Room-temperature superconductivity?

Point out news stories, on the net or in mainstream media, related to polywell fusion.

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johanfprins
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Post by johanfprins »

ladajo wrote:This link shows how the "orbit" of Juptier looks centered on Earth. There is a relativistic component to Juptiers position in Earth's gravity well as well as a speed component regarding Jupiters overall progression "around" Earth.

Scroll to the bottom to see the animation.

http://astronomy.nmsu.edu/geas/lectures ... ide01.html
I do not know why we are arguing about Jupiter or GPS sattelites since we all agree on this. The argument is about two clocks: One goes on a SR journey (NO GRAVITY CHANGES) and returns and is then compared to the clock that remained behind. In this case the clocks, after being synchronised at the beginning of the journey MUST srill show the same time for the whole journey or else Einstein's first postulate is wrong.

An experiment where one clock goes on a gravity free trip and returns to be compared, has not yet been done and until that time it is nonsensical for tomclarke to claim that Einstein's postulates are wrong.
Last edited by johanfprins on Wed Jan 11, 2012 9:41 am, edited 1 time in total.

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Post by johanfprins »

ben m wrote:Hello all,

Newbie here. I'm a regular on the James Randi Foundation forums (forums.randi.org). I just wanted to point out (and I apologize if this is inappropriate) that Johan F. Prins spent most of the past few months conducting exactly this argument on a forum over there---there was a thread on "room temperature superconductivity"; Johan showed up, and in the course of trying to explain his ideas, he had distinct trouble with the basics of energy conservation, acceleration, Gauss's Law, the definition of voltage, the Schrodinger equation, radiation from moving charges, kinetic energy, and of course relativity. He ended up accusing Hafele and Keating of fraud.

The relativity argument is uncannily familiar; I see a handful of patient and knowledgable posters giving Johan spacetime diagrams, detailed Lorentz transformations, and so on, and getting insults in return.

Over on JREF, about a week ago, Prins bailed out of the thread announcing that he had "better things to do". I'm greatly amused that his "better things to do" apparently includes doing exactly the same thing with a different audience.
If you read thuis thread you will see that I also wanted to bail out here for the same reason. It was easier to bail out from JREF since you, Sol, and Ziggurat are so cluieless when it comes to physics that it is a waste of time responding. On this thread tomclarke and many others have been raising good arguments and I felt that I had to answer. But I definitely would like to now spend my time on prototype room-temperature superconducting dervices.
Anyway: if anyone was sitting here thinking "Boy, Johan's error is so straightforward, I will be able to clear it up with one explanation and one more spacetime diagram"---well, take a look at the JREF thread (Google for it) to see how many such explanations have scrolled across Johan's browser.

Cheers!
Ben m is an anonymous troll who knows so little about physics that it is hilarious; Just two examples of many:

1. He claims that superconduction occurs when you inject a normal charge-carrier from a normal conductor so that this charge carrier coasts through the superconductor as if it has been injected into a perfect vacuum. Why the mainstream "experts" postulate Cooper Pairs when Ben's model is possible, I do not really know.

2. He also believes that charge-carriers within a metal are the same as free electrons within a vacuum. He has NEVER heard about "pseudo-particles" that show diffractive effects within a material.

It is worthless to even try and argue with such a person who obviously has no qualifications whatsoever when it comes to physics. No wonder he hides behind anonymity. At least tomclarke knows quite a bit of physics and do not need to hide his identity when he differs from me.

Not that a person must not use another name: But if a personn claims, like Ben m does, that he is an expert in physics and I am a crackpot, then he should back it up by revealing why he thinks that he is better qualified than I am. Though he did admit that he is incompetent when it comes to performing experiments.
Last edited by johanfprins on Wed Jan 11, 2012 9:45 am, edited 2 times in total.

