[PmbPhy (OP)]

Suppose a function f(x,y,z,t). f is zero in a region around (x0,y0,z0,t0). All derivatives of f are also zero. So, it is not possible all Christoffel symbols be zero while some of its derivatives escape this fate.
That is the reason for the Riemann tensor be zero in this case.

Why do you constantly keep insisting on the obvious? I know that better than anybody in the forums I go to. Why do you insist on ignoring what I told you, i.e. the Christoffel symbols become non-when changing from the space you keep talkin about to one that's accelerating??? Would you like me o post the proof?

[saspinski]

If I didn't get lost in the calculations, for the metric of an uniform gravitational field (mentioned in your link), R⁰₀₃₀ is not zero. So it is not a flat spacetime, even without tidal forces.

It is true that in the case of a uniform gravitational field, for the reference frame of an observer in free fall, the spacetime is flat. The same spacetime would be curved for an observer not in free fall.

That would be the case for a spaceship with uniform acceleration in outer space, and the RF of an astronaut jumping from the ship, compared to the others staying there.

But for a more conventional gravitational field, as the existing around the Earth, an observer in free fall (as the astronauts in the ISS) is not in a flat spacetime. It seems very flat indeed in the spatial range of that small ship, and if the period of observation is also small. After some minutes, looking through the window, the Earth will be rotating around. And geodesics should be straight lines for a flat spacetime.

So gravity is a curvature in spacetime, in the meaning that there is no gravitational field without that curvature, and no curvature without gravitational effects.

[PmbPhy (OP)]

If I didn't get lost in the calculations, for the metric of an uniform gravitational field (mentioned in your link), R⁰₀₃₀ is not zero. So it is not a flat spacetime, even without tidal forces.

You did get lost. Its not possible to have a non-zero Riemann tensor for a uniform gravitational field. Its actually the definition of a uniform field.

I'm letting this rest at this point. There's plenty of literature out there on the derivation such as in MTW (do you have that text?). IF anybody wants my help in finding one I'd be glad to help. Right now I just don't want to have to keep repeating myself. Especially when you made such a critical error here.

The best thing you can do is read one of those papers I mentioned at
http://www.newenglandphysics.org/physics_world/gr/uniform_field.htm

which so far you don't wish to read one of them for some odd reason. If you ever change your mind then read this one. Just click on this URL - http://booksc.org/dl/1945565/822468
Then click on the paper URL where it says "Download (pdf, 1.37 MB)"

I'd like to encourage you to stay on topic too. Whether its true or not has nothing to do with this thread. Its about who first said gravity= spacetime curvature. Understand?

[saspinski]

You did get lost. Its not possible to have a non-zero Riemann tensor for a uniform gravitational field. Its actually the definition of a uniform field.

Yes, I changed a γ by a δ of one of the Γ's. There are a lot of them. But now I checked everyone, and all components are really zero.

So, in the specific case of an uniform gravitational field, the spacetime is flat for any observer (being or not in free fall).

For conventional (non uniform) gravitational fields, the spacetime is curved for any observer.

[PmbPhy (OP)]

You did get lost. Its not possible to have a non-zero Riemann tensor for a uniform gravitational field. Its actually the definition of a uniform field.

Yes, I changed a γ by a δ of one of the Γ's. There are a lot of them. But now I checked everyone, and all components are really zero.

So, in the specific case of an uniform gravitational field, the spacetime is flat for any observer (being or not in free fall).

For conventional (non uniform) gravitational fields, the spacetime is curved for any observer.

The gravitational field of a vacuum domain wall has zero curvature everywhere off the wall itself. In this case the wall is repulsive. The field around a straight cosmic string is zero for both the curvature and there is no gravitational field around the string giving the space around it a conical solace but does not exert gravitational forces on things nearby.

[saspinski]

There is that tradicional picture of a sphere deforming a membrane as a representation of a gravitational field. It can spread some confusion if people take the curvature of this scalar field as a measure of its intensity. And by intensity I mean the fact that the same object would weight more in Jupiter than in the Earth.

As Jupiter is less dense, that curvature would be smaller on its surface.

So the idea that the greater the mass the greater the curvature, and stronger the gravity field is certainly wrong. 

But it is not wrong to say that the curvature of the metric tensor field, defined by the Riemann tensor, is related to its intensity. But intensity here meaning the tidal forces, because they are the same for an observer at rest in the planet, or in free falling, and more apropriate to a description of gravity following the relativity principle.

Using this criteria, a man free falling from a rocket in uniform acceleration would agree with the rest of the crew that there is no gravitational field, because there is no tidal forces.
 

[bluesXwinXtheXcup]

Ok, I still don't understand why gravity exists though. Why do two masses attract?

[PmbPhy (OP)]

So the idea that the greater the mass the greater the curvature, and stronger the gravity field is certainly wrong. 

I can't see how you got that idea. The greater the mass the greater the gravitational field and tidal forces (aka spacetime curvature) Ohanian's text shows the relationship between the two.

