# Thread: Time Dilation and Negative Coordinates in Moving Frame

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## Time Dilation and Negative Coordinates in Moving Frame

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

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

Do you apply Lorentz Transforms like this

t' = ( t - vx/cē )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) = 7/10ls.

Now apply transform. Since at origin, x = 0

t' = ( 7/10s - (3/5c)(0/cē) )γ

t' = (7/10s) γ

This means moving clock beats faster.

What's wrong with this?

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I don't think this is wrong. Remember negative coordinate vectors mean that the entity is moving slower relative to the center of gravity rather than faster for positive coordinates. You would then expect less time dilation and a faster progression of time. I have never seen negative coordinate vectors used concerning velocity, if I understand your model correctly.
Last edited by forrest noble; 2010-Dec-11 at 04:20 AM. Reason: added: I don't think and last sentence.

3. Originally Posted by chinglu1998
Apply length contraction to know when P' reaches the rest origin. Distance is (7/8)/γ = (7/8)(4/5) = 7/10ls.
Your problem is that you are mixing two approaches, the "intuitive" approach, where you use concepts like length contraction and time dilation, and the "mathematical" approach, where you simply apply Lorentz transformations and just think of everything as a coordinate rather than a physical length. Let's try it each way:
Intuitive:
Here we start out like you did, and infer that the moving observer reckons the distance to the observer at the origin to be 7/10 ls. However, the moving observer reckons the speed of the observer at the origin to be that same 3c/5, so the moving observer knows that they will require (7/10)/(3/5) = 7/6 seconds to reach the other observer. The observer at the origin reckons that time as being (7/8)/(3/5) = 35/24 s, which is a longer time, so the observer at the origin reckons the moving clock as running slow because the moving clock registers only 7/6 seconds for that motion.
Mathematical:
The observer at the origin reckons the coordinates of the displacement between the events to be (delta x, delta t) = (7/8, 35/24). Applying the Lorentz transformation to those displacement coordinates gives that for the moving observer they become (in units where c=1): (delta x', delta t') = (7/8 - 3/5*35/24, 35/24 - 3/5*7/8)*(5/4) = (0, 7/6).

So either way, we see that the observer at the origin reckons that the journey took 35/24 s on their own clock, and 7/6 s on the moving observer's clock. The moving observer agrees that it took 7/6 s on their own clock, but they don't agree that the observer at the origin's clock ticked off 35/24 s during that journey, because they don't agree with the other observer's assessment of what the clock at the origin read when the journey began. This can be traced to the fact that given the assumptions about the distances in the setup of the problem, the observer at the origin was taken to be inertial, whereas the other observer would have needed to accelerate to 3/5 c to use the same assumptions about the distance that you describe. In accelerating to 3/5 c, the moving observer would reckon a shift in the time read on the clock at the origin, accounting for the discrepancy.

4. Originally Posted by forrest noble
This is not wrong. Remember negative coordinate vectors mean that the entity is moving slower relative to the center of gravity rather than faster for positive coordinates.
That is not true.

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Originally Posted by chinglu1998

This means moving clock beats faster.

What's wrong with this?
You've got to be careful in how you interpret the Lorentz transform -- things depend on just whose coordinates "you're in".

Let's take both the t and t' transforms and put them in differential form:

dt = y(dt' - vdx'/c^2)
and
dt' = y(dt - vdx/c^2)

The first tells us that dt = y dt' *when dx' = 0*. dx' is zero for a clock stationary in the primed frame, which means that from the POV of the unprimed frame, that clock tick takes longer. That is dt'/dt < 1. The primed clock is ticking slow relative to the unprimed frame.

The second equation says just the reciprocal, dt' = y dt, but this time when dx = 0, meaning a a clock stationary in the unprimed frame. So this means the unprimed clock ticks slow relative to the primed frame.

We have to do a similiar thing with length contraction, this time using the space transforms:

dx = y(dx' - v dt')
and
dx' = y(dx - v dt)

The first tells us that dx = y dx', when dt' = 0. But dt' = 0 means we're talking about a length measurement in the primed frame, where dt' = 0. That is, these are simultaneous events in the primed frame. This means dx' = dx/y. The primed frame sees length contraction of the unprimed frame.

