Time dialtion tick tick skeptick

[Skeptick]

Please read the attached document and images. The other images can be found in the next post. Please feel free to critisise what I propose.

 

[Skeptick]

next images

here are the rest of the images for my original post.

 

[Skeptick]

revised document

Sorry folks I have made some minor amendments to the document as it contained some grammatical errors, it is it now easier to read and understand. The images all remian the same.

 

[Skeptick]

e_Freddo

The first person to find a flaw in my reasoning wins a zero calorie e_freddo :)

 

[JesseM]

The flaw is that you seem to be unaware of the relativity of simultaneity, which says that there is no frame-independent definition of what it means for clocks to be "synchronized"--if two clocks are synchronized in their own rest frame, they will be out-of-sync in a frame where they are moving. This is a natural consequence of the fact that each frame assumes light moves at c in that frame, so that clocks at rest in this frame can be synchronized using this assumption...if I am on board a rocket ship and I want to synchronize two clocks at the front and back of the ship, I can just set off a light flash at the midpoint of the two clocks, and then set each clock to read the same time at the moment the light from the flash reaches it. But if in your frame the rocket is moving forward, then naturally if you assume the light from the flash moves at c in both directions in your own frame, then you must say the light reaches the back clock before the front one, since the back clock is moving towards the point where the flash was set off, while the front clock is moving away from that point. So, my synchronization procedure is going to leave the two clocks out-of-sync in your frame.

A useful formula in SR is that if two clocks are synchronized in their mutual rest frame, and the distance between them in their rest frame is L, then in a frame where the clocks are moving at speed v along the axis between them, the time on the back clock will be ahead of the time on the front clock by vL/c^2. So although you're right that in fig. 2, the time as measured in this frame for the light to go from the green end to the red end will be L^/(v + c), where L^ is the Lorentz-contracted distance between the clocks (L^ would be equal to $$L * \sqrt{1 - v^2/c^2}$$) and v is the speed of the clocks in this frame, this does not mean that the time on the green clock when the light leaves it will differ from the time on the red clock when the light hits it by L^/(v + c)...instead you must take into account both the fact that the red clock was ahead of the green clock by vL/c^2 when the light departed from the green clock, and the fact that both of these clocks were ticking more slowly in this frame by a factor of $$\sqrt{1 - v^2/c^2}$$ due to time dilation, so that even though a time of L^/(v + c) passed in this frame the two clocks only ticked forward by $$\sqrt{1 - v^2/c^2}$$ * L^ /(v + c), and plugging in L^ = $$L * \sqrt{1 - v^2/c^2}$$ this means they ticked forward by L * (1 - v^2/c^2) / (v + c). Simplifying a little:

L * (1 - v^2/c^2) / (v + c) =
L * (1/c^2) * (c^2 - v^2) / (v + c) =
L * (1/c^2) * (c + v)*(c - v) / (c + v) =
L * (c - v) / c^2.

Now, if both clocks ticked forward by L * (c - v) / c^2 in the time between the moment the light left the green end and the moment it reached the red end, and the clock on the red end was ahead of the clock on the green end by vL/c^2, this means the time on the clock on the red end when the light reaches it will be greater than the time on the clock on the green end when the light departed it by:

[L * (c - v) / c^2] + [vL/c^2] =
[(cL - vL)/c^2] + [vL/c^2] =
cL/c^2 =
L/c.

Of course, this is exactly the same difference you'd expect if you calculated things from the perspective of the two clocks' rest frame, where the distance between them was L and the light moved at c between the green end and the red end.

A similar analysis could show that in fig. 3, where the green end is now positioned on the back end instead of the front end (relative to the direction of motion), although it will indeed take longer for the light to go from one end to the other in the frame where the clocks are moving, this frame will nevertheless predict that the difference between the time on the green clock when the light leaves it and the time on the red clock when the light arrives will still be L/c, because of the way that both clocks are ticking slowly and the clock on the green end is ahead of the clock on the red end.

 

[Dale]

Hi Skeptick,

Many people will not download .doc files or any other type of file that could contain a macro. I would recommend simply posting your comments directly rather than through a word file if you want a wider review.

That said the mistake or flaw is almost always neglecting the relativity of simultaneity.

