Is time universal? NO (and its proof)

[2inquisitive]

Let me rephrase question (1) to hopefully help understand what I am asking. Take those two light clocks I used as an example. Synchronize them in the same frame of reference, then send one away at a fraction of the speed of light. Will the light clocks remain synchronized? I assume you will say 'no' because Special Theory states time itself dilates due to relative motion. The 'tick' of the light clock is the time it takes the photon to move the 5 meters to the mirror, then 5 more meters back to the detector located beside the laser. In order for the relatively moving light clock to measure a longer 'tick', the speed of light has to be slower than the rest frame clock, or the total 10 meter travel of the photon has to increase relative to the rest frame. Does the ORIENTATION of the photon's direction of travel relative to the rest frame have an effect on the timekeeping of the moving clock?

Question (2). I meant to IGNORE SR velocity time dilations in question (2). Sorry if I was unclear. There is ample evidence that cesium clocks will beat faster the higher there are in the gravitational potential. When you state 'no', I assume you believe a light clock will beat faster when it is higher in the gravitational potential also (farther from Earth). Both light clocks are in the vacuum of space, yet they beat at different rates. Is the 10 meter distance transversed by the photon relatively shorter in one location? Which location has the relatively shorter meter?

 

[Billy T]

Let me rephrase question (1) to hopefully help understand what I am asking. Take those two light clocks I used as an example. Synchronize them in the same frame of reference, then send one away at a fraction of the speed of light. Will the light clocks remain synchronized? I assume you will say 'no' because Special Theory states time itself dilates due to relative motion. The 'tick' of the light clock is the time it takes the photon to move the 5 meters to the mirror, then 5 more meters back to the detector located beside the laser. In order for the relatively moving light clock to measure a longer 'tick', the speed of light has to be slower than the rest frame clock, or the total 10 meter travel of the photon has to increase relative to the rest frame. Does the ORIENTATION of the photon's direction of travel relative to the rest frame have an effect on the timekeeping of the moving clock?

Question (2). I meant to IGNORE SR velocity time dilations in question (2). Sorry if I was unclear. There is ample evidence that cesium clocks will beat faster the higher there are in the gravitational potential. When you state 'no', I assume you believe a light clock will beat faster when it is higher in the gravitational potential also (farther from Earth). Both light clocks are in the vacuum of space, yet they beat at different rates. Is the 10 meter distance transversed by the photon relatively shorter in one location? Which location has the relatively shorter meter?

Now that both (1) & b(2) are using your "light transit clocks" I will say some things that apply to both and let you see if that helps you sort it all out. BTW,. I am not the best person to ask, but here is what I think:
(1)Both frames have c constant and for them the 5m is unchanged by "the other guy's speed"
(2)Both notice that the photon launched toward the mirror on the other ship must travel farther (assuming mirror is more near the front of ship) or less (assuming mirror is more near the rear of ship) than after reflection because the ship was moving while the photon was in flight. These two reversed direction effects do not cancel out. Think of total time needed for rowing a boat up river and then back to the dock you left from. (Compared to rowing that distance in non flowing lake. Do some numbers if you need to.)
(3) For both, the other´s "5m" has contracted compared to his 5m. (This will tend to make the round trip required time less, in contrast to point (2).
(4) If you get someone good at this, like Pete, to do it correctly for you, not me, I think you will find that (2) dominates (3) because, if I have not overlooked something, that must be true for both to note that the others clocks are "running slow" (This is the SR view, but I am not sure of myself when making SRT calculations and too lazy now to even try.)

Sorry - that is about all I can say, but you have set up an interesting and thoughtfull question / thought experiment. If you get it well explaned, tell us.

 

[DaleSpam]

(2)Both notice that the photon launched toward the mirror on the other ship must travel farther (assuming mirror is more near the front of ship) or less (assuming mirror is more near the rear of ship) than after reflection because the ship was moving while the photon was in flight. These two reversed direction effects do not cancel out. Think of total time needed for rowing a boat up river and then back to the dock you left from. (Compared to rowing that distance in non flowing lake. Do some numbers if you need to.)
(3) For both, the other´s "5m" has contracted compared to his 5m. (This will tend to make the round trip required time less, in contrast to point (2).
(4) If you get someone good at this, like Pete, to do it correctly for you, not me, I think you will find that (2) dominates (3) because, if I have not overlooked something, that must be true for both to note that the others clocks are "running slow" (This is the SR view, but I am not sure of myself when making SRT calculations and too lazy now to even try.)

