SR and the earth, sun, and galaxy.

[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.

 

[Arcon]

Originally posted by 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.

Okay. I just wanted to make sure. I explained that I was very busy lately and that I don't find that I have the time to respond/read all of what you're posting. I didn't want you to get the impression that I was ignoring you or being rude. Some people never read e-mail/PM and thus I had no way of knowing if you read it or were one of those people. Thanks for clarifying. Much appreciated. I can see that you are not one of those people.

Arcon

 

[Sammywu]

Arcon,

Thanks. I did read that. I really appreciate your help.

 

[Sammywu]

Just try a sanity check on this GR effect as the Earth to the Sun.

-FP=(6.67*10^-11)*(2.0*10^30)/(150*10^9)= 1*10^10 m^2/sec^2.
-2FP/c^2=2*10^10/(9*10^16)=0.22*10^-6
sqrt(1-2FP/c^2) will be close to 1.

So, even if a spaceship escapes out of the gravity field of the Sun, the clock speed won't go too fast.

Try escaping from the Earth's surface relative to outside of Earth's gravity.

-FP=(6.67*10^-11)*(6.0*10^24)/(6.4*10^6)=6.8*10^7
-2FP/c^2=15.6*10^7/(9*10^16)=1.7*10^(-9)
sqrt(1-2FP/c^2) is even smaller than the one in the Sun's gravity field.

Can't find Milky way's data.

 

[Sammywu]

Just correct my item 3 formula.
T*SQRT(1-2GM/(R*c^2)-v^2/c^2).
There is an incorrect sign. Another possible formula is more likely
T*SQRT(1-2GM/(R*c^2))/SQRT(1-v^2/c^2).

So, for a free fall from far remote with initial speed=0, the denominator and numerator will be the same.

 

[Sammywu]

Try Sun surface's clock:
-FP=(6.67*10^-11)*(2.0*10^30)/(0.7*10^9)= 19*10^10 m^2/sec^2.
-2FP/c^2=38*10^10/(9*10^16)=4*10^-6
sqrt(1-2FP/c^2) will be close to 1.

This is trying to verify David's claim that muon's apparent life expectancy inflation is due to rather GR effect than SR effect. This small difference between the Sun's surface's clock and the Earth's clock will not produce the significant apparent time inflation in the muon phenomenon.

 

[Arcon]

Sammywu - I've just completed one item that I was working on. See

http://www.geocities.com/physics_world/gr/geodesic_deviation.htm

See Eq. (22). It represents the tidal acceleration of two particles in free-fall. That is to say that if there are two nearby particles in free-fall then that eqations tells you what the relative accelerations are. That relative accelertion cannot be transformed away, i.e. it exists in all frames of referance. Notice that its velocity dependant!

Arcon

 

[Sammywu]

Arcon,

Thanks. I will definitely take a look.

I just finished first part of your gravitational force. Trying too make clear.

t: proper time of a far remote observer.

tau: proper time of any objects , hereafter denoted as O, in the gravity.

bolded v: velocity of the object O from the far remote observer's view.

bolded r: Cartesian coondinate from far remote observer's view.

m: the apparent ( or relativistic ) mass of object O.

Spacetime event define all events in far remote observer's coordiante.

4-Velocity: O's 4-velocity as defined as the change from far remote observer's view but derivative of O's proper time. As we mentioned, it is a covariant.

The v(a) here is a 4-vector from of bold v adding time parameter.
Upper a or lower a seems to just denote a different form of the vector, either in horizotal or vertical.

4-Momentum and 4-force: the covariant form of the momentum and force . Just like 4-velocity.

Absolute Derivative is the subcomponent of 4-force. 4-force will be the total of the external force plus the adjustment from the connexion.

+++++++++++++++++++++++++++++++++++++++++++++++++++++

Another book I happened to read said the connexion is the partitial differentials of g(ij), the geometric attribute or the metric unit.

It also denote the form in a way that upper subscripts are the inverse of the lower scripts.

Just FYI.

++++++++++++++++++++++++++++++++++++++++++++++++++++++

 

[Sammywu]

Arcon,

Finshed your (1) thru (7) in the "Gravitational Force'.

