An “analytical-metaphysical” take on Special Relativity!

Doctordick · 06-07-2009

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Originally Posted by AnssiH

Ehh, at any rate, might be worthwhile to tidy it up in the OP, just in case

I have edited the OP. Not exactly what you suggested but I think it is better now.

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Originally Posted by AnssiH

Is there just a convention that they always use just the other part or something? (really just guessing here

Not really; they often use the exponential representation because of the convenience of representing some waves via complex amplitude particularly in electronics because of the ninety degree phase differences in voltage to current functions for different kinds of components (in a linear circuit, voltages in resistors and capacitors are ninety degrees out of phase when plotted as a function of the current). It provides a major simplification in the mathematics. A very similar thing occurs in quantum mechanics. A wave represented by $e^{bi}$ can be seen as having a constant amplitude everywhere (it is just a vector rotating in a complex space). I wouldn't worry about it as there is a lot of serious physics behind the representation. The only reason I bring that up is that your analysis is perfectly correct except that when you say, “that people build functions that exploit $e^{a+bi}$ within to come up with a wave” you sort of have things backwards.

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Originally Posted by AnssiH

I would like to understand how waves equations work so if you can provide more help with that, it would be good.

What you are missing is knowledge of differentiation of trigonometric functions. The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x). That being the case, the second derivative of either is a simple change of sign.

$\frac{d^2}{dx^2}sin(x)=-sin(x)$
and
$\frac{d^2}{dx^2}cos(x)=-cos(x)$

It follows that anytime you see a differential equation of the form

$\frac{\partial^2}{\partial x^2}\Phi(x,t)=a^2\frac{\partial^2}{\partial t^2}$

you immediately know that the solution is a wave function (a sine or cosine) with a simple argument of the form ax+t or ax-t (the sign of t is immaterial as when you take the second derivative the factor ends up squared). Furthermore, the second order differential equation can be written

$\left\{\frac{\partial^2}{\partial x^2}-a^2\frac{\partial^2}{\partial t^2}\right\}\Phi(x,t)=0$

which factors into

$\left(\frac{\partial}{\partial x}-a\frac{\partial}{\partial t}\right)\left(\frac{\partial}{\partial x}+a\frac{\partial}{\partial t}\right)\Phi(x,t)=0$

yielding two first order differential equations which will satisfy that equation.

$\left(\frac{\partial}{\partial x}-a\frac{\partial}{\partial t}\right)\Phi(x,t)=0$
and
$\left(\frac{\partial}{\partial x}+a\frac{\partial}{\partial t}\right)\Phi(x,t)=0$

The first is solved by a sine or cosine function with the argument (ax+t) and the second by a sine or cosine function with the argument (ax-t). These two functions are simple waves moving in opposite directions. If we call the argument of the sine and/or cosine function “z” then z=ax+t or z=ax-t. If z (the argument of the sine or cosine function) is a constant, we are talking about a specific point on that function. The question you need to ask yourself is, if t changes, how must x change in order for z (the argument) to remain constant? Obviously, if t increases by some amount, ax must either decrease by the same amount (for z=ax+t) or increase by the same amount (for z=ax-t). This can only be true if the change in x is identical to t divided by minus a or plus a (for the two cases). It should be clear to you that the change in x has to be given by x=x₀+vt so the velocity of the wave must be one over a.

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Originally Posted by AnssiH

Well after some head scratching, I could not understand that stuff above. Nor your further commentary about the issue. I'm guessing $\omega$ means angular frequency here, and I suppose that is essentially the rotation rate of the unit vector (the phase) or something like that.

Sorry about that. The $\omega$ just stands for a constant. What it means depends upon your specific definition of angles in the sine function and I was being very careless there as x is clearly in radians. Normally angles are expressed in degrees but in physics (for differential simplicity) angles are almost always expressed in radians. 360 degrees equal $2\pi$ radians. I am going to edit that post and replace $\omega$ with v. I really should have done that originally but you know I am getting senile and sometimes what pops into my head is just wrong. Getting old is a pain in the ass.

I hope I have made it a little clearer; I sincerely apologize for making this difficult for you.

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Originally Posted by AnssiH

Sorry I was slow, I wrote this reply over the course of many days, taking a hour from here and hour from there teaching myself the relevant wave function and power series stuff... I'll try to get around to continue from here soon...

You owe me no apology! It is I who owe you an apology as I was the one who made it difficult for you to understand. As I said, the only rational for my errors was pure senility.

It reminds me of a joke on work performance I heard a long time ago.

When you are young and you know nothing, you have to think everything out. As you spend time learning your job you discover things here and there that you don't have to think about; it's just routine. If you spend enough time at a specific job you will eventually reach the point where it is all routine and you don't have to think at all -- and that is called “senility”.

Have fun -- Dick

Bombadil · 06-10-2009

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Originally Posted by Doctordick

Any explanation of anything can be interpreted in a manner which makes it a solution to my fundamental equation.

A fact of little real use!

Schrödinger's equation (and thus Newtonian mechanics) constitutes an approximate solution to that equation.

Perhaps this is of some use; it sure justifies Newtonian mechanics.

The fact that the fundamental equation is essentially a wave equation with fixed velocity demands SR transformations.

That is interesting; it implies there cannot be an explanation which violates SR.

That is worth knowing.

And more will be developed here.

Firstly the fact that anything can be interpreted in a way that makes it a solution to the fundamental equation is of little real use because the fundamental equation is a n-body equation and as such it can’t be solved directly.

Now since you have derived the Schrödinger equation all that this says is that Newtonian mechanics can be used as an approximation to any explanation. We can construct objects, bounce things off of each other, things like that. Also it tells us that if we know the form of V(x,t) then we can explain the behavior of one element.

Finally that “The fact that the fundamental equation is essentially a wave equation with fixed velocity demands SR transformations.” means that if two observers are in different reference frames and can use the same explanation then their explanations must obey SR transformations.

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Originally Posted by Doctordick

No! They are trained scientists who are well aware of the supposed explanation of whatever phenomena they are investigating: i.e., they are both using exactly the same explanation. If they are not, then there is no reason to even dream there is any association between their experiments. If they are using exactly the same explanation then both their valid elements and their hypothesized elements are the same. I repeat, there is “NO” information in the fundamental equation; all information is in their explanation!

But even if there was a way of telling which frame was moving wouldn’t they still be using the same explanation in the example you put forward in that any explanation that would be considered useful must explain whatever property it was that separated the reference frames. Both being trained scientists trained in the same explanation must use an explanation that would explain this or the explanation would be so flawed that they would be forced to fined a new one. That or they will have to agree to ignore whatever information it is that doesn’t agree with their explanation.

Is this the point though, that no matter what explanation that they decide to use if it is a flaw-free explanation then it has an interpretation that satisfies the fundamental equation and in that interpretation all elements must have a constant speed as seen from any frame. Which leads us to the Lorenz transformation.

So my question is, is this one of the defenses for the two reference frames using the same explanation without any way of telling which one is moving? From what I can tell the only other defense lies in their observations (that is what they are explaining) and not in the fundamental equation.

There is no information about any particular explanation in the fundamental equation all that we know is that any flaw-free explanation can be interpreted in a way that makes it a solution to the fundamental equation.

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Originally Posted by Doctordick

You keep trying to use the fundamental equation to deduce something about the explanation. That is absolutely impossible because any explanation of anything can be interpreted in a manner which makes it a solution to my fundamental equation. That fact contains no information of any kind!

But didn’t you use the fundamental equation to deduce that there exists an interpretation of an explanation which must obey SR or is this considered a subtly different topic and not something that we deduced about the explanation?