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Post by johanfprins »

kcdodd wrote:All this argumentation has lead me to a new question for myself though. Why does vacuum energy not break the principle of relativity. I am definitely not familiar with quantum field theory. However, it seems to me that if one assigns an energy density to a space itself that seems to imply a preferred reference frame. If you write that energy density as a four-vector then there is a "rest vacuum energy" to be a Lorentz invariant, and singles out a particular frame as being special.
An excellent question
EDIT:

I think I answered this for myself. If the vacuum energy was actually only one component of a vacuum stress energy tensor, it could be invariant under lorentz transformations. The Minkowski metric is invariant by definition, and is a rank two tensor. So perhaps the vacuum stress-energy has the same form? Which would mean the energy would have to be accompanied by an isotropic pressure? Strange.
It is a good attempt to answer your own question, but I do not think that it does.

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Post by tomclarke »

To show the twins must have different ages, all you need is Doppler effect, and Einstein's 2nd postulate - which Johan has been silent about.

Apply "c is constant" postulate to light pulses sent between the two twins. It is about the level at which 16 year olds are taught physics. No Lorentz Transform is needed.

I gave this argument above, it was ignored, so here is a better version with pictures.

This is very similar to what Teahive posted, but maybe explained in more detail.


Image


[pic will expand if clicked, how can i make it bigger always?]


This picture plots the position (horizontal axis) versus time (vertical axis) of the two twins.

I am indebted to Johan for the idea of measuring twin age by requiring each twin to emit a pulse of light, say every second, which is received by the other twin. We can compare total numbers of pulses received when the twins get back together to get the ages.

There will be nothing weird in this argument except that one "c must be constant" rule.

Looking at the pictures, both are of the same situation, with a lefthand stationary twin and a righthand twin moving first away and then back with constant velocity v.

The only difference between the lefthand and righthand picture is that light rays between the two twins are shown in opposite directions. It could all be on one diagram but the light rays would then be difficult to see.

We note the following:

(1) Every light pulse emitted will be received by the other twin, so we can work out twin age by counting number of received pulses.

(2) Doppler effect alters the received frequency of the pulses. The frequency depends on whether the relative velocity of the twins is inward or outward.

(3) We need to be careful about what time to measure relative velocity. For a given light ray it is obviously the transmit-time velocity and the receive-time velocity which must be compared.

(4) The Left-hand stationary twin has no velocity. To work out Doppler we use the Receive velocity for rays going to the travelling twin, and the earlier time (because the rays are transmitted before received) Transmit velocity for rays coming from the travelling twin. This is a crucial difference.

(5) Looking at the pictures, you can see that the rays received by either twin are initially low frequency, and swap over to high frequency for the inbound journey.

(6) Looking at the pictures, it is clear that received rays in the two directions are different. Rays from the traveller will be lower frequency (outbound journey Doppler) for a longer time T1 than rays to the travelling twin, which are low frequency for only T1'. Conversely the high frequency time T2 is shorter than T2'.


(7) Here is where we get relativistic. Unlike sound, the speed of light is constant so the doppler effect due to light speed is the same for light pulses in both directions:

1 + v/c for the inbound journey (higher frequency).
1-v/c for the outbound journey (lower frequency).

This symmetry combined with the asymmetry from (4) will give us the surprising result. For sound rays, and a non-relativistic picture, the difference in wave velocity between the travelling and stationary twin would cancel with the asymmetry caused by the differing low and high frequency times.

(8) Doing the calculation for the number of pulses from each twin:

Stationary twin: f1T1 + f2T2
Moving twin: f1'T1'+f2'T2'
Let us also label total time:
T1 + T2 = T1' + T2' = T

(9)
f2 = f2' = (1+v/c)f
f1 = f1' = (1-v/c)f
[f is the frequency at which pulses are emitted. It will turn out we need to scale this by a time dilation factor. But that will be the result of our calculation, we do NOT assume this]

Difference in age is: f(1-v/c)(T1-T1') + f(1+v/c)(T2-T2') =
f[ (T - T) - v/c( T2' - T2) - v/c(T1- T1') ]

The first term cancels and (by inspection) T2' > T2, T1 > T1' so we must have the stationary twin older than the moving twin.


If we wanted we could go further and derive the gamma factor which is the (real) time dilation of the moving twin. That would need more care because once we have time dilation, we must check which reference frame each quantity is measured in. But without doing this we can see that the ages must be different.

All we have used to get this result is speed of light being constant, and physics remaining the same in any inertial frame.

