Negative moving coordinates and Time Dilation

[chinglu1998]

I have question. Want to know time dilation for clock at negative coordinate in moving frame to reach origin of rest frame.

Let's use P' = ( -7/8ls,0,0 ). Use v =3/5c, then γ = 5/4. ls is light seconds.

Now, do you apply Lorentz Transforms like this

t' = ( t - vx/c² )γ

But, we do not have t yet.

Apply length contraction to know when P' reaches the rest origin. Distance is (7/8)/γ = (7/8)(4/5) = 7/10l.

Now apply transform. Since at origin, x = 0

t' = ( 7/10s - (3/5c)(0/c²) )γ

t' = (7/10s) γ

This means moving clock beats faster.

What's wrong with this?

 

[ghwellsjr]

(7/8)(4/5) = 7/5l.

Shouldn't that be:
(7/8)(4/5) = 7/10?

 

[chinglu1998]

Shouldn't that be:
(7/8)(4/5) = 7/10?

Should, I fix many thanks.

 

[Dale]

I have question. Want to know time dilation for clock at negative coordinate in moving frame to reach origin of rest frame.

Let's use P' = ( -7/8ls,0,0 ). Use v =3/5c, then γ = 5/4. ls is light seconds.

Now, do you apply Lorentz Transforms like this

t' = ( t - vx/c² )γ

What are you trying to do here? If you already have coordinates P' specified in the primed frame then shouldn't you use the Lorentz transform in the other direction (from the primed frame into the unprimed frame). If something is at -7/8 ls in the primed frame and is moving at v=3/5 c in the primed frame then length contraction is not relevant.

 

[chinglu1998]

What are you trying to do here? If you already have coordinates P' specified in the primed frame then shouldn't you use the Lorentz transform in the other direction (from the primed frame into the unprimed frame). If something is at -7/8 ls in the primed frame and is moving at v=3/5 c in the primed frame then length contraction is not relevant.

Is relativity absolute and you are not allowed to calculate this from the view of the rest frame?

 

[Dale]

Is relativity absolute

No. I was just asking you to clarify your question.

 

[grav-universe]

The Lorentz transformations are only to be used to find what two observers view for the same event, not of each other. For instance, if an event takes place at coordinates (t, x, y, z) according to observer A, then the Lorentz transformations tell you at what coordinates (t', x', y', z') the event takes place for observer B. The time dilation that each observer measures of the other, however, is just t' = t / y. Using t and t' for both conditions can be confusing, there should really be different notations for each, but they are completely different situations.

 

[grav-universe]

I have question. Want to know time dilation for clock at negative coordinate in moving frame to reach origin of rest frame.

Let's use P' = ( -7/8ls,0,0 ). Use v =3/5c, then γ = 5/4. ls is light seconds.

Now, do you apply Lorentz Transforms like this

t' = ( t - vx/c² )γ

But, we do not have t yet.

Apply length contraction to know when P' reaches the rest origin. Distance is (7/8)/γ = (7/8)(4/5) = 7/10l.

Now apply transform. Since at origin, x = 0

t' = ( 7/10s - (3/5c)(0/c²) )γ

t' = (7/10s) γ

This means moving clock beats faster.

What's wrong with this?

Okay, let's work this out using the Lorentz transformation for time. Usually, in order to apply the Lorentz transformations, we originally have two observers that pass each other at the origin and synch their clocks to T=0 while passing, but we can still find the difference in times that pass upon the clocks for each between two events without synching to each other. For event 1, we have (0, -7/8 ls, 0, 0) for the observer at the origin in the rest frame. For event 2, we have ( (7/8 ls)/(3/5 c), 0, 0, 0), so t1 = 0 and t2 = (7/8 ls) / (3/5 c) = 35/24 sec, giving t = t2 - t1 = 35/24 sec. For the time that passes upon the clock of the traveling observer P', then, we have

t' = t2' - t1'

= γ ( t2 - v x2 / c² ) - γ ( t1 - v x1 / c² )

= γ (t2 - t1) - γ (x2 - x1) v / c^2

whereas v = (x2 - x1) / (t2 - t1), giving (x2 - x1) = v (t2 - t1), so

t' = γ (t2 - t1) - γ (t2 - t1) v^2 / c^2

= γ (t2 - t1) (1 - v^2 / c^2)

= γ (t2 - t1) / γ^2

= (t2 - t1) / γ

= t / γ

= (35/24 sec) / (5/4) = 7/6 sec

 

[grav-universe]

This actually requires slightly more explanation. As I mentioned before, the use of the parameters t and t' can be confusing since they are often used with the same notation for the Lorentz transformations as with time dilation and length contraction when there are really four parameters involved, not just two. So instead, with two observers A and B, we can use tA to denote what observer A reads upon A's own clock, tB for what observer B reads upon B's own clock, tA' for what observer B reads upon the clock of A, and tB' for what observer A reads upon the clock of B.