 

[robphy]

Hi Skeptick,

Many people will not download .doc files or any other type of file that could contain a macro. I would recommend simply posting your comments directly rather than through a word file if you want a wider review.

I wonder... are attachments uploaded to PF scanned for virii or other malware?

 

[JesseM]

I've already downloaded everything so I'll post the text myself, although if you want to see the diagrams you'll have to download them:
(edit: I uploaded the diagrams to imageshack so you could also see them that way--http://img119.imageshack.us/img119/122/fig1yg2.jpg , http://img134.imageshack.us/img134/323/fig2xf1.jpg , http://img223.imageshack.us/img223/1457/fig3qg7.jpg , http://img248.imageshack.us/img248/4862/fig4ka5.jpg , and http://img104.imageshack.us/img104/7885/fig5xd7.jpg )

By Ross Blenkinsop
Date 29 October 2007

_____________________
Time dilation
_____________________
Assumptions
_______________________________________

The speed of light is the same in all frames of reference and is C.
2 identical clocks located in a non accelerating frame of reference will keep exactly the same time.
2 identical clock located in a non accelerating frame of reference will be the same length, so long as they are oriented 180 degrees or 0 degrees in relation to each other.
A clock can be calibrated using the time taken for a photon to travel over a constant length.
The clocks in fig 1, fig 2 and fig 3 , fig 4 and fig 5 are all identical in the same frame of reference.

Fig 1
______________________________________

Depicts a clock of length L
At one end there is a detector (red) at the other end there is a laser (green)
At the red end there is also a clock that is perfectly synchronised with a clock positioned at the green end.
When a photon leaves the laser located at green the clock at green stops. When the photon eventually arrives at red the clock located at red stops. The time difference between the two clocks is noted, say 1 second if the clock is 300,000,000 meters long.
The time taken for a photon to travel from green to red in fig 1 will be t = L/C where C is the speed of light, L is the length of the clock and t is time difference.
A third clock is calibrated off the time difference between the two clocks. This third clock shows the true time.

In fig 1, if L = 300,000,000 meters, C = the speed of light 300,000,000 m/s then the third clock will be calibrated to tick once every second. T = 1 = 300,000,000/300,000,000

Fig 2
_______________________________________________

Now we have the same system of clocks within a moving frame of reference that is moving to the right at velocity V.

At some time the green laser will be located at point c and the red detector will be located at point a. At some time later the red detector will be located at point b and the green laser will be located at point d, so the entire system moves to the right.

The length of the clock will be L^ …L hat which will be contracted with respect to L as lengths contract in a moving frame of reference.

As time will be different in the moving frame of reference with respect to time in the stationary frame I will denote it a T^…T hat

From this the following analysis
Length L1 or the length from point a to point b ab = vt^
Length L2 or the length from point b to point c bc = ct^
L^ = L1 + L2
L^ = vt^ + ct^
L^ = t^(v + c)
L^/t^ = v + c
Equation 1 ….t^ = L^/(v + c)

From equation 1 as the velocity of the moving frame of reference increases the time taken decreases ie time goes by faster.

If the length L of the clock in a stationary frames of reference was 300,000,000 meters long then in the moving frame of reference as it will be contracted in length L ie L hat… it will be some value less than 300,000,000. Let's say v is 10,000 m/s then from the above. If v is small compared to C then the length L^ will be close to 300,000,000. As 10,000 m/s is way less than 300,000,000 Ill assume L^ is 300,000,000.

t^ = 300,000,000/(300,000,000 + 10,000)
t^ will be less than 1 and the third clock will be calibrated to tick at every say .99 seconds. The clock in the moving frame of reference is ticking faster than the stationary clock?

As is evident for any value of v, t^, time in the moving frame of reference will be less than T, time in the stationary frame of erefrence.

Fig 3
___________________________________________

Now we have the same system of clocks within a moving frame of reference that is moving to the right at velocity V.

At some time the green laser will be located at point a and the red detector will be located at point c. At some time later the red detector will be located at point d and the green laser will be located at point b, so the entire system moves to the right.

The length of the clock will be L^ same as in fig 2.