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Originally Posted by Doctordick

You don't seem to understand what relativity is all about. The central issue of relativity is that physics (the laws, equation and such) apply independent of your frame of reference. If you have the physics correctly specified and do all your calculations in one specified inertial frame then the issue of relativity does not even come up! You can use whatever frame you wish. In fact, that is the very central issue of relativity.

I suspect that the key words here are “If you have the physics correctly specified” and suspect that Newtonian mechanics is not the correct physics to use quite simply because using Newtonian mechanics it would be possible to construct an object that could accelerate to any speed and as far as I can tell if we obey the Lorenz transformation it is impossible to move an object faster then $V_0$ (the speed of light).

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Originally Posted by Doctordick

The effect is exactly as important as the probability that x_i=x_j. If that is not true, the impact of the Dirac delta function vanishes exactly. In our explanation of reality, our world view (which is the explanation we are working with), the probability that x_i=x_j for most of the elements making up our universe is so insignificant as to be non existent! So that two body relationship (Schrödinger's equation) is a very reasonable approximation. We are talking about that specific explanation and not the general implications of my equations (you should be well aware that there are none associated with my equation).

So we are only interested in explanations that will be equivalent to what we experience? That is when that two body equation (Schrödinger’s equation) is a close approximation in that no differences will be noticeable. Or the probability that $x_i=x_j$ is sufficiently small that it can be ignored.

Also aren’t there only no general implications because there is no general solutions and in order to get even a close approximation we must simplify it into something that we can solve (obtain implication of the explanation).

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Originally Posted by Doctordick

Again, you are confusing the fundamental equation with your explanation of reality. It is the physics (your explanation of reality) which must agree with the measures of both observers. Now, what does your world view say about a ruler you have in your office compared to that same ruler when you take it with you on a drive in your car. You want to get relativistic? If you get on a star cruiser and head for Alpha Centauri at 99% the speed of light and pull that same ruler out of your pocket. Does your world view suggest that you will find that ruler has changed its length? Or will it weigh down your pocket? Gee, if it did, you could use that fact to tell how fast you were moving (but that's a violation of relativity, the physics would be different). The observer on earth (who is using his Galilean inertial frame for his measurements) will look through the telescope and deduce the weight and length of the ruler. What will he say? My god, look how short that ruler has gotten and gee, it must weigh twenty pounds. Think about these things a little.

Firstly I can say that the laws of physics appear to be the same in every reference frame in so far as the laws that I can measure agree with each other. Further more there has not been one found that is not the same in any reference frame however this says nothing about why they are the same or what seems more important what are the requirements for this to be the case. The Lorenz transformation seems to be a consequence of this not a requirement.

This does however lead to what seems to be an interesting question which is, if the ruler was now shorter would we notice it. If it is shorter due to being in a moving reference frame won’t everything else now be shorter by the same amount so that, once again, we can’t tell if we are moving. If it is heaver, would it be possible to design a way to measure it so that we would be able to tell. Again I don’t think that we could, due to how things must be defined. What I’m saying is that even if the laws of physics do change if all that we are interested in is what is on the ship then I’m not sure that we can tell the difference, at least not if there exist a interpretation which obeys the fundamental equation. Again even with the speed of light (the oscillator) if it did change speed, could we tell. If we are truly using it to define simultaneity then there would be no way to time how long it takes to go one direction over the other (as we are timing it with its self) and so we would come to the same conclusions about how fast it moves.

Basically it seems that we have circularly defined the laws of physics in such a way that if one changes they all will change in such a way that it will not be noticeable, at least if there exist a interpretation which obeys the fundamental equation. That is, we can’t perform an experiment to find out which law has changed because it means that they all have changed.

Doctordick · 06-12-2009

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Originally Posted by Bombadil

Firstly the fact that anything can be interpreted in a way that makes it a solution to the fundamental equation is of little real use because the fundamental equation is a n-body equation and as such it can’t be solved directly.

Actually that is not quite true. There are some special circumstances where it can be but even then the amount of work required is beyond us.

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Originally Posted by Bombadil

Now since you have derived the Schrödinger equation all that this says is that Newtonian mechanics can be used as an approximation to any explanation.

What it actually says is that the elemental components upon which that explanation is based will approximately obey Newtonian mechanics. What you need to remember is that Newtonian mechanics includes the behavior of static structures. The issue there is that most explanations are like rocks and statues; their basic elements are presumed to be static.

The correct statement here is, if “we can construct objects, bounce things off of each other, things like that” there are conceivable entities whose hypothetical behavior will generate a V(x,t) such that Newtonian mechanics will approximately “explain the behavior of one element”. You have to keep your mind open to all the possibilities, not just presume the first one that pops into your head.

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Originally Posted by Bombadil

Finally that “The fact that the fundamental equation is essentially a wave equation with fixed velocity demands SR transformations.” means that if two observers are in different reference frames and can use the same explanation then their explanations must obey SR transformations.

Einstein proposed the axiom that the speed of light is the same in all frames and from that axiom deduced SR. My position is that his axiom could very well be “wrong” (and, down the road, I will demonstrate the power of dropping that axiom). I show that it is the fact that objects can be defined which requires SR. Objects are collections of elemental entities which can be seen as stable structures unto themselves: i.e., the rest of the universe can be ignored. If SR is not valid, you cannot define a structure such as clocks and rulers. Thus it is that any explanation which depends upon the existence of objects can not be valid. This is a much more powerful conclusion than Einstein’s SR.

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Originally Posted by Bombadil

But even if there was a way of telling which frame was moving ...

Try physically defining a frame of reference in a universe where objects can not exist.

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Originally Posted by Bombadil

That or they will have to agree to ignore whatever information it is that doesn’t agree with their explanation.

Ignoring information is one of the favorite moves of any unthinking person.

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Originally Posted by Bombadil

Is this the point though, that no matter what explanation that they decide to use if it is a flaw-free explanation then it has an interpretation that satisfies the fundamental equation and in that interpretation all elements must have a constant speed as seen from any frame. Which leads us to the Lorenz transformation.

You seem to overlook the fact that the fundamental equation is valid only in the rest frame of the universe. That fact was explicitly used in the derivation.

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Originally Posted by Bombadil

There is no information about any particular explanation in the fundamental equation all that we know is that any flaw-free explanation can be interpreted in a way that makes it a solution to the fundamental equation.

What you say is true but I wonder what is going on in your mind. Exactly what do you think such a statement means?

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Originally Posted by Bombadil

But didn’t you use the fundamental equation to deduce that there exists an interpretation of an explanation which must obey SR ...

I would have said that SR is embedded in the fundamental equation via that fixed velocity. Without it the universe cannot contain any objects and sans objects all we have is a manybody equation which can not be solved in general. How would you yourself exist in a universe without objects.

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Originally Posted by Bombadil

I suspect that the key words here are “If you have the physics correctly specified” and suspect that Newtonian mechanics is not the correct physics to use quite simply because using Newtonian mechanics it would be possible to construct an object that could accelerate to any speed and as far as I can tell if we obey the Lorenz transformation it is impossible to move an object faster then $V_0$ (the speed of light).

So you have pointed out a flaw in Newtonian mechanics. Isn’t it a much more powerful statement to say that if the elemental entities of your explanation do not obey SR, the explanation is flawed and just leave it there?

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Originally Posted by Bombadil

Basically it seems that we have circularly defined the laws of physics ...

That is indeed the apparent consequence of my deductions isn’t it.

Have fun -- Dick

Bombadil · 06-17-2009

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Originally Posted by Doctordick

What it actually says is that the elemental components upon which that explanation is based will approximately obey Newtonian mechanics. What you need to remember is that Newtonian mechanics includes the behavior of static structures. The issue there is that most explanations are like rocks and statues; their basic elements are presumed to be static.

