Most (if not all) of the so called 'Twin Paradox' thought experiments allow Relativists to use the 'asymmetry' argument in order to resolve it (e.g. one of the twins is accelerated or has to turn around etc). It is however clear that these arguments are in fact invalid as a time dilation should already be physically real before one of the twins turns around.
In order to clarify the true problem with any time dilation, I have thought up an experiment where no change of the state of motion of both systems occur whilst the clock is running

http://www.physicsmyths.org.uk/imgs/timedilation0.gif

One should note that the situation depicted above does not even involve any transmission of light signals at all and hence the basic assumption of Special Relativity does not apply in the first place. The clocks are started and stopped in each reference frame simultaneously and independently and both clocks will thus show identical times after having been stopped (after the clocks have been stopped it is obviously irrelevant if A or B (or both) turn around to compare the clocks).
The fact that it takes some time before the trigger signal reaches each clock does not produce any asymmetry here either, as the lengths are assumed identical and the signal travels (for instance with the speed of sound) within each rod separately.
Any assumption of a resultant time dilation is therefore ambiguous because of the symmetry of the system, i.e. time dilation is a logical impossibility.
The reason why Relativity mistakenly arrives at the conclusion of the existence of Time Dilation (and Length Contraction) are analyzed on my webpage http://www.physicsmyths.org.uk/timedilation.htm .

Note: I modified the graphics in the meanwhile so that the situation is now truly symmetric (see towards the bottom of Page 2 (http://www.sciforums.com/showthread.php?t=41636&page=2&pp=20). This should take account of some of the replies in this thread who saw the asymmetry in design as an argument to single out a preferred reference frame.


Thomas

Ah! but there IS asymmetry in your drawing. The one frame experiences switches that are a fixed uncontracted distance apart, yet the other frame experiences switches that are contracted distance apart.

I see you've tried to get around the light-speed postulate by dictating that the clocks are started and stopped mecanically.

But since there is some distance between the contact and each clock, there must be some delay between "mechanical contact" and "clock start".
No matter what method you use to transmit the start signal from the contact to the clock, it won't be instantaneous, and the delay will be different in each frame.

For example:
In Clock A's frame, the delay for the start signal to reach clock B is shorter than the delay for the start signal to reach clock A, so clock B starts first.

In Clock B's frame, things will probably be different, depending on the exact signal mechanism.

Pete:

You're right, but I don't think it is necessary to factor in delays in the mechanism to explain this. Even if the contacts start and stop the clocks instantaneously, time dilation is still there due to the contraction of the distance between the contacts of one of the clocks.

I don't know...
First, let me define some terms. In this post:
The term "active time" is used to mean the frame dependent time between a clock starting and stopping.
The term "proper active time" is used to mean a clock's proper time between starting and stopping.

Now, here's my reasoning.
If the signals are instantaneous, then:

In B's frame, Clock A and Clock B start and stop simultaneously, so their active time is identical.
In A's frame, Clock A and Clock B start and stop simultaneously, so their active time is still identical (but different to active times in frame B, as shown next).

In A's frame, the distance between the contacts is shorter than in B's frame.
Therefore, the active time for both clocks is less in A's frame than in B's frame.
Therefore, A's proper active time is less than B's proper active time.

It seems to me that this indicates:
Correct time dilation of clock A in clock B's frame (A's active time in B's frame is more than A's proper active time), but
Incorrect time dilation of clock B in clock A's frame (B's active time in A's frame is less than B's proper active time).

Please don't make me work through the formalism!

Ah! but there IS asymmetry in your drawing. The one frame experiences switches that are a fixed uncontracted distance apart, yet the other frame experiences switches that are contracted distance apart.But there is no preferred frame of reference as motion is only relative. I could as well draw the other frame as moving without changing anything about the kinematics, i.e. we have a twin paradox. This means a length contraction or time dilation of either frame is logically impossible.

Thomas

But there is no preferred frame of reference as motion is only relative. I could as well draw the other frame as moving without changing anything about the kinematics, i.e. we have a twin paradox. This means a length contraction or time dilation of either frame is logically impossible.

Thomas

The problem is not which frame you consider moving, the problem comes in when you say that both rods are identical in length. You fail to say under which conditions this is so. When the rods are at rest with respect to each other or when they are moving with respect to each other? If when they are moving with respect to each other, then from which frame are they measured as being identical in length?

Example, if they are moving with respect to each other at 0.866c and rod A measures itself as being 1 meter long and also measures rod B as being 1 meter long, That measurement of Rod B's length is due to length contraction.

Thus if the rods were placed side by side in the same frame, rod B would measure as being 2 meters long. This also means that in Frame B, rod B measures itself as being 2 meters long, and measures Rod A as being only 0.5 meters long. Thus While rod A measures the Rods as indentical in length, Rod B does not.

It does not matter whether you consider A or B as moving.

It would also be in your behoof to consider the fact that if time dilation could be shown to be "logically impossible" with such a simple example, Relativity would have never stood up for the 100 yrs it has.

Janus: Ok, but I think he already knows that. What is relevant is not merely that the rods contract per each other's viewpoints; what is essential is that the two switch activators are positioned on the rod ends of frame B, destroying any symmetry. B sees the switches activate with native (undistorted) timing, while frame A sees the switches activate in faster succession due to length distortion. See that.

The problem is not which frame you consider moving, the problem comes in when you say that both rods are identical in length. You fail to say under which conditions this is so. When the rods are at rest with respect to each other or when they are moving with respect to each other? If when they are moving with respect to each other, then from which frame are they measured as being identical in length?

Example, if they are moving with respect to each other at 0.866c and rod A measures itself as being 1 meter long and also measures rod B as being 1 meter long, That measurement of Rod B's length is due to length contraction.

Thus if the rods were placed side by side in the same frame, rod B would measure as being 2 meters long. This also means that in Frame B, rod B measures itself as being 2 meters long, and measures Rod A as being only 0.5 meters long. Thus While rod A measures the Rods as indentical in length, Rod B does not.
It is irrelevant for the present example what length A thinks rod B has and vice versa. It is only important that the clocks are stopped and started at the same time (which they are by design). It isn't even necessary that the length of the rods are measured identically within each rest-frame as any difference would affect only the time it takes the trigger signal to reach the clock. As this has nothing to do with any relative velocity of the two systems, this would result only in a constant (velocity independent) offset of the clock readings.

It would also be in your behoof to consider the fact that if time dilation could be shown to be "logically impossible" with such a simple example, Relativity would have never stood up for the 100 yrs it has.You can't really blame me that I have not presented the example already 100 years ago.
With your argument we would still be stuck in the Ptolemaic view of the universe which stood up for 2000 years until it was replaced by the Copernican view. Fact is that people simply tend to stick to views they have been taught without properly reflecting on them. A scientific view is not verified by history but in the first place by its logical consistency. As I have outlined on my webpage http://www.physicsmyths.org.uk/timedilation.htm and illustrated through this particular 'twin paradox', Relativity is not logically consistent and hence not acceptable as a theory.

Well A's rod is certainly irrelevant. In fact I think the picture should be like this (http://www.sciforums.com/attachment.php?attachmentid=3377&stc=1).

As Pete has pointed out B's rod has different lengths depending on the frame of reference.

This is an indication that time dilation has occurred. To whom the dilation occurs is what drives me squirrely.

So as far as I can tell you can only determine this by bringing the clocks together. Since the clocks have stopped, bringing them together will obviously not affect the readings that they have taken.

It seems to me that the result of the comparison implies another reference frame, I'll denote it as the "rest" frame. In which both A and B are accelerated from rest to their current respective velocities, whose velocity relative to each other is v.

So if they both accelerate to the same velocity, then the readings on the two clocks should be the same. If they accelerate to different velocities then each of the clocks should have different values, the clock that was travelling fastest relative to the "rest" frame will show a shorter amount of time than the clock travelling slower relative to the "rest" frame.

Of course this just sounds like I just specified a preferred frame of reference, but currently my brain cannot get around it.

Let the castigation begin. :D

Well A's rod is certainly irrelevant. In fact I think the picture should be like this (http://www.sciforums.com/attachment.php?attachmentid=3377&stc=1). As I said above, it is not really important what intrinsic length each rod has as long as one corrects the clock reading for the different delay times. In your diagram the length of rod a is simply 0 which means that there is no delay in frame A between triggering the relay and stopping the clock. In frame B there would still be a delay but this is constant (given by the distance between trigger point and the clock and the speed of sound) and you can just add this on to the clock reading later (or adjust the clock correspondingly in the first place).


