Post by: PmbPhy on 17/04/2018 08:13:42
I'm good on my word so when I said I'd post the original letter by Einstein o Lincoln Barnette I meant it so here it is. Please see attachment.
In essence Einstein stated in no uncertain terms that gravity should not be thought of as a curvature of spacetime. Spacetime curvature is just the relativistic term for tidal gradients. He also stated the relativity does not geometrize physics anymore that EM or the distance between two points.
Einstein_SR_GR.pdf (666.56 kB - downloaded 784 times)
Post by: geordief on 17/04/2018 11:10:17
It is the model that is geometric according to this letter?
A bit similar to the arguments as to why all physics can be reduced to maths and numbers....
EDIT:the geometric model has shown no signs of becoming less accurate at "higher definition" levels ,though has it? (ie measuring gravity/curvature at extremely small levels of mass/energy )
It is anticipated to break down at some point but that point seems not to have been reached or even "approached".
Post by: opportunity on 17/04/2018 12:05:37
I'm terrible at explaining this stuff, so hopefully it didn't offend.
I'd like to add more beyond this basic stuff......(?)
This is the physics section, not a new ideas symposium, so I look at the question and hesitate with an answer, such a good question.
Post by: jeffreyH on 17/04/2018 13:12:20
This is a leading question of course since I know the answer. I'm merely curious as to who believes it.
I'm good on my word so when I said I'd post the original letter by Einstein o Lincoln Barnette I meant it so here it is. Please see attachment.
In essence Einstein stated in no uncertain terms that gravity should not be thought of as a curvature of spacetime. Spacetime curvature is just the relativistic term for tidal gradients. He also stated the relativity does not geometrize physics anymore that EM or the distance between two points.
Somebody had to say it. Well done.
Post by: opportunity on 17/04/2018 15:55:49
< I should make that my call sign >
Post by: geordief on 17/04/2018 16:12:05
And the answer goes to.....the dude playing the flute we are making poetry to.....?Pan's People?
< I should make that my call sign >
https://en.wikipedia.org/wiki/Pan%27s_People
Post by: opportunity on 17/04/2018 16:13:54
Pan's pandemia.
I would like physics to have a few metaphors, but "math" seems to take centre stage, especially when billions of dollars and the investments of many are at stake......as a way to be serious about living. I love that. Couldn't love it less.
"Clearly" this response is in response to a question, yet please do not let this response, all members, detract you from the OP's question. Apologies if I answered outside of the OP's intent.
Who claimed gravity is a curvature of spacetime?
Einstein.
He didn't say spacetime is a curvature of gravity. Or did he, and is there a difference between saying gravity is a curvature of spacetime or spacetime is a curvature of gravity?
Post by: PmbPhy on 17/04/2018 17:09:07
Quote from: opportunity
Who claimed gravity is a curvature of spacetime?Nope. It was Max von Laue. Did you read the PDF file I posted in the OP?
Einstein.
Quote from: opportunity
He didn't say spacetime is a curvature of gravity. Or did he, and is there a difference between saying gravity is a curvature of spacetime or spacetime is a curvature of gravity?You can have spacetime with no curvature so no. They're not the same.
Post by: saspinski on 17/04/2018 19:02:37
Or, is it possible a Riemann tensor different from zero in a region and no gravity field there?
If the answer is no for both questions, then spacetime curvature <=> gravity.
Post by: PmbPhy on 17/04/2018 21:00:16
Quote from: saspinski
If the Riemann tensor is zero in a region of spacetime, is it possible a gravity field there?Yes. Without question. Its the affine connection that determines the presence of a gravitational field, no tidal forces.
Quote from: saspinski
Or, is it possible a Riemann tensor different from zero in a region and no gravity field there?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.
Quote from: saspinski
If the answer is no for both questions, then spacetime curvature <=> gravity.Wrong. And if I might add - Wow! You said all of that without reading the letter Einstein wrote. That's a problem in this thread and there's no excuse for it. In that letter Einstein wrote
Quote from: Albert Einstein
... 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 whyCase closed.
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.
