# Synchronised clocks anylsis help please?

## Main Question or Discussion Point

Synchronising clocks

Referring to fig 1

I want to synchronise two clocks.

I have a space ship travelling at v in the direction of the arrow. In the ship I have two identical clocks one at point A (clock1) and on a point D(clock 2). I also have a light source at point B. Length AB = length BD

I have observers at points E, F, G and H. E and F are at rest with the space ship. G and H are not at rest.

A flash of light is set of by the light source and if it hits both clock simultaneously then they are synchronised. As the light from the light source is travelling perpendicular to the direction of travel of the space ship the light will reach both clocks simultaneously and they’ll be synchronised.

I will look at what the observers see later.

Fig 2 shows how NOT to synchronise clocks as the clocks are equidistant and as they will change position before the flash of light reaches them then the flash will not reach them simultaneously and they will NOT be in sync

Does anyone disagree ?

So I sync two clocks using the method in fig 1

Referring to fig 3

I want to move the clocks, one to the front of the space ship and one to the back and I want them to still be synchronised from the perspective of the observer at E. The velocity of the space ship is V.

I move clock 1 from point A to point A hat at velocity V1.

I move clock 2 from point D to point D hat at velocity V2.

The length from point D to the point D hat is equal to the length from point A to point A hat.

There is an observer at E

Depending on the velocities that the clocks are moved at the observer will see the following:

1. If V1 = V and V2 = V
Clock 1 will be stationary, clock2 will be moving at 2V. therefore clock 2 will slow down clock1 will be ahead of clock 2

2. If V1 > 2V and V2 = V
Clock 1 will be moving faster than clock2. Clock 1 will slow down wrt clock2. Clock 2 will be ahead of clock 1

If V1 < V and V2 = V
Same as 1

If V1 = V and V2 < V
Same as 1

If V1 = V and V2 > V
Same as 1

If V1 = X and V2 = V - X
Both clocks moving at same speed therefore they are synchronised

If V1 = c and V2 = c
As both clocks are move at C both will stop and they will be synchronised when they make it to their final destination.

Does anyone disagree ?

I will do what the observer in the space ship sees next as this gets complicated.

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You create the same exact mistake you have yet to resolve in your last thread and start a new one without even so much as agreeing on where we disagree.

These clock are, again, all in the same ship. There is no velocity except what you imagined it might be. You can replace every V in every equation with 0 and you have your answer.

$$V_1 < V$$ cannot be because 0 < 0 is false. If you want to define another outside frame of reference that measures a velocity wrt the clocks then we can talk about the Relativity of Simultainiety, though I despair.

Your not just disagreeing with Einstein, you are even disagreeing with Newton. Your definition of velocity has been wrong since Galileo Galilei in 1632 and hasn't changed with any theory in modern times.
http://en.wikipedia.org/wiki/Galilean_relativity

the observer at E is not in the same frame of ref as the space ship.

So what you are saying is V1 cannot be different to V

you are the capitain of a ship and you look at the speedo of the ship and it says you are doing 20 kph. You walk from the front of the ship to the back at 8 kph so you are saying your velocity is zero and the velocity of the ship is zero...interesting

not only that you are saying someone watching from the dock will say the velocity of the ship is zero and so is the velocity of the captain...I must be missing something here

JesseM
Synchronising clocks

Referring to fig 1

I want to synchronise two clocks.

I have a space ship travelling at v in the direction of the arrow.
Are you claiming to know the ship is moving at v in some objective sense, or is it just moving at v relative to some observer? If it's just relative to an observer, then are you trying to synchronize the clocks in the observer's frame or the ship's frame? I'm sure by now you realize that different frames have different definitions of what it means for two clocks to be synchronized!

the observer at E is not in the same frame of ref as the space ship.
Oh ok. My mistake.

So what you are saying is V1 cannot be different to V.
Not if they are both sitting in the ship. Though E wasn't in the ship. My mistake.

you are the capitain of a ship and you look at the speedo of the ship and it says you are doing 20 kph. You walk from the front of the ship to the back at 8 kph so you are saying your velocity is zero and the velocity of the ship is zero...interesting
The velocity of the ship is 0 wrt the ship.
The velocity of the ship is 20 kph wrt the water.
The velocity of the ship is 8 mph wrt me walking.

My velocity is 0 wrt myself.
My velocity is 12 kph wrt the water.
My velocity is 8 mph wrt the boat.

The velocity of the dock person is 0 wrt himself.
The velocity of the dock person is 20 kph wrt the boat.
The velocity of the dock person is 0 wrt the water.

