Theory of Gravity - I can't resist [Archive]

ZenMaster

Mar 14th, 2011, 2:00 PM

That it is apparently impossible for the universe to be flat segues into the train of logic I used to come up with an idea for my own theory.

BE FOREWARNED: I followed a few blind alleys. Those who *actually* know the science are going to say to themselves "Well that's dead wrong" in some places. You're right, I was dead wrong. However, please follow me down these blind alleys, because they did lead to inspiration on the correct way to go.

When I first heard about Dark Matter, my knee-jerk reaction was to call bullshit. A mysterious particle we can't see or detect through any method except the circumstantial evidence of gravity? Like the crackpots, I decided immediately that General Relativity must be wrong, even though I knew nothing about it. Almost as a game I started to consider alternatives.

My first step was to decide that if spacetime was curved in a way different from that predicted by Einstein, the career astrophysicists would have figured it out, so I decided spacetime was probably flat. (Besides, I decided flat spacetime was easier to play with .. I know, another crackpot sign.)

This presented the immediate, obvious problem of what to think about gravitational lensing (or any light bending, really), when a beam of light bends around a massive object, like a galaxy. The puzzle: how do you bend a straight line in flat spacetime? It sounds like an impossible riddle, and I played the game for a long time. I eventually connected with Special Relativity, back when my understanding of Special Relativity was poor.

Blind alley, stay alert - It occured to me that maybe the path of light itself *is* straight, but that the spacetime between the massive object undergoes some sort of length contraction. In other words, the radial distance from the center of mass to an individual photon shrinks as the photon nears it, like varying the size of the graph paper while drawing a line.

I brought this idea to a physicist at a local university and he told me that Special Relativity didn't work this way. I studied SR more and discovered for myself how it really does work, and why it can't be applied this way.

ZenMaster

Mar 14th, 2011, 2:05 PM

Alright so, I started learning more about SR and eventually gave up trying to use it to explain how to make light appear to bend. SR really has zilch to do with empty spacetime, and is a statement about the relative relationship between physical objects.

I put it at the back of my head, but the idea of dark matter still bugged me. I sent an email to another physicist (who I have since forgotten, unfortunately) about some of my thoughts about it. He was the one who explained to me that trying to build a theory that explains the observations of gravity *without* assuming the existence of dark matter is, let's say, hard. The gaps left when we pretend dark matter is non-existent cannot be waved away, or at least, not easily. If you tinker with GR, gaps appear in other places instead. The physicist, anticipating my question, suggested that we can't just add corrective terms or small tinkers to GR to make dark matter superfluous. You'd have to add infinite corrections, and that's no theory at all. It's like saying all snowflakes are THIS one shape.. oh except the next ... and the next.. and the next is different too.

This is a long way of repeating what I said earlier. If you assume dark matter doesn't exist but our observations of it are the result of a new kind of gravity theory, then the gravity theory is very, very NOT "simple". Another thought I put at the back of my head: if I'm going to build a theory of gravity, it has got to have some way of being "not simple".

A relevant aside / metaphor - sorry, you need to know a bit about logic:
Some time later, I was taking a class in (of all things) basic computer design. One of the reasons I got into it is because I love rigid logic. A and B and not C etc. However, this one class blew my mind. I won't describe in detail, but we learned about how the computer is driven by a clock, and is prevented from spiralling into chaos by "edge-triggered" logic gates. The part of it that blew me away is that the LOGIC of the computer is circumvented by the PHYSICS of the circuits. You can't really use the output of a logical statement as the input to that same statement; it makes no sense. However, the very small but finite propagation delay of a current connected from the output of an electronic GATE to the input of that same gate makes edge-trigger possible.

I later connected this idea back to gravity. You could get some pretty messed up results from a simple theory, if you started feeding your "outputs" into your "inputs".

What would this mean in the context of cosmology? Let's say the "input" is the actual state of the universe, independent of our observations, and the "outputs" are our observations. If the latter feeds into the former, then our "cosmological state" has an effect on what we observe. I almost immediately thought back to Special Relativity. My observations of you are intimately dependent upon our relative states.

Here I ran into conflict. Special Relativity does NOT explain gravity, but it felt like it ought to.

Cartesiantheater

Mar 14th, 2011, 2:08 PM

Well, you better hurry up and get back to it before the real crackpots clutter your thread with useless crap.

ZenMaster

Mar 14th, 2011, 6:49 PM

Since I couldn't really justify Special Relativity in the context of gravity, again I brushed it aside and went on with my life.

I casually read science magazines, and I started looking at Quantum Physics articles. Time passed. Gravity still gnawed at my brain sometimes. (It became a quest for a new theory of gravity rather than an attempt to disprove dark matter, which are rather the same thing I suppose.)

Heisenberg's Uncertainty Principle (HUP) is something you've heard of, even if you know nothing else about quantum physics. Here, I had an idea.

For light, it is universally true that dx/dt = c. The speed is constant. However, the *actual* position of a given photon at any given time is bound by HUP. I wondered if, maybe, dp/dt = some universal constant, possibly zero.

It would imply that Special Relativity has a quantum twin brother. This brother would deal with changes of momentum. Maybe the reason we can't FIND a graviton or a Higgs-Boson is because they simply don't exist! Gravity could be a special type of relativity, previously unknown. Ironically, it would act like an unseen hand, a "spooky action at a distance"! It would not be limited by the speed of light, in precisely the same way that you can't really talk about Special Relativity being bound by the speed of light. (SR *defines* the speed of light!)

More along this train of thought: Einstein described Special Relativity in terms of x,y,z,t, and corresponding variables in the other frame of reference. Maybe we need variables that are sensitive to quantum physics? x,y,z,px,py,pz. Momentum implicitly includes a concept of time, but I've come to believe we can eliminate time as an independent variable. This would certainly agree with Quantum Physics, in which all of the operations of the fundamental particles are time-reversible. I'll explain later how we can get the apparent passage of time if the universe is built on time-reversible operations.

I don't have any logical backup to defend the idea that time is an emergent phenomenon. I just believe it to be so, because Particle and Wave Relativity would become more symmetric and, in fact, two heads of the theory, which I would call Quantum Relativity.

Here's basically what I've come to believe over the last 15 years:
- GR is wrong
- the universe is flat; I also believe this because I have since discovered that Quantum Physics allows a flat universe to come into existence from nothing
- Special Relativity is "wrong" in the same way the Ideal Gas Law is wrong; meaning, the model behind it is not technically correct but it's close enough that it produces good numbers
- "Particle Relativity", similar to SR but sensitive to quantum physics, would be a better version of Special Relativity. It will deal with the dx/dp of one frame of reference versus the dx/dp of another frame of reference, relative to the speed of light.
- "Wave Relativity" would be a slow, heavy twin brother theory to the quick and speedy Particle Relativity. It will deal with dp/dx of one object versus another travelling in a different "impulse frame" (i.e. a frame with a different change in momentum over a given distance), relative to the dp/dx of light which is probably but not necessarily zero.
- Both Particle and Wave Relativity are two heads of a grand Quantum Relativity theory; the two theories represent extreme cases of the grand theory

Stuff I'm not certain about, but which I'm entertaining as possible truths:
- light travels in a straight line at constant speed in all frames of reference and all "impulse frames of reference" (i.e. frames of constant momentum)
- gravitational lensing isn't a mirage; the duplicates are genuine twins in your frame of reference. (Don't ask me how this is possible yet. I need to do the math.)
- time is an emergent phenomenon; I didn't explain this above, but it is something I've come to believe through thinking on this stuff. I'll have to try and prove it.
- mass is an emergent phenomenon. Something I've just started to consider. It really has nothing to do with my theory, but I think it would be neat to prove since it would split the Standard Model in half.
- an interesting idea though I'm totally unsure if it has any meaning, because I haven't studied Quantum Physics to any meaningful depth. If my theory of Wave Relativity = Gravity is correct, it would work on *relative* mass and therefore exist down at the level of Quantum Particles. I'm not sure about this, but I think that if you abandon the assumption that electrons must exist at fixed orbitals, their orbital levels can be derived using Wave Relativity and given the fact that energy comes in discrete quanta.

I'm going to conclude here, because this is pretty much the jumping off point.

That's it. Thanks for reading, folks.

ZenMaster

Mar 14th, 2011, 8:16 PM

I decided to flesh this out a bit more, after all, since I left the conclusion so vague.

What's my theory? Actually, I don't have one yet, only the "roadmap" to get to the general area where I think the correct theories can be found. Really, the only actual assertions I'm making as-yet are those I mentioned in the above post; i.e. that the theories Particle Relativity, Wave Relativity, and their parent Quantum Relativity all await being discovered, but haven't been found yet. Now that I have a break in my life, I'm going to crack open that Quantum Physics textbook I bought (http://www.amazon.ca/Principles-Quantum-Mechanics-Second-Shankar/dp/0306447908 .. check it out)

I make no claims on the form of the theories except that Particle Relativity should be similar to, almost a clone of, Special Relativity. I do also strongly believe we'll find that time is an emergent variable, though this is more based on my suspicion that these final versions of relativity will be strongly symmetric, rather than on any good logic. I think Particle Relativity will be a statement of the dx/dp of one object versus the dx/dp of an object in another frame of reference. Similarly, I think Wave Relativity will be a statement of the dp/dx of one object versus the dp/dx of an object in another "impulse frame of reference", and thereby define the effect we call gravity. (A different "impulse frame of reference" is the frame of reference of an object moving with a different momentum.)

ZenMaster

Mar 15th, 2011, 8:09 PM

I also meant to talk about Newton's First Law. This is where the meat is, and where I'll explain how I believe gravity actually works.

