SR and the earth, sun, and galaxy.

[Sammywu]

DW, Would you mind elaborate a little more on the inertial force? What is it? How to calculate it? take a simple example is good enough. Also, are the inertial force and gravity the only two known ficticious forces?

Gravity is caused by the timespace curvature. How will the inertial force been considered? Does it appear as part of the stress energy tensor?

 

[Sammywu]

DW, I always thought Reimannian spacetime curvature is a description of gravity. From what you said, it's not. What is it then?

 

[Sammywu]

DW, Is the Riemann tensor the stress energy tensor? How does it look like?

 

[Arcon]

Originally posted by Sammywu

I know I have seen that strees energy tensor somewhere. I might ask question related to it later.

When you're ready then see
http://www.geocities.com/physics_world/sr/mass_tensor.htm

I call it the mass tensor in that page but the mass tensor and the energy-momentum tensor differ only by the multiplicative constant c². The reason for the namd "mass tensor" was only to make the point that the energy-momentum tensor is not a name that is fundamental to that tensor and that since mass is related to energy then one can speak of mass instead of energy. That fundamental fact seems to be lost on most students lately.
But the energy-momentum tensor is not something that is only used in GR. It's used in SR as well.

An inertial force is a force which is not present in an inertial frame of reference. For example: Suppose you're in an inertial frame of reference in a spaceship. A particle moving in a straight line at constant velocity in an inertial is said to be a free-particle. If the ship starts to accelerate then that same particle will now be an accelerating particle as viewed from your frame of reference. It behaves exactly as if it was in free-fall in a uniform gravitational field. That's Einstein's Equivalence Principle. Newton would have called that a fictitious force (hence dw is Newtonian in this respect). However Einstein would call it a real force. The reason Einstein referred to inertial forces as real is that they have the same nature as the gravitational force which Einstein considered to he real - almost by definition. In fact he did call it a real force. Many physicists do today as well.

Here's one such example: From Newtonian Mechanics, A.P. French, The M.I.T. Introductory Physics Series, W.W. Norton Pub. , (1971) , page 499.

From the standpoint of an observer in the accelerating frame, the inertial force is actually present. If one took steps to keep an object "at rest" in S', by tying it down with springs, these springs would be observed to elongate or contract in such a way as to provide a counteracting force to balance the inertial force. To describe such force as "fictitious" is therefore somewhat misleading. One would like to have some convenient label that distinguishes inertial forces from forces that arise from true physical interactions, and the term "psuedo-force" is often used. Even this, however, does not do justice to such forces experienced by someone who is actually in the accelerating frame of reference. Probably the original, strictly technical name, "inertial force," which is free of any questionable overtones, remains the best description.

The Coriolis force is a real force as well. As Einstein explained in the February 17, 1921 issue of Nature

Can gravitation and inertia be identical? This question leads directly to the General Theory of Relativity. Is it not possible for me to regard the Earth as free from rotation, if I conceive of the centrifugal force, which acts on all bodies at rest relatively to the earth, as being a "real" gravitational field of gravitation, or part of such a field? If this idea can be carried out, then we shall have proved in very truth the identity of gravitation and inertia. For the same property which is regarded as inertia from the point of view of a system not taking part of the rotation can be interpreted as gravitation when considered with respect to a system that shares this rotation. According to Newton, this interpretation is impossible, because in Newton's theory there is no "real" field of the "Coriolis-field" type. But perhaps Newton's law of field could be replaced by another that fits in with the field which holds with respect to a "rotating" system of co-ordiantes? My conviction of the identity of inertial and gravitational mass aroused within me the feeling of absolute confidence in the correctness of this interpretation.

 

[Sammywu]

Arcon,

Your statements are very interesting. I gradually figured that your arguments with DW seems to focus on whether the gravity mass is the same as inertial mass and the measured mass is a gravity mass.

Any way, I need some times to read them.

DW, you seem to believe that the inertial mass described by E/c^2 not the same as gravity mass. In one place, you said that g=GM/R^2 and the M is the rest mass. By the way, I am confused, proper time is the time of the mover's own time. How about proper mass? Straightly, it seems to be the rest mass. But you seem to say that's the E/c^2. Please help me with the terms. . Thanks.