But it is not wrong to say that the curvature of the metric tensor field, defined by the Riemann tensor, ..

You have it backwards. The metric defines the gravitational field and therefore defines the Christoffel symbols which in turn defines the Riemann tensor.

Using this criteria, a man free falling from a rocket in uniform acceleration would agree with the rest of the crew that there is no gravitational field, because there is no tidal forces.

Not true. The gravitational field is defined by the Christoffel symbols, no the tidal force tensor. A least not according most GR texts I've seen, to Albert Einstein, John Stachel (expert on Einstein and GR and the former editor of the Einstein papers project) and myself. I'm sure to find more when I asked the experts that I know personally.

[saspinski]

So the idea that the greater the mass the greater the curvature, and stronger the gravity field is certainly wrong. 

I can't see how you got that idea. The greater the mass the greater the gravitational field and tidal forces (aka spacetime curvature) Ohanian's text shows the relationship between the two.

I haven't written it clearly.     
The greater the mass the greater the gravitational field, OK. But not necessarly greater the tidal forces (curvature). The difference of gravity acceleration on the Jupiter surface (if it were possible be at rest there), in a given vertical lenght, is smaller than on the Earth surface:

da = GM/R² - GM/(R+d)². For d =1km, I have found a difference of 0,003 m/s² for Earth and 0,0007 m/s² for Jupiter.

The gravitational field is defined by the Christoffel symbols, no the tidal force tensor.

It is a matter of definition, but why to use GR concepts for something that is in the range of SR?

[PmbPhy (OP)]

I haven't written it clearly.     
The greater the mass the greater the gravitational field, OK. But not necessarly greater the tidal forces (curvature).

That is quite wrong when its for massive spherical bodies.
First off let's get something straight. The Riemann tensor is related by '

R^k_0lo  = Newtonian tidal force tensor
See Eq (5) at http://www.newenglandphysics.org/physics_world/cm/tidal_force_tensor.htm

Phi = GM/r where M is the mass of the body star. It's a constant and moves thoughout the equation for tidal forces so the larger the mass the larger the tidal force. Make no mistake about that.

The difference of gravity acceleration on the Jupiter surface (if it were possible be at rest there), in a given vertical lenght, is smaller than on the Earth surface:

da = GM/R² - GM/(R+d)². For d =1km, I have found a difference of 0,003 m/s² for Earth and 0,0007 m/s² for Jupiter.

Ae you now merely trying to find ways to make it different, Those are wrong  y the way. See my page on tidal forces.
Jupiter's tidal forces are significantly greater

It is a matter of definition, ..

Please remind us what the tiite and purpose of this thread is and why you made zero effort to answer it?   

quote author=saspinski]
...but why to use GR concepts for something that is in the range of SR?
[/quote]
Wrong. And you made no effort to support such an inalid claim. Why is that?

[/quote]

[saspinski]

Quote from: saspinski

    The difference of gravity acceleration on the Jupiter surface (if it were possible be at rest there), in a given vertical lenght, is smaller than on the Earth surface:

    da = GM/R² - GM/(R+d)². For d =1km, I have found a difference of 0,003 m/s² for Earth and 0,0007 m/s² for Jupiter.

Ae you now merely trying to find ways to make it different, Those are wrong  y the way. See my page on tidal forces.
Jupiter's tidal forces are significantly greater

No. It is right. It is the definition of tidal force, at least for weak fields. Another example is our ocean tides. The effect of the Moon is about twice that of the Sun. And the Sun’s gravitational field (gravity acceleration) is much greater here than the Moon’s one.

quote author=saspinski]
...but why to use GR concepts for something that is in the range of SR?

Wrong. And you made no effort to support such an inalid claim. Why is that?[/quote]

It is well known that SR can deal with uniformly accelerated frames of reference, see Rindler coordinates.

[PmbPhy (OP)]

Ae you now merely trying to find ways to make it different, Those are wrong  by the way. See my page on tidal forces.Jupiter's tidal forces are significantly greater.

saspinski - First off I want to apologize. I was becoming snippy and I'm taking every effort not to do that. Alas, in the end I'm only human.

Second, you are wrong. If you want to take the difference of the force F then its

Tidal force = dF = (@F/@z)dz

You neglected to put in the dz. Look at Eq, (3) at http://www.newenglandphysics.org/physics_world/cm/tidal_force_tensor.htm


No. It is right. It is the definition of tidal force, at least for weak fields. Another example is our ocean tides. The effect of the Moon is about twice that of the Sun. And the Sun’s gravitational field (gravity acceleration) is much greater here than the Moon’s one.

It is well known that SR can deal with uniformly accelerated frames of reference, see Rindler coordinates.