From http://www.physicsclassroom.com/class/newtlaws/u2l1a.cfm -

Newton's first law of motion is often stated as

An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

Let's reexamine this law, in light of my own idea on Wave Relativity, or in other words, the assertion that the path of light is straight in all reference frames of constant change in momentum with respect to position.

First I'll talk about why Newton's First Law is so very, very convincing.

The purpose of the law is to basically assert that the "natural movement" of an object, i.e. the mode of movement it would adopt in the absence of forces, is a straight line. All of the forces drop in power as you move away from the source and can be brought arbitrarily close to zero as the object gets farther and farther away. The path of the object, curved towards the source of the forces, gets less and less curved and so at the limit of infinity, the path becomes a straight line in space.

I won't debate this reasoning because I would fully agree with it .. IF gravity were a traditional force.

Here's the tricky subtle part where I diverge from Newton's First Law. Please Note: I'm going to build a special argument here. Debate the whole argument. Some of the premises may bother you so much that you'll complain just on the basis of incredulity. However, swallow your incredulity and debate rationally, please.

Alright, so, let's stop talking about non-local effects where the influence of gravity drops to near-zero. Let's put an observer and an object (observee?) close together, so that the forces could potentially influence the observer and object. Next, although the EM force, the weak force, and the strong force could cause an effect, let's choose the form of our observer and the object carefully so that they don't (or at least, too minimally to worry about). Let the observer and object be zero-dimensional, so that we don't worry about size or orientation.

The only remaining force is gravity and Newton's Laws. Next, let us assume that gravity does not exist. In this model, we just have the observer watching the object continue along a straight line on to infinity. (Maybe have the object flash at regular intervals. I won't discuss methods of observation, though I'll try to be very careful in setting up the models so such observational methods can in principle be set up without destroying the argument I'm trying to make.)

Now, a twist. Let us take the peculiar step of not assuming that the natural mode of movement is a straight line. (In other words, do not take Newton's First Law for granted.) In this model, although we can predict changes to a natural mode of movement if the other three forces were thrown out of balance, we have no natural mode of movement to begin with, so we're apparently at a loss.

ZenMaster

Mar 16th, 2011, 12:02 AM

Here I introduce the premise of Wave Relativity; it is a universal law that photons travel in a straight line in the impulsive frames of reference (i.e. those frames of reference of constant change in momentum relative to change in position) of any and all objects with mass. It may not seem obvious yet, but here we find that a natural mode of movement of non-light objects arises out of and hinges upon the fixed mode of movement of light.

Draw a point on some graph paper, name it "observer", then draw a straight line and name it "beam of light". Put yourself in the shoes of the observer and imagine you're observing another point travelling at a constant speed along that path. Let this photon start at the point on the line closest to you and then travel along the line off to infinity. Measure the distances between yourself and the photon at regular intervals, say, once per microsecond (or whatever is appropriate.. I haven't done the numbers). Clearly, the distance is shortest when the photon was closest to you and gets longer as the photon moves away, approaching the limit of infinity when your orientation is parallel to the direction the photon was travelling, at 90 degrees from your direction when you were looking at it in its starting position.

In this model, despite my graph-paper description, you have no sense of orientation since your only point of reference is the photon. Arbitrarily decide that the photon must be travelling in a straight line (per Wave Relativity) and the distances will magically create the line. (This is obvious though, since it is how we set up the model in the first place.)

Now, reintroduce the other object, the observee I mentioned. Let's start off with it being stationary to you, the observer. Put it on the graph paper in the diametrically opposite position from the photon, if the observer is the center of a circle and the starting point of the photon defines a radius. The line segment defined by <observer, object> should be exactly perpendicular to the path of light.

Ask yourself: how does the object view the path of the beam of light? You and the object have synchronized watches and decide to measure the distance to the travelling photon at the same regular times. (Your watches will remain synchronized over time since you're relatively stationary.)

You, the observer, can easily calculate the distance measurements that the object ought to be reading since you know the speed of light, the premise of Wave Relativity and the distances between you and the photon and between you and the object. You can also confirm the calculations during the experiment given the amount of time that has passed, and the angle (initially 180 degrees) between the photon and the object. Since spacetime is flat and you and the object are relatively stationary, there's no reason to think the object's measurements will differ from your calculations of what he ought to measure.

Note that if you and the object were both moving parallel to the path of light at the same speed, by the assumptions of Special Relativity, you couldn't tell the difference (and in reality there would BE no difference), so we'll skip this case.

Lets give the object a nudge in a direction parallel to the beam of light and in the same direction that the photon is travelling (or maybe you were the one who was nudged, again parallel, but in the opposite direction). We'll also hold firm to the rule that both you and the object must view the photon as travelling in a straight line at the same speed and work out the consequences.

ZenMaster

Mar 16th, 2011, 12:19 AM

Oh and, by "nudge", I mean that an acceleration is applied. Also, the nudge here must be "short", i.e. virtually instantaneous, so that it doesn't interfere with the consequences of the resulting natural movement. Saying that the object begins movement via a non-natural mode of movement is not cheating since we know from the previous scenario I set up that the natural mode of movement is for you and the object to remain relatively stationary today if you were relatively stationary yesterday.

In the scenario with the nudge, you still measure the photon to be moving along a straight path and, by the postulate I gave for Wave Relativity, so does the object from his own point of view. The real question is: given the postulate of Wave Relativity, how do you observe the object to move in response to a nudge?

Please assume for the next bit that I'm speaking in calculus terms, so measurements are all infinitesimal. (Just so I don't have to qualify everything I say.)

Due to the nudge, the object now has a momentum in the direction of the beam of light, relative to you. We're talking about fundamental modes of movement here, so let's keep Einstein in mind. Special Relativity suggests that an object with a relative velocity undergoes length contraction. As well as time dilation too.. oh that reminds me.

I vaguely associated time with momentum but forgot to mention: Particle Relativity would be almost a clone of Special Relativity, but backwards, in a way. With Special Relativity, a fast spaceship gets rotated out of length and into the dimension of time, so its clocks move more slowly. For Particle Relativity, a fast object gets rotated out of length AND out of momentum. Think of a grandfather clock on a spaceship: as time slows, the pendulum moves more slowly, and therefore its momentum is smaller. As the spaceship approaches the speed limit, it may be that time stretches out near-infinitely long, but it is equivalent to say that the momentum of the pendulum relative to the spaceship approaches zero.

Normally, Special Relativity is dealt with in non-calculus scenarios because it is only valid for uniform translatory movement.. i.e. no accelerations over long periods of time. However, just because it isn't normally used in a calculus treatment doesn't mean it can't be. (In fact, it would make no sense if it couldn't.) So, thanks to the nudge, over a tiny, tiny, time period, the object's relative velocity is non-zero. Its frame of reference suffers length contraction, in essence squeezing the graph paper in a direction parallel to the movement and the beam of light. For that tiny, tiny time period, the measurement you calculated is wrong. Taking Special Relativity into account over that time period, you would discover that the correct value is shorter, because the object's frame of reference is squeezed. If it were possible for the object to continue to undergo acceleration but never move, this effect would up over time and you will calculate that the path travelled by the beam of light from the perspective of the object will be a parabola wrapping down, around the two of you.

See Wikipedia's page http://en.wikipedia.org/wiki/Parabola. In the diagram of the parabola, you are the focus (assuming the initial distances between the photon, you, and the object were all equal.) The vertex is where the photon started. The actual shape of the parabola would depend on the final relative velocity achieved by the object after the short period of time.

ZenMaster

Mar 16th, 2011, 2:07 AM

When we last left the nudge model, with you, an object, and a photon which should travel according to the Wave Relativity premise, we ran into a problem. Left to its own devices, a nudge would cause light to travel in a parabolic arc, wrapping around you and the object, which violates the Wave Relativity premise. The parabolic path of the photon was calculated because we decided to take Special Relativity (Particle Relativity) into account. How do we fix the problem?

Before I continue, let me say that the path of the light beam is parallel to the X-axis, just to make it easier to talk about.

To solve the problem, we can't go back the way we came; i.e. have the object's space undergo length expansion and time contraction, since that really solves nothing. However, there is another path, very similar to Special Relativity. I'm rather proud of the trick, so excuse the smugness (though technically I cheated by gleaning inspiration off of Einstein's answer sheet.) We can have the object undergo length expansion, but this time, along the Y-axis, and a contraction of time (or equivalently, an increase of momentum).

To translate this: as the object is accelerated in this scenario, the space between you and it shrinks and it moves faster. This is getting damned close to looking like a theory of gravity, but we've only considered a very specific scenario with the object being accelerated by an external (non-natural) force, and you were all placed in specific spots.

The next question is: when does this effect apply? What are the preconditions? Let's start by switch the positions of you and the object. We still get the parabola effect, though it is lessened. This is interesting, because it means that we have to consider the space contractions along the axis from you to the object caused by Wave Relativity from two sources: your observations of it, and its observations of you. Next, let's throw our original line off at an arbitrary angle. We have to apply similar logic. Regardless of the angle of the line, we can drop a perpendicular from the line segment connecting you and the object, to the line defining the path of the photon. Though the photon may no longer start at the origin, its path must remain straight and our calculations will work out the same way.

We can draw lots of lots of lines. With the addition of these beams of light, the original calculations aren't complicated any more than defined above. We now have a workable explanation of how to build a theory of gravity, which I'm going to attempt soon. (Writing this all out on paper has made me discover that I might be able to build a rough theory by doing simple algebra .. maybe I won't yet need to study Quantum Physics after all.)

