How does a body "know" how to move??!!

[Strange]

 

Well, if the Universe is infinite in space, that would stand to reason that every possible outcome for every piece of matter is also possible.

Apart from those that are impossible.

[AbnormallyHonest]

 

Is "every piece" meaningfully labelled though? Especially if continually being destroyed and recreated as you said earlier...

 

Well, entanglement could be a form of "labeling". Our awareness is the "meaningful" because it is only aware of one probability over infinite possibility. Uncertainty allows for this, but not for coexistence for extended periods of time. That's why you and I must agree on the state of things... the joint waveform.

Apart from those that are impossible.

 

The only thing I could even imagine as being impossible, would be to take an object, and be able to move it outside of our perception. e.g. Outside of the observable Universe. Unfortunately, my mind only sees a gap in mathematical predictability, not possibility. Outside that boundary, everything is evenly distributed probability, and anything again becomes possible. Even the coexistence of the same "piece" of matter for an extended period of time, because it is not subject to our awareness.

[Strange]

The only thing I could even imagine as being impossible, would be to take an object, and be able to move it outside of our perception. e.g. Outside of the observable Universe.

 

 

Things move out of our observable universe all the time.

[AbnormallyHonest]

 

 

Things move out of our observable universe all the time.

 

They move out of it? Or do we lose perception of it? Our view of the Universe is shrinking, so yes, you are correct.

[Strange]

 

They move out of it? Or do we lose perception of it?

 

 

They move out of it.

 

 

 

Our view of the Universe is shrinking, so yes, you are correct.

 

Our view of the universe is not shrinking. Stop posting ignorant crap.

[AbnormallyHonest]

They move out of it.

 

 

 

Our view of the universe is not shrinking. Stop posting ignorant crap.

Honestly, if given a moment, you might find that it is far from ignorant. If space expands from all points evenly, the farther away two points are displaced from one another, the faster the expansion between them. Is it not as probable as an event horizon, that at a great enough distance, the rate of expansion between them would in fact exceed "c"? An "event horizon" of another sort. Edited April 19, 2017 by AbnormallyHonest

[Mordred]

The particle horizon or you may also be referring to the cosmological event horizon. Neither of which has anything to do with the OPs post

[steveupson]

I imagine the question in the OP could easily be restated to ask how all the particles in the universe "know" that they are supposed to expand. On-the-nose from my understanding of the OP.

[Mordred]

I imagine the question breaks down to particles and objects don't know...after all how can something with no consciousness know anything???? They simply respond to the environment via the principle of least action for movement.

 

The principle of least action has been mentioned before on this thread.

Edited April 19, 2017 by Mordred

[steveupson]

And somehow the particles know how to determine what least action is, and then they know to follow this principle. The question seems to be "by what mechanism" does this happen?

[Mordred]

I fail to understand the complication. A boat in water will follow the current and flow to the lower pressure eddies.

 

Principle of least action is similar in terms of potential and kinetic energy. Though the analogy isn't exact

[steveupson]

The boat knows the water is flowing because of friction. How does the particle know how spacetime is being bent? What could constitute the friction in your analogy?

 

Every particle occupies a position in spacetime and every position in spacetime has a future. The manifestation of force can most simply be seen as a change to these futures. Gravity changes the velocity (combination of speed and direction) of spacetime locally. This is no different from the way acceleration occurs. In either case, the futures of events are modified in a systematic process.

 

The process can be seen (mathematically and conceptually) as beginning with a change in the direction of the future of each event. This change occurs differently than the way we would normally view a change in position. We normally view a change in position as involving movement that can be expressed using the base quantity of length, in the base units of meters.

 

When we view a change in position in this manner, we have no way of understanding what happens instantly upon application of a force. All we can consider is what happens during an infinitesimal. This limitation is caused by the way we do the math. There has to be another method of doing the math where we can see how the force acts instantaneously, or synchronously.