Then even if an object is moving in our explanation we still use the same explanation of the object that we would if it was at rest, we just have to remember that it is now moving and transform the explanation of the object so that it still has a constant speed but is now moving partly along some x axis in addition to the $\tau$ axis. Which simply requires the Lorenz transformation?

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Originally Posted by Doctordick

The correct statement here is, if “we can construct objects, bounce things off of each other, things like that” there are conceivable entities whose hypothetical behavior will generate a V(x,t) such that Newtonian mechanics will approximately “explain the behavior of one element”. You have to keep your mind open to all the possibilities, not just presume the first one that pops into your head.

Are you saying that if we can construct objects then it is possible to choose elements in such a way as to create a V(x,t) that will explain the behavior of the elements as though they where obeying Newtonian mechanics?

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Originally Posted by Doctordick

Einstein proposed the axiom that the speed of light is the same in all frames and from that axiom deduced SR. My position is that his axiom could very well be “wrong” (and, down the road, I will demonstrate the power of dropping that axiom). I show that it is the fact that objects can be defined which requires SR. Objects are collections of elemental entities which can be seen as stable structures unto themselves: i.e., the rest of the universe can be ignored. If SR is not valid, you cannot define a structure such as clocks and rulers. Thus it is that any explanation which depends upon the existence of objects can not be valid. This is a much more powerful conclusion than Einstein’s SR.

So you have actually shown that rather then using the axiom, that the speed of light is the same in every reference frame to derive the Lorenz transformation. We can use the axiom that an object can exist in our explanation, where an object is simply a collection of elements that maintain their orientation and so can be explained separately of the rest of the universe.

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Originally Posted by Doctordick

Ignoring information is one of the favorite moves of any unthinking person.

But haven’t you already said that we have to ignore any information that will settle the question of what is the rest frame so that we can use the same explanation of an object that we would if the objects where at rest with the universe.

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Originally Posted by Doctordick

What you say is true but I wonder what is going on in your mind. Exactly what do you think such a statement means?

If we are using a flaw free explanation, meaning that whatever explanation it is that we are using, it is consistent with all of our observations, we certainly don’t have to be explaining things with a solution to the fundamental equation. There are in fact many ways to explain such information, however there must be a mapping between however we are explaining our observations and possible solutions to the fundamental equation. Whatever this maps our explanation into is an interpretation of our explanation. Or perhaps the mapping is the interpretation. (I’m not sure which one would be considered the interpretation) Either way what I am saying is that there is such a mapping that maps our explanation into a solution to the fundamental equation and whatever mappings satisfy this must depend on the observations, not on the possible solutions, to the fundamental equation.

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Originally Posted by Doctordick

I would have said that SR is embedded in the fundamental equation via that fixed velocity. Without it the universe cannot contain any objects and sans objects all we have is a manybody equation which can not be solved in general. How would you yourself exist in a universe without objects.

This question seems somewhat unanswerable as all of the explanations that I know any thing about require the use of objects, so I don’t even know how I would explain a universe where an object couldn’t exist., so I certainly cant say how I would exist in such a universe. More importantly (and perhaps more to the point you are trying to make) is that the word I or you would seem to imply that we are referring to an object which we are considering to not exist in the first place, So that we must find a new way to define such things before we can answer such questions.

Such explanations are perhaps for the time being of little interest as we can’t form objects, although it seems that it would be interesting to know what can be said about such explanations. While I may be wrong about this it seems that it may be outside of our current interests. It still seems that the Schrödinger equation could be used to explain a single element if we can choose the proper V(x,t) although perhaps the approximations that were made particularly setting $E=mc^2$ are too limiting to make this of much use in such a case.

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Originally Posted by Doctordick

So you have pointed out a flaw in Newtonian mechanics. Isn’t it a much more powerful statement to say that if the elemental entities of your explanation do not obey SR, the explanation is flawed and just leave it there?

Wouldn’t we be best off, though, to say that any explanation containing objects must obey SR or the explanation is flawed? or can any element be considered an object if objects can exist in our explanation?

AnssiH · 06-23-2009

Sorry about the delays, been quite busy, but should have some more time at my hands for a little while now...

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Originally Posted by Doctordick

What you are missing is knowledge of differentiation of trigonometric functions. The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x). That being the case, the second derivative of either is a simple change of sign.

$\frac{d^2}{dx^2}sin(x)=-sin(x)$
and
$\frac{d^2}{dx^2}cos(x)=-cos(x)$

Right, I see... I figure then that "second derivative" means "derivative of a derivative", and that that is exactly what you get by squaring the differential operator... Hmm, yeah, that bit seems to make perfect sense to me.

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It follows that anytime you see a differential equation of the form

$\frac{\partial^2}{\partial x^2}\Phi(x,t)=a^2\frac{\partial^2}{\partial t^2}$

you immediately know that the solution is a wave function (a sine or cosine) with a simple argument of the form ax+t or ax-t (the sign of t is immaterial as when you take the second derivative the factor ends up squared).

Hmm... okay.

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Furthermore, the second order differential equation can be written

$\left\{\frac{\partial^2}{\partial x^2}-a^2\frac{\partial^2}{\partial t^2}\right\}\Phi(x,t)=0$

which factors into

$\left(\frac{\partial}{\partial x}-a\frac{\partial}{\partial t}\right)\left(\frac{\partial}{\partial x}+a\frac{\partial}{\partial t}\right)\Phi(x,t)=0$

Yup.

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yielding two first order differential equations which will satisfy that equation.

$\left(\frac{\partial}{\partial x}-a\frac{\partial}{\partial t}\right)\Phi(x,t)=0$
and
$\left(\frac{\partial}{\partial x}+a\frac{\partial}{\partial t}\right)\Phi(x,t)=0$

The first is solved by a sine or cosine function with the argument (ax+t) and the second by a sine or cosine function with the argument (ax-t). These two functions are simple waves moving in opposite directions.

Right, so they are waves plotted against the change in t...

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If we call the argument of the sine and/or cosine function “z” then z=ax+t or z=ax-t. If z (the argument of the sine or cosine function) is a constant, we are talking about a specific point on that function. The question you need to ask yourself is, if t changes, how must x change in order for z (the argument) to remain constant? Obviously, if t increases by some amount, ax must either decrease by the same amount (for z=ax+t) or increase by the same amount (for z=ax-t). This can only be true if the change in x is identical to t divided by minus a or plus a (for the two cases).

"...identical to change in t divided by..." I suppose you meant to say.

One thing that puzzled me there for a moment. I understood that the argument of the sine/cosine function being a constant will satisfy the equations you laid down, but I thought, wouldn't it also be allowed that the partial derivative of x was equal or negative to "the partial derivative of t times a"... But it dawned on me that I guess that could only satisfy either one or the other of those first order differential equations, in which case the only possibility is if indeed the argument itself remains constant. I'm just laying my thoughts here just so you can make sure I'm understanding this correctly.

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It should be clear to you that the change in x has to be given by x=x₀+vt so the velocity of the wave must be one over a.

True.

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Sorry about that. The $\omega$ just stands for a constant. What it means depends upon your specific definition of angles in the sine function and I was being very careless there as x is clearly in radians. Normally angles are expressed in degrees but in physics (for differential simplicity) angles are almost always expressed in radians. 360 degrees equal $2\pi$ radians. I am going to edit that post and replace $\omega$ with v. I really should have done that originally but you know I am getting senile and sometimes what pops into my head is just wrong. Getting old is a pain in the ass.

Heh, I especially like your official "reason for editing: pure senility..."

And yeah, I know it's common human behaviour that we start making mistakes when we are good enough with something, as we become careless and start to overlook tiny mistakes without even realizing there ever was a mistake. I just view at it as a result of us interpreting the meaning of everything we see against our worldview, which includes our idea of the context of what we are seeing... I.e. we don't really look carefully but rather fill in most of the information with our expectations.

