# Time Dilation

Discussion in 'Pseudoscience Archive' started by chinglu, Dec 15, 2010.

1. ### chingluValued Senior Member

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Have coordinate at x' = -5/6ls.

What Lorentz transforms standard configuration say for elapsed time for x' to reach origin of unprimed frame at x = 0.

t' = ( t - vx/c² )γ

t' = ( t - 0 )γ = tγ.

So, moving clock coming toward rest origin beat not time dilated.

Is this wrong?

3. ### TachBannedBanned

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Yes, it is wrong but you are too stupid to figure out why.

5. ### arfa branecall me arfValued Senior Member

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Don't worry about Tach, he doesn't actually know anything.

Well, he does know how to type.

Last edited: Dec 15, 2010

7. ### James RJust this guy, you know?Staff Member

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31,079
That's half a set of spacetime coordinates. What is the t' value?

I don't understand this.

Which two events are you talking about? You have given one coordinate of one event, but then you start talking as if there are two events.

I don't understand this either. Perhaps your English isn't so good.

8. ### rpennerFully WiredRegistered Senior Member

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I think I have mastered the translation process. Please allow me to assist.
In a 1+1 space time, let us have two inertial frames S, with unprimed coordinates x and t, and S', with primed coordinates x' and t'.
Let A be the time-like world-line containing all events in this spacetime such that $x' = - \frac{5}{6} \, \textrm{light-seconds}$.
Assuming that the frames S and S' agree on event O as being the origin of their respective space and time coordinate systems, we call this standard configuration such that event O has the following coordinates:
$x_O = 0, \, t_O = 0, \, x'_O = 0, \, t'_O = 0$
Let B be the time-like world-line containing all events in this spacetime such that $x = 0$.
Obviously, unless the frame S and S' are not in relative motion, world-lines A and B share only one event in common. Following convention (established below) we call this event Q.
Let us call the motion of S' relative to S as v, so now we may compute the coordinates of Q in both frames:
$x_Q = 0, \, t_Q = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}}, \, x'_Q = - \frac{5}{6} \, \textrm{light-seconds}, \, t'_Q = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v}$

Applying the Lorentz transform from S to S' for Q, we see that
$\begin{pmatrix} t'_Q \\ x'_Q \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} & \frac{-v}{c^2 \sqrt{1- \frac{v^2}{c^2}}} \\ \frac{-v}{\sqrt{1- \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} t_Q \\ x_Q \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} \\ \frac{-v}{\sqrt{1- \frac{v^2}{c^2}}} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} \end{pmatrix} =\begin{pmatrix} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \\ - \frac{5}{6} \, \textrm{light-seconds} \end{pmatrix}$

The poster then asserts that a clock on world-line A, which is moving at speed v according to S, is not time-dilated.

It is wrong for a number of reasons. Just because at event O (which is on world-line B, and not on world-line A) $t_O = t'_O = 0$ does not imply that on world-line A (the world-line of the moving clock) that $t = t'$. Specifically, on world-line A, t=0 and t'=0 specify two different events.

To see this parameterize the events on A with the parameter a (with units of length):
$x_{A_a} = \frac{a}{\sqrt{1-\frac{v^2}{c^2}}}, \, t_{A_a} = \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{c^2-v^2}{v c^2}) + \frac{a}{v}}{\sqrt{1-\frac{v^2}{c^2}}}, \, x'_{A_a} = - \frac{5}{6} \, \textrm{light-seconds}, \, t'_{A_a} = \frac{(\frac{5}{6} \, \textrm{light-seconds}) + a}{v}$

So if you wanted to answer a question about time dilation of a clock moving (according to S) on world-line A and ending at event Q, you need to choose a starting event P on A. (Once again, O is not on A.)

$P_{t'=0} \, = \, P_{a = - \frac{5}{6} \, \textrm{light-seconds}}$ with coordinates:
$x = \frac{- \frac{5}{6} \, \textrm{light-seconds}}{\sqrt{1-\frac{v^2}{c^2}}}, \, t = \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{-v}{c^2})}{\sqrt{1-\frac{v^2}{c^2}}}, \, x' = - \frac{5}{6} \, \textrm{light-seconds}, \, t' = 0$

But $P_{t=0} \, = \, P_{a = - (1 - \frac{v^2}{c^2}) (\frac{5}{6} \, \textrm{light-seconds} )}$ has different coordinates:

$x = - \sqrt{1 - \frac{v^2}{c^2}} (\frac{5}{6} \, \textrm{light-seconds} ) , \, t = 0, \, x' = - \frac{5}{6} \, \textrm{light-seconds}, \, t' = \frac{v (\frac{5}{6} \, \textrm{light-seconds})}{c^2}$

So because $P_{t'=0} \neq P_{t=0}$ you have to be extra careful when talking about time dilation as a clock moves from P to Q, since
$t_Q - t_P$ depends on the choice of P.