 

[Sammywu]

DW, Arcon, If the argument focused on whether gravity mass is the same as inertial mass. I would like to ask another question.

If you noticed, some people are trying to capture a light in a confinement. Let's say if I am able to keep some photons bouncing inside a confinement, will this confinement exihibit any gravity from the mass by E/c^2 of the lights inside the confinement to the objects nearby?

What will your predict?

If I know a little bit about laser, maybe we can do some experiemnts with lasers.

 

[Arcon]

Originally posted by Sammywu

Your statements are very interesting.

We have Einstein to thank for that. :-)

By the way, the terms proper mass and rest mass are synonyms. I choose to use the term proper mass because the term rest mass can easily be interpreted to mean the relativistic mass when v = 0. However that is not always true since relativistic mass is also a function of the gravitational potential so you there can be a difference. I.e. for v = 0 relativistic mass, m, and proper mass, m₀ are related as

m = \frac {m_0}{\sqrt{1 + 2\Phi/c^{2} }}

Let's say if I am able to keep some photons bouncing inside a confinement, will this confinement exihibit any gravity from the mass by E/c^2 of the lights inside the confinement to the objects nearby?

Yes. But even a straight beam of light can generate a gravitational field sicne it has energy and energy has mass. John Archibald Wheeler once showed that it's possible for light was able to form an object and act as a gravitating body even with no confining walls. That object is called a geon. Its unstable though so it wouldn't last long.

 

[Sammywu]

Arcon, DW,

How about this one?

Since higher temperature means a group of molecules moving in higher average speed against us. Will they be measured with higher inertial mass and gravity mass?

 

[Sammywu]

DW, Arcon,

How about these two:

When an electron in a atom absorbed a photon and elevated to a higher state, will the atom show higher inertial mass and gavity mass?

If I spin a top to very high spin speed, not only now we have angular mometum also will it be measured with higher mI and mG?

 

[Arcon]

Sammywu - You do understand that there are two commonly used terms in relativity regarding mass right? These term terms are relativisitic mass and proper mass. In all cases when I use the term "mass" it means "relativisitic mass. It is in this sense of the term that I'm answering your questions. In the present case it makes no difference.

Originally posted by Sammywu

Since higher temperature means a group of molecules moving in higher average speed against us. Will they be measured with higher inertial mass and gravity mass?

Yes.

When an electron in a atom absorbed a photon and elevated to a higher state, will the atom show higher inertial mass and gavity mass?

Yes.

If I spin a top to very high spin speed, not only now we have angular mometum also will it be measured with higher mI and mG?

Yes. In fact I worked out a similar question this past week regarding a rotating cylinder. See
http://www.geocities.com/physics_world/sr/rotating_cylinder.htm

 

[Sammywu]

Arcon, DW, Actually I like the term proper mass. It's better than rest mass. Put it simple, there is no such thing as rest mass. An electron in an atom has different mass from a free electron as we already show, if Arcon is right and DW agreed. The elctron has a spin of 1/2, so its true rest-rest mass shall be smaller than our published figure if you can stop its spin. This is my stupid thought. Do not bother that much.

DW, can we show that stress energy tensor or the mass energe tensor changed due to the spin of the top with your modern SR? Arcon, when I have time, I will browse your explanation of spinning of cylinders.

Thanks.

 

[Sammywu]

DW, Arcon, About relativistic mass, I can't say what I agree. Let's examine another case. Let two objects A and B with the same mass m, moving toward each other in 2/3c relative to me, the observer. The threes are all in its own inertial reference frames. Agreed?

Now, two objects collide and lump together as one. What is the mass when they lump as one? What were they before lumped together, relative to me, A or B?

Can we apply this to the phenomenon of an electron and a positron lumped together as a gamma photon for a short period?

Is there a term. for apparent mass? Is it the same as relativistic mass?

By the way, I think DW does have good intention in puclishing the modern relativity formulae in another thread. I think we can show our difference in thinking and theory. Actually DW shall make his formulae even more clearer. Discussion of how they were derived will be even better.