Fortunately, I'm so unfamiliar with this math stuff that I don't know what to expect, and thus I am unable to overlook the mistakes and blunt shortcuts; I just stumble over immediately when something doesn't make sense.

I'm sure that that tendency to fill the gaps with our expectations has caused you a lot of grief in trying to explain your analysis to people who interpret it in terms of their idea of what "physics" is (as in seeing it as describing "real objects" as oppose to an excercise of defining what "object" is).

Anyhow;

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Originally Posted by Doctordick

...Put this together with the fact that the differential of the sine function is the cosine function (and vice versa) and one has the fact that

$\frac{\partial^2}{\partial x^2}\Phi (x \pm vt)=-\frac{1}{v^2}\frac{\partial^2}{\partial t^2}\Phi(x \pm vt)$

is the differential equation of a traveling wave. The shape of Phi can be a sine or cosine wave where a specific value is maintained at any point where $x=x_0 \mp vt$ (in other words, $x \pm vt = x_0$ : i.e., the shape of the wave is unaltered and only moved to a greater or lesser value as t increases. The solution has nothing to do with the wave length of the wave and thus a pulse can be created by summing a whole set of different wave lengths. That is what is displayed on the wikipedia entry for “Wave_equation”.

Right, that seems to correlate with what you just explained in the previous post...

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Notice further that the squared relationship can be factored into a product of two first order equations with solutions moving in opposite directions. A lot of people think of the first order equations as more fundamental than the squared expression.

As does that.

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It reminds me of a joke on work performance I heard a long time ago.

When you are young and you know nothing, you have to think everything out. As you spend time learning your job you discover things here and there that you don't have to think about; it's just routine. If you spend enough time at a specific job you will eventually reach the point where it is all routine and you don't have to think at all -- and that is called “senility”.

Heh, exactly, and I guess you could call it that

There is this old little trick which you've probably seen.

You are supposed to count the o's in:
"THE MISITRY OF UNAUTHORISED
CAR-OWNMENTSHIP OF MINORS
IN THE STATE OF WEST-VERGINIA"

If you are fluent with english and just a tiny bit careless, you may count less than there actually is, as you tend to skip some of the "of" words as irrelevant. Anyone not familiar with english would have to look much more carefully and wouldn't "know" how to see that without the "of"'s.

I wouldn't call it all just senility, it's just that you are "skipping a lot of of's" and other bits and pieces that are irrelevant when a person already knows what you are trying to say.

Back to the OP:

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Originally Posted by Doctordick

...the transformation from one coordinate system to the other can be no more complex than $x'=\alpha x -\beta t$ and $t'=\gamma x -\delta t$ .

...

The first thing I need to point out is that the position of the point, x'=0, being the origin of the primed coordinate system must be at x=vt in the unprimed coordinate system as that is the definition of the primed frame's movement in the unprimed coordinate system. That implies that $\alpha(vt)-\beta t = 0$ : i.e., that is exactly the transformation which yields the origin of the primed coordinate system which is, by definition x'=0.

Yup.

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From that we can immediately deduce that $vt=\frac{\beta}{\alpha}t$ or, dividing by t, that $v=\frac{\beta}{\alpha}$ . This is first of those four equations we are looking for.

Yup.

uuup, gotta go, I'll continue from here soon...

-Anssi

AnssiH · 06-25-2009

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Originally Posted by Doctordick

The first thing I need to point out is that the position of the point, x'=0, being the origin of the primed coordinate system must be at x=vt in the unprimed coordinate system as that is the definition of the primed frame's movement in the unprimed coordinate system. That implies that $\alpha(vt)-\beta t = 0$ : i.e., that is exactly the transformation which yields the origin of the primed coordinate system which is, by definition x'=0. From that we can immediately deduce that $vt=\frac{\beta}{\alpha}t$ or, dividing by t, that $v=\frac{\beta}{\alpha}$ . This is first of those four equations we are looking for.

Just to be sure, "v" (speed of primed coordinate system) is different from "v_?" (the constant speed of the elements of the explanation)?

I.e. we are NOT considering the coordinate system to move at the exact speed of the surface of the expanding sphere. (I guess if we did, we would run into infinities at the final transformation process)

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We now need to lay out three additional valid independent equations involving the unknown coefficients. We know that both coordinate systems must yield a spherical surface originally defined by $x^2+y^2+z^2+\tau^2=v_?^2t^2$ : i.e., that surface must transform exactly into the surface $x'^2+y'^2+z'^2+\tau'^2=v_?^2t'^2$ in the primed coordinate system.

So, in other words, we are looking for a transformation process from "unprimed view" to the "primed view", which only modifies the "x" and the "t" parameters in such a way as to maintain a constant velocity of the elements; i.e. in such a way that the "primed explanation" also yields a perfect sphere, while referring to the same elements as the unprimed explanation.

(I.e. we are examining the logical roots of Lorentz transformation)

Quote:

Simply performing the transformation defined above must yield exactly that result. When we use the proposed transformations perform the transformation (substitute the explicit forms for each primed coordinate) we get the following relationship: $(\alpha x -\beta t)^2+y^2+z^2+\tau^2=v_?^2(\gamma t -\delta x)^2$ which expands algebraically directly into

$\alpha^2x^2-2\alpha x\beta t+\beta^2 t^2+y^2+z^2+\tau^2=v_?^2[\gamma^2t^2-2\gamma t \delta x+\delta^2 x^2]$

Check.

Quote:

or, collecting terms related to the unprimed coordinates of interest, we get

$(\alpha^2-v_?^2\delta^2)x^2+y^2+z^2+\tau^2=v_?^2t^2\left(\gamma^2-\frac{\beta^2}{v_?^2}\right)+2xt(\alpha \beta -v_?^2\gamma \delta)$

Actually struggled with that part a bit, perhaps you can show some of the algebraic steps that gets us there. I can see though that the point of this step was to move all the terms having to do with position (x') onto the left side, and all the bits having to do with t' on the right side.

Quote:

which, as it must still yield that spherical surface as represented in the unprimed frame must be exactly $x^2+y^2+z^2+\tau^2=v_?^2t^2$ . This fact immediately yields three additional equations involving those four coefficients.

So we now have four equations in four unknowns:

$v=\frac{\beta}{\alpha}\quad ; \quad \alpha^2-v_?^2\delta^2=1 \quad ; \quad \gamma^2-\frac{\beta^2}{v_?^2} = 1\quad and \quad \alpha \beta -v_?^2\gamma \delta = 0 \quad which\quad is \quad\gamma=\frac{\alpha\beta}{v_?^2\delta}$ .

Right, substituting 1, 1 and 0 accordingly into the above equation would immediately yield $x^2+y^2+z^2+\tau^2=v_?^2t^2$ , so we know they must amount to 1, 1, and 0... Pretty clever.

Quote:

You can then eliminate $\beta$ by substituting $\beta=\alpha v$ which is obtained from that first equation. This reduces the set to three equations in three unknowns:

$\alpha^2-v_?^2 \delta^2=1\quad ; \quad\gamma^2-\left(\frac{v}{v_?}\right)^2\alpha^2=1\quad and \quad\gamma=\frac{\alpha^2v}{v_?^2\delta}$

Check.
And I guess that middle one also answers my first question also; "v" is indeed different from "v_?" (I did almost mix them up in my mind couple of times)

Quote:

Eliminating $\alpha$ via $\alpha^2=1+v_?^2\delta^2$ (obtained from the new first equation) reduces the set to two equations in two unknowns:

$\gamma^2-\left(\frac{v}{v_?}\right)^2(1+v_?^2\delta^2)=1\quad and \quad\gamma=\frac{(1+v_?^2\delta^2)v}{v_?^2\delta}$ .