Two observers at different points in a gravity well are not in a symmetric situation. The person lower in the well can drop a ball and see that it accelerates faster than the same ball dropped by the higher person. The higher person will see the lower person's clock running slow and the lower person will see the higher person's clock running fast. There is no ambiguity.

-Dale

 

[2inquisitive]

Now that both (1) & b(2) are using your "light transit clocks" I will say some things that apply to both and let you see if that helps you sort it all out. BTW,. I am not the best person to ask, but here is what I think:
(1)Both frames have c constant and for them the 5m is unchanged by "the other guy's speed"
(2)Both notice that the photon launched toward the mirror on the other ship must travel farther (assuming mirror is more near the front of ship) or less (assuming mirror is more near the rear of ship) than after reflection because the ship was moving while the photon was in flight. These two reversed direction effects do not cancel out. Think of total time needed for rowing a boat up river and then back to the dock you left from. (Compared to rowing that distance in non flowing lake. Do some numbers if you need to.)
(3) For both, the other´s "5m" has contracted compared to his 5m. (This will tend to make the round trip required time less, in contrast to point (2).
(4) If you get someone good at this, like Pete, to do it correctly for you, not me, I think you will find that (2) dominates (3) because, if I have not overlooked something, that must be true for both to note that the others clocks are "running slow" (This is the SR view, but I am not sure of myself when making SRT calculations and too lazy now to even try.)

Sorry - that is about all I can say, but you have set up an interesting and thoughtfull question / thought experiment. If you get it well explaned, tell us.

Thanks, Billy T, but you avoided my questions. I wasn't asking to intermix the two separate situations. Now I will respond to the statements you made.

(1) These examples are an attempt to explain HOW 'c' can remain constant using a light clock.
(2) You are mixing the examples I gave. There is relative velocity involved in my (1) question, but no gravity. There is gravity involved in my (2) question, but dilation due to relative velocity is ignored. But let's address your response involving relative velocity and the light clock. You state if the photon is traveling toward a mirror located in the forward direction of travel it will take a longer time to reach the mirror than a photon that is emitted toward a mirror located at the rear, opposite the direction of travel of the 'ship'. Do you remember how the light clock works? It emitts a photon toward a mirror, which is then reflected back to the clock. The total travel time is one 'tick', or 2.99792458 nano seconds. If the speed of light is a constant in both frames, the only way the clock can tick relatively SLOWER in the moving frame is if the photon has a greater distance to travel. From the rest frame of the distant clock, an 'observer' would determine a photon emitted orthogonal to the vector of the moving clock would have to travel a bigger distance, a 'v' shaped trajectory to the mirror and back to the clock. That is standard STR that slows the 'tick' on the moving clock as viewed from the distant observer.
The reason I asked about 'orientation' of the moving clock is because of this example. Suppose, as you said, the photon in the moving clock was emitted in the same direction as the direction of travel. According to an 'observer' in the rest frame of the distant clock, this photon has to 'catch up' to the moving mirror, increasing the travel time. The SAME photon is then reflected back to the detector, which is 'moving toward' the reflection event. This leg of travel decreases the travel time. The total of the two 'legs' of travel is the same 2.99792458 nano seconds. The moving clock does not beat slower in this orientation, according to the observer in the rest frame. So, according to an observer at rest, does a light clock beat at different rates according to the orientation of the clock?

(3) You state the '5 meters' have contracted according to a distant observer in both my examples. DaleSpam correctly stated the contraction is not symmetric in a gravity well. But he did not address my question of WHICH meter is relatively shorter in the gravity well. I already know what theory states, but let's see if we can make the light clock agree with theory.

I am addressing my questions to Billy T, but ANYONE is welcome to contribute. I have noticed James R has lately refrained from posting except in answering specific questions, mostly mathematical, but I would welcome his input if he feels like engaging in the discussion. I cannot make theory explain my examples, but maybe James R can with his much greater knowledge of theory. I don't get upset when I am shown to be wrong in my thinking, but I don't accept general statements like 'because GR says the meter is shorter deeper in the gravity well'.