As time will be different in the moving frame of reference with respect to time in the stationary frame I will denote it a T^…T hat

From this the following analysis
The distance the photon travels until it strikes the detector is the length from the point a to d which is given by the length from point a to point b plus the length from b to c , plus the length from point c to d
ab = vt
bc = L^ - vt
cd = vt
and the length from a to d is equal to t^C

therefore
t^C = vt + L^ - vt + vt
t^C = vt + L^

using similar reasoning as fig 2

t^ = L^/(C-v)

Intuitively as the velocity increases , time increases.

If the length L of the clock in a stationary frames of reference was 300,000,000 meters long then in the moving frame of reference as it will be contracted in length L ie L hat… it will be some value less than 300,000,000. Let's say v is 10,000 m/s then from the above. If v is small compared to C then the length L^ will be close to 300,000,000. As 10,000 m/s is way less than 300,000,000 Ill assume L^ is 300,000,000.

t^ = 300,000,000/(300,000,000 - 10,000)
t^ will be greater than 1 and the third clock will be calibrated to tick at every say 1.01 seconds. The clock in the moving frame of reference is ticking slower than the stationary clock?

As is evident for any value of v, t^ will be greater than T in the stationary frame of reference.

Fig 4
_______________________________________

In Einstein’s classic experiment he proposed that time in a moving frame of reference slows down. In his experiment he used a clock identical to the clock in fig 1, 2 and 3. His analysis for this clock in a stationary frames of reference is the same analysis as is in fig 1.

Fig 5
________________________________________

Einstein theorized if the clock depicted in Fig 4 was in a moving frame of reference then a photon emitted by the green laser would take the path as depicted in Fig 5 and eventually it would arrive at the red detector. Since the path taken by the photon in fig 5 is considerably longer than the path taken by the photon in fig4 he concluded that time in a moving frame of reference must slow down.

Conclusion
______________________________________

From my analysis I propose the following conclusions:
The measurement of time is dependent on the orientation and direction of travel of the clock, therefore time is a vector not a scalar quantity.

Depending on the orientation and direction of travel of a clock in a frame of reference time may increase or decrease and therefore I conclude there is no absolute reference point for the measurement of time.

Applications
______________________________________

This method of measurement may be used to determine if there is an ultimate frame of reference.

Einstein postulated that there was no experiment that could be conducted in a moving frame of reference (“MFR”) to determine if your are in a moving frame of reference. I conclude using this method you can determine if you are in a MFR.

Einstein postulated as you could not determine if your are in a MFR or not, by experiment, then the force due to gravity was identical to the force by acceleration. As I have demonstrated it is possible to determine if your are in a MFR and therefore the force due to gravity only may be the same as force from acceleration.

 

[Dale]

I wonder... are attachments uploaded to PF scanned for virii or other malware?

Probably. I'm just paranoid.

 

[robphy]

Probably. I'm just paranoid.

I hope so.
I'm paranoid too... but I opened the attachment anyway... [after making sure my anti-virus program is up to date].

Too bad something like http://view.samurajdata.se/ didn't work on that attachment.

 

[Skeptick]

virus

FYI

You cannot catch a virus from a bitmap file .bmp

You can catch a virus from a word doc but only if you have macros enabled. This word doc does not use a macro so you can view it with macros disabled.

Also you can catch a virus from carney so be wary of them as well

 

[robphy]

FYI

You cannot catch a virus from a bitmap file .bmp

You can catch a virus from a word doc but only if you have macros enabled. This word doc does not use a macro so you can view it with macros disabled.

Also you can catch a virus from carney so be wary of them as well

It's better to use standard web formats like .gif or .png or .jpg... which have better support for compression. 100kb is a lot for very simple graphics.

I'm thinking of use either http://www.winfield.demon.nl/" , or some online-service to read .doc files from the internet.

 

[jtbell]

It's better to use standard web formats like .gif or .png or .jpg... which have better support for compression. 100kb is a lot for very simple graphics.

I took "fig 1.bmp" from the first post and converted it to a GIF. Let's see how big it turns out to be after I attach it here... wow, only 1.9 KB!

The difference between 1.9 KB and 139.1 KB may not be important for people with broadband connections, but for someone like me who uses a dialup connection at home, it makes a real difference.

 

[Chris Hillman]

A Degree of Paranoia is Healthy (on the web)

Probably. I'm just paranoid.

As we all should be, unfortunately.

I wonder... are attachments uploaded to PF scanned for virii or other malware?

