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Re: A friendly request to Ken Seto from Uncle Ben
| harry | 27 Nov 2008 15:12 |
On Nov 26, 10:49 am, Uncle Ben <b...@greenba.com> wrote:
> On Nov 26, 9:29 am, kenseto <kens...@erinet.com> wrote: > [quoted text clipped - 169 lines] > the light flashes from the back and from the front are now not > simulotaneous.
: I thought t'1 and t'2 were the transit times from +x and -x locations. Wrong again: points in space-time are indicated by (x, t) and (x', t'). Thus the t' are NOT "transit times from locations" (= time intervals) but moments in time (in this case: the point in time of each strike).
Success! Harald
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| kenseto | 27 Nov 2008 14:05 |
> > > > > On Nov 25, 5:15 pm, "Dirk Van de moortel" > [quoted text clipped - 137 lines] > the light flashes from the back and from the front are now not > simulotaneous. I thought t'1 and t'2 were the transit times from +x and -x locations. I didn't know that they represent the transit times from the locations of the ends of the train after the lightning happened simultaneously. Your problem is that you can't comprehend that the locations of the ends of the train got nothing to do with how light propagates from the points of strikes at +x and -x locations.
Ken Seto
> You proved it yourself! > > Uncle Ben- Hide quoted text - > > - Show quoted text - |
| Uncle Ben | 26 Nov 2008 15:49 |
> > > > On Nov 25, 5:15 pm, "Dirk Van de moortel" > [quoted text clipped - 131 lines] > > - Show quoted text - I thought you had already shown the equations for t'1 and t'2 and had shown that they differed in proportion to (x2-x1). Now that you know that x1 and x2 are different, that shows that the starting times of the light flashes from the back and from the front are now not simulotaneous.
You proved it yourself!
Uncle Ben
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| kenseto | 26 Nov 2008 14:29 |
> > > On Nov 25, 5:15 pm, "Dirk Van de moortel" > [quoted text clipped - 115 lines] > occurred before t'=0 and t'1 occurred after t'=0. These are > coordinates, not transit times. There is no t'1 and t'2. There is t'. Why? Because once the flashes happened at the ends of the train the location of the ends become irrelevant. The only important thing is the transit time of the light fronts from the original points of the flashes (+x and -x) to arrive at M'.
Ken Seto
> Uncle Ben- Hide quoted text - > > - Show quoted text - |
| Uncle Ben | 26 Nov 2008 01:14 |
> > On Nov 25, 5:15 pm, "Dirk Van de moortel" > [quoted text clipped - 111 lines] > > - Show quoted text - And with regard to the times, if t'1 >0 and t'2 <0, it means that t'2 occurred before t'=0 and t'1 occurred after t'=0. These are coordinates, not transit times.
Uncle Ben
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| Uncle Ben | 26 Nov 2008 01:07 |
> On Nov 25, 5:15 pm, "Dirk Van de moortel" > [quoted text clipped - 98 lines] > > - Show quoted text - M <- -x -------x1----------------0++++++++++++++++x2+++++++++ +x -
M' tttttttttttttttttttttttttttttttttttt (train)
Can you see that x1 = -x2?
Uncle Ben
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| kenseto | 26 Nov 2008 00:27 |
On Nov 25, 5:15 pm, "Dirk Van de moortel" <dirkvandemoor...@nospAm.hotmail.com> wrote:
> kenseto <kens...@erinet.com> wrote in message > [quoted text clipped - 47 lines] > t'1 = - gamma v x1 / c^2 > t'2 = - gamma v x2 / c^2 Yes I did forgot the gamma factor.
> > Since x1 = x2 therefore the light fronts from events 1 and 2 will > > arrive at M simultaneously. [quoted text clipped - 10 lines] > meaning that > t'1 = - t'2 Right....This means that the transit time for the arrival of the light fronts from the strikes is the same in both the -x direction and the +x direction and thus the light fronts arrival at M' simultaneously.
>Meaning that 1 and 2 will *not* arrive at M simultaneously, Why? The same transit time for both directions.