As Pete has pointed out B's rod has different lengths depending on the frame of reference. a) Pete hasn't pointed it out, he alleged it.
b) As mentioned above already, only the intrinsic length within each frame is relevant for the clock reading and if this is different it amounts just to a constant correction time. A relative velocity of the two systems can not change anything at all about the clock readings.


To whom the dilation occurs is what drives me squirrely.That's why it is called a 'Twin Paradox'.


It seems to me that the result of the comparison implies another reference frame, I'll denote it as the "rest" frame. In which both A and B are accelerated from rest to their current respective velocities, whose velocity relative to each other is v.The thought experiment assumes that the two systems are in uniform motion whilst the clocks are running, i.e. they are accelerated to their respective velocities before the clocks start and decelerated only after the clocks have stopped.

Castigation, what's that?

mapsdnasggeyerg: I agree that your diagram is much more to the point!! :cool:

It is assumed that the clocks activate and later deactivate while the moving clock is coasting at its fixed speed, so acceleration doesn't factor in: only how much elapsed time each clock registers during the activated period. But the original poster has made a lousy asymmetrical arrangement, whose activation delays (after triggering) are dependent upon the length of B's rod, and only on the length of B's rod. It's gerry-rigged, to use the vernacular -- it's a rubbish concept all around.

Also, mapsdnasggeyerg, you cannot compare clocks after decelerating a moving one back to the lab frame, because that deceleration means that time dilation is compensated for by time hastening, FROM THE VIEWPOINT OF THE DECELERATED CLOCK ONLY. More asymmetry.

Trust me: tsmid is on to absolutely nothing.

tsmid:

One serious problem with the method you suggest in your first post to this thread is that you put the clocks in the centre of each device and assume that they correctly measure events which occur at some distance. But relativity tells us that events at each end of the device may happen at different times, especially in different frames of reference. You can't simply assign one clock for both ends.

But the original poster has made a lousy asymmetrical arrangement, whose activation delays (after triggering) are dependent upon the length of B's rod, and only on the length of B's rod. The apparent asymmetry of the location of the trigger points is completely irrelevant. I have just chosen the particular design because having trigger points on both sides would have complicated the situation as I would have to further specify which side starts and which side stops the clocks. It wouldn't make any difference if the trigger signal travels to the clock from the right or the left; the delay times are the same if the distances are identical (as mentioned the signal is assumed to travel with the speed of sound in thr rest frame of each rod).


Also, mapsdnasggeyerg, you cannot compare clocks after decelerating a moving one back to the lab frame, because that deceleration means that time dilation is compensated for by time hastening, FROM THE VIEWPOINT OF THE DECELERATED CLOCK ONLY. More asymmetry. As mentioned above already, there is no acceleration or deceleration involved. The two system are assumed to have constant velocity between starting and stopping the clocks.
Anyway, if you are still convinced that the clocks show different (velocity dependent) times when stopped, then you should be able to say which clock would appear to run slower , A or B (after all, this is supposed to be a Twin Paradox).

One serious problem with the method you suggest in your first post to this thread is that you put the clocks in the centre of each device and assume that they correctly measure events which occur at some distance. But relativity tells us that events at each end of the device may happen at different times, especially in different frames of reference. You can't simply assign one clock for both ends. If you read my replies to Janus58 and mapsdnasggeyerg above, it should be clear that Relativity is irrelevant here as any delay time is independent of the relative velocity of both systems. But if you still think that a time dilation exists, then you should be able to tell me which clock goes slower, A or B?

The problem presented by “tsmid” gives us a good possibility to illustrate the usage of Lorentz transformations (LT) to the solution of problems.

But, unfortunately, “tsmid” did one very crucial mistake at setting of its problem. There is no observer in The Nature that sees a picture, which was drawn by “tsmid”. Problem is that device 2 (with clock 2), even being constructed identical with device 1 (with clock 1) in the same workshop, at its motion in respect to device 1 never will give a picture shown by “tsmid”. Real picture (what actually sees the observer resting on device 1) is the following

. . . . . . . . . . . . 1
__________|__________
. . . . . . . . . . . . . . . . . . . . . . .|
. . . . . . . . . . . . . 2
. . . . . |_______|_______|---> v

(One should ignore all points on this picture - they are put in only to keep right positions of devices at posting)
This happens because of Lorentz contraction of the moving device. (I should say that this is a very common mistake of people who start recognizing the essence of SRT). With this improvement of the starting point I will call the problem initiated by “tsmid” as “Two clocks problem”, or TCP. There goes the full solution of TCP.

Let us base on the formulas that are given in my post “One Lecture on Twins Paradox” on the present Forum. IRF means “Inertial reference frame”.

1. As one can see, in TP we deal with two events:

Event A – the initial (the first) touching of contacts and it has the following 4D-coordinates:
(Xa, Ya, Za, ta) – in IRF-1 and (Xa’, Ya’, Za’, ta’) – in IRF-2.

Event B – the final (the second) touching of contacts and it has the following 4D-coordinates:
(Xb, Yb, Zb, tb) – in IRF-1 and (Xb’, Yb’, Zb’, tb’) – in IRF-2.

Attention: Subscript a and/or b indicate the events. And all variables marker with superscript ’ are measured in IRF-2, but without this superscript – in IRF-1.

2. One will immediately recognize that

* Events A and B are happening in the same 3D-point of IRF-1. Indeed, event A happens in 3D-point (Lo, 0, 0) at moment of time ta, and the distance between Clock 1 and right contact is Xa = Lo. Event B happens in the same 3D-point (Lo, 0, 0) but at another moment of time tb>ta, and the distance between Clock 1 and right contact is Xb = Lo = Xa
So, one has (Xa, Ya, Za, ta) => (Lo, 0, 0, ta) and
(Xb, Yb, Zb, tb) => (Lo, 0, 0, tb)

*Events A and B are happening in the two different 3D-points of IRF-2.
Indeed, event A happens in 3D-point (Xa’, 0, 0) at moment of time ta’, where distance between Clock 2 and right contact is La’ = Xa’. Event B happens in the 3D-point (Xb’, 0, 0) at another moment of time tb’>ta’, where distance between Clock 2 and left contact is Lb’ = Xb’.
So, one has (Xa’, Ya’, Za’, ta’) => (Xa’, 0, 0, ta’) and
(Xb’, Yb’, Zb’, tb’) => (Xb’, 0, 0, tb’).

3. Therefore, we have the following set of coordinates and times: non-zeroes Xb = Lo; ta; and Xb = Lo; tb; and Xa’; ta’ and Xb’; tb’, and all Y-s and Z-s are zeroes.
Using now the general formulas from “The practical usage of the Lorentz Transformations” of “One Lecture on Twins Paradox”, one will able to find that the observer in IRF-1 will obtain:

ta’=(ta-vXa/c^2)/g = (ta-vLo/c^2)/g and Xa’ = (Xa–vta)/g = (Lo–vta)/g
and ………………………………………………………………………...(1)
tb’=(tb-vXb/c^2)/g = (tb - vLo/c^2)/g and Xb’ = (Xb–vtb)/g = (Lo–vtb)/g

where g = (1 – v^2/c^2) and a^n means “a in power n”.

Therefore, one will have the following relations for
delay Tab = tb - ta and “distance” Xab = Xb - Xa = 0 and
delay Tab’ = tb’ – ta’ and “distance” Xab’ = Xb’ - Xa’:

Tab' = Tab/g and Xab'= vTab/g…………..(2)

As one can check out now, the Minkowski interval between events A and B
remains the same in both IRF. Indeed, one has

S = [c^2*( tb- ta^2 - (Xb– Xa)^2 -(Yb– Ya)^2- (Zb– Za)^2]^1/2=cTab
and
S’=[c^2*(tb' -ta')^2- (Xb'-Xa')^2- (Yb'-Ya')^2- (Zb'-Za')^2]^1/2=cTab

4. Let us notice that Xa' isn't taken as Xa'= Lo and Xb’ isn't taken as
Xb’= -Lo ether because of contraction of sizes along X-axis in the moving IRF-2. Let us also notice that without any restrictions we can chose ta = ta’ = 0 as the condition of synchronization of clocks in both reference frames: IRF-1 and IRF-2.

5. So, do not making any assumptions except obvious Xb = Xa and using general SRT formulas, we have came to relations (2).