FYI, this is likely chinglu1998 who recently appeared on various forums with contrived physical situations which he seeks to leverage into arguing that the Lorentz transform predicts absolute simultaneity or absence of time dilation for moving clocks.

Watch out for ambiguous problem statements that leave one or more endpoints of an interval ambiguously defined so that an apples and oranges comparison results or arguments that a Lorentz transformation of a uniformly moving particle may result in a frame where the particle has the same time dilation factor as the original speed. (This happens in 1+1 space time when you switch to a frame with twice the rapidity of the particle.)

9. ### chingluValued Senior Member

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1,637
This picture.
Code:

|---> v
---------x'---------|------------------
|
Origins of frames common here


Interval starts when origins are common. Interval ends when x' and unprime origin common.

Unprime, apply length contraction
dx = ( 0 - x'/γ )

dx = vdt

-x'/γ = vdt

dt = (-x'/γ)/v

prime
The unprime origin moves to x' coordinate in this frame.
dx' = -x'
dx' = vdt' = -x'
dt' = -x'/v.

dt = dt'/γ or dtγ = dt'. For x' to move to unprime origin prime clock beats faster than origin clock.

This my calculation.

10. ### chingluValued Senior Member

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1,637
Start of time interval is origins same.

End interval is when x' at same place as unprime origin.

This is well defined interval.

Next, when origins same, do not set t'=t=0. It is impossible to sync clocks in frames to 0 during motion. This only makes sense for light pulse.

So When origins same only record time on clocks as t0 and t0'.

Now simple question in unprimed frame how long it takes for x' to reach origin.

dx = vdt. It is that simple.

dx = -x'/γ. This applys lengths contraction. This is simplest motion possible in a frame.

So very simple dt = (-x'/γ)/v.

Now simple question in primed frame how long it takes for unprime origin to reach x'.

No length contraction.
dx' = vdt'.

dt' = dx'/v = -x'/v.

Check ratio dt'/dt.

11. ### chingluValued Senior Member

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Many thanks.
Since his post had no math I simply dismissed it out of hand.

12. ### TachBannedBanned

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If you want to refute relativity , you will need to learn it first. So far, despite of posting the same mistakes all over the internet, you haven't.

13. ### chingluValued Senior Member

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Here is thinking in a simple way.

Both frames can understand when origins same.
Both frames can understand when unprimed origin and coordinate (x',0,0) same.

dt = dx/v = (-x'/γ)/v.

dt' = dx'/v = -x'/v.

This does not refute relativity.

14. ### TachBannedBanned

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No, it only shows that after spamming multiple forums on the internet you still don't understand any.

15. ### chingluValued Senior Member

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1,637

Here is general problem. Let (x',y',z' ) coordinate in primed frame.

Let (x,y,z ) coordinate in unprimed frame. Assume standard configuration. this means motion along x-axis and time starts when origins same.

Question to answer how long it takes for (x',y,z ) to be at same location as (x,y,z ) from when origins same.

dx = vdt.

dx = x - x'/γ because of length contraction of x'.

x - x'/γ = vdt
x'/γ = x - vdt

x' = (x - vdt)γ Have you seen this equation before?

dx' = vdt'.

dx' = x/γ - x' because of length contraction of x.

x/γ - x' = vdt'
x/γ = x' + vdt'

x = (x' + vdt')γ Have you seen this equation before?

16. ### TachBannedBanned

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The "general problem" is that you don't know SR. You TRY to apply it, without knowing what you are doing. You have been refuted countless times on countless forums, yet , you persist. Why?

17. ### chingluValued Senior Member

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As any person can see in above post all I have been doing is a different way of rewriting Lorentz Transforms.

This mean if your thoughts are true, then Lorentz Transforms have been refuted.

I have read some your posts. I have see you use math when you have actual argument.

I see also you do not use math when responding to my posts.