Some similarities between the Wave Relativity theory and Einstein's gravity:
- weakens over the square of the distance between the two objects (since we had to calculate the radial space contraction contributed by both, and as the distance increases, the "special relativity causing a parabola" effect is diluted)
- depends on the masses of the objects (since, if I'm correct, what's important is not relative rates of time, but relative momentums)

Some differences between the Wave Relativity theory and Einstein's gravity:
- the math for the simple formula would be on the level of Newton (though the correct quantum formulation of it might be tricky as heck)
- no arbitrary universal gravitational constant, only the constant speed of light and zero change in momentum
- light cannot bend or change colour, except through universal expansion

ZenMaster

Mar 16th, 2011, 2:48 AM

Some other idea, which may be total crap but is interesting:

Wave Relativity would describe gravity based on the relative momentums, and therefore masses and velocities, of objects. Surely this would fail at the level of Quantum Particles?

As I have said earlier, if you first abandon Newton's First Law and do not assert the idea that, say, electrons must exist at fixed orbital distances from a nucleus, we simply need to allow Wave Relativity to take over. Since objects at that scale MUST have discrete amounts of energy (owing to the fact that energy comes in fixed-sized quanta), then each will have a particular, fixed momentum. Relative to a nucleus, an electron moving in a straight line would appear to be accelerating, so it must naturally travel in a cozy path defined by a zero relative change in momentum. The electrons must also occupy only well-defined orbital levels that match some whole number of wavelengths, not out of sheer quantum arbitrariness, but because if the electron travelled in a non-integral number of wavelengths, on average, it would appear to be travelling with a relative change in momentum from the perspective of the nucleus. This, again, would not be natural by the Wave Relativity postulate.

Finally, I'm fairly certain that my model will show that it could not (easily) spiral into the positive nucleus, despite the charge. In my model, the natural mode of movement is not a straight line, but a pattern where the relative change in momentum is zero, such as a circle. Any other movement is considered a relative acceleration, which will cause the electron to speed up in the direction it is travelling and scoot back to the shell.

At least, maybe. I'm FAR far far less certain about all this stuff that I am about the simple existence of Wave Relativity and Particle Relativity (and Quantum Relativity).

I think I'm finally done talking now.

ZenMaster

Mar 16th, 2011, 3:51 AM

(1) One quick thing: Inflation models and such have more to do with things like the absence of magnetic monopoles than GR.

(2) One more quick thing: there are one or two galaxies that DON'T behave like the other ones, and in fact seem to act exactly like GR and Newton say they should, indicating either dark matter is real, the laws of physics are NOT uniform throughout the universe, or something entirely different is going on.

http://www.newscientist.com/article/dn13280-galaxy-without-dark-matter-puzzles-astronomers.html

But do continue, unless you really are scared we'll steal your theory.

(also, as far as "crackpottery" goes, simply having a pet theory doesn't fit the category. It's when you have to do things like claim that evidence was fraudulent or that there is a grand conspiracy among academia to support your pet theory that you start becoming a crackpot. I have my own pet theory, but it is completely unjustified at this point since I lack the math skill to flesh it out).

Thanks CT, I didn't actually know about the monopoles thing. Funny, the monopoles mystery seems entirely unrelated :-.

Also, yes, I'm aware of those quirky galaxies which our current equations seem to explain precisely.

As I explained above, my Wave Relativity is "not simple", and in this case that means we must consider our own state in the universe as well as the state of what we're observing. We are on a random planet in a random place in a random galaxy, and therefore have a random average momentum relative to everything else in the universe.

Our equations were built upon the assumption that our place in it has no effect on those equations.

I'm claiming it does. I'm gambling that those places where our current equations happen to match exactly have, by sheer coincidence, the same local conditions (i.e. random average momentum relative to everything else in the universe).

Thanks for pointing it out. By the way, I'd love to hear your pet theory sometime if you're willing to share. (I can't help you with the math skills to flesh it out though.)

calliope

Mar 16th, 2011, 3:59 AM

ZenMaster

Mar 16th, 2011, 4:01 AM

Awesome thesis ZenMaster...i'm trying to work my way through it, but some of it's like a foreign language. I really loved my high school physics professor....he loved his science so much, the energy in his room was so calm, like a heightened sense, in there.

I really did try to keep up, because I found it all so beautiful and fascinating...like being in a different plane of existence.

That's how I feel in this thread. :cool14:

Wow, thank you Calliope :)

I wasn't kidding when I said I'd been building this in my head for 15 years (didn't I say that somewhere?) so it's become really personal. It means a lot to me that you said that. Thanks again.

By the way, I know I can ramble incoherently sometimes. Let me know if I can clarify anything.

ZenMaster

Mar 16th, 2011, 4:02 AM

I'mma put my name here, I think. Reveal my identity. Just in case there's something to my theory. Dan Blain

Nice to meet everyone. I'm REALLY going to bed now.

Cartesiantheater

Mar 16th, 2011, 7:58 AM

Thanks for pointing it out. By the way, I'd love to hear your pet theory sometime if you're willing to share. (I can't help you with the math skills to flesh it out though.)

Maybe one day when I really think about it like you have yours.

But there is one big problem with that: I like Brian Greene's ideas about brane theory a lot more than I like my own pet theory. ;) :noevl:

Ridiculous, no?

ZenMaster

Mar 16th, 2011, 3:48 PM

Maybe one day when I really think about it like you have yours.

But there is one big problem with that: I like Brian Greene's ideas about brane theory a lot more than I like my own pet theory. ;) :noevl:

Ridiculous, no?

lol, maybe not ridiculous. Despite the fact I'm firmly convinced of my own pet theories, I still love reading about alternate theories of gravity .. quantum loop gravity, and so on.

There are some absolutely fascinating theories out there.

ZenMaster

Mar 17th, 2011, 9:24 AM

I've realized two things from writing all of this down:

(1) I can make my earlier model better. If you make the nudge in a direction collinear with the line segment between you and the object (perpendicular to the beam of light), then you can restrict yourself to using only Special Relativity to come up with the same results. I no longer have to assume some sort of perpendicular length expansion is going on; I only need to stick to the rule that light travels in a straight line in all frames of reference. Less assumptions + same mathematical results == approaching bare bones truth, maybe.

(2) I really need to build the equation, now that I know it can be done. This algebraic version of gravity still won't be fundamentally correct (since I'll treat position and momentum as separate, in the same manner that Special Relativity ignores the HUP) but at least it should produce correct answers. I might be able to finally kill dark matter with this equation.

calliope

Mar 17th, 2011, 9:28 PM

Wow, thank you Calliope :)

I wasn't kidding when I said I'd been building this in my head for 15 years (didn't I say that somewhere?) so it's become really personal. It means a lot to me that you said that. Thanks again.

By the way, I know I can ramble incoherently sometimes. Let me know if I can clarify anything.

Very Nice, Master Zen. ::):

Thus far, you have a good way of expressing your ideas -- interesting and accessible. :2thumbs:

Cartesiantheater

Mar 18th, 2011, 5:39 AM

Let me honest and say that I HAVEN'T read all of this, and that honestly I don't really have the time to invest in seriously studying your theory since I have other theories that I'm graded on.

However, I have a question regarding Newton's first law.

It seems to me like your theory both abandons Newton's first law but also embraces special relativity. If I'm wrong, please correct me.

But if not, you have a sort of contradiction on your hands, because, if you weren't aware, one half of the two original postulates of special relativity IS IDENTICAL to Newton's first law.

The phrase is usually "the laws of physics are the same in all inertial reference frames," but it translates to this: "there is no physical test* you can do that will distinguish between inertial reference frames." Which is also identical to this: "there is no physical test that you can do that can distinguish between a frame moving with constant velocity and a frame that is at rest." If you really understand Newton's first law, you will realize that the above is equivalent to it. How? Because the heart of Newton's first law is that motion is relative. In fact, it is Galileo's relativity, originally, that led to Newton's first law. How can an object continue moving indefinitely without some propulsion behind it? Simple. The object itself is simply an inertial reference frame. From ONE perspective it is moving with constant velocity v. From another, it is moving with constant velocity u. From yet another it is as rest, with velocity = 0. And so on. It's constant velocity "motion" is frame dependent, and that is exactly what the cause of its uniform motion is. That is the heart of Newton's first law (hopefully, the readers will note that "inertial reference frame" and "law of inertia" have the word "inertia" in them for a reason).

Or let me put it this way: 1st law: A body in uniform motion will continue to be in that uniform motion and a body at rest will continue to be at rest unless acted on by some force. What does "at rest" mean? Say some object is moving with constant velocity. But we can choose a reference frame in which the object is at rest. Bam! The object is NOT in uniform motion, it is not moving! Whether an object is moving uniformly or at rest is frame dependent, which means that the form of any physical laws regarding the object will NOT depend on the object's motion <=> Newton's first law.

So, basically, you can't have special relativity without Newton's first law. Additionally, if Newton's first law does not hold, then space and time are neither homogeneous nor are they isotropic.

So, is that even relevant? As I said I have not read everything nor studied your theory very hard.

*obviously, I am excluding things like LOOKING at objects outside of your reference frame. The "lab" in which the tests are preformed is the inside of Galileo's boat.

ZenMaster

Mar 18th, 2011, 1:02 PM

CT,

*great* question. My response is wordy unfortunately. I apologize that I can't answer more simply. (Actually, the ideas are fairly straight-forward and maybe a YouTube video would make it all obvious.)

Note the subtle shift in emphasis during the translation of Newton's First Law.

(A) Newton - an object continues in its state of rest or uniform motion in the absence of forces.

(B) Einstein - the laws of physics are the same in all inertial reference frames.

In version A, you're only ever considering the universe from a single perspective; your own. THAT object over there continues in ITS state, etc. The relationship between you and the object is implied. This echoes the absolutist history of physics, so you can't blame Newton for it.