And finally, we can eliminate $\gamma$ via $\gamma^2=\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2$ (obtained by squaring the right hand equation of the two above). We thus arrive at a single equation with one unknown, “ $\delta$ ”:

$\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2-\left(\frac{v}{v_?}\right)^2 -\left(\frac{v}{v_?}\right)^2(1+v_?^2\delta^2)=1$ .

Hmm, I don't understand where that middle term comes from, I would have thought that at this point we'd have:

$\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2 -\left(\frac{v}{v_?}\right)^2(1+v_?^2\delta^2)=1$ .

Either way, I'm struggling with trying to understand the next step:

Quote:

If you multiply this equation through by $\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}$ , you will obtain

$(1+v_?^2\delta^2)\left(\frac{v}{v_?}\right)^2 -v_?^2\delta^2\left(\frac{v}{v_?}\right)^2=\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}$ .

I understand how to get the first term, but not the second one...
I think I'll pause here.

-Anssi

AnssiH · 06-26-2009

Even though I'm not through the first section, I can just take few things on faith and already start with the definitions and construction of a clock...

Quote:

Originally Posted by Doctordick

One very specific cavil still remains: in the derivation above, I set the velocity of a free element (that is neglecting interactions implied by the Dirac term, $\sum_{i \neq j} \delta(\vec{x}_i -\vec{x}j$ ) equal to v_? whereas the actual velocity is related to K , what seems, on first examination, to be a free parameter.

K is actually not a free parameter because we have not yet defined the actual measure of t. At this moment, t is an evolution parameter and is free to have any relationship with distances desired: i.e., velocities are essentially not defined. In order to relate that parameter to ordinary human perceptions, we have to design a mechanism to measure that parameter (essentially for reference purposes): i.e., it is required that a standard “clock” be defined before one can compare velocities as seen by different observers. In order to do that, one has to understand a few of the dynamic constraints implied by the model I have presented. In the design of my clock, for simple convenience, I will continue to use v_? as the fixed velocity implied by my fundamental equation.

...

I will define my standard clock to consist of two components: a mirror assembly and an oscillator. Both components are coherent macroscopic assemblies of elemental entities. The oscillator will have zero rest mass; therefore, every elemental entity which is part of the oscillator will have exactly zero momentum in the $\tau$ direction. The mirror assembly, on the other hand, will be massive: i.e., every elemental entity making up the mirror will have non zero mass. It follows that every event making up the mirror assembly must have significant momentum in the $\tau$ direction.

...

...

It follows from the above that, in macroscopic terms, although every elemental entity has exactly the same velocity, the mirror assembly is essentially an object moving parallel to the $\tau$ axis while the oscillator is an object (a coherent massless entity) moving parallel to the y axis. Since the entire assembly is infinite and uniform in the $\tau$ direction, motion in the $\tau$ direction yields utterly no changes in the structure of any part of our clock.

Check.

Quote:

If we now postulate that microscopic interactions (created by those Dirac delta interactions which we are essentially ignoring) between the mirror and oscillator are capable of reversing the sign of the oscillator's momentum upon contact with the mirror, the oscillator will bounce back and forth between the legs of the mirror assembly. Our standard clock will clearly have a period of 2L₀/v_?.

Yup.

Quote:

Since every event in the system described has non-negligible momentum only in the $(y,\tau)$ plane, we can display all important dynamic phenomena while considering only a cross section in that plane. Thus let us examine our standard clock as it appears in that cross section, paying particular attention to the associated velocity vectors.
...

[

It is interesting to note that T, the period of our standard rest clock, is identical to 1/v_? times the distance the mirror moves in the $\tau$ direction during one clock cycle.

So it is...

Quote:

Although actual position in the $\tau$ direction is a meaningless concept (as the entire object is infinite and uniform in that direction), our standard clock appears to be measuring the implied displacement of the mirror over time in that direction: i.e., we can infer that the mirror has moved a distance 2L₀ in the $\tau$ direction during one complete cycle.

Yup.

Quote:

This will turn out to be a very significant fact since the scale of the $\tau$ dimension is set by the form of the fundamental equation (setting the scale of any dimension sets the scale of all the others).

Hmmm... I'm struggling with this bit. I can take it on faith, but perhaps you could expand on it little bit.

Quote:

Now consider an identical standard clock in a moving reference frame: i.e., identical to the clock just described except for the fact that I will allow the momentum of the mirror assembly to be non negligible in the y direction.
...

Since all objects are uniform and infinite in the $\tau$ direction, it is reasonable to suppress actually drawing the objects themselves and, instead, deal entirely with the various displacement vectors. These displacement vectors are essentially v_?t where t is no more than a parameter of evolution: i.e., its scale is totally immaterial. It should be clear that these vectors contain all relevant information needed to predict the time evolution of the device. The only issue of great importance here is that, anytime the displacement vectors lead to identical (x,y,z) coordinates (which, in the $(y,\tau)$ plane which is being shown, means simply that two entities have identical y coordinates), microscopic interactions can occur between our macroscopic object anytime they lie on the same vertical line in these drawings (such a line specifies all points with the same y coordinate). This is important because all macroscopic objects are actually infinite and uniform in the $\tau$ direction, an issue which is no longer being explicitly shown in the drawing. Essentially, in the following drawings, x and z of every point in the picture is always identical so we need only concern our selves with a line at a y coordinate and the directions of the displacement vectors (essentially the angle $\theta$ they make with the tau axis).

All clear up to here.

Quote:

Note that the length of the moving clock is shown to be L'. This has been done because we know that the symmetry discussed in the previous section must require the Lorentz contraction to be a valid on any macroscopic solution if interactions with the rest of the universe may be neglected (up to this point the model was scale invariant): i.e., when we solve the problem in the moving clocks system we want the length of the clock as seen by the observer in that moving frame to be L₀.

So in this context, the length contraction is a consequence of any explanation having to accommodate for the fact that the speed of the probability waves must be isotropic across moving coordinate systems.

(And it is analogous to how relativistic length contraction is a direct consequence of having defined relativistic simultaneity, which is a direct consequence of having defined isotropic speed to light)

Btw, looking at your drawing, it's kind of interesting how simple it is to visualize L' through L₀ and the angle $\theta$ ... I guess someone used to math can see the same thing as easily from $L'=L_0\sqrt{1-sin^2(\theta)}$ :P

Quote:

We use the scale freedom in our model to set that length (as seen from the rest system) to be L'; then and only then can we seriously call the clocks identical. This will require $L'=L_0\sqrt{1-sin^2(\theta)}$ (the inverse of the relativistic transformation deduced earlier:

I.e. $\alpha= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}$

Quote:

i.e., in order to get the length of the moving clock in the primed coordinate system we have to multibly by $\alpha$ ). Note that $sin(\theta)$ is exactly the apparent velocity of the moving clock divided by the velocity of the elemental entities, v_?, which actually has nothing to do with time.

Hmmm, so it is...
$v = sin(\theta)v_?$
$sin(\theta) = \frac{v}{v_?}$

So in other words: $L'=L_0\sqrt{1-sin^2(\theta)}$ , just like you said.

Still haven't walked through the first section of the OP, but have no troubles taking some bits on faith as it appears to be quite analogous to standard relativity.

Quote:

Since all velocities are v_?, it follows directly that d₁ + d₂ = S. Please note that everything so far is being graphed as seen in the frame of the rest clock: i.e., S=v_?T_m, where T_m is the period of the moving clock as seen from the rest frame.

Notice that the following geometric figure is embedded in the previous diagram.