 

The other method expresses positions in Euclidean 3-space (events in spacetime) as a triplet of directions instead of a triplet of lengths. When anything changes in this model, then all the directions specifying every other position or event in the laboratory frame must also change in a synchronous manner, both mathematically and conceptually.

 

When you look at the illustration from nist, the diagram contains a derived quantity in the lower right hand corner that they call a radian, which is the unit of a plane angle. The correct representation would place the radian as the second SI base unit, and it would have its own dimension.

 

The way it's currently done, the radian is the cheater’s way of representing a turn. A turn in Euclidean 3-space is different than simply a ratio between two lengths. The same problem exists with representing volume as a product of three lengths. This assumes that the three directions are “orthogonal” to one another. What does that even mean without a base quantity of direction? Orthogonal, but without a unit of direction? That’s the non-sense that I see.

 

Maybe you understand this question better than I do, but I don't think that's the case here. I see things much too clearly to accept your argument. I know we disagree, but nothing that I've ever claimed to be true is in any conflict at all with state-of-the-art accepted scientific theory. As a matter of fact, it answers some questions that have been around for a long time.

[Mordred]

Think mass density then correlate that to the formulas used in relativity. If your on the right page this is what is described by spacetime curvature. The stress tensor in the Einstein field equations include mass density, flux and vorticity which are hydrodynamic terms.

 

If you think about it the main difference between water and what everyone calls spacetime is simply density. Though in the latter case it is field strength/energy density.

 

Throw a ball it will follow in essence the path of least resistance just like a boat in water or electric charge flowing through a circuit.

 

The ball must still have sufficient kinetic energy compared to the gravitational potential energy to maintain an elevation aka escape velocity.

 

As far as the infinitesimals this is important for accuracy. For example that thrown ball will not maintain a smooth arc. If you examine it close enough the path is actually jagged and only approximates the smooth flight path you see with the naked eye.

 

Here you may find this interesting its Arxiv.

 

"Relativistic hydrodynamics"

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=https://arxiv.org/abs/gr-qc/0603009&ved=0ahUKEwjX_p6jlbHTAhVY9GMKHdUaC0kQFggaMAA&usg=AFQjCNEtth45nhE3LhiqBB4d47GuHhLigA&sig2=UftLTG0IgJCijoIEFWSKIQ

 

an applicable line from that article.

 

"The equations of fluid motion are then deduced from the local conservation of energy and momentum in Sec. 5. They are given there in the standard form which is essentially a relativistic version of the Euler equations."

Edited April 19, 2017 by Mordred

[steveupson]

I’m pretty sure that I understand the principles (math and physics) that underlie this approach. The article is very good, and I think I understand the gist of how this approach is used. It is a much more straightforward explanation than what I’m used to. I feel as if we’re finally on the right track here.

 

When you use the term spacetime curvature, I am assuming you mean the gradient between adjacent vector spaces. The way the math is accomplished this gradient is expressed as a relationship between adjacent points of a manifold, or at least that’s how it looks to me.

 

The problem with this approach is that although it may connect space and time in a relativistic manner, Euclidean 3-space has a previously unknown relationship that exists between the spatial coordinates. This is due to them being mapped orthogonal to one another. The circumstance that is caused by them being mapped orthogonal is not differentiated in the method that is being used, and therefore it is lost. The space in spacetime that is being represented in this manner does not reflect real Euclidean 3-space.

 

If you were to open any book on spherical trigonometry, all of the mathematics is based on the function given in the above post. The function is what makes Euclidean 3-space work with time to create spacetime. It explains how the curvature actually manifests, and how it is able to wrap back on itself.

 

Since the relationship is expressed as a two-dimensional curve, I don’t know how it would be possible to differentiate it. I know that there has to be a way to show how Euclidean 3-space is slightly different than the way it’s being represented in relativistic terms. It has to be different than the approach of using 1-forms. I’m presently convinced that it involves calculating Bizarro vectors where coordinates are expressed as a triplet of directions. I don’t see any other way to proceed with this line of advancement, but then, I’m not a mathematician, either.