The second is again the reciprocal, dx' = y dx, but when dt = 0, meaning a length measurement in the unprimed frame, dx = dx'/y. The unprimed frame sees length contraction of the primed frame.

Note this is a bit tricky. For the clock rates, the t equation gives the POV of the unprimed frame (dx' = 0), but for lengths, the x equation gives the POV of the primed frame.

-Richard

6. I have already replied to this elsewhere but I will go ahead and reply here too.

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. Comparing what A and B observe according to their own clocks for the time that passes between two events that coincide with B's positions while travelling toward A, we have

ΔtB = tB2 - tB1

= γ ( tA2 - v xA2 / cē ) - γ ( tA1 - v xA1 / cē )

= γ (tA2 - tA1) - γ (xA2 - xA1) v / c^2

whereas v = (xA2 - xA1) / (tA2 - tA1), giving (xA2 - xA1) = v (tA2 - tA1), so

ΔtB = γ (tA2 - tA1) - γ (tA2 - tA1) v^2 / c^2

= γ (tA2 - tA1) (1 - v^2 / c^2)

= γ (tA2 - tA1) / γ^2

= (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.

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Originally Posted by Ken G
Your problem is that you are mixing two approaches, the "intuitive" approach, where you use concepts like length contraction and time dilation, and the "mathematical" approach, where you simply apply Lorentz transformations and just think of everything as a coordinate rather than a physical length. Let's try it each way:
Intuitive:
Here we start out like you did, and infer that the moving observer reckons the distance to the observer at the origin to be 7/10 ls. However, the moving observer reckons the speed of the observer at the origin to be that same 3c/5, so the moving observer knows that they will require (7/10)/(3/5) = 7/6 seconds to reach the other observer. The observer at the origin reckons that time as being (7/8)/(3/5) = 35/24 s, which is a longer time, so the observer at the origin reckons the moving clock as running slow because the moving clock registers only 7/6 seconds for that motion.
Mathematical:
The observer at the origin reckons the coordinates of the displacement between the events to be (delta x, delta t) = (7/8, 35/24). Applying the Lorentz transformation to those displacement coordinates gives that for the moving observer they become (in units where c=1): (delta x', delta t') = (7/8 - 3/5*35/24, 35/24 - 3/5*7/8)*(5/4) = (0, 7/6).

So either way, we see that the observer at the origin reckons that the journey took 35/24 s on their own clock, and 7/6 s on the moving observer's clock. The moving observer agrees that it took 7/6 s on their own clock, but they don't agree that the observer at the origin's clock ticked off 35/24 s during that journey, because they don't agree with the other observer's assessment of what the clock at the origin read when the journey began. This can be traced to the fact that given the assumptions about the distances in the setup of the problem, the observer at the origin was taken to be inertial, whereas the other observer would have needed to accelerate to 3/5 c to use the same assumptions about the distance that you describe. In accelerating to 3/5 c, the moving observer would reckon a shift in the time read on the clock at the origin, accounting for the discrepancy.
Here we start out like you did, and infer that the moving observer reckons the distance to the observer at the origin to be 7/10 ls

This I can not understand.

Moving observer measures distance to unprimed origin to be 7/8.

How you write this?

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Originally Posted by publius
You've got to be careful in how you interpret the Lorentz transform -- things depend on just whose coordinates "you're in".

Let's take both the t and t' transforms and put them in differential form:

dt = y(dt' - vdx'/c^2)
and
dt' = y(dt - vdx/c^2)

The first tells us that dt = y dt' *when dx' = 0*. dx' is zero for a clock stationary in the primed frame, which means that from the POV of the unprimed frame, that clock tick takes longer. That is dt'/dt < 1. The primed clock is ticking slow relative to the unprimed frame.

The second equation says just the reciprocal, dt' = y dt, but this time when dx = 0, meaning a a clock stationary in the unprimed frame. So this means the unprimed clock ticks slow relative to the primed frame.

We have to do a similiar thing with length contraction, this time using the space transforms:

dx = y(dx' - v dt')
and
dx' = y(dx - v dt)

The first tells us that dx = y dx', when dt' = 0. But dt' = 0 means we're talking about a length measurement in the primed frame, where dt' = 0. That is, these are simultaneous events in the primed frame. This means dx' = dx/y. The primed frame sees length contraction of the unprimed frame.

