Attention Paid To Accelerating Reference Frames Overthrows SR

[Severian596]

...but why on Earth I'm trying to do so on a message board is beyond me.

You failed to mention the horrible manner in which you present your arguments, and the circles in which you lead your "proofs."

 

[StarThrower]

"The only true wisdom is in knowing you know nothing."
~Socrates

"Wisest is he who knows what he does not know."
~Plato (428/7-348/7 B.C.E.)

"Wisdom is a poor substitute for knowledge."
~Author Unknown

 

[StarThrower]

You failed to mention the horrible manner in which you present your arguments, and the circles in which you lead your "proofs."

My arguments may be presented haphazardly, but I assure you there is nothing circular about my reasoning.

Kind regards,

The Star

 

[Severian596]

My arguments may be presented haphazardly, but I assure you there is nothing circular about my reasoning.

Kind regards,

The Star

I guess it's too bad that I can't base my beliefs on your own assurance of your methods.

 

[Michael F. Dmitriyev]

StarThrower,
This mathematics is incorrect.
You are ignoring the fact, that achievement of zero value is infinite process demanding infinite energy (time). Any value can come nearer to zero only (the speed too).
Do you have any own explanation of a constancy of the speed of light?

 

[StarThrower]

StarThrower,
This mathematics is incorrect.
You are ignoring the fact, that achievement of zero value is infinite process demanding infinite energy (time). Any value can come nearer to zero only (the speed too).
Do you have any own explanation of a constancy of the speed of light?

Micheal you have misunderstood something; my guess is that you think the speed of light is c in any inertial frame.

Kind regards,

The Star

 

[DrMatrix]

Are you saying that the speed of light depends upon the inertial frame? Because if you are, that is contrary to one of the fundamental assumptions of SR.

You are assuming the speed of light depends upon the reference frame, and (by trying to apply SR) assuming the speed of light is independent of the reference frame. It is no surprise that you can find a contradiction.

 

[Michael F. Dmitriyev]

Micheal you have misunderstood something; my guess is that you think the speed of light is c in any inertial frame.

Kind regards,

The Star

May be you have missed my reply on the previous pages. My point is:
all conclusions of SR are fair for all objects except of light itself.

Speed of light is unique and absolute speed.
It does not subjected to relativity at all.

 

[StarThrower]

Are you saying that the speed of light depends upon the inertial frame? Because if you are, that is contrary to one of the fundamental assumptions of SR.

You are assuming the speed of light depends upon the reference frame, and (by trying to apply SR) assuming the speed of light is independent of the reference frame. It is no surprise that you can find a contradiction.

Fundamental Postulate SR: The speed of a photon is c=299792458 meters per second in any inertial reference frame.

Here is what I am saying. I am saying that I can find at least one statement X such that:

If the fundamental postulate of SR is true then (X and not X).
At which point it follows that the fundamental postulate of SR is false.
At which point it follows that the following statement is true:

There is at least one inertial reference frame in which the speed of a photon isn't c=299792458.

So yes, the speed of a photon does depend upon what inertial frame you are in. In some inertial reference frames the speed of a photon will be 299792458 m/s, but in other inertial reference frames, the speed of the exact same photon is something other than 299792458 m/s.

I did not "assume" the speed of light depends upon inertial frame, in fact, I assumed the exact opposite (that it doesn't), and subsequently arrived at a contradiction.

Kind regards,

The Star

 

[StarThrower]

Let τ be the reading on a clock. Let t, x, y, and z be the temporal and spatial coordinates of the clock (according to a given inertial reference frame). Then, everywhere where the worldline of the clock is differentiable, (c dτ)² = (c dt)² - dx² - dy² - dz²

More precisely,

Let the parametrization (t(s), x(s), y(s), z(s)) (a <= s <= b) of the worldline of a clock through an inertial reference frame, and let τ(s) be the reading on the clock at (t(s), x(s), y(s), z(s)). Then, we have:

<br /> \left(c \frac{d\tau}{ds} \right)^2 = \left(c \frac{dt}{ds}\right)^2 -<br /> \left(\frac{dx}{ds} \right)^2 - \left(\frac{dy}{ds} \right)^2<br /> - \left(\frac{dz}{ds} \right)^2<br />

Or (assuming the clock's readout isn't going backwards as t(s) increases)

