Properties of a preferred frame, an inertial frame in SR and
in [Relativity]
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From: kenseto on 18 Jul 2010 18:33 On Jul 18, 11:05 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > kenseto says... > > >When you compare two clocks in relative motion the following > >possibilities exist: > >1. A is running faster than B then B is running slower than A. > >3. B is running faster than A then A is running slower than B. > > >At no time the bogus concept of mutual time dilation that: from A's > >point of view B is running slow and from B's point of view A is > >running slow. > > Strictly speaking, it's not that one clock runs slow relative to > another clock; one clock runs slow relative to a *coordinate* system. > A coordinate system is not a single clock, it is a system of *synchronized* > clocks. > > Let's go through an example: Suppose in one frame we have two synchronized > clocks at rest, C1 and C2, lined up left-to-right, with C1 to the left of C2. > In another frame, we also have two synchronized clocks at rest, D1 and D2, > that are also lined up left-to-right. Clocks D1 and D2 are traveling at 0..866c > relative to clocks C1 and C2, and their velocity is left-to-right. They > are traveling so that D1 & D2 pass very close to C1 and C2. > > So here are some events involving these two pairs of clocks: > > e1: D1 passes C1, and they both show time 12:00 OK....this is an assumption. > e2: D2 passes C2, and D2 shows time 11:50, while C2 shows time 12:10 No....at this event D2 must agree with what D1 read: 12:00. Why? Because D1 and D2 are in the same frame and they are synchronized. Also C2 must agree with what C1 read which is 12:00. > e3: D1 passes C2, and D1 shows time 12:10, while C2 shows time 12:20 At this event D1 has moved a distance of C1 to C2 at a speed of 0.866c....the C1/C2 observer said: his C1/C2 clock has elapsed 20 minutes when D1 passes C2. He uses the time dilation equation to determine that D1/D2 must have elapsed 20minute/gamma=20/2=10minutes. The rest of your arguement is based on some illogical assumptions. Ken Seto > > If we assume that C1 and C2 are synchronized, then we explain the > events this way: > (1) At time 12:00, D1 shows time 12:00, while D2 shows time 11:45; > D1 & D2 are not synchronized. > (2) At time 12:10, clock D2 has advanced 5 minutes, to 11:50, while > clocks C1 and C2 have advanced 10 minutes, to 12:10. So > it is running at 1/2 the speed of C1&C2. At this time, D1 shows > time 12:05. > (3) At time 12:20, clock D2 has advanced another 5 minutes, to 12:10. > Clock C2 has advanced another 10 minutes, to 12:20. > > So if we assume that C1 and C2 are synchronized, we conclude that > D1 and D2 are running slow, and that they are not synchronized > (D1 is 15 minutes ahead of D2). > > On the other hand, if we assume that D1 and D2 are synchronized, > we explain the events this way: > (1) At time 11:50, clocks D1 and D2 show time 11:50. Clock C2 > shows time 12:10, while clock C1 shows time 11:55. > (2) At time 12:00, clocks D1 and D2 have advanced 10 minutes, > to 12:00, while clock C1 has only advanced 5 minutes, to 12:00, > and C2 has also advanced 5 minutes, to 12:15. > (3) At time 12:10 clocks D1 and D2 have advanced another 10 > minutes, to 12:10. Clock C1 has advanced 5 minutes, to 12:05. > Clock C2 has advanced 5 minutes, to 12:20. > > So if we assume D1 and D2 are synchronized, we conclude that > C1 and C2 are running slow, and that they are not synchronized > (C2 is 15 minutes ahead of C1). > > So mutual time dilation is *not* a comparison of two clocks alone; > it is a comparison of 4 clocks, together with assumptions about > synchronization. Just comparing two clocks, you cannot possibly > say whether one clock is running slower than the other. > > -- > Daryl McCullough > Ithaca, NY
From: whoever on 18 Jul 2010 18:56 "JT" wrote in message news:8184e5eb-4594-494f-a73b-e9ab4388cc78(a)c10g2000yqi.googlegroups.com... > > Temporal order of spatial separated events is absolute Because you say so. Any proof other than you deciding how nature MUST work? --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: kenseto on 18 Jul 2010 19:57 On Jul 18, 6:33 pm, kenseto <kens...