Once again, since the triangles A and B are identical as are the triangles a and b, we discover that one clock cycle, rather surprisingly, measures exactly the length of time it takes the mirror to move the distance 2L₀ in the $\tau$ direction.

Now that took me a moment to dissect, and while I did find out that everything appears to be valid and without error, I walked through it in a reaaally topsy turvy way, and I really don't know in which order the logical steps should be carried out to end up with that graph.

Here's what I did to replicate your drawing:

- I laid down the mirrors in the (y,tau)-plane, setting $L_0$ between them.
- I drew velocity vectors for them (essentially $v_?$ ) so that they were moving partially in y-direction (just like in your diagram)
- I set the distance between mirrors to L' (Got that by rotating the vector $L_0$ so that it was orthogonal to those velocity vectors)

That allowed me to see the triangle A, and I could draw the triangle B, and see the position you have marked as "reflection must occur at this point", without really understanding how you originally found it out.

But I did set up an animation with specific constant velocities, and found out that that is indeed exactly the position where the reflection occurs.

Also I could from that point on just draw those $L_0$ vectors along tau-axis, and also see the triangles a and b, and once again by animating the whole thing I found out that the second reflection indeed does occur exactly where you are saying it would, and everything overall lines up just like you drew them...

...but still I have no idea how did you originally get the magnitudes of d₁ and d₂ (I mean, how did you find out the the first reflection point etc.)... And how did you arrive to $d_1 + d_2 = S$ ... I'm sure there is a better way to understand that than what I have in my mind right now.

btw, Arkain asked earlier if I could do some sorts of visualizations or animations that would be helpful in understanding this stuff, and I'm thinking this part might be something that should be relatively easy to visualize through animation... We'll see...

Quote:

Again, although our standard clock was originally designed to measure time, it appears that what is actually being measured here is inferred displacement in the $\tau$ direction. Once again, I assert that this is a very significant fact. One cycle of both the moving clock and the rest clock measure exactly an inferred displacement of 2L₀ in the $\tau$ direction; however, the time required for our moving clock to accomplish this feat is given by the length of the displacement vector S which is very clearly longer than 2L₀. It follows, from the fact that everything here moves at the velocity v_?, that S = v_?T_m or, the period of the moving clock (as seen in the rest observer's frame) is given by ,

$T_m=\frac{S_m}{v_?}= \frac{2L_0}{\sqrt{1-sin^2(\theta)}}\left(\frac{1}{v_?}\right)=\frac{T}{\sqrt{1-sin^2(\theta)}}$

Which happens to be exactly the result expected from the standard Lorentz relation: i.e., the moving clock appears to run slow by exactly the factor $\sqrt{1-sin^2(\theta)}$ .

Yes, so it appears...

I'll continue from here soon...

-Anssi

Doctordick · 06-28-2009

Quote:

Originally Posted by Bombadil

Then even if an object is moving in our explanation we still use the same explanation of the object that we would if it was at rest,

Somehow I get the impression that you are still missing the central issue. What we are talking about is a situation where we have an explanation of the universe as we see it. That means our explanation is not bothered by the fact that our reference frame (the coordinate system we use to represent our experiments) is moving with respect to the rest of the universe or not; the explanation includes the issues brought up by that circumstance. This has to do with absolutely any experiment we choose to examine. We choose to examine a few special circumstances because they point out, in detail, the problems with the old fashion Euclidean transformations used by Newton and, in fact, give us exactly what those transformations have to look like. The required transformations are a fact of life; your explanation must obey them and what kind of machinations you have to go through to achieve that is essentially beside the point.

Quote:

Originally Posted by Bombadil

Are you saying that if we can construct objects then it is possible to choose elements in such a way as to create a V(x,t) that will explain the behavior of the elements as though they where obeying Newtonian mechanics?

The “if we can construct objects then” does not need to be there. We don’t need the existence of “objects” (collections of elements which can be considered as universes unto themselves) in order to create a V(x,t) such that a single element will appear to approximately obey Newtonian mechanics. Given any collection of elements, it is always possible to imagine (or create fictional elements) such that we can see the impact of the rest of the universe (including those fictional elements) as causing an interaction of the form V(x,t). If we cannot create objects, then we cannot create rulers; this is another problem.

Quote:

Originally Posted by Bombadil

So you have actually shown that rather then using the axiom, that the speed of light is the same in every reference frame to derive the Lorenz transformation. We can use the axiom that an object can exist in our explanation, where an object is simply a collection of elements that maintain their orientation and so can be explained separately of the rest of the universe.

I essentially agree with what you are saying except of the fact that I don’t feel that the existence of objects can be considered an axiom. We could certainly conceive of a “universe” without objects and even talk about some of the constraints on explaining that universe; it could even be a subset of our universe which had negligible impact upon that portion of the universe we deal with on a day to day routine. Say the inside of stars (or perhaps the interior of nuclei) or the structure of those great voids between the clusters of stars. We just couldn’t hypothesize rulers or clocks in such a realm.

Quote:

Originally Posted by Bombadil

But haven’t you already said that we have to ignore any information that will settle the question of what is the rest frame so that we can use the same explanation of an object that we would if the objects where at rest with the universe.

There is a big difference between “having to ignore” and “ignoring”. If a decent approximation can be achieved by “ignoring” then, to the same extent that the approximation is decent, the Lorenz transformations must be decent. And an object (which by definition presumes there is no impact upon its fundamental structure from the rest of the universe) requires the Lorenz transformation to be a central characteristic of any flaw free explanation. And, in order for that explanation to be flaw free, it cannot be changed by moving into one of those regions where objects can not exist. The explanation is a flaw-free explanation of “the universe” not just part of the universe.

Quote:

Originally Posted by Bombadil

If we are using a flaw free explanation, meaning that whatever explanation it is that we are using, it is consistent with all of our observations, we certainly don’t have to be explaining things with a solution to the fundamental equation.

We cannot solve the general equation anyway so what difference does this make? What I am saying is that the elements of your explanation must obey that equation. I think what you are missing is the fact that most all common explanations are what is called “static”: i.e., none of the fundamental elements change in any way so obedience to the fundamental equation is a trivial issue. Just as the structure of my house obeys Newtonian mechanics: it just stands there without moving.

“Dynamic” explanations are a much more complex issue than are “static” explanations. In fact “physics” is perhaps the only field which makes much use of dynamic explanations of any kind at all. All the other fields are just far too unknown to even consider the existence of dynamic explanations. Chemistry and biology are now only beginning to think about dynamic explanation and those areas are often referred to as “physical chemistry” and “physical biology”. Physics sort of has ownership of dynamic explanations.

Quote:

Originally Posted by Bombadil

So that we must find a new way to define such things before we can answer such questions.

Well life is really not that complex; we have to think about what we want to know and how we might obtain it. Many many years ago, I was working on analyzing my fundamental equation (at a University) when an economist asked me what it had to do with economics. So I laid out the essentials of what economic knowledge was of interest and, using the variables as arguments of my fundamental equation, derived most of the common economic relationships taught in freshman economics. I gave it to him and failed to keep a copy for myself. I wish I had because some of it was rather interesting and might very well apply to situations where objects don't exist.

Quote:

Originally Posted by Bombadil

Wouldn’t we be best off, though, to say that any explanation containing objects must obey SR or the explanation is flawed? or can any element be considered an object if objects can exist in our explanation?

I think it is the issue of the existence of “rulers” without which we couldn’t talk about “coordinate frames of reference”. Without “frames of reference” transformation between frames of reference is a pretty meaningless concept.

Have fun -- Dick

Doctordick · 06-28-2009

Quote:

Originally Posted by AnssiH

"...identical to change in t divided by..." I suppose you meant to say.

