Attention Paid To Accelerating Reference Frames Overthrows SR

[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'}{\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'...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'}{\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 Δt is the time interval measured by clock A, and Δt' is the elapsed time in the coordinate frame F2, then

Δt = √(1 - (v/c)^2) Δt'

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

If you could also prove that

Δt' = √(1 - (v/c)^2) Δt

for the same Δt', Δt, 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.

 

[jdavel]

StarThrower, I finally see the source of your confusion.

You correctly say that the postulates of SR imply two equations,

Equation 1:

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

and Equation 2:

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

If these weren't both true then there would be something fundamentally different between moving to the left and moving to the right. The postulates say there is no such difference.

But then you mistakenly say that these equations together imply a third equation,

Equation 3:

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

This simply (and obviously) isn't true.

The reason you can't get to Eq 3 from Eqs 1 & 2 is that the variables delta-t and delta-t' don't represent the same thing in Eq 1 that they do in Eq 2. So the ratio, delta-t/delta-t' in Eq 1 isn't the same as the ratio delta-t/delta-t' in Eq 2. In fact, they are exactly the reciprocals of each other! So your Eq 3 should read:

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

which is true, but not very interesting, and certainly no threat to the validity of SR!

 

[Severian596]

I agree. The only way I can see StarThrower coming to the incorrect equation is by misinterpreting SR.

This is easy for him to fall into considering the following:
* He supports absolute time
* He supports absolute space (and therefore believes we live in Euclidian space)
* He adheres to classic Newtonian physics
* He believes he's brilliant and therefore justifies the previous three behaviors

Now I'm not saying he's ABSOLUTELY crazy because, well, that would be taking the same fanatic standpoint he's taking. I don't have all the answers, but I invest confidence in many of our past geniuses who advanced theory beyond Pythagoras and Euclid's day. I don't assume I have the luxury of being one of them myself...

 

[StarThrower]

I agree. The only way I can see StarThrower coming to the incorrect equation is by misinterpreting SR.

This is easy for him to fall into considering the following:
* He supports absolute time
* He supports absolute space (and therefore believes we live in Euclidian space)
* He adheres to classic Newtonian physics
* He believes he's brilliant and therefore justifies the previous three behaviors

Now I'm not saying he's ABSOLUTELY crazy because, well, that would be taking the same fanatic standpoint he's taking. I don't have all the answers, but I invest confidence in many of our past geniuses who advanced theory beyond Pythagoras and Euclid's day. I don't assume I have the luxury of being one of them myself...

Remember, if I can derive any contradiction whatsoever, then logically I can draw any conclusion whatsoever.

Look back at the equation Hurkyl arrived at:

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 />

This equation is valid for v=c is it not?

 

[Pergatory]

SR states that if we are in an inertial reference frame, v<c

 

[StarThrower]

SR states that if we are in an inertial reference frame, v<c

No, actually you just stated that... kind of ad hoc, if you know what I mean.

Kind regards,

The Star

 

[Severian596]

This equation

\Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;

is valid for v=c. Because of the flop you're no longer dividing by zero. The original equation was this:

\Delta t&#039; &amp;= \frac{\Delta t}{\sqrt{1 - \left( \frac{v}{c} \right)^2}}

This equation derives the time observed by a clock A at rest, of a clock B in motion, and denotes the observed time as \Delta t&#039;. Now if we look at the first equation again:

\Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;

This equation derives the time observed by clock B in motion, of a clock A at rest, and denotes the observed time as \Delta t, with respect to clock A's reference frame. This equation is in fact valid if clock B is traveling at v=c because the amount of time that B will observe passing on A is zero!

This equation tells us that photons do not observe time passing on any clocks because they travel at v=c.

EDIT:

I want to stress that the frame of reference for this equation did not change, therefore the symbols' meanings did not change. Thus there is no contradiction involved...all we did is multiply by the numerator. It doesn't support any contradiction (X and not X, for example) of special relativity. Deriving B's observation of A with respect to A's reference frame is completely valid, though a bit confusing if you don't keep the meaning straight.

 

[StarThrower]

This equation

\Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;

is valid for v=c. Because of the flop you're no longer dividing by zero. The original equation was this:

\Delta t&#039; &amp;= \frac{\Delta t}{\sqrt{1 - \left( \frac{v}{c} \right)^2}}

This equation derives the time observed by a clock A at rest, of a clock B in motion, and denotes the observed time as \Delta t&#039;. Now if we look at the first equation again:

\Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;

This equatin derives the time observed by clock B in motion, of a clock A at rest, and denotes the observed time as \Delta t, with respect to clock A's reference frame. This equation is in fact valid if clock B is traveling at v=c because the amount of time that B will observe passing on A is zero!

