SR and the earth, sun, and galaxy.

[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 the 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.