Not one of these velocities is any more or less real than any other.

So when you talk about the velocity of clocks just sitting on a ship with no other outside observer to measure velocity wrt then giving a velocity makes no sense. Like asking which way is down in the Universe.

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I want to move the clocks, one to the front of the space ship and one to the back and I want them to still be synchronised from the perspective of the observer at E. The velocity of the space ship is V.
The two clocks will not even stay synchronized before and after passing E (wrt E) even if you don't move them inside the ship. Even if you neglect the change in recessional velocity with position wrt E the two clocks must move to their new positions at the same speed.

If V1 = c and V2 = c
As both clocks are move at C both will stop and they will be synchronized when they make it to their final destination.

Does anyone disagree ?
I disagree. Even if we assume we can instantly move the clocks at the speed C so the time for themselves stops, time doesn't stop being measured by the other observers. E will not agree on how long they remained stopped.

I could give exact figures but it would be different for every observer and clock just like the boat, water, and dock situation I posted above. It would take far more of my time than it takes you to imagine these questions.

Are you claiming to know the ship is moving at v in some objective sense, or is it just moving at v relative to some observer? If it's just relative to an observer, then are you trying to synchronize the clocks in the observer's frame or the ship's frame? I'm sure by now you realize that different frames have different definitions of what it means for two clocks to be synchronized!
Yes the observer at E assumes he is stationary and that the space ship is moving. The analysis I have done is from the perspective of the observer at E.

As it says in my post I will do the same analysis from the perspective of an observer in the space ship (at rest with the space ship) next.

JesseM
Yes the observer at E assumes he is stationary and that the space ship is moving. The analysis I have done is from the perspective of the observer at E.
And so when you say you want to synchronize the clocks on the ship, you mean that this should be done in such a way that they'll be synchronized in the frame of the observer at E?

yes so E thinks that the clocks are synchronised

Actually looking at fig 99 when you are synchronising the clocks if you had observers at x, y , z and W and observers x and y assume they are stationary wrt the space ship, all observers would agree that the clocks are synchronised yeh?

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JesseM
Synchronising clocks

Referring to fig 1

I want to synchronise two clocks.

I have a space ship travelling at v in the direction of the arrow. In the ship I have two identical clocks one at point A (clock1) and on a point D(clock 2). I also have a light source at point B. Length AB = length BD

I have observers at points E, F, G and H. E and F are at rest with the space ship. G and H are not at rest.
You say here that E is "at rest with the space ship", but in subsequent posts you say that E sees the ship in motion, so I guess this is a mistake? You also don't seem to include any points F, G and H in your diagrams.
rab99 said:
A flash of light is set of by the light source and if it hits both clock simultaneously then they are synchronised. As the light from the light source is travelling perpendicular to the direction of travel of the space ship the light will reach both clocks simultaneously and they’ll be synchronised.
If E sees the ship moving, he won't see the light traveling perpendicular to the direction of motion...but if he sees the ship moving along the axis from tail to nose, perpendicular to the axis between the clocks, then he'll see the beams of light moving at symmetrical angles so that they both meet the clocks at the same time in his frame, so the clocks will be synchronized in his frame.
rab99 said:
I will look at what the observers see later.

Fig 2 shows how NOT to synchronise clocks as the clocks are equidistant and as they will change position before the flash of light reaches them then the flash will not reach them simultaneously and they will NOT be in sync
Yes, they won't be in sync in the frame of an observer who sees the ship moving forward in this case.
rab99 said:
Referring to fig 3

I want to move the clocks, one to the front of the space ship and one to the back and I want them to still be synchronised from the perspective of the observer at E. The velocity of the space ship is V.

I move clock 1 from point A to point A hat at velocity V1.

I move clock 2 from point D to point D hat at velocity V2.
You didn't say whether V1 and V2 are in the ship's frame or E's frame, but below you say that if V1 = V then E sees clock 1 at rest, so I guess they must be in the ship's frame.
rab99 said:
Depending on the velocities that the clocks are moved at the observer will see the following:

1. If V1 = V and V2 = V
Clock 1 will be stationary, clock2 will be moving at 2V. therefore clock 2 will slow down clock1 will be ahead of clock 2
Clock 2 won't be moving at 2V in E's frame if it moves at V in the ship's frame and the ship moves at V in E's frame--you have to use the relativistic velocity formula which tells you E will see clock 1 moving at $$\frac{2V}{1 + V^2/c^2}$$

