Dear Dr. Perez,
1. You wrote: "I performed the calculations based on SR step by step and I showed you that the time dilation equation (B) follows from the time transformation (A) and you still argue that I did the tricks found in textbooks. What is the trick? I am wrong (based on ST) tell me please were I am wrong".
I WILL SHOW YOU THE TRICK: When x = ct in SRT as a basic principle, x is substituted by x = vt
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2. I have copied below your step by step calculation and I have numbered each step as the alphabetical order for easy reference.
Let us take the step D. "Thus, the space interval as determined in the lab is
Dx=x_1-x_0=x_1. "
From the millions of experiments conducted in the last century, it has been confirmed that the displacement x_1 is not equal to v.t_1. (Where v is velocity that the particle of mass M is supposed to move relative to the EARTH, when the applied momentum is gamma.Mv).
It is equal to x'_1 = gamma (x_1 - ut) where u = 30 k/s and gamma = 1.000000005 for all experiments conducted on earth. You may verify this with any EXPERIMENTAL particle physicist.
When the displacement x' is measured relative to the lab-frame using the lab's 'metr rod', the time interval must be measured by the lab's clock. This time interval corresponds to t in the above equation. There is no t'.
3. In regard to the trick, I refer you to Rindler's derivation of the Lorentz transformations in 'Introduction to Special Relativity'.
The above equation is written as x' = gamma (x -vt) -------------(1)
but note here v is not directly referred to as the velocity of the particle as such but as the velocity of the reference frame relative to the observer's frame. c is the signal velocity in the moving frame (as well as the observer's frame).
Because in SRT co-ordinate and time are related by the signal velocity c, it is implicit that x'/c = t' and x/c = t and this finds confirmation when we divide (1) by c. We get,
t' = gamma(t - vt/c) --------------------(2a)
Now substitute in (2a) t = x/c and we get
t' = gamma( t - vx/c2) ----------------(2)
For the relationship of signal velocity and time c2t2 - x2 = 0, but Rindler takes a circuitous route to obtain the equations (1) and (2) because he takes c2t2 - x2 to be greater than 0 instead of the c2t2 - x2 = 0;
and then introduces the hypothesis (x - vt) = 0 which means not only that x' = 0 in (1) BUT MORE IMPORTANTLY v = c, since x = ct. Also v=c means that the particle is moving at velocity c relative to the lab frame which is a violation of SRT.
On the other hand if x = vt and v is considered to be less than c, then the basic principle of the relationship between the co-ordinate, signal velocity c and time is violated. The only condition when this principle is not violated is when v = c.
(Rindler gets out of this problem by a further hypothesis on the mathematical form of x' and t').
However, we must bear in mind that the condition that underlies the connection between the equations (1) and (2) is x = ct. Then when YOU in your correspondence with me use the same equation (2) the same condition must remain. And x cannot be substituted by vt in (2) unless v = c.
Then we can have two cases:
a) when v is less than c the vx/c2 term becomes vt/c (when ct substituted for x)
Then the equation (2) becomes t' = gamma t( 1 - v/c) ----------(2a)
b) When v = c , the vx/c2 terms becomes c2t/c2
Then the equation (2) becomes t' = 0 ----------------(2b)
Equations (1) and (2) as we saw are related by x = ct. Without violating the condition x = ct by illegitimately substituting x = vt, you cannot transform the Lorentz time equation to appear as the time dilation equation as you have done.
So you will see that your following statement does not hold:
L. "There is no conflict and relativity is in agreement with experience. By measuring time dilation we are testing the transformation expression t'=gamma(t-vx/c2)".
Best regards,
Viraj
Extract from your post of Oct 6 copied below:
A. Now assume that the muon is moving at speed v relative to the lab and we on the lab want to measure again the decay time Dt.
B. Suppose that we synchronize the clocks of both systems such when we start to measure, our clocks read t_0=t'_0=0 and our rulers read x'_0=x_0=0.
C. Then at a later time t_1 the muon decays. So the decay time Dt in the lab is Dt= t_1-t_0=t_1 and his position at the moment of decay is x_1.
D. Thus, the space interval as determined in the lab is Dx=x_1-x_0=x_1.
E. Remember that dt and Dt are the decay times of the muon at rest and in motion, respectively, as determined in the lab system.
F. So, when the muon is in motion we would expect (due to time dilation) that Dt would be greater than dt. Let us understand this.
G. The next step is to calculate the decay time in the system of reference of the muon, that is, the system of reference where now the muon is at rest.
H. For this we use the transformations equations: t'_0=gamma(t_0-vx_0/c2) and t'_1=gamma(t_1-vx_1/c2), where gamma=1/sqrt(1-v2/c2).
I. Hence the time interval in the system of the muon is: Dt'=t'_1-t'_0=t'_1=gamma(t_1-vx_1/c2), and since x_1=vt_1 we have that Dt'=gamma(t_1-v2t_1/c2).
J. Factoring out t_1=Dt we have Dt'=gammaDt(1-v2/c2); and considering that (1-v2/c2)=gamma^(-2) we finally arrive at the famous expression for time dilation: Dt'=gamma^(-1)Dt. Since gamma^(-1) is less than 1, Dt' is less than Dt as we expected.
K. This means that the decay time in the system where the particle is at rest is smaller than in the lab system in which the particle is in motion.
L. There is no conflict and relativity is in agreement with experience. By measuring time dilation we are testing the transformation expression t'=gamma(t-vx/c2).
