Exploring Space Expansion: Questions & Answers

[JimJast]

You are confusing two paradigms.

Einstein might not be explaining how it "really is" either.

All we can say is that here we have two theories of gravitation, one uses force-at-a-distance, the other geometry of space-time. When tested it so happens that Einstein's theory fits the data better than Newton.

The Einsteinian predictions are exact (as far as we know of course) the Newtonian predictions are wrong (which we know for sure). It is enough to reject Newtonian theory as physics. It is just an approximate math sufficient for many engineering applications (so Newton's is not physical but phenomenological approximate desription of gravitation). We even know what is missing in Newtonian physics: the curvature of space. It is only the math describing the time dilation part of physics of gravitation. The missing part is in Landau's Theory of fields desribing Einstein's gravitation. The (apparent) gravitational force is there as $$-dE/dx^i$$, where E is rest energy of the particle $$mc^2\sqrt{g_{00}}$$ where $$g_{00}$$ is time-time component of spacetime metric, and $$dx^i$$ is the displacement vector of the particle. So we have here the concrete rest energy of the particle and curvatures of spacetime and no forces, just enrgies and their conservation following from the fact that nature is unable to create enrgy out of nothing. If it were able, then our "gravitational force" as predicted by Einstein wouldn't be the same as Newtonian, yet it is. Which means energy conservation holds in Einstein's physics the same way as in Newtonian but it does not need a mysterious "potential energy" to be valid since it comes out of first principles through the relation between time dilation and space curvature.

In his day Newton was criticised for being a mystic because he did advocate 'force-at-a-distance'.

"I deduced that the forces which keep the Planets in their Orbs must reciprocally as the squares of their distances from the centers around which they revolve." Newton quoted by Barrow "The World within the World" 1988 pg 68.

It does not say anything about Newton believing in attraction coming through empty space from the Sun. It says only about the math of the phenomenon which according to Newton was the only legitimate conclusion that one can draw from observations without undue speculatons about the mechanism of the phenomenon.

[...] space curvature and time dilation are both part of the one united space-time curvature, are you saying here that they are equal parts? If you are doing so then in order to make such a statement the questions that have to be answered are:
1. "How do you measure space curvature and how do you measure time dilation independently of each other to make the comparison?"
2. "What units, what dimensions, are each measured in?"
As both the spatial and temporal components are frame dependent, then
3. "What frame of reference are both components to be measured into make this comparison?"

1. Use weak enough field that there are linear relations only (one does not influence the other) and measure deflection of light knowing that the Newtonian part is due to the time dilation, so the rest is due to the curvature of space (as Einstein's did with the bending of light rays in vicinity of the Sun when both part turned out to be equal).
2. The dimensionless relative changes.
3. The frame of any observer.

The debate between "action at a distance" and "spacetime curvature" sounds to me like the old religious debates about how many angels can dance on the head of a pin. As long as one accepts the mathematical results of the GR equations, there appears to be no meaningful distinction between which of the two physical mechanisms is more "real." And there won't be until the underlying physical mechanism of gravity is discovered.

But the underlying physical mechanism of gravity is discovered almost a century ago by Einstein. It is just not taught in high schools for some reason so it stays mostly unknown to anybody except doctoral students specializing in gravitation. While even most of them can't explain in simple English the reason for the illusion of gravitational force.

If gravity causes spacetime to curve, why is that "less spooky" than action at a distance (e.g., a force mediated by particles moving at the speed of light)? Curved spacetime is a damn weird concept, despite the elementary textbook explanations, which explain nothing. By what physical agent does spacetime become curved by being near a massive object? And in a physical, tangible sense, what is spacetime anyway? How do we know it goes beyond a mathematical concept and is actually physical?

The mechanism of rest energy of a particle diminishing along its displacement vector directed towards the source (of energy that for some reason bends the spacetime) is rather simple (see the top of this post). Why the energy bends the spacetime we don't know and it is not the subject of theory. It is a measurable conclusion since we can measure the amount of bending and it fits the theory through the principle of conservation of energy and the value of gravitational constant which origins we don't know yet just may try to guess. Maybe I'll do before I'm through with my phd project , then I might tell you. But there is no such potential in math of Newtonian gravitation so we may dismiss it as a candidate for delivering any understanding of the universe. We shouldn't even teach it to high school students since it only confuses them later in life, and the Einsteinian gravitation is also much simpler (the mystery of force acting through vacuum repalced by rather easy to understand inability of nature to make energy out of nothing).

If spacetime curvature is physically real, then it clearly is an ephemeral manifestation of instantaneous proximity and trajectory with respect to the source mass. An instantaneous change in spacetime curvature does not physically occur at a point near or far from the gravitational source any sooner than permitted by the travel speed of light. (Doesn't that suggest the need for a mediating particle or wave moving through space?)

Of course. And don't you know particles that carry energy from one atom to another? Just call this energy "gravitational", which it really is since numerically it is responsible for gravitational force (see top of this post) and you have the answer.

Spacetime attributes cannot reasonably be treated as an intrinsic attributes of any physical structure or substance ("space") located within the coordinate dimensions affected by the gravitational source. Empty vacuum is not viscous or malleable, nor can it assume or retain any definite shape, nor can it move or remain stationary, in any normal meaning of those words. How can empty vacuum itself be physically substantial enough to deflect a passing photon's worldline?

Of course it can't and it does not. A prticle taking a geodesic worldline is not deflected, unless some other particle pushes it away from its wordline. That's why gravitational force (the same as any other inertial force) can't be propagated through empty space and that's why the notion of "universeal gravitational attraction" is silly. Math without any physics.

One can, if one wants, treat any force or pseudo-force as a manifestation of spacetime curvature. When a constant wind blows, the "spacetime geodesic" of a flying arrow becomes curved, and the arrow's path curves off in the downwind direction. It must do so in order to conserve its energy. All moving objects follow the path of least resistance. Does this mean they are "moving in a straight line through curved space?" Does it "prove" the physical reality of spacetime curvature?

The concept of spacetime curvature seems like a "safe" place to mentally retreat to when one is confounded by the apparent mathematical complexity of gravitational action at a distance.

In addition to all of this, we (the lazy humans) try to choose a reference frame in which the calculations are the simplest (so we rather don't choose the frame of flying arrow) so we don't need to bother with everything what's there. But even if we choose the best frame we can't get rid of curvature of space which means it might be real. But we never observed action at a distance (especially instantaneous; all that looked like action at a distance turned out to be illusions) so we don't need to believe that it is real.

 

[jonmtkisco]

Hi JimJast,

The Einsteinian predictions are exact (as far as we know of course) the Newtonian predictions are wrong (which we know for sure). It is enough to reject Newtonian theory as physics. It is just an approximate math sufficient for many engineering applications (so Newton's is not physical but phenomenological approximate desription of gravitation).

I don't think you'll find anyone on this forum who claims that Newton's theory of gravity is mathematically better than or equal to Einstein's, or that it provides any useful physical description of the underlying physical mechanism of gravity. Of course it is a phenomenological approximation as you say. It just happens to be a particularly excellent approximation for a weak field. Newton was every bit as much a genius in his historical context as Einstein was in his.

But dismissing Newtonian theory answers nothing about the question whether spacetime curvature is physically "real" or just a superb mathematical tool.

