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Groups > sci.physics.relativity > #610289 > unrolled thread
| Started by | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| First post | 2023-05-09 23:40 -0700 |
| Last post | 2023-05-12 10:34 -0700 |
| Articles | 20 on this page of 87 — 19 participants |
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The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-09 23:40 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-10 07:21 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-10 04:52 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-10 23:22 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-10 18:38 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-11 12:38 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-11 06:37 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-11 23:59 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-11 19:22 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-12 13:12 +0000
Re: The Light Clock Thought Experiment - Questions "Dono." <eggy20011951@gmail.com> - 2023-05-10 19:15 -0700
Re: The Light Clock Thought Experiment - Questions Tom Roberts <tjoberts137@sbcglobal.net> - 2023-05-12 15:45 -0500
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-13 00:02 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-12 19:44 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-14 01:18 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-13 18:36 -0700
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-13 18:48 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-16 02:37 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-15 20:42 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-16 05:36 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-16 04:26 -0700
Re: The Light Clock Thought Experiment - Questions Tom Roberts <tjoberts137@sbcglobal.net> - 2023-05-14 11:06 -0500
Re: The Light Clock Thought Experiment - Questions Chad Stavropoulos <ssos@atoasoav.us> - 2023-05-14 20:08 +0000
Re: The Light Clock Thought Experiment - Questions whodat <whodaat@void.nowgre.com> - 2023-05-14 15:16 -0500
Re: The Light Clock Thought Experiment - Questions Chad Stavropoulos <ssos@atoasoav.us> - 2023-05-14 20:28 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-14 20:45 -0700
Re: The Light Clock Thought Experiment - Questions Tom Roberts <tjoberts137@sbcglobal.net> - 2023-05-15 02:15 -0500
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-15 04:15 -0700
Re: The Light Clock Thought Experiment - Questions Tom Roberts <tjoberts137@sbcglobal.net> - 2023-05-15 12:05 -0500
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-15 18:15 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-16 03:26 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-15 18:32 -0700
Re: The Light Clock Thought Experiment - Questions Paparios <mrios@ing.puc.cl> - 2023-05-20 08:34 -0700
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-20 12:13 -0700
Re: The Light Clock Thought Experiment - Questions nospam@de-ster.demon.nl (J. J. Lodder) - 2023-05-15 21:07 +0200
Re: The Light Clock Thought Experiment - Questions RichD <r_delaney2001@yahoo.com> - 2023-05-15 12:50 -0700
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-15 18:02 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-16 03:21 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-15 21:03 -0700
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-15 21:44 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-16 06:22 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-16 04:29 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-17 23:25 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-17 18:44 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-19 02:17 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-18 19:48 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-20 00:31 +0000
Re: The Light Clock Thought Experiment - Questions Tom Roberts <tjoberts137@sbcglobal.net> - 2023-05-18 22:32 -0500
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-20 00:45 +0000
Re: The Light Clock Thought Experiment - Questions Mikko <mikko.levanto@iki.fi> - 2023-05-20 12:56 +0300
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-21 03:39 +0000
Re: The Light Clock Thought Experiment - Questions Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2023-05-20 03:00 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-21 03:57 +0000
Re: The Light Clock Thought Experiment - Questions Paul Alsing <pnalsing@gmail.com> - 2023-05-18 21:55 -0700
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-18 22:37 -0700
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-18 23:10 -0700
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-18 23:15 -0700
Re: The Light Clock Thought Experiment - Questions Jane <Jane@home.com> - 2023-05-20 00:50 +0000
Re: The Light Clock Thought Experiment - Questions Errol Rooijakker <rjer@eioaljkk.ri> - 2023-05-20 09:30 +0000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-18 23:08 -0700
Re: The Light Clock Thought Experiment - Questions "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-05-18 11:33 -0700
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-12 19:42 -0700
Re: The Light Clock Thought Experiment - Questions nospam@de-ster.demon.nl (J. J. Lodder) - 2023-05-13 10:26 +0200
Re: The Light Clock Thought Experiment - Questions nospam@de-ster.demon.nl (J. J. Lodder) - 2023-05-10 10:04 +0200
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-10 01:42 -0700
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-10 05:00 -0700
Re: The Light Clock Thought Experiment - Questions nospam@de-ster.demon.nl (J. J. Lodder) - 2023-05-10 22:05 +0200
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-10 13:55 -0700
Re: The Light Clock Thought Experiment - Questions Tom Roberts <tjoberts137@sbcglobal.net> - 2023-05-11 00:04 -0500
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-11 03:44 -0700
Re: The Light Clock Thought Experiment - Questions nospam@de-ster.demon.nl (J. J. Lodder) - 2023-05-11 14:32 +0200
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-10 04:58 -0700
Re: The Light Clock Thought Experiment - Questions nospam@de-ster.demon.nl (J. J. Lodder) - 2023-05-10 22:05 +0200
Re: The Light Clock Thought Experiment - Questions carl eto <ccarleto4157990662@gmail.com> - 2023-05-10 14:12 -0700
Re: The Light Clock Thought Experiment - Questions Sylvia Else <sylvia@email.invalid> - 2023-05-11 15:37 +1000
Re: The Light Clock Thought Experiment - Questions "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> - 2023-05-11 03:45 -0700
Re: The Light Clock Thought Experiment - Questions Tom Roberts <tjoberts137@sbcglobal.net> - 2023-05-12 11:37 -0500
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-12 10:04 -0700
Re: The Light Clock Thought Experiment - Questions Python <python@invalid.org> - 2023-05-13 14:50 +0200
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-13 08:51 -0700
Re: The Light Clock Thought Experiment - Questions Python <python@invalid.org> - 2023-05-13 17:55 +0200
Re: The Light Clock Thought Experiment - Questions Volney <volney@invalid.invalid> - 2023-05-13 13:13 -0400
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-13 11:53 -0700
Re: The Light Clock Thought Experiment - Questions Chad Stavropoulos <ssos@atoasoav.us> - 2023-05-13 22:21 +0000
Re: The Light Clock Thought Experiment - Questions Maciej Wozniak <maluwozniak@gmail.com> - 2023-05-13 11:54 -0700
Re: The Light Clock Thought Experiment - Questions "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-05-13 12:11 -0700
Re: The Light Clock Thought Experiment - Questions "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-05-12 10:34 -0700
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-16 04:26 -0700 |
| Message-ID | <0ee99d88-1e1e-4828-8260-1dd790c56640n@googlegroups.com> |
| In reply to | #610912 |
On Tuesday, May 16, 2023 at 10:38:19 AM UTC+5, Jane wrote: > On Mon, 15 May 2023 20:42:35 -0700, gehan.am...@gmail.com wrote: > > > On Tuesday, May 16, 2023 at 7:38:59 AM UTC+5, Jane wrote: > >> On Sat, 13 May 2023 18:48:48 -0700, gehan.am...@gmail.com wrote: > >> > >> > On Sunday, May 14, 2023 at 6:18:53 AM UTC+5, Jane wrote: > >> > > >> > > >> >> > This is one way of looking at it. In effect the movement of the > >> >> > clock creates empty space between photons out of nothing. > >> >> The clock doesn't even move. It is simply observed in the moving > >> >> frame. > >> >> If it is plotted correctly all the beam elements remain vertically > >> >> aligned as they obviously must. The claim that the beam becomes > >> >> diagonal epitomizes the inferiority of the relavivist mentality. > >> >> > Creationex-nihilo, > >> >> > and that also asymmetrically. Only you and I understand this, and > >> >> > maybe Henry Wilson. > >> >> That name is prominent in my uncle's records. I will go through them > >> >> and see. > >> > > >> > Wilson had a nice graphic in BASIC called light exe or something. > >> > > >> >> > All we should do is cogently publish our peer reviewed books and > >> >> > be done with it. > >> >> I am writing a whole thesis on the history of the introduction of > >> >> SR. It contains some sensational discoveries...including a > >> >> completely new explanation of the cosmic redshift. > >> >> -- > >> >> -- lover of truth > >> > > >> > Maybe it us time to take out a piece of paper and do a 1915 type > >> > simulation of this. > >> > > >> > If you are writing anything please get it mercilessly peer reviewed > >> > at least for grammar. > >> > Your discoveries will have to be validated, agree? Or they are not. > >> No. I don't intend even trying to publish this as a formal scientific > >> paper. Nobody would touch it because what it reveals will destroy > >> thousands of reputations and millions of publications. It also has a > >> lot of sensational stuff that would be plagiarized by useless academics > >> like some who post here. It will be for sale in Ebook of PDF format, > >> copyrighted.. Some of it has already been checked for grammar but I > >> think I can handle that...being a science journalist for years. > >> -- > >> -- lover of truth > > > > There have been some papers published in peer- reviewed publications > > that do criticize aspects of Special Relativity. The widespread > > destruction you mentioned does not exist, so it is worth a try. > > > > https://www.researchgate.net/publication/ > 305807371_Criticism_of_the_Foundations_of_the_Relativity_Theory > I had a look....no good. > > > https://wiki.naturalphilosophy.org/index.php?title=Special_relativity > > This group has been trying to bring Einstein down for years but cannot. > The trouble is they don't realise how consistent SR is with its second > postulate. Yes, exactly, it is consistent with the second postulate. The Special Theory of Relativity will be proved wrong at a time and place of the scientific communities choosing, not before, and not before new discoveries come to light. I am saying that the explanations given to me do not make sense in some of their variations, and contain assumptions that are not clearly stated. >Also you have people like Tom Roberts who think they can > bamboozle people into submission with masses of 4D terminology and > associated jargon. One only needs 2D to analyse and understand SR. That is why avoid mathematics and stick to reason and basic algebra. I will not discuss what I do not understand. > > I would be willing to check the document and add my comments. I look > > forward to your publications, I am not afraid of looking at ideas I > > consider to be incorrect, after all I have been looking at Special > > Relativity for some time now. > I might let you...but like I said, it contains some sensational > discoveries and it suggest experiments to check them. > -- > -- lover of truth Good luck with your book.