Einstein takes the modern relativist perspective and says that the laws of physics are the same in all inertial reference frames. How do we know when two reference frames are mutually inertial? When they are relatively stopped or moving at a constant relative velocity. WAIT... there's a catch here that isn't immediately apparent. Let me bring it to light by building a model. I'll simplify the argument by considering a model where acceleration through the three forces (i.e. EM, weak nuclear, strong nuclear) is impossible. Let's build a universe with you (the observer), object A and object B. Define your view of the universe so that Object A is stationary relative to you and therefore you two are in the same inertial reference frame. Let's put object B moving initially with some fixed velocity relative to Object A, perpendicular to line segment created by <you, object A>. What is the velocity of Object B relative to you, if we assume that it continues in a straight line relative to the line segment created by you and Object A?

Answer: its velocity is suddenly non-constant, at least, from your point of view !! We started with a universe with supposedly strictly uniform motion, and forced out relative accelerations.

What does this say about Einstein's Special Relativity? I believe that it is still fundamentally correct, moreso than even Einstein imagined.

I don't know this for certain but I believe we've only ever tested Special Relativity across uniform reference frames, which would produce good answers regardless of whether I'm right or wrong. I'm also assuming that, from this, we've only ever extrapolated the idea that the theory is valid from frame A to frame B to frame C. I believe that this is, in general, not correct, and is only correct when frames A, B, and C are collinear.

Newton's First Law is therefore wrong because it fails to break out of the single perspective. If all acceleration-by-force is removed, of course everything moves in a straight line relative to you, but among the objects themselves, there may be (and likely is) relative accelerations.

Cartesiantheater

Mar 18th, 2011, 4:53 PM

Note: I'm not trying to be argumentative, but I don't think you've really shown me how your theory deals with my question. Elaboration below.

EDIT- let me put the bottom line at the top: What it appears you are giving up is not really Newton's first law particularly (this is the end result, not the fundamental cause), but rather the assumption that space is homogeneous and isotropic (that translation does not matter and that direction does not matter). Most of my post assumes homogeneity and isotropy (which is why I dispute your claim about relative velocities between object A, B, and C), until the last bit where I point out you must abandon them.

*note: you can not have an isotropic manifold without it also being homogeneous, but the converse is not true.

CT,

*great* question. My response is wordy unfortunately. I apologize that I can't answer more simply. (Actually, the ideas are fairly straight-forward and maybe a YouTube video would make it all obvious.)

Note the subtle shift in emphasis during the translation of Newton's First Law.

Yes, but consider the origin from which Newton's 1st law comes.

If you're locked inside a closed room in a boat, and the ride is smooth, you notice that you can jump, walk, etc, that fish in a fish bowl swim the same, and that throwing tennis balls, etc all work the exact same way they do when you're on the ground. Yet the boat is moving. But the earth is moving, and the solar system is moving, and the galaxy is moving, and all three would see that these things work for you (they must- physical reality must be the same in all frames even if measurements are not) ==> this motion is irrelevant to how the laws of physics work. Your "speed" simply doesn't matter (all acceleration is obviously negligible). But if the balls on your pool table suddenly start curving when you hit them straight (both you and all observers will see them curving with respect to you), you KNOW that you are accelerating (or have suddenly been zapped into a gravitational field or some force has suddenly started acting on you).

The first instance the law of inertia holds <=> it is an inertial reference frame. But since the experiment works, an observer on the sun would also see an inertial reference frame (disregarding long time intervals in which the sun's circular motion WRT the boat matters). Same for the galaxy. In other words, the boat is an inertial reference frame to BOTH of them, and in fact it must be for all observers in uniform motion with respect to the boat (otherwise different physical happenings would be observed based on different locations/velocities).

This can be shown mathematically:

Let there be a reference frame K so that the law of inertia holds good, and another reference frame K' moving at a constant speed u in the x direction (both x-axes are parallel) with respect to K in which the law of inertia holds as well. Then K and K' are inertial wrt to each other.

Assume there exists a frame K'' that is inertial to K', but is not inertial to K.

That means that K'' moves with constant velocity ω with respect to K' (parallel to x-axes), but does not move with constant velocity with respect to K.

So the transformations between K'' and K' are:

x'' = x' - ωt'

y'' = y'

z'' = z'

t'' = t' = t

The transformations between K' and K are:

x' = x - ut

y' = y

z' = z

t' = t

Notice these are linear equations. Since K'' and K are not inertial with respect to each other (we have assumed this), the transformation equations should NOT take the same form as they do between K'' and K', and K' and K.

But substitute x', y', z' and t' into the transformation between K'' and K':

x coordinates:

x'' = x' - ωt'

x'' = (x - ut) - ωt

x'' = x - ut - ωt

x'' = x - (ut + ωt)

x'' = x - (u + ω)t

Now, let u + ω = μ

x'' = x - μt

This means that K'' is moving at constant velocity μ with respect to K => K'' is inertial wrt to K, which contradicts the original assumption.

Therefore if K and K' are inertial reference frames, and K'' moves with constant velocity with respect to K', then it also moves with constant velocity with respect to K, and is also an inertial reference frame.

The point being that ALL inertial frames are equivalent in this way. That is, the transitive property holds for inertial reference frames (if K ~ K', and K' ~ K'', then K ~ K''). If they are not, the principle of relativity does not hold.

I will do the same thing with your example below (only this one will have perpendicular movement and vertical displacement).

In version A, you're only ever considering the universe from a single perspective; your own. THAT object over there continues in ITS state, etc. The relationship between you and the object is implied. This echoes the absolutist history of physics, so you can't blame Newton for it.

Einstein takes the modern relativist perspective and says that the laws of physics are the same in all inertial reference frames. How do we know when two reference frames are mutually inertial? When they are relatively stopped or moving at a constant relative velocity. WAIT... there's a catch here that isn't immediately apparent. Let me bring it to light by building a model. I'll simplify the argument by considering a model where acceleration through the three forces (i.e. EM, weak nuclear, strong nuclear) is impossible. Let's build a universe with you (the observer), object A and object B. Define your view of the universe so that Object A is stationary relative to you and therefore you two are in the same inertial reference frame. Let's put object B moving initially with some fixed velocity relative to Object A, perpendicular to line segment created by <you, object A>. What is the velocity of Object B relative to you, if we assume that it continues in a straight line relative to the line segment created by you and Object A?

Answer: its velocity is suddenly non-constant, at least, from your point of view !! We started with a universe with supposedly strictly uniform motion, and forced out relative accelerations.

I'm definitely not seeing why either object A or object C (you) should see object moving with a changing velocity (outside of the initial start of its motion). With an example:

Let A be reference frame K, C be reference frame K', and B be reference frame K''. A is separated from C by a distance of y = α in the vertical direction (no z, making this two dimensional; also, Galilean for simplicity, because the same principle should hold whether we're talking about Euclidean or Minkowski space). B moves in the x direction.

Then the transformation equations between K and K' are:

x' = x - 0t = x

y' = y - α = constant

t' = t

The velocity transformation is:

u' = u - v = u

u = u' + v = u'

since v is ZERO (v is the relative velocity between K and K')

Now object B moves with velocity ω with respect to object A (I'm assuming their x coordinates align, i.e. y'' - y' = 0, since you didn't specify. It will not matter much). So the transformation equations between K'' (object B) and K' (object A) are:

x'' = x' - ωt

y'' = y'

Velocities:

u'' = u' - ω = constant

u' = u'' + ω = constant

So what are the transformation equations between K'' and K? (object B and object C)

Well, x'' = x' - ωt, and x' = x - 0t = x. Also, ω = u' - u'', u'' = u' - ω, and u' = u - v = u. So substitute ALL that in, and you get:

x'' = x' - ωt

x'' = (x - 0t) - ωt

x'' = x - ωt

x'' = x - (u' - u'')t

x'' = x - (u - v - u''))t, but v = 0

x'' = x - (u' - u'')t

=> The velocity that K' sees K'' move is the same that K sees K'' move (this is because K and K' are AT REST with respect to each other)

This means that the velocity with which object B is moving with respect to object A is THE SAME as the velocity with which object B is moving with respect to object C.

So yeah, I don't see how you get a non-constant velocity out of it. As for the y direction, K'' will always be α units "up" from object C, no matter how far away it gets in the x direction => y'' = y' = y - α. Velocity in the y direction will be zero for all three.

What does this say about Einstein's Special Relativity? I believe that it is still fundamentally correct, moreso than even Einstein imagined.

I don't know this for certain but I believe we've only ever tested Special Relativity across uniform reference frames, which would produce good answers regardless of whether I'm right or wrong. I'm also assuming that, from this, we've only ever extrapolated the idea that the theory is valid from frame A to frame B to frame C. I believe that this is, in general, not correct, and is only correct when frames A, B, and C are collinear.

Newton's First Law is therefore wrong because it fails to break out of the single perspective. If all acceleration-by-force is removed, of course everything moves in a straight line relative to you, but among the objects themselves, there may be (and likely is) relative accelerations.

There cannot be. Not if you assume that there is nothing special about direction or translation in space. (i.e., you assume that (a) moving 300 yards away from one location does not suddenly change the laws of physics, and (b) turning 45 degrees does not suddenly change the laws of physics)

Also, if you give this up, you're going to lose conservation of both linear and angular momentum.

In fact, that is really what you're giving up. It is not really Newton's first law. It is actually the assumption that space is homogeneous and isotropic.

But if you do that, again, you'll lose conservation of momentum. Of course, that is why you lose Newton's law.