6. IRF-1-observer easily can find duration of work of Clock 1, P1
Indeed, if event A happened at tA, then the signal about that event will reach Clock 1 with delay T1 = Lo/c and Clock 1 will start running. It will run till the signal about event B will reach this Clock 1. It will happen with delay
T2 = Lo/c after period of time equal to Tab. Therefore, Clock 1 will work all together the following period of time:

P1 = Tab + T2 - T1 = Tab………………………………(3)

7. IRF-1-observer can go further and find evident expression for Tab.
Indeed, as a good student of SRT, he knows that distance between contacts of the moving device 2 will be equal to

L’ = 2gLo……………………………………………..(4)

(The Lorentz contraction) Therefore,

Tab = 2L’/v = 2gLo/v………………………………(5)

8. Substituting (5) in (2) one will have

Tab’ = 2Lo/v…………………………………………(6)
i.e.
P1 = Tab + T2 - T1 = Tab = 2gLo/v …………..(7)

9. All what we need for now is to find the duration of work of Clock 2, P2’. First of all we should find relation between P2’ and Tab’, just as we did it for P1 and Tab in formula (3). Because P2’ and Tab’ are quantities measured in IRF-2 we have to consider whole picture in that RF. It is easy task, and answer is

P2’ = Tab’ = 2Lo/v ..…………………..…………..(8)

Indeed, IRF-2-observer sees that device1, moving with velocity v, touches its right contact (event A) and then its left contact (event B), the distance to which is Lo (in his units of distance!!!). Therefore, clock 2 will starts at moment ta'+ Lo/c and will stops at moment tb’+ Lo/c, the duration of clock-2’s work will be (8).

10. It concludes the solution of TCP problem.

The problem presented by “tsmid” gives us a good possibility to illustrate the usage of Lorentz transformations (LT) to the solution of problems.

But, unfortunately, “tsmid” did one very crucial mistake at setting of its problem. There is no observer in The Nature that sees a picture, which was drawn by “tsmid”. Problem is that device 2 (with clock 2), even being constructed identical with device 1 (with clock 1) in the same workshop, at its motion in respect to device 1 never will give a picture shown by “tsmid”. Real picture (what actually sees the observer resting on device 1) is the following

. . . . . . . . . . . . 1
__________|__________
. . . . . . . . . . . . . . . . .|
. . . . . . . . . . . . . 2
. . . . . |_______|_______|--->

The only difference of rod 2 being shortened is that the time that elapses between starting and stopping the clocks will be less, but both clocks will still show the same time (after applying a constant correction for the different trigger- delay times in each system).

The only difference of rod 2 being shortened is that the time that elapses between starting and stopping the clocks will be less, but both clocks will still show the same time (after applying a constant correction for the different trigger- delay times in each system).
Nonsense! your original premise is utterly and completed devastated... it was all based on creating a lab experiment without inherent asymmetries, remember?? Here's you quoted: "Most (if not all) of the so called 'Twin Paradox' thought experiments allow Relativists to use the 'asymmetry' argument in order to resolve it."

As for your claim above that "both clocks will still show the same time, after applying a constant correction for the different trigger-delay times in each system", it is invalidated. One would actually have to perform the experiment to see the results. The results would doubtless exonnerate SRT. I leave it to grunts (no offense) to do the math.

Check out my Twin Paradox Without Accelerations (http://www.sysmatrix.net/~kavs/kjs/addend4.html) -- concise, elegant and doesn't waste words about accelerations or their attendant G-forces.

I wonder how many times tsmid has ignored exactly the same arguments in different arenas?

Four years (http://groups.google.com.au/groups?hl=en&lr=&threadm=8nj2qq%24ctv%241%40nnrp1.deja.com&rnum=1&prev=/groups%3Fq%3Dtsmid%2Btime%2Bdilation%26hl%3Den%26l r%3D%26selm%3D8nj2qq%2524ctv%25241%2540nnrp1.deja. com%26rnum%3D1) (at least).

Some people just don't learn.

As for your claim above that "both clocks will still show the same time, after applying a constant correction for the different trigger-delay times in each system", it is invalidated. One would actually have to perform the experiment to see the results. The results would doubtless exonnerate SRT. I leave it to grunts (no offense) to do the math.Dristam,
Why are you completely misrepresenting the situation? There is neither maths nor an experiment needed in order to find out if the concept of time dilation is logically consistent. The amount of time dilation would only be of interest *if* it is logically possible that both clocks show different times.


Check out my Twin Paradox Without Accelerations (http://www.sysmatrix.net/~kavs/kjs/addend4.html) -- concise, elegant and doesn't waste words about accelerations or their attendant G-forces. Elegant? Your argumentation (which is essentially the same as in the link above by the way), is about as convoluted as it gets.
Let's stick to the thought experiment suggested by me in this thread: no accelerations, no astronauts exchanging clocks, just two clock-systems that mutually and simultaneously start and stop their clocks and the question which clock would show the proposed time dilation, A or B?

Guys,
Now I really do not understand what for you really are on Forum? Simply to socialize or to discuss scientific issues?
I presented to you plain and easy to recognize mathematics of SRT in "Two clocks problem".
I have noticed initial mistake of tsmid in graphics, which destroys his initial proof, which was based on the “absolute symmetry of rods”.
I have shown you that clock 1 will read
P1 = 2gLo/v (formula 7 of my prior post) with g = [1-v^2/c^2]^1/2
and clock 2 will read
P2 = 2Lo/v (formula 8 of my prior post)
But even after that tsmid said “The only difference of rod 2 being shortened is that the time that elapses between starting and stopping the clocks will be less, but both clocks will still show the same time (after applying a constant correction for the different trigger- delay times in each system)”. Such statement assumes that tsmid is ready to show some proof that “both clocks will still show the same time”.
Instead of that you again started some discussion on the philological level. What is going on? What else you need to reveal calculations that prove your position? Do a real work, guys, and stop “socializing”. Whole word is waiting some concrete arguments, not prays… If you can not do Math, tell us what is stopping you in your calculations and we will help you to resolve your difficulties to give you a chance continue your research…

Dristam .. The amount of time dilation would only be of interest *if* it is logically possible that both clocks show different times.WHO'S logic?? If you had succeeded in designing an experiment utterly absent of ASYMMETRIES, then you would have made your point magnificently. But you designed no such experiment.
Let's stick to the thought experiment suggested by me in this threadSure, let's.

Dristam,

Let's stick to the thought experiment suggested by me in this thread: no accelerations, no astronauts exchanging clocks, just two clock-systems that mutually and simultaneously start and stop their clocks and the question which clock would show the proposed time dilation, A or B?


The real question is which clock accumulates less time after each has stopped.

Your error lies in assuming that both clocks do mutually and simultaneously start and stop.


Example: Assume that clock B's rod is 1 light second long, as measured by clock B. Also assume that the rods are made of a rigid material in which the speed sound is 0.1c. The rods have a relative velocity of 0.866c. We will also eliminate the rod for clock A, as it serves no real purpose. (If needed, we can later put back rod A to show that it would make no difference.)

First we will consider what happens according to clock B:

1. The contacts touch and clock A starts. Also the signal from this contact starts traveling to clock B at .1 sec.

2. 1.154... sec later, (the time it takes for clock A to traverse the length of Rod B at 0.866c) the second contact is touched and clock A stops. Clock A has experienced time dilation according to Clock B, so Clock A reads 0.577... sec when it stops. In addition the siganl form the second contact starts on its way to clock B

4. 5 sec after the first contacts touch. the signal arrives at clock B and clock B starts.

5. 5 sec After the second contact is touched the second signal arrives and clock B stops. The difference in time between the arrival of the two signals is equal to 1.154... sec, and this is the time on clock B when it stops.

Now consider what happens according to clock A.

1. The contacts touch and clock A starts, the signal begins its travels towards clock B

2. Rod B is length contracted so it is, according to clock A, .5 light sec long. Thus 0.577 secs later, Clock A has traversed this distance at 0.866c and stops. Clock A shows 0.577 sec when it stops.

3. The signal progressing along rod B travels at .1c relative to rod B as measured from rod B, but this does not hold as measured by clock A. To get the signals relative velocity to clock B as measured by clock A we must use the velocity addition formula. First we find the relative velocity of the signal to Clock A. this is equal to (.866c-.1c)/(1-(.1c)(.866c)) = 0.8386c (the minus sign is used because the signal travels in the opposite direction of Rod B. Then we take the difference between this and the 0.866c velocity of rod B to get .027c, for the relative velocity of the signal with respect to clock B as measured by clock A.
At this relative velocity, it takes 9.132 secs for the signal to travel the .25 light sec to clock B. Thus 9.132 secs after the contacts touch, clock B starts according to clock A

4. To find the relative velocity of the second signal to clock B we use the same procedure as before. This time however the siganal is traveling in the same direction as the rod, so we get (.866c+.1c)/(1+(.1c)(.866c))=0.889c for the relative velocity of the siganl to clock A. this gives a relative velocity of 0.023c to rod B as measured by clock A. Therefore by Clock A's reckoning, it takes 10.864 sec from the time the contacts touch until clock B stops.