>but, > assuming that x1 > 0 (on the right, so to speak) and thus x2 < 0 > (on the left), then 1 arrives first (before t' = 0), and 2 arrives later > (after t' = 0). This is a bogus assumption....x1 always equal to x2 (because the speed of light is isotropic). Once the flashes occurred simultaneously the motion of the ends of the train has no effect on how the light fronts will propagate in the M and M' frames. BTW that's why the speed of light is isotropic and independent of the motion of the source.
Ken Seto
> A transformation transforms coordinates. Not distances. > > Dirk Vdm- Hide quoted text - > > - Show quoted text - |
| Dirk Van de moortel | 25 Nov 2008 22:15 |
kenseto <kenseto@erinet.com> wrote in message b2924e9e-19bb-4351-a1a3-fa21040a2283@k8g2000yqn.googlegroups.com
>>>> Dear Ken, >> [quoted text clipped - 39 lines] > Event 2 arrive at M': > t'2 = -vx2/c^2 You forgot the gamma factors. Make that t'1 = - gamma v x1 / c^2 t'2 = - gamma v x2 / c^2
> Since x1 = x2 therefore the light fronts from events 1 and 2 will > arrive at M simultaneously. Remember that x1 and x2 and coordinates, not distances. So make that x1 = -x2 and you get t'1 = - gamma v x1 / c^2 t'2 = - gamma v x2 / c^2 and thus t'1 = - gamma v x1 / c^2 t'2 = + gamma v x1 / c^2 meaning that t'1 = - t'2 meaning that 1 and 2 will *not* arrive at M simultaneously, but, assuming that x1 > 0 (on the right, so to speak) and thus x2 < 0 (on the left), then 1 arrives first (before t' = 0), and 2 arrives later (after t' = 0).
A transformation transforms coordinates. Not distances.
Dirk Vdm
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| kenseto | 24 Nov 2008 18:19 |
> > > Dear Ken, > [quoted text clipped - 28 lines] > What you say may or may not be true, but I don't see any Lorentz > Transformation yet. What is the Lorentz Transformation? Event 1 t'1 = gamma(t1 - vx1/c*c) Lorentz Transformation, track -> train Event 2 t'2 = gamma(t2 - vx2/c*c) Lorentz Transformation, track -
> train When M and M' coincide with each other t1 = t2 =0 Therefore: Event 1 arrive at M': t'1 = -vx1/c^2 Event 2 arrive at M': t'2 = -vx2/c^2
Since x1 = x2 therefore the light fronts from events 1 and 2 will arrive at M simultaneously.
Ken Seto
- Hide quoted text -
> - Show quoted text - |
| Uncle Ben | 24 Nov 2008 17:34 |
> > Dear Ken, > [quoted text clipped - 25 lines] > > - Show quoted text - What you say may or may not be true, but I don't see any Lorentz Transformation yet. What is the Lorentz Transformation?
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| kenseto | 24 Nov 2008 14:36 |
> Dear Ken, > [quoted text clipped - 13 lines] > for all the years I have taught relativity, I will be very forever > grateful to you for correcting me. When M and M' are coincide with each other: The transit time for the light fronts from +x and -x arrive at M simultaneously is L/c from both directions. The light path length for the light fronts from +x and -x to arrive at M' is gamma*L Therefore the transit time for the light fronts from +x and -x to arrive at M' simultaneously is gamma*L/c.
Ken Seto
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| Uncle Ben | 23 Nov 2008 22:18 |
Dear Ken,
For weeks now you have been telling me, as we discuss elementary problems in Specialk Relativiy, that I have been executing the Lorentz Transformation incorrectly.
Maybe you are right, Ken. Educate me a little. Tell me, what is the Lorentz Transformation? In particular,
1. What is it for? It is a tool, I know, but what is it used for?
2. Show me a very simple example of how it is used for any problem you like.
If it turns out that I have bveen misusing the Lorentz Transformation for all the years I have taught relativity, I will be very forever grateful to you for correcting me.
Uncle Ben
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