SR and the earth, sun, and galaxy.

[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.

 

[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.