 

[Mordred]

Very good your getting it now, Spacetime is best treated as a vector field when you have an attraction or charge. However one can use scalar fields as well. The difficulty many have in picking up GR is the need for vector calculus. For example when you deviate from Euclidean flat geometry we can use techniques you may already be familiar with. Divergence, gradient and curl. This is used also in electrodynamics so you may already have some familiarity with it.

 

Rather than try to give a course into this I will supply a useful article

 

http://www.eecs.ucf.edu/~tomwu/course/eel6482/notes/Edition3_ch01.pdf

 

The matlab sections are useful if you happen to program if not just ignore those parts its the mathematical methods that are important. Other highly important study is Greenes, Stokes and Gauss theorem as they are involved in the above.

 

http://www.math.brown.edu/~deigen/labs35/lab9.pdf

Edited April 26, 2017 by Mordred

[steveupson]

Yes, exactly. The mathematical methods are important. The issue that I'm talking about has nothing at all to do with what you think it has to do with.

 

If you cannot understand me when I try and explain the math that I'm talking about then at least let me know what you think the math is telling you.

 

What do you make of the expression:

 

[latex] \alpha={\cot}^{-1 }(\cos\upsilon\tan{\sin}^{-1}(\frac{\sin\frac{\lambda}{ 2}}{ \sin\upsilon})) [/latex]

 

If this expression has no meaning to you, then you won't understand a word I've said.

 

It is impossible to understand unless you do the math.

 

on edit>>>>>

 

In my childish way of speaking, the expression is a characteristic of orthogonality. This characteristic is not captured by the vector calculus that you are trying to explain to me. In my childish way of speaking, the reason this characteristic is not captured by the vector calculus is because the characteristic relates all three orthogonal directions synchronously, in a single expression. It's very similar to the mathematical distinction between finite rotations and infinitesimal rotations. This is a distinctly different characteristic and it hasn't been noticed until now. You cannot find this in any of the literature because it's new.

Edited April 26, 2017 by steveupson

[Mordred]

How about not very useful in understanding the mathematical methods used in GR. That is what I am trying to show you. You wish to use directions instead of length but then the complication comes into play when you deal with coordinate systems and the symmetry group layout. I don't think you fully understand the problem.

 

Take your symmetry groups and examine them. There is a very useful and refined structure that is later incorperated into the S0(3,1) Lorentz group. Go ahead try to define your four momentum and four velocity with the line element

 

[latex]ds^2=(dx^0)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2[/latex]

 

(please note this equation is a direct translation of Pythagorous theory applied to our coordinates)

 

or better yet apply it to the equation of motion of a particle in a gravitational field with geodesic equation

 

[latex]\frac{d^2x^\mu}{ds^2}=\Gamma^\mu_{\lambda\nu}\frac{dx^\lambda}{ds}\frac{dx^\nu}{ds}=0[/latex]

 

there is an elegance in use of tensors and matrixes coupled with group theory

 

[latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex]

[latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex]

 

for example look at the above. the matrix literally follows the equation. and organizes it into a row column basis. Side note this is an orthogonal matrix you can identify it by its diagonal layout.

 

so how does this work with the boosts and rotations.

 

lets first show a boost in the x direction.

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex][latex]=\begin{pmatrix}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&o&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

now a boost in the y direction

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&-\beta\gamma&0\\0&0&1&0\\-\beta\gamma&0&\gamma&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

in the z direction

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&0&-\beta\gamma\\0&1&0&0\\0&0&1&0\\-\beta\gamma&0&0&\gamma\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

now stop and think about the coordinate ct which gives your length with the appropriate length contraction. remember the inertial frames will not have identical coordinate frames and we need to handle the frame transforms. We use pythagorous theory relations to reflect this in GR.

 

the above three boosts can be conpactly written as

 

[latex]\acute{x}=\Lambda(v) x[/latex]

 

the problem gets further complicated when you get acceleration which generates rapidity.