Please folks. Before making a claim on a paper/letter first read the paper/letter. Otherwise you come off quite poorly. Okay? :)
I should note something very important here. It's not me that who discovered this point. It was Dr. John Stachel, someone whom I became friends with over the last 18 years. He wrote a paper which touched on this subject and gave me a copy. He's a great guy who had a lot of interesting friends and acquaintances such as Karl Popper and Wolfgang Rindler (may he rest in peace).
https://en.wikipedia.org/wiki/John_Stachel
Post by: jeffreyH on 17/04/2018 21:29:22
Post by: opportunity on 17/04/2018 22:29:36
However the extract/letter you are referring to is central to "general" relativiity.
"Special" relativity has a differennt can of worms as Einstein would know.
Its is far easier to relate "special" relativity with geometry than "general" relatviity.......and I think this is the uniquue case of the letter you have provided.
Once again though, bravo for diigging this letter up.
Post by: saspinski on 18/04/2018 00:47:19
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.
Post by: PmbPhy on 18/04/2018 01:41:34
Quote from: saspinski
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.If you looked in MTW you'd see their example. But, sure. See:
http://www.newenglandphysics.org/physics_world/gr/grav_cavity.htm
Quote from: saspinski
If there are tidal forces, and the Riemann tensor is not zero, some Christoffel symbols must be non zero.Not true. In a locally inertial frame in a curved spacetime all of the affine connections vanish but the Riemann tensor doesn't. That's because the Riemann tensor is a function of both the Christoffel symbols as well as their derivatives. See "Coordinate expression" at
https://en.wikipedia.org/wiki/Riemann_curvature_tensor
Quote from: saspinski
What according to the Einstein text (..non-vanishing of the components of the affine connection...) => gravitational field.You're making it about Einstein when in fact many others agree with him and its in all of his books. Its simple. How do you tell if there's a gravitational field in your living room? Simple; hold an apple in your hand while its stretched out and then let it go. If it drops to the floor at a rate independent of its mass (drop others with different masses to check this out) then there's a gravitational field in that room. How do you tell if there are tidal forces? Use sensitive equipment and measure the small changes in the field with height. If they exist then there are tidal forces present.
You could read Einstein's gravitational field by Peter M. Brown which is online at:
https://arxiv.org/abs/physics/0204044
Yes. I wrote it. It should answer your questions. If not then your comments would be invaluable to me. :)
Post by: saspinski on 19/04/2018 02:23:52
Not true. In a locally inertial frame in a curved spacetime all of the affine connections vanish but the Riemann tensor doesn't. That's because the Riemann tensor is a function of both the Christoffel symbols as well as their derivatives.
The Christoffel symbols are functions of position and time. If in a point and its neighborhood they vanish, their derivatives also vanish at this point => Riemann tensor = 0.
About the example, it is too complicated to calculate the metric tensor and see if the Riemann tensor is zero. By the way, it is nomally said that if there are no tidal forces, the Riemann tensor is zero, but I don't know a proof from the tensor definition.
Post by: Bill S on 19/04/2018 20:59:07
Einstein Didn't Say That: Exposing the Common Sense in Relativity Theory.
Don Griffin
"Einstein Didn't Say That" is a lighthearted but respectful exploration of the everyday logic behind special and general relativity. The first person to use the term 'curved space' was not Einstein, and not a physicist, but a reporter for the New York Times, who needed a catchy headline. What did Einstein really say, or not say, about curved space, time travel, wormholes, traveling twins, extra dimensions, multiple universes, dark matter and more? Have we been sold more sizzle than steak? With the help of dozens of quotes from well-known physicists, but mainly from Einstein himself, the author tries to separate Einstein's original ideas from some of the speculation that those ideas inspired in others. This is not a textbook or a history of physics, but rather the personal story of an ordinary guy's delight in discovering
Post by: jeffreyH on 19/04/2018 23:34:52
Post by: PmbPhy on 20/04/2018 13:35:47
Quote from: saspinski
The Christoffel symbols are functions of position and time. If in a point and its neighborhood they vanish, their derivatives also vanish at this point => Riemann tensor = 0.That' incorrect. It's only when both al the first derivatives a ad the Christoffel symbols and tthey all vanish is the spacetime flat. That doesn't mean there's no gravitational filed. In fact Einstein said that a gravitational field can be "produced" by a change in the frame of reference. This is not about a time varying thing like you're implying.