Add the .557sec it took to travel between contacts, and subtract the 9.132 secs it took before clock B starts, and clock B ran for 2.31 sec by Clock A reckoning form the time it started and stopped. Clock B is time dilated according to clock A, and therefore should read according to clock A, half the time that Clock A recorded in that time, or 1.154 secs.

To recap:

According to clock B, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

According to clock A, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

Both clocks agree that twice as much time elaspes on clock B as A.

Most (if not all) of the so called 'Twin Paradox' thought experiments allow Relativists to use the 'asymmetry' argument in order to resolve it (e.g. one of the twins is accelerated or has to turn around etc). It is however clear that these arguments are in fact invalid as a time dilation should already be physically real before one of the twins turns around.
In order to clarify the true problem with any time dilation, I have thought up an experiment where no change of the state of motion of both systems occur whilst the clock is running

http://www.physicsmyths.org.uk/imgs/timedilation.gif

One should note that the situation depicted above does not even involve any transmission of light signals at all and hence the basic assumption of Special Relativity does not apply in the first place. The clocks are started and stopped in each reference frame simultaneously and independently and both clocks will thus show identical times after having been stopped (after the clocks have been stopped it is obviously irrelevant if A or B (or both) turn around to compare the clocks).
The fact that it takes some time before the trigger signal reaches each clock does not produce any asymmetry here either, as the lengths are assumed identical and the signal travels (for instance with the speed of sound) within each rod separately.
Any assumption of a resultant time dilation is therefore ambiguous because of the symmetry of the system, i.e. time dilation is a logical impossibility.
The reason why Relativity mistakenly arrives at the conclusion of the existence of Time Dilation (and Length Contraction) are analyzed on my webpage http://www.physicsmyths.org.uk/timedilation.htm .

Thomas

Tsmid, as you have drawn the schematics, the lower clock that is moving will not have its clock tuerned on at the same time as the signal from the ends of the ships are moving into a moving frame in one instance and in the other frame in the other drection is met head on. I didn't quite understand your argument regarding the lack of asymetry, but I see that all of that will fall away completely if you use just one switch for both clocks, then motion has nothing to do with synchroization at all. Actually I am on your side on this.

To repeat, you can repair this "defect" by simply having the trigger located such that both clocks get turned on at the same time by one switch.

Guys,
Now I really do not understand what for you really are on Forum? Simply to socialize or to discuss scientific issues?
I presented to you plain and easy to recognize mathematics of SRT in "Two clocks problem".
I have noticed initial mistake of tsmid in graphics, which destroys his initial proof, which was based on the “absolute symmetry of rods”.
I have shown you that clock 1 will read
P1 = 2gLo/v (formula 7 of my prior post) with g = [1-v^2/c^2]^1/2
and clock 2 will read
P2 = 2Lo/v (formula 8 of my prior post)
But even after that tsmid said “The only difference of rod 2 being shortened is that the time that elapses between starting and stopping the clocks will be less, but both clocks will still show the same time (after applying a constant correction for the different trigger- delay times in each system)”. Such statement assumes that tsmid is ready to show some proof that “both clocks will still show the same time”.
Instead of that you again started some discussion on the philological level. What is going on? What else you need to reveal calculations that prove your position? Do a real work, guys, and stop “socializing”. Whole word is waiting some concrete arguments, not prays… If you can not do Math, tell us what is stopping you in your calculations and we will help you to resolve your difficulties to give you a chance continue your research…

Yuriy, I didn't get your proof that time dilation will occur. I suspect that at some opint you use some relevant aspect of SR theory, is this correct? Then, is it not proper that you prove the SR construct that provides you your answer, that is if you use SR in your calculations.

The real question is which clock accumulates less time after each has stopped.

Your error lies in assuming that both clocks do mutually and simultaneously start and stop.


Example: Assume that clock B's rod is 1 light second long, as measured by clock B. Also assume that the rods are made of a rigid material in which the speed sound is 0.1c. The rods have a relative velocity of 0.866c. We will also eliminate the rod for clock A, as it serves no real purpose. (If needed, we can later put back rod A to show that it would make no difference.)

First we will consider what happens according to clock B:

1. The contacts touch and clock A starts. Also the signal from this contact starts traveling to clock B at .1 sec.

2. 1.154... sec later, (the time it takes for clock A to traverse the length of Rod B at 0.866c) the second contact is touched and clock A stops. Clock A has experienced time dilation according to Clock B, so Clock A reads 0.577... sec when it stops. In addition the siganl form the second contact starts on its way to clock B

4. 5 sec after the first contacts touch. the signal arrives at clock B and clock B starts.

5. 5 sec After the second contact is touched the second signal arrives and clock B stops. The difference in time between the arrival of the two signals is equal to 1.154... sec, and this is the time on clock B when it stops.

Now consider what happens according to clock A.

1. The contacts touch and clock A starts, the signal begins its travels towards clock B

2. Rod B is length contracted so it is, according to clock A, .5 light sec long. Thus 0.577 secs later, Clock A has traversed this distance at 0.866c and stops. Clock A shows 0.577 sec when it stops.

3. The signal progressing along rod B travels at .1c relative to rod B as measured from rod B, but this does not hold as measured by clock A. To get the signals relative velocity to clock B as measured by clock A we must use the velocity addition formula. First we find the relative velocity of the signal to Clock A. this is equal to (.866c-.1c)/(1-(.1c)(.866c)) = 0.8386c (the minus sign is used because the signal travels in the opposite direction of Rod B. Then we take the difference between this and the 0.866c velocity of rod B to get .027c, for the relative velocity of the signal with respect to clock B as measured by clock A.
At this relative velocity, it takes 9.132 secs for the signal to travel the .25 light sec to clock B. Thus 9.132 secs after the contacts touch, clock B starts according to clock A

4. To find the relative velocity of the second signal to clock B we use the same procedure as before. This time however the siganal is traveling in the same direction as the rod, so we get (.866c+.1c)/(1+(.1c)(.866c))=0.889c for the relative velocity of the siganl to clock A. this gives a relative velocity of 0.023c to rod B as measured by clock A. Therefore by Clock A's reckoning, it takes 10.864 sec from the time the contacts touch until clock B stops.

Add the .557sec it took to travel between contacts, and subtract the 9.132 secs it took before clock B starts, and clock B ran for 2.31 sec by Clock A reckoning form the time it started and stopped. Clock B is time dilated according to clock A, and therefore should read according to clock A, half the time that Clock A recorded in that time, or 1.154 secs.

To recap:

According to clock B, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

According to clock A, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

Both clocks agree that twice as much time elaspes on clock B as A.


Why not just use one switch to start and stop the clocks, and avoid all the problems with signals traveling sonically down rods. Does not this eliminate your basic objection of when clocks start and stop? I measn using one switch?

According to clock B, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

According to clock A, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

Both clocks agree that twice as much time elaspes on clock B as A. OK, but now assume that the situation is reversed, i.e. the rod is removed for clock B instead (which according to your own statement should be irrelevant). You would then have to conclude that clock A reads 1.154 sec and clock B 0.577 sec, i.e. we have a twin paradox!

To make the situation even clearer, you could actually remove all the rods and let the clocks start and stop themselves when colliding elastically with each other. You could confine them between two walls for instance and on the first collision the clocks would start and on the second they would stop. You would then have perfect symmetry from the outset.

I presented to you plain and easy to recognize mathematics of SRT in "Two clocks problem". Yuriy, with all respect, but your mathematical account of the problem is utterly incomprehensible, especially as some of the math symbols do not display.
The problem at hand is very trivial so a few simple arguments should really suffice to solve it. Everything else would not only be overkill but hide the central problem here and confuse people.

Why not just use one switch to start and stop the clocks, and avoid all the problems with signals traveling sonically down rods. Does not this eliminate your basic objection of when clocks start and stop? I measn using one switch?See my reply to Janus58 above, this might clarify the situation in your sense.