Cartesiantheater

Mar 18th, 2011, 5:41 PM

I vaguely associated time with momentum but forgot to mention: Particle Relativity would be almost a clone of Special Relativity, but backwards, in a way. With Special Relativity, a fast spaceship gets rotated out of length and into the dimension of time, so its clocks move more slowly. For Particle Relativity, a fast object gets rotated out of length AND out of momentum. Think of a grandfather clock on a spaceship: as time slows, the pendulum moves more slowly, and therefore its momentum is smaller. As the spaceship approaches the speed limit, it may be that time stretches out near-infinitely long, but it is equivalent to say that the momentum of the pendulum relative to the spaceship approaches zero.

Again, I'm skimming, but were you aware that in special relativity, momentum is NOT p = mu?

It's actually p = γmu, where γ is the Lorentz factor (the very same one that quantifies time dilation, as in t' = γt), but in the case of momentum,

γ = 1/√(1 - u²/c²),

where u is the velocity the observer at rest sees the object move.

But there is one problem: momentum has an INVERSE relation to time, since v = dx/dt, not dt/dx. That means as velocity increases and time slows, momentum should be observed by the guy at rest to be BIGGER in value, not smaller* (momentum is DIRECTLY proportional to velocity, and therefore INVERSELY proportional to the time differential).

This is actually an observed fact: fast moving objects in a particle accelerator experience time dilation, but they have a massive momentum (as measured by the observer at rest- in their own reference frame they have a momentum of zero because they are at rest with respect to themselves).

*NOTE: there is a HUGE point of common confusion with momentum in special relativity. See, there are two velocities to be considered: the velocity v between the two reference frames of the observers, and the velocity u of the moving object with respect to the moving observer. γ in this case uses the velocity v, not the velocity u. In order to measure the momentum the observer standing on earth measures, you will have to use the momentum transformation equation, which is:

px = γ(v) (p'x + vE'/c²)

γ(v) indicates that γ is a function of v now, not u.

The confusing part being that BOTH OBSERVERS consider themselves at rest and the object to be moving (it just also happens that both observers consider the other observer to be moving as well, but with a velocity magnitude smaller than the object).

Also, time dilation and length contraction are NOT something the moving observer notices (that is what the principle of relativity means- that he does not have a special frame of reference). While on the spaceship, all the laws of physics he is accustomed to must remain the same (from his perspective) as when he was "at rest" on a planet- because he IS at rest in his own reference frame. The differences arrive only when either (a) he looks out at another reference frame or (b) he compares his measurements with someone in another reference frame.

So even if there was the momentum change you are postulating, the observer on the spaceship would not notice it. Otherwise the principle of relativity is violated.

Generic special relativity question:

Assume an observer is moving at .99999 the speed of light, and is holding a mirror in front of his face. Does he see his own reflection being blue shifted or red shifted?

ZenMaster

Mar 18th, 2011, 6:15 PM

Hey CT,

don't apologize! X-) I'm actually loving this!

I've come to the stage now where I want to see if there's any merit to my theory (i.e. by beating the hell out of it ;-)

I need some time to digest what you've written. I'm not a fast thinker so I'll likely have to go over it several times but I get my head around it. I'm afraid you may have to be patient.

(Incidentally, I request that you don't let your studies suffer over this!)

Thanks again,
Daniel.

ZenMaster

Mar 19th, 2011, 1:45 AM

Cartesiantheater

Mar 19th, 2011, 2:32 AM

EDIT- Oh snap. You posted while I was writing this. Let me read your new post and then edit this.

Hey CT,

don't apologize! X-) I'm actually loving this!

I've come to the stage now where I want to see if there's any merit to my theory (i.e. by beating the hell out of it ;-)

Which is exactly how science works. (incidentally I knew you would want that, hence I dispute your claim at crackpottery)

I need some time to digest what you've written. I'm not a fast thinker so I'll likely have to go over it several times but I get my head around it. I'm afraid you may have to be patient.

Take your time. These few subjects are very subtle, and one of them is very heavy mathematically (the relation between homogeneity, isotropy, and conservation of momentum particularly).

But I would like you to see if you can figure out the Generic Special Relativity Question I posted at the end of my last post, if you don't know the answer already. Within the answer there is a lot of information about the nature of special relativity. So, if you don't know the answer just by reading the question, then figuring it out will definitely help your own understanding of traditional special relativity, which will no doubt help you in hatching out your own theory. Reposted for convenience:

Assume an observer is moving at .99999 the speed of light, and is holding a mirror in front of his face. Does he see his own reflection being blue shifted or red shifted?

Cartesiantheater

Mar 19th, 2011, 3:07 AM

CT,

I'm *extremely* interested in your response to my own model, i.e. A, B, C => K', K'', K. I submit that, indeed, my example doesn't show what I intended to express. (I don't know what I was smoking.) I think I might be able to pull it out of the scenario you built instead but I'm having trouble. The picture I constructed in my head is different a little bit from the one you've built due to my lack of specifications. Can you tell me what the velocity variables u and v refer to? I'd like to chew on your equations some more before I reply fully.

There's also an omega in that (in the FIRST of those consecutive posts). Now, u means something different in the TWO SECTIONS of that first post. (sorry. That was a stupid move, I know. Certainly only adds to the confustion /facepalm at self) In the first section it refers to the relative velocity of coordinate systems K and K'. In the second section (when I add A, B, and C), u refers to the velocity of any object within the coordinate system K.

In the second section, I actually made a slight fuckup, but it won't affect the argument. I would allow me to edit it and fix it before you read that SECOND part of the first post too deeply (although the physical result will be the same).

ω is the speed with which the coordinate systems K'' and K' move with respect to each other.

v is the speed with which the coordinate systems K' and K move with respect to each other.

u'' is the speed that an observer in K'' sees an object moving in K''.

u' is the speed an observer in K' sees an object moving in K'.

u is the speed an observer in K sees an object moving in K.

Quickly, to fix my screw up earlier, we want the transformation from K'' to K. The reason we want this is to see if K'' travels with the same velocity with respect to K as it does to K'. This is done by substitution, first from K'' to K', then from K'' to K.

x'' = x' - ωt

x'' = (x - 0t) - ωt

x'' = x - ωt

So what is ω?

We have to go back to K'' and find the velocity transformation between an object moving in K'' with velocity u'' to an object moving in K'. This is

u'' = u' - ω

Solving for ω:

ω = u' - u''

But we can put u' in terms of u, since A and C are at rest with respect to each other (that means v = 0), then u' = u - v = u. u' = u (this was actually my mistake). So we can just replace u' with u. So all the way back to before we solved for ω:

x'' = x - ωt

x'' = x - (u - u'')t

And that's the relation. And if you want to express it in terms of K to K'', you have:

x = x'' + (u - u'')t

and since u - u'' = ω

what we have is that the velocity that A sees B moving is the same as the velocity that C sees B moving: ω

I'm sure that doesn't help it all. Sorry.

Oh and .. it has occurred to me that I may have left a rather critical part of the argument out, and now I can't remember the name of the concept (if it in fact has a name). Google and Wikipedia aren't helping. The concept is that if you can make one scenario match another simply by changing the coordinate frame, then the two scenarios must actually be equivalent. E.G. having one particle move along the x-axis and another move along the y-axis is equivalent to saying that there is some single relative <x,y> movement between them. (Yes, I understand that is effectively your argument in a nutshell.) If they're the only particles in the universe, it is also equivalent to saying that they're not moving relative to each other, because you can grow and adapt the coordinate frame.

Thank you in advance :)

I don't follow you here on "grow and adapt."

Uniform motion is relative. You can say x is moving relative to y and y is at rest, or y is moving relative to x and x is at rest. That is what is equivalent. If you attach a coordinate system to each particle, then x would be at the origin of one system and y would be at the origin of the other, and yeah, they would be given a relative velocity (which is the velocity each measure the other moving WHILE ASSUMING THAT THEY ARE AT REST. This value MUST be identical if the typical assumptions about space are made- homogeneity, isotropy, etc. What I mean is, if x is at rest, then he sees the coordinate system y is at rest in moving with speed v. If y is considered at rest, then HE sees the coordinate system x is at rest in moving with speed v).

Any object that moves besides those two will be given a velocity by each that is frame dependent and related by the velocity transformation equations, and that new particles position and time will be related by the position and time transformation equations.

But I see no way to say that you can say they AREN'T moving relative to each other. If you take a new coordinate system and apply it to the whole of the space, x and y would not suddenly be at rest with respect to each other: x is moving to the "right" and y is moving "up," regardless.

If you want to be more confused, I'll drop this on you: angles are not invariant between coordinate systems, and direction of motion with respect to your chosen rest frame matters in this: y is not moving up with respect to the rest frame of x. It's moving up and in the negative x direction => diagonally with respect to the rest frame of x. This actually affects the shapes of objects (a moving object has length contraction ONLY in the direction of motion. If the base of a triangle shrinks but the height does not, the triangle will look skinnier, etc).

ZenMaster

Mar 20th, 2011, 4:29 AM

Thank you for the clarifications CT.

I've gone over your example a couple of times and my own again and changed my mind. Even before the clarifications, although I couldn't get what you were saying on the velocities, I picked up generally where your work was heading. I'm going to go over it again with the corrections applied so that I can understand the transformations for myself, but I don't expect I'll be arguing the logic.

I'm not going to do the whole ego-trip thing of back-pedalling and saying something like "Well .. but what I really meant was THIS." I think at some point I came up with the logic that justifies the theory, though I'm now less certain it was solid. Part of my reasoning was based on the example I gave, but I think if the example by itself doesn't lead to the conflict I'm thinking of, then there must have been something in the assumptions. I must have had either an unstated assumption or I was arguing against an assumption in mainstream physics. I'm going to try and go back through my path of logic to either rediscover the route to a new theory or else find the wrong turn I made.

