Is time universal? NO (and its proof)

No Pete, James R made it quite clear:

He said it does not matter who collects the paper read outs, BOTH observers will agree with the above readings.

That is not true according to special relativity, those are that thereadings that the stationary frame clocks will observe. The moving frame will not agree that his clock is the dilated clock from his own rest frame. That is not what special relativity predicts

This really is hilarious:
Aer: special relativity is nonsense.
James R: special relativity makes complete sense.
Aer: special relativity is nonsense.
James R: special relativity makes complete sense.
Aer: special relativity makes complete sense.
James R: special relativity is nonsense.

 

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Yes, that's correct.

No, that's not correct. Would you like to do the analysis?

 

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Do the following analysis: Assume our clock that was previously defined at rest in frame K' is stationary at x=0 and the other clocks move at .9c in the negative x direction. What will the markings be on the notepad?

 

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This is in fact the exercise that I suggested yesterday, so I will give you the solution
First of all, since I am a little bit lazy in typing, I shall use g instead of γ . I will also use c = 1., so that all velocities are given in fact in units of c, or if you prefer, instead of writting β i will use v. So that is g is defined to be g = 1/sqrt(1-v²).
In the following I will use the Lorentz transformation:

x' = g(x+vt)
t' = g(t+vx).

The inverse Lorentz transformation is

x = g(x'-vt')
t = g(t'-vx')

This assumes that the two referenxce frames have their origin x= 0 and t = 0 located at the same place when t=0 and t'=0.
This also assumes that the K' frames moves in the negative direction wrt the K frame.

We have now three events:
event A is clock 1 (the clock stationnary in the K frame) ticks 0.
event B is clock 2 (the clock stationnary in the K' frame) ticks 0
and event C is when both clocks reach each other.

Furthermore we assume (as a part of the exercise) that event A and event B are simultaneous in frame K (we shall see that they are not simultaneous in frame B). And finnally, we shall assume that the distance between the clocks when event A and B occur (simultaneously) in frame K, they are at a distance L (again, in frame K). Without loss of generality, we can assume that the events A and B occur at t = t' = 0. As you asked, since x'=0 for clock 2 at t'=0, at t=0, clock2 will be located also at x=0 in frame K (remember that the frames origin are located at the same position at t = 0 and t' = 0).
So that at t = 0, the location of clock1 will be at x = -L.
Let us see what happens in the K frame:
Clock 1 is stationnary so that its equation of motion is
x1(t) = -L
We can also write that for event A, xA = -L and tA = 0

Clock 2 moves to the negative direction with velocity -v, at t= 0 it is located at x=0 So that its equation of motion will be:
x2(t) = -vt (we assume that there is no acceleration and that for negative t the equation of motion is the same)
We can also write that for event B, xB = 0 and tB = 0.

Event C is when both clocks are located at the same position so that event C will occur when x1(t) = x2(t) ==> -L = -vt ==> t = L/v.
So that for event C we have xC = -L and tC = L/v.
tC = L/v will be the time written on the notepad by clock 1.
Due to time dilation, the time that will be shown by clock 2 will be tC/g = L/vg.

Let us go now to the K' frame.
In order to find tC' we have two ways, the first will be the easiest, the second will be more instructive, so let us look at both ways.

The first option is to say that we have an event C with coordinates xC and tC in the K frame and just do the Lorentz transformation in order to find xC' and tC'.

xC' will be given by:
xC' = g(Xc + vtC) = g(-L +v*L/v) = 0. That what we were supposed to find since the event C occur at the location of clock 2 which is allways at x'=0 in the K' frame.
Let us look now for tC'. Lorentz transformation will give:
tC' = g(tC + v xC) = g(L/v - vL) = gL/v(1-v²) = gL/vg² (since g = 1/sqrt(1-v², we have 1-v² -= 1/g²) so that tC' = L/gv. This is the same result as we have by just using the time dilation.