My understanding is that at PF, user-uploaded images (e.g. gif files to illustrate a post) are not made "live" until they can be checked for content and scanned for viruses by a moderator, but I've seen a number of practices I consider troubling, such as hiding the url of a link using VB code (some browsers can be configured to state the url and even do an ARIN lookup if one mouses over a live link) to a possibly unsafe external website. I am not sure whether all uploaded files are scanned at PF before being made live but they certainly should be. There was a recent discussion about security issues raised by a proposal to loosen restrictions for regular posters at the Homework Help forum in the Feedback forum.

I also caution PF users from believing non-authoritative (or even seemingly authoritative) statements of the form "it is not possible to catch a virus from..."). Unfortunately, closer examination often reveals that someone making such statements simply hasn't enough knowledge of the amazing variety of "exploits" used by criminal phishing gangs. As I wrote in the discussion just mentioned, it is also possible to go too far in the other direction; as always, its a matter of finding the right balance, which comes with knowledge and experience.

 

[Skeptick]

The flaw is that you seem to be unaware of the relativity of simultaneity, which says that there is no frame-independent definition of what it means for clocks to be "synchronized"--if two clocks are synchronized in their own rest frame, they will be out-of-sync in a frame where they are moving. This is a natural consequence of the fact that each frame assumes light moves at c in that frame, so that clocks at rest in this frame can be synchronized using this assumption...if I am on board a rocket ship and I want to synchronize two clocks at the front and back of the ship, .

Forget relavtivity this exeriment is taking place inside a snail

Sorry but I have thought of the synchronisation of the clocks. I would synchronise them thus:
I would place the two clocks as close together as possible. Depending on the physical construction of the clocks this means the actual timing mechnism could be only atoms apart. I would then synchronise the clocks thus ensuring the time difference between the two clocks was virtually zero, or the distnace between atoms divided by C. I would then move the clocks apart to their final positions. As they were synchronised before they were moved apart they will still be synchronised afterwards. All mopvements are as slow as you like say 0.00000000001 m/ 1 billion years.

In my scenario the speed of the MFR 10000 m/sec which is too slow for the effects of relativity to make a difference. Make the MFR's velocity as slow as you like 1 kilomteer per 1 million years if you like the maths remians the same. As the speed of the MFR is not relativisitic the effects of simutenaiety can be ignored the time diffrenece of the two clocks would be negligable.

Secondly if you look at fig 1 I have a clock at the green end and a clock at the red end. As explained the clock at the green end stops when a phton leaves the green end.

If the clock at the green end stopped when a photon from the red end struck it then simultenatoiety may be a problen assuming this whole gizmo is traveling very fast but as the clock at the green end stops when a photon leaves the green end and the clock at the green end is as close as you like to the green end simutenaiety effects can be ignored.

Same for the red end the clock at the red end stops when a ptohon arrives at the red end. The clock is as close as you like to the red end. Again negating simultenaiety.


...next

 

[Skeptick]

If the clock at the green end stopped when a photon from the red end struck it then simultenatoiety may be a problen assuming this whole gizmo is traveling very fast but as the clock at the green end stops when a photon leaves the green end and the clock at the green end is as close as you like to the green end simutenaiety effects can be ignored.

this should read ...

If the clock at the green end stopped when a photon from the red end struck it then simultenatoiety may be a problen assuming this whole gizmo is traveling very fast but as the clock at the green end stops when a photon leaves the green end and the clock at the green end is as close as you like to the green end and the whole gizmo is traveling at the pace of a snail simutenaiety effects can be ignored.

 

[JesseM]

Sorry but I have thought of the synchronisation of the clocks. I would synchronise them thus:
I would place the two clocks as close together as possible. Depending on the physical construction of the clocks this means the actual timing mechnism could be only atoms apart. I would then synchronise the clocks thus ensuring the time difference between the two clocks was virtually zero, or the distnace between atoms divided by C. I would then move the clocks apart to their final positions. As they were synchronised before they were moved apart they will still be synchronised afterwards. All mopvements are as slow as you like say 0.00000000001 m/ 1 billion years.

