# How Does Time Dilation Affect Clocks in Different Frames of Reference?

• rab99
In summary: L/(c+V) time to get from the red end to the blue end. Referring to fig 6I have a clock in the space ship. The spaceship is flying at constant velocity V.At the red end of the clock is a light source, at the blue end a detector.As the speed is constant the contraction is constant and the time rate is constant.The observer, point A...would see the light pulse traveling for L/(c+V) time to get from the red end to the blue end. In summary, in fig 5 the observer sees the light pulse traveling for a longer time than in fig 6. This disproves
rab99
Can someone please have a look at my analysis here and tell me if there are any mistakes
the rest of the pistures are in my other posts
___________________________________________________________________________
Einstein’s explanation of time dilation is given here http://en.wikipedia.org/wiki/Time_dilation

Referring to fig 1
In Einstein’s explanation you have 2 identical clocks. Clock 1 is at rest wrt the observer. Clock 2 is in a spaceship zooming along wrt the observer.

At the red end of the clock is a light source, at the blue end a detector.

If L is the length of the clock. The observer, observing the clock at rest wrt him would see that the light pulse would take L/c time to traverse the length of the clock.

Einstein postulated that as the observer would see the photons in clock 2 travel a longer path to move from the red end to the blue end when compared with length of the path the photons in the clock 1. Therefore clock 2 must tick slower (time passes slower) when compared to clock1.

Referring to fig 2
As in Einstein’s explanation again you have 2 identical clocks. Clock 1 and clock 2 are at rest wrt the observer. This is the trivial case. Again at the red end of each clock is a light source, at the blue end a detector.

If L is the length of the clock. The observer, observing both clocks would see that the light pulse would take L/c time to traverse the length of the clock.

Referring to fig 3
As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
Clock 1 is at rest wrt the observer, clock 2 is in a spaceship that is moving wrt the observer.

As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.

If L is the length of the clock.

The observer, observing the clock 1 would see that the light pulse would take L/c time to traverse the length of the clock.

T = L/c

As can be seen the path traveled by the photons in the clock2 is shorter than the path of the photons in the clock1. This is also what the observer would see.

The time the photons in clock2 would take would be at most

T = L/(c + V)

As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c + V).

I note the limit of the function t = L/(c + V) as V tends to infinity so t (time) goes to 0

Referring to fig 4
As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
Clock 1 is at rest wrt the observer, clock 2 is in a spaceship that is moving wrt the observer.
You should now note both clocks have been rotated through 180 degrees compared to fig 3. In Fig 3 the light source is at the right the detector at the left. In fig 4 the reverse.

As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.

If L is the length of the clock. The time taken in Clock 1 would be L/c time to traverse the length of the clock.

T = L/c

As can be seen the path traveled by the photons in the clock2 is longer than the path of the photons in the clock1. This is also what the observer would see.

The time the photons in clock2 would take would be at least

T = L/(c - V)

As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c - V).

I note the limit of the function t = L/(c - V) as V tends to infinity so t (time) goes to infinity

I have three questions

Is this analysis right?

How is it that the observer can get two different results simply by rotating the clock?

The theory is a moving clock ticks slower then a stationary clock. In fig 3 clock 2 is ticking faster than clock1 which refutes the accepted theory?

WHAT AN OBSERVER IN THE SPACE SHIP SEES

Referring to fig 5
I have a clock in the space ship.
The spaceship is flying at constant velocity V.
At the red end of the clock is a light source, at the blue end a detector.
As the speed is constant the contraction is constant and the time rate is constant.
The observer, point A and point B form an isosceles triangle.

The length of the clock is L which equals the length from B to the observer and also A to the observer.

I NOTE THE DRAWING IS NOT TO SCALE.

At some initial time the ends of the clock will be located at points A and B. At some time in the future as the spaceship is moving the ends of the clock will be located at points A hat and B hat.

At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.

By the time the pulse from the red end of the clock reaches the observer he will be located at point x.

When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.

By the time the pulse from the blue end reaches the observer the observer will be located at point z.

Analysis

Referring to fig 7
Using the cosine rule L3 can be calculated.
Times will be proportional to lengths as time is simply length/c

Angle g = 60 degrees
Angle e = 120 degrees
Cos 60 = 0.5
Cos 120 = -0.5
L3^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )

t1 is the time it takes for the pulse of light to travel from B hat to point z

Referring to fig 8
Using the cosine rule L1 can be calculated.