(a)erinet.com> wrote: > On Jul 18, 11:05 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > wrote: > > > > > > > kenseto says... > > > >When you compare two clocks in relative motion the following > > >possibilities exist: > > >1. A is running faster than B then B is running slower than A. > > >3. B is running faster than A then A is running slower than B. > > > >At no time the bogus concept of mutual time dilation that: from A's > > >point of view B is running slow and from B's point of view A is > > >running slow. > > > Strictly speaking, it's not that one clock runs slow relative to > > another clock; one clock runs slow relative to a *coordinate* system. > > A coordinate system is not a single clock, it is a system of *synchronized* > > clocks. > > > Let's go through an example: Suppose in one frame we have two synchronized > > clocks at rest, C1 and C2, lined up left-to-right, with C1 to the left of C2. > > In another frame, we also have two synchronized clocks at rest, D1 and D2, > > that are also lined up left-to-right. Clocks D1 and D2 are traveling at 0.866c > > relative to clocks C1 and C2, and their velocity is left-to-right. They > > are traveling so that D1 & D2 pass very close to C1 and C2. > > > So here are some events involving these two pairs of clocks: > > > e1: D1 passes C1, and they both show time 12:00 > > OK....this is an assumption. > > > e2: D2 passes C2, and D2 shows time 11:50, while C2 shows time 12:10 > > No....at this event D2 must agree with what D1 read: 12:00. Why? > Because D1 and D2 are in the same frame and they are synchronized. > Also C2 must agree with what C1 read which is 12:00. > > > e3: D1 passes C2, and D1 shows time 12:10, while C2 shows time 12:20 > > At this event D1 has moved a distance of C1 to C2 at a speed of > 0.866c....the C1/C2 observer said: > his C1/C2 clock has elapsed 20 minutes when D1 passes C2. He uses the > time dilation equation to determine that D1/D2 must have elapsed > 20minute/gamma=20/2=10minutes. > The rest of your arguement is based on some illogical >assumptions. I want to add: SR claims that from D1/D2 observer's point of view the C1/C2 clock must have elapsed 20 minutes/gamma=20/2=10 minutes. This gives rise to the bogus concept of mutual time dilation. In real life: the D1/D2 observer's prediction must agree with the C1/C2 observer's prediction. That means that the C1/C2 clock must read: gamma*10minutes=20 minutes when the D1/D2 clock read 10 minutes after 12:00. Ken Seto Ken Seto > > Ken Seto > > > > > > > If we assume that C1 and C2 are synchronized, then we explain the > > events this way: > > (1) At time 12:00, D1 shows time 12:00, while D2 shows time 11:45; > > D1 & D2 are not synchronized. > > (2) At time 12:10, clock D2 has advanced 5 minutes, to 11:50, while > > clocks C1 and C2 have advanced 10 minutes, to 12:10. So > > it is running at 1/2 the speed of C1&C2. At this time, D1 shows > > time 12:05. > > (3) At time 12:20, clock D2 has advanced another 5 minutes, to 12:10. > > Clock C2 has advanced another 10 minutes, to 12:20. > > > So if we assume that C1 and C2 are synchronized, we conclude that > > D1 and D2 are running slow, and that they are not synchronized > > (D1 is 15 minutes ahead of D2). > > > On the other hand, if we assume that D1 and D2 are synchronized, > > we explain the events this way: > > (1) At time 11:50, clocks D1 and D2 show time 11:50. Clock C2 > > shows time 12:10, while clock C1 shows time 11:55. > > (2) At time 12:00, clocks D1 and D2 have advanced 10 minutes, > > to 12:00, while clock C1 has only advanced 5 minutes, to 12:00, > > and C2 has also advanced 5 minutes, to 12:15. > > (3) At time 12:10 clocks D1 and D2 have advanced another 10 > > minutes, to 12:10. Clock C1 has advanced 5 minutes, to 12:05. > > Clock C2 has advanced 5 minutes, to 12:20. > > > So if we assume D1 and D2 are synchronized, we conclude that > > C1 and C2 are running slow, and that they are not synchronized > > (C2 is 15 minutes ahead of C1). > > > So mutual time dilation is *not* a comparison of two clocks alone; > > it is a comparison of 4 clocks, together with assumptions about > > synchronization. Just comparing two clocks, you cannot possibly > > say whether one clock is running slower than the other. > > > -- > > Daryl McCullough > > Ithaca, NY- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Daryl McCullough on 18 Jul 2010 20:38 kenseto says... > >On Jul 18, 11:05=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> Let's go through an example: Suppose in one frame we have two >> synchronized clocks at rest, C1 and C2, lined up left-to-right, >> with C1 to the left of C2. >> In another frame, we also have two synchronized clocks at rest, >> D1 and D2, that are also lined up left-to-right. Clocks D1 and D2 >> are traveling at 0.866c relative to clocks C1 and C2, and their >> velocity is left-to-right. They are traveling so that D1 & D2 >> pass very close to C1 and C2. >> >> So here are some events involving these two pairs of clocks: >> >> e1: D1 passes C1, and they both show time 12:00 > >OK....this is an assumption. > >> e2: D2 passes C2, and D2 shows time 11:50, while C2 shows time 12:10 > >No....at this event D2 must agree with what D1 read: 12:00. Why? >Because D1 and D2 are in the same frame and they are synchronized. >Also C2 must agree with what C1 read which is 12:00. You are confused. I didn't say that e1 and e2 are simultaneous! In frame D (the frame in which the D-clocks are at rest), e2 takes place 10 minutes before e1. In this frame, the D-clocks are correctly synchronized, but the C-clocks are not. In frame C (the frame in which the C-clocks are at rest), e2 takes place 10 minutes *after* e1. In this frame, the C-clocks are correctly synchronized, but the D-clocks are not. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 18 Jul 2010 21:33