 

[DaleSpam]

(2) You are mixing the examples I gave. There is relative velocity involved in my (1) question, but no gravity. There is gravity involved in my (2) question, but dilation due to relative velocity is ignored. But let's address your response involving relative velocity and the light clock. You state if the photon is traveling toward a mirror located in the forward direction of travel it will take a longer time to reach the mirror than a photon that is emitted toward a mirror located at the rear, opposite the direction of travel of the 'ship'. Do you remember how the light clock works? It emitts a photon toward a mirror, which is then reflected back to the clock. The total travel time is one 'tick', or 2.99792458 nano seconds. If the speed of light is a constant in both frames, the only way the clock can tick relatively SLOWER in the moving frame is if the photon has a greater distance to travel. From the rest frame of the distant clock, an 'observer' would determine a photon emitted orthogonal to the vector of the moving clock would have to travel a bigger distance, a 'v' shaped trajectory to the mirror and back to the clock. That is standard STR that slows the 'tick' on the moving clock as viewed from the distant observer.
The reason I asked about 'orientation' of the moving clock is because of this example. Suppose, as you said, the photon in the moving clock was emitted in the same direction as the direction of travel. According to an 'observer' in the rest frame of the distant clock, this photon has to 'catch up' to the moving mirror, increasing the travel time. The SAME photon is then reflected back to the detector, which is 'moving toward' the reflection event. This leg of travel decreases the travel time. The total of the two 'legs' of travel is the same 2.99792458 nano seconds. The moving clock does not beat slower in this orientation, according to the observer in the rest frame. So, according to an observer at rest, does a light clock beat at different rates according to the orientation of the clock?

That is a very interesting question. When we are transfering between frames we want to have 3 directions of space and 1 direction of time. So that is 4 degrees of freedom. However, by symmetry we cannot treat the two directions perpendicular to the direction of travel differently. So that is 3 degrees of freedom, time, direction parallel, and direction perpendicular.

Now, time cannot depend on the orientation or it would not be orthogonal, but we can have separate length contractions in the parallel or perpendicular directions. So the Lorentz transform dilates time and contracts distances in the parallel direction to account for exactly the effect you mention. In other words, time dilation explains the difference in tick rates of light clocks oriented perpendicular to the travel. As you noticed, once you do that it still doesn't correctly explain light clocks oriented parallel to the direction of travel, so parallel length contraction is used. Now any light clock in any orientation works correctly.

But that is only 2 degrees of freedom used, that means that there is one more degree of freedom left. In fact, the Lorentz transform is not the only transform that preserves c between frames. There is also the Voigt transform (http://en.wikipedia.org/wiki/Woldemar_Voigt) which was actually developed prior to the Lorentz transform. The Voigt transform is basically the Lorentz transform divided by gamma. It has length expansion in the orthogonal directions and twice as much time dilation. It can explain light clocks (oriented in any direction) just as well as the Lorentz transform, but experimental measures of time dilation agree with the Lorentz transform.

(3) You state the '5 meters' have contracted according to a distant observer in both my examples. DaleSpam correctly stated the contraction is not symmetric in a gravity well. But he did not address my question of WHICH meter is relatively shorter in the gravity well. I already know what theory states, but let's see if we can make the light clock agree with theory.

I am addressing my questions to Billy T, but ANYONE is welcome to contribute. I have noticed James R has lately refrained from posting except in answering specific questions, mostly mathematical, but I would welcome his input if he feels like engaging in the discussion. I cannot make theory explain my examples, but maybe James R can with his much greater knowledge of theory. I don't get upset when I am shown to be wrong in my thinking, but I don't accept general statements like 'because GR says the meter is shorter deeper in the gravity well'.

I'm sorry that I can't you here. My GR is very weak and the answer appears to require the full math of curved spacetime (which I don't yet know). So all I can say is 'because people who know the GR math say the meter is shorter deeper in the gravity well' Sorry. Perhaps one of the real experts will have a good explanation. I really wish my local library carried "Gravitation" but I am just not motivated enough to actually buy it.

-Dale

 

[Billy T]

...the photon in the moving clock was emitted in the same direction as the direction of travel. According to an 'observer' in the rest frame of the distant clock, this photon has to 'catch up' to the moving mirror, increasing the travel time. The SAME photon is then reflected back to the detector, which is 'moving toward' the reflection event. This leg of travel decreases the travel time. The total of the two 'legs' of travel is the same 2.99792458 nano seconds. The moving clock does not beat slower in this orientation, according to the observer in the rest frame. ...