Angle e = 60 degrees
Cos 60 = 0.5
L1^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )

t2 is the time it takes for the pulse of light to travel from point A to point x

from fig 5
L2 = L – V(t3)
t3 is the time it takes for the pulse of light to travel from point A to point B hat.

The time difference the observer will see between the first pulse of light he sees and the second is:

Time_Diff1 = (L1 – (L2 + L3))/c
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L2 = L – V(t3)

Referring to fig 6
I have a clock in the space ship.
The spaceship is flying at constant velocity V.
At the red end of the clock is a light source, at the blue end a detector.
As the speed is constant the contraction is constant and the time rate is constant.
The observer, point A and point B form an isosceles triangle.

The length of the clock is L which equals the length from B to the observer and also A to the observer.

I note the clock in fig 6 is rotated 180 degrees wrt the clock in fig 5 the red and blue ends have changed.

I NOTE THE DRAWING IS NOT TO SCALE.

At some initial time the ends of the clock will be located at points A and B. At some time in the future as the spaceship is moving the ends of the clock will be located at points A hat and B hat.

At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.

By the time the pulse from the red end of the clock reaches the observer he will be located at point x.

When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.

By the time the pulse from the blue end reaches the observer the observer will be located at point z.

Analysis

Referring to fig 7
Using the cosine rule L1 can be calculated.
Times will be proportional to lengths as time is simply length/c

Angle g = 60 degrees
Angle e = 120 degrees
Cos 60 = 0.5
Cos 120 = -0.5
L1^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )

t1 is the time it takes for the pulse of light to travel from point A to point x

Referring to fig 8
Using the cosine rule L3 can be calculated.

Angle e = 60 degrees
Cos 60 = 0.5
L3^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )

t2 is the time it takes for the pulse of light to travel from point B hat to point z

from fig 6
L2 = L + V(t3)
t3 is the time it takes for the pulse of light to travel from point A to point B hat.

The time difference the observer will see between the first pulse of light he sees and the second is:

Time_Diff2 = (L1 – (L2 + L3))/c
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L2 = L + V(t3)

Observations

Further observation:
L3 in fig 6 = L1 in fig 5
L1 in fig 6 = L3 in fig 5
V(t3) in fig 5 = V(t3) in fig 6

The time difference measured in fig 5 must equal the time difference measured in fig 6.

Time_Diff1 = (L1 – (L2 + L3))/c
Time_Diff2 = (L1 – (L2 + L3))/c
Time_Diff1 = Time_Diff2

Time_Diff2 = (L1 – (L2 + L3))/c
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L2 = L + V(t3)
Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – (L + V(t3) + sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) ) )
Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L - V(t3) - sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )

Time_Diff1 = (L1 – (L2 + L3))/c
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L3 = sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L2 = L – V(t3)
Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – (L – V(t3) + sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) ))
Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3) - sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )

Time_Diff1 = Time_Diff2

sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3) - sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
=

sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L - V(t3) - sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )

As can be seen Time_Diff1 does NOT equal Time_Diff2

As all frames of ref are completely contained within a larger frame I note for any observer, unless they are in the absolute frame of ref, Time_Diff1 will not equal Time_Diff2 as they will have some residual velocity due to the velocity of the larger containing frame of ref.

This analysis demonstrates that an experiment can be performed in a moving frame of reference to show the observer at rest with that frame of ref that they are in a moving frame of ref.

As all frames of ref are completely contained within a larger frame of ref. By using this apparatus it may be possible to determine if there is an absolute frame of ref. By using this apparatus and changing the direction and velocity of a frame of ref until Time_diff1 = Time_diff2 then the direction and velocity of that frame of ref would be the same as the absolute (or preferred) frame of ref.

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Rab99:

I tried to read your first post and stopped when I came to this

The time the photons in clock2 would take would be at most

T = L/(c + V)

As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c + V).
This made me unhappy because c+V implies something moving faster than light-speed, which I can't accept.