18. ### TachBannedBanned

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....while showing a total lack of understanding of what they represent.

No, it means you don't understand basic relativity. As proven by your threads being locked and you being banned for trolling.

There is no point , you would not understand it anyways.

19. ### chingluValued Senior Member

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This is respected poster in this forum. He agrees with my calculations.

physicsforums.com/showpost.php?p=3032483&postcount=23

physicsforums.com/showpost.php?p=3032528&postcount=24

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21. ### rpennerFully WiredRegistered Senior Member

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Recap:
• World-line B has events O and Q.
• World-line A has events $P_{t'=0}$, $P_{t=0}$ and Q.
• In frame S, world-line A moves at speed v and world-line B moves at speed 0.
• In frame S', world-line A moves at speed 0 and world-line B moves at speed -v.
• Every event has a x and t coordinate in S, and a x' and t' coordinate in S'.
• The Lorentz transformation connects the coordinates in S with the coordinates in S' for the same event.
$\begin{tabular}{l|cccc} \textrm{Event} & x & t & x' & t' \\ \hline \\ P_{t'=0} & \frac{- \frac{5}{6} \, \textrm{light-seconds}}{\sqrt{1-\frac{v^2}{c^2}}} & \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{-v}{c^2})}{\sqrt{1-\frac{v^2}{c^2}}} & - \frac{5}{6} \, \textrm{light-seconds} & 0 \\ O & 0 & 0 & 0 & 0 \\ P_{t=0} & - \sqrt{1 - \frac{v^2}{c^2}} (\frac{5}{6} \, \textrm{light-seconds} ) & 0 & - \frac{5}{6} \, \textrm{light-seconds} & \frac{v (\frac{5}{6} \, \textrm{light-seconds})}{c^2} \\ Q & 0 & \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} & - \frac{5}{6} \, \textrm{light-seconds} & \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \end{tabular}$
• There are therefore 3 time-like intervals in the forward direction that terminate at Q
• Which clock is considered moving depends on what world-line the time-like intervals describe.
• In all cases the time dilation appears based on v or -v, depending on in which frame the interval describes movement.
• So it is important both when and where your clock starts and when and where your clock ends to determine who sees it moving and who sees time-dilation
$\begin{tabular}{l|ccc|cccc|c} \textrm{Interval} & \textrm{World-line} & \textrm{Moving} & \textrm{Unmoving} & \Delta x_{Moving} & \Delta t_{Moving} & \Delta x_{Unmoving} & \Delta t_{Unmoving} & \frac{\Delta t_{Moving}}{\Delta t_{Unmoving}} \\ \hline \\ Q - P_{t'=0} & A & S & S' & \Delta x = \frac{5}{6} \, \textrm{light-seconds} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \Delta t = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}& \Delta x' = 0 & \Delta t' = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} & \frac{\Delta t}{\Delta t'} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ Q - O & B & S' & S & \Delta x' = - \frac{5}{6} \, \textrm{light-seconds} & \Delta t' = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} & \Delta x = 0 & \Delta t = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} & \frac{\Delta t'}{\Delta t} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ Q - P_{t=0} & A & S & S' & \Delta x = \frac{5}{6} \, \textrm{light-seconds} \sqrt{1 - \frac{v^2}{c^2}} & \Delta t = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} & \Delta x' = 0 & \Delta t' = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \left(1 - \frac{v^2}{c^2} \right) & \frac{\Delta t}{\Delta t'} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ \hline \end{tabular}$

Importantly, Special Relativity has a geometric interpretation, so that you can graph these events as points and intervals as line segments for either S or S' coordinates.

We are talking about a space-time interval, so you have to say when and where with clarity. Only at x'=0 does t=t'=0. At x' = -5/6 light-seconds t'=0 and t=0 are different events with different coordinates. See the tables above.
Event Q is the unique event in space-time when x' = -5/6 light-seconds and x = 0.

This is correct for $t_Q - t_{P_{t=0}}$.

This is correct for $t'_Q - t'_{P_{t'=0}}$ but importantly $P_{t'=0} \neq P_{t=0}$.
Your calculation mixes apples and oranges, and therefore in physically invalid and appears based on assuming absolute simultaneity in that you do not recognize that $P_{t'=0} \neq P_{t=0}$ since $x' \neq 0$ and that you have not considered the geometrical content of what is world-line A and B, and what is the the locus of all events where t = 0 and t' = 0.