Correct one item in the last paragraph:

Absolute Derivative is the subcomponent of 4-force. 4-force will be the total of the external force minus the adjustment from the connexion.

From (1) Thru (7) you show the total force from the far remote observer's view as 3-vector is the space componet of the 4-force/dilation factor plus G.

G is the space componet of the product of relativistic mass , the velocity of O measured by far remote oberver ( hereafter denoted OF ), connexion and O'svelocity again.

I need to think about your statements in explanation of (4) and (5).

 

[Sammywu]

Arcon,

By the way, you do not always need to respond to me unless you see it necessary to correct me.

Let me do a sanity check for myself. Let me assume a free fall object in the gravitational feld; its velocity is not constant from OF's view and so 4-U and 4-P are not constant either. dP(mu)/d(tau) will not be zero. F(mu) shall be zero because the affine connection adjustment l cancel out the dP(mu)/d(tau) in that the object is on its own geodisc.

Also, in an inertial frame, shall the affine connection be the zero 4x4x4 matrix?

Thanks

 

[Sammywu]

Arcon,

Following thru (14b), Your statements using Equivalence Principle to explain the gravity is "the force resulting entirely fromobserving the particles... frame" is crucial.

Up to (14b), you show the comparison between Loretz force against how the same concept to be applied to gravity. Even though I still had no idea how you do (11), the "more familar" form to me is unfamiliar, unfortunatelly. I will read the your appendix to see whether I got it there.

Any way, here you show gravitational charge as m, the relativistic mass, not the proper mass.

 

[Sammywu]

Arcon,

The mainpart that I have no idea is where the Christoffel'symbol is from. I can only assume that's true. I can follow you thru (16) now.

Also, I realized that your purpose is to prove the gravity mass is the relativistic mass.

I tried to use what you show me here to derive GR effect. It seems that you already assume GR and SR effct.

I also noticed your two threads about mass tensors.

Any way, remember I said in my model of GR effect, there is a problem. If mc^2-m0^c2=-EP=GMm0/R, then 1/2v^2 will not equal to -phi( I denote -FP). Now, if GMm0/R^2 needs to be adjusted to GMm/R^2, or with your mass tensor, will that correct this problem?

 

[Arcon]

Originally posted by Sammywu
Arcon,

The mainpart that I have no idea is where the Christoffel'symbol is from. I can only assume that's true. I can follow you thru (16) now.

There is a discussion of the meaning and derivation here -
http://www.geocities.com/physics_world/ma/chris_sym.htm

Also, I realized that your purpose is to prove the gravity mass is the relativistic mass.

No. That is not my purpose. It's just something that has always been true in general relativity. But there has been so much misinformation being passed around the internet on this topic that I've made things clear on my website for those who really want to learn it correctly. Mind you - this was done close to 100 years ago by Einstein. It's nothing new by any means. That equation is in Einstein's text The Meaning or Relativity. Problem is that people don't know that it its there. This misinformation was started in an article by Lev Okun on the concept of mass when he failed to comment on this. But in all fairness to Okun he didn't know Einstein did that in Einstein's book. The problem was that Okun made it appear as if Einstein never used relativistic mass and that's a false claim.

Any way, remember I said in my model of GR effect, there is a problem. If mc^2-m0^c2=-EP=GMm0/R, then 1/2v^2 will not equal to -phi( I denote -FP). Now, if GMm0/R^2 needs to be adjusted to GMm/R^2, or with your mass tensor, will that correct this problem?

You wrote so much that I was unable to read it all. But I don't know what EP is. The quantity "mc^2-m0^c2" is Kinetic energy so I take it that your EP is potential energy. What has happened is that you're using invalid approximations. I'll be getting to the correct one soon.

 

[Sammywu]

Arcon, Thanks. I am trying to gather some more info. from Sean M. Carroll's " Lecture Notes on General Relativity". Hopefully, that will resolve some questions about a few paradoxes I found in the GR effects.

 

[Sammywu]

I just read about this "Gravitational redshift". I never paid too much attention about redshift because I though that is always purely measurement issues from whose views.

But this seems to be related to the GR time dilation effect.