I skipped to the end and found

As all frames of ref are completely contained within a larger frame I note for any observer, unless they are in the absolute frame of ref, Time_Diff1 will not equal Time_Diff2 as they will have some residual velocity due to the velocity of the larger containing frame of ref.
What is the absolute frame within which all frames are contained ?

This analysis demonstrates that an experiment can be performed in a moving frame of reference to show the observer at rest with that frame of ref that they are in a moving frame of ref.
I don't believe this. I am convinced that this "it is not possible to distinguish a state of rest from a state of uniform motion" is true.

Take me for instance, sitting in front of my computer in a room on earth. Assuming there are observers around the cosmos on other planets or spaceships, they will each see me in a different state of motion. I don't have an 'absolute state of motion', only relative states of motion.

rab99 said:
At the red end of the clock is a light source, at the blue end a detector.
Hi rab, I am not even sure this apparatus is a clock. Usually a clock has some sort of recurring event at one location in its rest frame, this doesn't. How would you use this to time something?

Usually a light clock has a mirror on one side and then the emitter and detector on the same side. That way you are measuring the time between two events (emission and detection) at the same location in it's rest frame.

Mentz114 said:
This made me unhappy because c+V implies something moving faster than light-speed, which I can't accept.
No, it doesn't imply that. If you have a light signal emitted from the front end of a rod of length L in your frame which is moving at speed v in your frame, then in your frame it will take a time of L/(c+v) to reach the back end. For example, suppose the back end is at x=0 when the light is emitted at t=0, while the front end is at x=L; then the equation for the back end's position as a function of time will be x(t) = vt, while the equation for the light moving from front to back will be x(t) = L - ct. So, the light will meet the back end when vt = L - ct, which means t = L/(c+v).

JesseM,
OK, I wasn't paying attention. But what the OP is trying to do is false and hardly worth close attention. What do you think of his 'proof' that motion is absolute ?

rab99 said:
Can someone please have a look at my analysis here and tell me if there are any mistakes
the rest of the pistures are in my other posts
___________________________________________________________________________
Einstein’s explanation of time dilation is given here http://en.wikipedia.org/wiki/Time_dilation

Referring to fig 1
In Einstein’s explanation you have 2 identical clocks. Clock 1 is at rest wrt the observer. Clock 2 is in a spaceship zooming along wrt the observer.

At the red end of the clock is a light source, at the blue end a detector.

If L is the length of the clock. The observer, observing the clock at rest wrt him would see that the light pulse would take L/c time to traverse the length of the clock.
Remember that two different frames can only agree on the length of the clock if the clock is oriented at right angles to the two frames' relative motion. So in the setup of your first diagram they'll both agree on the length, but in the setup of your second diagram, if the clock's length is L in the ship's rest frame then it will be only $$L*\sqrt{1 - v^2/c^2}$$ in the frame where the ship is moving forward at speed c.
rab99 said:
Referring to fig 2
As in Einstein’s explanation again you have 2 identical clocks. Clock 1 and clock 2 are at rest wrt the observer. This is the trivial case. Again at the red end of each clock is a light source, at the blue end a detector.

If L is the length of the clock. The observer, observing both clocks would see that the light pulse would take L/c time to traverse the length of the clock.
Usually in this thought-experiment we talk about the time for the light to go from one end to the other and back, in which case the time would be 2L/c for the observer on the ship.
rab99 said:
Referring to fig 3
As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
Clock 1 is at rest wrt the observer, clock 2 is in a spaceship that is moving wrt the observer.

As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.

If L is the length of the clock.
Here is where what I said above about length contraction becomes important. If you take into account that the length is shrunk in the frame where the ship is moving, then you will find that the time for the light to go from one end to another and back is $$(2L/c)/\sqrt{1 - v^2/c^2}$$, exactly as predicted by the time dilation formula.

To calculate the one-way time for the clock to go from the left end to the right end is more complicated. It is different in different directions in the frame where the rocket is moving, but it will nevertheless be the same in both directions for the observer on the rocket if he places watches at each end; and notes the times the light hits each end; to understand this part you must take into account the relativity of simultaneity, which is based on the fact that each observer synchronizes watches in different locations using the "Einstein synchronization convention" which is based on the assumption that light moves at c in all directions in their own frame. So if I want to synchronize watches at the front and back of the rocket in the rocket's own frame, I can set off a flash at the midpoint of the two rockets, and set them to read the same time when the light from the flash reaches them. But this means that in the frame of an observer at rest in a different frame who sees the rocket moving forward, my watches will be out-of-sync, because the watch at the back of the rocket is moving towards the point where the flash was set off and the watch at the front is moving away from that point, so naturally if this observer believes light moves at c in both directions he'll say the light reached the back watch before the front watch.

