics that not only can change the energy/momentum of an object, but which also all satisfy F=ma.

In GR, gravity and all other forces can still change the energy/momentum of an object.  But gravity is modeled by Einstein's equations, which can be interpreted geometrically.  Non-gravitational forces all still satisfy something similar to F=ma. 

So gravity is special for reasons that have been discussed at length in this thread already.  What you choose to label it as isn't nearly as important as understanding that it's different.

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Offline MikeS

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« Reply #28 on: 11/09/2011 12:23:18 »
Quote Geezer
"It seems to me that there is no "force of gravity", although gravity clearly accelerates mass. What's wrong with this reasoning?
F = m.a"


Acceleration is a change in velocity over time.  So just by varying the passage of time you can have an acceleration.  Any gravitating body dilates the passage of time near its surface.

Gravity produces the effect of acceleration by altering the passage of time (by warping or bending space-time).
Normally one would expect a force to be required to produce an acceleration but in this instance it is brought about by the geometry of space-time.

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Offline Geezer

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« Reply #29 on: 11/09/2011 16:34:49 »

Gravity produces the effect of acceleration by altering the passage of time (by warping or bending space-time).


So, if I'm in free fall and observe my velocity using my wrist watch, I will find that my velocity is constant? I don't think so  [:D]
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline MikeS

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« Reply #30 on: 12/09/2011 09:31:30 »
Geezer you are obviously quite right.  The question is why?  Each of the above statements (in my last post) I believe to be true but the time dilation factor is no where near enough to account for the acceleration.

Added 19-09-11
Using a gravity calculator to check gravitational acceleration over 2000km, equating acceleration to the strength of gravity at any altitude and equating that to the time dilation feature, it can be estimated, time dilation is enough to account for the acceleration.
« Last Edit: 19/09/2011 11:00:23 by MikeS »

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Offline Geezer

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« Reply #31 on: 12/09/2011 21:13:41 »

The question is why? 


How the heck would I know  [;D]

Gravity remains a great mystery. The bendy space/time model seems to work really well in terms of describing and predicting the effects. I wonder it we will ever get much further than that.
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Offline JP

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« Reply #32 on: 12/09/2011 22:01:33 »
Gravity remains a great mystery. The bendy space/time model seems to work really well in terms of describing and predicting the effects. I wonder it we will ever get much further than that.

Is there a scientific theory that's can do more than being able to describe and predict effects?

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Offline Geezer

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« Reply #33 on: 13/09/2011 00:01:29 »
Gravity remains a great mystery. The bendy space/time model seems to work really well in terms of describing and predicting the effects. I wonder it we will ever get much further than that.

Is there a scientific theory that's can do more than being able to describe and predict effects?

Picky, picky! Some theories provide a bit more insight into the underlying mechanisms than others. The underlying mechanism of gravity remains mysterious (unless somebody figured it out while I wasn't paying attention.)

As I can see no harm in teaching you how to suck eggs, I should point out that "science" literally means "knowledge" (which you would know if you had ever taken the trouble to study a bit of Latin.) The fact that we promulgate a bunch of theories to explain "what" stuff does does not necessarily contribute to the "knowledge" of how the heck it does it. Ergo (another Latin word), a "theory" does not necessarily contribute to science.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline MikeS

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« Reply #34 on: 19/09/2011 10:32:16 »
Quote Geezer
"It seems to me that there is no "force of gravity", although gravity clearly accelerates mass. What's wrong with this reasoning?
F = m.a"


Acceleration is a change in velocity over time.  So just by varying the passage of time you can have an acceleration.  Any gravitating body dilates the passage of time near its surface.

Gravity produces the effect of acceleration by altering the passage of time (by warping or bending space-time).
Normally one would expect a force to be required to produce an acceleration but in this instance it is brought about by the geometry of space-time.


Gravity produces the effect of acceleration by altering the passage of time (by warping or bending space-time).


So, if I'm in free fall and observe my velocity using my wrist watch, I will find that my velocity is constant? I don't think so  [:D]

Geezer

An object free falling to Earth is seen to be accelerating.
An object free falling to Earth is experiencing time dilating more and more the closer it gets to the Earth.
Time dilating allows more distance to be covered per unit time.
This is acceleration due to time dilation.
This is an affect of gravity.

Your watch in free fall measures more distance being covered per unit time precisely because each unit of time is getting longer.  This effect is probably being diluted by other gravitating bodies like the solar system, our galaxy etc.

Time dilation due to the gravity of the Earth is a very small effect in the much larger picture.  To put this into context we must remember that gravity can make time stand still at the event horizon of a black hole or the passage of time could be infinite.  This gives a huge rage to the potential passage of time.
« Last Edit: 19/09/2011 10:37:21 by MikeS »

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Offline Geezer

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« Reply #35 on: 20/09/2011 01:12:35 »

Your watch in free fall measures more distance being covered per unit time precisely because each unit of time is getting longer.