This equation tells us that photons do not observe time passing on any clocks because they travel at v=c.

NO!

You are disobeying mathematical rules, namely you are dividing by zero in order to put the square root quantity in a denominator.

You absolutely are not permitted to do that. However, Hurkyl's formula

\Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;

certainly is valid.

Kind regards,

The Star

 

[Severian596]

NO!

You are disobeying mathematical rules, namely you are dividing by zero in order to put the square root quantity in a denominator.

You absolutely are not permitted to do that. However, Hurkyl's formula

\Delta t &amp;= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t&#039;

certainly is valid.

Kind regards,

The Star

Wait wait...first read my edit to my post just to make sure you get my meaning, then with a little less zeal and a little more annunciation tell me how I disobeyed mathematical rules. I'll examine my own post while you type.

 

[StarThrower]

EDIT:

I want to stress that the frame of reference for this equation did not change, therefore the symbols' meanings did not change. Thus there is no contradiction involved...all we did is multiply by the numerator. It doesn't support any contradiction (X and not X, for example) of special relativity. Deriving B's observation of A with respect to A's reference frame is completely valid, though a bit confusing if you don't keep the meaning straight.

Yes I know this. Let me state for the record:

Clock A is attached to the photon gun, and time measured by it is being denoted by \Delta t

Clock B was a clock moving at arbitrary speed v, relative to clock A. The Hurkelian result was then derived, where \Delta t^\prime is the time of the same event, but measured by clock B.

Up to this point you are good, so what are you trying to do now?

Kind regards,

The Star

 

[StarThrower]

Wait wait...first read my edit to my post just to make sure you get my meaning, then with a little less zeal and a little more annunciation tell me how I disobeyed mathematical rules. I'll examine my own post while you type.

I UNDERSTOOD YOU PERFECTLY.

The symbols meaning didn't change... of course the symbols meaning cannot change, that would be cheating just to derive a contradiction. Read what I wrote.

Kind regards,

The Star

 

[Severian596]

Yes I know this. Let me state for the record:

Clock A is attached to the photon gun, and time measured by it is being denoted by \Delta t

Clock B was a clock moving at arbitrary speed v, relative to clock A. The Hurkelian result was then derived, where \Delta t^\prime is the time of the same event, but measured by clock B.

Up to this point you are good, so what are you trying to do now?

Kind regards,

The Star

Please be careful of your terminology, because Delta t' is NOT an event, it's a measure of change in time between any two defined events. Please don't refer to it as the "time of the same event" unless you've first defined an event (ct,x,y,z) about which to speak. Furthermore the delta t of an event is not a delta at all, it's just t...

Up to this point you are good, so what are you trying to do now?

I'm trying to show you that in my opinion
1) v=c is valid for this equation, and
2) this fact does not contradict special relativity

Why? Well we set up two equations for A's rest frame, right? One solves for Delta t and the other solves for Delta t prime. All with respect to A. If we go ahead and solve each of them for v=c we get the following results:

\Delta t = 0
\Delta t&#039; = Undefined (division by zero)

You'll grant me the right to manipulate the equation BEFORE plugging in values, right? So I didn't divide by zero. But why oh why don't these results spell the downfall of Special Relativity??

 

[StarThrower]

Please be careful of your terminology, because Delta t' is NOT an event, it's a measure of change in time between any two defined events. Please don't refer to it as the "time of the same event" unless you've first defined an event (ct,x,y,z) about which to speak. Furthermore the delta t of an event is not a delta at all, it's just t...

I didn't say Delta t` is an event.

Let me define the 'event' I am talking about. The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

Clock B and clock C are in an inertial reference frame, and are at rest with respect to each other. Additionally, they have been synchronized. At the moment in time when clock A and clock B coincide, clock B reads x, and clock C also reads x (because clock C is synchronized with clock B by stipulation). Later, clock A coincides with clock C, and at this moment in time, clock C reads Y. Thus, the total time of the event in the inertial reference frame in which clock B and clock C are at rest, is equal to Y-X. In other words:

\Delta t^\prime = Y-X

The above quantity is the "time of the event" in reference frame F2.

I have now defined the 'event' I am talking about.

Kind regards,

The Star

P.S. I know why some verbal confusion has arisen.