1. Use weak enough field that there are linear relations only (one does not influence the other) and measure deflection of light knowing that the Newtonian part is due to the time dilation, so the rest is due to the curvature of space (as Einstein's did with the bending of light rays in vicinity of the Sun when both part turned out to be equal).

Please explain what you mean by "the Newtonian part [of light deflection] is due to time dilation." One doesn't typically see time dilation described as a Newtonian theory. Maybe it's just semantics. (I understand your point about space curvature.)

The mechanism of rest energy of a particle diminishing along its displacement vector directed towards the source (of energy that for some reason bends the spacetime) is rather simple (see the top of this post). ... It is a measurable conclusion since we can measure the amount of bending and it fits the theory through the principle of conservation of energy and the value of gravitational constant which origins we don't know yet just may try to guess.

The idea that GR allows a particle in freefall in a gravitational field to conserve its energy as it accelerates towards the gravitational source, and eliminates the Newtonian distinction between kinetic and potential energy, has always been a fundamental element of GR mathematics. I don't understand how a formulation derived from Landau is different from standard GR, or why you expect it to change our fundamental perception about gravitational action.

But even if we choose the best frame we can't get rid of curvature of space which means it might be real.

I disagree with this statement. In fact, in GR spatial curvature is ALWAYS frame dependent, and can be eliminated through a suitable change in coordinate systems. For example, an observer orbiting somewhat close to the event horizon of a black hole will measure the local space to be curved. But a plunging, freefalling observer who was initially at rest an infinite distance from the black hole and is accelerated towards the black hole by its gravity, and who falls through exactly the same local space, will measure that local space to be absolutely flat and Euclidian. This observer will continue to measure locally flat geometry even after plunging through the event horizon.

Curvature effects that are frame dependent aren't covariant, and arguably they are not universally "real" phenomena.

Jon

 

[MeJennifer]

Spatial curvature is a coordinate effect while spacetime curvature is coordinate independent.

 

[jonmtkisco]

Hi MeJennifer,

Yes I agree that spatial curvature and spacetime curvature are fundamentally different and distinct concepts. One should never treat the two as interchangeable for any purpose, or create ambiguity about which meaning is intended.

Jon

 

[JimJast]

Spatial curvature is a coordinate effect while spacetime curvature is coordinate independent.

Could you present some support for the idea that "Spatial curvature is a coordinate effect"? This is just opposite to what I'm accustomed to believe in. So if my belief is false I'd like to verify it and this way improve my understanding of Einsteinian gravitaton.

 

[JimJast]

I don't think you'll find anyone on this forum who claims that Newton's theory of gravity is mathematically better than or equal to Einstein's, or that it provides any useful physical description of the underlying physical mechanism of gravity. Of course it is a phenomenological approximation as you say. It just happens to be a particularly excellent approximation for a weak field. Newton was every bit as much a genius in his historical context as Einstein was in his.

Isn't it funny that one has to be a kind of genius to discover the inverse squares law of changing of some physical variables? What does it say about the rest of us? Aren't we, including famous philosophers, physicists and mathematicians who didn't find what Newton and Einstein found, just a bunch of idiots?

But dismissing Newtonian theory answers nothing about the question whether spacetime curvature is physically "real" or just a superb mathematical tool.

Please explain what you mean by "the Newtonian part [of light deflection] is due to time dilation." One doesn't typically see time dilation described as a Newtonian theory. Maybe it's just semantics. (I understand your point about space curvature.)

If you look a little deeper into gravitation then you see there a "tiny" effect of gravitational time dilation. Generally it is considered a "one more Einsteinian prediction that turned out to be right". It is "a tiny effect" because the Einsteinian gravitation is not taught in high schools and so physicists, if ever, learn about it only in their doctoral studies as a part of GR. This way they waste about 10 years of their best years and when they finally learn about the gravitational time dilation they think that first they have to understand tensor calculus before they can get involved in any thinking about the physics of gravitation. They are overwhelmed by the math and tend to think that the math is the key to understand Einstein (I live among them ). They are in no mood to pay much attention to physics and to particular curvatures. At least not that much as Einstein did. Since Einstein didn't have any tensor calculus yet to worry about. Just the plain physics.

So Einstein, by noticing that the Newtonian physics, if it is treated as physics and not math only, requires the gravitational time dilation, and knowing that nature must be consistent saw that this "tiny" effect is the basis of Newtonian gravitation. And it might have been causing the (apparent) gravitational force. Which in reality is an illusion caused by the time dilation. His genius was in the fact that the time dilation was not even observed then. So to postulate such an effect to explain the whole Newtonian gravitation was an act of faith in consistence of phisics that unlike the math must be consistent if it says something true about the nature. Afterwards he added the curvature of space, equal, as to its relative "strengh", to the time dilation, discovering this way all parts of gravitation (so far no other has been discovered, in nearly 100 years) justifying Einstein's leaving his theory to mathematicians to polish it after he solved his equations for the universe and discovered in the process the cosmological constant that mathematician tried to dismiss since it didn't appear "elegant" tho them. Then Einstein said "If you are out to describe the truth, leave elegance to the tailor." Now we know that Einstein's gravitation makes no sense without the gravitational constant, we don't agree yet to "Einstein's value" of it since it collides with the present model of the universe which is surely not Einstein's.

So as you may see, the Newtonian part of gravitation, is the time dilation being half of the reason for the deflection of light ray near the Sun. There is equal second part which has to be the space curvature (at least according to Einstein) and if Einstein is wrong you need to develop a new gravitational theory in which space can be flat. Then I should wait with my work until you do since my present work is based strictly on Einstein's gravitation and the curvature of space being the same as the time dilation. If it is wrong I don't need to go on and should find better use for my free time.

The idea that GR allows a particle in freefall in a gravitational field to conserve its energy as it accelerates towards the gravitational source, and eliminates the Newtonian distinction between kinetic and potential energy, has always been a fundamental element of GR mathematics. I don't understand how a formulation derived from Landau is different from standard GR, or why you expect it to change our fundamental perception about gravitational action.

Of course you are right. I learned it rading Landau, that's why I describe Landau. I could learn it from any other textbook. Why do you think that Landau is not standard GR?

In fact, in GR spatial curvature is ALWAYS frame dependent, and can be eliminated through a suitable change in coordinate systems.

You may draw a flat map (or a set of maps) of the Earth surface. Do you think it elininates the curvature of the Earth surface by "a suitable change in coordinate systems"? You wouldn't advise "flat Earth" even if to some people it is "impossible to understand" how come people in Australia don't fall off the Earth (and so they suspect the government fooling them for ideological reasons). Even worse, knowing that according to Einstein there is no "attractive gravitational force" in nature?

 

[MeJennifer]

You may draw a flat map (or a set of maps) of the Earth surface. Do you think it elininates the curvature of the Earth surface by "a suitable change in coordinate systems"? You wouldn't advise "flat Earth" even if to some people it is "impossible to understand" how come people in Australia don't fall off the Earth (and so they suspect the government fooling them for ideological reasons). Even worse, knowing that according to Einstein there is no "attractive gravitational force" in nature?

A false analogy.

Curved spacetime is intrinsically curved, independent of how it is charted but whether 3-surfaces in spacetime are curved entirely depends on how they are "sliced" from spacetime.

 

[jonmtkisco]

Aren't we, including famous philosophers, physicists and mathematicians who didn't find what Newton and Einstein found, just a bunch of idiots?