The second possibility is to transform the events A and B to the K' frame and to transform the equation of motions

First let us look at clock 1.
Its equation of motion is x1(t) = -L, So using the inverse Lorentz transformation we have g(x1' - vt') = -L, from this we get x1'(t') = vt'-L/g.
The equation of motion of clock 2 in K is given by x2 = -vt so that again by using the inverse Lorentz transformation we have
g(x2'-vt')= -vg(t'-vx2'), g cancels on both side and we are left with x2' - vt' = -vt' +v²x2'.
-vt' also cancel so that we are left with x2' = v²x2' ==> x2'(t) = 0.Let us calculate the time of event C. Event C will occur when both clocks are at the same location so that x1'(t) = x2'(t)
==> vt' - L/g = 0 ==> t' = L/gv. So that tC' will be tC' = L/gv.

Now (and this is for you MacM, maybe this will help you understand SR), what about time dilation?
Let us calculate at what time event A occurs in reference frame K'.
Since A and B are simultaneous in frame K, we shall see that they are not simultaneous in reference frame K'. Time dilation says that if we take the time from event A to event B in reference frame K' and divide by g, this is what time dilation will give.
since for event A we have tA = 0 and xA = -L' tA' and xA' will be related by Lorentz transformation.
Let us first calculate xA':
xA' = g(xA + tA) = -gL.
Now if we take this location and put it in the equation of motion of clock 1 in the K' frame, (I remind you that it is x1' = vt'- L/g) we get -gL = vt'-L/g ==> vt' = L/g-gL = Lg(1/g² - 1), but 1/g² = 1-v², so that we get vtA' = Lg(1-v²-1) = -Lgv² ==> tA' = -Lgv. We see that event A occured before event B in the K' frame.
Of course, we could just have taken the Lorentz transformation of tA in order to get:
tA' = g(tA + vxA) and use tA = 0 and xA = -L to get tA' = -gvL, which is of course the same result.

Now in order to use the time dilation formula, we must calculate the elapsed time between event A and event c in the K' frame.
This elapsed time will be given by Dt' = tC' - tA' = L/gv -(-vgL) = L/gv + vgL = gL/v(1/g² + v²), again by using the fact that 1/g² = 1-v², one gets that Dt' = gL/v(1-v² + v²) = gL/v.

Time dilation in the K' frame POV will state that the time between the two events A and C is gL/v, so that in the K frame the time between the two events should be gL/vg = L/v.

So that according to reference frame K', on the notepads there should be L/gv as the time given by clock 2 and L/v as the time given by clock 1.

All "accumulated" times are consistent in all frames.

 

See my previous post. All frame agrees that what it sees is the dilated time of the moving clock. This is what relativity predicts

 

Minor correction - the clock at rest in K' is stationary at x'=0, not x=0.

You haven't done this analysis yourself, have you? Isn't it funny that whenever I ask you to do an analysis, you never bother? Are you incapable? Lazy? What?

Perhaps you did a quick mental analysis in which you assumed that the clocks and notepads spaced 1 light-second apart in frame K have the same spacing in frame K'? Oops!


Here's one way to do it properly:

The clock and notepad fixed at x=0 are at x'=0 when t'=0.
According to SR, the distance between the x=0 notepad and the x=1 notepad will be contracted in K' by a factor of 0.436.
So according to SR, the notepad at x=1 must be at x'=0.436 when t'=0. Do you agree?

Now, since that notepad is moving at v=-0.9c in K', it must arrive at x'=0 when t' = -0.436/-0.9 = 0.484 seconds, so...

SR predicts in the K' frame that the x=1 notepad will be marked with t'=0.484.


Figuring the value of t stamped on the notepad is a bit trickier. We know that the x=1 clock reads t=0 simultaneously in frame K with the x=0 clock reading t=0, but what about in frame K'?

SR says that in frame K', the x=1 clock reads zero at x'=2.29, t'=-0.9γ=-2.06 seconds. Check this.
We need to know what the x=1 clock reads when it reaches x'=0 at t'=0.484, which is 2.55 seconds later in frame K'.
SR says that the clock in question is dilated by a factor of 0.436 in frame K', so in those 2.55 seconds in K', the clock goes from zero to 0.436x2.55 = 1.11 seconds, so...

SR predicts in the K' frame that the x=1 notepad will be marked with t=1.11.


