# Invariant spacetime and time dilation

Discussion in 'Physics & Math' started by Confused2, Jun 11, 2020.

1. ### Confused2Registered Senior Member

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607
I'm channelling someone else (rpenner) here:
You can't distinguish between relative and absolute space using maths - the maths is the same for both*.
So, absolute or relative space is an interpretation. Most people (myself included) seem to accept relative space as a fact - but it doesn't have to be.

*proof is beyond my ability.

3. ### HalcRegistered Senior Member

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227
I thought of a way to argue against the abolute interpretation: No known coordinate system foliates all of spacetime, and that seems to be a requirement for absolute time. In other words, there is no known single method that objectively orders all events. Notable problem events are distant ones (beyond the event horizon 16 BLY away), and those in the interior of geometries like black holes. So somebody who has fallen into a black hole cannot ask what time it is on Earth right now, but if there was absolute time, there'd be an answer to that question even if there is no way to empirically know it.

5. ### phytiRegistered Senior Member

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492
Neddy Bate;

Then you are misunderstanding me. But that's ok, because it's reciprocal!

The conclusions are always relative to the frame of the observer, and any inertial frame may serve as a reference for the purpose of comparison.
The motion induced phenomena of td and lc alter all processes in a moving frame. One of those processes is perception, the working of the mind. Therefore the moving observer cannot be aware of td and lc of their own frame, only outside observers moving with a different velocity.
If the moving observer was aware, he could determine his motion in space relative to light and that of other frames. The constant independent speed of light is equivalent to a fixed frame, and why Lorentz preferred the luminiferous ether. If events (light emission) moved, there would be no invariant intervals.

A comparison is made for the elapsed interval of time, from departure to return. The moving B clock shows a smaller time interval than the static A clock. The difference is real/actual. The clocks are identical as to function, so the explanation is, the B clock rate was slower than the A clock rate while away. This is the conclusion after the motion occurred, i.e. verification of the SR prediction. It's not about which times are real vs apparent, but why they are different.

7. ### Neddy BateValued Senior Member

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2,121
Well, that's good, because if it turns out that you are not trying to tell me that there is any time dilation (td) or length contraction (lc) in any clock's/object's own reference frame, (i.e. the reference frame in which that clock's/object's velocity is zero), then I will be very happy, and we should be able to come to an agreement very soon.

Unfortunately that sounds to me like you are again trying to tell me that there is time dilation and length contraction in their own frame, but they just cannot be aware of it, due to something to do with the working of the mind. Have you ever considered that there might not be any time dilation or length contraction in their own frame to begin with? Because that is what I have been trying to get you to consider.

Yes, in the twin scenario, SR predicts that in the inertial reference frame of the say-home clock A, the rate of the traveling clock B would be slowed by time dilation. And since clock A never accelerates, it never changes reference frames, and so we know the end result must be that clock B will show a smaller time interval than clock A.

However, SR also predicts that during the inertial portions of the voyage of clock B, it would be the rate of clock A which must be slowed by time dilation, according to clock B's own inertial reference frame. When considered from the inertial reference frames of clock B, (the outbound leg and inbound leg can both be inertial), we have to accept that SR predicts that it is the rate of clock A which must be slowed by time dilation. That is why it is was considered to be a "paradox" because the explanation could not be as simple as, "The explanation is, the B clock rate was slower than the A clock rate," as you stated. At least not when considered from the inertial reference frames of clock B, where the outbound leg and inbound leg can both be inertial.

If you would like to know the better explanation for the twin scenario, we can talk about that more, but I would rather leave the twin scenario out of this discussion until you and I can come to an agreement on the other points made in this post.

Last edited: Jul 4, 2020
8. ### phytiRegistered Senior Member

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492
Neddy Bate;

NO. The home twin is also moving and thus affected by td, but can be used for comparison purposes based on relative speed. Standards are arbitrarily defined for an intended purpose. Motion is relative, so we are not looking for anything absolute.

NIST, who's business is precision timekeeping, knows from experiment, all clocks lose time depending on their motion and gravity. That variation is reason to issue the world time standard as an average of a few atomic clocks.

NO. There is td and lc in all frames, none of which can be detected within the frames. Observer A with the A-clock is describing effects for the B-frame resulting from its (B-frame) motion relative to A. The B-clock could be in a speeding spacecraft, and would not be part of the A-frame.
If the ship is lc, then all its contents are lc, and why the anaut in the ship cannot detect any lc in his own frame. It's only logical, his speed is the same as that of his ruler!
The moving observer cannot detect his own constant velocity in space. He can only detect a change (speed or direction).

Keep the observer separate from what he observes.

A quote from the author:"From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by t(v/c)^2, t being the time occupied in the journey from A to B."
OTEOMB, A.Einstein, 1905, par.4
Here K is the rest frame, and clock-2 is moving relative to clock-1, thus clock-2 is not part of the K frame, has relative motion, and runs slower.

