On The Non Relativity Of Simultaneity

[master_coda]

Originally posted by TheAtheistKing
Ok I really really need to know how to make those math symbols, then this will be simple, can you explain that to me?

I can't explain it any better than the "Introducing LaTeX Math Typesetting" thread in the General Physics forum.

 

[TheAtheistKing]

We are going to consider the following event:

Initial state
R1: A...B
R2:_____A`...C`...B`

Intermediate state
R1:_______A...B
R2:_______A`...C`...B`

Final state
R1:__________________A...B
R2:_______A`...C`...B`

Let the total time of the event in this frame be denoted by


\Delta t'

This time corresponds to the amount of time it takes B to move from A` all the way to B`. The distance traveled in this frame is L0, which is the rest length of the ruler. Thus, the speed of b relative to b` is

v(bb') = \frac{L_0}{\Delta t'}

Assumption 1:

L= L_0 \sqrt{1-\frac{v^2}{c^2}

Now consider things from ruler one's perspective. Suppose there is a stationary clock located at B. It will note when A` coincides with B, and later this same clock will note when B` coincides with B. Thus, this clock can make an inertial time measurment of this event which I will denote by

\Delta t

Now, the distance that the point B` traveled in ruler one's frame is L. Thus, the speed of B` relative to B is

v(b'b) = \frac{L}{\Delta t}

And since the speed is relative, we must have

v(b'b) = v(bb')

Hence, we must have

\frac{L_0}{\Delta t'} = \frac{L}{\Delta t}

From which the time dilation formula follows. I will pick this up later, I have to go right now.

 

[Hurkyl]

In other words:


Diagram as drawn in the primed reference frame (that of ruler 2):

Time t'0:
         A  B
Ruler 1: |--|
Ruler 2:    |--|---|
            A' C'  B'


Time t'1:
            A  B
Ruler 1:    |--|
Ruler 2:    |--|---|
            A' C'  B'

Time t'2:
                A  B
Ruler 1:        |--|
Ruler 2:    |--|---|
            A' C'  B'

The two rulers, AB and A'B' have the same proper length, L_0.

At (primed) time t'_0, B and A' are at the same event of space-time.

At (primed) time t'_1, A and A' are at the same event.

The point C' is defined to be the point on ruler 2 which is at the same event as B at (primed) time t'_1.

At (primed) time t'_2, A and B' are at the same event.


You've defined \Delta t' := t'_2 - t'_0.

The velocity, v of ruler 2 (as measured in the primed system) must be v = L_0 / \Delta t'.

By length contraction, we know the length L of ruler 1 as measured in the primed coordinate system; it is L = L_0 \sqrt{1 - v^2/c^2}.


Now, we consider things in the unprimed reference frame (that of ruler 1):

Time t0:
         A      B
Ruler 1: |------|
Ruler 2:        |--|
                A' B'


Time t2:

         A      B
Ruler 1: |------|
Ruler 2:     |--|
             A' B'

You define \Delta t := t_2 - t_0.

The velocity of ruler 2 (as measured in the unprimed system) must be v = L / \Delta t.

By (insert favorite reason here), these two velocities must be the same, thus:

\frac{L}{\Delta t} = \frac{L_0}{\Delta t'}

Okay...

 

[TheAtheistKing]

Hey Hurkyl, your diagrams are better than mine so I will change, but please if you can, edit your third state diagram. You have the event ending when A coincides with A`, but the event I am analyzing ends in a state where B coincides with B`. As for your usage of the word 'event' in spacetime, in state theory the word 'state' is used, to clearly differentiate between which theory is being used. In state theory, a state is a moment in time, a point of time, and instant, and as such has no duration. In contrast, in state theory an event must have a duration, an event must last for some amount of time. An event has a beginning, and an ending. The beginning of an event is a state, and the end of an event is a state, and there is a binary relation on the set of states 'before' such that the following statement is true in any reference frame:

Let [X,Y] denote an event. X denotes the beginning of the event, and Y denotes the end of the event. The following binary relation holds:

X before Y

Undefined binary relation on the set of states: before
In state theory, this binary relation is not defined logically, instead an operational definition is used. That means that some experiment serves to give meaning to the term. For example, consider the dropping of a stone and a feather in the Earth's gravitational field. Suppose the two are released simultaneously from the same height above the Earth's surface. The experimental result is that the stone hits the ground BEFORE the feather hits the ground every time. Hence, such an experiment can be used to 'operationally define' the binary relation 'before'. Now, on the moon both objects would hit the ground simultaneously because there is no air resistance, but this need cause us no concern, because on the moon the experiment would be that we release the feather BEFORE the stone, from the same height, and the result will always be that the feather hits the ground before the stone hits the ground.
In state theory there are two defined binary relations on the set of states they are:

Let X denote a state, and let Y denote a state.
Definition: X simultaneous to Y if and only if (not(x before Y) and not (y before x))
Definition: X after y if and only if y before x

Some axioms are:

Axiom 1 (Non circularity):
For any state x, and any state y, not (X before y and y before x)

Axiom 2 (Transitivity):
For any states x,y,z: If x before y and y before z then x before z.

Axiom 3 (Quantum Hypothesis QH):
For any state x, there is at least one state y, such that x before y AND not (there is at least one state z, such that x before z and y before z)

Axiom 4 (First state):

There is at least one state A, such that for any state B, if not (A simultaneous to B) then A before B.

There are others, but you get the idea. Basically, state theory models time using the natural number system. In other words, in state theory there is an absolute order to the states of the universe, which is totally independent of reference frame, inertial or otherwise. At any rate, unless relativity is overthrown, I don't see state theory being seriously worked on by the physics community, even though a whole host of intuitively true statements about time can be derived from a very small set of axioms.

The quantum hypothesis is the statement that there is at least one possible next state of the universe (from which it follows that time cannot end, or equivalently that relative motion cannot cease, which fact is obvious via the senses, thermodynamics, or even derivable from Newton's laws of motion). The statement that there is at most one possible next state is called the determinism conclusion DC, and is one of the theorems of state theory. In order to reach the determinism conclusion, the mathematical theory of probability is used. I don't wish to discuss state theory further, because I consider it more important to address whether or not simultaneity is absolute. It isn't until after you are certain that simultaneity is absolute and not relative, that you should begin seriously considering the temporal modal logic that is used in state theory.

 

[TheAtheistKing]

Consider the uniform motion of C' from A to C, as illustrated below:

Initial State
         C'
Ruler 1: |
Ruler 2: |------------|
         A            C

Intermediate state

              C'
Ruler 1:      |
Ruler 2: |------------|
         A    B       C

Final State

                      C'
Ruler 1:              |
Ruler 2: |------------|
         A    B       C

Suppose that there are four clocks, one at each of the four locations A,B,C,C`, and that clocks A,B,C are not moving relative to one another, and tick at the same rate. Now consider time readings. When C' coincides with A, clocks C' and A read something, when C' coincides with B, clocks C' and B read something, and when C' coincides with C clocks C' and C read something. Let us presume that clocks A,B,C are synchronized at the beginning of the event. Therefore, when C' coincides with A, clocks A,B,C all read X. When C' coincides with B, clock B will read somewhat greater than X, say it reads X+ \Delta t_1. Therefore, the total time of travel of C' from A to B can be computed by subtracting the reading of clock A when C' coincided with it, from the reading of clock B when C' coiincided with it. Thus:

(X+\Delta t_1) - X = \Delta t_1

When C' coincides with C, clock C will read somewhat greater than X +\Delta t_1, say X + \Delta t_1 + \Delta t_2. Therefore, the total time of travel of C' from B to C is computed by subtracting the reading of clock B when C' coincided with it, from the reading of clock C when C' coincided with it. Thus:

(X+\Delta t_1 + \Delta t_2) - (X+\Delta t_1) = \Delta t_2

The total time of travel of C' from A to C in the unprimed frame is

\Delta t = \Delta t_1 + \Delta t_2


Let L_0 denote the distance from A to C in the unprimed frame. The speed of C' relative to C is

V(C'C) = \frac{L_0}{\Delta t}

Now consider things from the point of view of clock C'. Suppose that when C' coincides with A, clock C' reads X', and when C' coincides with B, clock C' reads X' + \Delta t_1 '. Therefore the total time of travel of C' from A to B in the primed frame is given by:

(X' + \Delta t_1 ') - X' = \Delta t_1 '

Now when C' coincides with C, clock C' will read an amount \Delta t_2 ' more than what it read when it coincided with B. Therefore, the total time of travel of C' from A to C in the primed frame is given by:

\Delta t' = (X' + \Delta t_1 ' + \Delta t_2 ') - X' = \Delta t_1 ' + \Delta t_2 '

Let L denote the distance between A and C in the primed frame. The speed of C with respect to C' is:

V(CC') = \frac {L}{\Delta t'}

Let it be stipulated that clock C' is of identical construction to clocks A,B, and C. Therefore clock C ticks at the same rest rate as they do. In other words, when clock C' and clock A are both at rest with respect to each other, one tick of clock A corresponds to one tick of clock C'. Now, assuming that the rate of clock C' does not change when clock C' is in relative motion to clock A, \Delta t_1 = \Delta t_1 ' and \Delta t_2 = \Delta t_2 '. In which case, \Delta t = \Delta t', from which it follows that the relative velocity computations will be equal if and only if L=L_0. If clock C' ticks at a constant rate \gamma which is different from the rate that clocks A,B,C tick at when C' is in relative motion to them then:

\Delta t_1 = \gamma \Delta t_1 '
and
\Delta t_2 = \gamma \Delta t_2 '

In which case:

<br /> <br /> \Delta t = \Delta t_1 + \Delta t_2 = \gamma \Delta t_1 &#039; + \gamma \Delta t_2 &#039; = \gamma (\Delta t_1 &#039; + \Delta t_2&#039;) = \gamma \Delta t&#039;

Thus, the relative speed calculations will be equal if and only if

L = \frac {L_0}{\gamma}

Notice how the readings of the clocks fall out of the analysis, and only 'amounts of time' in a frame are left. A reading on a clock corresponds to some particular state of the universe, it is differences in readings that are used to make computations of 'amounts of time'.

Also notice that the previous conclusion is completely independent of the theory of relativity as well as the Lorentz Fitzgerald length contraction formula. The previous conclusion would hold for any formula for L, not just the Lorentz Fitzgerald one.

Mathematical Analysis Of Simultaneity

Let us consider an event where two rulers of identical rest length L_0 move past each other in uniform relative parallel motion, as shown in the state diagrams below. Let the relative speed be denoted by v, and let c denote the speed 299792458 meters per second. Let it be supposed that there are four clocks of identical construction, at the ends of both rulers. Thus, the rest rate of all the clocks is the same. It can be stipulated that at the beginning of the event all four clocks have the same reading without loss of generality, since clock readings inevitably fall out of the analysis. What we end up doing is comparing the amount of time of an event [X,Y] in one reference frame to the amount of time of the same event [X,Y] in another reference frame.

Initial State (B coincides with A' in all reference frames)
         A      B
Ruler 1: |------|
Ruler 2:        |--|
                A' B'

Final State (B coincides with B' in all reference frames)
         A      B
Ruler 1: |------|
Ruler 2:     |--|
             A' B'

Stipulation 1: The relative speed between the two reference frames is constant, hence if either reference frame is inertial, so is the other.
Stipulation 2: Neither ruler is subjected to any external forces, hence each ruler is in an inertial reference frame. Hence clocks in a frame tick at their rest rate.
Stipulation 3: All four clocks are synchronized at the beginning of the event, in other words they all read the same number, for simplicity let all four clocks read zero at the beginning of the event. There is nothing objectionable to this stipulation, since clock readings fall out of the analysis.
Stipulation 4: The rest rate of all four clocks is identical, thus clocks in a frame remain synchronous at all moments in time.
Stipulation 5: The two rulers, AB and A'B' have the same proper length, L_0.