The second is again the reciprocal, dx' = y dx, but when dt = 0, meaning a length measurement in the unprimed frame, dx = dx'/y. The unprimed frame sees length contraction of the primed frame.

Note this is a bit tricky. For the clock rates, the t equation gives the POV of the unprimed frame (dx' = 0), but for lengths, the x equation gives the POV of the primed frame.

-Richard
dt' = y(dt - vdx/c^2)

This part get me confused.

Is it the case that dx = (7/8)/y ?

Then what you get for dt?

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Originally Posted by grav
I have already replied to this elsewhere but I will go ahead and reply here too.

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. Comparing what A and B observe according to their own clocks for the time that passes between two events that coincide with B's positions while travelling toward A, we have

ΔtB = tB2 - tB1

= γ ( tA2 - v xA2 / cē ) - γ ( tA1 - v xA1 / cē )

= γ (tA2 - tA1) - γ (xA2 - xA1) v / c^2

whereas v = (xA2 - xA1) / (tA2 - tA1), giving (xA2 - xA1) = v (tA2 - tA1), so

ΔtB = γ (tA2 - tA1) - γ (tA2 - tA1) v^2 / c^2

= γ (tA2 - tA1) (1 - v^2 / c^2)

= γ (tA2 - tA1) / γ^2

= (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.
OK, unprimed frame concludes clock at ( -7/8,0,0) beats time dilated.

Is this what you say?

11. Originally Posted by chinglu1998
Here we start out like you did, and infer that the moving observer reckons the distance to the observer at the origin to be 7/10 ls

This I can not understand.

Moving observer measures distance to unprimed origin to be 7/8.

How you write this?
Let's say the stationary observer at the origin has a rigid iron bar of length 7/8 ls, and reckons the moving observers motion starting from the end of that bar until they reach the origin, all at speed 3/5 c. The moving observer will reckon that the rigid iron bar has length 7/10 ls, that is length contraction. The moving observer can also reckon their motion starting at the edge of that bar (which to them is a moving bar) until they reach the origin (which to them is just the arrival of the other end of the bar). If they do that, they will get the results I described.

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Originally Posted by Ken G
Let's say the stationary observer at the origin has a rigid iron bar of length 7/8 ls, and reckons the moving observers motion starting from the end of that bar until they reach the origin, all at speed 3/5 c. The moving observer will reckon that the rigid iron bar has length 7/10 ls, that is length contraction. The moving observer can also reckon their motion starting at the edge of that bar (which to them is a moving bar) until they reach the origin (which to them is just the arrival of the other end of the bar). If they do that, they will get the results I described.
Let's say the stationary observer at the origin has a rigid iron bar of length 7/8 ls, and reckons the moving observers motion starting from the end of that bar

Where did you account for length contraction?

The moving coordinate is at (-7/8,0,0)

Is this length contraction now false in relativity?

13. Originally Posted by chinglu1998
Let's say the stationary observer at the origin has a rigid iron bar of length 7/8 ls, and reckons the moving observers motion starting from the end of that bar

Where did you account for length contraction?
When I said the moving observer reckons the bar as being 7/10 ls long.
The moving coordinate is at (-7/8,0,0)
I have no idea what a "moving coordinate" is. However, I do know that the moving observer reckons their own starting coordinate, at the end of the rod, as (0,0), say, and their final coordinate, when they are reached by the other observer (at the other end of the rod), is (0, 7/6). All you need to get that is length contraction-- nothing else. That's also what publius was saying, and he pointed out that if you do it for the other observer, all you need is time dilation. Or you can do it with the Lorentz transformation, in which case you apply no manual corrections of any kind-- just let the transform do all the work. Your mistake above was mixing these approaches.
Is this length contraction now false in relativity?
I hope I have clarified under what circumstances one needs to invoke length contraction, and why not understanding that is the source of your problem. If you cannot get it from the above, I doubt there is much more I can say that will help.

14. Basically chinglu wants to move the origin (in this whole discussion) to (now) -7/8, that is all. And then because he does not take into account he has shifted the origin, and thus should make more translations comes up with a "nice" example of numerology, where, when you put in numbers you will find either 1=1 or you find 1=2.

I fail to see the reason for this thread, it is just a rehash of the ATM that was produced.