johanfprins wrote: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.
No I am just saying i do not need the reference frame concept for this proof.
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?
If you read through my post's steps, which clearly you have never done, you would see how.
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.
Sigh. The "stationary" twin is so called because his movement is inertial and therefore there exists an inertial frame in which he is always stationary. The moving twin has no such frame. It is a shorthand to call them stationary and moving.
You may if you like consider my whole proof to be carries out in the rest frame of the stationary twin.
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.
That is clearly not true, look at the diagrams to see why, or my many explanations: the "return journey" effect on received pulses starts at different times for each twin.
Thus the total number of pulses received by both twins is still the same once they meet up.
See above.
The mathematics derived from the Lorentz transformation confirms that this is so.
You apply the LT assuming that the inward and outward segments are the same at both ends, hence your wrong answer. maths of LT is irrelevant to this.
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?
Johan. I have stated clearly my arguments. I have also stated what is the (initial and basic) defect in yours. Therefore instead of asking me to say what is wrong with your argument (I already have) try to engage with mine. If you are correct there will be some step in it which you can question.
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.
That is a silly statement, see above.
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.
Physically the turnaround is asymmetrical. Relativity applies to inertial frame, not to frames which suddenly reverse direction. that is one of your fundamental mistakes.
The turnaround alters the Doppler effect,
Correct, it does: My impeccable mathematics using the Lorentz equations prove that it makes no difference.
See above
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.
Ahhh... Most revealing. You admit this is true. You see that it proves the asymmetry. You have however no answer to this except to illogically assert that it makes no difference to your stated (without reasons) result. Of course it does, because it chnages the number of pulses received by the stationary twin at the lower frequency, and the number at the higher frequency.
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.
For example, you could go through the steps of my original post saying which is the first one you disagree with, and why. That is the normal way in which arguments are resolved.
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.
I don't see that can be when your equations do not address the issue, obvious from my diagrams, that the length of pulses received at higher frequency is shorter on the stationary twin than the moving twin. In your derivation you have just assumed the situation is symmetrical.
Your stated reasons, a distorted diagram in which the change in velocity of one twin cannot be seen, and an apparent change in velocity of the other twin is added which does not exist, have been explained by me.
This distorted diagram does not contain the information needed to determine whether a given light ray from the stationary to the moving twin is doppler shifted up or down, because you cannot see on it where the moving twin chnages velocity.
My reasons, the two diagrams I give and the points in my post, have not been addressed by you.
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.
No, such diagrams are exactly what is used to prove the Lorentz transform from Einstein postulates, which I thought was what you would want, given that you claim "first principles" theory superior to that others derive.