SR and the earth, sun, and galaxy.

[Sammywu]

Arcon,

Thank you for the lesson. It's of great help.

As you know, My Physics is just high school grade. It will take me a while to understand all those formulas.

One thing I do'nt know is what is tidal force. I have read it many places. Would you mind explain a little more about it?

Also, if applying this 4-dimention vector formula to my imaginary experiments, does its outcome fit my calculation?

 

[DW]

Originally posted by Arcon
The gravitational force on a particle in free fall is given by

$$G_{\mu} = \frac {1}{2} m g_{\alpha \beta, \mu} v^{\alpha} v^{\beta}$$

Which you should mention is a fictitious force because it is not a four vector and can be transformed away.

The Newtonian gravitational potential is related to g₀₀ as

$$\Phi = \frac{c^2}{2} (g_{00}-1)_$$

Only in a linearized weak field approximation.

The mass of the particle isw hose proper mass is m₀ is given by

$$m = \gamma m_{0} = m_{0} \frac {dt}{d\tau}$$

This is bad terminology for special relativity and is just plain wrong for arbitrary spacetimes in general relativity. In general the conserved energy parameter is not always even proportional to $$\frac{dt}{d\tau}$$

The mass is not only a function of speed but a function of the gravitational potential as well. For detailed defintions and proof of these relations see
http://www.geocities.com/physics_world/gr/grav_force.htm...

This is your own site and is wrong. The word mass of a particle free of a vector potential referrs to the positive root for m in the equation
$$(mc)^2 = g_{\mu}_{\nu}P^{\mu}P^{\nu}$$
which is manifestly invariant.

However the tidal force may not be zero as is the case of tidal forces, which cannot be transformed away.

Yes they can. You just can't tranform away both the affine connections and the tides "at the same time" when the gravitational spacetime curvature field described by the Riemann tensor is not zero.

 

[Arcon]

Originally posted by Sammywu

One thing I do'nt know is what is tidal force. I have read it many places. Would you mind explain a little more about it?

The definition of Tidal force is that which causes gravitational tides i.e. tidal forces are what cause ocean tides.

In GR it is defined in hte more precise way as is the difference in the gravitational force as measured in a freelly falling frame of referance.

Also, if applying this 4-dimention vector formula to my imaginary experiments, does its outcome fit my calculation?

Which calculation is that again?

 

[Sammywu]

Arcon,

Thank you for the lesson.

There are three simple experiments that could be used to examine your theory:
The experiment I proposed about two spaceships A, B and the station in the remmote space assuming no gravity force.

Another is an experiment that actually similates the movement of free falling bodys and orbiting objects caught by a gravity.

The last one was the two parallel moving electrons.

Once I digested your formulas, maybe I can propose some more complicate experiments for you to demonstrate how your formula can be applied to a problem.

DW, thank you for pointing out it's Arcon's own website.
I guess when you know more, there are more differences that people have to straight out. But, this shall be focused on a basis that we are all looking for the truth.

I have browsed his website. Did you? Any formulas shall be corrected to reflect a current Physics points of view.

 

[Arcon]

A note on the low speed limit of these equations.

Note - I just noticed an error in the influential article The Concept of Mass, Lev B. Okun, Physics Today, June 1989. This is the article which has been highly influential in bad mouthing the concept of relativistic mass. But the problem is that it has several errors in it. There is an expression in this paper for the gravitational force on a moving particle. The equation number is (16) and is on page 34. For the case of radial fall as I did it his expression and my expression are identical.

I have never seen Okun's work in the derivtion of that equation so I didn't know know the meaning of the terms he used. Therefore, before today at least, I was not in a position to comment on it. However. since I took the time recently figure out how he derived this I can now see his errors of which there are two. Okun claims that when v << c the equation reduces to (using my symbols for mass)

$$F_{g} = \frac {GMm_{0}}{r^{2}}$$

However that is an invalid statement. That is not the limit. The correct limit is

$$F_{g} = \frac {GMm}{r^{2}}$$

The massm, m, this second equation equals

$$m = \frac {m_{0}}{\sqrt{1 + 2\Phi/c^{2} - \beta^{2}}}$$

only attains the value m = m₀ when

$$2\Phi/c^{2} = \beta^{2}$$

For v << c m is not m₀. However if v << c and Phi = 0 then m = m₀. For example: If the particle is at rest in the field (or moving slowly) then m becomes

$$m = \frac {m_{0}}{\sqrt{1 + 2\Phi/c^{2}}}$$

In his article Okun incorreclty implies that Einstein did not use the concept of relativistic mass in Einstein's text The Meaning of Relativity. However Einstein does utilize this expression in a derivation regarding Mach's principle. Notice that that mass is a function of the gravitational potential. Therefore if the particle is in a gravitational field its inertia is altered. For this reason Einstein stated in his text (page 100,102)

(page 100) The inertia of a body must increase when ponderable masses are piled up in its neighboorhood.