Alternately for the interval Q-O (which is on world-line B) your calculations are correct, but world-line B is not moving in the primed coordinates. So the physical meaning of t and t' swap places and time dilation of the moving clock is still exactly as predicted.

Actually, we have shown that at x' = -5/6 light-seconds, t' = 0 is not the same time as t = 0. You need a time and a place for the start or end of an interval.
The event Q has both a time and a place and has x' = -5/6 light-seconds and x = 0.

Not until you pick t' = 0 or t = 0, because you can't have both when x' = -5/6 light-seconds.

t'=t=0 only at the event where x'=x=0 i.e. at event O. You were the one who insisted on "standard configuration". t and t' are coordinate time not synchronized clock time -- we don't care about the synchronized time in this example, we only care about the time elapsed between two events in space time, so synchronization is not an issue. The Lorentz transformation is homogeneous, so it works perfectly fine with coordinate differences, provided you are correctly working with well-defined intervals.

That's fine for a clock on world-line B, but the clock you seem to want to talk about is on world-line A. So you have to choose, t = 0 or t' = 0.

That's starting at $P_{t=0}$ and ending at $Q$.

The length contraction formula is only valid when the t coordinate of the ends is the same, and since you are having problems with keeping your events straight, you should probably stick with the full Lorentz transforms.

But this is starting from $P_{t'=0}$ and ending at $Q$.

There's no pedagogically valid reason to, since you have impermissible mixed apples and oranges and due to your error of assuming absolute simultaneity.

Your posts have bad and misapplied math, which we do not dismiss out of hand, but attempt to engage you and determine where your geometrical and physical reasoning has gone awry so that you might learn how to apply these concepts correctly.

O and Q are well-defined events in space-time. They have a well-defined place and a well-defined when. But for Q, $x_Q \neq x'_Q$ and $t_Q \neq t'_Q$ so obviously someone should be careful when talking about $t = 0$ or $t' = 0$ since they only both happen at O.

Last edited: Dec 15, 2010
22. ### chingluValued Senior Member

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First, the coordinate ('x,0,0) is not an event. It is a space coordinate. It translates to ('x/γ,0,0) when origins are same.

There are two events in this problem.
1) The event of both origins being same.
2) The event of x' and x being same.

Next you cannot set t'=t=0 when time-like intervals. It is impossible to Einstein synchronize the clocks during the motion and get meaningful results.

So all you can deal with is time intervals as relativity produces.

So you can only record the frame times when origins same and work with time intervals.

This is simple question and if you refute it you refute lorentz transforms.

Is below true or false.
If false please give specific reasons on these simple equations.

Here is general problem. Let (x',y',z' ) coordinate in primed frame.

Let (x,y,z ) coordinate in unprimed frame. Assume standard configuration. this means motion along x-axis and time starts when origins same.

Question to answer how long it takes for (x',y,z ) to be at same location as (x,y,z ) from when origins same.

dx = vdt.

dx = x - x'/γ because of length contraction of x'.

x - x'/γ = vdt
x'/γ = x - vdt

x' = (x - vdt)γ Have you seen this equation before?

dx' = vdt'.

dx' = x/γ - x' because of length contraction of x.

x/γ - x' = vdt'
x/γ = x' + vdt'

x = (x' + vdt')γ Have you seen this equation before?

Maybe you should look at Galilean transformations and how they function.

They answer two questions in a frame if time start when origins are same.
1) If t in a unprimed frame elapsed, what x' will at same place as x.
2) If given x' and x what is elapsed time these coordinates be at same place.

Lorentz no different except must use length contraction of the other frames coordinates to get relativity answer.

23. ### chingluValued Senior Member

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We also went through this in other forum. We assumed your answer was correct that all time intervals for moving frame time dilated.

We then had clock at negative coordinate x'. We had clock at origin of moving frame and rest frame.

We then start time intervals when origins same. Record time t0 and t0'. All clocks in moving frame have t0' when origins same and all clock in rest frame have t0 in rest frame. These apply only to frames themselves.

We then assume when x' clock reach rest origin and rest origin elapsed t and rest clock has t0 + t as its time.

We said clock at x' elapsed t/γ and has t0' + t/γ to agree with your calculations.

But clock at moving origin elapsed t/γ also and has t0' + t/γ on its clock.

This mean rest frame concludes these clock remain synchronized which contradict relativity of simultaneity.