Well as they say, it's all relative, so compared to Newton and Einstein we could perhaps be considered idiots, but compared to clams we're all geniuses. Don't feel bad, no individual's intelligence comes within even an order of magnitude of the genius of perfect hindsight. In 2028 we're all going to feel really dumb about all the things we failed to comprehend in 2008 that will seem so obvious in 2028. And by 1955, Einstein in retrospect had any number of reasons to kick himself for missing many discoveries in his field made by others in the prior 40 years that were within the capabilities of his genius and subject matter expertise to discover himself.

If you look a little deeper into gravitation then you see there a "tiny" effect of gravitational time dilation.
...
So Einstein, by noticing that the Newtonian physics, if it is treated as physics and not math only, requires the gravitational time dilation, and knowing that nature must be consistent saw that this "tiny" effect is the basis of Newtonian gravitation. And it might have been causing the (apparent) gravitational force. Which in reality is an illusion caused by the time dilation.
...
So as you may see, the Newtonian part of gravitation, is the time dilation being half of the reason for the deflection of light ray near the Sun.

JimJast, after reading this I feel no closer to understanding why you say that Newtonian gravity is nothing but time dilation. I don't recall reading this anywhere else, and I assume I'm not the only one. Can you please give us a quote from a textbook or other source that describes how this conclusion is derived?

Why do you think that Landau is not standard GR?

I don't, but it was my impression that you claimed in an earlier post that mainstream cosmologists don't understand GR as you have conceptually derived it from Landau, and your PhD thesis is going to enlighten them about it. Maybe I misunderstood.

Jon

 

[yuiop]

The idea that GR allows a particle in freefall in a gravitational field to conserve its energy as it accelerates towards the gravitational source, and eliminates the Newtonian distinction between kinetic and potential energy, has always been a fundamental element of GR mathematics.

Are you suggesting that the rest mass energy of a falling particle changes as it falls so that the potential energy of the gravitational field does not have to be invoked to conserve energy?

If that is the case then I have derived this equation for Total Energy (TE) that satisfies those requirements with the assumption that the initial velocity of the particle at infinty is zero:

$$TE = \sqrt{m_0^2 c^4 \left( 1- \frac{2GM}{c^2R} \right) +m_o^2 v^2 c^2 }$$

Is that the sort of thing you had in mind?

Also note that energy is conserved without requiring potential energy of the field in the coordinate measurements of an observer that remains at infinity, but not from the point of view of local non inertial observers. For example local observers see the frequency (and therefore the energy) of a falling photon as increasing while the frequency in coordinate measurements remains constant.

 

[JimJast]

JimJast, after reading this I feel no closer to understanding why you say that Newtonian gravity is nothing but time dilation. I don't recall reading this anywhere else, and I assume I'm not the only one. Can you please give us a quote from a textbook or other source that describes how this conclusion is derived?

Jon, I found a quote for you in Wolfgang Rindler's Relativity, Special, General and Cosmological, chapter 9.4 Newtonian support for the geodisic law of motion. On p. 189 Rindler writes (as a conclusion about geodesic lines for slow motion in weak fields):

"A deformation of the time dimension therefore has a first-order effect on the length of those lines, wherease a deformation of the spatial dimensions has only a second-order effect. [...]".

And then comes the metric with Newtonian potential in time-time component and flat spatial part, as a sufficient approximtion of gravitation for velocities < 50km/s as producing error < 3x10^{-8} (as propotional to v/c squared).

 

[JimJast]

A false analogy.

Curved spacetime is intrinsically curved, independent of how it is charted but whether 3-surfaces in spacetime are curved entirely depends on how they are "sliced" from spacetime.

Of course you are right if no "suitable change in coordinate systems" is applied. Then certain ways of slicing can produce zero curvatures and others, non zero curvatures.

With a "proper scaling" of coordinates one can make any slice flat. This is done for the maps of the surface of the Earth (which are usually orthogonal to the time coordinate and so they are surely of a curved surface) so the analogy with a "suitable change in coordinate systems" is still exact.

 

[mysearch]

First, I would like to get clarification of Kev’s post #44, as I know he has spent some considerable time looking that the issue of relativity. Which observer perceives the conservation of energy, the distant observer or onboard observer?

[1] Distant Observer $$v_R = c\left( 1+ \frac {Rs}{r}\right)* \sqrt {\frac {Rs}{r}}$$

[2] Onboard Observer $$v_R = c\sqrt {\frac {Rs}{r}}$$

Now equation [1] suggests that the observed radial velocity goes to zero, while the onboard radial velocity goes to [c] at [r=Rs]. I would have thought only [2] was consistent with the conservation of energy?

#36: The Einsteinian predictions are exact (as far as we know of course) the Newtonian predictions are wrong (which we know for sure).

If possible I would like to try and get a better idea as to the scope and scale of the ‘errors’ in Newtonian predictions in comparison with GR. As an initial statement, it was my understanding that the discrepancies between Newtonian physics and Relativity only become apparent under the extremes of gravitational potential, i.e. space curvature, and/or the velocity of an observed frame approaching [c].

#43: JimJast, after reading this I feel no closer to understanding why you say that Newtonian gravity is nothing but time dilation. I don't recall reading this anywhere else, and I assume I'm not the only one.

I would also like to get a better understanding of the details of the premise on which the original statement was based. Have only just seen #46, which may address this question.

#36: So we have here the concrete rest energy of the particle and curvatures of spacetime and no forces, just energies and their conservation following from the fact that nature is unable to create energy out of nothing. If it were able, then our "gravitational force" as predicted by Einstein wouldn't be the same as Newtonian, yet it is. Which means energy conservation holds in Einstein's physics the same way as in Newtonian but it does not need a mysterious "potential energy" to be valid since it comes out of first principles through the relation between time dilation and space curvature.

Classically, while we refer to many forms of energy, they are said to all reduce to just 2 forms, kinetic and potential. If I remove `potential energy` from my vocabulary:

- How do I semantically describe the conservation of energy?

- Is the statement above implying that the conservation of kinetic energy is balanced by time dilation and space curvature?

Accepting the tenet of GR that gravitational force does not exist, but rather the path of an mass object with kinetic energy [Ek] is described by a geodesic path:

- Does this prevent an equivalent model being constructed based on the concept of force and potential energy?

- If the kinetic energy is conserved by space curvature, is there any implication that this curvature must require a source of energy and, if so, might we call it – potential energy?

Finally, I wanted to see if there was a comparative yardstick by which we could judge the scope and scale of the results provided by Newtonian & GR physics. In Newtonian physics, the total energy [Et], excluding rest energy, can be derived based on the conservation of energy:

[3] $$E_T = 1/2 m\left(v_r^2 + v_o^2\right) +\left (-\frac{GMm}{r} \right)$$

Where [vr] is the radial velocity and [vo] is the orbital velocity with the middle term corresponding to the kinetic energy, whilst the right term corresponds to the potential energy of gravitation. I believe the next equation is an equivalent form of [3] derived from the Schwarzschild metric, when divided thorugh by $$[d\tau]$$, which has its roots in GR.