 

[latex]\epsilon^\phi=\gamma(1+\beta=1+v/c[/latex]

 

which gives Lorentz transforms as

[latex]ct-x+\epsilon^{-\phi}(\acute{ct}-\acute{x})[/latex]

[latex]ct=x+\epsilon^{-\phi}(\acute{ct}+\acute{x})[/latex]

[latex]y=\acute{y}[/latex]

[latex]z=\acute{z}[/latex]

 

which is a hyperbolic rotation of your IF frames.

 

[latex]\gamma=cosh\phi=\frac{\epsilon^\phi+\epsilon^{-\phi}}{2}[/latex]

[latex]\beta\gamma=sinh\phi=\frac{\epsilon^\phi-\epsilon^{-phi}}{2}[/latex]

and therefore

[latex]\beta=tanh\phi\frac{\epsilon^\phi-\epsilon^{-\phi}}{\epsilon^\phi-\epsilon^{-\phi}}[/latex]

 

which in matrix form is

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}cosh\phi&-sinh\phi&0&0\\-sinh\phi&cosh\phi&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

 

Are you starting to see the problem with trying to use just directions instead of distance?

1) the length of your ct coordinate defines your length contraction

2) the tensor organization system is a series of boosts and rotations that are organized according to vector symmetry relations including the inner outer dot products and tensor forms.

3) we are dealing with a coordinate independent system these relations work equally well in any coordinate system. Not just Euclidean flat but also with curvature

4) acceleration is handled as a rotation while velocity is your boosts (Lorentz transforms are under constant velocity coupled with relativity of simultaneaty)

5) GR is a coordinate metric the tensors allow us to be independent upon a particular set of coordinates

 

[latex]G_{\mu\nu}[/latex]

 

one can easily translate all of the above to polar, Cartesian or cylindrical coordinates the above relations will work in all three.

 

Hows that for proper mathematical methodology?

Edited April 26, 2017 by Mordred

[steveupson]

Why don't you think it's useful? Can you explain why I'm wrong? The math is either correct or incorrect, isn't it? Show me the error. Explain the error to me.

 

And, I don't know how to do any productive calculations using this new information. All I can do is show how and why the geometry is not correct when you try to do the math the way you are describing. It's an order of magnitude more complicated than that what you are doing.

 

[latex] ds^2=(dx^0)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2 [/latex]

 

In this expression, the last three coordinates no longer contain the relationship that I'm talking about. The reason is obvious to me, but for some reason you cannot see it. Basically, when there is no curvature, the sine function is the dominate (or operative) setting for expressing physical phenomena that involve Euclidean 3-space, and so the expression is fine for describing Newtonian spacetime because in that case the first element doesn't vary. When curvature is maximum then the hyperbolic function dominates. This has to happen when you mathematically start to vary all of the components. Of course you have to understand the equation to understand why this is true.

 

I'm sure you must have had basic trig somewhere in your studies. How do you explain a smooth function that maps the angles from 0°as a sine to 90° as a hyperbola? If you haven't seen this done before, why haven't you? And if you don't think it is significant that there is a way to map spherical excess as a curve, then why do you think it's not important?

 

The physics we're trying to model is nature. The model that I'm talking about will do everything you want, and more. I just don't have the algebra skills necessary to communicate the math.

 

You may recall that when we first started this discussion we were at odds because I couldn't show the algebra. Now I've shown the algebra. It was an effort. It would be nice if someone, anyone, would take the time to do the math and comment on the math.

 

 

 

 

on edit >>>> Of course I see the problem. And I don't know the complete answer to it yet. The thing is, just as using a triplet of lengths yields a direction, a triplet of directions yields a length. http://www.scienceforums.net/topic/101726-formalizing-length-as-an-antivector-algebra-help-requested-once-again/page-2#entry986027

 

 

 

second edit >>>> Not to be snarky or anything, the mathematical rigor that I'm looking for is someone to look at the equation I've presented and then comment on what they think the equation shows. For me, I see the equation as a partial invalidation of your statement 3) in that if the coordinate system is to remain orthogonal then it doesn't work the way you think it does. If orthogonality isn't a requirement, then sure, your way works fine, but it isn't doing what you think it's doing.