Also, if they are both moving the same distance in the ship's frame, then in E's frame clock 1 which is moving towards the tail will reach its destination D-hat before clock 2 which is moving towards the nose reaches its destination A-hat. This is because the tail is moving towards the position clock 1 started at, while the nose is moving away from the position clock 2 started at. Of course they have different velocities in E's frame so this isn't totally obvious, but if they both move at V in the ship's frame then they must reach their destinations simultaneously in the ship's frame, and we know clocks at the nose and tail which are synchronized in the ship's frame will be out-of-sync, with the clock at the tail ahead, so clock 1 must reach the clock at the tail first in E's frame. and clock 1 both have the same time dilation for the same amount of time in the ship's frame, so they must also show the same time as one another when they reach their destinations, so clock 1 must be ahead of clock 2 in E's frame, by exactly the same amount that the clock which is fixed at the tail is ahead of the clock which is fixed at the nose (vL/c^2, where L is the distance between them in the ship's frame).
rab99 said:
2. If V1 > 2V and V2 = V
Clock 1 will be moving faster than clock2. Clock 1 will slow down wrt clock2. Clock 2 will be ahead of clock 1
Why 2V? Even if you are thinking of the Newtonian velocity addition formula, if V1 was 2V in the ship's frame that would make it only V in E's frame, while if V2 is V in the ship's frame then it is 2V in E's frame. Perhaps you meant meant V1 > 3V? Anyway, like I said, the Newtonian velocity addition formula isn't really correct...it would be easier to just specify that clock 1's speed is greater than clock 2's speed in the frame of E, but even in this case you couldn't guarantee that clock 2 is ahead, because you have to take into account that in E's frame clock 2 takes longer to get to its destination and thus spends more time with a time dilation factor that's greater than the time dilation factor both of them share once they are at rest relative to the ship.

The rest of your comments seem to have similar mistakes, but I'll comment on one more:
rab99 said:
If V1 = c and V2 = c
As both clocks are move at C both will stop and they will be synchronised when they make it to their final destination.
This isn't physically possible--you can't have clocks moving at c. Even if you could, it's sort of weird to talk about two stopped clocks being "synchronized" just because they permanently show the same time.

This is advice from a fellow 'layman': Read Richard Feynman's book "6 Not-So-Easy Pieces". If your questions deal with spaceships and light clocks, then no matter whether you imagine a spaceship moving at a constant velocity (meaning no acceleration) OR you imagine a spaceship accelerating, your answers should be in that book and very clear. that's what Feynman does, after all... explain stuff clearly. And to be fair about the confusion regarding your post, you did initially say E and F are at rest and G and H are not at rest, then switched it to E being an observer moving with the ship. That's pretty darn confusing.

yes I must agree that was confusing sorry to all

got a few questions to answer here

JesseM

but if he sees the ship moving along the axis from tail to nose

yes this is how the clocks are synchronised so the observer at E thinks they are synchronised. Observer E is initially positioned along the axis of the flight of the space ship he then moves to a second position where he is perpendicular to the flight of the space ship. As no other parameters change the clcoks must still be synchronised from his perspective.

My original drawing was going to have observers at efgh but it became too confusing so i wittled it down to one but forgot to remove the reference...sorry all... and thank you for using the word "darn" its one of my favorites :)

v, v1 and v2 are from the perspective of the observer at E after all if I get
If V1 = V and V2 = V
Clock 1 will be stationary

this could only be if v1 was from the perspective of the observer at E

I beleive the ships frame has no relevance here at all. It is from the perspective of E who cares what someone in the ship sees ....yet

Why 2V? Even if you are thinking of the Newtonian velocity addition formula, if V1 was 2V in the ship's frame that would make it only V in E's frame, while if V2 is V in the ship's frame then it is 2V in E's frame. Perhaps you meant meant V1 > 3V? Anyway, like I said, the Newtonian velocity addition formula isn't really correct...it would

here yu are plain wrong from the perspective of E if v2 = v then v1 must be greater than (2 x V) if v1 is to be travelling faster than V2

OK so looking

The obser

ok so looking at the figures again lets say V is 10 kph as measured by E

the clocks are already synchronised by E being along the aixs of flight of the space ship and then E moves so he is perpendicular to the flight is that is OK? I cant see how it is not as the observer moving position is not going to alter the synchronisation of the clocks in the space ship unless you believe in magic

lets say v1 = .99c and v2 = 0.01 kph
second and further lets say v1 = 0.01 kph and v2 = .99c

you can try a few more combinations here if you like

V1 and V2 and V3 are as measured by E

now E knows the follwowing,
E knows the velovcity of the space ship and therefore the time dialation and contraction coefficients of the Space ship

now from the perspective of E by varying the velocities of the positioning, and if you like the position, of the clocks, can the clock at the back of the space ship (SS) be:
1) ahead of the front clock ?
2) synchronised with the front clock ?
3) behind the front clock ?