The equation to remember for the relativity of simultaneity is that if you have two clocks at rest relative to each other which are synchronized in their own frame and a distance L apart in that frame, then in the 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 the amoung vL/c^2. If you take this into account, and also take into account length contraction, you'll find that the observer on the ship still measures a time of L/c between the light leaving one end of the light clock and the light arriving at the other end, even if you analyze the situation from a frame where the ship is moving.

If you want an example along these lines, take a look at the one I gave in post #6 from this thread.

Last edited:
DaleSpam said:
Hi rab, I am not even sure this apparatus is a clock. Usually a clock has some sort of recurring event at one location in its rest frame, this doesn't. How would you use this to time something?

Usually a light clock has a mirror on one side and then the emitter and detector on the same side. That way you are measuring the time between two events (emission and detection) at the same location in it's rest frame.

this does have two events... the time the photon left the light source and thr time that it arrives at the detector. The time interval between these two events could be timed and used to set a clock

JesseM
Remember that two different frames can only agree on the length of the clock if the clock is oriented at right angles to the two frames' relative motion. So in the setup of your first diagram they'll both agree on the length, but in the setup of your second diagram, if the clock's length is L in the ship's rest frame then it will be only in the frame where the ship is moving forward at speed c.

The clock at rest wrt the observer is an identical clock to the one in the space ship. Why is the length of the clock even relevant they are identical. Just assume the clock in the spaceship is shorter wrt the one at rest with the observer?

rab99 said:
this does have two events... the time the photon left the light source and thr time that it arrives at the detector. The time interval between these two events could be timed and used to set a clock
These two events take place at different locations in the rest frame of the apparatus.

rab99 said:
this does have two events.
Yes, but the interval between the two events is lightlike, not timelike. This means that there does not exist any frame where the events measure two points in time at a single point in space. The interval between two clock tick events needs to be timelike.

This was extraordinarily difficult to follow but I think I got the point.

rab99 said:
Einstein postulated that as the observer would see the photons in clock 2 travel a longer path to move from the red end to the blue end when compared with length of the path the photons in the clock 1. Therefore clock 2 must tick slower (time passes slower) when compared to clock1.

No this was not a postulate, this is one of the consequences of two postulates.
1) The speed of light is the same for all observers.
2) The laws of physics is the same regardless of motion.
Below you have taken issue with the second postulate.

rab99 said:
As all frames of ref are completely contained within a larger frame I note for any observer, unless they are in the absolute frame of ref, Time_Diff1 will not equal Time_Diff2 as they will have some residual velocity due to the velocity of the larger containing frame of ref.

I'm not sure what you mean by "contained within a larger frame" but I suspect it has something to do with your setup in "fig 6". You have the observers contained in the same frame as point A and B. You then proceed with an analysis showing length contraction. In what frame of reference does this time dilation take place? There is no motion within that frame to say that A or B changed position at all so there is no velocity V to use in your analysis.

By assuming it has a velocity at all you implicitly defined a third frame of reference that you never actually specified. You just assumed from an intellectual perspective, your minds eye. This make you the one in motion wrt the spaceship observer, A, and B, and must be treated as such. It is you, which is in motion separately from the system you defined, that is the only defined frame for which "Time_Diff1 does NOT equal Time_Diff2". It cannot be said of the observers in the spaceship.

rab99 said:
JesseM
The clock at rest wrt the observer is an identical clock to the one in the space ship. Why is the length of the clock even relevant they are identical. Just assume the clock in the spaceship is shorter wrt the one at rest with the observer?

The observer has to agree that the clock in the spaceship is shorter than his clock...full stop.

rab99 said:
The observer has to agree that the clock in the spaceship is shorter than his clock...full stop.