Since this thread was diverted from is purpose I'm brining it back on track. Einstein wrote
Quote from: Albert Einstein
... 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 whyPlease focus on the subject at hand, i.e. What Einstein actually said, regardless of whether you think that he was right opr wrong. Okay?
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.
Quote from: Albert Einstein
About the example, it is too complicated to calculate the metric tensor and see if the Riemann tensor is zero.QuoteIt's rehire easy. See: http://www.newenglandphysics.org/physics_world/gr/uniform_force.htm
Post by: PmbPhy on 20/04/2018 16:09:13
"Einstein Didn't Say That" is a lighthearted but respectful exploration of the everyday logic behind special and general relativity. The first person to use the term 'curved space' was not Einstein, and not a physicist, but a reporter for the New York Times, who needed a catchy headline.
[/quote]
Awesome. Can you tell me what news paper it was in and how I can get a copy of it:?
I really appreciate this bill. I love you! :) That ..doesn't mean we'll be taking warm showers in the wee hours of he morning - Clint Eastwood in "Heartbreak Ridge". I love that saying! :)
Post by: saspinski on 20/04/2018 16:26:57
That is the reason for the Riemann tensor be zero in this case.
Post by: PmbPhy on 20/04/2018 17:39:51
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.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?
That is the reason for the Riemann tensor be zero in this case.
Post by: saspinski on 21/04/2018 15:50:31
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.
Post by: PmbPhy on 21/04/2018 17:20:24
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? :)
Post by: saspinski on 22/04/2018 22:39:27
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.
Post by: PmbPhy on 24/04/2018 07:52:36
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.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.
Post by: saspinski on 25/04/2018 02:08:40
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.
Post by: bluesXwinXtheXcup on 25/04/2018 06:07:07
Post by: PmbPhy on 25/04/2018 16:31:43
Quote from: 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.
Quote from: saspinski
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.
Quote from: saspinski
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.
Post by: saspinski on 26/04/2018 02:04:13
Quote from: saspinskiSo 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?
Post by: PmbPhy on 26/04/2018 13:04:39
Quote from: saspinski
I haven't written it clearly.That is quite wrong when its for massive spherical bodies.
The greater the mass the greater the gravitational field, OK. But not necessarly greater the tidal forces (curvature).
First off let's get something straight. The Riemann tensor is related by '
Rk0lo = 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.
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:Ae you now merely trying to find ways to make it different, Those are wrong y the way. See my page on tidal forces.
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.
Jupiter's tidal forces are significantly greater
Quote from: saspinski
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]
Post by: saspinski on 26/04/2018 23:30:08
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]Wrong. And you made no effort to support such an inalid claim. Why is that?[/quote]
...but why to use GR concepts for something that is in the range of SR?
It is well known that SR can deal with uniformly accelerated frames of reference, see Rindler coordinates.
Post by: PmbPhy on 27/04/2018 05:44:42
Quote from: pmbphy
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.
Quote from: saspinski
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
Quote
... 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.
Post by: PmbPhy on 27/04/2018 06:14:04
Quote from: saspinski
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: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.
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.
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
Post by: PmbPhy on 27/04/2018 10:46:14
Post by: Petrochemicals on 28/04/2018 03:11:47
Is space curved around a cube ? Like the borg ships out of startrek ?
Post by: PmbPhy on 28/04/2018 10:56:27
https://www.pdf-archive.com/timer.php?id=351738
Post by: Bill S on 28/04/2018 13:02:11
Quote from: Petrochemicals
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?
Post by: PmbPhy on 28/04/2018 14:24:36
Quote from: Bill S
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.Howdy Bill. Do you know how I can get my hands on that article you mentioned about the NY Times?
Wouldn't that be in line with Einstein's concept; without suggesting that Einstein actually said spacetime was curved?
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
Quote
... 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.
Post by: mad aetherist on 18/10/2018 14:52:47
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.
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.
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.
Post by: alancalverd on 18/10/2018 15:06:00
Post by: mad aetherist on 18/10/2018 15:21:30
Quote from: saspinski
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.If you looked in MTW you'd see their example. But, sure. See:
http://www.newenglandphysics.org/physics_world/gr/grav_cavity.htm
Comment. I had a look. That link does not refer to an infinite plane or plate. It refers to a hole. And here i am thinking that the equations shown might be wrong because i feel sure that the equations dont explain the well known borehole (gravity) anomaly.