Part I made bold in your text is not correct.
You did not do a numerical example, as I suggested, to understand that the extra time on, “forward leg” is not compensated for by less time leg on the rearward leg, i.e. the round trip takes longer, so I do it for you:

Assume ship moving at 0.9c and that if it were stationary it would take 10 units of time to make the one-way trip (20 time units for one of your round trip “ticks.”)

Thus for the observer NOT on the ship, it takes 10 time units of the ship for photon to get to where the mirror was when photon was launched. Lets call the launch point x=0 then after 10ship time units the photon is at x=10, but moving at 0.9c the mirror is now at x=19.

For the photon to get to x=19. will take another 9 ship time units, during which time the mirror movers to x= 19+0.9x9 = 27.1 the photon still has 27.1-19=8.1 length units to go to catch the mirror. I.e. it has already used up 7.1 more time units than the total round tip time of 20time units and still has more than 80% of the initial forward distance to travel.

(Hint: Use an infinite geometric series summation to compute the total time it takes to get to the forward moving mirror.)

Now in the observer’s frame the distance from source is not the 5m, but less, due to SRT’s “length contraction effects.” So time lapsed on observer's own clock is not as great as above ananlyis suggests, but it still, even in the observer’s frame clock time, takes much more than 10 time units, so just as SRT predicts, the rate of ticks is slower in the moving frame than in the local clocks.
_________________________________________
There is a usually ignored quadratic term in the SRT theory that is significant when the photon’s path is orthogonal to the velocity (and the linear term is thus zero.) You can see this in that from the observer’s POV the photon, although returning to the detector near the source still, does not return to the same point in the observer’s reference frame. (Makes a V path as you state.) Both theses paths are longer than the 5m as they are the hypotenuse of two right triangle with one leg of 5m and the other leg directly proprtional to the product of speed. and time to get to the mirror. So once again the moving clock ticks slower as the photon must travel the longer path of the hypotenuse twice, not 5 + 5 = 10m as it does on the moving ship.

 

[2inquisitive]

"by Billy T:

So the Lorentz transform dilates time and contracts distances in the parallel direction to account for exactly the effect you mention. In other words, time dilation explains the difference in tick rates of light clocks oriented perpendicular to the travel. As you noticed, once you do that it still doesn't correctly explain light clocks oriented parallel to the direction of travel, so parallel length contraction is used. Now any light clock in any orientation works correctly."
===============================================================

I haven't much time now as I have to leave on a trip. Billy T, think your expaination through again, please. How does contracting the length in the direction of travel cause the moving clock to tick SLOWER. That would cause a shorter distance the photon has to travel and a relatively FASTER beating clock in that direction. I get back to the other stuff later.

 

[Billy T]

...How does contracting the length in the direction of travel cause the moving clock to tick SLOWER. That would cause a shorter distance the photon has to travel and a relatively FASTER beating clock in that direction. I get back to the other stuff later.

No. you read what I demonstrated in numertical example again (assuming you did at least skim it.)

summary of what was shown: The on board observer notes 20 time units for the round trip. Not yet including the contraction or time dilation, the distant observer notes that 27.1 ship time units have passed and yet more that 80% of the 5m trip only to the mirror still remains! (Because the mirror is running ahead of the advancing light at 0.9c the light is gaining on the mirror at only 0.1c so it takes 100 time units to catch the mirror instead of just 10)

Now when contraction is included, but still not assuming time dilation as that is what we are proving, it takes less than 27.1 time units to travel to the same (less than 20% waypoint) but still much more that the 20 time units to reach the mirror and reflect. Thus even if it took zero time to return, the round trip takes longer than 20 time units. I.e. the moving clock is running slow for the distant observer, as SRT states.

 

[2inquisitive]

Sorry, Billy T, I was in a hurry and pasted DaleSpam's comments under your name.
Let's go to your comments now.

"Assume ship moving at 0.9c and that if it were stationary it would take 10 units of time to make the one-way trip (20 time units for one of your round trip “ticks.”)

Thus for the observer NOT on the ship, it takes 10 time units of the ship for photon to get to where the mirror was when photon was launched. Lets call the launch point x=0 then after 10ship time units the photon is at x=10, but moving at 0.9c the mirror is now at x=19."