I don't think I'm proposing a non-homogeneous and anisotropic model. I have come across the ideas in studying Special Relativity, in that I'd heard it mentioned that these properties are implicitly assumed in SR. I know them on a rough level, i.e. that in a homogeneous universe, experiments don't depend on location, and in an isotropic universe, experiments don't depend on direction. However, I've never studied them so alternatives to a homogeneous, isotropic universe aren't something I've explicitly considered.. but who knows? Maybe what I'm thinking of is such an alternative universe.

<..>

But I would like you to see if you can figure out the Generic Special Relativity Question I posted at the end of my last post, if you don't know the answer already. Within the answer there is a lot of information about the nature of special relativity. So, if you don't know the answer just by reading the question, then figuring it out will definitely help your own understanding of traditional special relativity, which will no doubt help you in hatching out your own theory. Reposted for convenience:

Assume an observer is moving at .99999 the speed of light, and is holding a mirror in front of his face. Does he see his own reflection being blue shifted or red shifted?

Hrm, well, I may be biased unfortunately. I don't specifically remember hearing the question but I probably have in my readings. It seems to be one of those questions where the knee-jerk answer is wrong. Regardless, I'll try to answer it by just sticking to the statement of Einstein's postulate as I can remember it: the speed of light is independent of the speed of the source and the observer. I can forget the speed of light as it left the observer, because it simply gets absorbed and re-emitted by the mirror. Therefore, the distance over which the beam of light travels is from where it was when it left the mirror and the location of the observer when he sees the image. Since the observer is moving towards that location, the total distance is shorter than the relatively-at-rest distance. My final answer - blue-shifted? (Now that I've written this out .. yeah I've heard the question before. The only question now is whether I also remembered the reasoning correctly.)

Incidentally, I have also been reading more on your second post, regarding momentum. I was actually aware of the relativistic version of momentum (i.e. p = γmu) but I don't believe my own theory conflicts with it since relativistic momentum (and the standard equation for momentum itself) all depend on the time variable. I'm using a different version of momentum that, literally as I'm writing this sentence, I'm beginning to wonder if I should specially name. It does not depend on the existence of time per se, only on a reference clock. You can slip this new quantity, call it relational momentum, invisibly into mainstream physics as a replacement for standard momentum IF you assume that the ticking of the reference clock really does occur at regular intervals in time. I expect this sounds like I'm drawing a distinction where none exists. If I can get to it, I'll try and delve into it deeper in my other thread "The Emergence of Time from a quantum space" (http://forums.armageddononline.org/emergence-time-quantum-t28806.html). Here I'll build an equation, for no other reason than to clearly and unequivocally define it, though I've never actually written it down or used it before so it might be crap:

Relational momentum of an object relative to an observer -
P = γm • (∆x / ∆(∆Φ) )

This is similar to the normal equation of linear momentum but deserves clarification. Normally all the stuff in brackets would be replaced by the relative velocity of the object, which is ∆x / ∆t. In my equation, ∆t is translated to something I claim is analogous to time under the "regular ticking" condition I mentioned. ∆Φ is the relative change in position of a standard clock in our frame of reference; say, the distance between notches on a sundial. ∆(∆Φ) is the duration measured this way, but only if ∆Φ occurs over constant time. Let ∆Φ = 1 notch on the sundial, then from your perspective, the relational momentum of a turtle might be P = 0.8 kg • (1.2 km / (3.4 • ∆Φ) )

Hrm .. maybe I should stop getting *more* complex and sort out the root stuff with my theory, but I hate to delete what I've written up so far.

Anyway, I hadn't yet seen the momentum transformation equation but it looks very similar to the transformation equations for space and time so I'll buy it. I am also aware that the effects of Special Relativity are invisible inside the relatively moving frame of reference.. it would be difficult to believe otherwise! Astronauts don't see themselves getting flatter during launch. (Well, I suppose the effects are too small anyway, but whatever.)

I don't follow you here on "grow and adapt."

Uniform motion is relative. You can say x is moving relative to y and y is at rest, or y is moving relative to x and x is at rest. That is what is equivalent. If you attach a coordinate system to each particle, then x would be at the origin of one system and y would be at the origin of the other, and yeah, they would be given a relative velocity (which is the velocity each measure the other moving WHILE ASSUMING THAT THEY ARE AT REST. This value MUST be identical if the typical assumptions about space are made- homogeneity, isotropy, etc. What I mean is, if x is at rest, then he sees the coordinate system y is at rest in moving with speed v. If y is considered at rest, then HE sees the coordinate system x is at rest in moving with speed v).

Any object that moves besides those two will be given a velocity by each that is frame dependent and related by the velocity transformation equations, and that new particles position and time will be related by the position and time transformation equations.

<... etc. ...>

Haha! Awesome, I think you're helping me reconstruct my path of logic.

Okay, yup, totally with you on the Galilean relativity between the two frames, but it is quite different from what I was referring to. I'm glad you brought this up. Put my theory aside for a moment. Instead, I want to describe a mathematical tool I've been making heavy use of on the way to coming up with the theory.

Take this into a purely mathematical realm, such that the speed of light is no longer a universal constant. To quote you:

If you attach a coordinate system to each particle, then x would be at the origin of one system and y would be at the origin of the other, and yeah, they would be given a relative velocity (which is the velocity each measure the other moving WHILE ASSUMING THAT THEY ARE AT REST.

You're attaching coordinate systems quite freely. In discussions on Special Relativity, this is valid because of the constancy of the speed of light. However, with no universal constant in this mathematical diversion, the coordinate systems are entirely arbitrary. Take a 2D space where the only two reference points that exist are the zero-dimensional particles x and y. Adopting x's point of view, we can arbitrarily define the movement of y by playing with our coordinate system. Define line segment L to connect x and y, and define a rectangular coordinate system using L as the horizontal and vertical distance between perpendicular gridlines. We can make y approach us by stretching L on a whim. y moves away from us if we shrink L. We can create funkier movement by changing the coordinate system in weirder ways, or even by changing our own position relative to the coordinate system (by asserting the we are moving away from the origin).

Although this "feels" like movement, I think you can see that ultimately all coordinate changes like this are isomorphic (is that the word I mean?) and so there is no change in terms of the relative relationships between the particles (i.e. relative distance and orientation). This means that if I can show that one scenario is isomorphic to another, then they must be equivalent.

As I mentioned earlier, in reality we can use the speed of light as a sort of truth-teller in regards to distances and times, so the isomorphism trick doesn't really seem to work. However, in coming up with my own theory, I remember depending on the isomorphism trick very heavily. I believe I had found a loophole.

After I study your example some more, if I'm still convinced I'm not wrong, I'll try and come up with a demonstration of my own which actually works.

Cartesiantheater

Mar 20th, 2011, 8:03 AM

Since early in the morning here and I've been up all night doing stupid crap, I'm not really in the mood for deep thinking, so I'll come back to the rest of your post later, but the "Generic question" is one I can discuss without doing that.

It actually comes from a physics textbook.

Disregard parts that you understand, as always, and do not take offense if I try to explain something you already understand, or seem to go about it too slowly or whatever.

Hrm, well, I may be biased unfortunately. I don't specifically remember hearing the question but I probably have in my readings. It seems to be one of those questions where the knee-jerk answer is wrong. Regardless, I'll try to answer it by just sticking to the statement of Einstein's postulate as I can remember it: the speed of light is independent of the speed of the source and the observer. I can forget the speed of light as it left the observer, because it simply gets absorbed and re-emitted by the mirror. Therefore, the distance over which the beam of light travels is from where it was when it left the mirror and the location of the observer when he sees the image. Since the observer is moving towards that location, the total distance is shorter than the relatively-at-rest distance. My final answer - blue-shifted? (Now that I've written this out .. yeah I've heard the question before. The only question now is whether I also remembered the reasoning correctly.)

The correct answer is actually that he sees his reflection NEITHER blue-shifted NOR red-shifted. It is a trick question. The observer sees his reflection in EXACTLY the same way if he were standing on Earth, or in a boat, or on some spaceship, etc.

I do find it interesting, however, that you also said this in the same post:

I am also aware that the effects of Special Relativity are invisible inside the relatively moving frame of reference.. it would be difficult to believe otherwise! Astronauts don't see themselves getting flatter during launch. (Well, I suppose the effects are too small anyway, but whatever.)

So it's clear you know what's going on intellectually, yet you missed this very concept with respect to the guy looking at his reflection. Very interesting.

Anyway, the answer gets to the very heart of what the principle of relativity is (also, remember the principle of relativity predates Einstein, so I'm not just talking about SR): the laws of physics are the same in every inertial reference frame. What that literally equates to is that whatever things you see happening in your reference frame (assuming you are closed off from the rest of the world, say locked in a room) are the same no matter how fast you are moving with respect to someone else.

If you haven't thought through it yet to see why, here is what made me understand it: I'm on Earth, but Earth is moving in a big ellipse around the sun. But the sun (the entire solar system) is moving in a big ellipse around the galaxy. And in fact, the entire galaxy is moving. But what's more, the Earth is moving at RIDICULOUS velocities around the sun, and the solar system is moving MUCH FASTER than that.

Yet I feel NONE of it. Why?

(remember the pathways are so long that the acceleration from curvature is not something we can feel in our daily lives, so I'm ignoring it)

As for homogeneity or isotropy, these aren't just implicitly assumed in special relativity. They're also assumed in Newtonian relativity (Galilean invariance). They are required for relativity in general, whether it follows Galilean transformations or Lorentz transformations, etc. It is also assumed that time is homogeneous (in both pre-Einstein and post-Einstein physics, because energy is conserved in Newtonian relativity and special relativity).