<br /> \begin{equation*}\begin{split}<br /> \tau(s) &amp;= \tau(a) + \int_a^s \frac{d\tau(u)}{du} \, du \\<br /> &amp;= \tau(a) + \int_a^s \frac{1}{c} \sqrt{<br /> ( c t&#039;(u) )^2<br /> - x&#039;(u)^2<br /> - y&#039;(u)^2<br /> - z&#039;(u)^2<br /> } \, du<br /> \end{split}\end{equation*}<br />
[/size]

Hurkyl, you have made an error in all of this, and it is too hard to find with the symbolism you are using. Really, you should do what I asked, which is to derive the time dilation formula from the assumption that the speed of light is c in any inertial reference frame.

Kind regards,

The Star

 

[Hurkyl]

So you're not going to try and do it yourself, then? I think I'm going to have a heart attack and die from that surprise! What's wrong with the symbolism? Looks like ordinary calculus to me...

 

[StarThrower]

So you're not going to try and do it yourself, then? I think I'm going to have a heart attack and die from that surprise! What's wrong with the symbolism? Looks like ordinary calculus to me...

Well you are still missing the error, so rather than put the blame on you, I blamed your needless parametrization of the worldline curve.

Regards,

The Star

 

[StarThrower]

So you're not going to try and do it yourself, then? I think I'm going to have a heart attack and die from that surprise! What's wrong with the symbolism? Looks like ordinary calculus to me...

You don't want to derive the time dilation formula do you.

Maybe I can find the derivation I am thinking of on the web already, and then just paste the link. It is of an experiment on measuring the speed of light using a mirror.

Kind regards,

The Star

P.S. I too can do melt your brain mathematics, but I don't find that kind of flowerly display of intelligence necessary.

 

[Hurkyl]

Don't complain about a statement being vague if you don't want to see it written more precisely.

Shall I try again? &tau; is the time on the clock under analysis, and t is the time given by the coordinate system.

You don't want to derive the time dilation formula do you.

Are you serious that you would like to see a derivation, or are you trying to deflect attention from the fact that you offered to try and prove something first, and have yet to make that attempt?

 

[StarThrower]

Don't complain about a statement being vague if you don't want to see it written more precisely.

Shall I try again? τ is the time on the clock under analysis, and t is the time given by the coordinate system.




Are you serious that you would like to see a derivation, or are you trying to deflect attention from the fact that you offered to try and prove something first, and have yet to make that attempt?

I really am serious. I know what has to be done to accomplish the proof. The first step is going to be to derive the time dilation formula. What is going to happen, is that you will set up two coordinate systems in relative motion at speed v. Then we can constantly refer to the way you set up the coordinate systems in order to make comments about the times in either system.

I tried to search the web for the exact derivation I am referring to, so that neither of us would have to repeat the work, here is a site that uses the derivation I am referring to, but I don't like the guy's symbolism.

Derivation Of The Time Dilation Formula

Here is the specific experiment (an experiment to measure the speed of light) we will need analyzed:

You are going to make a measurement of the speed of a photon. The photon is going to be fired from a photon gun, straight to a mirror, and then be reflected back to you. So you have a clock (clock A) at rest in this lab frame F1 which will mark the precise moment the photon is emitted, and the precise moment the photon returns. Denote this time by \Delta t. Now, you have a ruler set up just sitting at rest in the lab frame. It runs from the photon gun to the mirror. Let the distance from the photon gun to the mirror be measured to be D. Thus, the total distance the photon travels is 2D. Let c denote the speed of the photon in this frame. Thus, the speed of the photon in F1 is by definition:

c = \frac{2D}{\Delta t}

Now, consider this event as viewed from a reference frame F2 which is moving at speed v relative to F1. In this frame, the path of the photon is not a straight line, but rather the path is triangular (that is how the pythagorean theorem enters the derivation). Let it be the case that if you are at rest in F2, that you see the photon gun moving from left to right. So you see the photon travel diagonally upwards until it strikes the mirror, and then diagonally downwards until it strikes clock A(which is attached to the photon gun). There is a second clock (clock B) at rest in F2, and let the time it takes the photon to move from the photon gun to the mirror and back to the photon gun in F2 be measured by clock B to be \Delta t^\prime