kenseto says... I'm going to try again to explain what's going on with mutual time dilation. Let's assume that there is a frame C such that: There are two clocks, C1 and C2 at rest in this frame. They are lined up left-to-right, with C1 to the left of C2. Initially, clocks C1 and C2 are set to the same time, 12:00. There are two more clocks, D1 and D2 which are lined up left-to-right, and are moving to the right. They are traveling at the same rate of speed such that it takes 20 minutes to travel between clocks C1 and C2. Initially, D1 is set to 12:00 and initially D2 is set to 11:45 (why this discrepancy? I can explain later, but for now, let's just assume that whoever was setting the clocks set them that way, for whatever perverse reason). Clocks D1 and D2 run at 1/2 the rate of C1 and C2. Initially, D1 is lined up with C1, while D2 is halfway between C1 and C2. This is all just assumptions. Surely it is *possible* to arrange things so that all of the above is true. You can set clocks to whatever times you like. You can adjust the rates on clocks. It is certainly possible for the above description to be true. Right? If you think otherwise, then let's stop here and discuss it further. Of *course* it is possible to set things up this way. Now, at 12:00, we have the following situation: 1. D1 is lined up with C1. They both show time 12:00. 2. D2 is halfway between C1 and C2. It shows time 11:45. 3. C2 shows time 12:00 At 12:10, we have the situation: 1. C1 shows time 12:10 2. D1 is halfway between C1 and C2. It shows time 12:05. 3. D2 is lined up with C2. D2 shows time 11:50, while C2 shows time 12:10. At 12:20, we have the situation: 1. C1 shows time 12:20. 2. D1 is lined up with C2. D1 shows time 12:10, while C2 shows time 12:20 3. D2 is past C2, a distance equal to 1/2 the distance between C1 and C2. It shows time 11:55. Surely you agree that it is *possible* to set things up so that all the above are true? Right? Now, from all the facts above, let's extract those facts that are independent of any coordinate system, that everyone agrees on: e1: When clocks C1 and D1 are lined up, they both show time 12:00. e2: When clocks C2 and D2 are lined up, clock C2 shows time 12:10, while D2 shows time 11:50. e3: When clocks C2 and D1 are lined up, clock D1 shows time 12:10, while C2 shows time 12:20. Do you agree with these coordinate-independent facts? These facts were all derived in frame C, in which the C-clocks are at rest, and are synchronized, and the D-clocks are traveling, are time-dilated, and are not synchronized. But there is a frame D consistent with facts e1, e2, and e3 such that: the D-clocks are at rest, and are synchronized, the C-clocks are traveling, are time-dilated, and are not synchronized. Here's the same story, told from the point of view of frame D: At 11:50: 1. Clock D1 is to the left of C1. D1 shows time 11:50 2. Clock C1 is halfway between D1 and D2, and shows time 11:55. 3. Clock C2 is lined up with D2. C2 shows time 12:10, while D2 shows time 11:50. At 12:00: 1. Clock C1 has moved to the left to line up with D1. They both show time 12:00. 2. Clock C2 has moved to the left, and is now half-way between D1 and D2. C2 shows time 12:15. 3. Clock D1 shows time 12:00. At 12:10: 1. Clock C1 has moved to the left of D1. It shows time 12:05. 2. Clock C2 has moved to the left, and is now lined up with clock D1. C2 shows time 12:20, while D2 shows time 12:10. These two wildly different stories for what goes on have *exactly* the same coordinate-independent facts: e1: When D1 and C1 are lined up, they both show time 12:00 e2: When D2 and C2 are lined up, D2 shows time 11:50, while C2 shows time 12:10. e3: When D1 and C2 are lined up, D1 shows time 12:10, while C2 shows time 12:20. But in the first story, it is the D-clocks that are unsynchronized, and running slow, while in the second story, it is the C-clocks that are unsynchronized and running slow. Both stories are consistent with all the coordinate-independent facts, but they disagree about which clocks run slower than which other clocks. -- Daryl McCullough Ithaca, NY |