If you read my replies to Janus58 and mapsdnasggeyerg above, it should be clear that Relativity is irrelevant here as any delay time is independent of the relative velocity of both systems. But if you still think that a time dilation exists, then you should be able to tell me which clock goes slower, A or B?
Please excuse my blundering into your post. I was too hasty in reading what was going on. Let me see if I understand you completely. Anyway, I find that The B clock has more time on it but both clocks run the same rate and the difference in time is not a rime dilation factor in the slightest, but contrary to your assumption tha the relative velocity is irrelevant I must add the following:. The relative velocity is the reason for the dfference in times on the clocks, but adds no time dilation implications..

When the A clock is triggered, assumed stationary, then the moving clock gets turned on at different times. Lets us assume that the signal in the A frame moves a distance ct(a) and the lower signal moves ct(b) - vt(b), the distance the B frame has moved to get turned on, before the A clock.

Now the other "off" signal. The signal in the A frame is the same as before. Let us break the problem down such:
When the lower signal has just the same amount of distance to move as the upper signal, i.e. vt, both signals have the same distance to go right?

No. When the upper signal has just reached the A clock the lower signal has covered the same distance, but the B frame is moving, therefore there is a little bit more the B signal must move through in order to stop the B clocks.

Therefore, there will be more time on the lower clock, but this says nothing about time dilatiion as the difference in times can be calculated knowing the appropriate distances and frame velocity, and the times between events , we see:

ct(a) = ct(b) +vt(b)
and therefore t(b) = Ct(a)/ (C + V)

Now when turning the clocks off we see that the A signal must move the same distance as before. Lets look at the signals just as the B signal has reached the point ct - vt(b). Both signals have to travel the same vt distance to shut off the clocks, right?

Wrong (Irepeat myself). The B clock has to cover the same vt ground that the A signal must cover, but, the B frame is moving, therefore it takes a little extra time to reach this point, I call t'.
or ct' = ct(b) + vt(b) + vt'
or t' = Ct(b)(C + V)/(C - V), or substituting t(b) above, and if all my algebra is correct,

t' = ct(a)/(C - V).



This says the signal in the A frame, travels the complete distance in a time less than the lower signal needed to catch the clock in the B frame. But this number being known, there is no SR implications or time dilation consequences, or twin paradox. But that little delta t' is the cause of all the SR bally hoo.

..To recap:

According to clock B, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

According to clock A, clock B reads 1.154 sec when it stops and clock A reads .577 sec when it stops.

Both clocks agree that twice as much time elaspes on clock B as A.

I gotta hand it to you, Janus58; Bravo! Kudos! The differential will always be equal to the gamma factor, because it boils down to being solely a reflection of the fact that the rod is contracted (by the gamma factor) per Frame A's reckoning, yet uncontracted per Frame B's reckoning. All else equals out, ie. cancels out!

It's EXACTLY like I've insisted all along in this thread, beginning with my terse post, the very first reply of the thread, wherein I pointed out that tsmid's precious SYMMETRY is naught but a PIPE DREAM. There's no perfectly symmetrical lab experiment that can be devised, so tsmid's original premise is disproven -- plain and simple.

..To make the situation even clearer, you could actually remove all the rods and let the clocks start and stop themselves when colliding elastically with each other. You could confine them between two walls for instance and on the first collision the clocks would start and on the second they would stop. You would then have perfect symmetry from the outset.
You are as stupid as stupid gets; because then you've added accelerations and G-forces; and under Relativity, those will compensate for relative time dilation with relative time hastening, just so. Get savvy!

And check out my Sorting Out a Paradox (http://www.sysmatrix.net/~kavs/kjs/addend5.html)

Dristam, read my post a couple above your latest. You haven't comented on that have you? Is here some reason for avoiding the post? Its right there.
Geistkiesel.

Dristam, read my post a couple above your latest. You haven't comented on that have you? Is here some reason for avoiding the post? Its right there.
Geistkiesel.
I didn't answer it because it didn't say anything of value. The simple answer -- my original answer -- is the definitive answer, as Janus58 has so aptly demonstrated: Clock A will always read less time -- less by the gamma factor -- because the rod span is reckoned as contracted by Frame A, yet uncontracted for Frame B. And the rod of Frame B is where the trigger span is located, so that's what throws it one-sided.
http://www.sysmatrix.net/~kavs/stg/redrawn.JPG

then you've added accelerations and G-forces; and under Relativity, those will compensate for relative time dilation with relative time hasteningYes there would in practice be accelerations, but these would appear completely identical in both frames and you could not single out one frame in any way.

Yes there would in practice be accelerations, but these would appear completely identical in both frames and you could not single out one frame in any way.
That's right, sir -- and that's why the clock readings will be identical; but that only proves Relativity -- it DISproves nothing! In an utterly symmetrical scenario with accelerations, the clock hastening distortion will exactly counter the clock slowing distortion. Read my Sorting Out a Paradox (http://www.sysmatrix.net/~kavs/kjs/addend5.html)

Clock A will always read less time -- less by the gamma factor -- because the rod span is reckoned as contracted by Frame A, yet uncontracted for Frame B. And the rod of Frame B is where the trigger span is located, so that's what throws it one-sided. Why would this single out one of the frames as a preferred reference frame? The usual argument is that one of the twins is not in an inertial system all the time (i.e. either accelerated or changes frame by passing the clock to another traveller) and this is not the case here at all.

But if the asymmetry in the design still bothers you, consider a slight modification of my original drawing
http://www.physicsmyths.org.uk/imgs/timedilation2.gif
You have now perfect symmetry here with the length of both rods serving as the trigger span.

In an utterly symmetrical scenario with accelerations, the clock hastening distortion will exactly counter the clock slowing distortion. Read my Sorting Out a Paradox (http://www.sysmatrix.net/~kavs/kjs/addend5.html) Why would that be so? The amount of time lost due to time dilation would continuously increase the further the observers travel apart, but the time gained due to the time hastening would always be the same as it just depends on the moment of acceleration when turning around. In general I can't see the two therefore being identical.

tsmid,
1. You said: "especially as some of the math symbols do not display".
Please, tell me which symbols do not display on your PC? How you found out that?
2. Is my notice, that there is no observer in Nature, which could see your picture with two identical rods as the rods with the same length if one is moving in respect of other, also "incomprehensible". Why you still continue drawing such pictures?

.. But if the asymmetry in the design still bothers you, consider a slight modification of my original drawing .. You have now perfect symmetry here with the length of both rods serving as the trigger span.
That's good; that's GREAT! NOW you have symmetry, and so the clocks will read the same, following execution of the lab experiment. Congratulations!

But you have NOT proven any obvious fallacy to time dilation, which was your intent. You have not proven any obvious contradiction because that could only happen in an experiment devised so that the clocks start and stop simultaneously. Here above, they do not: because of length contraction of the other frame's rod, the signal does not arrive at the two clocks simultaneously. The start & stop signals travel an UNcontracted half-rod length to activate one's native clock, but a contracted half-rod length to activate the other.

But you have NOT proven any obvious fallacy to time dilation, which was your intent. You have not proven any obvious contradiction because that could only happen in an experiment devised so that the clocks start and stop simultaneously. Here above, they do not: because of length contraction of the other frame's rod, the signal does not arrive at the two clocks simultaneously. The start & stop signals travel an UNcontracted half-rod length to activate one's native clock, but a contracted half-rod length to activate the other.Yes, but native clock for frame A is A and the native clock for frame B is B i.e. the frames would disagree about which clock will show a time dilation when you compare them after the clocks have been stopped.


Anyway, it is not the 'symmetry' which is the the crucial point here but the logical consistency of the statement 'moving clocks run slower'. It can obviously have no effect on the momentary clock rate if one of the twins decides to turn around in 100 years time. Corresponding arguments to resolve the twin paradox are therefore already flawed due to violation of causality.

Why would that be so? The amount of time lost due to time dilation would continuously increase the further the observers travel apart, but the time gained due to the time hastening would always be the same as it just depends on the moment of acceleration when turning around. In general I can't see the two therefore being identical.
You're dead wrong my friend: the time hastening distortion is proportional to the thrust at turnaround AND proportional to the distance apart at turnaround. So the more time they spend moving apart, the more time contraction distortion results at turnaround.

.. Anyway, it is not the 'symmetry' which is the the crucial point here but the logical consistency of the statement 'moving clocks run slower'. It can obviously have no effect on the momentary clock rate if one of the twins decides to turn around in 100 years time. Corresponding arguments to resolve the twin paradox are therefore already flawed due to violation of causality.
You are wrong here because the turnaround after 100 years indeed has NO effect on the Earthbound twin's appraisal of the astronaut twin's clock. The Earthbound appraisal will encorporate only simple time dilation. But the G-force thrust at turnaround affects the thrusting twin's appraisal of Earth's clock.