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| From | Tom Roberts <tjoberts137@sbcglobal.net> |
|---|---|
| Date | 2023-05-14 11:06 -0500 |
| Message-ID | <X96dnTWmKt9xm_z5nZ2dnZfqlJxh4p2d@giganews.com> |
| In reply to | #610543 |
On 5/12/23 7:02 PM, Jane wrote:
> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote:
>> On 5/10/23 6:22 PM, Jane wrote:
>>> The proof is that if the positions of points in a vertical light
>>> beam are plotted against time in any horizontally moving frame,
>>> it becomes obvious that the only light beam present remains
>>> vertical in all such frames. There are NO 'diagonal light beams'
>>> moving at c.
>>
>> NONSENSE! Just making stuff up and pretending it is true is no
>> "proof" of anything except that you are ignorant and delusional.
>>
>> Use the usual (x,y,z,t) coordinates in inertial frame S, and
>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x
>> direction relative to the S; z and z' are vertical. Vacuum.
>>
>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0)
>> and reaches a point vertically a distance L above the origin in S
>> (x,y,z,t) = (0,0,L,L/c)
>>
>> [The speed of this pulse relative to S is clearly c.]
>>
>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second
>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g =
>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this
>> light pulse CLEARLY moved diagonally relative to S'.
>>
>> [Calculate the speed of this pulse relative to S' and one obtains
>> c, the same as its speed relative to S.]
>>
>> What if instead of the Lorentz transform we used the Galilean
>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and
>> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second
>> point in S' is not x'=0, so this light pulse CLEARLY moved
>> diagonally relative to S'. Here the only difference from the
>> Lorentz transform is the absence of g in the coordinates of the
>> second point.
>>
>> Your claim is false in Newtonian mechanics and in relativistic
>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing
>> whatsoever to do with "Einstein indoctrination" (whatever that
>> is), it is just basic physics.
>>
>> You really need to learn basic physics before attempting to write
>> about it.
>
> A vertical light beam when plotted in the frame of a horizontally
> moving observer remains the same vertical beam but simply moving
> sideways.
This is not true. The beam is moving diagonally in that frame (my S').
> Consecutively emitted elements remain vertically aligned.
Finally you have said something that is true. To see this, let's extend
my discussion quoted above.
In S, consider a light beam that leaves the spatial origin moving
vertically (along z) for all t>0. The light pulse considered above is
the start of the beam. At any time t=T, with T>0, the beam extends up to
z=cT, and all portions of the beam are located at
x=0, y=0, 0<z<cT, t=T.
In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to
z'=cT with all portions of the beam located at:
x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#]
That is, the leading edge of the beam clearly traces a diagonal line in
the x',z' plane, and the entire beam is vertically aligned (along z').
[#] This is just the Lorentz transform of the extent of
of the beam in S.
Similarly, every infinitesimal region of the beam is moving diagonally
in S'. In particular, at time t=T the infinitesimal region emitted at
t=T0 (0<T0<T) is located at
x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0)
So it is clear that the entire beam is moving diagonally in S'.
Summary: In S', at any given value of t' (>0), all portions of the beam
have the same value of x', which is moving in the -x' direction with
speed v; the beam's extent along z' is increasing, and the entire beam
is moving diagonally in the x',z' plane.
Tom Roberts
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| From | Chad Stavropoulos <ssos@atoasoav.us> |
|---|---|
| Date | 2023-05-14 20:08 +0000 |
| Message-ID | <u3rf34$2mr2k$2@dont-email.me> |
| In reply to | #610757 |
Tom Roberts wrote: > On 5/12/23 7:02 PM, Jane wrote: >> A vertical light beam when plotted in the frame of a horizontally >> moving observer remains the same vertical beam but simply moving >> sideways. > > This is not true. The beam is moving diagonally in that frame (my S'). > >> Consecutively emitted elements remain vertically aligned. > > Finally you have said something that is true. To see this, let's extend > my discussion quoted above. I come to think that if the *_future_exists_*, it must have been *_existent_in_the_past_*. You can't see the past, since we always are *_in_the_present_*. But the future if it exists, as it does, it must have *_been_existent_in_the_past_*.
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| From | whodat <whodaat@void.nowgre.com> |
|---|---|
| Date | 2023-05-14 15:16 -0500 |
| Message-ID | <kcctspFmc6aU1@mid.individual.net> |
| In reply to | #610771 |
On 5/14/2023 3:08 PM, Chad Stavropoulos wrote: > Tom Roberts wrote: >> On 5/12/23 7:02 PM, Jane wrote: >>> A vertical light beam when plotted in the frame of a horizontally >>> moving observer remains the same vertical beam but simply moving >>> sideways. >> >> This is not true. The beam is moving diagonally in that frame (my S'). >> >>> Consecutively emitted elements remain vertically aligned. >> >> Finally you have said something that is true. To see this, let's extend >> my discussion quoted above. > > I come to think that if the *_future_exists_*, it must have been > *_existent_in_the_past_*. You can't see the past, since we always are > *_in_the_present_*. But the future if it exists, as it does, it must have > *_been_existent_in_the_past_*. No wonder the Russians are so far behind...Their future is their past.
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| From | Chad Stavropoulos <ssos@atoasoav.us> |
|---|---|
| Date | 2023-05-14 20:28 +0000 |
| Message-ID | <u3rg8n$2mr2t$1@dont-email.me> |
| In reply to | #610773 |
whodat wrote: >>>> Consecutively emitted elements remain vertically aligned. >>> >>> Finally you have said something that is true. To see this, let's >>> extend my discussion quoted above. >> >> I come to think that if the *_future_exists_*, it must have been >> *_existent_in_the_past_*. You can't see the past, since we always are >> *_in_the_present_*. But the future if it exists, as it does, it must >> have *_been_existent_in_the_past_*. > > No wonder the Russians a̶r̶e̶ s̶o̶ f̶a̶r̶ b̶e̶h̶i̶n̶d̶...Their future is their past. your mother's liver, seen through her vagina.
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-14 20:45 -0700 |
| Message-ID | <c28fe715-07a0-4d4d-8340-43c8829e3857n@googlegroups.com> |
| In reply to | #610757 |
On Sunday, May 14, 2023 at 9:06:17 PM UTC+5, Tom Roberts wrote: > On 5/12/23 7:02 PM, Jane wrote: > > On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: > >> On 5/10/23 6:22 PM, Jane wrote: > >>> The proof is that if the positions of points in a vertical light > >>> beam are plotted against time in any horizontally moving frame, > >>> it becomes obvious that the only light beam present remains > >>> vertical in all such frames. There are NO 'diagonal light beams' > >>> moving at c. > >> > >> NONSENSE! Just making stuff up and pretending it is true is no > >> "proof" of anything except that you are ignorant and delusional. > >> > >> Use the usual (x,y,z,t) coordinates in inertial frame S, and > >> (x',y',z',t') in inertial frame S' moving with velocity v in the +x > >> direction relative to the S; z and z' are vertical. Vacuum. > >> > >> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) > >> and reaches a point vertically a distance L above the origin in S > >> (x,y,z,t) = (0,0,L,L/c) > >> > >> [The speed of this pulse relative to S is clearly c.] > >> > >> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second > >> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = > >> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this > >> light pulse CLEARLY moved diagonally relative to S'. > >> > >> [Calculate the speed of this pulse relative to S' and one obtains > >> c, the same as its speed relative to S.] > >> > >> What if instead of the Lorentz transform we used the Galilean > >> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and > >> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second > >> point in S' is not x'=0, so this light pulse CLEARLY moved > >> diagonally relative to S'. Here the only difference from the > >> Lorentz transform is the absence of g in the coordinates of the > >> second point. > >> > >> Your claim is false in Newtonian mechanics and in relativistic > >> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing > >> whatsoever to do with "Einstein indoctrination" (whatever that > >> is), it is just basic physics. > >> > >> You really need to learn basic physics before attempting to write > >> about it. > > > > A vertical light beam when plotted in the frame of a horizontally > > moving observer remains the same vertical beam but simply moving > > sideways. > This is not true. The beam is moving diagonally in that frame (my S'). > > Consecutively emitted elements remain vertically aligned. > Finally you have said something that is true. To see this, let's extend > my discussion quoted above. > > In S, consider a light beam that leaves the spatial origin moving > vertically (along z) for all t>0. The light pulse considered above is > the start of the beam. At any time t=T, with T>0, the beam extends up to > z=cT, and all portions of the beam are located at > x=0, y=0, 0<z<cT, t=T. > > In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > z'=cT with all portions of the beam located at: > x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > That is, the leading edge of the beam clearly traces a diagonal line in > the x',z' plane, and the entire beam is vertically aligned (along z'). > > [#] This is just the Lorentz transform of the extent of > of the beam in S. > > Similarly, every infinitesimal region of the beam is moving diagonally > in S'. In particular, at time t=T the infinitesimal region emitted at > t=T0 (0<T0<T) is located at > x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > So it is clear that the entire beam is moving diagonally in S'. > > Summary: In S', at any given value of t' (>0), all portions of the beam > have the same value of x', which is moving in the -x' direction with > speed v; the beam's extent along z' is increasing, and the entire beam > is moving diagonally in the x',z' plane. > > Tom Roberts Assume that the light clock has a light bulb at the bottom, emitting light in a spherical wavefront. In that case, light will reach equal distances from the light source in time t, since the speed of light is constant. This is in the rest frame. However, no measurement in the rest frame will record light travelling a greater distance than ct in any direction. If you want to assume that what happens in the light clock is that light goes up and down in the inertial frame of reference of the light clock, and you want to translate those movements back into your frame of rest, you are free to do so using any formula or numerical method in existence. Saying that the ray of light travels a greater distance in one direction than the in the same time violates the second postulate, and everyone should see that.