Thus, the total distance traveled by the photon in F2, is the sum of the two sides of an isosceles triangle. Let the side length of this triangle be denoted by S. Thus, the total distance traveled by the photon in F2 is 2S. Let c` denote the speed of the photon in F2. Thus, the speed of the photon in F2 is by the same definition as before:


c^\prime = \frac{2S}{\Delta t^\prime}

Now, the base of the isosceles triangle is the product of the relative speed v between the two frames F1,F2, and the total time it takes the photon to strike the mirror and return to the photon gun, as measured by clock B. Thus, the base of the isosceles triangle is:

v \Delta t^\prime

This is certainly enough information for you to use the pythagorean theorem to derive the time dilation formula. The height of the isosceles triangle is D, hence the following equation is true:

S^2 = D^2 + (\frac{v\Delta t^\prime}{2})^2

In the derivation of the time dilation formula, the fundamental postulate of the special theory of relativity is invoked the moment you equate c with c`. In classical mechanics they are unequal, in SR they are equal by hypothesis. The exact equation you get (by hypothesizing that c=c`) is:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

Once that is done, all quantities in the formula are clear. Once you have carried out the derivation, more discussion will be necessary. Let me show you the logic I am after by asking you to do this for everyone:

Fundamental postulate of SR: The speed of a photon in any inertial reference frame is c=c`= 299792458 meters per second. (Keep in mind that in the derivation both F1, AND F2 are inertial reference frames. F1 is inertial by stipulation, and since F2 isn't accelerating relative to F1 and F1 is inertial, it follows that F2 is inertial.

Here is what is being perfectly proved:

Theorem(contingent knowledge): If the fundamental postulate of SR is true then (\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}})

Now, at this point reason alone cannot yet determine whether or not the fundamental postulate of SR is true. More reasoning will have to be done.

In order to prove that the fundamental postulate is false, you now have to prove the following theorem:

Theorem(contingent knowledge): If the fundamental postulate of SR is true then NOT(\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}).

At that point you will have proven the following:


There is at least one statement X, such that:

Theorem(contingent knowledge): If the fundamental postulate of SR is true then (X and not X).

At this point, you can now prove the following:

Theorem(Absolute knowledge): The fundamental postulate of SR is false.


The main purpose of this excercise, is so that we can both have the same exact thought experiment in mind simultaneously, using the exact same letters to denote the variables and constants which are crucial to the time dilation formula.

Kind regards,

StarThrower

P.S. Here is a possible enstasis(enstasis is a greek word and means objection). For the appearance of mathematical terms like postulate, lemma, enstasis see Heath's translation of Euclid.

Enstasis:

Clock B is at a fixed location in F2, and the photon gun only passes near to clock B once during the event.


Solution: Let there be a third clock (clock C) which is also at rest in F2. Let the photon leave the photon gun at the precise moment when the apparatus passes by clock B, and let the photon return to the gun at the precise moment that the apparatus passes by clock C. Thus, observers at rest in F2 will know the clock readings of clocks B,C at the moment the photon gun passed them. Let clocks B,C have been synchronized. Thus, the time \Delta t^\prime can be computed by subtracting the reading of clock B from the reading of clock C.

Additionally, let it be the case that all three clocks are of identical construction, therefore they all tick in the same unit (let the unit of time on these clocks be the second). Since all clocks are of identical construction, we can be certain that they have the same rest rate (the rate they tick when they are at rest with respect to one another). Thus, if any two of these three clocks are at rest with respect to one another and have been synchronized, then they will remain synchronous.

In order to synchronize clocks B,C, we can imagine that a rigid ruler connects them, and that we have located the midpoint, and then a spherical light pulse goes off there, and travels to both clocks, setting them to zero simultaneously in reference frame F2. They therefore remain in sync forever in reference frame F2.

Incidentally, here is a link to Euclid's proof of the Pythagorean Theorem

Euclids Proof Of The Pythagorean Theorem

The above site is excellent.

There should be no objection to the use of the Pythagorean theorem in the derivation, because it is certainly true of rulers at rest with respect to each other in an inertial frame. My favorite proof of the Pythagorean theorem can be seen at the following site:

Simple Proof Of Pythagorean Theorem

 

[Hurkyl]

I have seen that derivation, but I would normally take a totally different approach to the proof that just looks at what sort of coordinate transformations preserve straight lines and lines with coordiante velocity c.