I agree with you up to this point, by using your example of .9c. Light travels 10 meters in 2.99792458 ns. However, in my example the mirror was 5 meters from the laser, but we will use your new distance. The mirror traveled another 9 meters while the photon was in flight, according to the distant observer. So, the photon will contact the mirror after it has traveled 19 meters. There is no additional distance to travel, contracting the frame only makes the photon reach the mirror sooner. Remember, I am using LIGHT to time my clock. This is what throws a monkey wrench in your example. The speed of light does not slow, according to the ground observer, so you can't state the clock is already ticking slower than the distant observer's clock and add additional time because the colck is ticking slow. Don't you get the reason WHY I used the light clock? I know the standard SRT exercise, this is my example to show you where the STR 'trick' is pulled. I have known it for a long time, but it is very hard to get those that simply overlook the 'trick' to see it. I once compared it to a magician's slight-of-hand. The return trip for the photon is, of course, much shorter because the laser/detector is approaching the photon-in-flight.

 

[geistkiesel]

Let me rephrase question (1) to hopefully help understand what I am asking. Take those two light clocks I used as an example. Synchronize them in the same frame of reference, then send one away at a fraction of the speed of light. Will the light clocks remain synchronized? I assume you will say 'no' because Special Theory states time itself dilates due to relative motion. The 'tick' of the light clock is the time it takes the photon to move the 5 meters to the mirror, then 5 more meters back to the detector located beside the laser. In order for the relatively moving light clock to measure a longer 'tick', the speed of light has to be slower than the rest frame clock, or the total 10 meter travel of the photon has to increase relative to the rest frame. Does the ORIENTATION of the photon's direction of travel relative to the rest frame have an effect on the timekeeping of the moving clock?

2inquisitive,
If you look at the photon moving in the direction of the frame and compare the motion with a photon moving in the oppositie direction of the 4emitted left moving photon then the picture willclear. In the time the left photon moves a distance ct the frmae move a distance vt and the left mirror and photon have met. The righ moving photon, also moved a distance ct yet is a distrance 2vt from the right moving morror. The photon will cross the distance ct' equal to 2vt plus a distance the frame moves in the interim, vt', or ct' = 2vt + vt', or t' = t(2v)/(c - v). This is not SRT of course, but you can see if the frame is at rest wrt the vacua, the embankment, the round trip is of a shorter duration. the time difference for the round trip motion is t', which is not equal to the time dilation calculated using SRT.

If the mirrors are opriented tranzverse to the forection of motion you have a different situation. There are two views on this, 1). (most popular) is that the reflected light is "carried along" by the moving frame giving the SRT explanation for the time dilation. This view is seen in most diescripotions of the MM experiment. 2).back the light beams reflect 180 degrees along the incoming trajectory, therefore the moving frame runs away from the reflected beam that maintains its position in real soace (rememeber the SOL is independent of the motion of the light source). In the second case the time of flight of the outgoing and reflected lights are identical.

This runnuing away from the beam, suspended in space, can be compensated for. As the wall in the rear of the compartment is about to absorb the at rest light trajectory, the beam is reemitted at the forward end of the compartment. It seems to move to the rear of the compartment (the frame is actually in motion) and opposite to the direction of motion of the frame, always keeping a "live" reflected beam in which to measure time.

Question (2). I meant to IGNORE SR velocity time dilations in question (2). Sorry if I was unclear. There is ample evidence that cesium clocks will beat faster the higher there are in the gravitational potential. When you state 'no', I assume you believe a light clock will beat faster when it is higher in the gravitational potential also (farther from Earth). Both light clocks are in the vacuum of space, yet they beat at different rates. Is the 10 meter distance transversed by the photon relatively shorter in one location? Which location has the relatively shorter meter?

I understand your gravity potential and have a conditional agreement. Only when acceleration affects can be eliminated experimentally can I join with gravity potential as a time dilation explanation. In order to reach "higher orbits" a larger acceleration time is required, that also changes the gravity potential, so the jury is still out on this one, at least for me.

In your last part of the question are you asking about the two directions of motions of the light maintianing symmetry where SRT claims that time dilation and frame contraction account for the lights (emitted in oppositie directions)arriving simultaneously at equally spaced clock/mirrors(regarding srt)? If this is so, I was wondering about this myself -w here did the symmetry go?

Geistkiesel​

 

[2inquisitive]

I'm not sure what you are getting at, geistkiesel. There are two separate questions, question (1) is pertaining to time dilation due to RELATIVE VELOCITY (SR). Question (2) is a question pertaining to gravitational potential effects ONLY (GR), so I advised to ignore SR velocity effects. Is that what you were questioning? Billy T mixed the two examples together, I assume because I used a light clock in both questions.