if not why not ?

the clocks are already synchronised by E being along the aixs of flight of the space ship and then E moves so he is perpendicular to the flight is that is OK? I cant see how it is not as the observer moving position is not going to alter the synchronisation of the clocks in the space ship unless you believe in magic

the clocks are synchronised by E being along the axis of the flight of the space ship and then E moves so he is perpendicular to the flight is that is OK? I cant see how this would not be OK as the observer simply changing position is not going to alter the synchronisation of the clocks in the space ship, unless you believe in magic

Give it up. You will never demonstrate that you can detect 'absolute motion' because it can't be done. Better minds than ours have thought about this for decades.

I wouldn't mind if he kept trying if he didn't just ignore issues and those that point out those issues with his physical assumptions. Assumptions that have consistantly been wrong since Galileo.

ok so looking at the figures again lets say V is 10 kph as measured by E

the clocks are already synchronised by E being along the aixs of flight of the space ship and then E moves so he is perpendicular to the flight is that is OK? I cant see how it is not as the observer moving position is not going to alter the synchronisation of the clocks in the space ship unless you believe in magic

lets say v1 = .99c and v2 = 0.01 kph
second and further lets say v1 = 0.01 kph and v2 = .99c

you can try a few more combinations here if you like

V1 and V2 and V3 are as measured by E

now E knows the follwowing,
E knows the velovcity of the space ship and therefore the time dialation and contraction coefficients of the Space ship

now from the perspective of E by varying the velocities of the positioning, and if you like the position, of the clocks, can the clock at the back of the space ship (SS) be:
1) ahead of the front clock ?
2) synchronised with the front clock ?
3) behind the front clock ?

if not why not ?

JesseM
JesseM said:

but if he sees the ship moving along the axis from tail to nose
yes this is how the clocks are synchronised so the observer at E thinks they are synchronised. Observer E is initially positioned along the axis of the flight of the space ship he then moves to a second position where he is perpendicular to the flight of the space ship. As no other parameters change the clcoks must still be synchronised from his perspective.
"Positioned"? Of course position is irrelevant in questions about simultaneity and time dilation and so forth, all that matters is velocity. If E draws his x axis along the axis from nose to tail of the rocket, with the nose further in the +x direction than the tail, then does E see the rocket moving purely in the +x direction, or not?
rab99 said:
v, v1 and v2 are from the perspective of the observer at E after all if I get
If V1 = V and V2 = V
Clock 1 will be stationary
It makes no sense for you to say v1 is v "from the perspective of the observer at E" and also that "clock 1 will be stationary", unless v = 0 (i.e. the rocket is at rest in E's frame, not moving). Are you sure that V1 and V2 aren't meant to be in the frame of the rocket?
rab99 said:
this could only be if v1 was from the perspective of the observer at E

I beleive the ships frame has no relevance here at all. It is from the perspective of E who cares what someone in the ship sees ....yet
If v1 is clock 1's velocity in E's frame, and clock 1 is stationary in E's frame, then v1 = 0. Is that really what you meant?
rab99 said:
JesseM said:
Why 2V? Even if you are thinking of the Newtonian velocity addition formula, if V1 was 2V in the ship's frame that would make it only V in E's frame, while if V2 is V in the ship's frame then it is 2V in E's frame. Perhaps you meant meant V1 > 3V? Anyway, like I said, the Newtonian velocity addition formula isn't really correct...it would
here yu are plain wrong from the perspective of E if v2 = v then v1 must be greater than (2 x V) if v1 is to be travelling faster than V2
I was thinking v1 and v2 were in the ship's frame, and v was in E's frame. So if v1 is in the direction from nose to tail while v2 is in the opposite direction, then using the Newtonian velocity addition formula, in E's frame clock 1's velocity would be v - v1 while clock 2's velocity would be v + v2. So, this is why I said that if v2 = v and v1 = 3v, the clocks would have equal speeds in E's frame. But, obviously I may have misunderstood what frame v1 and v2 were supposed to be in.
rab99 said:
ok so looking at the figures again lets say V is 10 kph as measured by E