Quote from: saspinski
If there are tidal forces, and the Riemann tensor is not zero, some Christoffel symbols must be non zero.Not true. In a locally inertial frame in a curved spacetime all of the affine connections vanish but the Riemann tensor doesn't. That's because the Riemann tensor is a function of both the Christoffel symbols as well as their derivatives. See "Coordinate expression" at
https://en.wikipedia.org/wiki/Riemann_curvature_tensor
Quote from: saspinski
What according to the Einstein text (..non-vanishing of the components of the affine connection...) => gravitational field.You're making it about Einstein when in fact many others agree with him and its in all of his books. Its simple. How do you tell if there's a gravitational field in your living room? Simple; hold an apple in your hand while its stretched out and then let it go. If it drops to the floor at a rate independent of its mass (drop others with different masses to check this out) then there's a gravitational field in that room.
Comment. I dont agree. If your living room is a part of a rotating artificial g space station with no gravity field then your apple test there would-might give results no different to tests in a living room sitting on Earth.
How do you tell if there are tidal forces? Use sensitive equipment and measure the small changes in the field with height. If they exist then there are tidal forces present.
Comment. I dont understand the concept of tidal forces, why are tidal forces a different thing to plain old gravity forces. A small change in the field with height shows gravity is present -- & gravity gives tidal effects (or can). Have i missed something?
Just remembered. There are two tidal forces. On the Earth's near side to the moon the tidal force is the moon's gravity partly negated by centrifugal force (arising from Earth orbiting the barycenter). On the Earth's far side from the moon the tidal force is a combination of the reduced gravity from the moon plus a largish centrifugal force. In between the near side & the far side we have the in-between bits of the Earth where there is in effect no tidal force, yet in all 3 sections sensitive equipment would find small changes in the field with height. Or am i missing something?
You could read Einstein's gravitational field by Peter M. Brown which is online at:
https://arxiv.org/abs/physics/0204044
Yes. I wrote it. It should answer your questions. If not then your comments would be invaluable to me. :)
Post by: mad aetherist on 18/10/2018 16:00:37
I reckon that the gravitational field around an infinitely long straight wire (or a long straight string) is not zero --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.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 field varies as per 1/R (the field around a ball varies as per 1/RR). How could it be zero?
Post by: mad aetherist on 18/10/2018 16:08:07
Ok, I still don't understand why gravity exists though. Why do two masses attract?Dont expect an answer from Einsteinians. I am told that according to Einstein's field equations (which i dont understand) two stationary masses do not attract. Alltho "stationary" i think has no meaning in SR & GR.
Post by: Bill S on 19/10/2018 14:03:31
Quote from: Mad Aetherist
I reckon that the gravitational field around an infinitely long straight wire (or a long straight string) is not zero –
One trouble with thoughts about anything “infinitely long” is that, whatever the maths/theory might establish, you will never be able to provide physical “proof”.
Tipler had some interesting ideas along these lines, even postulating time travel using an infinitely long cylinder.
Come to think of it; he had some flamboyant ideas about omega-point theory, that Teilhard de Chardin might never have intended. :)
Post by: mad aetherist on 20/10/2018 01:10:02
Yes & no. U can get a useful result by simply bringing the test particle up very close, eg if the wire/string is only 1000 m long (instead of infinite m) u can place the test particle 1 mm away (& get a 0.999999 result).Quote from: Mad AetheristI reckon that the gravitational field around an infinitely long straight wire (or a long straight string) is not zero –One trouble with thoughts about anything “infinitely long” is that, whatever the maths/theory might establish, you will never be able to provide physical “proof”.
Post by: mad aetherist on 20/10/2018 02:42:35
I dont think i have heard of 1/R deriving from Newton. My own idea is simply based on aether flowing in to the wire-string to replace aether annihilated in the wire-string, such inflow streamlines being 2 dimensional (giving 1/R) instead of the 3D inflow streamlines for a ball (giving 1/RR).That a linear mass (wire/rod) has gravity dropping off as an inverse of radius (instead of inverse squared) derives directly from Newton's equations, and M-A gives a valid finite way to test this. It doesn't require infinite length.Quote from: Mad AetheristI reckon that the gravitational field around an infinitely long straight wire (or a long straight string) is not zero –One trouble with thoughts about anything “infinitely long” is that, whatever the maths/theory might establish, you will never be able to provide physical “proof”.