No, this method won't synchronize them in any universal sense. Are you familiar with the concept of "limits" in calculus? In the limit as the speed of the two clocks relative to one another approaches zero, there will be some frame in which the velocity of both clocks is approaching zero as well (the rest frame of whichever clock is not being changed in the limit), and in this frame and this frame only your method will result in the clocks being arbitrarily close to being perfectly synchronized in the limit as their velocity relative to one another approaches zero. But in some other frame--say, a frame where the first clock is moving at 0.8c and the second clock is moved away from it at 0.800000000000000000000000000000000000000001c, they will not remain close to synchronized if they are given enough time to move some significant distance apart...in fact, if you move them a distance of L apart from one another, in this frame your method will result in them being out-of-sync by an amount that's arbitrarily close to the amount they'd be out-of-sync if you use the conventional Einstein synchronization method involving light-signals. I gave a proof of something basically identical in post #41 of this thread, where someone suggested that you could synchronize two clocks C1 and C2 at a fixed distance apart by moving a third clock C3 very slowly between them, with C1 and C2 set to read the same time as C3 at the moment C3 passed each one. Here's what I said there:

Presumably you want C3 to move very slowly relative to C1/C2 so there is little difference in the time dilation factor between C3 and C1/C2, but the difference in time dilation factors cannot be eliminated entirely. Assume the distance between C1 and C2 in their rest frame is 10 light-second, and the velocity of C3 in their rest frame is v. In this case, in their frame it takes a time of 10/v seconds for C3 to move between the two clocks. But C3 is slowed down by a factor of $$\sqrt{1 - v^2/c^2}$$, so it advances forward by

$$\frac{10 \sqrt{1 - v^2/c^2}}{v}$$

in that time. So, what's the difference between this, the time on C3 as it reaches C2, and the time that C1 reads at the same moment in this frame, namely 10/v? Well, it'd just be:

$$\frac{10(1 - \sqrt{1 - v^2/c^2})}{v}$$

If we take the limit of this quantity as v approaches zero, it will work out to zero (use L'Hospital's rule along with the chain rule of calculus to prove this), which means C2 will get arbitrary close to being synchronized with C1 in this frame as the velocity of C3 becomes very slow compared to c.

But that's just for the C1/C2 rest frame; now consider what happens in another frame where C1 and C2 move at some nonzero velocity, say 0.6c, and C3 moves with some slightly different velocity (0.6c + v)--again, we're going to be taking the limit as v approaches zero. In this frame the distance between C1 and C2 is shrunk to 8 light-seconds.

So if C3 and C1 start at x=0 light-seconds at time t=0 seconds, C3's position as a function of time x(t) is:

x(t) = (0.6c + v)*t

And if C2 starts at x=8 l.s. at time t=0 s, C2's position as a function of time is:

x(t) = 0.6c*t + 8 l.s.

So to figure out when C3 catches up with C2, set them equal:

0.6c*t + vt = 0.6c*t + 8 l.s.

...and solving this for t gives t=(8 l.s.)/v. So, this is the time it takes for C3 to go from C1 to C2 in this frame. Now, since C1 is slowed down by a factor of 0.8 in this frame, it will have elapsed a time of (0.8)*(8 l.s.)/v in this time. And C3 is slowed down by a factor of $$\sqrt{1 - (0.6c + v)^2 /c^2}$$ in this frame, so it will have elapsed a time of:

$$\frac{\sqrt{1 - (0.6c + v)^2 /c^2} * 8 \, l.s.}{v}$$

So, the difference between the reading of C1 and the reading of C3 at the moment C3 reaches C2 will be:

$$\frac{(0.8 - \sqrt{1 - (0.6c + v)^2 /c^2}) * 8 \, l.s.}{v}$$

Now we want to know what this difference will approach in the limit as v approaches zero. Since both the numerator and the denominator of this fraction individually approach zero in the limit as v approaches zero, we can again use L'Hospital's rule, taking the derivative of both the top and bottom and seeing what the new fraction approaches in the limit as v approaches zero...to take the derivative of the numerator we must also use the chain rule again. This gives us the following complicated-looking fraction:

$$\frac{8 \, l.s. * [-(1/2) * (1 - (0.6c + v)^2 / c^2)^{-1/2} * (\frac{-1.2c - 2v}{c^2})]}{1}$$