The bottom line is this: Every inertial frame's time continuum is a proprietary one. When the astronaut twin returns to Earth, he will be importing with him the accrued effects that alien time continua have had on his person and on his timepiece. Relativity never violates causality. Relativity has been approved by the World's TOPMOST minds. YOU have not found a flaw in THEIR reasoning. Yet even so, the particulars need to be hashed out in order to satisfy you... if we relativists can in any way accomodate.

Yes, but native clock for frame A is A and the native clock for frame B is B i.e. the frames would disagree about which clock will show a time dilation when you compare them after the clocks have been stopped.
The scheme is 100% symmetrical, so, according to Relativity, all of the Lorentz transforms cancel each other out and the clocks will read the same when later compared. This can either be taken as proof of the Lorentz transform, or proof of nothing whatsoever, since the clock readings don't differ.

In order to invalidate time dilation, one would have to devise a scheme where moving clocks start simultaneously and later stop simultaneously. That cannot be done. I DO APOLOGIZE however for misleading you a bit back there, when I indicated that symmetry alone would suffice. No, simultaneity of triggering is needed also, in order to show Relativity flawed. You saw what Janus58 did to your first attempt... the math checks out, sorry.

tsmid,
1. You said: "especially as some of the math symbols do not display".
Please, tell me which symbols do not display on your PC? The symbols that apparently do not display are for instance ?T1 , ?P1 , ?Tab and others (the question mark appears as a square in Internet Explorer).

Is my notice, that there is no observer in Nature, which could see your picture with two identical rods as the rods with the same length if one is moving in respect of other, also "incomprehensible". Why you still continue drawing such pictures?Sure I can draw these pictures. If you are bothered about the rods both standing still, just imagine that they have been photographed with an infinitely short exposure time which freezes their motion.
Otherwise, as I said before, it is immaterial that I have drawn the rods with equal length. The crucial point is that no frame can claim to be preferred because it is an inertial frame and the other not.

You're dead wrong my friend: the time hastening distortion is proportional to the thrust at turnaround AND proportional to the distance apart at turnaround. So the more time they spend moving apart, the more time contraction distortion results at turnaround.It is news to me that time dilation is assumed to depend not only on relative velocity and G-force but also on distance (and it is neither mentioned on your webpage http://www.sysmatrix.net/~kavs/kjs/addend5.html ).
Do you have a reference for this if you can't make it more plausible?

Dear tsmid,
1. ? mark is exactly the square which means "delta T, or difference of t". I use it only because this Forum does not contain font "Symbol" (Greece letters). Now you will be not confused, I guess, to use right Math of SRT from my post.
2. But it is more important that you wrote: "Sure I can draw these pictures. If you are bothered about the rods both standing still, just imagine that they have been photographed with an infinitely short exposure time which freezes their motion. Otherwise, as I said before, it is immaterial that I have drawn the rods with equal length. The crucial point is that no frame can claim to be preferred because it is an inertial frame and the other not." That means you still do not accept the fact of Nature that the length of your both robs will be seen for any observer at any exposure as different ones, as far as those rods are in mutual movement.And namely that is the Reality, and namely any rejection of this fact of Nature is a fantasy.

It is news to me that time dilation is assumed to depend not only on relative velocity and G-force but also on distance (and it is neither mentioned on your webpage Do you have a reference for this if you can't make it more plausible?
You will HAVE TO read my posts more carefully (or write yours more carefully), because it is time contraction that is proportional to both g-force of acceleration AND distance of remote timepiece -- not SRT time dilation. BUT HEY! MY ERROR! From the standpoint of an accelerating entity, time contraction occurs in one direction and dilation in the opposite direction; yet both phenomena are compounded upon SRT time dilation. You'll see.. keep reading.

You're right that I do not cover the precise formula on my web pages. The formula is found in abundance at reference web sites and in books: the rate attributed to a distant clock, per the reckoning of an accelerating entity (in unencumbered space) is one's OWN RATE TIMES (1+gx)/gamma, where g is the G-force of acceleration and x is the distance (sign significant!!) and gamma is the usual and denotes the time dilation that is concurrently attributed based on relative speed (if any).

I've seen the units worked out for that formula before but now I am at a (momentary?) loss to resolve the units. CAN ANYONE PLEASE ASSIST??

So you see, it is as I originally stated.

Dear dristam,
to make your formula in consistence with "the same dimensions rule" you have to use term like gx/mc^2 or "product of distance and force/per unit of mass" divided square of speed of light.
BTW. If you will use U/c^2 instead your gx/mc^2, where U is the potential of the external force, you will get exactly the Einstein's formula from his first work on GRT.

Dear dristam,
to make your formula in consistence with "the same dimensions rule" you have to use term like gx/mc^2 or "product of distance and force/per unit of mass" divided square of speed of light.
Please! How can that be?? There is no mass involved in my calculation. Perhaps I got something wrong. Is not the term "g" expressed as meters per second squared? If so, then gx is in meters squared per second squared and a simple division by c^2 remedies that, ie. makes it a pure number.

1.I do not know what you ment, but your words: "where g is the G-force of acceleration " "Force of acceleration" and acceleration itself have different dimensions, as you know, I guess. May be you are using term "Force" in some non-conventional sense? Anyway, if you divide product of acceleration (which is a force per unit of mass) and x on c^2 you will fix your initial formula.
2. Please, pay attention to my remark about Einstein's formula!

Many thanks, Yuriy!

.. I guess. May be you are using term "Force" in some non-conventional sense? ..
Again, many thanks to Yuriy. I checked with another source and the formula is as we both suspected: (1 + gx/c^2)/gamma, where g is the strength of the acceleration. If you use units where c=1, such as acceleration in units of light-years per year per year... THEN I think that the c^2 term is understood and need not be written. But either way, the matter is now fully resolved.

And YES, that formula is derived (a differentiation mostly) from the basic Lorentz transform.

So, tsmid... put that in your pipe and smoke it!

dristam,
everything indeed will be OK, if you would mentioned that this formula works (and can be proven) only if g =const (i.e. only in case of the constant external force. BTW, Einstein derived this formula exactly from his famous "falling lift" gedunken experiment, i.e. at g = const or for gravity-acceleration equivalence). At arbitrary external force (but potential one!) gx/c^2 should be changed by U/c^2, where U is a local potential of the external force.

the rate attributed to a distant clock, per the reckoning of an accelerating entity (in unencumbered space) is one's OWN RATE TIMES (1+gx)/gamma, where g is the G-force of acceleration and x is the distance (sign significant!!) and gamma is the usual and denotes the time dilation that is concurrently attributed based on relative speed (if any).As Yuriy said already, I think you misinterpret the equation: x has to be seen as the distance the twin travels under the influence of the gravitational acceleration and not as the distance he travels before or after. The former can be assumed as constant however as it is just the distance travelled by the spacecraft during turnaround. Also, you said that the effect depends on the sign of x, so the effect should actually cancel to 0 because x changes sign during the turnaround.
Also, should a G-force not slow down a clock according to GR?

But anyway, the reason why I suggested the particular thought experiment in this thread was to avoid the acceleration issue in the first place.

As Yuriy said already, I think you misinterpret the equation: x has to be seen as the distance the twin travels under the influence of the gravitational acceleration and not as the distance he travels before or after. The former can be assumed as constant however as it is just the distance travelled by the spacecraft during turnaround. Also, you said that the effect depends on the sign of x, so the effect should actually cancel to 0 because x changes sign during the turnaround.
Also, should a G-force not slow down a clock according to GR?

But anyway, the reason why I suggested the particular thought experiment in this thread was to avoid the acceleration issue in the first place.

I have misinterpreted NOTHING, sir.

You said, "should a G-force not slow down a clock according to GR?". Under Relativity, one's OWN clocks NEVER distort. Learn that and you're halfway home!

The formula I supplied tells how much distortion is attributed to distant clocks, per the reckoning of an accelerating entity. The term x is the distance to the remote clock... positive in one direction yet negative off in the opposite direction. So, an accelerating entity ascribes time slowing to remote/distant clocks that are off in the direction opposite to the vector of acceleration, while ascribing time hastening to remote/distant clocks that are off in the direction congruent with the vector of acceleration. The term g is the degree of acceleration.