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| From | Tom Roberts <tjoberts137@sbcglobal.net> |
|---|---|
| Date | 2023-05-15 02:15 -0500 |
| Message-ID | <K9-dnazxn9OJQfz5nZ2dnZfqlJxh4p2d@giganews.com> |
| In reply to | #610816 |
On 5/14/23 10:45 PM, gehan.am...@gmail.com wrote: > On Sunday, May 14, 2023 at 9:06:17 PM UTC+5, Tom Roberts wrote: >> On 5/12/23 7:02 PM, Jane wrote: >>> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: >>>> On 5/10/23 6:22 PM, Jane wrote: >>>>> The proof is that if the positions of points in a vertical light >>>>> beam are plotted against time in any horizontally moving frame, >>>>> it becomes obvious that the only light beam present remains >>>>> vertical in all such frames. There are NO 'diagonal light beams' >>>>> moving at c. >>>> >>>> NONSENSE! Just making stuff up and pretending it is true is no >>>> "proof" of anything except that you are ignorant and delusional. >>>> >>>> Use the usual (x,y,z,t) coordinates in inertial frame S, and >>>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x >>>> direction relative to the S; z and z' are vertical. Vacuum. >>>> >>>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) >>>> and reaches a point vertically a distance L above the origin in S >>>> (x,y,z,t) = (0,0,L,L/c) >>>> >>>> [The speed of this pulse relative to S is clearly c.] >>>> >>>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second >>>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = >>>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this >>>> light pulse CLEARLY moved diagonally relative to S'. >>>> >>>> [Calculate the speed of this pulse relative to S' and one obtains >>>> c, the same as its speed relative to S.] >>>> >>>> What if instead of the Lorentz transform we used the Galilean >>>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and >>>> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second >>>> point in S' is not x'=0, so this light pulse CLEARLY moved >>>> diagonally relative to S'. Here the only difference from the >>>> Lorentz transform is the absence of g in the coordinates of the >>>> second point. >>>> >>>> Your claim is false in Newtonian mechanics and in relativistic >>>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing >>>> whatsoever to do with "Einstein indoctrination" (whatever that >>>> is), it is just basic physics. >>>> >>>> You really need to learn basic physics before attempting to write >>>> about it. >>> >>> A vertical light beam when plotted in the frame of a horizontally >>> moving observer remains the same vertical beam but simply moving >>> sideways. >> This is not true. The beam is moving diagonally in that frame (my S'). >>> Consecutively emitted elements remain vertically aligned. >> Finally you have said something that is true. To see this, let's extend >> my discussion quoted above. >> >> In S, consider a light beam that leaves the spatial origin moving >> vertically (along z) for all t>0. The light pulse considered above is >> the start of the beam. At any time t=T, with T>0, the beam extends up to >> z=cT, and all portions of the beam are located at >> x=0, y=0, 0<z<cT, t=T. >> >> In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to >> z'=cT with all portions of the beam located at: >> x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] >> That is, the leading edge of the beam clearly traces a diagonal line in >> the x',z' plane, and the entire beam is vertically aligned (along z'). >> >> [#] This is just the Lorentz transform of the extent of >> of the beam in S. >> >> Similarly, every infinitesimal region of the beam is moving diagonally >> in S'. In particular, at time t=T the infinitesimal region emitted at >> t=T0 (0<T0<T) is located at >> x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) >> So it is clear that the entire beam is moving diagonally in S'. >> >> Summary: In S', at any given value of t' (>0), all portions of the beam >> have the same value of x', which is moving in the -x' direction with >> speed v; the beam's extent along z' is increasing, and the entire beam >> is moving diagonally in the x',z' plane. >> >> Tom Roberts > > [... irrelevant diversion] In S at time t=T the leading edge of the light beam has moved from (x,y,z,t)=(0,0,0,0) to (x,y,z,t)=(0,0,cT,T); (distance traveled in S)/(time duration in S) = c. In S' at time t'=gT the same leading edge has moved from (x',y',z',t')=(0,0,0,0) to (x',y',z',t')=(-gvT,0,cT,gT); (distance traveled in S')/(time duration in S') = c. So in S' it traveled a longer distance in a longer time; the (vacuum) speed of light is the same value as in S: c. This does NOT violate Einstein's second postulate (as you wrongly fantasize), and everyone should see that. Tom Roberts
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-15 04:15 -0700 |
| Message-ID | <3942a7e2-48c6-4851-8ca0-787a2d882b88n@googlegroups.com> |
| In reply to | #610831 |
On Monday, May 15, 2023 at 12:15:46 PM UTC+5, Tom Roberts wrote: > On 5/14/23 10:45 PM, gehan.am...@gmail.com wrote: > > On Sunday, May 14, 2023 at 9:06:17 PM UTC+5, Tom Roberts wrote: > >> On 5/12/23 7:02 PM, Jane wrote: > >>> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: > >>>> On 5/10/23 6:22 PM, Jane wrote: > >>>>> The proof is that if the positions of points in a vertical light > >>>>> beam are plotted against time in any horizontally moving frame, > >>>>> it becomes obvious that the only light beam present remains > >>>>> vertical in all such frames. There are NO 'diagonal light beams' > >>>>> moving at c. > >>>> > >>>> NONSENSE! Just making stuff up and pretending it is true is no > >>>> "proof" of anything except that you are ignorant and delusional. > >>>> > >>>> Use the usual (x,y,z,t) coordinates in inertial frame S, and > >>>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x > >>>> direction relative to the S; z and z' are vertical. Vacuum. > >>>> > >>>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) > >>>> and reaches a point vertically a distance L above the origin in S > >>>> (x,y,z,t) = (0,0,L,L/c) > >>>> > >>>> [The speed of this pulse relative to S is clearly c.] > >>>> > >>>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second > >>>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = > >>>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this > >>>> light pulse CLEARLY moved diagonally relative to S'. > >>>> > >>>> [Calculate the speed of this pulse relative to S' and one obtains > >>>> c, the same as its speed relative to S.] > >>>> > >>>> What if instead of the Lorentz transform we used the Galilean > >>>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and > >>>> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second > >>>> point in S' is not x'=0, so this light pulse CLEARLY moved > >>>> diagonally relative to S'. Here the only difference from the > >>>> Lorentz transform is the absence of g in the coordinates of the > >>>> second point. > >>>> > >>>> Your claim is false in Newtonian mechanics and in relativistic > >>>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing > >>>> whatsoever to do with "Einstein indoctrination" (whatever that > >>>> is), it is just basic physics. > >>>> > >>>> You really need to learn basic physics before attempting to write > >>>> about it. > >>> > >>> A vertical light beam when plotted in the frame of a horizontally > >>> moving observer remains the same vertical beam but simply moving > >>> sideways. > >> This is not true. The beam is moving diagonally in that frame (my S'). > >>> Consecutively emitted elements remain vertically aligned. > >> Finally you have said something that is true. To see this, let's extend > >> my discussion quoted above. > >> > >> In S, consider a light beam that leaves the spatial origin moving > >> vertically (along z) for all t>0. The light pulse considered above is > >> the start of the beam. At any time t=T, with T>0, the beam extends up to > >> z=cT, and all portions of the beam are located at > >> x=0, y=0, 0<z<cT, t=T. > >> > >> In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > >> z'=cT with all portions of the beam located at: > >> x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > >> That is, the leading edge of the beam clearly traces a diagonal line in > >> the x',z' plane, and the entire beam is vertically aligned (along z'). > >> > >> [#] This is just the Lorentz transform of the extent of > >> of the beam in S. > >> > >> Similarly, every infinitesimal region of the beam is moving diagonally > >> in S'. In particular, at time t=T the infinitesimal region emitted at > >> t=T0 (0<T0<T) is located at > >> x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > >> So it is clear that the entire beam is moving diagonally in S'. > >> > >> Summary: In S', at any given value of t' (>0), all portions of the beam > >> have the same value of x', which is moving in the -x' direction with > >> speed v; the beam's extent along z' is increasing, and the entire beam > >> is moving diagonally in the x',z' plane. > >> > >> Tom Roberts > > > > [... irrelevant diversion] > > In S at time t=T the leading edge of the light beam has moved from > (x,y,z,t)=(0,0,0,0) to (x,y,z,t)=(0,0,cT,T); (distance traveled in > S)/(time duration in S) = c. An observer in S sees the edge of the leading edge of a stationary clock traveling L in time t. When the clock is moving, how far does S see the edge of the beam of light moving? In time t it has moved a distance, as measured in S, of L or a distance greater than L? > > In S' at time t'=gT the same leading edge has moved from > (x',y',z',t')=(0,0,0,0) to (x',y',z',t')=(-gvT,0,cT,gT); (distance > traveled in S')/(time duration in S') = c. So in S' it traveled a longer > distance in a longer time; the (vacuum) speed of light is the same > value as in S: c. > Fictions about what happens inside the moving clock prove nothing here. In S' it has traveled a longer distance relative to what? I thought in S' the light is simply going up and down. > This does NOT violate Einstein's second postulate (as you wrongly > fantasize), and everyone should see that. > > Tom Roberts