 

[geistkiesel]

I'm not sure what you are getting at, geistkiesel. There are two separate questions, question (1) is pertaining to time dilation due to RELATIVE VELOCITY (SR). Question (2) is a question pertaining to gravitational potential effects ONLY (GR), so I advised to ignore SR velocity effects. Is that what you were questioning? Billy T mixed the two examples together, I assume because I used a light clock in both questions.

The reference is to a paper where using a Sagnac correction the MM experiment was looked at differently.
http://qem.ee.nthu.edu.tw/f1b.pdf

Here no orbital motion could be detected using GPS systems. The author concluded that a local "gravity ether" explains the results, which is where you are heading (or are already there). Taking both clocks to deep space, the at rest clock will measure a shorter time of flight to reach the mirror than the moving frame, but the difference can be easily calculated and you already have determined the difference in the one way time of flight. The difference in the two round trip times can lead directly to the measurment of the velocity difference, or if known, then can lead to a measurement of the SOL.

Untill I can experimentally exclude acceleration as the cause for apparent time dilations I remain on the fence post regarding those claiming the gravity potential explanation. Without any attempt to compromise the differences of opinion, it seems intuitive that both affects are possible.
Geistkiesel​

 

[2inquisitive]

Yes, you are correct that I suspect something that could be called a 'local gravity ether'.
I posted befor you made the edit, of course, with just references to my post. I'll read the paper. Thanks, geistkiesel.

 

[2inquisitive]

Oh, by the way, it has been shown in experiments that acceleration alone, at least the kind of acceleration produced by a centrifuge, does not slow clocks, even when the acceleration forces are several magnitudes of g. I have seen nothing on straight-line acceleration results, though. I have 'heard' that clocks on rocket sleds lose time, but I have found no confirmation.

 

[geistkiesel]

Part I made bold in your text is not correct.
You did not do a numerical example, as I suggested, to understand that the extra time on, “forward leg” is not compensated for by less time leg on the rearward leg, i.e. the round trip takes longer, so I do it for you:

Assume ship moving at 0.9c and that if it were stationary it would take 10 units of time to make the one-way trip (20 time units for one of your round trip “ticks.”)

Thus for the observer NOT on the ship, it takes 10 time units of the ship for photon to get to where the mirror was when photon was launched. Lets call the launch point x=0 then after 10ship time units the photon is at x=10, but moving at 0.9c the mirror is now at x=19.

For the photon to get to x=19. will take another 9 ship time units, during which time the mirror movers to x= 19+0.9x9 = 27.1 the photon still has 27.1-19=8.1 length units to go to catch the mirror. I.e. it has already used up 7.1 more time units than the total round tip time of 20time units and still has more than 80% of the initial forward distance to travel.

(Hint: Use an infinite geometric series summation to compute the total time it takes to get to the forward moving mirror.)

Billy T,
You do not need any ifinite geometrical series. Look at the problem from the lights point iof view. The light sees 10 meters it must cross to get to the opposite mirror, plus a short distance the frame moves in the interim. The light moves ct = 10m + vt, or t = 10m/(v - c). If v = .9c, then t = 10/.3 x 10^8 = 33.33 x 10^-8. This must be added to the reflected time of flight, t1 = 10/(c + v)= 10/1.9x3 X10^8 = 1.754 10^-8 and added together = 35.087 x 10^-8.

For the clock at rest 20/3 = 6.667 x 10^-8
or a difference of 28 x 10^-8 seconds.
I USED 10 METERS BETWEEN THE SOURCE AND THE MIRROR DUE TO 2INQ'S REVISION.

Now in the observer’s frame the distance from source is not the 5m, but less, due to SRT’s “length contraction effects.” So time lapsed on observer's own clock is not as great as above ananlyis suggests, but it still, even in the observer’s frame clock time, takes much more than 10 time units, so just as SRT predicts, the rate of ticks is slower in the moving frame than in the local clocks.
_________________________________________
There is a usually ignored quadratic term in the SRT theory that is significant when the photon’s path is orthogonal to the velocity (and the linear term is thus zero.) You can see this in that from the observer’s POV the photon, although returning to the detector near the source still, does not return to the same point in the observer’s reference frame. (Makes a V path as you state.) Both theses paths are longer than the 5m as they are the hypotenuse of two right triangle with one leg of 5m and the other leg directly proprtional to the product of speed. and time to get to the mirror. So once again the moving clock ticks slower as the photon must travel the longer path of the hypotenuse twice, not 5 + 5 = 10m as it does on the moving ship.