Thanks

 

[Arcon]

Originally posted by Sammywu
DW, Arcon, About relativistic mass, I can't say what I agree. Let's examine another case. Let two objects A and B with the same mass m, moving toward each other in 2/3c relative to me, the observer. The threes are all in its own inertial reference frames. Agreed?

Im sorry but I don't understand what you mean by "The threes". Please clarify. Do you mean you and the other two particles are each in an inertial frame of referance?

Now, two objects collide and lump together as one. What is the mass when they lump as one?

Let the proper mass of each particle before collision be m₀. Let the proper mass of the final lump be M₀. The mass of each particle before collision is

m = \gamma m_{0}

The total mass, m_total, before the collision is the sum of the two masses

m_{total} = 2\gamma m_{0}

Since this is a closed system mass is conserved and therefore

M_{0} = m_{total} = 2\gamma m_{0}

Can we apply this to the phenomenon of an electron and a positron lumped together as a gamma photon for a short period?

No. If you did then the momentum of the system would not be conserved. Look at this from the viewpoint of the zero momentum frame of reference. When there were the electron and positron then there was no momentum. When these particles appear and only one photon appears there will now be momentum since one single photon always has momentum. But you can have them anihilate and yield two photons moving in opposite directions as obeserved in the zero momentum frame of reference.

 

[Sammywu]

Arcon, Yes. When I say the three, I mean looking from the three different objectss points of view. Can we show the 4 vector momentum conservation and energy conservation from the three different observers.

About the other statement, I oversimplified that, there needs to be a subvelocity 2/3c in both particles toward the screen. so, their true velocity will be c*sqrt(8/9) with an angle toward the middle of teh seceren.

Sorry about that.

 

[Sammywu]

Arcon, Any way this went far away from our original topic. Let's try to resolve my question about a free falling object in a gravity field in your equations.

Your gamma is dt/dt'. Since dt/dt' is a not constant here, this gamma is not constant either. OK, I stuck here and don't know how to continue. Want to help me. Thanks.

My object is try to derive the clock difference when the object returns to the initial falling point in your formulae.

 

[Sammywu]

Arcon, This is the test case for you in case you do not know what test I was up to.

Making a device like a donut with a connecting tunnel through one of the diameters. With an open end at the joint of the tunnel and the donut, you can shoot an object in with certain speed. Put a comparable significant mass in the middle; now this is a artificial Sun experiment that you can let objects orbiting through the donut and objects falling through the tunnel. You can perform a true comparison of clocks between free falling objects and orbiting objects.

YOu can assume the middle mass is a evenly densed sphere of M0 with a radius of RM. Apparently your formula will only deal with detail gravity density.

 

[Arcon]

Sammywu - Please read your PM

Arcon

 

[Janus]

Originally posted by Sammywu
You can perform a true comparison of clocks between free falling objects and orbiting objects.

Orbiting objects are free falling objects.

 

[Arcon]

Originally posted by Janus
Orbiting objects are free falling objects.

Good point. Even objects moving away from the Earth, even at escape velocity, are in free-fall.

 

[Sammywu]

Arcon, Janus, Well. We can agree on that. But here I am just trying to understand how Arcon's formulae shall be applied to resolve a problem. My free falling object refers to the one that was held by some thing at one end of the tunnel before the orbiting object was shot into the donut. If you like you can assume there is a clock at this initial point and compare the moving object( the oscillating object from a Newtonian point-of-view. ) to this static clock. You don't even need to bother a comparison between the obiting object, hereafter as A, and the object in question here, hereafter as B.

I have pointed out that since the relative velocity will be changing, the time dilation factor is not a constant rather a function of Ts ( for Time of static ) or Tb.

My poitn is let's test Arcon's formulae to a case and just want to see how it can be used to solve a practical problem. If a Physic theory is just a whole bunch of formulae but unable to predict or soleve some problems, what is the use of these formulae?

I tried to skip a few evaluation in Arcon's formulae and come to a point that Fex=0, so f(total)= 0+G. Now Arcon' (21) is a denotation I am not familiar with. How would you put the G here as a function of r or Tb?

 

[Sammywu]

DW, By the way, who is pmb, when you said you do not agree with his Newtonian's approach?