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| From | Tom Roberts <tjoberts137@sbcglobal.net> |
|---|---|
| Date | 2023-05-15 12:05 -0500 |
| Message-ID | <SeudnVoasaLp-__5nZ2dnZfqlJxh4p2d@giganews.com> |
| In reply to | #610839 |
On 5/15/23 6:15 AM, gehan.am...@gmail.com wrote: > On Monday, May 15, 2023 at 12:15:46 PM UTC+5, Tom Roberts wrote: >> On 5/14/23 10:45 PM, gehan.am...@gmail.com wrote: >>> On Sunday, May 14, 2023 at 9:06:17 PM UTC+5, Tom Roberts wrote: >>>> On 5/12/23 7:02 PM, Jane wrote: >>>>> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: >>>>>> On 5/10/23 6:22 PM, Jane wrote: >>>>>>> The proof is that if the positions of points in a vertical light >>>>>>> beam are plotted against time in any horizontally moving frame, >>>>>>> it becomes obvious that the only light beam present remains >>>>>>> vertical in all such frames. There are NO 'diagonal light beams' >>>>>>> moving at c. >>>>>> >>>>>> NONSENSE! Just making stuff up and pretending it is true is no >>>>>> "proof" of anything except that you are ignorant and delusional. >>>>>> >>>>>> Use the usual (x,y,z,t) coordinates in inertial frame S, and >>>>>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x >>>>>> direction relative to the S; z and z' are vertical. Vacuum. >>>>>> >>>>>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) >>>>>> and reaches a point vertically a distance L above the origin in S >>>>>> (x,y,z,t) = (0,0,L,L/c) >>>>>> >>>>>> [The speed of this pulse relative to S is clearly c.] >>>>>> >>>>>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second >>>>>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = >>>>>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this >>>>>> light pulse CLEARLY moved diagonally relative to S'. >>>>>> >>>>>> [Calculate the speed of this pulse relative to S' and one obtains >>>>>> c, the same as its speed relative to S.] >>>>>> >>>>>> What if instead of the Lorentz transform we used the Galilean >>>>>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and >>>>>> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second >>>>>> point in S' is not x'=0, so this light pulse CLEARLY moved >>>>>> diagonally relative to S'. Here the only difference from the >>>>>> Lorentz transform is the absence of g in the coordinates of the >>>>>> second point. >>>>>> >>>>>> Your claim is false in Newtonian mechanics and in relativistic >>>>>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing >>>>>> whatsoever to do with "Einstein indoctrination" (whatever that >>>>>> is), it is just basic physics. >>>>>> >>>>>> You really need to learn basic physics before attempting to write >>>>>> about it. >>>>> >>>>> A vertical light beam when plotted in the frame of a horizontally >>>>> moving observer remains the same vertical beam but simply moving >>>>> sideways. >>>> This is not true. The beam is moving diagonally in that frame (my S'). >>>>> Consecutively emitted elements remain vertically aligned. >>>> Finally you have said something that is true. To see this, let's extend >>>> my discussion quoted above. >>>> >>>> In S, consider a light beam that leaves the spatial origin moving >>>> vertically (along z) for all t>0. The light pulse considered above is >>>> the start of the beam. At any time t=T, with T>0, the beam extends up to >>>> z=cT, and all portions of the beam are located at >>>> x=0, y=0, 0<z<cT, t=T. >>>> >>>> In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to >>>> z'=cT with all portions of the beam located at: >>>> x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] >>>> That is, the leading edge of the beam clearly traces a diagonal line in >>>> the x',z' plane, and the entire beam is vertically aligned (along z'). >>>> >>>> [#] This is just the Lorentz transform of the extent of >>>> of the beam in S. >>>> >>>> Similarly, every infinitesimal region of the beam is moving diagonally >>>> in S'. In particular, at time t=T the infinitesimal region emitted at >>>> t=T0 (0<T0<T) is located at >>>> x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) >>>> So it is clear that the entire beam is moving diagonally in S'. >>>> >>>> Summary: In S', at any given value of t' (>0), all portions of the beam >>>> have the same value of x', which is moving in the -x' direction with >>>> speed v; the beam's extent along z' is increasing, and the entire beam >>>> is moving diagonally in the x',z' plane. >>>> >>>> Tom Roberts >>> >>> [... irrelevant diversion] >> >> In S at time t=T the leading edge of the light beam has moved from >> (x,y,z,t)=(0,0,0,0) to (x,y,z,t)=(0,0,cT,T); (distance traveled in >> S)/(time duration in S) = c. > > [...] Everything you wrote is garbled and appears to be unrelated to the scenario above (e.g. you ask about "clocks", but there are no clocks in this scenario; also there is no "up and down") -- you REALLY need to read more carefully. I described this carefully and precisely, and your questions show that you did not actually read and understand what I wrote. There's no point in continuing until YOU learn to read carefully and accurately, and reflect that in your questions and writing. Tom Roberts
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-15 18:15 -0700 |
| Message-ID | <40db639d-b58a-44dd-93cf-4c8270b0c3f6n@googlegroups.com> |
| In reply to | #610857 |
On Monday, May 15, 2023 at 10:07:16 PM UTC+5, Tom Roberts wrote: > On 5/15/23 6:15 AM, gehan.am...@gmail.com wrote: > > On Monday, May 15, 2023 at 12:15:46 PM UTC+5, Tom Roberts wrote: > >> On 5/14/23 10:45 PM, gehan.am...@gmail.com wrote: > >>> On Sunday, May 14, 2023 at 9:06:17 PM UTC+5, Tom Roberts wrote: > >>>> On 5/12/23 7:02 PM, Jane wrote: > >>>>> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: > >>>>>> On 5/10/23 6:22 PM, Jane wrote: > >>>>>>> The proof is that if the positions of points in a vertical light > >>>>>>> beam are plotted against time in any horizontally moving frame, > >>>>>>> it becomes obvious that the only light beam present remains > >>>>>>> vertical in all such frames. There are NO 'diagonal light beams' > >>>>>>> moving at c. > >>>>>> > >>>>>> NONSENSE! Just making stuff up and pretending it is true is no > >>>>>> "proof" of anything except that you are ignorant and delusional. > >>>>>> > >>>>>> Use the usual (x,y,z,t) coordinates in inertial frame S, and > >>>>>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x > >>>>>> direction relative to the S; z and z' are vertical. Vacuum. > >>>>>> > >>>>>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) > >>>>>> and reaches a point vertically a distance L above the origin in S > >>>>>> (x,y,z,t) = (0,0,L,L/c) > >>>>>> > >>>>>> [The speed of this pulse relative to S is clearly c.] > >>>>>> > >>>>>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second > >>>>>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = > >>>>>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this > >>>>>> light pulse CLEARLY moved diagonally relative to S'. > >>>>>> > >>>>>> [Calculate the speed of this pulse relative to S' and one obtains > >>>>>> c, the same as its speed relative to S.] > >>>>>> > >>>>>> What if instead of the Lorentz transform we used the Galilean > >>>>>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and > >>>>>> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second > >>>>>> point in S' is not x'=0, so this light pulse CLEARLY moved > >>>>>> diagonally relative to S'. Here the only difference from the > >>>>>> Lorentz transform is the absence of g in the coordinates of the > >>>>>> second point. > >>>>>> > >>>>>> Your claim is false in Newtonian mechanics and in relativistic > >>>>>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing > >>>>>> whatsoever to do with "Einstein indoctrination" (whatever that > >>>>>> is), it is just basic physics. > >>>>>> > >>>>>> You really need to learn basic physics before attempting to write > >>>>>> about it. > >>>>> > >>>>> A vertical light beam when plotted in the frame of a horizontally > >>>>> moving observer remains the same vertical beam but simply moving > >>>>> sideways. > >>>> This is not true. The beam is moving diagonally in that frame (my S'). > >>>>> Consecutively emitted elements remain vertically aligned. > >>>> Finally you have said something that is true. To see this, let's extend > >>>> my discussion quoted above. > >>>> > >>>> In S, consider a light beam that leaves the spatial origin moving > >>>> vertically (along z) for all t>0. The light pulse considered above is > >>>> the start of the beam. At any time t=T, with T>0, the beam extends up to > >>>> z=cT, and all portions of the beam are located at > >>>> x=0, y=0, 0<z<cT, t=T. > >>>> > >>>> In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > >>>> z'=cT with all portions of the beam located at: > >>>> x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > >>>> That is, the leading edge of the beam clearly traces a diagonal line in > >>>> the x',z' plane, and the entire beam is vertically aligned (along z'). > >>>> > >>>> [#] This is just the Lorentz transform of the extent of > >>>> of the beam in S. > >>>> > >>>> Similarly, every infinitesimal region of the beam is moving diagonally > >>>> in S'. In particular, at time t=T the infinitesimal region emitted at > >>>> t=T0 (0<T0<T) is located at > >>>> x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > >>>> So it is clear that the entire beam is moving diagonally in S'. > >>>> > >>>> Summary: In S', at any given value of t' (>0), all portions of the beam > >>>> have the same value of x', which is moving in the -x' direction with > >>>> speed v; the beam's extent along z' is increasing, and the entire beam > >>>> is moving diagonally in the x',z' plane. > >>>> > >>>> Tom Roberts > >>> > >>> [... irrelevant diversion] > >> > >> In S at time t=T the leading edge of the light beam has moved from > >> (x,y,z,t)=(0,0,0,0) to (x,y,z,t)=(0,0,cT,T); (distance traveled in > >> S)/(time duration in S) = c. > > > > [...] > > Everything you wrote is garbled and appears to be unrelated to the > scenario above (e.g. you ask about "clocks", but there are no clocks in > this scenario; also there is no "up and down") -- you REALLY need to > read more carefully. I described this carefully and precisely, and your > questions show that you did not actually read and understand what I > wrote. There's no point in continuing until YOU learn to read carefully > and accurately, and reflect that in your questions and writing. > > Tom Roberts Sorry, re-reading