Thisw method is in serious doubt. If the reflection is supposed to be 180 degrees, the the 'v shapped trajectory' is impossible. The light reflecting 180 degrees will use the same outgoing trajectory as the incoming trajectory with no distortion in the taraveled distance. To use the 'v shapped' trajectory one has imposed a momentum comoponent in the direction of travel of the frame which increases the SOL considerably to a value greater than c.
Geistkiesel​

Why aren't the [sup]8[/sup] working?

 

[2inquisitive]

Oh, I've read that paper before. It is actually useless, as the author starts his paper from the "synchronized atomic clocks" premis. It offers no explaination of HOW the clocks were synchronized, only that the Sagnac effect is noted from that prospective.
All GPS literature I have read recognizes the Sagnac effect has to be taken into account. He also makes a mistake in referring to the 'ECI' frame as the one where the Sagnac effect is evident. I will cut & paste from the paper and explain:
"Thereby, the wave propagation in GPS can
be viewed in a classical way, if an ECI frame, rather than the ECEF or any other frame, is
selected as the unique propagation frame. Thus the wave propagation in GPS depends on
the Earth’s rotation, but is entirely independent of the Earth’s orbital motion."

In the Earth Centered Inertial frame, THE EARTH DOES NOT ROTATE. The satellite constellation orbits around the Earth, but the Earth is 'assumed' stationary in this frame. I know it sounds counter to what I am about to say, but the Earth rotates in the Earth-centered Earth-fixed (ECEF) frame. 'Earth-centered' in both frames simply means a frame with the Earth at its center. 'Earth-fixed' refers to a LOCATION on Earth's surface that rotates with the Earth, not that the Earth is 'fixed' and does not rotate. In the 'ECI' frame, the Earth's surface is not even represented, just the satellite constellation rotating around a point, the center of the Earth. The Sagnac effect CANNOT appear in that frame since the surface is not represented. Both the ECI frame AND the ECEF frame have to be combined to represent orbiting satellites traveling over a rotating Earth. The article had to have been written by a student not familiar with GPS. By the way, I have been doing some more reading of papers related to the new Galileo satellite positioning system being put into service by the European nations.

 

[DaleSpam]

I haven't much time now as I have to leave on a trip. Billy T, think your expaination through again, please. How does contracting the length in the direction of travel cause the moving clock to tick SLOWER. That would cause a shorter distance the photon has to travel and a relatively FASTER beating clock in that direction.

Here is a concrete example. I am sure you have seen similar diagrams in the past (click the tumbnails to see the full images), but let's go step by step through them anyway. In this example I am using 2 light clocks that are 4 sc (lightseconds) long in our frame. The light clocks are moving at .6 c relative to our frame; one is oriented parallel and one perpendicular to the motion. So now let's look at the operation of these two clocks in our frame and in the frame of the clocks themselves. As I mentioned previously we have 3 degrees of freedom, time, parallel distance, perpendicular distance. So lets take the Lorentz approach, assume that perpendicular distances are unaffected, and start with the perpendicular clock.


In this one the light clock is oriented perpendicular to the direction of motion. Due to the motion of the clock the light travels 10 sc in our frame. Since perpendicular distances are unaffected it travels 8 sc in its proper frame. So the time of the clock has dilated to 10 s per tick in our frame instead of 8 s per tick in its proper frame. The ratio of our time to the clock's time is 10/8 = 1.25


In this one the light clock is oriented parallel to the direction of motion. Due to the motion of the clock the light travels 12.5 sc in our frame. We know from the perpendicular clock that time in the clock's frame is running slow by a factor of 1.25, so our 12.5 s is equal to 10 s of proper time for the clock. If a light clock ticks once every 10 s then it's proper length is 5 sc. So the parallel length of the clock has contracted from 5 sc in its proper frame to 4 sc in our frame. The ratio of our parallel length to the clock's parallel length is 4/5 = 1/1.25

I did my best to lay this out clearly and correctly, but I may have made an error somewhere. I hope it shows how time dilation and length contraction are both required in order to properly explain the function of light clocks. As I mentioned previously Voigt started with the parallel clock and assumed that parallel distances were unchanged. This lead to a greater amount of time dilation and a perpendicular length expansion, but experimental measures of time dilation agree with the Lorentz transform.