LEARN about Relativity someday, tsmid. The primary thing one must understand is that clock and ruler distortions never occur to ONESELF -- NEVER! they are always attributed to OTHERS !!

TSMID: I give you credit for making a BOLD effort.

Guys,
I'm afraid that your discussion will go in wrong direction if you will keep in mind the different sense of the Einstein's formula for correction of the Lorentz-factor 1/gamma in the presence of the gravitational field (and at acceleration without gravitational field, if they locally effect the body in the same way. (Rrecall the Principle of Equivalency of local G-field and acceleration).
The rule 1/gamma --> (1+U/c^2)/gamma works locally. So, if some body moves under influence of the external gravitational field (or acceleration), then to reckon distortion of some characteristic (like time delay) this body experienced, one has to integrate usual (pure SRT) infinitesimal distortion with factor (1+U(x)/c^2) along body's trajectory. So, quantitatively one will get results that are not the same as a vulgar substitution of 1/gamma by (1+U/c^2)/gamma into usual formulas of SRT.
BTW, factor U/c^2 can not be delivered by SRT, its discovery is the one more grate achievement of Einstein and namely it had signified the beginning of GRT.

Guys .. if some body moves under influence of .. acceleration, then to reckon distortion of some characteristic (like time delay) this body experienced, one has to integrate usual ..
The body itself, the body that is accelerating in flat (unencumbered) space -- THAT body does NOT "experience" any distortions whatsoever! I am trying to impress upon tsmid the fact that the formula, (1 + ax/c^2)/gamma, describes a clock rate distortion that must be ascribed to any remote (ie. DISTANT) timepiece -- never one's OWN!!

So, Yuriy, YOU are not helping matters by implying that a body under acceleration "experiences" distortions of its own... as that is simply NOT TRUE.

The formula (1 + ax/c^2)/gamma is a direct derivation (differentiation mostly) of the basic Lorentz Transform, YES! I have the derivation!

Dear dristam,
Please, stop right there. Because if you can derive Einstein's formula directly from LT it will have a lot more important meaning for physics than all you have already said in this Forum.
So, please, do delivery of such formula from LT. We will see it with a huge attention and interest.

Dear dristam,
Please, stop right there. Because if you can derive Einstein's formula directly from LT it will mean a lot more important meaning for physics than all you have already said in this Forum.
So, please, do delivery of such formula from LT. We will see it with a huge attention and interest.

Wrong; there's nothing Earth-shattering about this know-how... you should double check your science history. Here is a link to the derivation I wangled out of David Waite: CLICK HERE. (http://www.sysmatrix.net/~kavs/stg/derivegx.html)

Have a ball!

Dear dristam,
I read your post, I was on page you asked me to go, and I read book of Mr. D.Waite.

Actually, the “derivation” of term (1+U/c^2)/gamma, we are talking about, you got from Mr. D. Waite in response on your inquiry: “The formula for reckoning of a remote clock's rate relative to one's own, while accelerating in unencumbered space, is the well-known (1+gx)/gamma. I'd like to understand, can this be derived somehow from the rudimentary Lorentz transform?”Mr. Waite answered: “Have a look at problem
http: //www.geocities.com/zcphysicsms/chap5.htm#BM65 part c.
Your expression is mine only with c = 1 and g in place of alpha. To get it start with my Lorentz like transformation equation 3.3.8.
......ct'
ct = ∫ γdct'' + γβx' ( ∫ means integral)

.............ct'
x = γx' + ∫ γβdct'' ”

(Please, ignore dots - they are only to make a right positionning of other symbols on the screen)

where β = v/c and γ = 1/ (1- v^2/c^2)^1/2 = 1/ (1- β^2)^1/2

(a^n means a in power n)

1. First of all, do you recall what we were talking about and what you promised to show? You promised to show how rule “1/gamma replaced by (1+gx)/gamma“ can be derived from Lorentz transformation. And what you are trying to give us instead of that? You are giving us some derivation of some very similar formula from Mr. D. Waite’s invented transformations (“with my Lorentz like transformation equation”).

2. As you know Lorentz transformations have the following form:

ct = γct' + γβx'

x = γx' + γβct'

and have nothing to do with case when v is depending upon t, and/or x, and/or t' and/or x', no matter does x depend on t and/or x' depend on t'.
And these transformations after trivial differentiation give (at v = const, i.e. γ = const and β = const as the real Lorentz transformations state):

dt/dt' = γ (1+ βdx'/dt') and dx/dt' = γdx'/dt' + γβc

As you can see these equations have nothing to do neither with Einstein’s formula

"γ replaced by γ (1+gx/c^2)"

nor with Mr. Waite’s formula

dt/dt' = (1 + alpha*x'/c2)/γ where αlpha = c (dβ/dt')γ^2

3. You even did not noticed that you and Mr. Waite are talking about absolutely different cases: in your formula we have a value of x, in Waite’s formula we have a value of x'; in Einstein's formula γ stays in the numerator, in Waite’s formula γ stays in the denominator. (Because your “gamma" is 1/γ !)
4. Conclusion: you did not provided derivation of Einstein's formula from the Lorentz transformations.
5. But the good news is that the Einstein’s formula is right. The more good news is that it can not be derived from the Lorentz transformations and requires much more powerful Principles of General Relativity.

.. the “derivation” of term (1+U/c^2)/gamma, we are talking about .. in Einstein's formula γ stays in the numerator, in Waite’s formula γ stays in the denominator.

Look again. The term, gamma is in the denominator in both formulae.

In short, I am 100% satisfied that the derivation D.Waite supplied is correct and pertinent. In short, it is very obvious to me that the clock rate distortion that a constantly thrusting entity (in flat space) must ascribe to distant clocks is equal to

(1 + gx/c^2)/gamma

.. where g is the acceleration, x is the distance (sign significant) and gamma is the usual. I can see clearly that the above formula follows directly from the LT. I am totally satisfied to have seen it derived right before my very eyes. What you said about gamma being in the numerator one place and in the denominator another place is dead wrong... please look back and you will see that gamma resides in the denominator in both formulae.

In conclusion, you are making some comprehension errors and you are completely mistaken about the derivation being invalid.

General Relativity may have a lot more complicated and more profound implications than just this one formula, but the formula above for describing distant clock rate distortions in a uniform pseudo-force field (ie. gravitational potential gradient) along a single linear axis is indeed precisely as shown and is exactly equivalent to what is shown, and the formula is derived directly from LT and that isn't even shocking.

How is it that you failed to notice that in Mr. Waite's final formula, gamma is raised to the power negative one, hmmm?? Raising a term to the negative one power means moving it into the denominator. Didn't that superscript get rendered properly in your browser?? If NOT, then perhaps some other symbols also failed to render correctly in your lousy browser. But in my browser (MSIE 6.0) it is crystal clear, and I see gamma raised to the negative one power in the derived formula. Please have another look!

Just to make sure that you and I are looking at the exact same derivation, I have uploaded the derivation as an image CLICK HERE. (http://www.sysmatrix.net/~kavs/stg/derivegx.jpg) In order to see it clearly, you may have to Save Target to your own computer and then use image Zoom-in. It's an 85kb file.

And Yuriy, or anyone: do you know with certainty what is meant by the term, proper acceleration?? I can only imagine that it refers to acceleration that has attendant G-forces felt, as opposed to acceleration in freefall relative to another entity. For example, if one fell off a building, then one would be "accelerating" toward the Earth, per any Earthling's view. That kind of acceleration has no attendant onboard G-forces felt (you'd feel weightless, ignoring air drag of course).

But I wish to know with absolute certainty that I have guessed the meaning of the term correctly.

Dear dristam,
1. If you assert that your "gamma" is square root of (1- v^2/c^2) than your formula is nothing to do with Einstein's rule for Lorentz gamma-factor in the presence of the external field, which is
"1/square root of (1- v^2/c^2) should be changed by
(1 + U/c^2)/square root of (1- v^2/c^2)".
In that case, please forget all what I told you about your formula, because it simply is a WRONG one.

2. You somehow forget notice that Waite's formula and your one contain a different x: yours has x, Waite's has x'. I hope you understand a huge difference in consequences arising of that fact.

3. Please, do not substitute necessity of presenting of a straight math prove with philology like "I believe", "I know" etc. Mr.Waite had delivered his formula from his own "Lorentz-like transformations" that are not used nor by Einstein, nor by Lorentz. Hence even Wait's formula can not be useful for us in our dispute: can you or can not derive your formula from Lorentz transformations?