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| From | Jane <Jane@home.com> |
|---|---|
| Date | 2023-05-16 03:26 +0000 |
| Message-ID | <175f81e1f5b2c2d2$181$2806839$fd385da@news.newsgroupdirect.com> |
| In reply to | #610886 |
On Mon, 15 May 2023 18:15:03 -0700, gehan.am...@gmail.com wrote: >> >> Everything you wrote is garbled and appears to be unrelated to the >> scenario above (e.g. you ask about "clocks", but there are no clocks in >> this scenario; also there is no "up and down") -- you REALLY need to >> read more carefully. I described this carefully and precisely, and your >> questions show that you did not actually read and understand what I >> wrote. There's no point in continuing until YOU learn to read carefully >> and accurately, and reflect that in your questions and writing. >> >> Tom Roberts > > Sorry, re-reading Make sure you read my latest esponse to Tommy before you post a reply. -- -- lover of truth
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-15 18:32 -0700 |
| Message-ID | <9054198b-dc32-4fe6-8971-f83cfdf54dd6n@googlegroups.com> |
| In reply to | #610839 |
On Monday, May 15, 2023 at 4:15:48 PM UTC+5, gehan.am...@gmail.com wrote: > On Monday, May 15, 2023 at 12:15:46 PM UTC+5, Tom Roberts wrote: > > On 5/14/23 10:45 PM, gehan.am...@gmail.com wrote: > > > On Sunday, May 14, 2023 at 9:06:17 PM UTC+5, Tom Roberts wrote: > > >> On 5/12/23 7:02 PM, Jane wrote: > > >>> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: > > >>>> On 5/10/23 6:22 PM, Jane wrote: > > >>>>> The proof is that if the positions of points in a vertical light > > >>>>> beam are plotted against time in any horizontally moving frame, > > >>>>> it becomes obvious that the only light beam present remains > > >>>>> vertical in all such frames. There are NO 'diagonal light beams' > > >>>>> moving at c. > > >>>> > > >>>> NONSENSE! Just making stuff up and pretending it is true is no > > >>>> "proof" of anything except that you are ignorant and delusional. > > >>>> > > >>>> Use the usual (x,y,z,t) coordinates in inertial frame S, and > > >>>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x > > >>>> direction relative to the S; z and z' are vertical. Vacuum. > > >>>> > > >>>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) > > >>>> and reaches a point vertically a distance L above the origin in S > > >>>> (x,y,z,t) = (0,0,L,L/c) > > >>>> > > >>>> [The speed of this pulse relative to S is clearly c.] > > >>>> > > >>>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second > > >>>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = > > >>>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this > > >>>> light pulse CLEARLY moved diagonally relative to S'. > > >>>> > > >>>> [Calculate the speed of this pulse relative to S' and one obtains > > >>>> c, the same as its speed relative to S.] > > >>>> > > >>>> What if instead of the Lorentz transform we used the Galilean > > >>>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and > > >>>> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second > > >>>> point in S' is not x'=0, so this light pulse CLEARLY moved > > >>>> diagonally relative to S'. Here the only difference from the > > >>>> Lorentz transform is the absence of g in the coordinates of the > > >>>> second point. > > >>>> > > >>>> Your claim is false in Newtonian mechanics and in relativistic > > >>>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing > > >>>> whatsoever to do with "Einstein indoctrination" (whatever that > > >>>> is), it is just basic physics. > > >>>> > > >>>> You really need to learn basic physics before attempting to write > > >>>> about it. > > >>> > > >>> A vertical light beam when plotted in the frame of a horizontally > > >>> moving observer remains the same vertical beam but simply moving > > >>> sideways. > > >> This is not true. The beam is moving diagonally in that frame (my S'). > > >>> Consecutively emitted elements remain vertically aligned. > > >> Finally you have said something that is true. To see this, let's extend > > >> my discussion quoted above. > > >> > > >> In S, consider a light beam that leaves the spatial origin moving > > >> vertically (along z) for all t>0. The light pulse considered above is > > >> the start of the beam. At any time t=T, with T>0, the beam extends up to > > >> z=cT, and all portions of the beam are located at > > >> x=0, y=0, 0<z<cT, t=T. > > >> > > >> In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > > >> z'=cT with all portions of the beam located at: > > >> x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > > >> That is, the leading edge of the beam clearly traces a diagonal line in > > >> the x',z' plane, and the entire beam is vertically aligned (along z'). > > >> > > >> [#] This is just the Lorentz transform of the extent of > > >> of the beam in S. > > >> > > >> Similarly, every infinitesimal region of the beam is moving diagonally > > >> in S'. In particular, at time t=T the infinitesimal region emitted at > > >> t=T0 (0<T0<T) is located at > > >> x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > > >> So it is clear that the entire beam is moving diagonally in S'. > > >> > > >> Summary: In S', at any given value of t' (>0), all portions of the beam > > >> have the same value of x', which is moving in the -x' direction with > > >> speed v; the beam's extent along z' is increasing, and the entire beam > > >> is moving diagonally in the x',z' plane. > > >> > > >> Tom Roberts > > > > > > [... irrelevant diversion] > > > > In S at time t=T the leading edge of the light beam has moved from > > (x,y,z,t)=(0,0,0,0) to (x,y,z,t)=(0,0,cT,T); (distance traveled in > > S)/(time duration in S) = c. > An observer in S sees the edge of the leading edge of a stationary clock traveling L in time t. > > When the clock is moving, how far does S see the edge of the beam of light moving? In time t > it has moved a distance, as measured in S, of L or a distance greater than L? > > > > In S' at time t'=gT the same leading edge has moved from > > (x',y',z',t')=(0,0,0,0) to (x',y',z',t')=(-gvT,0,cT,gT); (distance > > traveled in S')/(time duration in S') = c. So in S' it traveled a longer > > distance in a longer time; the (vacuum) speed of light is the same > > value as in S: c. Some questions: >>So in S' it traveled a longer > > distance in a longer time; In S' the leading edge of the light travels a distance 2L. Why should anything happening inside S' change? As viewed from the stationary frame, at time T, the light is seen to travel a distance ct from the light source. This cannot change. The entire experiment is framed in terms of what the stationary observer sees, but what does he see? Whatever he sees, he cannot see light move a greater distance than cT in a time T. > > > Fictions about what happens inside the moving clock prove nothing here. > In S' it has traveled a longer distance relative to what? I thought in S' the light is simply going up and down. > > This does NOT violate Einstein's second postulate (as you wrongly > > fantasize), and everyone should see that. > > > > Tom Roberts
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| From | Paparios <mrios@ing.puc.cl> |
|---|---|
| Date | 2023-05-20 08:34 -0700 |
| Message-ID | <741839c4-fd2f-4012-8914-60838434c06an@googlegroups.com> |
| In reply to | #610887 |
El lunes, 15 de mayo de 2023 a las 21:32:08 UTC-4, gehan.am...@gmail.com escribió: > On Monday, May 15, 2023 at 4:15:48 PM UTC+5, gehan.am...@gmail.com wrote: > Some questions: > >>So in S' it traveled a longer > > > distance in a longer time; > In S' the leading edge of the light travels a distance 2L. Why should anything happening inside S' change? As usual, you can't read in English. There are two observers (in frame S inside the train, and frame S' on the embankment). The observer in S just sees light signals moving up and down and calculates the light photons take dt=2L/c seconds in the path, where L is the distance between the mirrors A and B. The observer in the embankment frame S', on the other hand, sees the train moving at speed v (to the right). So the observer sees each photon moving up, but when the photon emitted from A reaches the top mirror B, mirror B is no longer at the same coordinate as mirror A was when the photon was emitted. Therefore, the light photons, as seen from S', follow a diagonal path (of length D up and down). Hence, the total time for the light pulse to trace its path is given now by dt'=2D/c , where D=sqrt((1/2(vdt')^2+L^2). From these two equations you get dt'=dt/sqr(1-v^2/c^2), implying dt' is longer than dt. Simple geometry and logic and the consideration that (according to the second principle) light signals move at c in both frames.
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| From | Maciej Wozniak <maluwozniak@gmail.com> |
|---|---|
| Date | 2023-05-20 12:13 -0700 |
| Message-ID | <3238d6d8-2a94-43f1-a1ca-0749d5259442n@googlegroups.com> |
| In reply to | #611231 |
On Saturday, 20 May 2023 at 17:34:50 UTC+2, Paparios wrote: > El lunes, 15 de mayo de 2023 a las 21:32:08 UTC-4, gehan.am...@gmail.com escribió: > > On Monday, May 15, 2023 at 4:15:48 PM UTC+5, gehan.am...@gmail.com wrote: > > > Some questions: > > >>So in S' it traveled a longer > > > > distance in a longer time; > > > In S' the leading edge of the light travels a distance 2L. Why should anything happening inside S' change? > > As usual, you can't read in English. There are two observers (in frame S inside the train, and frame S' on the embankment). The observer in S just sees light signals moving up and down and calculates the light photons take dt=2L/c seconds in the path, where L is the distance between the mirrors A and B. And in the meantime in the real world, forbidden by your bunch of idiots improper clocks keep measuring improper t'=t in improper seconds.