Assumption 1:
Let L denote the length of the moving ruler, as measured in a rest frame. Let us assume the Lorentz Fitzgerald length contraction formula is a true statement about the length of a moving ruler. Therefore,

L = L_0 \sqrt{1 - v^2/c^2}.

From the previous analysis:

L = \frac {L_0}{\gamma}

Hence, we must have

\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}.

Let us define g such that:

g = \frac{1}{\gamma} = {\sqrt{1 - v^2/c^2}}


(I am still working on this, ignore everything below)





Diagram as drawn in the primed reference frame (that of ruler 2):

Time t'0:
         A  B
Ruler 1: |--|
Ruler 2:    |--|---|
            A' C'  B'


Time t'1:
            A  B
Ruler 1:    |--|
Ruler 2:    |--|---|
            A' C'  B'

Time t'2:
                A  B
Ruler 1:        |--|
Ruler 2:    |--|---|
            A' C'  B'

The two rulers, AB and A'B' have the same proper length, L_0.

At (primed) time t&#039;_0, B and A' are at the same event of space-time.

At (primed) time t&#039;_1, A and A' are at the same event.

The point C' is defined to be the point on ruler 2 which is at the same event as B at (primed) time t&#039;_1.

At (primed) time t&#039;_2, A and B' are at the same event.


You've defined \Delta t&#039; := t&#039;_2 - t&#039;_0.

The velocity, v of ruler 2 (as measured in the primed system) must be v = L_0 / \Delta t&#039;.

By length contraction, we know the length L of ruler 1 as measured in the primed coordinate system; it is L = L_0 \sqrt{1 - v^2/c^2}.


Now, we consider things in the unprimed reference frame (that of ruler 1):

Time t0:
         A      B
Ruler 1: |------|
Ruler 2:        |--|
                A' B'


Time t1:

         A      B
Ruler 1: |------|
Ruler 2:     |--|
             A' B'

You define \Delta t := t_1 - t_0.

The velocity of ruler 2 (as measured in the unprimed system) must be v = L / \Delta t.

By (insert favorite reason here), these two velocities must be the same, thus:

\frac{L}{\Delta t} = \frac{L_0}{\Delta t&#039;}

Okay... [/B][/QUOTE]

 

[Hurkyl]

Stipulation 3: All four clocks are synchronized at the beginning of the event, in other words they all read the same number, for simplicity let all four clocks read zero at the beginning of the event.

In which reference frame are you making this stipulation?

 

[mathfeel]

To me, this is not a paradox at all. If you consider the Lorentz invariant for the events A/A' coincide and B/B' coincide in either frame, it turns out to be
ds^2=-L_0^2 \frac{1}{\beta}(1-\sqrt{1-\beta^2}) + L_0^2

One can simply prove (just plot it) that \frac{1}{\beta}(1-\sqrt{1-\beta^2}) is less than 1 for 0\leq\beta\leq1
So the interval is spacelike. So there is no causal relation between the two events. So it is perfectly legitimate to have the order of two events reversed in two different frame.

 

[pallidin]

A horizontal hard, wooden dowel, of 1-foot length is placed against a vertical platform; this portends non-compressabilty.
Underneath, attached to that same platform, is a spring with a weight on the end and a minor support dowel, total mass equaling that of the upper dowel; this portends compressability.
Force is applied to the platform, in this case, left to right.
The upper dowel will move left to right at the velocity of platform movement.
The lower arrangement will be compressed and so delayed, yet building potential energy will spring forward, balancing all forces involved.
So, what's your point?