Thanks

 

[Sammywu]

Arcon,

Back to your Gravtitational Force, I Think I could decipher some of your notations now. Your v with either alpha on the upper or lower right is the velocity in vector form but in a vertical or horizon writing.

What is the reverse L? Is it a matrix, or it denotes a matrix calculation?

Is it correct?

 

[Arcon]

Originally posted by Sammywu
Arcon,

Back to your Gravtitational Force, I Think I could decipher some of your notations now. Your v with either alpha on the upper or lower right is the velocity in vector form but in a vertical or horizon writing.

What is the reverse L? Is it a matrix, or it denotes a matrix calculation?

Is it correct?

I'm not sure which symbols you're referring to. Please post the equation number. Thanks

Arcon

 

[Sammywu]

Arcon,

In equation (1), you describe the 4-force to be big D differential of 4-momentum to the proper time. Then it equals to the sum of two components. The first part is a formula of sum of partial differential multiplied by the dx(b)/d(tau). The second part is a product of a so-called connexion and the two different forms of 4-momentum and 4-vector. The product of 4-momentum and the 4-vector shall be a 4x4 matrix. The conection is also a 4x4 matrix. their matrix product shall be a 4x4 matrix. But this special symbol must mean you sum a row of the 4x4 matrix to make a subcomponet of the vector.

In equation (4), you said the total of the two parts is the external force.

maybe I shall buy a modern SR textbook to decipher your notation here.

 

[Arcon]

Originally posted by Sammywu
In equation (1), you describe the 4-force to be big D differential of 4-momentum to the proper time.

Yes. That is the absolute derivative (aka derivative along the curve) of the covariant 4-momentum.

Then it equals to the sum of two components. The first part is a formula of sum of partial differential multiplied by the dx(b)/d(tau).

Yes. Notice that the chain rule is used, i.e.

\frac {\partial P_{\mu}}{\partial x^{\beta}} \frac {dx^{\beta}}{d\tau} = \frac {dP_{\mu}}{d\tau}}

The second part is a product of a so-called connexion and the two different forms of 4-momentum and 4-vector. The product of 4-momentum and the 4-vector shall be a 4x4 matrix. The conection is also a 4x4 matrix. their matrix product shall be a 4x4 matrix. But this special symbol must mean you sum a row of the 4x4 matrix to make a subcomponet of the vector.

If you wish to view that last part in matrix form then note that the affine connection has 3 indices and as such is not a 4x4 matrix. A 4x4 matrix has 16 components. The affine connection has 4x4x4 = 64 components. If you want to form a matrix equation then view the connetion as a 4x4 matrix which is a function of "u" (mu). Then note that you have to rearrange it to read

P_{\alpha} \Gamma^{\alpha}_{\mu \beta} U^{\beta}

This then reads in matrix form

P\Gamma(\mu)U

In equation (4), you said the total of the two parts is the external force.

Eq.(4) has on the left side dP/dt while on the right side it has F_external + Gravitational Force

maybe I shall buy a modern SR textbook to decipher your notation here.

An SR text won't have this material. You need a text either on GR or on differential geometry. There is a nice text by D'Inverno that I like. It's called Introducing Einstein's Relativity, Ray D'Inverno, Oxford Univ. Press, (1992)

I'm working on writing tutorials at the moment. Here is one on an intro to tensors in general

http://www.geocities.com/physics_world/ma/intro_tensor.htm

This one gives you an idea of the geometrical meaning of the Christoffel symbols (affine connection)

Arcon

 

[Sammywu]

Arcon, Thanks. You did not comment on my part of connexion. was the guess correct?

Before I can get to a book store and buy a modern SR book, I can only limp along your formulae.

By the way, in my experiment of two objects hiting each other in 2/3c relative to me and lumping together, where is the inertial energy?

 

[Sammywu]

Arcon, Thanks. I just saw your reply of connexion. I was trying to guess thatis a 4X4X4 matrix or different things. You definitely pointed me to correct path.

 

[Arcon]

Originally posted by Sammywu
Arcon, Thanks. You did not comment on my part of connexion. was the guess correct?

Which part was that? All I saw was that you said it was in the second part and was a 4x4 matrix. The gamma part is the connection but it is not a 4x4 matrix.

By the way, in my experiment of two objects hiting each other in 2/3c relative to me and lumping together, where is the inertial energy?