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| From | nospam@de-ster.demon.nl (J. J. Lodder) |
|---|---|
| Date | 2023-05-15 21:07 +0200 |
| Message-ID | <1qarlv3.ggw4lzuxa02xN%nospam@de-ster.demon.nl> |
| In reply to | #610816 |
gehan.am...@gmail.com <gehan.ameresekere@gmail.com> wrote: > On Sunday, May 14, 2023 at 9:06:17?PM UTC+5, Tom Roberts wrote: > > On 5/12/23 7:02 PM, Jane wrote: > > > On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: > > >> On 5/10/23 6:22 PM, Jane wrote: > > >>> The proof is that if the positions of points in a vertical light > > >>> beam are plotted against time in any horizontally moving frame, > > >>> it becomes obvious that the only light beam present remains > > >>> vertical in all such frames. There are NO 'diagonal light beams' > > >>> moving at c. > > >> > > >> NONSENSE! Just making stuff up and pretending it is true is no > > >> "proof" of anything except that you are ignorant and delusional. > > >> > > >> Use the usual (x,y,z,t) coordinates in inertial frame S, and > > >> (x',y',z',t') in inertial frame S' moving with velocity v in the +x > > >> direction relative to the S; z and z' are vertical. Vacuum. > > >> > > >> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) > > >> and reaches a point vertically a distance L above the origin in S > > >> (x,y,z,t) = (0,0,L,L/c) > > >> > > >> [The speed of this pulse relative to S is clearly c.] > > >> > > >> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second > > >> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = > > >> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this > > >> light pulse CLEARLY moved diagonally relative to S'. > > >> > > >> [Calculate the speed of this pulse relative to S' and one obtains > > >> c, the same as its speed relative to S.] > > >> > > >> What if instead of the Lorentz transform we used the Galilean > > >> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and > > >> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second > > >> point in S' is not x'=0, so this light pulse CLEARLY moved > > >> diagonally relative to S'. Here the only difference from the > > >> Lorentz transform is the absence of g in the coordinates of the > > >> second point. > > >> > > >> Your claim is false in Newtonian mechanics and in relativistic > > >> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing > > >> whatsoever to do with "Einstein indoctrination" (whatever that > > >> is), it is just basic physics. > > >> > > >> You really need to learn basic physics before attempting to write > > >> about it. > > > > > > A vertical light beam when plotted in the frame of a horizontally > > > moving observer remains the same vertical beam but simply moving > > > sideways. > > This is not true. The beam is moving diagonally in that frame (my S'). > > > Consecutively emitted elements remain vertically aligned. > > Finally you have said something that is true. To see this, let's extend > > my discussion quoted above. > > > > In S, consider a light beam that leaves the spatial origin moving > > vertically (along z) for all t>0. The light pulse considered above is > > the start of the beam. At any time t=T, with T>0, the beam extends up to > > z=cT, and all portions of the beam are located at > > x=0, y=0, 0<z<cT, t=T. > > > > In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > > z'=cT with all portions of the beam located at: > > x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > > That is, the leading edge of the beam clearly traces a diagonal line in > > the x',z' plane, and the entire beam is vertically aligned (along z'). > > > > [#] This is just the Lorentz transform of the extent of > > of the beam in S. > > > > Similarly, every infinitesimal region of the beam is moving diagonally > > in S'. In particular, at time t=T the infinitesimal region emitted at > > t=T0 (0<T0<T) is located at > > x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > > So it is clear that the entire beam is moving diagonally in S'. > > > > Summary: In S', at any given value of t' (>0), all portions of the beam > > have the same value of x', which is moving in the -x' direction with > > speed v; the beam's extent along z' is increasing, and the entire beam > > is moving diagonally in the x',z' plane. > > > > Tom Roberts > > Assume that the light clock has a light bulb at the bottom, emitting light > in a spherical wavefront. In that case, light will reach equal distances > from the light source in time t, since the speed of light is constant. > This is in the rest frame. However, no measurement in the rest frame > will record light travelling a greater distance than ct in any direction. Then you should take a spherical mirror, and you have a spherical light clock. (already discussed by Ehrenfest, shortly after 1905) Excercise: describe what happens when it moves. Will you still see all the light passing through the centre, periodically? Jan
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| From | RichD <r_delaney2001@yahoo.com> |
|---|---|
| Date | 2023-05-15 12:50 -0700 |
| Message-ID | <a2c6200d-28af-4197-b439-e6e260d8df35n@googlegroups.com> |
| In reply to | #610865 |
On May 15, J. J. Lodder wrote: > Then you should take a spherical mirror, > and you have a spherical light clock. > > Excercise: describe what happens when it moves. > Will you still see all the light passing through the centre, > periodically? https://www.amazon.com/OMIRO-Folding-Compact-Mirror-Magnification/dp/B0799JCCFF/ref=asc_df_B0799JCCFF/ She holds the mirror at eye level, at a few inches distance, with hinge vertical, bent 90*. She winks her right eye. What does she see? What if this mirror is used in Einstein's mirror light clock? -- Rich
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-15 18:02 -0700 |
| Message-ID | <ad0d19ce-78a1-46aa-8da1-0e49337407d6n@googlegroups.com> |
| In reply to | #610865 |
On Tuesday, May 16, 2023 at 12:09:22 AM UTC+5, J. J. Lodder wrote: > gehan.am...@gmail.com <gehan.am...@gmail.com> wrote: > > On Sunday, May 14, 2023 at 9:06:17?PM UTC+5, Tom Roberts wrote: > > > On 5/12/23 7:02 PM, Jane wrote: > > > > On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: > > > >> On 5/10/23 6:22 PM, Jane wrote: > > > >>> The proof is that if the positions of points in a vertical light > > > >>> beam are plotted against time in any horizontally moving frame, > > > >>> it becomes obvious that the only light beam present remains > > > >>> vertical in all such frames. There are NO 'diagonal light beams' > > > >>> moving at c. > > > >> > > > >> NONSENSE! Just making stuff up and pretending it is true is no > > > >> "proof" of anything except that you are ignorant and delusional. > > > >> > > > >> Use the usual (x,y,z,t) coordinates in inertial frame S, and > > > >> (x',y',z',t') in inertial frame S' moving with velocity v in the +x > > > >> direction relative to the S; z and z' are vertical. Vacuum. > > > >> > > > >> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) > > > >> and reaches a point vertically a distance L above the origin in S > > > >> (x,y,z,t) = (0,0,L,L/c) > > > >> > > > >> [The speed of this pulse relative to S is clearly c.] > > > >> > > > >> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second > > > >> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = > > > >> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this > > > >> light pulse CLEARLY moved diagonally relative to S'. > > > >> > > > >> [Calculate the speed of this pulse relative to S' and one obtains > > > >> c, the same as its speed relative to S.] > > > >> > > > >> What if instead of the Lorentz transform we used the Galilean > > > >> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and > > > >> the second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second > > > >> point in S' is not x'=0, so this light pulse CLEARLY moved > > > >> diagonally relative to S'. Here the only difference from the > > > >> Lorentz transform is the absence of g in the coordinates of the > > > >> second point. > > > >> > > > >> Your claim is false in Newtonian mechanics and in relativistic > > > >> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing > > > >> whatsoever to do with "Einstein indoctrination" (whatever that > > > >> is), it is just basic physics. > > > >> > > > >> You really need to learn basic physics before attempting to write > > > >> about it. > > > > > > > > A vertical light beam when plotted in the frame of a horizontally > > > > moving observer remains the same vertical beam but simply moving > > > > sideways. > > > This is not true. The beam is moving diagonally in that frame (my S'). > > > > Consecutively emitted elements remain vertically aligned. > > > Finally you have said something that is true. To see this, let's extend > > > my discussion quoted above. > > > > > > In S, consider a light beam that leaves the spatial origin moving > > > vertically (along z) for all t>0. The light pulse considered above is > > > the start of the beam. At any time t=T, with T>0, the beam extends up to > > > z=cT, and all portions of the beam are located at > > > x=0, y=0, 0<z<cT, t=T. > > > > > > In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > > > z'=cT with all portions of the beam located at: > > > x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > > > That is, the leading edge of the beam clearly traces a diagonal line in > > > the x',z' plane, and the entire beam is vertically aligned (along z'). > > > > > > [#] This is just the Lorentz transform of the extent of > > > of the beam in S. > > > > > > Similarly, every infinitesimal region of the beam is moving diagonally > > > in S'. In particular, at time t=T the infinitesimal region emitted at > > > t=T0 (0<T0<T) is located at > > > x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > > > So it is clear that the entire beam is moving diagonally in S'. > > > > > > Summary: In S', at any given value of t' (>0), all portions of the beam > > > have the same value of x', which is moving in the -x' direction with > > > speed v; the beam's extent along z' is increasing, and the entire beam > > > is moving diagonally in the x',z' plane. > > > > > > Tom Roberts > > > > Assume that the light clock has a light bulb at the bottom, emitting light > > in a spherical wavefront. In that case, light will reach equal distances > > from the light source in time t, since the speed of light is constant. > > This is in the rest frame. However, no measurement in the rest frame > > will record light travelling a greater distance than ct in any direction. > Then you should take a spherical mirror, > and you have a spherical light clock. > (already discussed by Ehrenfest, shortly after 1905) > > Excercise: describe what happens when it moves. > Will you still see all the light passing through the centre, > periodically? > > Jan Is the answer to that question to take a spherical mirror? For any non moving light source, light will travel equal distances in a specific time. If the light source such as a light bulb is moving, what will the stationary observer see? Will he see light travel greater distances in one direction in a specific time. I would be happy to see an answer.