You refer to a single moment in time as an 'event'. That is the standard terminology yes. But there is another word for this, which is used in philosophical circles, and that word is 'state'.

You represent a state using the following symbolism: (ct,x,y,z)

But you have no precise notion of the "time of an event" which can only be defined using two states, one which occurred before the other, in universal time.

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Regards,

The Star

 

[Severian596]

I thought maybe you'd answer me, or try to tell me that it does in fact lead to a contradiction in SR.

This situation (where v=c) just doesn't make sense. Distance = rate*time. No matter what your rate, if time is zero, distance is zero. But for v=c, no matter what amount of time you try to assert has passed at clock A denoted by \Delta t, clock B traveling at v=c will measure \Delta t = 0, which is why we cannot define \Delta t&#039; with respect to A. We just wouldn't understand it! No matter how much time we experience traveling with clock A, clock B never measures any change in our clock.

Your initial argument that two photons traveling parallel to each other are traveling relative to each other at velocity zero is classic Newtonian velocity addition. This is an incorrect application of Newtonian physics, just like the application of Newtonian physics to explain the "almost insignificant" change in Mercury's perihelion did not apply (but was precisely explained by Einstein's equations). So, if I can prove that one of your assumptions is incorrect, I can call your entire conclusion incorrect.

I won't mention that you assume that we're in Euclidian space (not proven), and that space is the same for a photon as it is for us (not proven), AND that you assume time is absolute (which I'm sure you know carries implications of variable c and the ether, as well as absolute space, which you've admitted you agree with).

I don't know why I bother. Think what you like, but your thoughts aren't as original as you believe they are.

 

[Severian596]

The event I am talking about begins at the moment...and ends later...

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

 

[Severian596]

I didn't say Delta t` is an event.

Let me define the 'event' I am talking about. The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

Clock B and clock C are in an inertial reference frame, and are at rest with respect to each other. Additionally, they have been synchronized. At the moment in time when clock A and clock B coincide, clock B reads x, and clock C also reads x (because clock C is synchronized with clock B by stipulation). Later, clock A coincides with clock C, and at this moment in time, clock C reads Y. Thus, the total time of the event in the inertial reference frame in which clock B and clock C are at rest, is equal to Y-X. In other words:

\Delta t^\prime = Y-X

The above quantity is the "time of the event" in reference frame F2.

I have now defined the 'event' I am talking about.

Kind regards,

The Star

Alright, this is convoluted at best, and a diagram may help or it may not, but

The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

I guess I'll assume that, in order to synchronize all three clocks a light source is placed in the center of a Circle (or sphere) in an extremely local spatial area with respect to clocks A, B, and C. A pulse of light is emitted from the light source, and the clocks begin ticking the moment the light reaches them. They are equidistant from the source and not in relative motion to each other or the light source, and therefore we should be confident that they are synchronized. So, in this reference frame F1 (notorious label by now) the lines of simultaneity for clocks A, B, and C are parallel to their spatial axis x with respect to their time axis t.

Clock B and clock C are in an inertial reference frame, and ...

Whoa! When did they change reference frames? You just synchronized them for crying out loud! So we must conclude that now clocks B and C are in a different reference frame F2 than A, then some force must have acted upon them for them. So an acceleration took place. Bad for SR...very bad...you're getting out of special relativity's jurisdiction.

Additionally, they [clocks B and C] have been synchronized.

As long as they've had the same force applied to them and are now in the same reference frame I assume this was not necessary. I'll assume also that you're just reiterating the fact.

At the moment in time when clock A and clock B coincide, clock B reads ...

Okay that's it. Now you're proposing a situation where special relativity does not apply. Because A and B are in different reference frames their lines of simultaneity are NOT parallel to each other. A sees B's lines of simultaneity at an angle, and B sees A's lines of simultaneity at an angle. Therefore there exists no "moment in time" that A and B coincide unless both A and B are in uniform motion with respect to some OTHER reference frame F3. This other reference frame (with respect to which A and B must have been synchronized before hand, then had equal forces applied to each of them to put them in different reference frames than F3 itself) could possibly measure "coinciding" times on A and B, but ONLY RELATIVE TO ITSELF.

Therefore because A and B will never coincide with respect to each other, the rest of your statements never occur.

 

[StarThrower]

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

See my PS which I added to my previous post. It was/is:

P.S. I know why some verbal confusion has arisen.

You refer to a single moment in time as an 'event'. That is the standard terminology yes. But there is another word for this, which is used in philosophical circles, and that word is 'state'.