4. Last time I'm trying to attract your attention to the fact that Einstein's formula comes from redefinition of the metric tensor of 4D-space under influence of the external field - the basic idea of GRT.

5. And last comment. I'm trying to help you to manage your ideas. I do not want to spend our time for philological discussion. So, either you will represent a straight and clear derivation of your formula from Lorentz transformation, or this is a LAST TIME I'm mentioning it.

Dear dristam,
1. If you assert that your "gamma" is square root of (1- v^2/c^2) than your formula is nothing to do with .

But I never ever said that anywhere, anytime! When you check back at my posts you will not find a single instance of my saying that! Gamma is the reciprocal of what you show above.

As for the rest of your comments, you are doubtless confused. David's starting formula is just a fancier, more precise, more rigorous way of stating the very same LT that you put forth. Clear the haze my friend; clear out the cobwebs.

And of course... LET'S SEEK A THIRD PARTY'S OPINION.

I repeat, there is no doubt whatsoever about it: (1+gx/c^2)/gamma derives directly from LT. Of course it can! if it couldn't, then the Equivalence Principle would be in grave doubt!

Dear dristam,

1. Let me ones again underline your possibilities in our discussion:
Either you assert that your "gamma" is square root of (1- v^2/c^2) than your formula is nothing to do with Einstein's rule for Lorentz gamma-factor in the presence of the external field, or you agree that your
“gamma”=1/1/square root of (1- v^2/c^2)” and then your formula is a particular case of well-known Einstein’s formula at special condition U =gx with g =const. But in that case your formula has nothing to do with Waite’s formula, at all.

2. You said: “David's starting formula is just a fancier, more precise, more meticulous way of stating the very same LT that you put forth.” Sorry, I do not have interest in Mr. Waite’s “inventions” in classic Lorentz transformations no matter how fancier, more precise, more meticulous they look for you. If you want my personal opinion – all what I read in Mr. Waite’s web site is either well-known facts or absolute nonsense; particularly, his “invention” of substitution of the classic Lorentz transformations by his own formulas to make them applicable to the non-inertial observers are absolute nonsense. But this Forum is not a place where I can discuss this matter.

3. In any case, Waite’s formula has nothing to do with Einstein’s one.

4. Remember, you still own us derivation of your formula from the classic Lorentz transformations.

5. You do not accept my argumentation – wait for a third, fourth, etc, etc opinion. For me case becomes boring.

DW's notation is fancier, more precise, more rigorous. I stand by all my aforeposted claims and the derivation.

It seems patently obvious to me that the Equivalence Principle leads to it being no surprise that Einstein's formula for distant clock rate reckonings along the axis of potential from within a (real) gravitational field... is absolutely identical to a derived member of the Lorentz Group that denotes the equivalent reckoning from within a pseudo-force field, ie. within a fictitious force field, ie. within the grav'l potential gradient engendered by a thrusting/proper acceleration. I stand by this all!

.. But if the asymmetry in the design still bothers you, consider a slight modification of my original drawing .. You have now perfect symmetry here with the length of both rods serving as the trigger span.

TSMID: Are you still subscribed to this thread? I am working fervently on your latest challenge and the new symmetrical drawing of your device. I hope to have a new post on that, the original intent of this thread!

That back & forth with Yuiry about the derivation was divergent.

http://www.physicsmyths.org.uk/imgs/timedilation2.gif

Assume that each rod is 1 light-second long, as measured by a native. Also assume that the rods are made of a rigid material in which the speed of sound is 0.1c. The rods have a relative velocity of 0.866c.

Consider what happens according to an observer in Frame A.

1. The Start contacts touch and the signals begin their travels toward stopwatches A and B.

2. 5 seconds later, the signal activates stopwatch A.

3. 1.154 seconds elapse in A's frame while the right-hand edge of rod B advances across the 1 light-sec length of A's rod. And then the contracted span of rod B must thereafter traverse, until the final Stop contacts meet. That will require an additional .577 secs, as the velocity is still .866c but the span is contracted in half. So the Stop signal occurs after 1.731 seconds, per A's reckoning. And once again, another 5 seconds is required to transmit the signal along the half of A's rod. But those delay times are symmetrical, a late start and a late stop; so they cancel, and 1.731 seconds total is what A's stopwatch will record.

4. The Start signal progressing along rod B travels at .1c relative to rod B as measured from rod B, but this does not hold as measured by an observer in Frame A. To get the signal's relative velocity along rod B, as measured by Frame A, we must use the velocity addition formula. First we find the relative velocity of the signal per Frame A's metric; this is equal to (.866c-.1c)/(1-(.1c)(.866c)) = 0.8386c (the minus sign is used because the signal travels in the opposite direction of Rod B). Then we take the difference between this and the 0.866c velocity of rod B to get .027c, for the relative velocity of the signal with respect to rod B as measured by an observer in Frame A.
At this relative velocity, it takes 9.132 secs for the signal to travel along rod B the .25 light-sec to stopwatch B. Thus 9.132 secs after the contacts touch, stopwatch B starts -- according to Frame A. [ALL of this entire sequence is according to how an observer native to Frame A perceives events.]

5. To find the relative velocity of the later Stop signal to stopwatch B, we use the same procedure as before. This time however the signal is traveling in the same direction as the rod, so we get (.866c+.1c)/(1+(.1c)(.866c))=0.889c for the relative velocity of the signal per Frame A. This gives a relative velocity of 0.023c to rod B as measured in Frame A. Therefore by Frame A's reckoning, it takes 10.864 sec from the time the Stop contacts touch until stopwatch B shuts off.
Add the 1.731 secs it took to travel between Start and Stop contacts (see #2 above), and subtract the 9.132 secs it took before stopwatch B started, and stopwatch B ran for 3.463 sec, per Frame A's reckoning, between the time it started and stopped. Stopwatch B is time dilated according to A, and therefore it should read, according to Frame A, only half the time that elapsed in Frame A for that period, or 1.731 seconds.

To recap:

According to an observer in Frame A, stopwatch B records 1.731 seconds total duration, and stopwatch A also records 1.731 seconds. The readings figure to be equal, even allowing for time dilation and whatnot.

And since the device is perfectly symmetrical, Frame B's reckoning will be identical to A's. By either frame's reckoning, the stopwatches will show identical times upon completion of the trial run, so no contradiction is detected and nothing is truly proven -- but nothing discredited either!

dristam has asked me the following question: "What is the meaning of term a proper accelleration ?" I think the answer will be useful for many of us.
1. The term "proper" is used in SRT to emphasize a value of some quantitative physical feature that was measured by the inertial observer, which sees the physical carrier of this feature in the mechanical state of rest as a whole. Example: a proper time, a proper length, a proper angle, a proper volume, etc.
2. I never used term "proper" in respect to acceleration, but following to explanation 1 one can make conclusion that consistent sense of application of that term to acceleration should be the following one:
a proper acceleration is the acceleration that could be measured by the inertial observer instantly seeing the physical carrier (body) in the state of the mechanical rest as a whole.
3. But it immediately raises a question: "what about motion of the source of this acceleration?" If this source is in the instant rest with accelerating body everything is OK. But if it is not? We certainly know that mutual motion of the interacting bodies causes changes of force between them (In respect with case when both interacting partners are in the state of mutual resting.)
Here we are touching the most mysterious problem of whole contemporary physics: what is the energy of interaction? (Because before notion "force" we have to clarify notion "the potential energy")

before notion "force" we have to clarify notion "the potential energy")That's not possible because potential energy is defined as the path integral over force.

No, tsmid, you are mistaken.
The path integral over force defines the work. For instance, path integral over force of friction defines the work that goes in work against that force; the path integral over force of propulsion defines the work that goes in the kinetic energy of rocket, etc. In all these cases mentioned forces are not potential ones and there does not exist any potential energy. So, these examples should convinced you that notion potential energy and path integral over force are independent notions - second one is more general than the first one. Yes, path integral over potential force can measure how much potential energy, defining that force, was changed between initial and final positions of body, but it does not defines potential energy of body in the given position. We can use path integral over potential force to measure potential of body in the given point only in the academic sense, using the following exact definition of potential:
"the potential of body in the given point is the minimal work that should be done to bring this body from infinity in that point".
And I do not think we should change the subject of this thread. If you are indeed interested in discussion on potential energy - open a new thread.

I do not think we should change the subject of this thread. If you are indeed interested in discussion on potential energy - open a new thread.I have replied to your post above under
http://www.sciforums.com/showthread.php?t=41076&page=2&pp=20