You can't have them if the following is not true:

if an object moves uniformly with respect to an inertial reference frame, then it moves uniformly with respect to all inertial reference frames (they won't all measure the same magnitude of the velocity, but all will measure a constant one).

The reason is because the above is a requirement for Newton's third law to be true (see link below), and Newton's third law is a requirement for conservation of momentum to be true (see link below), but the conservation of momentum is true only if space is homogeneous (see Noether's theorem).

More slightly related interesting information here:

http://www.mathpages.com/home/kmath386/kmath386.htm

A simpler way to look at it:

A reference frame is called an inertial frame if Newton's laws are indeed valid in that frame

So if we have an inertial reference frame, then Newton's second law is valid: F = mẍ

That involves the second time derivative of x, the position vector, as you know.

So what happens if you do a coordinate transformation involving ẋ? NOTHING. a change of coordinates involving ẋ, which is velocity, affects NOTHING in F = mẍ.

Can you see why? It's because F = mẍ is not a function of ẋ.

If Newton's laws are valid in one reference frame, then they are valid in any reference frame in uniform motion (i.e. not accelerated) with respect to the first system. This is a result of the fact that the equation F = mẍ involves the second time derivative of x: a change of coordinates involving a constant velocity does not influence the equation. This result is called Galilean invariance or the principle of Newtonian relativity.

*used x's instead of r's, which the textbook uses, because I can't find an r with two dots on it.

But if no coordinate transformation involving ẋ will change anything, that means something deeper: it means a velocity coordinate transformation does not change the law.

With the same reasoning, it is clear that simple translational changes in coordinate transformations will not change it either (if the integral of ẍ -velocity- has no bearing on the law, neither does the integral of ẋ-position).

The same holds true for Newton's first and third laws under the assumption that space is homogeneous and isotropic: neither position nor velocity translations change anything about the laws themselves.

You can see how thoroughly connected physics is, no doubt. It is very interesting how it is so internally consistent.

After skimming through your post, I am concerned about one thing regarding coordinate systems. We can define them however we like, but how we define coordinate systems doesn't have anything to with the universe itself. I can "decide" to label the units on my axis between me and Tokyo so that Tokyo is less units away from me, but I can't arbitrarily cause Tokyo to come closer or move away from me by "stretching" or "shrinking" my coordinate system. I suppose I didn't quite follow you there.

I'll check out your momentum later, maybe after I sleep for a couple of days.

ZenMaster

Mar 22nd, 2011, 3:30 AM

CT,

I'll check out your momentum later, maybe after I sleep for a couple of days.

Before you take a look at my definition of momentum, I'd like to stay on the K, K', K'' example. I've figured out how I broke it the first time, and I want some time to formalize my response; i.e. clearly define the relative orientations of my reference frames and double-check my equations, so that I can ensure it communicates what I want. (I'll try and post the model before the weekend.) I'm going to cheat and mostly use your equations towards my own evil, evil ends, but I think I'm going to have to come up with a few of my own in order to explain the concept properly.

Note that, from going over your response, I've figured out that my model is indeed either not homogeneous and/or anisotropic. I will be debating against the claim that non-collinear reference frames can ever be "relatively inertial" when travelling rectangularly (i.e. in a straight line.) I know that this is raising warning flags in your head. Sorry .. the model is coming. You also mentioned that most of your responses were based on the assumption that space is both homogeneous and isotropic, so I no longer believe they're applicable to my model.

I'm also going to establish some terminology here, to make it easier to express my ideas.

coordinate frame - any system for measuring relative positions
inertial reference frame - a system for measuring relative positions in which all of Newton's three laws are valid
isomorphism - a configuration of points in a frame that is equivalent to another under certain transformations

A coordinate frame is strictly mathematical, and if you add a dimension of time, the definition is significantly looser than the definition of an inertial reference frame. For the purposes of this discussion, I'm only going to be talking about coordinate frames that include time implicitly (i.e. I won't be graphing events in time, I'll just talk about velocities and accelerations).

I'm going to talk about isomorphism in regards to coordinate systems and inertial reference frames. I'll restrict my attention to isomorphisms where the only allowable transformations are:

translations
rotations
scales that maintain angle measures (and therefore scale equally along the x and y axes)

This concept of isomorphism is important for my argument. Special Relativity necessarily requires scaling along one arbitrary axis of movement but not another; however, I will step carefully so as not to require the invocation of Special Relativity.

After skimming through your post, I am concerned about one thing regarding coordinate systems. We can define them however we like, but how we define coordinate systems doesn't have anything to with the universe itself. I can "decide" to label the units on my axis between me and Tokyo so that Tokyo is less units away from me, but I can't arbitrarily cause Tokyo to come closer or move away from me by "stretching" or "shrinking" my coordinate system. I suppose I didn't quite follow you there.

Precisely. Two isomorphisms may look very different, but they are mathematically equivalent and do not change the reality of the situation. Please keep that in mind when you're going over the model to come.

dedanoe

Mar 22nd, 2011, 11:39 AM

a system of weights can have only one true referential point. entire frame is over sufficient but it needs only one point as its origin. all weights in the system carry their own axis or set of axes and they are applied in addition to the referent point. this point of origin must be the equilibrium point cause the point of equilibrium is equally the global representor for all the weights of the system. you are not free to move the frame across the space cause weight (as magnitude) is deeply related with distance (as magnitude). referring to the equilibrium point is assurance that the representation of the system is the simplest possible. and one more thing screw the physics after newton. the universe knows of one law only -- the lever law -- and to be the universe is to have yin verse that is completely reciprocal with the verse out. i don't care how genius were all those stomach driven for salary physicians from newton to hawking cause none of them was able to define universe. yin verse over uni verse equals uni verse over verse out -- that universe is me alone.

rJso8iJdPrQ

and now if you'll excuse me, i have to clean the floors in the psycho asylum.

dedanoe

Mar 22nd, 2011, 11:46 AM

imagine weight in uniform motion traveling with constant velocity. don't it mean that the weight continuously is distancing away from its equilibrium point? and won't then the price for the distance gained be the lost of weight? or else how does the weight conserve its imbalance evaluated as norm of the array[weight, distance]

I = sqrt(weight^2 + distance^2) = const, but how if only distance changes?

ZenMaster

Mar 24th, 2011, 2:09 PM

Cartesiantheater

Mar 24th, 2011, 3:27 PM

Dedanoe,
I think your ideas would be fascinating to learn about but they really don't have to do with the original post, so I'd like to request that you start a different thread. It's like bringing up Roman Catholicism in the middle of a discussion of whether Islam is inherently unfair to women. Sure, they're both religions but the former really has nothing to do with the debate. (I sincerely look forward to a new thread if you're willing to post one. As it stands, I couldn't penetrate your logic. 1*1*1*1*.. ??? What are you saying??)

So anyway, back on topic.

CT, I'll direct this to you since you're the only one really addressing the content of my posts. To quote myself:

In my initial attempts at the mathematics, I've avoided using Special Relativity altogether and had absolutely no luck in breaking out of the Newtonian universe. However, it finally occurred to me that I can't avoid using Special Relativity, despite my initial comments. I made the claim that reference frames moving non-collinearly relative to a light source will fail to be relatively inertial, but if I never mention Special Relativity, I lose the magic trick.

This is a long way of saying "D'uh!" :blink:

Back to the drawing-board. Model still to come but the time I've spent thus far is wasted so I'll make good use of the weekend.

Problem is that the same thing applies with special relativity, except the math gets more complex (instead of using the simple Galilean transformations you have to use the more complex Lorentz transformations- that's the only thing that changes). Here is the same reasoning, but with Lorentz covariance (in fact, the phrase "Lorentz covariance" means exactly this: the feature of nature that says experimental results are independent of the orientation or the boost velocity of the laboratory through space.)

Note: γ(ω) = 1/√(1 - ω²/c²) for any velocity ω.

We want the transformation from K'' to K. The reason we want this is to see if K'' travels with the same velocity with respect to K as it does to K'. This is done by substitution, first from K'' to K', then from K'' to K.

x'' = γ(ω) (x' - ωt)

x'' = γ(ω) [γ(v)(x - vt) - ωt]

*noting that v = 0, and γ(0) = 1

x'' = γ(ω) [γ(0)(x - 0t) - ωt]

x'' = γ(ω) (x - ωt)

So what is ω?

We have to go back to K'' and find the velocity transformation between an object moving in K'' with velocity u'' to an object moving in K'. This is

u'' = (u' - ω)/(1 - u'ω/c²)

Solving for ω:

ω = u' - u'' + (u''u'ω)/c²

*notice that velocity transformations are not linear (one difference between Newtonian relativity)

But we can put u' in terms of u, since A and C are at rest with respect to each other (that means v = 0), then u' = (u - v)/(1 - uv/c²), v = 0, => u' = u . So we can just replace u' with u. So all the way back to before we solved for ω:

x'' = γ(ω) (γ(v)[x - vt] - ωt] from above

x'' = γ(ω) (x - ωt)

*substituting ω

x'' = γ(ω) (x - [u' - u'' + (u''u'ω)/c²]t)

*remembering that u' = u since v = 0 and γ(0) = 1

x'' = γ(ω) (x - [u - u'' + (u''uω)/c²]t)

Notice that the only thing that is different is the Lorentz factor for ω, which is denoted γ(ω), and is equal to 1/√(1 - ω²/c²), and the term added to u - u'' which multiplies t in the Newtonian relativity, (u''uω)/c².

If you want to express it in terms of K to K'', you have:

x = γ(ω)(x'' + [u - u'' + (u''uω)/c²]t)

and since u - u'' + (u''uω)/c²' = ω

we again have that the velocity that A sees B moving is the same as the velocity that C sees B moving.