You represent a state using the following symbolism: (ct,x,y,z)

But you have no precise notion of the "time of an event" which can only be defined using two states, one of which must occur before the other, in universal time.

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Regards,

The Star

 

[Severian596]

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Thank you for this. I don't participate in philosophical circles, and we're not in one right now. Your 'event' is actually a duration, I'm sure you know that. Read my previous post to see why your proposed system is invalid...you make assumptions, leave out details, and change frames of reference willy nilly. You know better than that.

Oh and remember that your "Universal Time" you referred to when defining the philosopher's circle 'event' is actually just another frame of reference which is subject to relativity itself. In my post, that "Universal Time" could be in terms of F3, but that means you're not presenting all the necessary facts. And remember that there is no preferred frame of reference, and if there is one (in your opinion) how do YOU personally define that?

 

[StarThrower]

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

I apologize for that, though I knew it would happen. I simply prefer the word 'state' to the word 'event' when I am discussing points of time, or moments in time, and I like to assign quantities of time to something that I wish to call 'events'.

You use the word 'event' to mean what I call 'state', and I use the word 'event' for something you don't have a word for.

Kind regards,

The Star

(P.S. It is easier for you to change than me.)

EDIT: Correction, you do have a word for it... duration. However, duration isn't a noun, the way an event is. Philosophers would have a field day with you.

To familiarize yourself with the philosophical use of the terms 'state' and 'event', peruse the following reputable site which discusses temporal logic.

Temporal Logic

Keep in mind that these guys are just dying to talk with physicists.

 

[Severian596]

duration isn't a noun, the way an event is.

Duration is a noun...and I'm not going to debate about it. Here's dictionary.com's definition (beware an annoying popup).

 

[matt grime]

And linguists would have a field day with you. Not to mention psychologists, given your egoism.

 

[StarThrower]

Duration is a noun...and I'm not going to debate about it. Here's dictionary.com's definition (beware an annoying popup).

Absolutely no point in arguing semantics. I can switch back and forth between your terminology and mine, I am intelligent enough for that. I presume you have the same capability?

Kind regards,

The Star

 

[StarThrower]

And linguists would have a field day with you. Not to mention psychologists, given your egoism.

In the time it takes them to comprehend me, I will have died. Hence, I conclude their efforts would be in vain. Better they should become physicists.

Kind regards,

The Universe

 

[Severian596]

To suppliment my post about your incorrect use of Special Relativity, peruse http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

I admire your apparent understanding of physics (both classical and electromagnetic), and your mathematical ability. I'm just a beginning student to the topic(s) yet I spot your error so easily. Why do you persist in thinking you can prove an error in SR where SR does not apply?

 

[StarThrower]

To suppliment my post about your incorrect use of Special Relativity, peruse http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

I admire your apparent understanding of physics (both classical and electromagnetic), and your mathematical ability. I'm just a beginning student to the topic(s) yet I spot your error so easily. Why do you persist in thinking you can prove an error in SR where SR does not apply?

Firstly, I didn't make the error someone accused me of, which was to reverse the meanings of \Delta t and \Delta t^\prime

I understand which quantity of time is being measured by which clock. That wasn't/isn't the contradiction I am reaching for. As for your new question about the twin paradox, let me read the article at the site first, and then I will come back and give you my comments. As for my persistence in thinking I have spotted an error in SR... the reason for that is because I have. I don't really believe I am the first to ponder it either. Nonetheless, the error to me seems obvious, yet it hasn't caught on to mainstream physics yet. And that is what I really cannot understand.

If the relative speed in this problem is given by v=c, then we have two photons moving at right angles to each other. The clocks B,C are thus in a photonic frame (see other threads on this). The conclusion I reach, is that if the fundamental postulate of the theory of special relativity is true, then photons cannot move relative to each other. That cannot possibly be true, based upon simple facts that come from sensory perception.

Now let me give this site a look, and I will come back and tell you what I think.

Kind regards,

StarThrower


EDIT: Correction, the clocks B,C are in a photonic reference frame (reference frame in which photons are at rest)

 

[matt grime]

So, you're an expert, a genius able to see to the very inherent contradictory nature of relativity, and you've not come across the twin paradox?

 

[Severian596]

That cannot possibly be true, based upon simple facts that come from sensory perception.

This is the biggest pitfall you could've fallen into. You cannot argue anything based "upon simple facts that come from sensory perception," especially mathematics. Sensory perception is not an axiom. In fact that this phrase even escaped your keyboard surprised the heck out of me. How atrocious.