The nice thing about this proof is that it can be done, and shows the relativistic effect, entirely without frames of reference, Lorentz transform, or any other complex math.
Last edited by tomclarke on Wed Jan 11, 2012 1:52 pm, edited 7 times in total.

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Post by tomclarke »

ben m wrote:Hello all,

Newbie here. I'm a regular on the James Randi Foundation forums (forums.randi.org). I just wanted to point out (and I apologize if this is inappropriate) that Johan F. Prins spent most of the past few months conducting exactly this argument on a forum over there---there was a thread on "room temperature superconductivity"; Johan showed up, and in the course of trying to explain his ideas, he had distinct trouble with the basics of energy conservation, acceleration, Gauss's Law, the definition of voltage, the Schrodinger equation, radiation from moving charges, kinetic energy, and of course relativity. He ended up accusing Hafele and Keating of fraud.

The relativity argument is uncannily familiar; I see a handful of patient and knowledgable posters giving Johan spacetime diagrams, detailed Lorentz transformations, and so on, and getting insults in return.

Over on JREF, about a week ago, Prins bailed out of the thread announcing that he had "better things to do". I'm greatly amused that his "better things to do" apparently includes doing exactly the same thing with a different audience.

Anyway: if anyone was sitting here thinking "Boy, Johan's error is so straightforward, I will be able to clear it up with one explanation and one more spacetime diagram"---well, take a look at the JREF thread (Google for it) to see how many such explanations have scrolled across Johan's browser.

Cheers!
Thanks for this. I guess I think Johan must surely be able to see the error of his ways: but as you say it is of course probably an illusion and the above (very neat) proof of twins paradox ages being different is I promise my last attempt!

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Post by johanfprins »

tomclarke wrote:To show the twins must have different ages, all you need is Doppler effect, and Einstein's 2nd postulate - which Johan has been silent about.
Not at all. It should really not be a separate postulate since it can be considered a direct consequence of Einstein's first postulate that the laws of physics are the same within each and every inertial reference frame. This first potulate is in turn just a reconfirmation of Galileo's postulate that when you are within an enclosed space you cannot do any physics-experiment to determine whether you are moving with a speed v or not moving at all. Einstein realised that if the speed of light is not the same relative to different inertial reference frames then it will be possible to do a light experiment within your enclosed space to determine whether you are moving with a speed v or not moving at all (just like Michelson-Morley tried to do). In effect, this means that all inertial refrence frames are "stationary" and that motion is an "illusion" like Zeno has concluded already 3000 years ago.
Apply "c is constant" postulate to light pulses sent between the two twins. It is about the level at which 16 year olds are taught physics. No Lorentz Transform is needed.
If you do not require the Lorentz transform you are not doing SR. It is equivalent to claiming that you can model a dynamic force without Newton's second law.
I gave this argument above, it was ignored, so here is a better version with pictures.
Thanks. I sincerely appreciate your effort and your time. But before discussing your proof futher, I have some questions which I need to have cleared up. How did you calculate the distance between the two twins? Is it correct to assume that in the first diagram the distance at any time T is v*T? And in the second diagram the distance at time T(prime) is given by v*T(prime)?

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Post by tomclarke »

Johan wrote: How did you calculate the distance between the two twins? Is it correct to assume that in the first diagram the distance at any time T is v*T? And in the second diagram the distance at time T(prime) is given by v*T(prime)?
I do not calculate or use the distance between the twins. I rely only on the Doppler effect, which is a factor of:
1+v/c or 1-v/c
That determines the frequencies f1,f2,f1',f2'.

Since, by symmetry, v has constant magnitude, and by postulate, c is constant, f1=f1', f2=f2'. Clearly f2>f1. That is all I need.

This proof avoids many of the issues that would otherwise complicate the analysis. If you add them in, you have to get them right (in a relativistic context) which perhaps we will argue about. Luckily, they are not needed to prove the twins relative ages unequal.

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Post by tomclarke »

Johan wrote: If you do not require the Lorentz transform you are not doing SR. It is equivalent to claiming that you can model a dynamic force without Newton's second law.
That is clearly untrue, because the Lorentz transform can be derived from Einstein's postulates.

I am doing the same type of derivation here, but because it is a specific problem and I am only interested in whether the twins have the same age, not what is the magnitude of the difference, the derivation is much simpler.