Edited April 26, 2017 by steveupson

[Mordred]

Orthogonal matrix.

 

https://en.m.wikipedia.org/wiki/Orthogonal_matrix

 

please I know what I am talking about and I know your definition on orthogonal axis. ie at right angles to each other. The problem is GR rotates those axis.They lose orthogonality when you apply length contraction in the direction of motion.

 

Yes I do recognize your trigonometric identities. Its not as useful as you think when the very geometry deviates from Pythagoras norm.

 

Or for example a commoving volume coordinate system ie expansion in the FLRW metric.

 

Tbe point your missing is you need to show the changes in the metric geometry itself as well as the particle path.

Yes I am familiar with what you have been attempting from this post.

 

http://www.scienceforums.net/topic/101726-formalizing-length-as-an-antivector-algebra-help-requested-once-again/page-1

Edited April 26, 2017 by Mordred

[steveupson]

A right angle isn't simply a condition where Pythagoras holds true. I mean, of course that's all there is to it in 2-space. In 3-space it's more than that because we can express the orthogonal angles as a interrelationship between all of the angles that make up the space. This description or definition will hold true both locally and globally.

We don't need any length or what you're calling a metric in order to define orthogonality. The relationship that is expressed in the equation defines all angles in the system as a function of the orthogonal axes. And it makes all of the angles in the entire coordinate system relate to one another. This is independent of length or metric. It uses angles to relate all the angles to one another.

In other words, the x, y, and z coordinates of a position should be able to be expressed as three directions and thereby represent a length that has no particular orientation (other than being positive or negative.) The radius of a sphere is a length like the one I'm talking about that has no particular direction. It's the exact inside-out version of using three lengths to define a direction that has no particular length.

One more time, the best way to state this is to say that the equation expresses the relationship of all angles relative to orthogonality at once, or synchronously. I've spent some time studying the implications and I sure wish you would take a closer look.

[Mordred]

I will take a closer look at your Bizarto vector but the fact remains it has little to do with the Op of this thread.

 

The thread is specifically about "How does a body know how to move."

 

Trying to utilize your formula into this has no bearing I can see on that topic. There is a logical reason why the related formulas I have been posting work. They are all used in describing kinematics.

 

For example why would you express a coordinate point as three directions?

Where is the possible advantage of doing so when the coordinate system itself contains all the information on the system state.?

 

For example I could just as easily use a direction and a radius from origin to describe the new coordinate but then I lose the details on the geometry change itself.

 

Ie the change of coordinate scale in the direction of motion. aka length contraction.

 

Does your Bizarro vectors work in 4d? is the other problem. Our coordinate system is

 

ct,x,y,z. It is not 3d. We need the time coordinate to keep track of variable time.

 

How does your Bizzaro vector describe this?

Edited April 26, 2017 by Mordred

[steveupson]

To explain how this pertains to the OP, it is necessary to return to first principles. Beginning with Newton, I'd direct your attention to the paper on the Origins of Torque that was linked in post # 97.

http://www.scienceforums.net/topic/99258-how-does-a-body-know-how-to-move/page-5#entry950396

This is important because it tries to provide an explanation of how force germinates. Using their approach, they show that when we try and model a rigid two-body it becomes difficult to make the math work properly, based on our normal use of the definitions of vectors and geometry and such.

"... We conclude that rigid bodies are inconsistent with

classical (non-relativistic) physics. When we allow for nonrigidity,

the model becomes consistent with all of Newtons

laws, and the correct equation of motion [Eq. (10)] is then

obtained straightforwardly. Moreover, considering the

effect of non-rigidity on the tangential components of the

internal forces makes evident the physical importance of

the torque."

In order to resolve the conflicts in the math, they assume that there must be some deep connection that doesn't allow for rigid bodies to occur in nature. This is a solution to their problem, and it's probably the most obvious, although I don't believe that it's the most elegant. There's another way to look at this question.