I discussed it with my friend. So he asked this question. If a light beam was shot from the tower down to the groud, the ground observer shall se a blueshift instead of redshift. Is this right?

 

[Sammywu]

I followed through that Gravitational Redshift actually leads to the algorithm of GR effect. Do a sanity check here.

If I continuously send a light beam to the tower for one second: wave length * number of wave packet = 1 second. The tower observer will see longer wave length * number of wave packet which means more than one second. So, the tower's clock needs to be faster than mine.

 

[Sammywu]

The analogy between two elevators in the same frame and the tower and the ground in the Earth could be extended to add another elevator by the side of the higher elevator but with lower acceleration; this shall give a comparison between the two objects in the gravitational field but different acceleration.

 

[Sammywu]

This analogy can explain the difference between two standing clocks at different alttudes. Note the elevator's acceleration is upward just in reverse to gravity. It's like the supporting forces for standing clocks.

This shall not straightly applied to any free fall or orbiting objects.

The only thing certain here is the initial clock of a free fall object shall be the same as the standing clock at where it was released.

Note if I released a clock to free fall, it will oscillate between two points in the imaginary experiment I proposed as digging a tunnel through a significant mass. We shall be able to compare its clock against many standing clock along the tunnel. Since it is on its own geodesic, its clock shall be faster than the standing clock which is always at its original release point. We can actually compare their clocks without ambiguity when the free fall clock oscillate back to the point. No need for redshift or anything else. True clock comparison.

 

[Sammywu]

My impressions from the first 70 pages of Sean's lecture notes.

1. Event space is a manifold.
2. At each point of the event space, there is a metric tensor, which shall be viewed an unique one at this point from all objects' worldlines that pass thru this event point. Even though all objects have their own time-space coordinate, the metric tensor is the same one, so the metric tensor seen from different coordinates follows the rule of tensor transformation equation by different coordiantes.
3. All objects moving around this event space will leave a world-line like a time-like curve in the manifold.
4. The connexion is 4 4x4 matrixes, which is related to metric unit in the EQ (3.21), that is needed to make partial derivatives to become tensors.
5. In the page 26 thru 30, the 4-velocity is not the covariant 4-velocity.

 

[Sammywu]

I still see a flaw in this clock formula used to adjust GPS clock. This formula basically adjust the clock based on GR effect and then on SR effect.

Since there are experiments proving the GR effect is real, I think this effect is true without doubt. But applying the SR effect right after the GR effect actually conflicts with the underlying assumption that Gravity needs to be traeted differently from a regular external force.

SR effect is a relative effect, not absolute effect. GR effect is an absolute effect because the two clocks are relatively fixed.

When I compare a clock Ta standing at the GPS's orbit, ( we can build a very high tower ), while the GPS is orbiting around the earth, its clock Tg shows a SR difference from Ta. After the GPS makes a circle, the clock can be compared. The point here is SR effect is a relative effect, Tg is slower than Ta from Ta's view, Ta is slower Tg from Tg's view. Which one will be truly slower after a circle?

Gravity does not do any work on the GPS. ( Is this correct? ) The supporting force to the standing clock does not do any work either.

The GPS is on its own geodesic. Ta is the one pushed off its geodesic.

From an analogy of this to a sample experiment created as artificial gravity, the one on the artificial gravity shall have a slower clock. This will implys the standing clock has a slower clock.

This is different from the formula applied to GPS clock setting.

Apparently I must be wrong.

 

[Sammywu]

Before I can further provide a calculation from this tensors and SR/GR, I thought of some analogies that show these interesting facts and underlying implications:

Let us send two rockets to height H from the earcth surface. One we will gradually curve it so as it will evently orbit the Erach at H. Once it gain its height and its escape velocity, it will orbit the Earth without further energy expenditure. Another one will be sent staright up and kept afloat by continuously pump out energy to gain the exact acceleration to counter the gravity GMm/R^2.

We will see very clear that it's actually more difficult to keep a rocket standing than orbiting at the same height.

Another one is the one we send straight out to the same hwight and let it free fall back to the Earth. We can see this one expends the least energy.

A side finding: This seems to imply that by holding us at the Earth's surface, The Earth has to expend some energy continuously.