[4] $$E_T =1/2 m\left(v_r^2 + v_o^2\right) - \frac{GMm}{r} \left( 1+ \frac{v_o^2}{c^2}\right)$$

The similarities of [3] and [4] seem obvious, except for the term on the right, which now has an extra component that corresponds to the orbital velocity. If this equation is a valid assumption, it would suggest that the radial free-fall energy of both equations is comparable in Newtonian physics and GR, at least, to an onboard observer, as only the orbital kinetic energy changes as [vo] approaches the speed of light. As such, the discrepancy appears, in this specific example, to be associated with the orbital conservation of energy, only when [vo] approaches [c].

I would appreciate any clarification of the points raised. Thanks

 

[Garth]

There are a few ground rules to get cleared up.

First GR does NOT in general conserve energy, it conserves energy-momentum instead, which is different.

Energy is a frame dependent concept. When observers in inertial 'freely falling' frames are accelerating relative to each other, because they are at different positions in a gravitational field, then they will measure energy differently to each other, particularly the kinetic energy of other bodies. However they will all agree on the energy-momentum of a third body.

Emmy Noether showed that GR belongs to a class of theories she called "Improper Energy Theories" that do not in general conserve energy.

You can try to construct a gravitational theory that does conserve energy but that would be an alternative theory, not GR, and the challenge would be to get it to pass all the observational tests that GR has passed.

However, there are special cases in GR when energy is conserved, when there is a temporal or time-like Killing vector. One such case would be the static field of a spherical body, such as the Earth, ignoring other gravitational bodies in the universe, as observed from a frame co-moving with the Earth.

In this case for example you (on the Earth's surface with a clock and ruler) can watch an apple fall to the Earth and observe that its kinetic energy, and therefore total energy, increases as its falling velocity increases. This is so even though there are no forces operating, no work is being done, because the Earth and the apple are simply following their geodesic paths through curved space-time.

However this increase in kinetic energy is compensated by the increasing time dilation the apple experiences as it falls further into the Earth's field. The clock and the gravitational field are 'carried' by the Earth.

On the other hand we can transpose ourselves into the apple's frame of reference with clocks and rulers of measurement now being carried by the clock.

In this frame of reference the Earth is seen to fall towards the apple, and constantly increase in its velocity and kinetic energy.

Now the clock is being carried by the apple while the field is being carried by the Earth. As a consequence the time dilation operates in the opposite sense and does not compensate for the increasing kinetic energy, but the reverse.

I hope this helps.

Garth

 

[JimJast]

I would appreciate any clarification of the points raised. Thanks

I'll do it from the beginning, and only for Einstein's gravitation: Total energy of a particle is $$E = \frac{m}{\sqrt{1-\frac{v^2}{c^2}}}c^2\sqrt{g_{00}}$$ (Landau's Theory of fields, p. 285). Being total this energy is necessarily conserved. You may check it by taking derivative with respect to displacement along its free fall path to see that after a few transformations it comes out as $$\frac{dE}{dx}=0$$. By putting $$v = 0$$ (no kinetic energy) you get also a force $$F = -\frac{dE(v=0)}{dx}=mg$$ that used to be called "universal gravitational attraction" (not anymore though since it is plainly an inertial reaction resulting from forcing $$v=0$$). Ain't Einstein's gravitation neat?

 

[Garth]

Being total this energy is necessarily conserved.

Why is it necessarily conserved? Are you assuming a time-like Killing vector?

And yes Einstein's gravitation is 'neat' but if it wasn't it would not have worked and stood up so long under so many experimental tests.

Garth

 

[yuiop]

First, I would like to get clarification of Kev’s post #44, as I know he has spent some considerable time looking that the issue of relativity. Which observer perceives the conservation of energy, the distant observer or onboard observer?

[1] Distant Observer $$v_R = c\left( 1+ \frac {Rs}{r}\right)* \sqrt {\frac {Rs}{r}}$$

[2] Onboard Observer $$v_R = c\sqrt {\frac {Rs}{r}}$$

Now equation [1] suggests that the observed radial velocity goes to zero, while the onboard radial velocity goes to [c] at [r=Rs]. I would have thought only [2] was consistent with the conservation of energy?

Hi mysearch,

First I should make it clear that I am comparing distant measurements with local measurements made by non-inertial observers that are stationary in the gravitational field and not "onboard" measurements of free falling observers.

The Total Energy (E_T) equation from Special Relativity relates Rest Energy (R_E) to Momentum Energy (M_E) by the relationship:

$$E_T = \sqrt{E_R^2 + E_M^2} = \sqrt{m^2c^2 + m^2(1-v^2/c^2)^{-1} v^2 c^2}$$

To put this into the context of a free falling object in the Schwarzschild metric that falls from infinity with an initial velocity of zero, the relativistic gamma factor of sqrt(1-v^2/c^2) can be replaced by the gravitational gamma factor sqrt(1-R_s/R) because the local falling velocity v/c = sqrt(R_s/R) to give:

$$E_T = \sqrt{M^2 C^4 + M^2(1-R_s/R)^{-1} V^2 C^2}$$

I have also taken the liberty of using uppercase M, V and C to indicate measurements made by a distant observer and reserved lower case m, v and c for local measurements .

From other threads we know that

$$C = c\left( 1 - \frac {Rs}{R}\right)$$

and this applies to all velocities so

$$V = v\left( 1 - \frac {Rs}{R}\right)$$

The total energy equation can now be written as:

$$E_T = \sqrt{M^2 c^4 (1-R_s/R)^4 + M^2(1-R_s/R)^{-1} (v^2 c^2) (1-R_s/R)^4}$$

By assuming the Total Energy of the particle at rest at infinity is mc^2 and by further assuming that the Total Energy is constant for a falling particle in coordinate measurements, we can now say

$$mc^2 = \sqrt{m^2 f^2 c^4 (1-R_s/R)^4 + m^2 f^2 v^2 c^2 (1-R_s/R)^3}$$

where f is some factor that relates M to m.
By substituting c*sqrt(R_s/R) for v we can solve for f to give:

$$f = (1-R_s/R)^{-1.5}$$ so that

$$M^2 = m^2 f^2 = m^2(1-R_s/R)^{-3}$$ and also

$$mc^2 = \sqrt{m^2 c^4 (1-R_s/R) + m^2 v^2 c^2}$$

It can be seen that the expression is now in terms of local velocity and that when v=c AND R_s=R the total energy is mc^2 and also when v+0 AND R=infinity that the total energy is still mc^2

Solving the last equation for the local falling velocity gives the result v/c = sqrt(R_s/R) which should not come as too much of a surprise as that was assumed in deriving the equation.

I originally derived the equation without assuming v/c = sqrt(R_s/R) by simply assuming M = m(1-R_s/R)^(-1.5) because in SR parallel relativistic mass is also related by m = mo(1-v^2/c^2)^(-1.5)

Have you noticed that in SR the parallel kinetic transformations

L = Lo (1-v^2/c^2) ^(0.5)
T = To (1-v^2/c^2) ^(-0.5)
M = Mo (1-v^2/c^2) ^(-1.5)

are analogous to the vertical gravitational transformations

L = l (1-R_s/R) ^(0.5)
T = t (1-R_s/R) ^(-0.5)
M = m (1-R_s/R) ^(-1.5)

and that the transverse SR kinetic transformations

L = Lo
T = To (1-v^2/c^2) ^(-0.5)
M = Mo (1-v^2/c^2) ^(-0.5)

are analogous to the horizontal gravitational transformations ?