You can do the same analysis for the other notepads if you're willing and able. I have, and they all check out.

Your turn!

 

Hi 1100f,
That's a very dense post... I was exhausted by the time I got to transforming the equations of motions, but I'll check it properly later.

I spotted a couple of issues so far:

Note that this is slightly different to James's problem, in which the K' frame moved in the positive direction wrt the K frame.

Event B only. I assume this is a typo, since I don't think you've actually used the stated assumption that event A occurs at t'=0.

 

I don't understand the problem here. Two clocks start ticking at the same spactime point and then meet up again later at another spacetime point for comparison. Each clock reads its own elapsed proper time since the first meeting. Every frame agrees on the elapsed proper time of each clock since proper time is an invariant. Either clock A is ahead of clock B or it isn't.

 

Of course not. Everything I said is correct.

You mean the clock that was fixed at x=1 (but now the moving clock..).

And here you see that the clocks starting times are not simultaneous.

Now that you have taken the time to see that distances create what is referred to as the relativity of simultaneity, you might want to try to explain how satellites in orbit are synchronized with ground clocks.

 

Aer and MacM appear to think that SR predicts differently.

 

No, I've correctly told you that the moving frame is always dilated compared to the stationary frame and that switching frames necessarily means there is a shift in simultaneity.

 

Yes... x=1 is a moving point in the K' frame. The clock that is fixed at x=1 is moving in the K' frame.

Of course... what did you expect?

Yes, that is true. The wrong part was when you said that both observers would not agree on the notepad record.

As the analysis shows (both mine and 1100f's), the notepad readings are frame invariant.

Aer, do you agree that both observers agree on the notepad readings or not?

You seem to think that if they do, it implies that the K' clock is dilated in the K' frame? Why do you think that?

One step at a time.

 

Usually true, but there is an exception:
When two events simultaneous in one frame are in the same location in the direction of motion of a second frame, then the two events are also simultaneous in the second frame. I'm really surprised that you haven't caught on to this yet... it slaps you in the face if you look at the Lorentz transform. Do you know what the Lorentz transform is?

Do you agree with PhysicsMonkey's post? Do you agree that SR says that every frame agrees on the elapsed proper time of each clock since proper time is an invariant?

 

Of course time is universal: has no beginning and no end and is experienced by all. But not experienced by all simultaneously during the exact same event. Why can we not postulate a multi-dimensional theory, rather than arguing about different frames of reference? This thread seems to be going around in circles.

 

That is the simultaneity issue that I've brought up countless times. "When" each clock starts is dependent on the frame you choose.

That is the model in which all things are simultaneous, the local ether frame. I don't necessarily "believe" either.

Are you suggesting that the distance between the two does not matter?

Pete, you have suggested to me that all frames will give the same result, which is not true for the reasons I've stated. The origin of the ground frame for instance never even intersects the satellite frame.

 

Aer also asserts that elapsed time for an accelerating clock is not well defined. He hasn't explained why.

 

Yes I have, because of the break in simultaneity inherent in a noninertial frame, or rather slipping through various inertial frames.

Also, neither the satellite frame, the ground frame, or the ECI frame are inertial frames as they are all all accelerating in orbits (or rotations).

 

I think we still consider the speed of light as a constant, no? So what's all the fuss about? If the speed of light is a constant than so is time. So what if our frame of reference differs. Think outside the box.

 

And I agree! ...Except if they start in the same place... as I've brought up countless times.

What rot. Please explain why you think that "both observers agree on Notepad markings" implies absolute time.

I strongly suspect you've misunderstood what I've suggested.

I've suggested two things:

1. SR says that if two events happen in the same place at the same time, then they happen in the same place and time in all frames.
2. SR says that when two events simultaneous in one frame are in the same location in the direction of motion of a second frame, then the two events are also simultaneous in the second frame.

The first statement is a consequence of the second.
The second statement is a consequence of the Lorentz transform.

Do you agree with them as they are written? Please don't read anything else into them.

I don't think it's constructive to talk specifically about the satellite situation if we can't agree on these general statements or the simpler notepad situation.