User:Jh17710

A hidden fact in LT

Let's study Lorentz Transformation (LT) step by step. If we follow the "ruler contraction" we will find the hiding equation t'=γt after the following steps.

1. In LT, there is no assigned observers, however, to make sure the equation x'=γ(x-vt) is established solidly, we do need some more details as followings: 1-1. We need an event eA at the location A(x.y.z) in the system (t,x,y,z), named S, with coordinate A(x',y',z') in the system (t’,x’,y’,z’), named S', so that we know what are x' and x stand for. 1-2. The event time of eA recorded in S is t and the event time of eA recorded in S' is t'. 1-3. The velocity of S' is v as measured in S and we assume that the velocity of S is -v as measured in S'. We then let v in the equation be the speed, not the velocity. 1-4. We may name the origin point of S as O(0,0,0) and of S' as O'(0,0,0) just for convenience purpose.

2. To understand the equation x'=γ(x-vt) clearly, we need to think about the locations of three points A, O, and O': 2-1. To make things easier, we let the project point of A on x-axis be Ax(x,0,0). 2-2. Let x+ point to the right, then the relation of Ax and O can be Ax-O when x<0, Ax=O when x=0, and O-Ax when x>0. 2-3. At the event time t', in S', the location of O' can have 9 possible relationship of locations with Ax and O. They are: O'-Ax-O, O'=Ax-O, Ax-O'-O, Ax-O'=O, Ax=O'=O, O=O'-Ax, O-O'-Ax, O-O'=Ax, and O-Ax-O'.

3. We will figure out how do we derive the equation x'=γ(x-vt) base on the above 9 scenarios. Let the distance of O and O', measured in S’, be represented by the symbol (OO')', then in S’, base on the hypothesis of “ruler contraction”: 3-1. When O'=O, we have t'=t=0, x'=γx. 3-2. When O'-O, we have x'-(OO')'=γx. Since it takes total time of |t'|=-t' for O' to reach O under the speed of v, (OO')'=-t'v. We have x'=γx-vt'. 3-3. When O-O', we have x'+(OO')'=γx. Since O' moves from O for a total time of |t'|=t' under the speed of v, (OO')'=t'v. We have x'=γx-vt' too. 3-4. We established that for all 9 scenarios, we have x'=γx-vt'.

4. If we try to get the equation in LT, x'=γ(x-vt), from the equation we just established x'=γx-vt', we must have γvt=vt', t'=γt. That means x'=γ(x-vt) is based on t'=γt.

There is another way to derive t'=γt by time-period. I will list the procedure next time. Please comment. Thanks.

Regards, John Huang Jh17710 (talk) 02:55, 22 October 2010 (UTC)

To establish t'=γt in LT

Let's find the hiding equation t'=γt in LT by the following steps.

1. In LT, there is no assigned observers, however, to make sure we will have a solid communication, we do need some more details as followings: 1-1. We need an event eA at the location A(x.y.z) in the system (t,x,y,z), named S, with coordinate A(x',y',z') in the system (t’,x’,y’,z’), named S', so that we know what are x' and x stand for. 1-2. The event time of eA recorded in S is t and the event time of eA recorded in S' is t'. 1-3. The velocity of S' is v as measured in S and we assume that the velocity of S is -v as measured in S'. We then let v in the equation be the speed, not the velocity. 1-4. We may name the origin point of S as O(0,0,0) and of S' as O'(0,0,0) just for convenience purpose.

2. To establish the hiding equation t'=γt in LT we select a point B(b,0,0) in S: 2-1. Let x+ point to the right, then the relation of B and O can be B-O when b<0, B=O when b=0, and O-B when b>0. 2-2. Let the event be the motion of O’ from B to O or from O to B. In S, we have distance between B and O measured as OB and event-time-period measured as △t, so that v=OB/△t. Then, in S’ we have event-time-period measured as △t’. 2-3. Now, we can calculate the distance between O and B in S’ by the “ruler contraction” hypothesis, (OB)’=γ(OB), so that the speed of O’ as calculated in S’ should be v=(OB)’/△t’=γ(OB)/△t’, since the v measured in S should be the same as the v calculated for S', OB/△t=γ(OB)/△t’, so that △t’=γ△t.