L = l
T = t (1-R_s/R) ^(-0.5)
M = m (1-R_s/R) ^(-0.5)

My assumption of M = m (1-R_s/R) ^(-1.5) from the above observations happily gives us v/c = sqrt(R_s/R)

Anyway, going back to your original concern about whether we are talking about velocity measured by a local or distant observer, the total coordinate energy equation can be expressed in terms of the velocity V measured by the distant observer as:

$$E_T = \sqrt{m^2 c^4 (1-R_s/R) + m^2 V^2(1-R_s/R)^{-2} c^2}$$ . . . . . . [3]

Substituting c*sqrt(R_s/R)*(1-R_s/R) for V we get:

$$E_T = \sqrt{m^2 c^4 (1-R_s/R) + m^2 c^2(R_s/R) c^2}$$

which simplifies to

$$E_T = \sqrt{m^2 c^4 (1-R_s/R+R_s/R)}$$

which is obviously constant for all R.

We can also solve equation [3] for V to obtain

$$V = c*\sqrt{(R_s/R)}*(1-R_s/R)$$

which agrees with your equation [1] (assuming you made a typo with the sign)

Please note that I have not proved that coordinate energy is conserved because I assumed that in the derivations, but it seems a reasonable assumption because we already know that the coordinate energy of a falling photon is constant. It should also be clear that energy is not conserved according to measurements by local observers and Garth has also indicated it is not conserved from the point of view of the falling particle. However, I hope I have shown that IF the coordinate total energy of a falling particle is conserved then the rest energy of the particle diminishes as the particle falls.

 

[Nereid]

[...]

If possible I would like to try and get a better idea as to the scope and scale of the ‘errors’ in Newtonian predictions in comparison with GR. As an initial statement, it was my understanding that the discrepancies between Newtonian physics and Relativity only become apparent under the extremes of gravitational potential, i.e. space curvature, and/or the velocity of an observed frame approaching [c].

[...]

What follows may be a list of things that you already know very well (in which case I have misunderstood your question)...

If by "extremes of gravitational potential" you include the surface of the Earth and the solar system from the Sun's photosphere to the orbit of Jupiter (say), then indeed "the discrepancies between Newtonian physics and Relativity only become apparent" there.

However, such environments are not, I submit, usually so regarded.

And yet it is just these in which "the discrepancies between Newtonian physics and Relativity only become apparent"! (as long as we add "first").

Example: the anomalous advance of the perihelion of Mercury - known for ~half a century before the publication of GR

Example: the bending of light as it passes the Sun's 'surface' - known since the 1919 total eclipse (and now a standard correction in contemporary astrometry, for example, for GAIA this deflection needs to be factored in across the whole sky, and all planets and the Moon will produce a 'weak lensing' signal that will be easily measurable by GAIA)

Example: gravitational redshift (as in Pound-Rebka).

 

[mysearch]

General Response to #48, #49, #51, #52

Thanks for all the helpful comments, especially the detailed explanation in #51. I wanted to initially respond by trying to explain my general approach and then respond to specific points after I have had a chance to work through the details provided.

- First, apologises for the incorrect sign in equation [1] #47, as picked up by kev #51.

- Second, while appreciating Nereid’s comments, my reasons for wanting to compare the approximation of Newtonian physics with GR were mainly in response to the Jim-Jon exchanges. However, as a general comment, while the issues surrounding the perihelion of Mercury and bending of light support GR, the effects in absolute terms are quite small. Therefore, I will try to better explain my rationale below.

From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction. Hence my correlation of GR theory to Newtonian physics, but please note I am not advocating a Newtonian preference. However, I did want to test Jim’s statement that `Newtonian predictions are wrong` rather than the usual implication that they were still a good approximation under most circumstances. Therefore, the point of my original equations (1-4) in #47 was to try to narrow the overall complexity, as alluded to in Garth’s comments in #48, to a very specific, and possibly over-simplistic, example in the hope that some physical interpretation might be more obvious. For example, equation [2] is the free-fall velocity as perceived by both the onboard observer and any local observer at [r].

[2] $$v_R = c\sqrt {\frac {Rs}{r}}$$

It is possibly worth highlighting that [2] is also the perspective derived from Newtonian physics. As such, what are the real physical implications of equation [#47:1], if the velocity, as perceived by the distant observer, slows under free-fall gravitational acceleration to zero?

Based on what appears to be the most ‘real’ physical interpretation, and I realize that some may question the applicability of the word ‘real’, I then tried to correlate the implications of equation #2 on equations [3,4]. In this respect, I was focusing on the onboard observer, while Kev in #51 appears to have focused more on the distant observer. Equation [3] being the Newtonian form, while [4] is derived from Schwarzschild’s metric and therefore assumed consistent with GR under the general caveats assumed by this solution. However, the implication on [3] and [4] for an onboard free-falling observer is that orbital velocity [vo=0] and, as such, both equations appear to collapse to the form in [5], i.e.

[3] Newtionian: $$E_T = 1/2 m\left(v_r^2 + v_o^2\right) +\left (-\frac{GMm}{r} \right)$$

[4] GR: $$E_T =1/2 m\left(v_r^2 + v_o^2\right) - \frac{GMm}{r} \left( 1+ \frac{v_o^2}{c^2}\right)$$

[5] $$E_T = 1/2 mv_R^2 - \frac{GMm}{r}$$

Substituting for $$[v_R]$$ and $$Rs=2GM/c^2]$$ leads to the form:

[6] $$E_T = \frac{GMm}{r} - \frac{GMm}{r} = 0$$

Note, this is only valid for the radially solution, not an orbiting one; also this equation appears not to account for any relativistic effects, i.e. velocity or gravitation as normally associated with $$[\gamma]$$. However, the form of equation [6] has transpose the kinetic energy associated with $$[v_R]$$ into an equivalence amount of potential energy via equation [2]. In this form, the only quantity that would be subject to relativistic effects is [m], so any value of $$[\gamma]$$ would cancel out in [6].

As such, energy would seem to be conserved based on potential energy being converted to kinetic energy, noting that classically, potential energy is negative ranging from a maximum of zero at an infinite [r] to a minimum, i.e. maximum negative, as the centre of mass [M] is approach. It is noted that free-falling into a black hole raises an anomaly when [v=c] at [r=Rs]. However, generally, the conclusion being forwarded seems to contradict the statement made in #51and so I would like to better understand if I have made a wrong assumption or misinterpreted what was being said in this specific case?

#51: It should also be clear that energy is not conserved according to measurements by local observers and Garth has also indicated it is not conserved from the point of view of the falling particle.

Referencing `Exploring Black Holes` by Taylor and Wheeler p3-12/section-5:

The fact that E/m is constant for a free particle yields a great simplification in describing the motion of a radially plunging particle

Equation [10] in this reference also gives an equation of the form:

[7] $$\frac{E}{mc^2} = \left(1-Rs/r\right) \frac{dt}{d\tau} = 1$$

From which it might be assumed that:

[8] $$\frac{dt}{d\tau} = \left(1-Rs/r\right)^{-1}$$

This raised the question in my mind as to whether the relativistic factor $$[\gamma]$$ could be physically interpreted as follows:

[9] $$\gamma = \frac{1}{\sqrt(1-v^2/c^2)}*\frac{1}{\sqrt(1-Rs/r)}$$

However, in the specific case of a free-falling observer, velocity is also proportional to radius such that:

[10] $$\frac{1}{\sqrt(1-v^2/c^2)} = \frac{1}{\sqrt(1-Rs/r)}$$

I believe this is consistent to Kev’s statement in #51. However, in #49, Jim presented the following equation and I would therefore like to clarify that the use of the metric tensor [g00] is essentially equivalent to equation [7] given the assumption of this specific case?

$$E = \frac{m}{\sqrt{1-\frac{v^2}{c^2}}}c^2\sqrt{g_{00}}$$

This post is already too long, and possibly drifting off the main thrust of this thread, so I will terminate at this point, but will continue to review the detailed points already raised. Many thanks

 

[JimJast]

Why is it necessarily conserved? Are you assuming a time-like Killing vector?