Erm, are you sure about that? If that were the case, you wouldn't observe any acceleration at all because the time to travel any distance would be proportional to the distance traveled, which would mean you had constant velocity.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline MikeS

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« Reply #36 on: 20/09/2011 07:59:07 »

Your watch in free fall measures more distance being covered per unit time precisely because each unit of time is getting longer.


Erm, are you sure about that? If that were the case, you wouldn't observe any acceleration at all because the time to travel any distance would be proportional to the distance traveled, which would mean you had constant velocity.


Geezer
Yes, when observed from a great distance this is true.  The geometry of space-time produces a constant velocity.  I 'believe' it is in accordance with relativity.


Normally one would expect a force to be required to produce an acceleration but in this instance it is brought about by the geometry of space-time.


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Offline imatfaal

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« Reply #37 on: 20/09/2011 10:17:56 »
Mike - do the maths!  If I drop a ball out of my office window (well my old office) after a second the ball is doing 10 metres per second - after two seconds it is doing twice that speed; please don't tell me that this effect is due to time dilation.

And if you read the thread on falling into black holes you will find that time doesn't actually come to a stop a the Event horizon of a black hole - that is an artefact of using the wrong form of coordinates for the job; whilst schwarzchild coordinates are great for outside the black hole near the eh they create a mathematical singularity (ie you get undefined/infinite answers) - better choices of coordinates remove this anomaly.
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Offline MikeS

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« Reply #38 on: 21/09/2011 16:06:48 »
Mike - do the maths!  If I drop a ball out of my office window (well my old office) after a second the ball is doing 10 metres per second - after two seconds it is doing twice that speed; please don't tell me that this effect is due to time dilation.

And if you read the thread on falling into black holes you will find that time doesn't actually come to a stop a the Event horizon of a black hole - that is an artefact of using the wrong form of coordinates for the job; whilst schwarzchild coordinates are great for outside the black hole near the eh they create a mathematical singularity (ie you get undefined/infinite answers) - better choices of coordinates remove this anomaly.

imatfaal
Yes, I think it is due to time dilation and I agree it sounds impossible but, I suspect, that may just be a matter of scale and our arbitrary way of measuring the passage of time.
"gravity can make time stand still at the event horizon of a black hole" "from the perspective of a distant observer" I should have added. 

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Offline Pmb

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« Reply #39 on: 22/09/2011 02:52:52 »
I forgot to mention that Newton defined force as F = dp/dt, not F = ma. The former is in the beginnning of the Principia and the later is Euler's expression for the force/mass/acceleration relastionship. The expression F = dp/dt is relativistically correct, where generally F = ma is not.

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Offline Geezer

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« Reply #40 on: 22/09/2011 08:22:18 »
I forgot to mention that Newton defined force as F = dp/dt, not F = ma. The former is in the beginnning of the Principia and the later is Euler's expression for the force/mass/acceleration relastionship. The expression F = dp/dt is relativistically correct, where generally F = ma is not.

What's the difference?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Pmb

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« Reply #41 on: 27/09/2011 19:08:37 »
I forgot to mention that Newton defined force as F = dp/dt, not F = ma. The former is in the beginnning of the Principia and the later is Euler's expression for the force/mass/acceleration relastionship. The expression F = dp/dt is relativistically correct, where generally F = ma is not.

What's the difference?
They don't always have the same value, e.g. for v << c the mass of a rocket which is accclerating from its engines being turned on, and when the velocity is relativistic. Its best to define F = dp/dt and when v << c and m = constant.

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Offline Geezer

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« Reply #42 on: 27/09/2011 22:33:46 »
I forgot to mention that Newton defined force as F = dp/dt, not F = ma. The former is in the beginnning of the Principia and the later is Euler's expression for the force/mass/acceleration relastionship. The expression F = dp/dt is relativistically correct, where generally F = ma is not.

What's the difference?
They don't always have the same value, e.g. for v << c the mass of a rocket which is accclerating from its engines being turned on, and when the velocity is relativistic. Its best to define F = dp/dt and when v << c and m = constant.

Can rockets achieve relativistic velocities?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Pmb

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« Reply #43 on: 27/09/2011 22:51:35 »
Quote from: Geezer
Can rockets achieve relativistic velocities?
Theoreticaly, yes. But that would be to impractical to build.

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Offline imatfaal

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« Reply #44 on: 28/09/2011 10:57:57 »
Quote from: Geezer
Can rockets achieve relativistic velocities?
Theoreticaly, yes. But that would be to impractical to build.

Theoretically - Not really.

[attachment=15308]

[attachment=15310]


It is clear from the above (MR is mass ratio of Massfull over Massempty) that to accelerate a 1000kg rocket with a exhaust velocity of 10000 m/s (ie double what is currently achievable to make maths simpler) to a tenth of the speed of light you would need the mass of the full rocket to be 1000e^3000kg  - That is the mass of many many universes. Even taking the velocity of exhausted gases up to 10^6 m/s would require a rocket that outweighs an asteroid (10^17kg) - and that's at maximal efficiency and with make-beleive technology

Realistically - it is doubtful that even .05c is on the cards, even with the exotic tech envisaged for the future such as ion drive and fusion drive



There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n