One question: Is this observer in the spaceship or moving with it?

my_wan said:
I'm not sure what you mean by "contained within a larger frame" but I suspect it has something to do with your setup in "fig 6". You have the observers contained in the same frame as point A and B. You then proceed with an analysis showing length contraction. In what frame of reference does this time dilation take place? There is no motion within that frame to say that A or B changed position at all so there is no velocity V to use in your analysis.

What I mean by two frames are always contained within a larger frame is this
you are walking on the Earth and a car passes you by. your walking is in a frame which can be descirbed by a vector, direction and velocity, say vector X. The car can be described by say vector y. The Earth is spinning on its axis with vector w. The vector for the car is now the resultant of Vector W, x and y. The Earth is also moving around the sun with vector B the car and your vectors are now the resultant of vectors b, x, y and W. The Earth is within the solar system which is moving by vector D which adds the all of the aformentioned vectors. The solar system which is within the milky way, moving by vector U. The milky way is within the galaxy moving at vector Q. The galaxy is within within some larger struscture moving at vector K. And if you believe in big bang theorey this whole maras of vectors is within the universe probably descirbed by vector H. Now the vector decribing the car and you is a resultant of all the above vectors. I would call this vector that is the reultant of all the vectors up to the largest structure the preferred vector or frame

the observer and the clock next to the observer is in a diff frame of ref to the space ship

two identical clocks one in the spaceship one at rest wrt the observer

what do you think of the fact that time1 does not equal time 2 or is it too poorly communicated?

rab99 said:
What I mean by two frames are always contained within a larger frame is this
you are walking on the Earth and a car passes you by. your walking is in a frame which can be descirbed by a vector, direction and velocity, say vector X. The car can be described by say vector y.

Here's the thing: In the walking frame of reference the vector has no direction and a velocity of is 0. In the cars frame the vector has no direction and a velocity of is 0. In the walking frame the Earth, the car, and the rest of the Universe has a vector. In the car frame the walking frame, the Earth, and the rest of the Universe has a vector.

Within a spaceship all observers and clocks inside that spaceship has a 0 vector to everything in that spaceship. The only way you can describe a vector in that spaceship is to add another observer that is moving wrt that spaceship. Only another observer that is moving wrt that spaceship can you define a spaceship vector.

rab99 said:
the observer and the clock next to the observer is in a diff frame of ref to the space ship

If the clock stays next to the observer it is not another frame of reference to any observer.

rab99 said:
two identical clocks one in the spaceship one at rest wrt the observer
rab99 said:
what do you think of the fact that time1 does not equal time 2 or is it too poorly communicated?

Yes you keep using frame of reference but you have yet to define what observers are actually moving differently from the other observers and clocks. You appear to assume that the frame of reference is different just because they aren't in the same place.

In my anaylsis of fig 5 and fig 6 I am purely concerned with what the guy in the spaceship sees and he is at rest wrt the clock he is observing. I am not interested in any other frame. Just the guy in the spaceship observing the clock in the spaceship and a mathematical analysis of what he sees.

With the clock in one oreintation he sees two flashes of light separated by a time delay
With the clock rotated by 180 degrees he sees two flashes of light separated by a time delay that is not the same as the first time delay...how difficult can it be?

the weirdness lies in the fact that time delay one is not equal to time delay two ...why not?

no one has given me an answer much less a satisfactory answer yet

rab99 said:
In my anaylsis of fig 5 and fig 6 I am purely concerned with what the guy in the spaceship sees and he is at rest wrt the clock he is observing. I am not interested in any other frame. Just the guy in the spaceship observing the clock in the spaceship and a mathematical analysis of what he sees.

With the clock in one oreintation he sees two flashes of light separated by a time delay
With the clock rotated by 180 degrees he sees two flashes of light separated by a time delay that is not the same as the first time delay...how difficult can it be?

That's what I thought. Then there is absolutely nothing moving and V=0 in all cases. Defining V as anything but 0 is not valid and doing so you must "concern" another observer outside the spaceship.

rab99 said:
the weirdness lies in the fact that time delay one is not equal to time delay two ...why not?

It does always equal as I have restated several times. The mistake is giving a velocity when in fact there is no velocity between any of the clocks or observers.

1. the guy in the spaceship may not be aware of it but the reality is at time 1 he will be at a position say position 1 and some time in the future he will be at a different position ..is that agreed?

