Attention Paid To Accelerating Reference Frames Overthrows SR

[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