With these notations, the coordinates of an event according to A are (tA, xA, yA, zA) and (tB, xB, yB, zB) according to B for the same event. The Lorentz tranformation for time now becomes tB = γ (tA - xA v / c^2). The time dilation B observes of A, however, is ΔtA' = ΔtB / γ. Likewise, the time dilation A observes of B is ΔtB' = ΔtA / γ. To see this, let's go back to what we found before with

ΔtB = tB2 - tB1

= ... = (tA2 - tA1) / γ

= ΔtA / γ

This is so far still in terms of what each observer views of their own clocks only. However, observers in all frames must agree about what a clock will read when it coincides in the same place as an event. According to A, an event takes place at (tA1, xA1, 0, 0). B coincides with that event and B's clock reads tB1. Since observers in all frames must agree with what B's clock reads while coinciding with the event, then tB1' = tB1 as both A and B observe the reading upon clock B at the same place that the event takes place. B also coincides with the second event at the origin and observer A can directly read B's clock there as well, so tB2' = tB2. Therefore

ΔtB' = tB2' - tB1'

= tB2 - tB1

= ΔtA / γ

giving the time dilation formula as observer A views the time that passes upon clock B between the events as compared to the time that passes upon A's own clock between the events.

 

[rpenner]

The original poster of this thread originally titled another thread Time dilation is false.

Event P is not well agreed upon. The OP seems to have chosen t = 0 and x' = -7/8 light seconds. While grav-universe prefers t' = 0 and x' = -7/8 light seconds.
Event Q is well agreed upon with x = 0 and x' = -7/8 light seconds.

Both camps can be satisfied with parameterizing P on x = -a c and x' = -7/8 light seconds.

Event P has
t = 7/6 seconds - (5/3) a
x = -a c
t' = 35/24 seconds - (4/3) a
x' = -7/8 light-seconds

Event Q has
t = 7/6 seconds
x = 0
t' = 35/24 seconds
x' = -7/8 light-seconds

Q - P has
Δt = (5/3) a
Δx = a c
Δt' = (4/3) a
Δx' = 0

Obviously, for all positive a, Δx/Δt = (3/5) c, Δx'/Δt' = 0, and Δt' = Δt/γ < Δt.

Also, since the Lorentz transform preserves it, (cΔt)² - (Δx)² = (16/9)² a² c² = (cΔt')² - (Δx')², which is a quick way to compute the proper time for any time-like interval.

 

[chinglu1998]

The original poster of this thread originally titled another thread Time dilation is false.

Event P is not well agreed upon. The OP seems to have chosen t = 0 and x' = -7/8 light seconds. While grav-universe prefers t' = 0 and x' = -7/8 light seconds.
Event Q is well agreed upon with x = 0 and x' = -7/8 light seconds.

Both camps can be satisfied with parameterizing P on x = -a c and x' = -7/8 light seconds.

Event P has
t = 7/6 seconds - (5/3) a
x = -a c
t' = 35/24 seconds - (4/3) a
x' = -7/8 light-seconds

Event Q has
t = 7/6 seconds
x = 0
t' = 35/24 seconds
x' = -7/8 light-seconds

Q - P has
Δt = (5/3) a
Δx = a c
Δt' = (4/3) a
Δx' = 0

Obviously, for all positive a, Δx/Δt = (3/5) c, Δx'/Δt' = 0, and Δt' = Δt/γ < Δt.

Also, since the Lorentz transform preserves it, (cΔt)² - (Δx)² = (16/9)² a² c² = (cΔt')² - (Δx')², which is a quick way to compute the proper time for any time-like interval.

I do not understand why you are doing P and Q.

LT is for events after the co-location of the origins in the standard configuration. Does that mean this all built into LT? does LT calculate the elapsed times since the origins were co-located. If not, then what is purpose if LT? In standard configuration.

But, your way OK with x' = -7/8 and then want to know when origin of unprimed frame reaches (x',0,0). Is this t' = -x'/-v = (-7/8)(-5/3) = 35/24s

Now, I have space-time coordinate [ 35/24s, -7/8,0,0 ] and apply LT matrix for standard configuration. Is this correct?