Which simplifies to:

$$\frac{8 \, l.s. * (\frac{0.6c + v}{c^2})}{\sqrt{1 - (0.6c + v)^2 / c^2}}$$

And if we take the limit of this as v approaches zero, it just turns out to be:

$$\frac{8 \, l.s. * (\frac{0.6c}{c^2})}{0.8}$$

Which is 6 seconds. So, even in the limit as the velocity of C3 relative to C1 and C2 gets arbitrarily small, this method will still leave C1 and C2 6 seconds out-of-sync in this frame (and you're free to imagine that this frame in which C1 and C2 move at 0.6c also happens to be the absolute rest frame of Lorentzian relativity). And not-so-coincidentally, it turns out that if you had synchronized C1 and C2 using the Einstein synchronization convention, they would also be 6 seconds out-of-sync in this frame. So again, this method is useless for absolute synchronization, all it does is to replicate the same type of synchronization as the Einstein convention, which will cause clocks that are in-sync in one frame to be out-of-sync in another (or, if you prefer, will cause clocks that are moving relative to the absolute rest frame to be absolutely out-of-sync).

Incidentally, if you don't trust my calculus, feel free to take the equation I gave earlier for the difference between the reading of C1 and C3 at the moment C3 reaches C2:

$$\frac{(0.8 - \sqrt{1 - (0.6c + v)^2 /c^2}) * 8 l.s.}{v}$$

...and instead of taking limits, just plug in some very small v like v=0.000001c, you should end up with an answer very close to 6 seconds.

In my scenario the speed of the MFR 10000 m/sec which is too slow for the effects of relativity to make a difference. Make the MFR's velocity as slow as you like 1 kilomteer per 1 million years if you like the maths remians the same. As the speed of the MFR is not relativisitic the effects of simutenaiety can be ignored the time diffrenece of the two clocks would be negligable.

But if the velocity v of the clocks in the MFR is negligible, then the difference of L^ /(v+c) from L/c will be negligible too (because the Lorentz contraction will be negligible, and the difference between c and v+c will be negligible). To the degree there is any small difference, it will be exactly compensated for by the small Lorentz contraction, time dilation and lack of simultaneity between the two clocks, in exactly the right way to ensure that, if the green clock stops when the light leaves it and the red clock stops when the light arrives at it, then the difference in readings between the two stopped clocks will end up being precisely L/c.

Secondly if you look at fig 1 I have a clock at the green end and a clock at the red end. As explained the clock at the green end stops when a phton leaves the green end.

If the clock at the green end stopped when a photon from the red end struck it then simultenatoiety may be a problen assuming this whole gizmo is traveling very fast but as the clock at the green end stops when a photon leaves the green end and the clock at the green end is as close as you like to the green end and the whole gizmo is traveling at the pace of a snail simutenaiety effects can be ignored.

The difference in simultaneity means that in the MCR, if the clock at the green end reads a time of T at the exact moment the light departs from the green end (the moment when the clock is stopped), then at that precise moment in the MCR the clock at the red end will read a time of T + vL/c^2, where L is the distance between the clocks in their rest frame and v is the speed of the two clocks in the MCR (this is assuming the clocks have been synchronized using a method that gives the same definition of 'synchronized' as the Einstein clock synchronization convention, which would include your proposal of synchronizing them at a common spot and moving them apart with an arbitrarily small relative velocity).

You are free to assume the the velocity of the two clocks v in the MCR is negligible compared to see, in which case T + vL/c^2 will differ negligibly from T, but the result will be the same--if the clock on the red end also stops when the light from the green end reaches it, and we examine the readings on both clocks after they have stopped, their readings will differ by precisely L/c.

 

[Skeptick]

Lorentz contraction LC

from wikipedia

Length contraction, according to the special theory of relativity, which was formulated in the early twentieth century through the seminal work of Einstein, Poincaré and Lorentz, is the physical phenomenon of a decrease in length detected by an observer in objects that travel at any non-zero velocity relative to that observer

In my example the observer is in the MFR there is no relative movemenet hence no LC so that can be diregarded. The clock is fig 2 would be exactly the same length as the clock in fig three they are both in the same MFR moving at the same velocity the only difference is the clock in fig 3 is 180 degree rotation to clock fig 2

Time dilation same deal ... again wikipedia
Time dilation is the phenomenon whereby an observer finds that another's clock which is physically identical to their own is ticking at a slower rate as measured by their own clock. This is often taken to mean that time has "slowed down" for the other clock, but that is only true in the context of the observer's frame of reference. Locally (i.e., from the perspective of any observer within the same frame of reference, without reference to another frame of reference), time always passes at the same rate. The time dilation phenomenon applies to any process that manifests change over time.