A flat sheet has gravity that doesn't drop off at all. If Earth was flat and large enough, gravity would be 1 g all the way up.Here i say no, the gravity would be 0.00 g for a distance, then as u got further away the g would gradually rise, & when u were far enuff away (& could see the whole Earth below) the g would be nearnuff its full value, say 1.00 g.
This is based on g near an infinite plate being 0.00 g at all distances. This is due to the fact that the inflow streamlines must all be parallel, ie there will be no acceleration of aether flowing in to the plate, & if no acceleration of aether then no gravity (& no mass)(depending on how u define mass). All of this is Aether 101.
If u accept that then u must accept that g near a spiral galaxy aint g. There are Nobel prizes waiting for this stuff, but Einsteinians are happy to die ignorant.
Just realized. Einsteinians must surely say that spacetime is flat near an infinite plate, in which case 0.00 g at all distances is pure Einsteinian 101 also.
Post by: CPT ArkAngel on 20/10/2018 04:00:18
Post by: mad aetherist on 20/10/2018 04:31:16
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.I missed this posting. Yes a plane has no mass. I wonder why the plane came up in the discussion, must be re some technical aspect of looking up Einstein's quoit with a microscope.
Yes gravity is additive (in which case it is also subtractive).
But i might not say infinitesimal masses. I reckon that there are only two types of mass, free-photons (eg light), & confined-photons (elementary particles). And a third type of mass if u like, which is electromagnetic fields (these being made up of photinos (my name) which are a part of every photon, & emanate out to infinity from every photon (if indeed photinos have mass)(i havnt made up my mind)(so i include them here just in case)(i think that Einstein said that they did have mass).
Einstein said that gravitational fields have mass. I dont agree (so i dont include it).
Einstein said that energy had a mass equivalence in some circumstances, but i dont think that that meant that mass increased with speed (so i dont include relativistic mass).
Post by: CPT ArkAngel on 20/10/2018 05:06:41
Post by: mad aetherist on 20/10/2018 07:41:02
Even though gravity is invisible apart from its effect on relative motion of bodies, gravitational energy is real. Gravitational waves prove beyond any reasonable doubt that GR is right on that specific topic. All energy produces a gravitational potential, gravity itself included. Many experiments proved the gravity-inertia equivalence, none proved the contrary. When you throw an object, you transfer some of your mass-energy to the object and the object increases its relativistic mass-energy by the same amount. Gravity included, relative to any inertial frame (or non accelerating observer).Nah i dont believe any of that. GWs that travel at c dont exist. Ligo's GWs are fake (or an error)(praps a harmonic of the calibration signal). What we have are gravity pulses, that travel at over 20 billion c (Van Flandern). Plus we have gravity turbulence, which travels at 500 kmps south to north throo Earth (Cahill).
Yes gravity energy is of course real, but that doesnt mean that it has mass or wt (it doesnt have mass or wt).
Yes all energy produces a gravity potential (ie due to having a gravity field), except that gravity energy (ie a gravity field) does not produce a gravity field. That would be silly, a field producing a field which produces a field which produces a field which produces a field which produces a field which produces a field etc etc.
Yes gravitational mass equals inertial mass, because gravitational mass is inertial mass, because of the way we measure gravitational mass, we measure it by measuring its inertial mass, & then they trumpet about how they have proven them to be equal to umpteen decimals, when all they have proven is that inertia equals inertia (give that man a Nobel).
Yes, when u throw an object u increase its energy, but no u dont increase its mass. Its ok for an object to have an energy relative to a frame or something. But its not ok for an object to have a mass relative to a frame or something -- Einsteinians have been saying that sort of thing & getting away with it for so long that they dont even think about it any more. Mass is absolute, it has one value, however i am ok with the notion that apparent mass depends on the frame of reference, ie that it depends on gamma (affecting apparent time, & affecting apparent length), that is a possibility (i will say for sure after i think it throo one day).
Post by: yor_on on 23/10/2018 05:47:02
affine connections
Maybe this will give a intuitive idea of what it is
https://www.physicsforums.com/threads/affine-connection.216136/