[ 35/24s, -7/8,0,0 ] LT = [7/8s, 0, 0, 0] If this false, then what good LT is for this? In short, I want to translate the event/coordinate of [ 35/24s, -7/8,0,0 ] in the primed to the unprimed frame using LT.

 

[chinglu1998]

Okay, let's work this out using the Lorentz transformation for time. Usually, in order to apply the Lorentz transformations, we originally have two observers that pass each other at the origin and synch their clocks to T=0 while passing, but we can still find the difference in times that pass upon the clocks for each between two events without synching to each other. For event 1, we have (0, -7/8 ls, 0, 0) for the observer at the origin in the rest frame. For event 2, we have ( (7/8 ls)/(3/5 c), 0, 0, 0), so t1 = 0 and t2 = (7/8 ls) / (3/5 c) = 35/24 sec, giving t = t2 - t1 = 35/24 sec. For the time that passes upon the clock of the traveling observer P', then, we have

t' = t2' - t1'

= γ ( t2 - v x2 / c² ) - γ ( t1 - v x1 / c² )

= γ (t2 - t1) - γ (x2 - x1) v / c^2

whereas v = (x2 - x1) / (t2 - t1), giving (x2 - x1) = v (t2 - t1), so

t' = γ (t2 - t1) - γ (t2 - t1) v^2 / c^2

= γ (t2 - t1) (1 - v^2 / c^2)

= γ (t2 - t1) / γ^2

= (t2 - t1) / γ

= t / γ

= (35/24 sec) / (5/4) = 7/6 sec

This equation

t' = γ ( t2 - v x2 / c² ) - γ ( t1 - v x1 / c² )
I can not understand as you have it.

Since in view of unprimed frome, x1=0, t1=0, is this not correct?

Also, x2 = x'2/γ by length contraction, is this not correct?

All measurements in moving frame length contracted along motion of travel (x-axis) in view of rest frame.

 

[grav-universe]

This equation

t' = γ ( t2 - v x2 / c² ) - γ ( t1 - v x1 / c² )
I can not understand as you have it.

Since in view of unprimed frome, x1=0, t1=0, is this not correct?

Also, x2 = x'2/γ by length contraction, is this not correct?

All measurements in moving frame length contracted along motion of travel (x-axis) in view of rest frame.

Let's say that P' carries two balloons. According to P, P' pops one balloon at x1 = -7/8 ls at t1. P' pops the other balloon upon reaching P, so at coordinates x2 = 0 at t2. Since P' carries the balloons, both events occur at x1' = 0 and x2' = 0 and we also have the relation v = (x2 - x1) / (t2 - t1), so we can find t2' - t1' as before to determine the time dilation. x1, x2, t1, and t2 themselves, however, are just the coordinates, a specific marking upon a ruler for the distance of an event and a specific reading upon a clock for when each event takes place. The time dilation is a ratio of some time that passes for each observer, so we must use a difference in times between two events. That is to say, for example, if the time dilation is 1/2, then P will read 2 seconds passing between two readings upon the clock of P as compared to 1 second that P views to have passed between two readings upon the clock of P'. It doesn't matter how the clocks are synched in that case. If, for instance, the difference between clock readings for P is 3.5 sec - 1.5 sec and for P' is 4.21 sec - 3.21 sec, then the time dilation for the ratio of time that passes on each clock according to P is 1 sec / 2 sec = 1/2 when finding the ratio by taking a difference between the readings at each event. If the difference for P is 7.4 sec - 3.4 sec and the difference for P' is 2.1 sec - .1 sec, then the time dilation is still 2 sec / 4 sec = 1/2. Likewise, if say P and P' have identical ships as each measures it in their own frame, the length contraction is the difference in markings upon a ruler of P as P simultaneously measures each end of the ship of P' as compared to the length, or the difference between markings upon the same ruler of P, for the ship of P.

 

[rpenner]

I do not understand why you are doing P and Q.

I thought it was physically obvious, since you want to talk about an elapsed time you need two events, and since you are talking about the event where a clock at x' = -7/8 light-seconds intersects something else at x = 0, that gives Q as the unique intersection.

Since you insist on calling the thing at x = 0 "the origin" (as if the "rest frame" was privileged) we can O to label its world-line, and like I did with P, I can with a single parameter on t, t' or x' express all the events on that worldline.