Velocity = distance over time

2. But photons are completely separate from the frame they always travel at C regardless of the direction or velocity of the frame. Photons are not ballistic. The photon at time 1 will be at a position and at some time in the future guess what its in a diff position.

Velocity = distance over time

if you agree statement 1 then you also must agree there is a velocity..there is a change in position and there is a time differential. The guy in the spaceship is not aware of this velocity but none the less he has a velocity and his velocity is completely disconnected from the velocity of the photon in the space ship. The two move independently. The photon has no intelligence and doesn't know its in a moving frame of ref.

we now have two velocities the veocity of the photon and the velocity of the clock and the observer. If you agree statement 1 then you must agree with my analysis. if you agree statement 1 but not my analysis how?

If you don't agree statement 1 then how?

rab99 said:
1. the guy in the spaceship may not be aware of it but the reality is at time 1 he will be at a position say position 1 and some time in the future he will be at a different position ..is that agreed?

No! He is at the same place. The rest of the Universe moved. You cannot under any circumstances tell the difference.

rab99 said:
If you don't agree statement 1 then how?

How what? How does he have a velocity? He don't. How does his vector have a direction? It don't. The rest of the Universe does.

Look at it this way. You have A and B and they are going 100 m/hour wrt each other. Now A says B has a vector magnitude of 100 m/h. B says A has a vector magnitude of 100 m/h. What happens if they are both right. The vector adds up to 200 m/h! That can't be right because they are both only going 100 m/h wrt each other.

Ahh, but A thinks his vector magnitude is 0 and B vector magnitude is 100 m/h.
Also, B thinks his vector magnitude is 0 and A vector magnitude is 100 m/h.
Now suddenly the vectors add up properly.

JesseM said:
To calculate the one-way time for the clock to go from the left end to the right end is more complicated. It is different in different directions in the frame where the rocket is moving, but it will nevertheless be the same in both directions for the observer on the rocket if he places watches at each end; and notes the times the light hits each end; to understand this part you must take into account the relativity of simultaneity, which is based on the fact that each observer synchronizes watches in different locations using the "Einstein synchronization convention" which is based on the

JesseM i forgot to mention the space ship's velocity is 1 meter per sec and the clock is 600,000,000 meters long.

the clock has moved 2 meters by the time the light strikes the detector.

As the clock is moving so slow the effects of simutenaiety are virtually zero, same for contraction and time dilation.

rab99 said:
The clock at rest wrt the observer is an identical clock to the one in the space ship. Why is the length of the clock even relevant they are identical. Just assume the clock in the spaceship is shorter wrt the one at rest with the observer?
But if the light clock is shorter than L in the frame of the guy on the ship, then of course it will no longer be true that he'll measure the time for the light to go from one end of the clock to the other and back is 2L/c. The point is that whatever the length L' is in the frame of the ship, the time for the light to go from one end to the other and back is 2L'/c as measured by a stopwatch the observer on the ship is holding, so the experiment doesn't help at all in determining whether the ship is "really" in motion (in relativity this question has no single right answer).

rab99 said:
JesseM i forgot to mention the space ship's velocity is 1 meter per sec and the clock is 600,000,000 meters long.

the clock has moved 2 meters by the time the light strikes the detector.

As the clock is moving so slow the effects of simutenaiety are virtually zero, same for contraction and time dilation.
Again, the point is that whatever the length of the light clock L' in the ship's own frame, the time for light to go from one end to the other as measured by watches at either end of the clock which have been synchronized in the ship's frame (and are at rest in this frame) will be exactly L'/c, and the time for light to go from the back end to the front end and back as measured by a watch at the back end and at rest in the ship's frame will be 2L'/c. Even if you do the analysis in the frame where the ship is moving, you'll still make this prediction about the time as measured by these watches. Do you disagree?

I knew I should have split that proposition into two posts.

In his clock in the spaceship mind experiment (the link is in the post) Einstein looked at the length of the path that the photons would take as compared to the length of the path for the clock at rest wrt the observer. Einstein made a simple observation, the path is longer in the moving clock so it ticks slower. In my post it looks to me as tho the path is shorter. Make the speed of the spaceship 1 meter an hour so the effects of simutenaitey are negligable the result remains the same the path taken by the photons as observed will be shorter than the clock at rest wrt the observer

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