Before I go too far into this, I need to reiterate that a change in position can be viewed as a change in either distance or direction, or both. This is important because if we were to disconnect distance and direction from each other, then we can come up with a plausible scenario where F=ma can be explained a little differently than using this non-rigid body approach.

The acceleration is the rate of change of velocity, and the velocity is the rate of change of position, or, the rate of change of direction or distance, or both. Again, if we can break the mathematical connection between direction and distance, and if we can then tease out the direction from the distance, then we can have a change in position that is solely a change in direction, without any changes in any lengths. I know, it sounds odd, but it's only math. We've always seen direction as a purely ratiometric quantity, but that's only because its the way we've always done the math.

What makes this relevant is that direction does parse out differently when we use ct,x,y,z coordinates. If we have a point mass at xyz at ct, and its still in the same location in space at ct' (or, in other words, if xyz = x'y'z'), then the point mass is moving through spacetime in a defined direction. If a force causing an acceleration occurs, or if gravity acts to change the magnitude of t, then the direction that the point mass is moving through spacetime will also change, regardless of what happens to xyz. The least action mathematically is to have the force (or gravity) cause only the change in t, and then this change in t is what results in the acceleration.

This way of viewing how a body in motion knows to stay in motion leads to an observation that everything has a future that lies in a precise direction, and that this direction is all that needs to change, mathematically, in order to achieve the relationship F=ma. What makes this look like nonsense is only the fact that we have always dealt with direction as a ratio and not a quantity. Once it is recognized as a base quantity then it is a simple matter (mathematically and conceptually) to separate direction out from distance. This distinction has no relevance at all until it is applied to the spacetime construct. Directional relationships in Euclidean 3-space take on a different importance when they are combined with the direction of another quantity, which in this case is the quantity time.

In another discussion at another site, the following explanation was produced, utilizing the rotational equivalent of the third law. I hope it's self explanatory as to how this relates to this question.

Note that if you work under the rotational equivalent of Newton's third law, which is "if a torque is applied to an object and its rotational acceleration doesn't change, there exists an equal and opposite torque" we can assert that the force [LATEX]T[/LATEX] at [LATEX]m_1[/LATEX] gives a different torque than the force [LATEX]T[/LATEX] at [LATEX]m_2[/LATEX]. To resolve the issue we can either rescale the forces so that the torques match (which is effectively achieved in the paper with the relation [LATEX]T_1r_1 = T_2r_2[/LATEX]) or we can redefine the net torque as:



[LATEX]\tau = \tau_1 + \tau_2[/LATEX]



then



[LATEX]\tau = r_1 T + r_2 (F-T)[/LATEX]

[LATEX] = r_1 T + r_2 F - r_2 T[/LATEX]

[LATEX] = (r_1 - r_2) T + r_2 F[/LATEX]

[LATEX] = (r_1 - r_2) m_1 a_1 + r_2 (m_1a_1 + m_2 a_2)[/LATEX]

[LATEX] = (r_1 - r_2) m_1 r_1 \alpha + r_2 (m_1r_1 \alpha+ m_2 r_2 \alpha) [/LATEX]

[LATEX] = \alpha(m_1r_1^2 - m_1r_1r_2 + m_1r_1r_2 +m_2r_2^2)[/LATEX]

[LATEX] = \alpha(m_1r_1^2 + m_2r_2^2)[/LATEX],



which is the correct equation of motion. However, the rotational equivalent of Newton's third law is derived from the linear form. By stating the above I am actually dodging the perfect-rigid body assumption. To be consistent, we must recognize that in doing the above, we have implicitly assumed that the object is a rigid-body that is capable of transmitting force along its length. -- Benit13

http://mymathforum.com/physics/72236-physical-property-direction-7.html


A close look at the method that was used in order to dodge the rigid body assumption reveals:

[LATEX] = (r_1 - r_2) m_1 r_1 \alpha + r_2 (m_1r_1 \alpha+ m_2 r_2 \alpha) [/LATEX]

Since this introduces a change in angular velocity, and since angular velocity is the derived quantity that determines direction, can angular acceleration be viewed as a derivative of direction? We understand that the direction of the angular velocity vector doesn't change with a change in angular velocity, but is that really the case? Can the direction become more so? I don't think that there's any proof one way or the other on this. There's simply the fact that we've always done it this way.