Event O(b) has
t = 7/6 seconds - (4/3) b
x = 0
t' = 35/24 seconds - (5/3) b
x' = a c -7/8 light-seconds

The choice b = 7/8 seconds gives the conventional origin:

Event O(7/8 seconds) has
t = 0
x = 0
t' = 0
x' = 0

Event P(a) has
t = 7/6 seconds - (5/3) a
x = -a c
t' = 35/24 seconds - (4/3) a
x' = -7/8 light-seconds

The choice a = 7/10 seconds gives the t=0 choice:

Event P(21/30 seconds) has
t = 0
x = -7/10 light seconds
t' = 21/40 seconds
x' = -7/8 light-seconds

The equally valid choice a = 35/32 seconds made by grav-universe gives the t'=0 choice

Event P(35/32 seconds) has
t = -21/32 seconds
x = -35/32 light-seconds
t' = 0
x' = -7/8 light-seconds

But Q = O(0) = P(0) is where "the origin" meets "the clock".

So the elapsed time on "the clock" is given by the Δt' of Q - P(a) which has

Δt = (5/3) a
Δx = a c
Δt' = (4/3) a
Δx' = 0

LT is for events after the co-location of the origins in the standard configuration.

The homogenous Lorentz transformation works for all events in the space-time, event if they happen "before" the origin or with a space-like separation. If you want you can even compose Lorentz transformation with a translation in space and time to get the inhomogenous Lorentz transformation, so co-location of the origins is not strictly necessary. Since the Lorentz transformation is a linear transformation of coordinates, it is also a valid transformation of coordinate differences:

LT(t2) - LT(t1) = t2' - t1' = Δt' = LT(Δt)

And this means that the location of the origin is of no physical significance. There is no preferred origin or rest frame.

Does that mean this all built into LT? does LT calculate the elapsed times since the origins were co-located.

The Lorentz transformation does indeed calculate elapsed time between any two events, but you are responsible for clearly identifying the two events in both space and time. Here the co-located origin is an event O(7/8 seconds), but does not have physical relevance to the clock which is on the line P(a).
Nor does the t or t' coordinate of the starting point of the clock have physical significance -- they are choices that have to be made by the one asking the question.

But, your way OK with x' = -7/8 and then want to know when origin of unprimed frame reaches (x',0,0). Is this t' = -x'/-v = (-7/8)(-5/3) = 35/24s

The clock at x' = -7/8 light seconds meets the "origin" at x = 0 at the event Q = O(0) = P(0).

"When" has no meaning until you specify whose clock you are using and when that clock was started. Choose P(21/30 seconds) with t = 0 or P(35/32 seconds) with t' = 0.

Q - P(35/32 seconds) gives:
Δt = 175/96 seconds
Δx = 35/32 light-seconds
Δt' = 35/24 seconds
Δx' = 0

So the proper time (the time measured by a clock moving from P to Q) is √((Δt)²-(Δx/c)²) = √((Δt')²-(Δx'/c)²) = 35/24 seconds.

Now, I have space-time coordinate [ 35/24s, -7/8,0,0 ] and apply LT matrix for standard configuration. Is this correct?

That is the coordinates of Q in the moving frame.

[ 35/24s, -7/8,0,0 ] LT = [7/8s, 0, 0, 0] If this false, then what good LT is for this? In short, I want to translate the event/coordinate of [ 35/24s, -7/8,0,0 ] in the primed to the unprimed frame using LT.

Then you need to apply the Lorentz transform with v = -3/5 c.

As it turns out, it's [7/6 s, 0, 0, 0] which you could have also figured out with:

t² = (t²-(x/c)²) = ((t')²-(x'/c)²) = ((35/24 seconds)² - (-7/8 seconds)²) = 49/36 seconds²

In 1908 it was shown that Special Relativity makes as much sense as Euclidean geometry. But you have to think about space and time at the same time.

 

[chinglu1998]

I thought it was physically obvious, since you want to talk about an elapsed time you need two events, and since you are talking about the event where a clock at x' = -7/8 light-seconds intersects something else at x = 0, that gives Q as the unique intersection.

Since you insist on calling the thing at x = 0 "the origin" (as if the "rest frame" was privileged) we can O to label its world-line, and like I did with P, I can with a single parameter on t, t' or x' express all the events on that worldline.

Event O(b) has
t = 7/6 seconds - (4/3) b
x = 0
t' = 35/24 seconds - (5/3) b
x' = a c -7/8 light-seconds

The choice b = 7/8 seconds gives the conventional origin:

Event O(7/8 seconds) has
t = 0
x = 0
t' = 0
x' = 0

Event P(a) has
t = 7/6 seconds - (5/3) a
x = -a c
t' = 35/24 seconds - (4/3) a
x' = -7/8 light-seconds

The choice a = 7/10 seconds gives the t=0 choice:

Event P(21/30 seconds) has
t = 0
x = -7/10 light seconds
t' = 21/40 seconds
x' = -7/8 light-seconds

The equally valid choice a = 35/32 seconds made by grav-universe gives the t'=0 choice

Event P(35/32 seconds) has
t = -21/32 seconds
x = -35/32 light-seconds
t' = 0
x' = -7/8 light-seconds

But Q = O(0) = P(0) is where "the origin" meets "the clock".