Of course, I could have used LT, but that would be a less strong proof in the sense that it could only be fully understood by those who fully understand the LT. That is not all on this thread.

The core of this proof is in identifying what is symmetric and what asymmetric about the twins problem. Once you see that there is an asymmetry you understand why the ages must be different. Many people analysing this problem don't see the key Doppler effect asymmetry (on this thread however teahive at least does).

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Post by johanfprins »

tomclarke wrote:
Johan wrote: How did you calculate the distance between the two twins? Is it correct to assume that in the first diagram the distance at any time T is v*T? And in the second diagram the distance at time T(prime) is given by v*T(prime)?
I do not calculate or use the distance between the twins. I rely only on the Doppler effect, which is a factor of:
1+v/c or 1-v/c
That determines the frequencies f1,f2,f1',f2'.

Since, by symmetry, v has constant magnitude, and by postulate, c is constant, f1=f1', f2=f2'. Clearly f2>f1. That is all I need.

This proof avoids many of the issues that would otherwise complicate the analysis. If you add them in, you have to get them right (in a relativistic context) which perhaps we will argue about. Luckily, they are not needed to prove the twins relative ages unequal.
I am, however afraid that you are inserting an asymmetry where there is none. The symmetry must be done from the inertial reference frame of each twin since the signals are sent out from both inertial refrence fames on both the clocks which (as you agreed) are keeping the same time rate within their respective reference frames: The two diagrams that must be compared are where the light arrows go from the vertical time axes to the other reference frame that moves relative to the reference frame that you choose for the vertical time axis. Thus, from the vertical time axis T of twin 1 to the moving reference frame of twin 2; and then from the vertical time axis Tp of twin 2 to the moving reference frame of twin 1. The situation is then totally symmetric as it must be.

Let me do the calculation by using the Lorentz transformation:

After a time T(E) after synchronisation, twin 1 sends out a light pulse towards twin 2. According to the Lorentz transformation this light pulse appears within the reference frame of twin2 at a position

Xp(Ep)= -(gamma)*v*T(E)....................................[1]

and at a time

Tp(Ep)=(gamma)*T(E)..........................................[2]

And it thus proceeds by light speed c to cover the distance -Xp(Ep) in order to reach twin2 at a time Tp(Ap) on the clock of twin2 given by:

Tp(A)=Tp(Ep)+(MOD)Xp(Ep)/c..............................[3]

It is then simple algebra to derive that:

Tp(A)=(1+v/c)*(gamma)*T(E)...............................[4]

The corresponding position X(Ap) and time T(Ap) within the reference frame of twin1 follows from the LT as:

X(Ap)=(gamma)*vTp(A)=(1+v/c)*[(gamma)^2]*v*T(E).....[5]
and
T(Ap)=(gamma)*Tp(A)=(1+v/c)*[(gamma)^2]*T(E)...........[6]

Doing a bit of algebra, it is easy to calculate

(delta)T(E,Ap)=T(Ap)-T(E)=[(v/c)*T(E)]/(1-v/c) ..............[7]

and the speed of the light as measured by twin1, while it travelled from twin1 to twin2, is thus given by
SPEED OF LIGHT=X(Ap)/(delta)T(E,Ap) and it is c as it must be.

Now let twin2 send out a light pulse after a time Tp(E) towards twin 1.
According to the Lorentz transformation this light pulse appears within the reference frame of twin1 at position X(Ep) and at a time T(Ep) where:

X(Ep)=(gamma)*v*Tp(E)...............................................[8]
and
T(Ep)=(gamma)*Tp(E)..................................................[9]

It then rushes with light speed towards twin1 to reach twin1 at a time T(A) on his clock where we have that:

T(A)=T(Ep)+X(Ep)/c= (1+v/c)*(gamma)*Tp(E)..................[10]

The corresponding position Xp(Ap) and time Tp(Ap) within the reference frame of twin2 are thus:

Xp(Ap)=(gamma)*v*T(A)=(1+v/c)*[(gamma)^2]*v*Tp(E].....[11]
and
Tp(Ap)=(1+v/c)*[(gamma)^2]*Tp(E)

As above, one can calculte (delta)Tp(E,Ap)=Tp(Ap)-Tp(E) and prove that the speed of light as measured by twin2 is alo eqaual to c.