 

[Severian596]

If the relative speed in this problem is given by v=c, then we have two photons moving at right angles to each other. The clocks A,B are thus in photonic frames (see other threads on this).

Note that this example should be stricken from the record unless StarThrower is willing to better qualify "this problem" and exactly what A and B have to do with any frames, much less "photonic" frames.

 

[StarThrower]

So, you're an expert, a genius able to see to the very inherent contradictory nature of relativity, and you've not come across the twin paradox?

Of course I have, I am merely looking at this site for the first time... then I will give my comments. Immediately upon looking at it I thought to express the population as follows:

P(t) = P_0 e^r^t

Give me a moment here, I'm reading.

Regards,

Star

 

[Severian596]

Nonetheless, the error to me seems obvious, yet it hasn't caught on to mainstream physics yet. And that is what I really cannot understand.

You need to understand relativity better. It never caught on because actual physicists (who know more on this topic than I do) evaluated it, pondered it, and dismissed it. Why? Because they know more than you.

 

[StarThrower]

This is the biggest pitfall you could've fallen into. You cannot argue anything based "upon simple facts that come from sensory perception," especially mathematics. Sensory perception is not an axiom. In fact that this phrase even escaped your keyboard surprised the heck out of me. How atrocious.

My apologies, allow me to be more professional.

Consider the following experiment performed by an ancient caveman:

He started a fire for the women to sit, directly on his left, so he didnt have to listen to them squalk, and then started a fire directly in front of him, for the men to sit around. He noticed that not only was the front of his body warm, but so was the left side of his body too. He concluded that photons can move relative to each other.

P.S. Or was this a thought experiment? Geeze, I forget. Maybe he just thought he was warm?


Regards,

Star

 

[StarThrower]

You need to understand relativity better. It never caught on because actual physicists (who know more on this topic than I do) evaluated it, pondered it, and dismissed it. Why? Because they know more than you.

They cannot make me believe that which is false. They can confuse me about it for awhile, but they can never make me believe it.

Regards,

StarThrower

 

[Severian596]

He concluded that photons can move relative to each other.

This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.

 

[Severian596]

And...I shall exit stage left, as I should've done from square one.

Goodbye, StarThrower

 

[StarThrower]

To suppliment my post about your incorrect use of Special Relativity, peruse http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

Ok I've looked at the site. The first thing to say, is that I understood Dolan's question.

If you have ever been to the carnival, there is a ride where you are spun around really fast, and then the floor drops out, but you don't fall to the earth. In this ride, you are pinned to the wall by a force which has arisen because you are in uniform circular motion.

Principle of equivalence: It is impossible to tell if you are at rest on the surface of a body exerting a gravitational force F on you, or accelerating uniformly through space.

Dolan asks about the following experiment.

You have two vials of some radioactive substance, with a precisely known half life. For example, suppose that the half life is one hour. The samples are identical, in the sense that they contain the same initial population. So for example, suppose that P0 = 1,000. Thus, if the vials are left at rest, then in one hour, there will be (approximately) 500 radioactive particles left in both samples.

Now, perform this experiment, but this time put one of the samples in a device like the carnival ride described before.

The particles in the centrifuge are obviously subjected to a force which the particles in the vial on the bench are not subjected to.

Dolan's question then, is whether or not the experiment has ever been done, and whether or not any time dilation was observed.

 

[StarThrower]

This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.

Umm but they are moving relative to each other, even if they cannot measure the relative speed.

 

[Severian596]

Umm but they are moving relative to each other, even if they cannot measure the relative speed.

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

 

[StarThrower]

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

I thought you exited stage left.

Self-contradiction? Oh now you are totally coming unglued huh. I find a flaw in the theory of SR, and you come apart.

 

[StarThrower]

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star

 

[DrMatrix]

moment in Universal time

If you're going to assume SR (or GR), you must drop the notion of Universal time. Otherwise, you will run into a contradiction.

 

[Severian596]

Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star

Ignoring the impossibility of a string (which has mass) "keeping up" with the photon traveling at c, the compression wave triggered by the snapping string would travel at roughly the speed of sound, just as any compression wave. For example if you and your friend stand with thighs against a table and he bumps it, how much time passes before you feel the table move? The speed is not infinite, and in fact travels at the speed of sound through the medium of the solid table. The photon would never learn of the "universal time" at which the string snapped.

I will discuss the theory of SR with you no further, but am happy to educate you on some basic physical phenomenon.