Yes, you are absolutely correct.

Quote:

Originally Posted by AnssiH

And yeah, I know it's common human behavior that we start making mistakes when we are good enough with something, as we become careless and start to overlook tiny mistakes without even realizing there ever was a mistake.

...

I wouldn't call it all just senility, it's just that you are "skipping a lot of of's" and other bits and pieces that are irrelevant when a person already knows what you are trying to say.

On reading this whole section, I am moved to tell you about a book I bought in Golden, Colorado. I saw it in a “used book” store near where my wife was visiting a quilt store. (She is very much into quilting and, wherever we go, if there is a quilt store around, she wants to see it. New patterns and such.)

At any rate, the book I bought was “Consciousness Explained” by Daniel Dennett. I read the whole book in about three days. It was extremely interesting as he expressed a great number of very rational insights which had not occurred to me; but I don’t think he really “explained” consciousness for reasons I don’t have time to go into. I would very much recommend the book.

I googled Dennett and discovered two sites you might find interesting. They are videos which talk about some of the stuff covered in the book but are very limited as compared to the book itself.

On false memories

and

On memes

I think he would find a lot of my work very applicable to some of the issues he brings up in his book.

Quote:

Originally Posted by AnssiH

(I.e. we are examining the logical roots of Lorentz transformation)

Absolutely correct!

Quote:

Originally Posted by AnssiH

I can see though that the point of this step was to move all the terms having to do with position (x') onto the left side, and all the bits having to do with t' on the right side.

Not quite accurate. The point was (now that we have eliminated the primed coordinates) to rearrange the terms such that we reproduced the correct form of the equation in the unprimed coordinates. Thus we must collect all the terms proportional to x² on the left and all the terms proportional to v_?t² to the right. Where we go with the remaining terms (those proportional to 2xt; the only other terms) is actually immaterial as terms like 2xt do not exist in the equation in the unprimed coordinates. In detail, the algebra looks like this after one multiplies out the parenthesis on the right

$\alpha^2x^2-2\alpha x\beta t+\beta^2t^2+y^2+z^2+\tau^2=v_?^2\gamma^2t^2-v_?^22\gamma t\delta x+v_?^2\delta^2x^2$

then rearranging the coefficient of each term, we have

$\alpha^2x^2-\alpha \beta 2xt+\beta^2t^2+y^2+z^2+\tau^2=\gamma^2 v_?^2t^2-2xtv_?^2\gamma \delta+v_?^2\delta^2x^2$

Moving the x² term on the right to the left and the t² term on the left to the right (which changes the sign of both) we have

$\alpha^2x^2 - v_?^2\delta^2x^2-\alpha \beta 2xt+y^2+z^2+\tau^2=\gamma^2 v_?^2t^2-\beta^2t^2-2xtv_?^2\gamma \delta$

Moving the 2xt term on the left to the right and factoring out the appropriate terms this may be written

$(\alpha^2 - v_?^2\delta^2)x^2+y^2+z^2+\tau^2=(\gamma^2 v_?^2-\beta^2)t^2+(\alpha\beta-v_?^2\gamma \delta)2xt$

All that is left to do is divide the coefficient of the first term on the right by v_? squared in order to factor that same term and one has

$(\alpha^2 - v_?^2\delta^2)x^2+y^2+z^2+\tau^2=\left(\gamma^2-\frac{\beta^2}{v_?^2}\right)v_?^2t^2+(\alpha\beta-v_?^2\gamma \delta)2xt$

The rest you seem to have understood.

Quote:

Originally Posted by AnssiH

And I guess that middle one also answers my first question also; "v" is indeed different from "v_?" (I did almost mix them up in my mind couple of times)

Context is often as important in mathematics as it is in any language. We tend to lack a sufficient number of symbols to make everything different (at least we don’t like to use that many).

Quote:

Originally Posted by AnssiH

Hmm, I don't understand where that middle term comes from, I would have thought that at this point we'd have:

$\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2 -\left(\frac{v}{v_?}\right)^2(1+v_?^2\delta^2)=1$ .

You are absolutely right once again. That middle term just shouldn’t be there. When typing in the LaTex, I apparently retyped the coefficient. Anything to generate a little confusion right??? Sorry about that. I have edited out the error. Thank you again.

Quote:

Originally Posted by AnssiH

Either way, I'm struggling with trying to understand the next step:

Forgetting that added middle term is the most important thing here. Working from your (correct) expression, performing the indicated multiplication one has

$\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2 -\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}\left(\frac{v}{v_?}\right)^2(1+v_?^2\delta^2)= \frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}$ .

In the first term, the denominator gets rid of the square of the $(1+v_?^2\delta^2)$ term and the numerator just cancels out. In the second term, the $(1+v_?^2\delta^2)$ term cancels out and, finally the right hand side consists of exactly what we are multiplying by.

It appears that, once again, your only confusion arises from my errors. Sorry about that. It is quite clear to me that you are the only one reading this stuff at all carefully.

Quote:

Originally Posted by AnssiH

Hmmm... I'm struggling with this bit. I can take it on faith, but perhaps you could expand on it little bit.

The requirement that the probability of seeing an element at a future time (when it started at a specific point and interactions with the rest of the universe can be ignored) must be an expanding sphere requires all scale of all dimentions to be established if one is established. That is all there is to it.

Quote:

Originally Posted by AnssiH

Now that took me a moment to dissect, and while I did find out that everything appears to be valid and without error, I walked through it in a reaaally topsy turvy way, and I really don't know in which order the logical steps should be carried out to end up with that graph.

Self consistency is self consistency. I am not bothered at all by the way you stepped through the problem. Maybe there were a few steps which presumed the correctness of the diagram but they could have been reached by other methods. The fact that the total is internally self consistent is the single most important issue.

Quote:

Originally Posted by AnssiH

...but still I have no idea how did you originally get the magnitudes of d₁ and d₂ (I mean, how did you find out the first reflection point etc.)... And how did you arrive to $d_1 + d_2 = S$ ... I'm sure there is a better way to understand that than what I have in my mind right now.

Well, first of all, I am somewhat proficient at plane geometry: i.e., I pretty well know where these lines will intersect. The fact that d₁+d₂ has to equal S comes directly from the idea that time (as I have defined it) is an interaction parameter (time being the same is equivalent to “interactions can take place”). The pulse of photons (that would be the oscillator) interacts twice with the left hand mirror. Once when the experiment begins and again at the end of the experiment. Once again, presuming all other interactions can be ignored (and presuming that interaction with the right hand mirror face does nothing except to reverse the travel direction of the oscillator), the position of the mirror and the oscillator must change by exactly v_?t where t is the specified interaction time. Since the elapsed interaction time for both elements must be the same and v_? must be the same, d₁+d₂ has to equal S.

Quote:

Originally Posted by AnssiH

...btw, Arkain asked earlier if I could do some sorts of visualizations or animations that would be helpful in understanding this stuff, and I'm thinking this part might be something that should be relatively easy to visualize through animation... We'll see...

that would be quite nice.

You seem to be picking up on everything quite well. Thanks for your attention.

Have fun -- Dick

AnssiH · 07-01-2009

Quote:

Originally Posted by Doctordick

Not quite accurate. The point was (now that we have eliminated the primed coordinates) to rearrange the terms such that we reproduced the correct form of the equation in the unprimed coordinates. Thus we must collect all the terms proportional to x² on the left and all the terms proportional to v_?t² to the right. Where we go with the remaining terms (those proportional to 2xt; the only other terms) is actually immaterial as terms like 2xt do not exist in the equation in the unprimed coordinates. In detail, the algebra looks like this after one multiplies out the parenthesis on the right

$\alpha^2x^2-2\alpha x\beta t+\beta^2t^2+y^2+z^2+\tau^2=v_?^2\gamma^2t^2-v_?^22\gamma t\delta x+v_?\delta^2x^2$

....