I was reminded already a few times, that such problems can't be discussed in regular threads that may spoil the minds of innocent students who are coming here for knowledge and not for some crazy ideas. Those crazy ideas (like global conservation of energy) have to be placed in IR section of this forum. IR section is provided specially for those who propose (as eg. Feynman did in his Feynmans lectures...) that energy is conserved separately from momentum and not together, as Wheeler's momenergy, as the present form of GR requires. If you want to discuss a time-like Killing vector and consequences of its existence you should open a thread in IR section. We can't discuss it here since this thread is about space expansion which necessarily excludes the global conservation of energy.

My remark about energy necessarily conserved regards Einstein's theory from 1916, and so from before it was assumed that the universe is expanding that could happen only after 1929 when Hubble proposed the dependence of recession velocities of galaxies on distance. Before that it was universally accepted (also by Einstein, the patent office clerk) that energy and momentum are conserved separately, and that no theory that contradicts this principle can be consdered a scientific theory (the patent office didn't even allow applications for devices proposing perpetual motion machines; maybe they allow them now).

But since you mention this subject you might know the answer to the question how energy can be converted into momentum and vv. to keep the both of them being conserved together? I was banned from astronomy forum before learning the answer to this question (and possibly just for asking it) and so I'm still curious.

 

[Garth]

As I have said now several times, if you are talking about GR then you need a time-like Killing vector to conserve energy.

If you are not talking about GR, but some unpublished theory of your own, then you need to post on the IR thread.

Garth

 

[JimJast]

As I have said now several times, if you are talking about GR then you need a time-like Killing vector to conserve energy.

If you are not talking about GR, but some unpublished theory of your own, then you need to post on the IR thread.

You're (obviously) right about a time-like Killing vector and its necessity to conserve energy. You are wrong though that Einstein's theory of 1916 is my own unpublished theory even if it requires posting it in IR thread to be discussed. I wouldn't call this theory neither mine (dispite that I believe it is true) nor unpublished and I would even go as far as calling it a GR theory, distinctive from the mainstream GR by the fact that it has the conservation of energy built in (and therefore it is admitting a time-like Killing vector). Since I don't want to be banned from this forum for discussing Einstein's theory in the mainstream forum I submited my post to IR thread hoping that the discussion of Einstein's theory will be allowed there. We just need to wait for the decision of moderators.

 

[Garth]

But the Einstein's static model as a possible GR cosmological solution has been falsified.

1. By the observation of Hubble red shift, which not be there in a static GR model.

2. By the fact that such a solution would be unstable wrt small perturbations. Any perturbation would increase and kick the model into an ever increasing expanding or contracting solution.

3. Therefore, apart from the other observations and tests it would have to pass that I mentioned in post #28, any resurrection of the model would have to include new physics and therefore be an unpublished new theory.

Garth

 

[yuiop]

Hi,


In General Relativity can we always find a coordinate system or reference frame where the total energy of an evolving system is conserved over time independently of momentum?


I ask, because in Special Relativity the total energy of an evolving system is always conserved independtly of momentum from the point of view of an inertial observer even if the observer is not at rest with respect to the centre of momentum frame of the system. The only time energy appears not to be conserved in Special Relativity is when switching from one reference frame to another.

I am trying to make clear the difference between energy not being conserved over time and energy not being conserved when switching reference frames or coordinate systems.

 

[JimJast]

But the Einstein's static model as a possible GR cosmological solution has been falsified.

1. By the observation of Hubble red shift, which not be there in a static GR model.

2. By the fact that such a solution would be unstable wrt small perturbations. Any perturbation would increase and kick the model into an ever increasing expanding or contracting solution.

3. Therefore, apart from the other observations and tests it would have to pass that I mentioned in post #28, any resurrection of the model would have to include new physics and therefore be an unpublished new theory.

This new theory is a published already theory of Einstein known as general relativity. If you admit the conservation of energy (which is built into Einstein gravitation in a form of vanishing divergence of stres-energy tensor) then for a static space you get the Hubble resdshift roughly as observed for the density of matter in space roughly as observed. You may calculate it easily and exactly with Newtonian approximation of GR since it may be done for space with weak gravitational field and slow matter. If you do you'll see how it simulates the accelerating expansion, with observed acceleration, and that the dynamical friction of photons was never before calculated exactly. It was just assumed to be negligible by inventors of the Big Bang hypothesis to provide for their idea of expanding universe. Similarly the stuff about stability (note that we don't know how to solve 3 body problem, and the universe contains even more than 300 bodies, so how we can tell that this system is unstable?)

Regretable we can't write about Einstein's GR in a mainstream thread because certain calculations were neglected and now we need to pretend that we all believe that they were done and Einstein's universe was falsified. Have you ever seen the proof that there is no Hubble redshift in Einstein's universe (while energy is conserved) and that photons fly in it without any redshift while everything else, including space probes (eg. Pioneers 10 ans 11) have this redshift?

 

[Garth]

This new theory is a published already theory of Einstein known as general relativity. If you admit the conservation of energy (which is built into Einstein gravitation in a form of vanishing divergence of stres-energy tensor)

Have you understood nothing about what we have been saying about GR?

You cannot admit the conservation of energy in GR because GR is an Improper Energy Theory that does not in general conserve energy.

The vanishing covariant divergence of the stress-energy tensor, otherwise known as the energy-momentum tensor, $T_{\mu \nu}$, conserves energy-momentum, not in general energy, which is different.

This is basic 'Einstenian' GR, which you had better get under your belt if you are going to base a PhD thesis on it. Einstein himself realized that energy was not conserved in general and worried about it until Noether explained that this was how a theory such as GR should behave.

then for a static space you get the Hubble resdshift roughly as observed for the density of matter in space roughly as observed. You may calculate it easily and exactly with Newtonian approximation of GR since it may be done for space with weak gravitational field and slow matter.

No, no, no...

The static GR cosmological solution has a spatial curvature constant k = +1, it has to.
Newtonian approximations do not generally work in curved space over cosmological distances, even if the field is weak and the non-Newtonian effects are locally insignificant nevertheless the accumulation of the effect over cosmological distance is not insignificant.

You have to work consistently in GR.

This is where I find your logic baffling; in the GR static solution, because the field is static, there is a time-like Killing vector and energy is conserved! The galaxies are not moving apart and there is no Doppler shift, the energy of the photon is conserved in the frame of all static representative observers in such a universe and the frequency is constant from emission to observation. There would be no red shift.

If you do you'll see how it simulates the accelerating expansion, with observed acceleration, and that the dynamical friction of photons was never before calculated exactly.

What the heck is "dynamical friction of photons" this is pure crackpottery...