Of course, the reason that we assume that there isn't a change in direction of the angular velocity vector could be caused by the way we apply Euclidean 3-space to spacetime. I don't think that there can be any difference between the two constructs in Newtonian spacetime, but it appears that there certainly could be a difference in relativistic spacetime. Most of the stuff that I obsess over has no real meaning in flat spacetime, but it could have significance when applied to curved spacetime.

As for your suggestion about using a direction and a radius for identifying a position, there's a more interesting use for combining a direction and a radius. We can use a direction and a radius to specify a volume, and this raises some other issues that have to be explained somehow.

The SI derived unit for volume is [latex]{m^3}[/latex]. But that's not the only way to specify a volume. If we know the radius of a sphere we can calculate the volume inside the sphere using a simple formula. But what happens if we take a partial area of the surface of the sphere and use this to calculate volume? The most straightforward method is to use the steradian (symbol [latex]sr[/latex]) as the partial area of the surface of the sphere. The steradian (see the SI chart in the above post) is referred to as the derived quantity for solid angles, but in this case its just as appropriate to look at it as the area of [latex]{r^2}[/latex] on the curved surface of a sphere.

The procedure is to take the volume of the space that lies within the steradian and the sphere center. To make this into a volume we'll take the product of the angular quantity of the [latex]sr[/latex] and the length quantity of the meter. Well call this unit of volume the steradian meter, or [latex]srm[/latex]. There are [latex]{4}\pi[/latex] steradians in a complete sphere, so there must also be [latex]{4}\pi[/latex] units of volume that will be equal to the volume of the complete sphere.

The first thing to notice is that the [latex]srm[/latex] can use either direction or length as the unit quantity. A quantity of [latex]3sr[/latex] where [latex]r=1m[/latex] is different than the quantity of [latex]1sr[/latex] where [latex]r=3m[/latex]. Mathematicians will probably know of a convention that can be used in order to capture this difference. If we put the number in the appropriate location to show which quantity is the unit quantity and which quantity the number is associated with, then the first case would be written as [latex]3srm[/latex] and the second case would be written as [latex]sr3m[/latex]. A little arithmetic reveals that:

[latex]1srm = sr1m = \frac{1}{3}m^3[/latex] and also, for example:

[latex]sr8m = 512srm = \frac{512}{3}m^3 [/latex]

If space is simply an empty volume, these three examples of expressing a volume should have identical meanings. They should all be identical, both mathematically and conceptually. The problem is that there is a difference. The expression [latex]\frac{512}{3}m^3[/latex] has no spherical excess, [latex]sr8m[/latex] has a spherical excess of [latex]\frac{\pi}{2}[/latex], and [latex]512srm[/latex] has a spherical excess of [latex]256\pi[/latex].

I guess it would be possible to pretend that this difference has no relevance at all. Once again, I think were up against the same mathematical quandary. The difference between using unit length or unit direction to express a volume has no impact at all on Newtonian spacetime where ct is a constant. But it does raise a whole universe of issues once we try to use these two different methods to express a volume in relativistic spacetime.

Some of this explanation (like the stuff about the direction of the angular velocity vector) is merely speculation at this time. The algebra hasn't been worked out for most of that yet. But enough has been accomplished to know that it won't be as simple as some of the stuff that has already been done. It would be necessary to show that gravity has an impact on the direction of the angular velocity vector and it looks as though this cannot be accomplished using conventional geometry. This is one of the reasons why a coordinate system that uses directions instead of lengths is being suggested.

It appears that there's a portal that exists between Euclidean 3-space expressed as a system of lengths and Euclidean 3-space expressed as a system of directions. We should walk through it.


On edit>>>> removed rant

Edited May 2, 2017 by steveupson