The important issue for our discussion is to compare equations [4] and [10] and to note that when the two twins send out their signals after a time T(E)=Tp(E)=1 (second), which they can do since their clocks have the same time rate within their respective reference frames, the time when twin 2 will receive the pulse from twin 1 will be (from Eq. [4]

Tp(A)=(1+v/c)*(gamma)

And the time that twin 1 will receive the pulse coming from twin2 will (from Eq. 10) be:

T(A)=(1+v/c)*(gamma)

The times are exactly the same on their clocks. A similar calculation can be done for the return leg of the trip. The Doppler factor v/c then enters as (1-v/c) but the pulses are still received at the exact same times on the two clocks. Thus if twin 1 received N pulses from twin2 during the whole trip, twin 2 MUST also receive N light pulses from twin 1. Their clocks MUST thus show the same elapsed time at the end of the journey.

Do you now see how dangerous it is to ignore the Lorentz transformation and to use a so-called Minkowski time-diagram?

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Post by tomclarke »

johanfprins wrote:
tomclarke wrote:
Johan wrote: How did you calculate the distance between the two twins? Is it correct to assume that in the first diagram the distance at any time T is v*T? And in the second diagram the distance at time T(prime) is given by v*T(prime)?
I do not calculate or use the distance between the twins. I rely only on the Doppler effect, which is a factor of:
1+v/c or 1-v/c
That determines the frequencies f1,f2,f1',f2'.

Since, by symmetry, v has constant magnitude, and by postulate, c is constant, f1=f1', f2=f2'. Clearly f2>f1. That is all I need.

This proof avoids many of the issues that would otherwise complicate the analysis. If you add them in, you have to get them right (in a relativistic context) which perhaps we will argue about. Luckily, they are not needed to prove the twins relative ages unequal.
I am, however afraid that you are inserting an asymmetry where there is none. The symmetry must be done from the inertial reference frame of each twin since the signals are sent out from both inertial refrence fames on both the clocks which (as you agreed) are keeping the same time rate within their respective reference frames: The two diagrams that must be compared are where the light arrows go from the vertical time axes to the other reference frame that moves relative to the reference frame that you choose for the vertical time axis. Thus, from the vertical time axis T of twin 1 to the moving reference frame of twin 2; and then from the vertical time axis Tp of twin 2 to the moving reference frame of twin 1. The situation is then totally symmetric as it must be.
Sorry, what inertial reference frame? I am not involving reference frames here. All I consider is the Doppler effect on light emitted by one twin and received by the other.

Why is it asymmetric? Just look at the diagrams! Consider the light-ray sent from the moving twin at turaround. This is received by the stationary twin AFTER turnaround and marks the start of T1.

By (time) symmetry, T2' = T1'
therefore T1 < T2 (by the time it takes light to travel between the twins when they are furthest separated).

These equations remain true in all frames, which is why I do not need to use them, and why you need to anwer this question of the inherent asymmetry simply.

Intuitively, the asymmetry derives from the fact that one twin turns around, the other does not. The turnaround alters the Doppler effect, but it does so unevenly. Light going to the twin which turns around will have different doppler effect immediately after the turnaround. Light coming from it will have different Doppler effect at time of turnaround + light travel time.
Let me do the calculation by using the Lorentz transformation:

...snip...

Do you now see how dangerous it is to ignore the Lorentz transformation
Wow. You were the one saying I misused mystical LTs instead of good-old einstein's postulates! I'll correct your incorrect use of LTs here after you have dealt with the (simpler) issue of Doppler asymmetry.
and to use a so-called Minkowski time-diagram?
I have not used Minkowski anything. My diagrams are simple time/position diagrams such as are commonly used to illustrate moving objects. You are the one using (without proof direct from Einstein's postulates) Lorentz transforms, frames, etc.
Last edited by tomclarke on Wed Jan 11, 2012 4:00 pm, edited 3 times in total.

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Post by happyjack27 »

tomclarke wrote:To show the twins must have different ages, all you need is Doppler effect, and Einstein's 2nd postulate - which Johan has been silent about.