It was just assumed to be negligible by inventors of the Big Bang hypothesis to provide for their idea of expanding universe. Similarly the stuff about stability (note that we don't know how to solve 3 body problem, and the universe contains even more than 300 bodies, so how we can tell that this system is unstable?)

de Sitter showed that the static model was unstable in the 1920's.

If you want to learn about it I suggest you read this recent paper by Barrow, Ellis et al. http://www.iop.org/EJ/article/0264-9381/20/11/102/q311l2.html which looks at the model as possible primordial states for the present universe. The conclusions of that paper:

There is considerable interest in the existence of preferred initial states for the universe and in the existence of stationary cosmological models. So far this interest has focused almost entirely upon the de Sitter universe as a possible initial state, future attractor, or global stationary state for an eternal inflationary universe. Of the other two homogeneous spacetimes, the Einstein static provides an interesting candidate to explore whether it could play any role in the past evolution of our universe. It is important to know whether it can provide a natural initial state for a past eternal universe, whether it allows the universe to evolve away from this state, and whether under any circumstances it can act as an attractor for the very early evolution of the universe. We might also ask whether it is not possible for it to provide the globally static background state for an inhomogeneous eternal universe in which local regions undergo expansion or contraction, manifesting an instability of the Einstein static universe. With these questions in mind we have investigated in detail the situations under which the Einstein static universe is stable and unstable.

We have shown that the Einstein static universe is neutrally stable against inhomogeneous vector and tensor linear perturbations, and against scalar density perturbations if $c_s^2 > \frac{1}{5}$ extending earlier results of Gibbons for purely conformal density perturbations. However, we find that spatially homogeneous gravitational-wave perturbations of the most general type destabilize a static universe. We pointed out the link that can be forged between this homogeneous instability and the behaviour of the inhomogeneous gravitational wave spectrum by choosing modes with imaginary wave number. Our results show that if the universe is in a neighbourhood of the Einstein static solution, it stays in that neighbourhood, but the Einstein static is not an attractor (because the stability is neutral, with non-damped oscillations). Expansion away from the static state can be triggered by a fall in the pressure of the matter. Typically, expansion away from the static solution will lead to inflation. If inflation occurs, then perturbations about a Friedmann geometry will rapidly be driven to zero. The nonlinear effects (which will certainly be important in these models because of the initial infinite timescale envisaged) will be discussed in a further paper, as will other aspects of the spatially homogeneous anisotropic modes.

Regretable we can't write about Einstein's GR in a mainstream thread because certain calculations were neglected and now we need to pretend that we all believe that they were done and Einstein's universe was falsified. Have you ever seen the proof that there is no Hubble redshift in Einstein's universe (while energy is conserved) and that photons fly in it without any redshift while everything else, including space probes (eg. Pioneers 10 ans 11) have this redshift?

Yes I have given the 'proof' above; energy is conserved in a static GR universe there is no motion and therefore in a GR theory there is no red shift in a theory where particle masses are constant.
This is because red shift measurement is the comparison the ratio (energy of the photon)/(mass of the atom) at emission with the ratio (energy of the photon)/(mass of the atom) at observation. In the static GR model it would be the same.

As I have said you have to do it consistently - I have published in peer reviewed journals a theory that does what you seem to be getting at, but it is a modification of GR, not GR itself.

Garth

 

[Garth]

Hi,In General Relativity can we always find a coordinate system or reference frame where the total energy of an evolving system is conserved over time independently of momentum?I ask, because in Special Relativity the total energy of an evolving system is always conserved independtly of momentum from the point of view of an inertial observer even if the observer is not at rest with respect to the centre of momentum frame of the system. The only time energy appears not to be conserved in Special Relativity is when switching from one reference frame to another.

I am trying to make clear the difference between energy not being conserved over time and energy not being conserved when switching reference frames or coordinate systems.

Hi kev.

You have to define how you measure energy and then how to compare that measurement between different coordinate systems.

The problem in GR is introduced, by space-time curvature.

Energy is the time component of a particle's energy-momentum vector, how do you transport a vector across curved space time?

Consider the surface of a cylinder with a vector in the tangent plane at one point on that surface.

Now transport the vector over the surface.

If you go around the curved surface the tangent plane and the vector being parallel transported in it will turn, so although the vector's overall length does not change a coordinate based measurement of its components will in general change.

However it we move up the surface parallel to the axis of the cylinder, where there is a symmetry, the vector will not turn. In this case the components will not change, they will be conserved under the translation.

If the vector is energy momentum and the axis of the cylinder represents the time axis of a space-time surface then there is a time-like Killing vector and the 'axis' component that is conserved is energy.

I hope this helps.

Garth

 

[JimJast]

What the heck is "dynamical friction of photons" this is pure crackpottery...

not to astronomers for whom this effect is an observable and who are able to calculate it easily in clouds of dust. Eg. in a static cloud of dust of density $$6\times10^{-27}kg/m^3$$ it produces a redshift corresponding to the recession velocity of the source of light $$70km/s/Mpc$$ (looks familiar?). Those astronomers are always surprised that the cosmologists despite their sophistication and ease of handling of tensors of GR never heard about the effect that necessarily follows the simple Newtonian physics with its hypothetical conservation of energy, the same effect that is described in popular articles on http://en.wikipedia.org/wiki/Dynamical_friction" .

 

[Garth]

not to astronomers for whom this effect is an observable and who are able to calculate it easily in clouds of dust. Eg. in a static cloud of dust of density $$6\times10^{-27}kg/m^3$$ it produces a redshift corresponding to the recession velocity of the source of light $$70km/s/Mpc$$ (looks familiar?). Those astronomers are always surprised that the cosmologists despite their sophistication and ease of handling of tensors of GR never heard about the effect that necessarily follows the simple Newtonian physics with its hypothetical conservation of energy, the same effect that is described in popular articles on http://en.wikipedia.org/wiki/Dynamical_friction" .

I asked what you meant by the "dynamical friction of photons".

From the same wikipedia article that you referred to: (Note, Wikipedia articles have to be used with much caution and are not always a good reference)

Fritz Zwicky proposed in 1929 that a gravitational drag effect on photons could be used to explain cosmological redshift as a form of tired light.[6] However, his analysis had a mathematical error, and his approximation to the magnitude of the effect should actually have been zero, as pointed out in the same year by Arthur Stanley Eddington. Zwicky promptly acknowledged the correction,[7] although he continued to hope that a full treatment would be able to show the effect.

It is now known that the effect of dynamical friction on photons or other particles moving at relativistic speeds is negligible, since the magnitude of the drag is inversely proportional to the square of velocity.

Any attempt to resurrect this discredited idea must therefore be 'new' physics, which must either be tested and published, in which case please give the reference to the published paper, or simply falls into the category of 'crackpottery'.

Garth

 

[Nereid]

Thanks for all the helpful comments, especially the detailed explanation in #51. I wanted to initially respond by trying to explain my general approach and then respond to specific points after I have had a chance to work through the details provided.

- First, apologises for the incorrect sign in equation [1] #47, as picked up by kev #51.

- Second, while appreciating Nereid’s comments, my reasons for wanting to compare the approximation of Newtonian physics with GR were mainly in response to the Jim-Jon exchanges. However, as a general comment, while the issues surrounding the perihelion of Mercury and bending of light support GR, the effects in absolute terms are quite small. Therefore, I will try to better explain my rationale below.