Did you read the PM message I sent you?

 

[Sammywu]

Arcon, I saw your complete response after I wrote that response. So don't bother with it.

What do you mean by the PM mark?

 

[Sammywu]

Arcon, I am sorry. What do you mean by the PM message?

Did you send me an Email?

 

[Sammywu]

Arcon, I am reading your "Introduction to Tensors". I believe that will be very helpful.

Thanks

 

[Arcon]

Originally posted by Sammywu
Arcon, I am sorry. What do you mean by the PM message?

Did you send me an Email?

Take a look below this post where you normally click on "quote" to respond. To the left there are other buttons to click. One says "pm" - when you click that you can send the person a "Private Message (PM)"

To read PM's take a look at the top of this web page. There is a button that says "user cp" which stands for "User Control Panel"

I sent you a few PM's. I guess that you don't know about this function. Click on PM and you can see what I sent you.

Arcon

 

[Sammywu]

Arcon, I finished your tensor introduction. A contravariant is a matrix whose values will change with the coordinates we use. A covariant is a matrix whose value will not change with the coordinates.

Your equation (31) might have messed up between j and k. Either k shall be j or k shall be j.

In equation (16), shall it be rank 1 or rank zero?

Hopefully I am right.

The 4-momentum and 4-force shall all be covariant.

How about the connexion? Is it also a covariant?

 

[Sammywu]

Arcon, I am sorry. Take back part of the first statement. An invariant is a matrix or vector that won't change with any coordinates we use. A product of covariant and its own unit cubicle will not change with any coordinates we use.

Thanks

 

[Sammywu]

Acon, In equation (31), is it a covariant tensor? Or, a tensor is a covariant? Your numerators and denumerators in the fraction of the partial differentials probably are in reversed places.

Thanks

 

[Arcon]

Originally posted by Sammywu
A contravariant is a matrix whose values will change with the coordinates we use. A covariant is a matrix whose value will not change with the coordinates.

To be precise - a covariant vector is a geometrical object whose components transform from one coordinate system to another as shown in the notes. The components may be represented using matrix notation.

Your equation (31) might have messed up between j and k. Either k shall be j or k shall be j.

Thanks. It seems I did make an error. I'll look into it and correct it if neccesary. All those little damn numbers can look confusing huh?

In equation (16), shall it be rank 1 or rank zero?

A tensor of rank 1. There is one index so the rank is one.

The 4-momentum and 4-force shall all be covariant.

There are both covariant and contravariant forms of all vectors. A covariant vector is said to the the dual of a contravariant vector since there is a one to one relationship between the two.

How about the connexion? Is it also a covariant?

The term "covariant" as used there refers to a type of tensor. Since the affine connection is not a tensor the term does not apply.

 

[Arcon]

Originally posted by Sammywu
Acon, In equation (31), is it a covariant tensor? Or, a tensor is a covariant? Your numerators and denumerators in the fraction of the partial differentials probably are in reversed places.

Thanks

That is called a mixed tensor since it is covariant with respect to some indices and contravariant with respect to other indices.

A tensor is said to be covariant if and only if all indices are covariant.

A tensor is said to be contravariant if and only if all indices are contravariant.

 

[Sammywu]

Arcon, Reading your article "Invariant", most important concept is the metric tensor. It's not a constant matrix. Its value will change with the coordinates we choose. Take (ct, x, y, z) as the coordinate of an inertial frame ( or flat spacetime ), it shall be the matrix with the diagonal value as (-1,1,1,1) and the rest subcomponent is 0.
Let me denote it as Gs.

In a gravity field, with (cTf, Rf, yf, zf ) as coordinates ( make Rf for the x ) for a free fall observer, it shall be Gs as well, since the free fall observer will see straight light beam ( Something seem to need to be adjusted here, it must be related to its initial field potential if this person was held steady in the gravity field before released. ) When we replaced Rf, yf, zf with R, y,z , the coordinate observed by the static infinite observer, what would this metric matrix look like? hmm, I need to think.

 

[Sammywu]

Actually, I found I have to redo some thinkings done for the two imaginary experiment I brought into examine the SR and GR effect.

