tomclarke wrote: I'm not even a physicist, just enjoying learning SR on this thread, but I have not found any questions recently I could not answer And answer consistently.
Yes you gave answers but they are all wrong. I had a student who always had an answer but it was never really an answer. You remind me of him. I think that you do not understand what consistency means: This in addition to the fact that you refuse to do the simple algebra I have posted on Wednesday; or maybe you just are not competent enough to follow simple mathematical logic?
[You know, being able to answer all questions does not necessarily mean correctness.
Exactly and you have been CONSISTENTLY WRONG on this thread.
So let me try again: Twin1 in K and twin2 in Kp; They synchronise their clocks at the beginning of the journey and then move away from one another with a RELATIVE SPEED v. Both have atomic clocks, and the transitions on which these clocks are based MUST according to Einstein's first postulate be the same in both reference frames. Thus within both reference frames the clocks MUST keep the same time with respect to each reference frame within which each clock is stationary. This means that after a time T(E) on the clock of twin1 there is a time Tp(E) on the clock of twin2 and that T(E)-Tp(E). This also means that at that instant in time twin2 is a distance D=v*T(E) from twin 1 and similarly twin1 is a distance D=-v*Tp(E) away from twin2. The actual distance between them, as measured from either twin, is thus the same; namely D.
As already pointed out on Wednesday, if twin 1 now sends out a light pulse at the time T(E)=Tp(E) then, although this light pulse moves with speed c relative to twin1, it must also move with speed c relative to twin2. Thus to find out what twin2 sees within his inertial reference frame one must transform the coordinates X=0, and T=T(E) of this appearance of the light pulse within K, into Kp. The coordinates within Kp are thus Xp(Ep) and Tp(Ep) given by the Lorentz transformation as:
Xp(Ep)= -(gamma)*v*T(E)=-(gamma)*D....................................[1]
Thus the distance relative to twin1 at which the light pulse appears IS LARGER than the ACTUAL DISTANCE D between twin1 and twin2. This is counterintuitive but is what the Lorentz transformation gives: And when you ignore this increase in distance, you reach the wrong conclusions that tomclarke has posted repeatedly, and with false confidence, on this thread.
Furthermore
Tp(Ep)=(gamma)*T(E)..........................................[2]
In addition to the fact that the light pulse appears at a longer distance away from twin2 than the ACTUAL DISTANCE D between the twins, twin2 also sees it appearing at a later time Tp(Ep) than the actual time T(E) that twin1 sends the pulse. If you do not take BOTH these dilations into account you draw the silly, and wrong, time-space diagrams that tomclake has posted.
Now as I have also have posted it is simple to derive at which time Tp(A) on twin2’s clock that the light pulse will reach twin2. Since I have already shown how the derivation must be done I will just post the result: is longer than the actual
Tp(A)=(1+v/c)*(gamma)*T(E)...............................[4]
But twin2 also sends out a light pulse at time Tp(E)=T(E) when the two twins are an ACTUAL DISTANCE D apart 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]
And as already derive the time T(A) on the clock of twin1 that this light pulse reaches twin1 is:
T(A)=T(Ep)+X(Ep)/c= (1+v/c)*(gamma)*Tp(E)..................[10]
Since T(E)=Tp(E) then according to Eqs. [4] and [10] one must have that T(A)=Tp(A) Thus both twins receive the other twin’s light pulse when their clocks show the exact same time on their respective cocks.
Now let us slightly change the argument at this point, and assume that the twins agreed that after each has received the first light pulse they will both send out their second light pulse at the instant when the receive each others incoming light pulse. So they send out their second light pulses at T(E)2=T(A) and Tp(E)2=Tp(A): Thus in this case T(E)2=Tp(E)2, and if you go through the mathematics again the twins will now simultaneously receive the second light pulses at time T(A)2 and Tp(A)2, where:
T(A)2=(1+v/c)*(gamma)*Tp(E)2=(1+v/c)*(gamma)*Tp(A)
And from Eq. [4] this becomes:
T(A)2=(1+v/c)*(gamma)*[(1+v/c)*(gamma)*T(E)]=
{[(1+v/c)*(gamma)]^2}*T(E)……………...................................[11]
Similarly for
Tp(A)2={[(1=v/c)*(gamma)]^2}*T(E)………………………….(12)
And again T(A)2=Tp(A)2
Thus after N pulses twin1 receives the Nth pulse at time T(A)N on his clock where:
T(A)N={[(1+v/c)*(gamma)]^N}*T(E)…………………………..(13)
And
Tp(A)N={[(1+v/c)*(gamma)]^N}*T(E)………………………….(14)
And of course again we have that T(A)N=Tp(A)N. The times on the clocks are still exactly the same when the two twins receive the Nth light pulse simultaneously.
To simplify the mathematics, we can assume without any loss in generality, that at the instant that each receive the Nth pulse simultaneously, they start to move back towards on another, and at the same instant each one sends out the (N+1)th pulse at the time T(A)N=Tp(A)N. Since they are now moving towards each other the Doppler factor changes from (1+v/c) to (1-v/c) so that the signals reach the twins at times so that the pulse from twin1 will reach twin2 at a time Tp(A)(N+1) which similar to Eq. [4] is now given by:
Tp(A)(N+1)=(1-v/c)*(gamma)*Tp(A)N…………………………(15)
And the pulse from twin2 reaches twin1 at the time T(A)(N+1) which similar to Eq. [10] is given by:
T(A)(N+1)=(1-v/c)*(gamma)*T(A)N……………………………(16)
Thus since Tp(A)N=T(A)N, the twin’s clocks still read the same when they each receive the (N+1)th pulse from the other twin.
After M pulses during the return journey the times on the clocks when the twins receive and send out pulses are Tp(A)(N+M) and T(A)(N+M)
Where Tp(A)(N+M)={[(1-c/v)*(gamma)]^M}*Tp(A)N……….(17)
And
T(A)(N+M)={[(1-c/v)*(gamma)]^M}*T(A)(N)…………………(18)
An since Tp(A)N=T(A)N one must have that T(A)(N+M)=Tp(N+M). Thus if the two twins reach one another after each one of them has received M pulses on the return journey, their clocks will read exactly the same time. Although, owing to the Doppler effect, the number of pulses N on the outward leg is not the same as the the number of pulses M on the return leg the total number of pulses received by each twin is during the whole journey is exactly the same namely (N+M): And since the clocks of the twins read exactly the same time when each receives the others pulse, their clocks will also show exactly the same time after the whole journey is completed.
I think the derivation is now so clear that even my goldfish will understand it. I hope that tomclarke will now do the honourable thing and admit that he was wrong all the time and apologise to all of us for having wasted so much of our time on this thread.