From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction. Hence my correlation of GR theory to Newtonian physics, but please note I am not advocating a Newtonian preference. However, I did want to test Jim’s statement that `Newtonian predictions are wrong` rather than the usual implication that they were still a good approximation under most circumstances. Therefore, the point of my original equations (1-4) in #47 was to try to narrow the overall complexity, as alluded to in Garth’s comments in #48, to a very specific, and possibly over-simplistic, example in the hope that some physical interpretation might be more obvious.

[...]

Thanks for the clarification.

Not to derail the line of discussion you wish to pursue, I think there are some important clarifications to be made.

However, as a general comment, while the issues surrounding the perihelion of Mercury and bending of light support GR, the effects in absolute terms are quite small.

If, for purposes of teaching and learning, you choose to start with something like Newton (and history) and work your way to Einstein and GR, then this is fine. However, as GR is more general and as it has passed all the experimental and observational tests (to date), it is just as sensible to start with GR and derive Newton as a good approximation in the appropriate limit.

From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction.

Hear hear!

For some, the math and the abstractions are the side-show; the experimental results and observations must always been seen as the main show.

 

[mysearch]

Response to #64

Nereid: Thanks for the response. You raise some useful points against which I would like to make some general comments, as I assume that many, like me, must come to the PF to get some help in understanding the science that underpins the standard models.

#64: If, for purposes of teaching and learning, you choose to start with something like Newton (and history) and work your way to Einstein and GR, then this is fine.

In practical terms, most people become acquainted with the principles of physics through Newton’s laws of motion and gravitation. Therefore, this is an implicit starting point, for most people, which normally precedes any introduction to SR/GR. I also assume that it is an implicit postulate of relativity that in the absence of any near light speeds or massive gravitational presence, its equations should reduce to the Newtonian approximations. Finally, there is also a sense of Pareto’s (80/20) rule that suggests that you can learn 80% of something with 20% effort. By this rule GR plus all the maths will demand 80% of my time, in order, to come closer to some overall understanding of the detailed arguments supporting the standard model(s) of cosmology.

#64: However, as GR is more general and as it has passed all the experimental and observational tests (to date), it is just as sensible to start with GR and derive Newton as a good approximation in the appropriate limit.

As stated, I am not suggesting a preference for a Newtonian model and fully accept the weight of authority that supports GR. However, for the reasons above, I not sure that getting to grips with GR and its apparent dependency on differential geometry and tensors always seems the most “sensible” way forward, although I now accept the necessity and have started to work my way into these topics.

#53: From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction

#64ear hear! For some, the math and the abstractions are the side-show; the experimental results and observations must always been seen as the main show.

I fully concur with your statement, i.e. #64, in the following sense. I accept the need that has led to the abstraction of the mathematical notation used in GR, however, the conclusions via this approach should still be expressible in plain English and correlated to “experimental results and observations”. By way of example, I have tried to follow many of the threads discussing the curvature and expansion of spacetime. However, I am still struggling to get a clear physical interpretation of the various arguments being put forward, i.e.

- Measurements suggest that k = 0 implying little to zero space curvature. Does this not suggest that space is flat? If so, what other curvature is implied?

- On the very large scale, the universe is said to be homogeneous. As such, it is assumed that the universe has no overall centre of gravity. If so, does this not suggest that any net gravitational effects must be localised anomalies of mass-density, e.g. galaxies?

- As a very simplistic reduction of relativity, might we look at the effects of the relativistic factor $$[\gamma]$$ due to velocity and/or gravity? Do the following equations not suggest little overall effect until the relative velocity approaches [c] or a mass object approaches the Schwarzschild radius of mass-density corresponding to a black hole?

[1] velocity $$\gamma_v = \frac{1}{\sqrt(1-v^2/c^2)}$$

[2] gravity $$\gamma_g = \frac{1}{\sqrt(1-Rs/r)}$$

- Again, if the homogenous mass-density, inclusive any dark matter, is aligned to current estimates and there is no overall centre of gravity, how does gravity, and therefore GR, affects the net curvature of the universe?

- There appears to be conflicting comments over the real nature of any expansion of space. I have include a quote from Garth from another thread purely as any example, as he is also contributing to this thread and may wish to explain the level of “contention”:

The only reason for saying that "space expands", which is itself a contentious expression, is that the FRW predicts it to do so, that is cosmic expansion is one solution of that metric when GR is used to determine the scale factor a(t) and curvature parameter k.

- If the expansion of space is not to violate SR, how does $$[\gamma_v]$$ have any overall affect on the curvature of space or spacetime?

- On the basis of k=0, the FRW metric appears to reduce to the form below. However, while this metric alludes to an expansion, does it make any other statement about curvature other than [k]?

$$- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2}\right) = - c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2$$

However, I fully accept that I still have a lot to learn on this subject, so will get back to my teach-yourself maths courses. However, would welcome any other comments not encoded in tensors

 

[MeJennifer]

Mysearch you cannot apply gamma to curved spacetimes except locally. Note also that special relativity is not compatible with spacetimes that have a non-zero cosmological constant.

 

[mysearch]

Mysearch you cannot apply gamma to curved spacetimes except locally.

Now if you could just explain the physical interpretation of your definition of curved spacetime. I am really not tying to be facetious.

 

[MeJennifer]

Frankly I do not see how physical interpretations of spacetime (curved or not) are going to be relevant here. Fact is gamma does not apply to curved spacetimes except locally.

 

[jonmtkisco]

Hi Kev,

Are you suggesting that the rest mass energy of a falling particle changes as it falls so that the potential energy of the gravitational field does not have to be invoked to conserve energy?

GR treats a particle's energy as a unified whole, and it is not possible to divide that energy into the Newtonian components of rest mass, kinetic and potential energy. If the particle begins at rest at an infinite distance from a planet, its total GR energy is simply equal to its rest mass. Then as the particle begins to accelerate gravitationally toward the planet, that total energy does not change, as judged by a distant observer. The falling particle does not gain energy from the gravitational source. Energy of mass and motion is an invariant in GR. When the particle collides with the planet, its rest mass is simply added to the planet's rest mass.

As you mention, it is important to recognize that a stationary local observer (say, standing on the surface of the gravitating planet, or orbiting around it), does NOT measure the plunging particle's energy to be invariant. This stationary observer's clock runs at a different rate from both the distant observer and an observer falling alongside the particle, so the calculated energy of the particle is different. If she doesn't adjust for the clock difference, the stationary observer will judge that the total energy of the particle increases as it falls.

I'm not going to comment on your equation because math is not a strong suit for me!

Jon

 

[MeJennifer]

If the particle begins at rest at an infinite distance from a planet, its total GR energy is simply equal to its rest mass. Then as the particle begins to accelerate gravitationally toward the planet, that total energy does not change, as judged by a distant observer. The falling particle does not gain energy from the gravitational source. Energy of mass and motion is an invariant in GR. When the particle collides with the planet, its rest mass is simply added to the planet's rest mass.

That is actually not true under GR. Two masses of 1kg closing in on each other due to gravitation will slowly decrease their total rest mass. Except when they are separated infinitely far from each other their total mass will be less than 2kg.

Also the "stacking order" or matter matters. For instance if you have a set of Lego blocks in space far removed from any gravitational influences then their total mass depends on how they are stacked.