-Dale

 

[2inquisitive]

Thanks, DaleSpam, for your obvious effort in making this post. I, of course, have seen similar digrams before. But they don't describe my gedankin. I am not aware of a method to illustrate my gedankin with diagrams. The problem with the diagrams you posted is the don't illustrate the proper POSITION of the distant (at rest) observer. When a mathematician uses the Lorent transforms to calculate time dilation and length contraction, where is the observer located? He is at a rest point 'A' viewing a 'clock' (B) receed from him in a straight line, directly away from him. A photon from the light clock would appear to be traveling left-to-right for a light clock oriented perpendicular to the direction of travel. He is viewing the clock from the 'end' of the frame while the clock is receeding from his viewing location. The result would be the same as your first diagram. I don't think it is possible to illustrate this in a diagram.

The photon from the light clock oriented in the direction of travel would appear to receed from observer 'A' while traveling toward the mirror, and then travel back toward the observer after the photon was reflected from the mirror, both motions in-line with the observer's line-of-sight. Why do I insist on these orientations to illustrate a problem with the Lorentz transforms? Because these are the positions the Lorent transforms are calculated from. Your first diagram places the observer BESIDE the direction of travel of the light clock, time dilation will be evident, but NO length contraction will be seen from that perspective.

by DaleSpam:

"We know from the perpendicular clock that time in the clock's frame is running slow by a factor of 1.25, so our 12.5 s is equal to 10 s of proper time for the clock. If a light clock ticks once every 10 s then it's proper length is 5 sc."
==============================================================

This quote is in reference to the second diagram. No, we do not know time in the clock's frame is running slow. If this were the ONLY orientation of the light clock viewed, how would you determine that? Besides, the only thing shown in the first diagram was that a photon has to tranverse a bigger distance because of THE SAGNAC EFFECT. The laser/detector moved while the photon was in flight. Since the first diagram was using a PHOTON to measure time, if you state 'time slowed', then you are stating the speed of light is slower in that frame. Again, this is why I use a light clock. I'll stop at this point for now.

 

[2inquisitive]

Let me give a little exercise that I read online that is similar to the 'time dilation' gedankins often given in Special Theory.

Let's start with the basic premise of a plane flying from New York to San Francisco and back, a distance of about 3000 miles each way. The plane travels at an air speed of 500 miles per hour. Assuming no wind at the planes altitude, it will take the plane 6 hours to travel each 'leg' of the trip, for a total of 12 hours both ways.

Now, assume a wind speed of 100 miles per hour, the wind blowing from the west to the east. From New York to San Fran, the plane will be traveling in a head wind, making its speed relative to the ground 500 - 100 mph, or 400 mph. It will take the plane 7.5 hours to get to San Fran. Traveling back to New York, the plane will travel 500 + 100 mph relative to the ground, or 600 mph. This 'leg' will take 5 hours to complete. 7.5 plus 5 equals 12.5 hours.

Does the .5 hour difference in travel time mean 'time' was dilated in the second example with the 100 mph wind?

 

[MacM]

No I'm not. What part of my words "different cars" and "his group of Mustangs is different from mine" don't you get (or don't you think I get). I understand your idea. The two Mustangs that me and my buddy see are completely different Mustangs. One is blue and one is red, but even if I were colorblind they would still be different Mustangs. Too bad it is hard to read those VIN's zipping by at 50 mph, that and the windows are all tinted so we can't see the interior. Your idea is basically that the fact that they are in more or less the same spot is just confusing other people into inventing something silly like the special color-changing paint instead of being reasonable like you and realizing that they are different Mustangs.

Frankly I am at a loss for words. I didn't bring up colors. You did. You have specifically made them red and blue, which I thinks suggest doppler shift.

For the last time forget colors. Forget VIN's. The Mustang going 80 Mph (50 relative to you at 30 Mph) does not exist at all to the car going 50 Mph (another Mustang is going 100 Mph (50 Mph to it) and vice versa.

These Mustangs to not exist in but one frame and that is the frame where they are going 50ph relative to any observer.

Why not? Again relativity is not based on any properties of the identity of light, just its motion. You aren't changing the motion so you won't be changing the conclusions of relativity.

-Dale

False. Relativity is based on the assumption that these are the same mustangs and to have a constant relative veloicty to different observers that is where relativity comes in. Since these are not the same Mustangs there is no need for relativity.