In my example the observer is in the MFR there is no relative movement/Frame hence no Time Dialation so again that can be diregarded


The person in the MFR synchronises the clocks, runs the experiments as shown in fig 1, 2, 3, 4, 5
when experiment in fig 2 is run the observer in the MFR will see time speed up. When experiment 3 is run the observer in the MFR will see time slow down. explain this without time being a vector


An observer external to the MFR would see the same as the person internal to the MFR.

In the experiment depicted by fig 2 an external observer would percieve time to speed up
In the experiment depicted by fig 3 an ext observer would percieve time to slow down

Thankyou for your post it has given me food for thought

 

[JesseM]

from wikipedia

Length contraction, according to the special theory of relativity, which was formulated in the early twentieth century through the seminal work of Einstein, Poincaré and Lorentz, is the physical phenomenon of a decrease in length detected by an observer in objects that travel at any non-zero velocity relative to that observer

In my example the observer is in the MFR there is no relative movemenet hence no LC so that can be diregarded.

I thought you were using the "MFR" to refer to the frame in which the two clocks are in motion, ie the frame where you calculated the time to be L^ /(v + c). If you intend the MFR to be the frame where the two clocks are at rest, then of course if the distance between the clocks is L and light moves at c in this frame, the time for the light to go from the green end to the red end will be L/c. But my analysis shows that if you analyze the situation from the point of view of the frame where the two clocks are in motion--whatever you want to call this frame--then even though the time measured in this frame will be L^ /(v+c), the difference in readings between the two clocks after they both stop will still be predicted to be L/c, because of the way the clocks are ticking slowly and the way the distance between them is less than L and the fact that they are out-of-sync in this frame.

The person in the MFR synchronises the clocks, runs the experiments as shown in fig 1, 2, 3, 4, 5

Again, I'm not entirely clear on what you mean the "MFR" to be--is it the clocks' own rest frame, or the frame in which they are in motion? Your figures 2-5 seem to be drawn from the perspective of a frame where the clocks are in motion, since they show the position of the green and red end being different at the moment the light departs the green end and the moment the light arrives at the red end.

In the experiment depicted by fig 2 an external observer would percieve time to speed up
In the experiment depicted by fig 3 an ext observer would percieve time to slow down

If by "external observer" you mean the observer in the frame where the clocks are moving, I've already shown that this observer will predict the difference in the two clocks' readings will be L/c (although the time as measured in this frame will be different from L/c).

Would you like to go through a numerical example? You can tell me the velocity v of the clocks in the external observer's frame, the distance L between the clocks in their own rest frame, and I can show that because of the way the clocks are slowed down and out-of-sync in the frame where they're moving, the difference in their readings after both stop (assuming the clock on the green end stops at the moment the light departs that end, and the clock on the red end stops at the moment the light arrives at that end) will work out to L/c. This will be true regardless of whether the clocks are moving left-to-right or right-to-left in the external observer's frame, I can do the calculation both ways.

 

[Skeptick]

Ill attempt to clear up one problem at a time

I am in a space shipthat is moving, can I sunchronise two clocks in the spaceship Yes / No?

 

[JesseM]

Ill attempt to clear up one problem at a time

I am in a space shipthat is moving, can I sunchronise two clocks in the spaceship Yes / No?

It depends what you mean by "synchronize"--you can synchronize them relative to a particular frame's definition of simultaneity (like the rest frame of the ship), but you can't synchronize them in any objective frame-independent sense. Every inertial reference frame in relativity has its own separate definition of simultaneity, so if the event of one clock reading 3:00 PM happens simultaneously with the event of another clock at a different position reading 3:00 PM in one frame (meaning that the clocks are synchronized in that frame), then these events are not simultaneous in all other frames--there might be some frame where the event of the first clock reading 3:00 PM was simultaneous with the event of the other clock reading 2:59 PM, for example. This is what is meant by "the relativity of simultaneity".