(page 102) The inert mass is proportional to 1 + q, and therefore increases when ponderable masses approach the test body

Where Einstein's q is the negative of (gravitational potential)/c² and is positive. Einstein's relation is an approximation to the exact relation which I gave above. When the field is weak and the particle is movig slowly then

$$m \approx (1 - \Phi/c^{2})m_{0} = (1 + q)m_{0}$$

Okun also makes another error in this paper. His formula has an E in it which he claims is energy and his use implies that his E has the value

$$E = m_{0}c^{2}\frac {dt}{d\tau}$$

However this is not how the energy for such a particle is defined in general relativity. Okun assumes that the energy E is related to the time component of the 4-momentum P as E = cP⁰. However this only holds on special relativity. The correct expression in general relativity is P as E = cP₀.

Long story short - While Okun't goal was to claim that m can be called E which is the time component of the 4-momentum and doing otherwise causes confusion what he actually did was to prove the opposite. His thinking of m as E has led to make the errors above!

 

[Sammywu]

DW, I will be busy for a while. Any way, if you do not agree with Arcon, is that possible that you show how you will handle these problems.

Thanks a lot.

 

[DW]

Originally posted by Sammywu
DW, I will be busy for a while. Any way, if you do not agree with Arcon, is that possible that you show how you will handle these problems.

Thanks a lot.

I totally dissagree with pmb's rather Newtonian approach to relativistic problems. As I've mentioned just about every view he has outlined above is wrong in the context of modern general relativity. If you have a specific problem you would like me to show the correct relativistic treatment for just give me the details. Otherwise I'll just give a brief account of the modern relativistic paradigm for gravitation. The equation of motion for general relativity takes a form similar to Newton's second law.
This general relativistic equation is
$$F^\lambda = \frac{DP^\lambda}{d\tau}$$
There are 3 important differences between this and Newton's second law. First the vectors represented here contain a fourth element corresponding to a timelike coordinate. Second, the time derivative here is not a derivative with respect to your coordinate frame, but is with respect to time for the "particle" in question. Third this derivative is not an ordinary total derivative, but is a covariant total derivative. It is a sum of an ordinary derivative and an affine connection term which is a description of the curvature of the spacetime coordinates being used. The capitalization of the D in the numerator is not a typo. This force $$F^\lambda$$ is a real force and is sometimes qualified and called the four vector force because it is a four vector and is different than the ordinary force defined by
$$f^i = \frac{dP^i}{dt}$$, and sometimes including $$f^0 = 0$$, which is still sometimes used at a basic level in the context of "special" relativity.
The four vector force is real and as such can not be transformed away. In the case that there is only gravitation and no real forces, $$F^\lambda = 0$$
and
$$F^{\lambda};_\mu = 0$$.
The first is sufficient to guarantee that a frame exists in which the inertial forces also called fictitious forces or pseudo forces including the gravitational force (a sum of terms proportional to affine connections) is "locally" transformed away.
The second is sufficient to guarantee that a frame exists in which the tidal force(a sum of terms proportional to first order derivatives of affine connections) is "locally" transformed away.
The tricky part is that when there is Riemannian spacetime curvature these two things can not be transfomed away going to a "single" frame.
You can transform away one or the other, but when Reimannian spacetime curvature is present at least one of those is always present. In short this is because the Riemann tensor which is the expression for this kind of curvature is a sum of first order derivatives of affine connections and products of the affine connections. A nonzero tensor is nonzero according to every frame, so one set of terms can be transformed away, but never both when this tensor isn't zero.
The Newtonian concept of the gravitational field was an acceleration field. The relativistic analog of the acceleration field is the affine connections. However the affine connections are not tensors and can be transformed away. Therefor by themselves they can not represent a complete real physical entity. For this reason in modern relativity a different quantity all together has become the field concept in gravitation and that is the field of Riemannian spacetime curvature described by the Riemann tensor. This is because when the Riemann tensor is not zero one can never have a frame in which the gravitational force can be "globally" transformed away and when the Riemann tensor is zero one always has a frame according to which the gravitational force is globally transformed away and the fact that the Riemannian spacetime curvature is described by a tensor which can not be transformed away is sufficient that it can be taken to stand alone as a complete real physical entity.
Up to this point I've referred to the gravitational force several times, but as I specifically mentioned it is a fictitious force corresponding to $$F^\lambda = 0$$. In fact when doing modern general relativity it is rare that one would even make reference to a "gravitational force". One instead thinks of motion in terms of geodesic motion Vs nongeodesic motion. When there is a real force present the motion is nongeodesic. In the absence of the four vector forces motion is geodesic. This concept replaces Newtons first law of motion. Newton's first law of motion can be expressed as unacted on a particle will have constant motion. The general relativistic law corresponding to this is unacted on a particle will follow geodesic motion. If the four vector force is zero then one can show that the equation of motion for general relativity reduces to what is called the geodesic equation. This equation describes optimal paths between events in a curved four dimensional spacetimes called geodesic paths named after paths between points along the curved surface of the Earth that optimise travel distance along the surface. The geodesics of general relativity are also paths of extremal proper time. Consider two events or two different points in space and time. You have an observer that is to start at the location of the first event and to arive at the second. The most time will be recorded by that observers watch if he is to follow the geodesic that connects them. This is the principle of maximal proper time.
Theres a lot more detail I could get into, but I'm getting tired now so I'll just tie it together with the short paragraph following.