Ultimately, this applies to either Newtonian or Einstein/Lorentzian relativity (special relativity): any inertial reference frame will be seen to be an inertial reference frame by ALL inertial reference frames <=> given any inertial reference frame, all other inertial reference frames will move with a constant velocity with respect to that inertial reference frame, and all reference frames moving with respect to that inertial reference frame are themselves inertial reference frames.

That works in special relativity just as it does in Newtonian relativity.

I taken you wish to modify special relativity to change that?

ZenMaster

Mar 28th, 2011, 3:38 AM

Nope, my math is too rusty and I've gotten stuck.

Anyone who happens to be reading .. willing to lend a hand in the calculation at the bottom? CT, don't neglect your studies for this. :pirate: (I'm rather embarrassed actually. The problem should be hideously simple for anyone who's taken university level mathematics. Oh well.)

My goal is to challenge Newton's concept of inertia though I know I haven't given anything to support an alternative yet. Below I'm starting construction of a bridge to an alternative, beginning from a Newtonian perspective.

To reiterate the variables, and add a few more:
α – not used here and repurposing variables would be confusing
K, K’ – two separate coordinate frames; each has an independent object fixed at its origin
K’’ – an object that may or may not be the origin of its own coordinate frame
θ is the angle between the line segments <K, K’’> and <K’, K’’>
ω is the speed with which K’’ and K’ move with respect to each other
v is the speed with which K’ and K move with respect to each other
W is the speed with which K’’ and K move with respect to each other
ϒ(ω) = 1/√(1-ω²/c²) for any velocity ω (i.e. Lorentz contraction)

My goal is to reconstruct the coordinate frames over again a few times, but all constructions will be isomorphically equivalent to each other. Therefore, the reader can picture what's going on more easily since they can get one scenario from any other just by changing the origin, angle, and scale of the graph paper underneath.

************************************************** **

Scenario #1 – I’ll start with the scenario that was constructed earlier. Given some arbitrary rectangular reference frame K, the origin of K’ is some distance in the negative direction along the y-axis and object K’’ is moving with velocity ω along the positive x-axis of K’ (i.e. y’=y’’). Coordinate frame K’ has the same units as K, with axes parallel. We can calculate the velocity transformation of object K’’ from coordinate frame K’ to coordinate frame K. Start with an equation of the velocity of K’’ in the frame of K’.

ω = x’/t’
x’ = ωt’ where x’ = ϒ(v)(x – vt) and t’ = ϒ(v)(t – v/c²∙x) by Lorentz
Substitute and cancel the duplicate term ϒ(v):
x – vt = ω(t – v/c²∙x)
x – vt = ωt – vω/c²∙x
x + vω/c²∙x = vt + ωt
(1 + vω/c²)∙x = (v + ω)∙t
x/t = (v + ω) / (1 + vω/c²)
W = x/t
W = (v + ω) / (1 + vω/c²)
And when v=0; W = ω

Conclusion: When coordinate frames K and K’ are relatively stationary, the object K’’ is moving along the +x axis with velocity ω in both.

************************************************** **

Scenario #2 – This scenario is similar to #1, except I’m giving the whole thing a spin. Relative velocities of the coordinate frames are not modified. With K as the origin, rotate around the z-axis counterclockwise until K’’ is along the +x axis. In this isomorphic configuration, object K’ is now at some location a distance of <+x, -y> and is moving with velocity v = v[x] + v[y] relative to the coordinate frame of K. Similarly, object K’’ has velocity ω = ω[x] + ω[y] relative to K’. To figure out the velocity of object K’’ relative to K, break the vector for the velocity into components that align with the new x and y axes:

ω = ω[x] + ω[y] where ω[x] = ω∙cos(θ) and ω[y] = ω∙sin(θ)
v = v[x] + v[y] where v[x] = v∙cos(θ), v[y] = v∙sin(θ)

The variables with square brackets [x] are supposed to be subscripts but there doesn't seem to be that feature here.

To get the velocity of K’’ in the coordinate frame of K, we can perform similar calculations as before except that we have to do it component-wise so it gets a bit complicated. It's ONLY the second scenario but I've been working on it all weekend and can't seem to grunt my way through it.

I believe I can start with:
W[x] = (v[x] + ω[x])/(1 + v[x]ω[x]/c²)
W[y] = (v[y] + ω[y])/(1 + v[y]ω[y]/c²)
W = W[x] + W[y]

And end up with
W = (v + ω) / (1 + vω/c²)
which, when v = 0, W = ω

Ultimately I want to show that the velocity ω = ω[x] + ω[y] is observed relative to both coordinate systems K and K' when v = 0.

Come to think of it: if CartesianTheater or anyone else is interested, I'll keep posting but the hot-potato-in-my-brain has cooled now that I have some concrete math on paper (even though it only confirms Newton, so far). If no one cares, I'll just keep working on my own.

Cartesiantheater

Mar 28th, 2011, 7:39 PM

*note: technically neither one of us is doing it right, since we're only giving average velocities here. But of course, who wants to use differential notation? Screw that. x/t = dx/dt for the purposes of this portion of the discussion, right?

One thing that might help:

You correctly derived the velocity transformation along the axis of which the coordinate frames are moving with respect to each other (i.e. you correctly derived the velocity transformation in the x direction).

Here are the velocity transformations for y and z (for reference frames moving relatively in the x direction, this is the velocity of an object in the y or z direction):

First of all, the Lorentz coordinate transformation equations:

x' = γ(v) (x - vt)

y' = y

z' = z

t' = γ(v) (t - vx/c²)

Therefore, the y and z velocity transformations are y'/t' and z'/t' respectfully. (I use small font for subscripts):

u'y = y/γ(v) (t - vx/c²)

= (y/t) /γ(v) ((t/t) - vx/tc²)

= uy/γ(v) (1 - vux/c²)

It's the same thing for the z direction:

u'z = uz/γ(v) (1 - vux/c²)

That might be of use to you, if you didn't have that information already.

Things of interest: (note the second one may not be to your mathematical liking. It gets into abstract algebra a tiny bit and, if you go far enough, it gets pretty heavy into vectors)

(1)

Both velocity in the y and z direction of an object depend on the velocity of the object in the x direction, but velocity in the x direction does not depend on either.

(2)

Now, I'm not particularly fond of the direction this is going to go. It gets hairy. But here is some more general information that you might like (if you can stand the vector notation and abstract algebra (just in the fact that they define an operation and use notation for it)):

http://en.wikipedia.org/wiki/Velocity-addition_formula#General_case_.28engineering_units .2C_replaced_with_v_.2F_c_.29

As you can see, things get quite tricky when you move into three dimensions.

Now, you can take the above results (the velocity transformation equations), and you will find out that if the velocity vector u only has components parallel to the velocity it is added to, the addition will be commutative. But if the vector u has perpendicular components to the velocity it is added to, it will NOT be commutative. If you want a good math exercise, you might want to show this to yourself by doing the following two operations TWICE (two cases, described below):

v ⊕ u

and then

u ⊕ v

where u = ux + uy,

but in the first case uy = 0 and in the second where it does not. Do both operations with uy = 0, and then both with uy ≠ 0.

Note:

Operation definition:

a ⊕ b = (a + b‖+ αab⊥)/(1 - ab/c²)

where αa is the reciprocal of the Lorentz factor γ(a)

You will see that in the first case the two operations are equal, but in the second case they are not. (that's because when you do the u ⊕ v operation, v has no y component, so when you reverse all the symbols as you do that operation, vy will be zero in both cases, and uy will not come into play)

Or more compactly:

u ⊕ v = v ⊕ u only when u is parallel to v.

It is an interesting thing... of course, sorry, that was all slightly off topic. The wikipedia link shows how you can take this to where you have z components as well. As I said, it gets hairy. I think the coordinate notation (column vectors) is the easiest to visualize (shown a little down in the link - Everyone has their own notation preferences, I suppose). It is also pretty much the most general equation for velocity addition.

ZenMaster

Mar 29th, 2011, 2:27 AM

*note: technically neither one of us is doing it right, since we're only giving average velocities here. But of course, who wants to use differential notation? Screw that. x/t = dx/dt for the purposes of this portion of the discussion, right?

Yes, that's fine. It would have come to a point where I was going to break everything up into infinitesimals and use calculus anyway. Ditto your "screw that".

That might be of use to you, if you didn't have that information already.
<..>
Things of interest: (note the second one may not be to your mathematical liking. It gets into abstract algebra a tiny bit and, if you go far enough, it gets pretty heavy into vectors)

Hrm, as a matter of fact, I wasn't aware. I had come to believe that the effects of Special Relativity along a single axis were independent of the effects along perpendicular axes so yes, thank you for bringing this up. I've somehow managed to convince myself that the logic of the new premises is good, but if the consequences of these premises don't even look like gravity on a subjective level, I'm hooped.

It is an interesting thing... of course, sorry, that was all slightly off topic. The wikipedia link shows how you can take this to where you have z components as well. As I said, it gets hairy. I think the coordinate notation (column vectors) is the easiest to visualize (shown a little down in the link - Everyone has their own notation preferences, I suppose). It is also pretty much the most general equation for velocity addition.

Not off-topic at all, actually. I'm going to be taking a closer look at this since my own theory hinges critically on a good understanding of Special Relativity in more than one dimension. I was tempted to skip the rotation scenario altogether in my construction, but now I'm glad I didn't!

******
CartesianTheater, I'm tremendously grateful your input. Thank you for all your help! I'm going to be lurking on this thread for awhile because I want to go over your math, but I won't be actively posting until I can play with Special Relativity in 3D. (I really need to brush up on my vectors too, blah.)

Thanks again,
ZM