So the elapsed time on "the clock" is given by the Δt' of Q - P(a) which has

Δt = (5/3) a
Δx = a c
Δt' = (4/3) a
Δx' = 0

The homogenous Lorentz transformation works for all events in the space-time, event if they happen "before" the origin or with a space-like separation. If you want you can even compose Lorentz transformation with a translation in space and time to get the inhomogenous Lorentz transformation, so co-location of the origins is not strictly necessary. Since the Lorentz transformation is a linear transformation of coordinates, it is also a valid transformation of coordinate differences:

LT(t2) - LT(t1) = t2' - t1' = Δt' = LT(Δt)

And this means that the location of the origin is of no physical significance. There is no preferred origin or rest frame.

The Lorentz transformation does indeed calculate elapsed time between any two events, but you are responsible for clearly identifying the two events in both space and time. Here the co-located origin is an event O(7/8 seconds), but does not have physical relevance to the clock which is on the line P(a).
Nor does the t or t' coordinate of the starting point of the clock have physical significance -- they are choices that have to be made by the one asking the question.

The clock at x' = -7/8 light seconds meets the "origin" at x = 0 at the event Q = O(0) = P(0).

"When" has no meaning until you specify whose clock you are using and when that clock was started. Choose P(21/30 seconds) with t = 0 or P(35/32 seconds) with t' = 0.

Q - P(35/32 seconds) gives:
Δt = 175/96 seconds
Δx = 35/32 light-seconds
Δt' = 35/24 seconds
Δx' = 0

So the proper time (the time measured by a clock moving from P to Q) is √((Δt)²-(Δx/c)²) = √((Δt')²-(Δx'/c)²) = 35/24 seconds.

That is the coordinates of Q in the moving frame.

Then you need to apply the Lorentz transform with v = -3/5 c.

As it turns out, it's [7/6 s, 0, 0, 0] which you could have also figured out with:

t² = (t²-(x/c)²) = ((t')²-(x'/c)²) = ((35/24 seconds)² - (-7/8 seconds)²) = 49/36 seconds²

In 1908 it was shown that Special Relativity makes as much sense as Euclidean geometry. But you have to think about space and time at the same time.

This is a simple problem. Still use v = 3/5c.

Standard configuration so both frames start when origins are same.

Both frames can agree when origins are same.

There is clock at (-7/8,0,0) in primed coordinates.

Question: how long it take for this coordinate to reach unprimed origin when unprimed frame taken as stationary.


Next Question: how long it takes for unprimed origin to reach (-7/8,0,0) when primed frame taken as stationary.

Both frames can agree when unprimed origin and (-7/8,0,0) are same.

This is all i am asking.

 

[rpenner]

This is a simple problem.

This is a deceptive problem.

Both frames can agree when origins are same.

The event O has t = t'.

There is clock at (-7/8,0,0) in primed coordinates.

Question: how long it take for this coordinate to reach unprimed origin when unprimed frame taken as stationary.

The starting event is when t = 0 and x' = -7/8 and the ending event is when x = 0 and x' = -7/8.

Next Question: how long it takes for unprimed origin to reach (-7/8,0,0) when primed frame taken as stationary.

The starting event is when t' = 0 and x' = -7/8, which is a different event than in the first question. Comparing the two measures of time tells you nothing about time dilation because due to the relativity of simultaneity, you are comparing apples and oranges.

Both frames can agree when unprimed origin and (-7/8,0,0) are same.

FALSE! For event Q, t does not equal t'.

 

[chinglu1998]

This is a deceptive problem.The event O has t = t'.The starting event is when t = 0 and x' = -7/8 and the ending event is when x = 0 and x' = -7/8. The starting event is when t' = 0 and x' = -7/8, which is a different event than in the first question. Comparing the two measures of time tells you nothing about time dilation because due to the relativity of simultaneity, you are comparing apples and oranges.
FALSE! For event Q, t does not equal t'.

1) Both frame can agree that the event of the co-location of unprimed origin and primed coordinate is same event.
2) Both frame can agree that the event of the co-location of unprimed origin and primed origin is same event. I not say they agree on the elapsed time between events.
Is that not nature of relativity? Each assigns different space and time coordinates to same event?

Is the above true?