 

[StarThrower]

I have seen that derivation, but I would normally take a totally different approach to the proof that just looks at what sort of coordinate transformations preserve straight lines and lines with coordiante velocity c.

I don't doubt you have seen it, but this is the derivation I want us to discuss. I have my reasons, humor me. I know it's just a bit of algebra, but it will help if we both are certain the other guy is using the exact same symbol we are, to denote the exact same thing.

Kind regards,

The Star

 

[Hurkyl]

For the record, I'd like to point out that you have assumed D is the same in both frames. I imagine (based on the rest of your post) you intended that the F2 be moving in a direction perpendicular to the ruler, though you never stated as such.


Anyways, I'll agree with you up until

<br /> S^2 = D^2 + \left(\frac{v \Delta t&#039;}{2}\right)^2<br />.

because from this point, we have:

<br /> \begin{equation*}\begin{split}<br /> S &amp;= \frac{1}{2} c \Delta t&#039; \\<br /> D &amp;= \frac{1}{2} c \Delta t \\<br /> \left(\frac{1}{2} c \Delta t&#039;\right)^2 &amp;=<br /> \left(\frac{1}{2} c \Delta t\right)^2 + \left(\frac{v \Delta t&#039;}{2}\right)^2 \\<br /> c^2 \Delta t&#039;^2 &amp;= c^2 \Delta t^2 + v^2 \Delta t&#039;^2 \\<br /> \Delta t^2 &amp;= \left( 1 - \left(\frac{v}{c}\right)^2\right) \Delta t&#039;^2 \\<br /> \Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;<br /> \end{split}\end{equation*}<br />

 

[StarThrower]

For the record, I'd like to point out that you have assumed D is the same in both frames. I imagine (based on the rest of your post) you intended that the F2 be moving in a direction perpendicular to the ruler, though you never stated as such.


Anyways, I'll agree with you up until

<br /> S^2 = D^2 + \left(\frac{v \Delta t&#039;}{2}\right)^2<br />.

because from this point, we have:

<br /> \begin{equation*}\begin{split}<br /> S &amp;= \frac{1}{2} c \Delta t&#039; \\<br /> D &amp;= \frac{1}{2} c \Delta t \\<br /> \left(\frac{1}{2} c \Delta t&#039;\right)^2 &amp;=<br /> \left(\frac{1}{2} c \Delta t\right)^2 + \left(\frac{v \Delta t&#039;}{2}\right)^2 \\<br /> c^2 \Delta t&#039;^2 &amp;= c^2 \Delta t^2 + v^2 \Delta t&#039;^2 \\<br /> \Delta t^2 &amp;= \left( 1 - \left(\frac{v}{c}\right)^2\right) \Delta t&#039;^2 \\<br /> \Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;<br /> \end{split}\end{equation*}<br />

Interesting.

 

[Severian596]

I'm definitely riveted.

 

[StarThrower]

Do you agree that if I can now show that:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

SR is finally overthrown?

Kind regards,

The Star

 

[Severian596]

Wait a tick...you just lost me. You want to prove that one of special relativity's formulas is correct, and then determine that SR is overthrown because its formula is correct?

I'm thinking out loud. This formula comes about because of the "hypothesis" that c' = c no matter what reference frame you're in. So you want to prove that the formula is correct, but also assert that c' != c (please excuse the programming convention of NOT EQUAL)?

 

[StarThrower]

Wait a tick...you just lost me. You want to prove that one of special relativity's formulas is correct, and then determine that SR is overthrown because its formula is correct?

I'm thinking out loud. This formula comes about because of the "hypothesis" that c' = c no matter what reference frame you're in. So you want to prove that the formula is correct, but also assert that c' != c (please excuse the programming convention of NOT EQUAL)?

Hurkyl proved this:

If the fundamental postulate of SR is true then:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c^2}}

When a theory self-contradicts, you cannot finish reasoning after the derivation of a single implication. In other words, when a theory self-contradicts, you can derive two implications as follows:

Let T denote a theory of physics.
Thus, T is the conjunction of many statements. Suppose that theory T is the result of conjoining exactly 8 uncertain statements, with 1000 certain statements. Consider just that part of the theory, which is uncertain.