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| From | Jane <Jane@home.com> |
|---|---|
| Date | 2023-05-16 03:21 +0000 |
| Message-ID | <175f819dee089fb7$180$2806839$fd385da@news.newsgroupdirect.com> |
| In reply to | #610757 |
On Sun, 14 May 2023 11:06:03 -0500, Tom Roberts wrote: > On 5/12/23 7:02 PM, Jane wrote: >> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: >>> On 5/10/23 6:22 PM, Jane wrote: >>>> The proof is that if the positions of points in a vertical light beam >>>> are plotted against time in any horizontally moving frame, it becomes >>>> obvious that the only light beam present remains vertical in all such >>>> frames. There are NO 'diagonal light beams' >>>> moving at c. >>> >>> NONSENSE! Just making stuff up and pretending it is true is no "proof" >>> of anything except that you are ignorant and delusional. >>> >>> Use the usual (x,y,z,t) coordinates in inertial frame S, and >>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x >>> direction relative to the S; z and z' are vertical. Vacuum. >>> >>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) >>> and reaches a point vertically a distance L above the origin in S >>> (x,y,z,t) = (0,0,L,L/c) >>> >>> [The speed of this pulse relative to S is clearly c.] >>> >>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second >>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = >>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this light >>> pulse CLEARLY moved diagonally relative to S'. >>> >>> [Calculate the speed of this pulse relative to S' and one obtains c, >>> the same as its speed relative to S.] >>> >>> What if instead of the Lorentz transform we used the Galilean >>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and the >>> second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second point in >>> S' is not x'=0, so this light pulse CLEARLY moved diagonally relative >>> to S'. Here the only difference from the Lorentz transform is the >>> absence of g in the coordinates of the second point. >>> >>> Your claim is false in Newtonian mechanics and in relativistic >>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing whatsoever >>> to do with "Einstein indoctrination" (whatever that is), it is just >>> basic physics. >>> >>> You really need to learn basic physics before attempting to write >>> about it. >> >> A vertical light beam when plotted in the frame of a horizontally >> moving observer remains the same vertical beam but simply moving >> sideways. > > This is not true. The beam is moving diagonally in that frame (my S'). I said the beam remains vertical in the moving frame. >> Consecutively emitted elements remain vertically aligned. > > Finally you have said something that is true. To see this, let's extend > my discussion quoted above. Finally you have understood some basic physics. > In S, consider a light beam that leaves the spatial origin moving > vertically (along z) for all t>0. The light pulse considered above is > the start of the beam. At any time t=T, with T>0, the beam extends up to > z=cT, and all portions of the beam are located at > x=0, y=0, 0<z<cT, t=T. > > In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > z'=cT with all portions of the beam located at: > x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > That is, the leading edge of the beam clearly traces a diagonal line in > the x',z' plane, and the entire beam is vertically aligned (along z'). > > [#] This is just the Lorentz transform of the extent of of the beam > in S. That is the error. > Similarly, every infinitesimal region of the beam is moving diagonally > in S'. In particular, at time t=T the infinitesimal region emitted at > t=T0 (0<T0<T) is located at > x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > So it is clear that the entire beam is moving diagonally in S'. Silly Tom. Every infinitesimal element of the beam is moving diagonally. Clearly, the paths of those elements represent just the loci of infinitesimal points....an infinite number of unique imaginary lines on a graph. ...and you are so naive you claim they are light... and therefore must move at c!! How blatantly stupid! Those lines are obviously 'nothing' and the infinitesimal elements are not light. They move at sqrt(c^2+v^2) in the moving frame and the beam takes exactly the same time to go up and down in all frames. > Summary: In S', at any given value of t' (>0), all portions of the beam > have the same value of x', which is moving in the -x' direction with > speed v; the beam's extent along z' is increasing, and the entire beam > is moving diagonally in the x',z' plane. Tom, to emphasize the error in your claims, let the moving observer shine a real laser beam along one of the diagonal paths. Do you see that it is different from the main beam? I hope so....Its 'axis' is not vertical and it clearly does not get to the top in the same time as the diagonally moving points. Obviously its speed is not the same as that of the infinitesimal points. Note, if an Ether really existed, the beam of a light clock would have to be angled diagonally at arctan(v/c) in order to return to the source. It would then genuinely take longer than if v = 0. The Lorentz factor would then apply. However the fact that TWLS is always very constant proves that no ether exists...and therefore LET and SR are both nothing but SciFi. > Tom Roberts -- -- lover of truth
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-15 21:03 -0700 |
| Message-ID | <25a832a9-374f-402f-9504-e356966cd3e6n@googlegroups.com> |
| In reply to | #610897 |
On Tuesday, May 16, 2023 at 8:21:44 AM UTC+5, Jane wrote:
> On Sun, 14 May 2023 11:06:03 -0500, Tom Roberts wrote:
>
> > On 5/12/23 7:02 PM, Jane wrote:
> >> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote:
> >>> On 5/10/23 6:22 PM, Jane wrote:
> >>>> The proof is that if the positions of points in a vertical light beam
> >>>> are plotted against time in any horizontally moving frame, it becomes
> >>>> obvious that the only light beam present remains vertical in all such
> >>>> frames. There are NO 'diagonal light beams'
> >>>> moving at c.
> >>>
> >>> NONSENSE! Just making stuff up and pretending it is true is no "proof"
> >>> of anything except that you are ignorant and delusional.
> >>>
> >>> Use the usual (x,y,z,t) coordinates in inertial frame S, and
> >>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x
> >>> direction relative to the S; z and z' are vertical. Vacuum.
> >>>
> >>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0)
> >>> and reaches a point vertically a distance L above the origin in S
> >>> (x,y,z,t) = (0,0,L,L/c)
> >>>
> >>> [The speed of this pulse relative to S is clearly c.]
> >>>
> >>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second
> >>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g =
> >>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this light
> >>> pulse CLEARLY moved diagonally relative to S'.
> >>>
> >>> [Calculate the speed of this pulse relative to S' and one obtains c,
> >>> the same as its speed relative to S.]
> >>>
> >>> What if instead of the Lorentz transform we used the Galilean
> >>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and the
> >>> second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second point in
> >>> S' is not x'=0, so this light pulse CLEARLY moved diagonally relative
> >>> to S'. Here the only difference from the Lorentz transform is the
> >>> absence of g in the coordinates of the second point.
> >>>
> >>> Your claim is false in Newtonian mechanics and in relativistic
> >>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing whatsoever
> >>> to do with "Einstein indoctrination" (whatever that is), it is just
> >>> basic physics.
> >>>
> >>> You really need to learn basic physics before attempting to write
> >>> about it.
> >>
> >> A vertical light beam when plotted in the frame of a horizontally
> >> moving observer remains the same vertical beam but simply moving
> >> sideways.
> >
> > This is not true. The beam is moving diagonally in that frame (my S').
> I said the beam remains vertical in the moving frame.
> >> Consecutively emitted elements remain vertically aligned.
> >
> > Finally you have said something that is true. To see this, let's extend
> > my discussion quoted above.
> Finally you have understood some basic physics.
If you consider the light clock to consist of light pulses (not sure how to handle the interactions with the mirror and related delays)
but this is an useful idea.
> > In S, consider a light beam that leaves the spatial origin moving
> > vertically (along z) for all t>0. The light pulse considered above is
> > the start of the beam. At any time t=T, with T>0, the beam extends up to
> > z=cT, and all portions of the beam are located at
> > x=0, y=0, 0<z<cT, t=T.
> >
> > In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to
> > z'=cT with all portions of the beam located at:
> > x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#]
> > That is, the leading edge of the beam clearly traces a diagonal line in
> > the x',z' plane, and the entire beam is vertically aligned (along z').
> >
> > [#] This is just the Lorentz transform of the extent of of the beam
> > in S.
> That is the error.
What is this 'trace?' I am a little more free to ask questions from Jane _E since he seems to see things from the same point of view. Questions asked from other persons, many times, in my view, result in a cloud of mathematics that do no answer the question direclty.
> > Similarly, every infinitesimal region of the beam is moving diagonally
> > in S'. In particular, at time t=T the infinitesimal region emitted at
> > t=T0 (0<T0<T) is located at
> > x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0)
> > So it is clear that the entire beam is moving diagonally in S'.
If the beam is made up of photons, and not continuous, or light pulses which definitely consist of groups of photons,
then it is clear that the projected movement of the apparatus results in increased spacing between light pulses, in that
arrangement, which may not represent reality.
> Silly Tom. Every infinitesimal element of the beam is moving diagonally.
> Clearly, the paths of those elements represent just the loci of
> infinitesimal points....an infinite number of unique imaginary lines on a
> graph. ...and you are so naive you claim they are light... and therefore
> must move at c!!
A trace is not light. I find it useful to think of the light clock filled with smoke. The stationary observer will see the light reflect off the smoke, and ignoring viewing delays, they will see a moving set of pulses, in case of the light pulse clock, a series of dotted lighted lines or points. With reference to the light clock, these lines are not moving at all even as seen by a stationary observer. This is all conjecture, though.
__________
|
|
|__________
>How blatantly stupid! Those lines are obviously 'nothing'
> and the infinitesimal elements are not light. They move at sqrt(c^2+v^2)
> in the moving frame and the beam takes exactly the same time to go up and
> down in all frames.
> > Summary: In S', at any given value of t' (>0), all portions of the beam
> > have the same value of x', which is moving in the -x' direction with
> > speed v; the beam's extent along z' is increasing, and the entire beam
> > is moving diagonally in the x',z' plane.