3) I believe this is all with the context of the relativity of simultaneity. Let us say there was clock at primed origin. I will agree you correct that clock at primed origin will beat time dilated compared to clock at unprimed origin. We agree on that.

When (-7/8,0,0) moves to unprimed origin the primed origin clock still beats time dilated compared to unprimed origin clock. But you say clock at (-7/8,0,0) also beats time dilated between co-location of origins and co-location of (-7/8,0,0) and unprimed origin while primed origin clock beat time dilated for same unprimed frame time interval.

So these two moving clocks are both time dilated in the view of the unprimed frame and thus are in sync in the view of the unprimed frame contradict relativity of simultaneity.

Is this correct?

 

[Dale]

This is a simple problem. Still use v = 3/5c.

Standard configuration so both frames start when origins are same.

Both frames can agree when origins are same.

There is clock at (-7/8,0,0) in primed coordinates.

Question: how long it take for this coordinate to reach unprimed origin when unprimed frame taken as stationary.Next Question: how long it takes for unprimed origin to reach (-7/8,0,0) when primed frame taken as stationary.

Both frames can agree when unprimed origin and (-7/8,0,0) are same.

This is all i am asking.

Your question is unclear. "How long it take" implies a starting event and an ending event. You are clear on the ending event, it ends when the worldline of the (primed frame) clock at r'=(-7/8,0,0) intersects with the worldline of the (unprimed frame) clock at r=(0,0,0). But what is the starting event?

 

[chinglu1998]

Your question is unclear. "How long it take" implies a starting event and an ending event. You are clear on the ending event, it ends when the worldline of the (primed frame) clock at r'=(-7/8,0,0) intersects with the worldline of the (unprimed frame) clock at r=(0,0,0). But what is the starting event?

Starting event is same for all standard configurations. It is when both origins are same.


Also assume that the origins of both coordinate systems are the same. If all these hold, then the coordinate systems are said to be in standard configuration.

http://en.wikipedia.org/wiki/Lorent...ormation_for_frames_in_standard_configuration

 

[Dale]

Starting event is same for all standard configurations. It is when both origins are same.

Both origins being the same is an event on the worldline of the (unprimed frame) clock at (0,0,0). It is not an event on the worldline of the (primed frame) clock at (-7/8,0,0). Therefore I surmise that you are asking about the coordinate time in each frame, i.e. from t=0 and from t'=0 respectively.

I will use j to represent the position of the primed frame clock in the unprimed coordinates, I will use j' to represent the position of the primed frame clock in the primed coordinates, I will use k to represent the position of the unprimed frame clock in the unprimed coordinates, and I will use k' to represent the position of the unprimed frame clock in the primed coordinates. I will use the usual four-vector notation (ct,x,y,z) and choose my units such that c=1. L is the Lorentz transform matrix from unprimed to primed coordinates and L' is the inverse (i.e. k'=L.k and k=L'.k').

In primed coordinates:
j'(t')=(t',-7/8,0,0)
k'(t')=(t',-3/5 t',0,0)
These intersect at -3/5 t' = -7/8 or t' = 35/24
j'(35/24)=k'(35/24)=(35/24,-7/8,0,0)
So it takes 35/24 units of time in the primed frame

In unprimed coordinates
j=k=L'.j'(35/24)=L'.k'(35/24)=(7/6,0,0,0)
So it takes 7/6 units of time in the unprimed frame

 

[chinglu1998]

Both origins being the same is an event on the worldline of the (unprimed frame) clock at (0,0,0). It is not an event on the worldline of the (primed frame) clock at (-7/8,0,0). Therefore I surmise that you are asking about the coordinate time in each frame, i.e. from t=0 and from t'=0 respectively.

I will use j to represent the position of the primed frame clock in the unprimed coordinates, I will use j' to represent the position of the primed frame clock in the primed coordinates, I will use k to represent the position of the unprimed frame clock in the unprimed coordinates, and I will use k' to represent the position of the unprimed frame clock in the primed coordinates. I will use the usual four-vector notation (ct,x,y,z) and choose my units such that c=1. L is the Lorentz transform matrix from unprimed to primed coordinates and L' is the inverse (i.e. k'=L.k and k=L'.k').

In primed coordinates:
j'(t')=(t',-7/8,0,0)
k'(t')=(t',-3/5 t',0,0)
These intersect at -3/5 t' = -7/8 or t' = 35/24
j'(35/24)=k'(35/24)=(35/24,-7/8,0,0)
So it takes 35/24 units of time in the primed frame

In unprimed coordinates
j=k=L'.j'(35/24)=L'.k'(35/24)=(7/6,0,0,0)
So it takes 7/6 units of time in the unprimed frame

Yes, I understand and this.

t' = 35/24

t = 7/6

γ = 5/4

Hence (7/6)(5/4) = (35/24)

That mean t' = γt

That mean, in the view of the rest/unprimed frame, the clock at the primed coordinate )=(-7/8,0,0) beats opposite of time dilation.