Wrong. By definition SR is relativity in inertial frames. Just because yu see something in acceleration in an SR text it doesn't mean its an SR subject. For example: Use SR to show what the speed of light is in an accelerating frame. What are the Christoffel symbols in such a frame? MTW do this iand call it GR as do most texts = As does Einstein, regardless of how hard you avoid the issue its what Einstein held to be true and for good reasons. Einstein wrote

... what characterizes the existence of a gravitational field from the empirical
standpoint is the non-vanishing of the components of the affine connection],
not the vanishing of the [components of the Riemann tensor]. If one does
not think in such intuitive (anschaulich) ways, one cannot grasp why
something like curvature should have anything at all to do with gravitation.
In any case, no rational person would have hit upon anything otherwise. The
key to the understanding of the equality of gravitational mass and inertial
mass would have been missing.

You can find this definition of SR in Schutz's text.

[PmbPhy (OP)]

The greater the mass the greater the gravitational field, OK. But not necessarly greater the tidal forces (curvature). The difference of gravity acceleration on the Jupiter surface (if it were possible be at rest there), in a given vertical lenght, is smaller than on the Earth surface:

da = GM/R² - GM/(R+d)². For d =1km, I have found a difference of 0,003 m/s² for Earth and 0,0007 m/s² for Jupiter.

It's not clear at all to me where you're getting these numbers from. What is the value of R you're using in both expressions, i.e. for Earth and for Jupiter. If the value used is less than the radius than the gravitational force is inside the body where the force is linear in r.

There's a reason I said that you're tidal force was wrong other than what said and that's because its not a tensor. A tensor gives a complete form of the forces involved and that includes the inwardly directed forces from the sides as well as the outward forces along the direction of the field lines. Also tidal forces fall of as 1/r^3 not 1/r^2 as you have them.

Do you disagree with the derivation of tidal forces I derived at:
http://www.newenglandphysics.org/physics_world/cm/tidal_force_tensor.htm

[PmbPhy (OP)]

I have to stop here. I myself am contributing to  diverting my own thread from its stated purpose. Please feel free to start another thread on your subject.

[Petrochemicals]

I thought einstein said space was straight and so where the orbits ? It was merly the curve of the body represented ?

Is space curved around a cube ? Like the borg ships out of startrek ?

[PmbPhy (OP)]

By the way,. Gravitation by Misner, Thorne and Wheeler is online in PDF format for free at

https://www.pdf-archive.com/timer.php?id=351738

[Bill S]

Is space curved around a cube ? Like the borg ships out of startrek ?

If matter is gravitationally attracted towards Borg ships; which, presumably it would be; then a directionality would be present, which could best be described in terms of spacetime curvature.

Wouldn't that be in line with Einstein's concept; without suggesting that Einstein actually said spacetime was curved?

[PmbPhy (OP)]

If matter is gravitationally attracted towards Borg ships; which, presumably it would be; then a directionality would be present, which could best be described in terms of spacetime curvature.

Wouldn't that be in line with Einstein's concept; without suggesting that Einstein actually said spacetime was curved?

Howdy Bill. Do you know how I can get my hands on that article you mentioned about the NY Times?

No. Tidal forces would be much worse in describing the field than the affine connection. The connection coefficients, aka the Christoffel symbols, play the same part as the gravitational field vectors do in Newtonian gravity. The Riemann tensor plays the same role as the tidal force tensor in Newtonian gravity. That means they don't all point in the direction of the cube, some point in the direction perpendicular to it.

By the way. This is the part of that letter I was talking about. The one I posted first was an error

... what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the components of the affine connection], not the vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive (anschaulich) ways, one cannot grasp why
something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality of gravitational mass and inertial mass would have been missing.

[mad aetherist]

 
Yes. Without question. Its the affine connection that determines the presence of a gravitational field, no tidal forces.

Do you have an example? I can imagine a gravitational field generated by an infinite plane. In that case the field is uniform and there are no tidal forces. But infinite planes were not detected by astronomers until now.

Yes. In fact if you have a gradiometer in free fall while in orbit of Earth then in that frame the gravitational field is zero but there are still tidal forces present.

If there are tidal forces, and the Riemann tensor is not zero, some Christoffel symbols must be non zero. What according to the Einstein text (..non-vanishing of the components of the affine connection...) => gravitational field.

The gravitational field of an infinite plane (or an infinite plate if u like) is zero. It is zero above the plate, & it is zero below. Calling it uniform is sort of wrong. Zero is for sure uniform, but uniform is not zero.

Therefore two infinite plates will not attract each other (at least not gravitationally).
However, an ordinary object, eg a ball, will probly attract an infinite plate. However i am not sure of this. The answer puts a severe strain on what exactly is gravity, what exactly happens when two objects attract. I spend a lot of time thinking about this sort of thing.
 
Anyhow the next question should be whether 2 very large but not infinite plates attract according to GMm/RR. I dont think they do [edit 19.10.2018][if very close together]. Do u see any relationship to a flat spiral galaxy here? I do.

[alancalverd]

Difference between an infinite plane, which, having no thickness, has no mass, and an infinite plate, which has mass and therefore a gravitational field since it is an infinite array of infinitesimal masses, each of which has a field, and gravitation is additive. Indeed it is exactly this additivity that makes it special.