The stress energy tensor is the gravitational source for the field of Riemannian spacetime curvature. This curvature determines what forms the differential geometry of spacetime expressed by the metric tensor can take. Things unacted on by four forces then follow geodesics or paths of maximal proper time between events which are paths descriptive of this geometry. Due to the curvalinear nature of the spacetime an observer's coordinates will not be globally rectilinear or inertial which leads to his experiencing of inertial forces which he then calls the gravitational force. Locally transforming this away is as simple as going into free fall.

 

[Sammywu]

Arcon, DW, Sorry, I haven't even time to really read your stuff. Just on the train, checking Arcon's formula populated in his website. The first part of Gravitational force.

One question: What is t and tow? Is t the time of the moving observer, or the infinite observer ( the center observer in textbook maybe ), assuming here t and tow are all in a gravity influencing area?

 

[Arcon]

Originally posted by Sammywu
Arcon, DW, Sorry, I haven't even time to really read your stuff. Just on the train, checking Arcon's formula populated in his website. The first part of Gravitational force.

One question: What is t and tow? Is t the time of the moving observer, or the infinite observer ( the center observer in textbook maybe ), assuming here t and tow are all in a gravity influencing area?

t = time as measured by distant observer
Tau = time as measured by wristwatch strapped to falling particle

Arcon

 

[Sammywu]

Arcon, What is the 4-velocity? Why did you divide the space measurement by the remote observer by the proper time of the falling obhect?

 

[Arcon]

Originally posted by Sammywu
Arcon, What is the 4-velocity? Why did you divide the space measurement by the remote observer by the proper time of the falling obhect?

Relativity implies that the laws of physics must be expressable in a form which is independant of the coordinate system. I.e. they must be covariant. Such laws may expressed mathematically using tensors. When expressed as tensors then they are said to be manifestly covariant. For example: In Newtonian dynamics the equation

$$F = \frac {dP}{dt}$$

is said to be manifestly covariant, i.e. it does not depend on the coordinate system. However the equation

$$F_{x} = \frac {dP{x}}{dt}$$

which is one of the components of the first equation, is not manifestly covariant since it is not a vector equation. However it is still covariant since it has the same form in all coordinate systems. I.e. in another, primed, coordinate system

$$F'_{x} = \frac {dP'{x}}{dt}$$


A 4-vector is an example of a tensor, i.e. it is said to be a tensor of rank 1. The proper time is also a tensor. It's called a tensor of rank 0, also known as a scalar or an invariant. As such the laws of general relativity must be expressible in the form of tensors. The mathematics of general relativity is of course geometry just as it is in Newtonian dynamics. However in general relativity one deals with differential geometry, the equations of which are often tensor equations. In order to describe motion one starts with 4-vectors and then squeezes out the desired quantities. In this case the 4-vector of velocityh (i.e. 4-velocity) must be obtained by division by a proper time interval rather than a coordinate time interval since the former gives a 4-vector while the later does not.

This is a bit different than one finds in Newtonian dynamics. For example, vectors in Newtonian vectors have components which all have the same physical meaning. However the same cannot be said of 4-vectors. Thus when we have the final result we want to pick off the components to obtain the physical stuff that we're interested in. For example: in special relativity one might desire to work a problem using 4-vectors and yet they might not even be seeking a 4-vector as their goal. They might want to know the the energy of something is and that is a component of a 4-vector. However there is a subtle relationship between components and scalars. Since it takes some space to describe it I wrote it up and posted it in the internet at

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


If you have the stomach for it then you might want to take a gander at something I wrote up as an quicky intro to tensors. It's located at
http://www.geocities.com/physics_world/ma/intro_tensor.htm

It describes a few different classes of tensors, namely general tensors, affine tensors, Cartesian tensors and Lorentz tensors.