Given clock at primed origin beat time dilated, this mean relativity of simultaneity is preserved for these two clocks.

Any thought that clock at (-7/8,0,0) for the interval beats time dilated make relativity of simultaneity false.

 

[Dale]

That mean t' = γt

I always recommend to not use this formula or the length contraction formula, and I never use them myself.

That mean, in the view of the rest/unprimed frame, the clock at the primed coordinate )=(-7/8,0,0) beats opposite of time dilation.

Given clock at primed origin beat time dilated, this mean relativity of simultaneity is preserved for these two clocks.

Any thought that clock at (-7/8,0,0) for the interval beats time dilated make relativity of simultaneity false.

There is some sort of language barrier. I cannot make sense of what you are asking.

Are you surprised that asking a question about coordinate time gives you a frame-dependent answer?

 

[chinglu1998]

I always recommend to not use this formula or the length contraction formula, and I never use them myself.

I have much agreement. I feel good to find someone that calculates the way I do.

There is some sort of language barrier. I cannot make sense of what you are asking.

Are you surprised that asking a question about coordinate time gives you a frame-dependent answer?

No I am feel good you calculate rules consistent with the relativity of simultaneity.

I can not find others that use Lorentz transforms as should.

 

[Dale]

I can not find others that use Lorentz transforms as should.

I wouldn't worry too much about that. It was a little difficult to understand the question, so I am sure that is the source of most of the disagreements.

 

[chinglu1998]

I wouldn't worry too much about that. It was a little difficult to understand the question, so I am sure that is the source of most of the disagreements.

Many thanks for meeting you.

Hope have nice day.

 

[chinglu1998]

The original poster of this thread originally titled another thread Time dilation is false.

Event P is not well agreed upon. The OP seems to have chosen t = 0 and x' = -7/8 light seconds. While grav-universe prefers t' = 0 and x' = -7/8 light seconds.
Event Q is well agreed upon with x = 0 and x' = -7/8 light seconds.

Both camps can be satisfied with parameterizing P on x = -a c and x' = -7/8 light seconds.

Event P has
t = 7/6 seconds - (5/3) a
x = -a c
t' = 35/24 seconds - (4/3) a
x' = -7/8 light-seconds

Event Q has
t = 7/6 seconds
x = 0
t' = 35/24 seconds
x' = -7/8 light-seconds

Q - P has
Δt = (5/3) a
Δx = a c
Δt' = (4/3) a
Δx' = 0

Obviously, for all positive a, Δx/Δt = (3/5) c, Δx'/Δt' = 0, and Δt' = Δt/γ < Δt.

Also, since the Lorentz transform preserves it, (cΔt)² - (Δx)² = (16/9)² a² c² = (cΔt')² - (Δx')², which is a quick way to compute the proper time for any time-like interval.

[Moderator: This post is so deeply confused that we have to work backwards to extract any meaning.

For observer one, we have two events separated by a certain amount of space, Δx, and a certain amount of time, Δt. Here, Δx = 0.2 light-seconds, and Δt = 0.6 seconds. Because cΔt > |Δx|, or equivalently (cΔt)²−(Δx)² > 0, the separation between the two events is called time-like and in theory all observers agree in the order the two events occur and that a massive particle could move from the earlier event to the latter event with a speed less than c. That speed, for observer one, is w = Δx/Δt = c/3.]
http://www.physforum.com/index.php?showforum=30


Do you understand when you pointed folks to your website there was no logic at all from me where w = Δx/Δt = c/3. You made this up. This cause your entire long winded argument to fall apart and be false.


Also, do you understand yet your thoughts in this forum on time dilation of the moving coordinate (-7/8,0,0) contradicts the relativity of simultaneity?

 

[rpenner]

It does not harm me that you are howlingly ignorant about the mathematics of space-time. But, it does bothers me that you have brought your broken understanding to three different forums, in a language which you have not mastered, and sought to convince people by misleading problems that are not part of sound pedagogy that the Lorentz transformations predict no time dilation and absolute simultaneity.

Your questions are not formulated by a teacher who wishes to teach physics or a student that has mastered the underlying concepts of relativity or the need for space and time coordinates for every event in space time.