> Tom, to emphasize the error in your claims, let the moving observer shine
> a real laser beam along one of the diagonal paths. Do you see that it is
> different from the main beam? I hope so....Its 'axis' is not vertical and
> it clearly does not get to the top in the same time as the diagonally
> moving points. Obviously its speed is not the same as that of the
> infinitesimal points.
>
This is a good point. In the stationary frame, shine a laser light so that it coincides exactly with the angle of the 'trace' of the 'light'.
Will this light reach the 'top' of the clock before the light inside the light clock?
How is the light clock different from a long corridor with mirrors at the top and the bottom, with a laser light shining at the same angle as the 'trace' of the 'diagonal light'?
_____________________________________________________________________________________________________________________
/
/
/_____________________________________________________________________________________________________________________
>
>
>
> Note, if an Ether really existed, the beam of a light clock would have to
> be angled diagonally at arctan(v/c) in order to return to the source. It
> would then genuinely take longer than if v = 0. The Lorentz factor would
> then apply.
If the Aether existed in the frame of the stationary observer, right?
>
> However the fact that TWLS is always very constant proves that no ether
> exists...and therefore LET and SR are both nothing but SciFi.
>
>
> > Tom Roberts
> --
> -- lover of truth
There is a way to use the light clock to illustrate time dilation, and I will come up with it later.
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| From | "gehan.am...@gmail.com" <gehan.ameresekere@gmail.com> |
|---|---|
| Date | 2023-05-15 21:44 -0700 |
| Message-ID | <ed12f933-cf4b-4a19-afd4-f610d9918785n@googlegroups.com> |
| In reply to | #610904 |
On Tuesday, May 16, 2023 at 9:03:27 AM UTC+5, gehan.am...@gmail.com wrote: > On Tuesday, May 16, 2023 at 8:21:44 AM UTC+5, Jane wrote: > > On Sun, 14 May 2023 11:06:03 -0500, Tom Roberts wrote: > > > > > On 5/12/23 7:02 PM, Jane wrote: > > >> On Fri, 12 May 2023 15:45:59 -0500, Tom Roberts wrote: > > >>> On 5/10/23 6:22 PM, Jane wrote: > > >>>> The proof is that if the positions of points in a vertical light beam > > >>>> are plotted against time in any horizontally moving frame, it becomes > > >>>> obvious that the only light beam present remains vertical in all such > > >>>> frames. There are NO 'diagonal light beams' > > >>>> moving at c. > > >>> > > >>> NONSENSE! Just making stuff up and pretending it is true is no "proof" > > >>> of anything except that you are ignorant and delusional. > > >>> > > >>> Use the usual (x,y,z,t) coordinates in inertial frame S, and > > >>> (x',y',z',t') in inertial frame S' moving with velocity v in the +x > > >>> direction relative to the S; z and z' are vertical. Vacuum. > > >>> > > >>> Consider a light pulse that leaves the origin (x,y,z,t) = (0,0,0,0) > > >>> and reaches a point vertically a distance L above the origin in S > > >>> (x,y,z,t) = (0,0,L,L/c) > > >>> > > >>> [The speed of this pulse relative to S is clearly c.] > > >>> > > >>> In S', the first point is (x',y',z',t') = (0,0,0,0) and the second > > >>> point is (x',y',z',t') = (g(0-v*L/c),0,L,g(L/c-0)) here g = > > >>> 1/sqrt(1-v^2/c^2) The second point in S' is not x'=0, so this light > > >>> pulse CLEARLY moved diagonally relative to S'. > > >>> > > >>> [Calculate the speed of this pulse relative to S' and one obtains c, > > >>> the same as its speed relative to S.] > > >>> > > >>> What if instead of the Lorentz transform we used the Galilean > > >>> transform? In S' the first point is (x',y',z',t') = (0,0,0,0) and the > > >>> second point is (x',y',z',t') = (0-v*L/c,0,L,L/c) The second point in > > >>> S' is not x'=0, so this light pulse CLEARLY moved diagonally relative > > >>> to S'. Here the only difference from the Lorentz transform is the > > >>> absence of g in the coordinates of the second point. > > >>> > > >>> Your claim is false in Newtonian mechanics and in relativistic > > >>> mechanics, AND THIS OUGHT TO BE OBVIOUS. This has nothing whatsoever > > >>> to do with "Einstein indoctrination" (whatever that is), it is just > > >>> basic physics. > > >>> > > >>> You really need to learn basic physics before attempting to write > > >>> about it. > > >> > > >> A vertical light beam when plotted in the frame of a horizontally > > >> moving observer remains the same vertical beam but simply moving > > >> sideways. > > > > > > This is not true. The beam is moving diagonally in that frame (my S'). > > I said the beam remains vertical in the moving frame. > > >> Consecutively emitted elements remain vertically aligned. > > > > > > Finally you have said something that is true. To see this, let's extend > > > my discussion quoted above. > > Finally you have understood some basic physics. > If you consider the light clock to consist of light pulses (not sure how to handle the interactions with the mirror and related delays) > but this is an useful idea. > > > In S, consider a light beam that leaves the spatial origin moving > > > vertically (along z) for all t>0. The light pulse considered above is > > > the start of the beam. At any time t=T, with T>0, the beam extends up to > > > z=cT, and all portions of the beam are located at > > > x=0, y=0, 0<z<cT, t=T. > > > > > > In S', at any time t'=gT, with T>0, the beam extends from z'=0 up to > > > z'=cT with all portions of the beam located at: > > > x'=g(0-v*T), y'=0, 0<z'<cT, t'=gT [#] > > > That is, the leading edge of the beam clearly traces a diagonal line in > > > the x',z' plane, and the entire beam is vertically aligned (along z'). > > > > > > [#] This is just the Lorentz transform of the extent of of the beam > > > in S. > > That is the error. > What is this 'trace?' I am a little more free to ask questions from Jane _E since he seems to see things from the same point of view. Questions asked from other persons, many times, in my view, result in a cloud of mathematics that do no answer the question direclty. > > > Similarly, every infinitesimal region of the beam is moving diagonally > > > in S'. In particular, at time t=T the infinitesimal region emitted at > > > t=T0 (0<T0<T) is located at > > > x'=-gvT, y'=0, z'=c(T-T0), t'=gT (valid only for T>T0) > > > So it is clear that the entire beam is moving diagonally in S'. > If the beam is made up of photons, and not continuous, or light pulses which definitely consist of groups of photons, > then it is clear that the projected movement of the apparatus results in increased spacing between light pulses, in that > arrangement, which may not represent reality. > > Silly Tom. Every infinitesimal element of the beam is moving diagonally. > > Clearly, the paths of those elements represent just the loci of > > infinitesimal points....an infinite number of unique imaginary lines on a > > graph. ...and you are so naive you claim they are light... and therefore > > must move at c!! > A trace is not light. I find it useful to think of the light clock filled with smoke. The stationary observer will see the light reflect off the smoke, and ignoring viewing delays, they will see a moving set of pulses, in case of the light pulse clock, a series of dotted lighted lines or points. With reference to the light clock, these lines are not moving at all even as seen by a stationary observer. This is all conjecture, though. > > __________ > | > | > |__________ > >How blatantly stupid! Those lines are obviously 'nothing' > > and the infinitesimal elements are not light. They move at sqrt(c^2+v^2) > > in the moving frame and the beam takes exactly the same time to go up and > > down in all frames. > > > Summary: In S', at any given value of t' (>0), all portions of the beam > > > have the same value of x', which is moving in the -x' direction with > > > speed v; the beam's extent along z' is increasing, and the entire beam > > > is moving diagonally in the x',z' plane. > > Tom, to emphasize the error in your claims, let the moving observer shine > > a real laser beam along one of the diagonal paths. Do you see that it is > > different from the main beam? I hope so....Its 'axis' is not vertical and > > it clearly does not get to the top in the same time as the diagonally > > moving points. Obviously its speed is not the same as that of the > > infinitesimal points. > > > This is a good point. In the stationary frame, shine a laser light so that it coincides exactly with the angle of the 'trace' of the 'light'. > Will this light reach the 'top' of the clock before the light inside the light clock? > > How is the light clock different from a long corridor with mirrors at the top and the bottom, with a laser light shining at the same angle as the 'trace' of the 'diagonal light'? > > _____________________________________________________________________________________________________________________ > / > / > /_____________________________________________________________________________________________________________________ > > > > > > > > Note, if an Ether really existed, the beam of a light clock would have to > > be angled diagonally at arctan(v/c) in order to return to the source. It > > would then genuinely take longer than if v = 0. The Lorentz factor would > > then apply. > If the Aether existed in the frame of the stationary observer, right? > > > > However the fact that TWLS is always very constant proves that no ether > > exists...and therefore LET and SR are both nothing but SciFi. > > > > > > > Tom Roberts > > -- > > -- lover of truth > There is a way to use the light clock to illustrate time dilation, and I will come up with it later. The light clock illustration can be used to describe the concept of time dilation. Refer to the Wikipedia article on time dilation and the light clock described there: https://en.wikipedia.org/wiki/Time_dilation The light clock illustrated consists of a light pulse that travels up and down in the rest frame of the clock. The time taken for the light to complete one cycle is 2L/c. If the light clock is moving left to right relative to a stationary observer, if the the classical addition of velocities is used we get a speed for the light pulse relative to the stationary frame as being sqrt(v^2 + c^2) where v is the speed of the light clock. This value is of course greater than c. The calculated speed of the light pulse in the moving clock frame projected onto the stationary frame has to be reduced to c using a suitable correction factor.
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