 

[Creator]

Originally posted by Arcon
[In his article Okun incorreclty implies that Einstein did not use the concept of relativistic mass in Einstein's text The Meaning of Relativity. However Einstein does utilize this expression in a derivation regarding Mach's principle. Notice that that mass is a function of the gravitational potential. Therefore if the particle is in a gravitational field its inertia is altered. For this reason Einstein stated in his text (page 100,102)
quote:
--------------------------------------------------------------------------------

(page 100) The inertia of a body must increase when ponderable masses are piled up in its neighboorhood.

(page 102) The inert mass is proportional to 1 + q, and therefore increases when ponderable masses approach the test body

--------------------------------------------------------------------------------
[/B]

Acron;
Though I haven't read Okun's article, I believe you may have incorrectly assumed that Okun was unaware of Einstein's use of 'relativistic' mass.

In reality, he probably simply chose to ignore it. The reason?:

Well, you may be unaware of it but...
The quote you gave above (pg. 100 & 102 from The Meaning of Rel.) is the first of three Machian ideas Einstein initially thought to be consistent with GR.
However, it is well known since then and quite well established (proved in 1962 by Carl Brans) that this first statement is in fact not true of GR.
So although this statement was initially used by Einstein, it has long
ago been recognized as inaccurate due to papers by Carl Brans and others...
In other words, the statement that "The inertia of a body must increase when ponderable masses are piled up in its neighboorhood"
has long ago been shown to be invalid,;
And Okun's statements are simply reflecting his knowledge of that fact.
(I have not the Brans reference ; I'm sure it is locatable).

Creator

 

[Sammywu]

Arcon, David, I can't find my SR book. But I remember it did mention this 4 vector force and momentum, even though I am not sure how to practically use it.

I designed an experiment so as we can see how these four vector momentum and force can be of any pratical use.

Please also list the formula for gamma. I believe that's the contraction factor. But I can not even memorize that formula.

Let's even keep the gravity away from this experiment.

Here, We have one object C whose rest mass is m0. I have two observers A and B. C is moving relative to A in 2/3c northward and B is moving relative to A in 2/3c southward. A , B and C coincide each other at the initial event point.

Now, At the time 0, there is a external force ( not gravity ) acting on C westward. The force is measured as m0*g to C. This force will be acting on C for 10 seconds of C's time.

Actually this experiment will be a little difficult to execute. Any thruster will involve certain mass change. If we put it into an electric field, there will be other issue. Let's just assume this is possible.

Tell me what is the 4 vector force and mometum relative to A, B and C. Will it be fixed or it become a function of tC?

At any time of tC, what is the relationship between tB, tA and tC.

When tC = 10, that is After 10 seconds of C's time, what are tA and tB?

What is the velocity of C relative to A and B at any time of this 10 second interval?

 

[Arcon]

Originally posted by Creator
Acron;
Though I haven't read Okum's article, I believe you may have incorrectly assumed that Okum was unaware of Einstein's use of 'relativistic' mass.

No. He was most definitely unaware of it. He told me that in e-mail. I.e. I was curious as to his comments given Einstein's text so I e-mailed him and asked him. Until I mentioned it to him he was unaware of it.

In reality, he probably simply chose to ignore it. The reason?:

Well, you may be unaware of it but...
The quote you gave above (pg. 100 & 102 from The Meaning of Rel.) is the first of three Machian ideas Einstein initially thought to be consistent with GR. However, it is well known since then and quite well established (proved in 1962 by Carl Brans) that this first statement is in fact not true of GR.

That is incorrect. The statement of Einstein's is by no means false and has never been proven wrong. What you're most likely thinking about is a different use of the term "mass"

Okun referred to Einstein's text in order to prove that Einstein did not use the idea in his book. The problem was that he didn't look closely enough. I guess if you don't expect to find something then its not surprising that you don't find it.

But please note - Einstein is by no means wrong. That statement is by no means wrong. If it was wrong then Einstein's GR would be wrong since all Einstein assumed there was that the quantity M = m_o*dt/dT (m_o = proper mass) was a function of the gravitational potentials (note - for a coordinate system in which the metric is time-orthogonal, i.e. g_0k = 0, metric its a function only of Phi = (g_oo - 1)c²/2) as well as a function of speed of the particle.

Thank you for your response though. If you find that reference to the 1962 article by Carl Brans please pass it on. Thanks.

 

[Sammywu]

Arcon, DW,

I am digesting your statements. Arcon seems to indicate that the object under gravity does lose mass. DW, I like your explanation of geodisc and ficticious force.

I know I have seen that strees energy tensor somewhere. I might ask question related to it later. Any way, need some times to digest these. I also read your formula in another thread. I like it. But I still have problems reading it. The denotation is still not clear enough. I think that's common to you, but they are not to us, the laymen. If you can make clear which is what mesaured by whom, I will catch them quicker. Sorry, DW, My previous sample experiment was for Arcon and you. Trying to locate where exactly your arguments are.
I wrote David because somebody else was in my mind.

 

[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?