An example from your most recent post (which is an objection to moderation on another forum):

If you are OK with the interpretation of your first post in that a particle moves with uniform speed from x = 0 to x = 1/5 light seconds. And if you are OK with the interpretation of your confused presentation that this takes 3/5 second, how in the world do you object to the calculation of the particle's speed as w = Δx/Δt = c/3 ?

Furthermore, you look at the same two events in a frame moving at a speed of v = (3/5) c. But according to the Einstein velocity addition formula, there is an exact relation between w (the particle in frame one) and v (the speed of observer two in frame one).

$$v = \frac{w + w}{1 + \frac{w^2}{c^2}}$$
or
$$\tanh^{-1} \frac{v}{c} = 2 \tanh^{-1} \frac{w}{c}$$

That is v represents twice the hyperbolic angle of w, so when you transform to the frame moving at speed v, the particle which was moving at speed w is now seen as moving at speed -w,
$$\frac{w - v}{1 - \frac{wv}{c^2}} = \frac{w - \frac{2 w}{1 + \frac{w^2}{c^2}}}{1 - \frac{w \frac{2 w}{1 + \frac{w^2}{c^2}}}{c^2}} = \frac{w + \frac{w^3}{c^2} - 2 w}{1 + \frac{w^2}{c^2} - \frac{2 w^2}{c^2}} = \frac{\frac{w^3}{c^2} - w}{1 - \frac{w^2}{c^2}} = -w$$
or
$$\tanh^{-1} \frac{w}{c} - \tanh^{-1} \frac{v}{c} = \tanh^{-1} \frac{w}{c} - 2 \tanh^{-1} \frac{w}{c} = -\tanh^{-1} \frac{w}{c}$$
and the time dilation factor is the same for both.
$$\gamma(w) = \frac{1}{\sqrt{1-\frac{w^2}{c^2}}} = \frac{1}{\sqrt{1-\frac{(-w)^2}{c^2}}} = \gamma(-w)$$
or
$$\cosh \tanh^{-1} \frac{w}{c} = \cosh \tanh^{-1} \frac{-w}{c}$$

But just because the time dilation is the same, that doesn't mean that there is no time dilation, since you haven't looked at the particle in its comoving frame.

Both with this example from that other forum and this thread, you take simple geometrical problems and mangle important concepts like elapsed time verses time dilation or the relativity of simultaneity, and from that flawed conceptual background seek to claim that they "contradict relativity of simultaneity" or "proof time dilation false.".

So you are 105 years out-of-date. Stop trying to drag the world backwards.

 

[myuncle]

The original poster of this thread originally titled another thread Time dilation is false.

We should then link some of the threads in your forum where you show clearly that you are a joke as a moderator, for the simple fact that you don't allow any debate.

http://www.physforum.com/index.php?showtopic=28593
http://www.physforum.com/index.php?showtopic=28580
http://www.physforum.com/index.php?showtopic=28579

 

[Dale]

I think the three of you should take your arguments back to the other forum. It is rather impolite to register on a new forum for the purpose of continuing an argument that was already banned or moderated in another forum. I don't think it is explicitly against the rules, but it is certainly not a way to make a good impression.

 

[Beer w/Straw]

I think the three of you should take your arguments back to the other forum. It is rather impolite to register on a new forum for the purpose of continuing an argument that was already banned or moderated in another forum. I don't think it is explicitly against the rules, but it is certainly not a way to make a good impression.

It happens a lot.

 

[chinglu1998]

I think the three of you should take your arguments back to the other forum. It is rather impolite to register on a new forum for the purpose of continuing an argument that was already banned or moderated in another forum. I don't think it is explicitly against the rules, but it is certainly not a way to make a good impression.

Agreed, I should not brought moderator's comments. That is my fault.

But, they followed me here and I just want to discuss this narrow topic of OP in this forum to explore further.

 

[chinglu1998]

If you are OK with the interpretation of your first post in that a particle moves with uniform speed from x = 0 to x = 1/5 light seconds. And if you are OK with the interpretation of your confused presentation that this takes 3/5 second, how in the world do you object to the calculation of the particle's speed as w = Δx/Δt = c/3 ?

It is my apologies I brought your comment here.

However, a moving clock is at (-1/5,0,0) in the moving coordinates and moves to (1/5,0,0) in the stationary coordinates. So, your statement is not what I wrote.

Anyway, if you would like to continue this discussion of this topic in your forum, simply remove the suspension and we can continue. We still more to discuss since you can now know w = Δx/Δt = c/3 is false based on (-1/5,0,0) moving coordinate which moves to stationary coordinate (1/5,0,0) all in view of stationary system.

Again, apologies that I brought your comments here.