the clocks are already synchronised by E being along the aixs of flight of the space ship and then E moves so he is perpendicular to the flight is that is OK? I cant see how it is not as the observer moving position is not going to alter the synchronisation of the clocks in the space ship unless you believe in magic
Do you just mean E changes position, or does he change velocities so that the ship is moving perpendicular to the nose-tail axis in his frame? If he actually changes velocities, then this will of course change the synchronization of the clocks relative to the definition of synchronization of his current rest frame, because each frame has its own definition of what it means for two clocks to be synchronized (different definitions of simultaneity).
rab99 said:
lets say v1 = .99c and v2 = 0.01 kph
second and further lets say v1 = 0.01 kph and v2 = .99c

you can try a few more combinations here if you like

V1 and V2 and V3 are as measured by E
What is v3? And how long does each clock spend moving at these velocities? If each clock is moving from the middle of the ship to the front/back and then changing velocities so they now are moving at the same speed relative to the ship, then E may see them moving at v1 and v2 for different amounts of time before they reach their end of the ship, and this influences how far the clocks get out-of-sync.
rab99 said:
now E knows the follwowing,
E knows the velovcity of the space ship and therefore the time dialation and contraction coefficients of the Space ship
If you already gave the velocities of the two clocks in E's frame, then the time dilation of the ship is irrelevant, a clock's time dilation in E's frame depends only on that clock's velocity in E's frame.
rab99 said:
now from the perspective of E by varying the velocities of the positioning, and if you like the position, of the clocks, can the clock at the back of the space ship (SS) be:
1) ahead of the front clock ?
2) synchronised with the front clock ?
3) behind the front clock ?
You haven't given enough information--you gave two different possible velocities for clock 1 and clock 2, and you didn't specify how much time each clock spends moving at its velocity. Certainly by picking the right combination of answers to these questions, any of the three options above could become true.

start from the beginning referring to fig 1

a space ship with some clocks in it are flying past the observer at E. E is standing on the ground the space ship is flying past in the air. E is stationary the space ship is moving at velocity V as measured by observer E.

E also measures the velocities that the clocks are moved at V1 and V2.

in scenario 1 lets say v1 = .99c and v2 = 0.01 kph
in scenario 2 lets make v1 = 0.01 kph and v2 = .99c

you can try a few more combinations here if you like

V1 and V2 and V are as measured by E

now E knows the follwowing,
E knows the velovcity of the space ship and therefore the time dialation and contraction coefficients of the Space ship

now from the perspective of E by varying the velocities that the two clocks are positioned at, can the clock at the back of the space ship (SS) be:
1) ahead of the front clock ?
2) synchronised with the front clock ?
3) behind the front clock ?

Dear whoever when you reply if you dont understand one thing or you find an error please only give me that one thing or error, NOT one hundred things or error.

If there is an error in the first sentence and you do this fantastically long anaylysis based on the error then all the analysis after the error is most likely irrelevant cos its based on an error :) ......... thankyou

It will take longer but it will be more productive

JesseM
start from the beginning referring to fig 1

a space ship with some clocks in it are flying past the observer at E. E is standing on the ground the space ship is flying past in the air. E is stationary the space ship is moving at velocity V as measured by observer E.

E also measures the velocities that the clocks are moved at V1 and V2.

in scenario 1 lets say v1 = .99c and v2 = 0.01 kph
in scenario 2 lets make v1 = 0.01 kph and v2 = .99c
You still haven't told me how long each clock spends at these speeds--eventually they'll hit the front or back of the ship and have to come to rest in the ship's frame (i.e. change to speed V in E's frame), right?
rab99 said:
V1 and V2 and V are as measured by E

now E knows the follwowing,
E knows the velovcity of the space ship and therefore the time dialation and contraction coefficients of the Space ship
Do you understand that these things are irrelevant to predicting the time on clock 1 and clock 2 if E knows their own velocities in his frame? Clock 1's rate of ticking in E's frame is determined solely by clock 1's velocity in E's frame, and likewise for clock 2.

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I quote my initial post

___________________________________________________________
Referring to fig 3

I want to move the clocks, one to the front of the space ship and one to the back and I want them to still be synchronised from the perspective of the observer at E. The velocity of the space ship is V.

I move clock 1 from point A to point A hat at velocity V1.

I move clock 2 from point D to point D hat at velocity V2.

The length from point D to the point D hat is equal to the length from point A to point A hat.

___________________________________________________________

the time will be the distance divided by the velocity as seen by the stationary observer. The clocks are moved equidistant from thier initial positions.

Each clock moves the same distance from the inital position say length D it i only the velocities V1 and V2 that alter.