Denote the uncertain part of a theory of physics using the following symbol:

\Im

Thus,

\Im = S_1 \wedge S_2 \wedge S_3 \wedge ... S_8

\wedge = AND

Let

\rightarrow = if...then

So when the uncertain part of a theory of physics is wrong, you can derive two implicative statements from the theory of the following form:

\Im \rightarrow X
\Im \rightarrow notX

In other words, you can find a statement X, such that:

\Im \rightarrow (X \wedge notX)

At this point in your reasoning, you will know that there is an error in the theory, although you won't know which of the 8 statements is false, all you will know is that at least one of the 8 statements is false.

Now, consider a theory which is based upon only one uncertain statement. For example, the theory of relativity certainly uses all statements of algebra, but such statements are true by stipulation.

Let us define the theory of relativity to be a theory with just one uncertain postulate, namely the following postulate:

Fundamental postulate SR: The speed of a photon in any inertial reference frame is c.

SR will be overthrown if we can now find a statement X, such that:

If the fundamental postulate of SR is true then X
If the fundamental postulate of SR is true then not X.

Hurkly just derived the following implication:

If the fundamental postulate of SR is true then:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c^2}}

Suppose that I can now derive the following implication:

If the fundamental postulate of SR is true then:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

It will thus follow that:

If the fundamental postulate of SR is true then:

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

From which it will follow that:

1 = 1 - v^2/c^2

From which it will follow that V=0.

But not (v=0) by stipulation, since F1 and F2 are in relative motion.

Thus, it will follow that:

If c=c` then (v=0 and not (v=0)).

It will therefore follow that:

not (c=c`)

And c is the speed of light in inertial reference frame F1, and c` is the speed of light in inertial reference frame F2.

Thus, it is not the case that the speed of light is 299792458 meters per second in all inertial reference frames. In other words, the fundamental postulate of the theory of special relativity is false.

So, what I said was, that if I can now derive the second implication above, then the theory of SR is wrong.

Kind regards,

The Star

 

[Severian596]

You made it clear. I understood it that way, but wanted to make sure I was not confusing the prime character...the two formulae are very similar.

So following soon will be your proof of the proposed equation?

 

[Severian596]

This is strange though. This equation:


\Delta t = \frac{\Delta t&#039;}{\sqrt{1-v^2/c^2}}


is true, and is used to calculate the dilation of the clock at rest with respect to the clock in motion...so I want to make sure you're not switching frames. You will attempt to show that relative to the same clock in the same reference frame, both equations are true?

 

[Pergatory]

Hurkly just derived the following implication:

If the fundamental postulate of SR is true then:

\Delta t^\prime = \frac{\Delta t}{\sqrt{1-v^2/c2}}

Suppose that I can now derive the following implication:

If the fundamental postulate of SR is true then:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c2}}

It will thus follow that:

If the fundamental postulate of SR is true then:

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

Could you please explain to me how you arrived at the last quoted conclusion?

 

[Severian596]

Could you please explain to me how you arrived at this last quoted conclusion?

Yes, please. You cannot substitute 1 for both \Delta t and \Delta t&#039;...if that was in fact what you did. This would imply absolute time, and you can't use this implication in an attempt to prove just that.

 

[StarThrower]

This is strange though. This equation:


\Delta t = \frac{\Delta t&#039;}{\sqrt{1-v^2/c^2}}


is true, and is used to calculate the dilation of the clock at rest with respect to the clock in motion...so I want to make sure you're not switching frames. You will attempt to show that relative to the same clock in the same reference frame, both equations are true?

Yes, be patient, I am awaiting Hurkyl's response.

Kind regards,

The Star

 

[StarThrower]

Could you please explain to me how you arrived at the last quoted conclusion?

I didn't arrive at that conclusion yet. That conclusion is contingent upon me showing within the framework of this problem that:

\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}

Kind regards,

The Star

 

[Hurkyl]

If &Delta;t is the time interval measured by clock A, and &Delta;t' is the elapsed time in the coordinate frame F2, then

&Delta;t = &radic;(1 - (v/c)^2) &Delta;t'

(sanity check: clock A is indeed running slower than coordinate time)

If you could also prove that

&Delta;t' = &radic;(1 - (v/c)^2) &Delta;t

for the same &Delta;t', &Delta;t, and v, then there would indeed be a contradiction in SR.


Furthermore, if one could build a model of SR in some mathematical theory (say... by using linear algebra to construct Minowski geometry), then you will also have proven that mathematical theory to be inconsistent.