The orbiting object, the free fall object released from a point inside the gravity field and the standing person in the gravity field might all see a different degree of light bent.

I have to assume a far remote observer who is in a tre flat spacetime and a free fall object coming from this far remote place: in other words, EP=0. These two seem to have the same clock and they will see all light beams straight.

 

[Sammywu]

In order to make the pont clearer, I would like to build a closed box, julst like the elevator in Einstein's elvator, but much bigger. I will put three holes, to let light thru, on one sidewall in different altitude, or height. On the opposite wall, I will mark the corresponding spots at the opposite wall. I will lock myself at the same latitude as the middle hole. In a true inertial frame, no gavity, all three light beams will hit the corresponding spots. When I put the closed box in a gravity field, either free fall or standing, all of three light beams will hit different spots to reflect different acceleration at different altitude.

To me, the observer, all the light beams will probably look straight, because light beams are the only perception for guiding a straight line. Only by the spots shift we can tell the lights actually bend because this box was built in a environemnt without gravity.

 

[Sammywu]

Actually, I found it interesting to draw how the light beams will travel in this box in different environments: far away, free fall from far away, free fall released from one point in the gravity field, standing rest in the gravity field.

My gut feeling is that the standing rest observer in the gravity field has the same clock as the clock released from this point. Though I am wondering how I can prove that.

I also got this formula from the site that Marcus showed me at another thread.

This formula describes the clock standing at distance r from the Earth center and a geocentric latitude A.

(delta)v/v= [ V(r,A) - I^2*r^2*cosA^2/2 - ( V(a1,0) - I^2a1^2/2)]/c^2

delta)v/v is the fractional frequency, a way to show clock difference in fraction, I believe. No time to check the detail.

V(r,a) is the gravitational potntial.

I: Earth's angular rotation rate.

a1: Earth's quatorial radius.

 

[Sammywu]

Just make it clearer.

(delta)v/v= [ V(r,A) - I^2*r^2*cosA^2/2 - ( V(a1,0) - I^2*a1^2/2)]/c^2

Something was wrong here, or the document did not make it clear. This delta fraction is definitely compared with a clock at the sea level high in equator. If you replace r as a1 and A as 0, you will see this turned to 0. Even though the document claimed it's compared to Earth centered clock. Maybe I did not carefully read the document.

 

[Sammywu]

Arcon, Janus,

I am confused. This document apparently was written by some authoritative sources.

But its model contradicts what I thougt here.

Its formula actually says that the higher you are, ao as the higher your field potential ( Since -GM/R is negative, the higher you are, the smaller absolute field potentail you have, the higher field potential after you apply the negative sign.), the faster your clock will run.

It also has a sample to make that clear.

By my model, it will be reversed. The higher we are, we will be more close to the infitely far away clock, which is the slowest clock.

This is really true clock difference, not a measurement issue. See the two experiments I proposed.

Please help. What's wrong with my model?

 

[Sammywu]

Janus, Arcon,

I checked a little further. There is a 1976 rocket experiment that proved the GR time dilation effect. This further bothered me. Doesn't this contradict to the twin paradox? We put energy to push this rocket up and let it fall back going thru certain brake accelaration. Its clock turned out faster. Isn't this a twin paradox experiment?

So, is the twin paradox just some balloney? It seemed to show that the astronauts sent out will be older rather younger.

How do we reconcile these two theory?

 

[russ_watters]

Originally posted by Sammywu
I checked a little further. There is a 1976 rocket experiment that proved the GR time dilation effect. This further bothered me. Doesn't this contradict to the twin paradox? We put energy to push this rocket up and let it fall back going thru certain brake accelaration. Its clock turned out faster. Isn't this a twin paradox experiment?

So, is the twin paradox just some balloney? It seemed to show that the astronauts sent out will be older rather younger.

How do we reconcile these two theory?

There are quite a number of experiments that confirm the SR and GR time dilation effects. They involve clocks on the ground compared to clocks on towers, clocks in space, clocks in planes, etc. The GPS system is my favorite example: the satellites are launched with their clocks calibrated to run slower than identical clocks on earth. When they reach orbit, they stay synchronized with their twins on earth.