....

(8) Doing the calculation for the number of pulses from each twin:

Stationary twin: f1T1 + f2T2
Moving twin: f1'T1'+f2'T2'
Let us also label total time:
T1 + T2 = T1' + T2' = T

(9)
f2 = f2' = (1+v/c)f
f1 = f1' = (1-v/c)f
[f is the frequency at which pulses are emitted. It will turn out we need to scale this by a time dilation factor. But that will be the result of our calculation, we do NOT assume this]

Difference in age is: f(1-v/c)(T1-T1') + f(1+v/c)(T2-T2') =
f[ (T - T) - v/c( T2' - T2) - v/c(T1- T1') ]

The first term cancels and (by inspection) T2' > T2, T1 > T1' so we must have the stationary twin older than the moving twin.
this is all fine and good math, but which of the two twins is the "stationary" one? you are sure to get a different answer from each twin!

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Post by tomclarke »

Johan wrote: I am, however afraid that you are inserting an asymmetry where there is none. The symmetry must be done from the inertial reference frame of each twin since the signals are sent out from both inertial refrence fames on both the clocks which (as you agreed) are keeping the same time rate within their respective reference frames: The two diagrams that must be compared are where the light arrows go from the vertical time axes to the other reference frame that moves relative to the reference frame that you choose for the vertical time axis. Thus, from the vertical time axis T of twin 1 to the moving reference frame of twin 2; and then from the vertical time axis Tp of twin 2 to the moving reference frame of twin 1. The situation is then totally symmetric as it must be.
this appears to be your only rebuttal of my (simple) argument.

Even if you do as you say here, the situation is not symmetric. Consider the inbound (second) part of your two diagrams. The light rays received by the moving twin during the inbound journey are (obviously) all blue-shifted. The light rays received by the stationary twin during the first part of the inbound journey will have been transmitted by the moving twin during the previous outbound journey, and so be red shifted.

The journey is symmetrical on time reversal, so any effects due to frame change (like LT time dilation etc) do not alter the fact that half the time the moving twin received rays are blue-shifted, less than half the time the stationary twin received rays are blue shifted.

The situation is symmetric if both twins move away from each other equally at v/2, of course. Your diagram redrawn so that the top and bottom reference frame are different and the moving twin appears stationary is misleading, and that is where you go wrong.

The redrawn diagram incorrectly makes you think the Doppler effect changes at the time of the apparent change in velocity of the stationary twin (kink on diagram) - when it must be at the real change in velocity of the moving twin (not shown on your diagram).

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Post by tomclarke »

happyjack27 wrote:
tomclarke wrote:To show the twins must have different ages, all you need is Doppler effect, and Einstein's 2nd postulate - which Johan has been silent about.

....

(8) Doing the calculation for the number of pulses from each twin:

Stationary twin: f1T1 + f2T2
Moving twin: f1'T1'+f2'T2'
Let us also label total time:
T1 + T2 = T1' + T2' = T

(9)
f2 = f2' = (1+v/c)f
f1 = f1' = (1-v/c)f
[f is the frequency at which pulses are emitted. It will turn out we need to scale this by a time dilation factor. But that will be the result of our calculation, we do NOT assume this]

Difference in age is: f(1-v/c)(T1-T1') + f(1+v/c)(T2-T2') =
f[ (T - T) - v/c( T2' - T2) - v/c(T1- T1') ]

The first term cancels and (by inspection) T2' > T2, T1 > T1' so we must have the stationary twin older than the moving twin.
this is all fine and good math, but which of the two twins is the "stationary" one? you are sure to get a different answer from each twin!
The left-hand twin is stationary (the two pictures are the same situation, drawn twice to show light rays in different directions).

I do get a different answer for the stationary and moving twin, as I show.

You can't reverse the argument because it depends on one twin turning round. This turnaround point alters the Doppler shift for the moving twin immediately, for the stationary twin only after light has travelled to it from the moving twin. Obviously, only the moving twin turns around!

Of course, if both twins turnaround (and move at v/2) the problem is symmetric and there is no difference.

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Post by tomclarke »

Anyone who wants to see the mistake in Johan's incredibly complex Lorentz Transform post above should note only the sign of the Doppler effect.