Okay, I was able to follow that whole thing now. (And spotted a small typo, the last $v_?$ isn't squared there, but it was correct in the subsequent steps)

Btw, I found a very handy LaTex editor, should make life easier (and less error-prone

LaTeX Equation Editor for the Internet
It's exactly what I was dreaming of earlier, you can just type in some LaTex code and it is attempting to render it all the time. Or, you can just click on the buttons above to create whatever you need to, and paste the code into your post.

It seems to have some clever features in that you can just paint parts of the code and press parenthesis or fraction button, and it will put the painted parts inside the added code.

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You are absolutely right once again. That middle term just shouldn’t be there...

...Working from your (correct) expression, performing the indicated multiplication one has

$\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2 -\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}\left(\frac{v}{v_?}\right)^2(1+v_?^2\delta^2)= \frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}$ .

In the first term, the denominator gets rid of the square of the $(1+v_?^2\delta^2)$ term and the numerator just cancels out. In the second term, the $(1+v_?^2\delta^2)$ term cancels out and, finally the right hand side consists of exactly what we are multiplying by.

Right, okay, that seems pretty trivial now.

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It is quite clear to me that you are the only one reading this stuff at all carefully.

I was just thinking about the same... Hmmm... :I

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The requirement that the probability of seeing an element at a future time (when it started at a specific point and interactions with the rest of the universe can be ignored) must be an expanding sphere requires all scale of all dimentions to be established if one is established. That is all there is to it.

Oh, right, of course.

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Self consistency is self consistency. I am not bothered at all by the way you stepped through the problem. Maybe there were a few steps which presumed the correctness of the diagram but they could have been reached by other methods. The fact that the total is internally self consistent is the single most important issue.

Yeah, I was thinking about the same thing, and certainly I was presuming the correctness, otherwise I would not have known which steps to take. I'm quite confident it's valid, and part of the reason I asked about that was to get a better ideas for how to represent the issue as clearly as possible in an animation. Anyway, I think I know what I'll do now...

But first, back to OP, the Lorentz transformation part:

Quote:

Originally Posted by Doctordick

And finally, we can eliminate $\gamma$ via $\gamma^2=\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2$ (obtained by squaring the right hand equation of the two above). We thus arrive at a single equation with one unknown, “ $\delta$ ”:

$\frac{(1+v_?^2\delta^2)^2}{v_?^2\delta^2}\left(\frac{v}{v_?}\right)^2 -\left(\frac{v}{v_?}\right)^2(1+v_?^2\delta^2)=1$ .

If you multiply this equation through by $\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}$ , you will obtain

$(1+v_?^2\delta^2)\left(\frac{v}{v_?}\right)^2 -v_?^2\delta^2\left(\frac{v}{v_?}\right)^2=\frac{v_?^2\delta^2}{(1+v_?^2\delta^2)}$ .

So that's all clear now.

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The left hand side clearly reduces to $(v/v_?)^2$ . Thus if we multiply through by $(1+v_?^2\delta^2)$ we obtain a very simple result:

$\left(\frac{v}{v_?}\right)^2 +v_?^2\delta^2\left(\frac{v}{v_?}\right)^2=v_?^2\delta^2\quad or \quad\left[1-\left(\frac{v}{v_?}\right)^2\right]v_?^2\delta^2=\left(\frac{v}{v_?}\right)^2$

which is easily solved for $\delta$ (just divide through by the coefficient of $\delta$ and take the square root of both sides of the equation. The final result is:

$\delta= \left(\frac{v}{v_?}\right)\frac{1}{v_?\sqrt{1-\left(\frac{v}{v_?}\right)^2}}$ .

That step was throwing me off a bit, but I finally realized that, if I just focused on the equation on the right, I could just take the square root of it:
$\sqrt{\left[1-\left(\frac{v}{v_?}\right)^2\right]v_?^2\delta^2}=\sqrt{\left(\frac{v}{v_?}\right)^2}$

$\sqrt{1-\left(\frac{v}{v_?}\right)^2}v_?\delta=\left(\frac{v}{v_?}\right)$

And move everything but the $\delta$ to the right side, to arrive at:

$\delta= \left(\frac{v}{v_?}\right)\frac{1}{v_?\sqrt{1-\left(\frac{v}{v_?}\right)^2}}$

So, I guess I just don't know what you meant by "divide through by the coefficient of $\delta$ ...", as when I tried that (however I interpreted it), I wasn't finding my way to your result.

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Since $\alpha^2=1-v_?^2\delta^2$ ,

Oop, looks like another typo there at the OP. I think that should be: $\alpha^2=1+v_?^2\delta^2$

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we know that

$\alpha^2= 1+\left(\frac{v}{v_?}\right)^2\frac{1}{\left[1-\left(\frac{v}{v_?}\right)^2\right]}$ .

Hmmm... Okay, this once again involves algebraic steps that I'm not familiar with but have to guess, so let me know whether this is valid route. Just substituting $\delta$ first:

$\alpha^2=1+v_?^2 \left(\left(\frac{v}{v_?}\right)\frac{1}{v_?\sqrt{1-\left(\frac{v}{v_?}\right)^2}}\right)^2$

$\alpha^2=1+v_?^2 \left(\frac{v}{v_?}\right)^2 \frac{1}{v_?^2\left[{1-\left(\frac{v}{v_?}\right)^2}\right]}$

$\alpha^2=1+ \left(\frac{v}{v_?}\right)^2 \frac{1}{{1-\left(\frac{v}{v_?}\right)^2}}$

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Use “common denominators” to add the two terms above and you will discover that the square root of the result is:

$\alpha= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}$ .

Okay, once again new stuff to me, but after some head scratching I think it goes little bit something like this:

$\alpha^2=1+ \left(\frac{v}{v_?}\right)^2 \frac{1}{{1-\left(\frac{v}{v_?}\right)^2}}$

$\alpha^2= \frac{1}{1}+ \frac{\left(\frac{v}{v_?}\right)^2}{{1-\left(\frac{v}{v_?}\right)^2}}$

$\alpha^2=\frac{1-\left(\frac{v}{v_?}\right)^2}{1-\left(\frac{v}{v_?}\right)^2}+ \frac{\left(\frac{v}{v_?}\right)^2}{{1-\left(\frac{v}{v_?}\right)^2}} = \frac{1-\left(\frac{v}{v_?}\right)^2 + \left(\frac{v}{v_?}\right)^2}{1-\left(\frac{v}{v_?}\right)^2} = \frac{1}{1-\left(\frac{v}{v_?}\right)^2}$

So:

$\alpha= \sqrt{\frac{1}{1-\left(\frac{v}{v_?}\right)^2}}$

And I suppose that can then be written:

$\alpha= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}$

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Finally, since $\beta=\alpha v$ and $v_?^2\delta=\alpha v$ , it is quite obvious that $\gamma=\alpha \frac{\beta}{v_?^2\delta}$ clearly implies $\gamma=\alpha$ .

Hmmm, I remember $\beta=\alpha v$ , but I don't remember (nor spot from the OP) where did we establish $v_?^2\delta=\alpha v$ ... Also could not figure out how you got the $\gamma=\alpha \frac{\beta}{v_?^2\delta}$

Perhaps you could expand on that a bit...?

Phew, the Lorentz-part almost done already!

I'd have a few things to say to lurkers, about what all this implies, but I'll save it for little bit later... I hope these explicit baby-steps make it easier for all the lurkers to follow the thing though, and perhaps think about the implications themselves.

-Anssi

An “analytical-metaphysical” take on Special Relativity!

Hypography?

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