The twins paradox is a mental exercise. It only seems like a paradox. If it were a real paradox, SR and GR would be invalid.

 

[Arcon]

Originally posted by Sammywu

Gravity is caused by the timespace curvature. How will the inertial force been considered? Does it appear as part of the stress energy tensor?

Please note that spacetime curvature is not a cause of gravity. The term spacetime curvature is merely a geometric analogy of purposes of description and should not be taken as a literal truth and should, by no means, not be thought of as a cause. It is an unfortunate truth that this geometric interpretation has become predominant in general relativity because it makes people stop thinking. Especially since spacetime curvature is not always present when there is a gravitational field present.

Regarding this analogy Steven Weigberg comments on this very point in his GR text, which is a main staple of any GRist. From page 147

... the geometric interpretation of the theory of gravitation has dwindled to a mere analogy, which lingers in our language in terms like "metric, "affine connection," and "curvature," but is not otherwise very useful.
[...]
(The reader should be warned that these views are heterodox and would meet with objections from many general relativists).

Good ole Weinberg! Good stuff as is expected from a Nobel Laureate

 

[Sammywu]

Russ, Arcon, Janus,

So do we all agreed that an astronaut going out of the Earth and made a trip back to the Earth will be older rather than younger than his twin brother? He most likely will be making trips inside gravity fields of some major masses: the Earth, the Sun, and the Galaxy.

 

[Sammywu]

Russ, Arcon, Janus,

Were any known experiments doone to check a clock that was brougt to the international space station and back to the Earth by astronauts?
That shall confirm this faster clock on higher ground also. Correct?

 

[Arcon]

Originally posted by Sammywu
Russ, Arcon, Janus,

Were any known experiments doone to check a clock that was brougt to the international space station and back to the Earth by astronauts?
That shall confirm this faster clock on higher ground also. Correct?

Sammywu - I've asked you a direct question and you've refused to answer it. Until you have the courtesy to respond then I will no longer respond to any further questions. I asked if you got my PM message I was met with stone silence. If you got it and don't acknowledge it and the content I'm going to assume that this is something you will always do and therefore not respond to anything else from you

 

[Sammywu]

Arcon,

I am sorry. I have read your PM messages, after you told me how to. I assumed that you should know that. I don't remember there was any other questions you asked me.

Did I skip other questions you asked me? Please let me know which one.

You can send me another PM.

If you referred to the question of relativistic mass, I would say the concept of relativistic mass is useful. I tends to agree with it more and more now. But I still have questions on how gravity and EM energy will be included.

By the way, I am still reading your gravity force document, I found it is very useful to me. But I still have problems truly able to apply it or totally comprehend it.

That's why I came back to revisit my concept of GR effect.

Any way, I tried another model to resolve this FR effect.

Assumptions:
1. an object on its geodisc will run fastest clock because it is not affected by any true external forces.
2. An object free fall from far remote runs a clock in sync with the far remote clock because it is on its own geodisc. Let's denote its time as T for this paragraph.
3. When this object passes by a standing object, the standing object has a slower clock because it undergoes the support force, an external force.

This will derive a GR effect if 1/2*v^2=-FP; FP stands for field potential and equals to -GM/R.

This model will match the formula I got from this document.

Just one problem, if applying m*c^2-m0*c^2=-EP, 1/2*v^2 will not be the same as -FP.

Let me threw this problem aside, and check the model against the three observers I have mentioned:

1. For a standing clock supported by a ground support force, its time will be T*SQRT(1-2GM/(R*c^2)).
2. For a free fall clock with initial speed 0, it will depend on where this clock was released. It's on its own geodisc, but related to where it was originally released. Its clock is the same as the standing clock where it was released.
3. For an orbiting clock, its clock seems to be T*(1-2GM/(R*c^2)+v^2/c^2).
4. In general any objects in the gravity field with certain speed v, will follow the same formula used for item 3. It does nto matter where the speed direction is. This at first bothered me, but I realized in the free fall object from far remote will pass thru the midtunnel, assuming we build a midtunnel that it can fly thru, and fly away from the mass center in a decreasing speed v toward far remote. So, the direction of its speed apparently does not matter.

This model match most of experiments that Russ mentioned and the formula published in this authoritative document.