Johan is only considering the (1+v/c) component of the Doppler (inbound journey) when he has to consider both inbound and outbound. Look through all his equations. See any (1-v/c)?

But it is clear from the diagrams (or personal observation of say train whistles) that Doppler effect is different outbound and inbound.

If he did that, he would have to deal with the fact that between the two twins the received inbound part would start at a different time, which is the source of the asymmetry.

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Post by johanfprins »

tomclarke wrote: Sorry, what intertial reference frame?
Both twins are each within an inertial reference frame and they each send out light pulses from their reference frames.
I am not involing reference frames here. All I consider is the Doppler effect on light emitted by one twin and received by the other.
So you are saying that the twins are not each within an inertial reference frame and that the two light sources are not each stationary within each of the twin's inertial refrence frames. The hwole theory of SR is based on the concept on inertial refrence frames moving relative to one another. How can you make any concluions by not involving refrence frames?
why is it asymmetric? Just look at the diagrams! Consider the light-ray sent from the moving twin at turaround.
Both twins are moving relative to the other twin and they both send light pulses to one another at turnaround. There is not a unique stationary twin. This is the whole point of Special Relativity that there is not a unquely stationary reference frame.
This is received by the stationary twin AFTER turnaround and marks the start of T1.
The times at which each twin receives the other twin's signal after the other twin has sent the signal is shorter on the return journey owing to the Doppler shift, but for each pulse that twin1 receives twin2 also receives a pulse at the same exact time. Thus the total number of pulses received by both twins is still the same once they meet up. The mathematics derived from the Lorentz transformation confirms that this is so.
By (time) symmetry, T2' = T1'
therefore T1 < T2 (by the time it takes light to travel between the twins when they are furthest separated).
Look at equation [4] and [10] above. They do not agree with your assertion here at all. Where do you say have I made a mistake when I derived these equations?
These equations remain true in all frames, which is why I do not need to use them, and why you need to anwer this question of the inherent asymmetry simply.
As I have proved by using the Lorentz transformation, there is no asymmetry invloved when you do the mathematics correctly.
Intuitively, the asymmetry derives from the fcat that one twin turns around, the other does not.
Their speed is relative: You can only conclude that one turns around as seen from a third reference frame. But this is nonsensical, since one twin sees the other twin to "turn around" and other twin sees the first twin as turning around. The turn around is symmetrical.
The turnaround alters the Doppler effect,
Correct, it does: My impeccable mathematics using the Lorentz equations prove that it makes no difference.
Light going to the twin which turns around will have different doppler effect immediately after the turnaround. Light coming from it will have different Doppler effect at time of turnaround + light travel time.
This does NOT change the fact as proved above, that the two twins must receive exactly the same number of pulses during the whole journey.
Let me do the calculation by using the Lorentz transformation:

...snip...

Do you now see how dangerous it is to ignore the Lorentz transformation
Wow. You were the one saying I misused mystical LTs instead of good-old einstein's postulates!
You are distorting what I have said. I said that when you derive a result, for example that the twins will age at diffrent rates, you are violating Einstein's postulates since according to these postulates and the Lorentz transformation derived from it, the twins cannot age at different rates.
I'll correct your incorrect use of LTs here after you have dealt with the (simpler) issue of Doppler asymmetry.
My equations have dealt with it. What eklse must I do. Why wait? "Correct" me it if you can: Which I doubt. You cannot correct which is already correct. The Doppler effect is part of the Lorentz transformation and does appear within my equations as (1+v/c) on the outbound leg and as (1-v/c) on the return leg but the equations still prove compellingly that the two twins will each receive the SAME number of pulses during the whole journey. Therefore tgeir respective clopcks MUST show the time for the whole journey
I have not used Minkowski anything. My diagrams are simple time/position diagrams such as are commonly used to illustrate moving objects.
Exactly! And such a diagram is nonsense when the Lorentz transformation applies.
You are the one using (without proof direct from Einstein's postulates) Lorentz tranforms, frames, etc.
The Lorentz transformation is a direct result of Einstein's postulates. Thus, by using these equations I am proving my case directly from Einstein's postulates. Your conclusion directly cotradicts Einstein's postulates as well as the LT

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