3. So far, we have established △t’=γ△t for any given point B(b,0,0) on x-axis. 3-1. For b<0, O’ moves from B to O so that △t’=0-t’ and △t=0-t. We have –t’=γ(-t), so that we have derived t’=γt for the situation of b<0. 3-2. For b>0, O’ moves from O to B so that △t’=t’-0 and △t=t-0. We have t’=γt for the situation of b>0. 3-3. For b=0, we have t’=t=0 so that t’=γt. We have established that in LT, to any given point B on x-axis, for the motion of O' between O and B, we always have time equation t’=γt when O' is at B. That means, for any location of O', t’=γt.

4. That means, LT should have time equation t’=γt. Then, why do we have the popular time equation t’= γ(t-(xv/c^2)) in LT? I will explain it next time.

Please comment. Thanks.

Regards, John Huang Jh17710 (talk) 03:14, 23 October 2010 (UTC)

Why t’= γ(t-(xv/c^2))?

First of all, we should understand that Lorentz assumed LT and Einstein tried to derive LT, at least twice, but all failed. The first one is on the paper dated 1905-6-30 and the second one is the appendix I (Supplementary to section XI) of Relativity: The Special and General Theory, 1920. That history reveals the reason why Lorentz finalized LT like that t’= γ(t-(xv/c^2)) ---(1) and x'= γ(x-vt) ---(2).

I think, we should blame on the equation of Galilean Transformation (GT), x'=x-vt ---(3). Since Lorentz just assumed "ruler contraction", so that he could base on the arrangement of GT and assume x'= γ(x-vt) directly. If GT is listed as x'=x-vt' ---(4), then Lorentz might have found the time equation hiding in the motion of O', the moving origin point, which is t'=γt ---(5).

The equation (4) is the true spatial equation for GT when S' is the moving system, because when observers in S' record event time t', the distance between origin points O and O' is always v|t'| and x' must go by that location of O'. That means at time t', when t'<0, x'=x+OO'=x-vt' and when t'>0, x'=x-OO'=x-vt', we always have equation (4). Since in GT we also have t'=t to make (3) correct, I will say it was a pure bad luck that Lorentz did not find the time equation (5) at the beginning.

John Huang Jh17710 (talk) 18:18, 23 October 2010 (UTC)

How t’= γ(t-(xv/c^2)) works?

Assume that we mark equations t’= γ(t-(xv/c^2)) ---(1) and x'= γ(x-vt) ---(2). There is nothing wrong to establish (1), however, what people should know is that when (1) was derived from combination of two equations, then, the variable domain for (1) may reduce to the common domain of two variable domains for each of the original two equations.

Lorentz started from (2) as we learned from the history, then, he derived the inverse spatial equation for moving S at constant velocity -v as x=γ(x'+vt') ---(6), then he replaced x' in (6) by the right side of (2) to get (1). Even if he did not do so, base on LT, when we replace the vt in (2) by the vt=(vt'/γ)+(v^2/c^2)x from (1) we will derive (6). That means, (2) and (6) will always coexist in LT. The variable domain for (1) is the common domain of variable domains for (2) and (6).

To see restrictions to variables in (2) clearly, we can get help from the true spatial equation for LT, x'=γx-vt' ---(7). Under the standard configuration (SC-LT) of GT and LT, if we assume "ruler contraction" then we will derive (7) and (5), t'=γt, follow logic reasonings. From (7) and (5) we can get (2). Now, it looks clear that variables in the variable domain for (2) are variables satisfied all of (SC-LT), (7), and (5). It is quite the same reasonings that when S is moving, at inverse LT, we will find x=γx'+vt ---(8) and t=γt' ---(9) first, then, from (8) and (9) we can derive (6) so that variables in the variable domain for (6) are variables satisfied all of inverse (SC-LT), (8), and (9).

That means variables in the common domain of variable domians for (2) and (6) must satisfy both of (5), t'=γt, and (9), t=γt'. Do you see the result? We find out variables in that common domain for (2) and (6) must satisfy (γ^2)=1. When c>|v|, γ>0, so that γ=1 when c>|v|. That means v=0 when c>|v|. That means (1) is almost useless with the restriction that v=0 is the only possible variable for v in (1), t’= γ(t-(xv/c^2)).

Now, we all understand how (1) works, for all |v|<c, the equation (1), t’= γ(t-(xv/c^2)), will reduce to t'=t by the restriction of v=0.

John Huang Jh17710 (talk) 16:09, 23 October 2010 (UTC) Jh17710 (talk) 19:15, 23 October 2010 (UTC)