SR and the earth, sun, and galaxy.

[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.

 

[Sammywu]

My first try to resolve a SR/GR clock question with this modern SR theory.

The easiest question is the spaceship flying in a circle about a statis observer in a flat spacetime.

The most important issue here seems to choose a stable reference frame. It's very difficult to use the spaceship's reference frame because it actually change with time.

Using the flat spacetime as the reference frame, (t,x,y,z) for the world line of the spaceship, (t, R*cos(2*3.14*t/T), Rsin(2*3.14*t/T), 0) denotes the path using t as the parameter.

Replacing t by t', the proper time of the spaceship, d(t,x,y,z)/dt'= dt/dt'*(1, -2*3.14*R/T*sin(2*3.14*t/T), 2*3.14*R/T*cos(2*3.14*t/T), 0). dt/dt' can be assumed to be a constant as 1/sqrt(1-v^2/c^2) since the spaceship is kept in constant velocity.

When integrating this from t =0 thru T or t'= 0 thru T*sqrt(1-v^2/c^2), we will get T'=T*sqrt(1-v^2/c^2).

The whole process actually is redundent. When dt/dt' = sqrt*(1-v^2/c^2) was determined, the outcome is basically completely based on this.

From the spaceship's view, by SR, dt'/dt shall be also sqrt*(1-v^2-c^2).

The only reason this at last shows that spaceship has a slower clock is because the flatspace has a referencible coordinates.

 

[Sammywu]

By people's help from here, especially Arcon ( pmb_phy ), I read a few documents on tensor algebra and modern SR/GR, I got enought to come back to solve this porblem, even though I still do not understnd all the informations contained in thes documents.

1). We need a stable coordinates to start with. The assumed coordinate is (T(lab), X, Y Z ). T(lab) is also T(world) or T(coordinate). You can see Okun's paper about Gravity Redshift for a better description of it.

2). In this coordinate, g(00) , the 00 component of metric tensor, is -(1+2Phi). I perfer to write it as -(1-2|Phi|), since Phi is negative. In this formula, we can easier see 1-2|Phi| is smaller than 1.

3). For a standing clock in the gravitational potential Phi, Its time can be denoted as T(stand). We can establish another coordinate as (T(stand), X, Y Z ).

4). T(stand) is the same as a clock at thecomoving ( in this case, velocity as zero ) local inertial frame, denoted as T(loc1), to the order of the first differentials.
You can see Okun's paper for better description of this frame. I personally think T(stand) and T(loc1) does have difference when we integrate them over the world line, but to the order of second differentials, ignored by most treatment and for now.

T(loc1) is really the clock held at the same area and just released to free fall. Just at this moment dT(loc1) is the same as dT(stand).

When the object falled to a diffrent area and with aspeed not zero any more, its ticking rate might be different.

5). Based on ds^2=g(ij)dx(i)dx(j), dT(loc1) is the ds and dT(lab) is the dx(0). dx(1), dx(2) and dx(3) are all zero because the clock never move. -(dT(Loc1)^2)=-(1-2|Phi|)*dT(lab)^2. So dT(Loc1)=sqrt(1-2|Phi|)dT(lab).

6). dT(stand) is very close to dT(loc1). So dT(stand)=sqrt(1-2|Phi|)dT(lab). The deeper you are in the gravitational field, the higher is 2|Phi|. dT(stand) is smaller and it means T(stand) is slower.

Once we established T(stand)'s relationship with T(lab), we can now use a different coordinates (T(lab), X, Y, Z). This cordinates can be unchangely describing the orbit of an orbiter. Now let's see how to calculate the T(orbit), the time of an orbiter around the center mass.

7). First, we can establish another T(loc2) for T(orbit), T(loc2) is the free fall object going on the same speed as the orbiter. So, the orbiter is in the local inertal frame. T(loc2)=T(orbit), no doubt.

8). From T(loc1)'s view, T(loc2)=T(loc1)*gamma. Here beta is v/c and gamma is sqrt(1-beta^2).

Why don't we look from T(loc2)'s view? It's the most important question. The coordinates we can establish with clear relationship between time and space is the coordinate (T(stand), X, Y Z ). The rotating of the orbiter makes it diffecult to relate X,Y,Z to its T(orbit).

The selection of the coordinate again determines how the resulit will be. Anyway, I will think more on this.

9) T(orbit)=T(loc2)=gamma*T(loc1)=gamma*T(stand)=gamma*sqrt(1-2|Phi|)*T(lab). So, the orbiter is slower than a standing clock.

Note all I did above, I simplified that taking T as c*T. The coordinates established assuming c=1.


My question is:

How much confidence will we have about the existence of T(lab) and the relationship established between T(lab), X, Y and Z? T(lab) is the same as the far remote obserevr free of gravity I have imagined in the first place.

 

[Sammywu]

Using EQ (3.48) in the "Lecture Notes on General Relativity" written by Sean M. Carroll, with the same coordinate labeled (T(lab), X, Y, Z ) as before, g(00)=-(1+2Phi), g(11), g(22) and g(33) are close to 1. T(orbit)=integration from 0 to 2Pie of SQRT((1-2|Phi|)*(T/2Pie)^2-R^2), where T is the lab. time for the orbiter to make a circle. R is the raius of the orbit. The item R^2 comes from dX/dA^2+dY/dA^2, where dA is the angular velocity of the orbiter. X=R*cosA and Y=R*sinA, so dX/dA=R*(-sinA) and dY/dA=R*cosA.

Any way, 2Pie*R=T*beta, where beta=v/c as SR people know. R=T*beta/2Pie.

So, T(orbit)= integration from 0 to 2Pie of SQRT((1-2|Phi|)*(T/2Pie)^2-(T/2Pie)^2*beta^2).

SQRT((1-2|Phi|)*(T/2Pie)^2-(T/2Pie)^2*beta^2)= (T/2Pie)*SQRT(1-2|Phi|-bate^2).
T(orbit)=T*SQRT(1-2|Phi|-beta^2). Note T=T(lab), so
T(orbit)=T(lab)*SQRT(1-2|Phi|-beta^2).
We already know T(stand)=T(lab)*SQRT(1-2|Phi|).

This shows a match with what we know as GPS time formula.
Also this shall match the H&K experiment.
Note the formula can be applied to an airplane flying around the Earth with smaller amount of v than the so-called escape velocity that drives an orbiter, where the airplane will need to rely on air pressure beneath the wing to support it from not falling down.