9. ### Neddy BateValued Senior Member

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2,121
phyti;

Yes, I agree that observer A is at rest in the A frame, and he is describing the B frame which is moving relative to the A frame. Observer A can say something like, "From my reference frame, (frame A), I can measure that the B-clock runs at a slower rate, due to my measuring the velocity of clock B to be NON-ZERO." I can agree with that.

I can also agree with the observer A saying, "From my reference frame, (frame A), I can measure that the biological functions of observer B are also running at a slower rate, by the same factor as the slower rate of clock B, due to my measuring the velocity of observer B to be that same NON-ZERO velocity." That is fine by me.

The only thing I would not agree with is if observer A said, "From the reference frame of the traveler, (frame B), the slow rate of observer B's biological functions will cause him to be unable to detect that his own clock is running at a slower rate." I find that unnecessary and wrong, because it is mixing two different effects from two different frames. I think now that you and I can agree on this, hopefully?

It would be much better if he would instead say, "From the reference frame of the traveler, (frame B), the rate of observer B's biological functions and the rate of clock B would both be running at their proper rate, not at a slower rate, due to him measuring his own velocity and the velocity of clock B to both be ZERO."

When you say things like this...
...it still sounds to me that you are saying that there is LC in the astronaut's own frame, because you even use the words, "in his own frame." You also use the words "cannot detect" as if the LC is really there, but just cannot be detected. Compare that carefully to the three different statements I made above, and you will see that the one I object to uses the words, "unable to detect" which is yet again very similar to what you keep saying.

But I think we can agree now, that you are not quite saying that there is LC in his own frame, I just think you are explaining it weirdly.

Last edited: Jul 8, 2020
10. ### phytiRegistered Senior Member

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492
Neddy Bate;

From this, it sounds like you don't think td or lc is physically real.
When the B clock departs from the A clock, after both are synchronized (to simplify calculations), then rejoins the A clock, there is a real difference in clock readings.
Each observer will say their clock ran at 'normal' speed, but they couldn't have, if there is a difference.

How do you explain the difference?

Relativity was and is weird compared to the accepted theories of the day.
Reciprocity of lc is only possible if it is real for at least one of the pair of observers.

11. ### Neddy BateValued Senior Member

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2,121
No, I am saying that TD and LC only exist when there is relative motion between two inertial frames. So for the A observer, the real TD and LC exist in the B frame, but there is absolutely no TD or LC in the A observer's own frame. Likewise, for the B observer, the real TD and LC exist in the A frame, but there is absolutely no TD or LC in the B observer's own frame.

I already answered that in post #39 in reply to your question, "Without the biological clock change, how do you explain the difference in aging for the 'twin scenario'?"

My answer: There is a biological clock change, but it is not what you seem to be describing. You make it sound like the "real" amount of time that passed during the 'twin scenario' is the amount of time that the stay-home twin experienced, and therefore the traveling twin must have been aging slow in his own frame, but unaware of it. No, the traveling twin aged at the normal rate in his own frame, and the smaller amount of time he experienced is just as "real" of a measurement of how much time passed during the 'twin scenario'.

The word reciprocity should explain it all, because it is truly reciprocal. For the A observer, the real TD and LC exist in the B frame, but there is absolutely no TD or LC in the A observer's own frame. And the reciprocal of that is that for the B observer, the real TD and LC exist in the A frame, but there is absolutely no TD or LC in the B observer's own frame.

I'm glad you kept arguing your point instead of just agreeing with me, because now I can see that you really have been saying that you think that TD and LC exist in one's own frame, but that one is just unable to detect it. That is not the case. Your mistake is in taking the results of one frame, (Observer A finds there is TD and LC happening in frame B), and then and assuming that those results also apply to the other frame, (So now we must explain why the B observer does not detect his own TD and LC in his own frame.) WRONG!! As soon as you start explaining the point of view of observer B, you are switching to the B frame. In the B frame, the B observer is just as capable of detecting TD and LC as anyone else, it's just that for observer B, the TD and LC are happening in frame A, not in B's own frame.

Last edited: Jul 10, 2020
12. ### Confused2Registered Senior Member

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607
If Clock B is moving at velocity v relative to Clock A then isn't Clock A moving at velocity -v relative to Clock B? (My thanks to Halc for making this point earlier).

With clock A 'stationary' and clock B moving at velocity v followed by Clock B 'stationary' and Clock B moving at velocity v I'm not sure you are any further forward.

Last edited: Jul 11, 2020
13. ### phytiRegistered Senior Member

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492
Neddy Bate;
Length contraction is real, and required for reciprocal observations.
g=gamma
Left:
Bill moves past Ann at relative speed v. They are anauts in identical capsules of length d, with identical clocks.
A measures the B capsule as length d, on his axis of simultaneity Ax
B measures the A capsule as length d/g^2, on his axis of simultaneity Bx.
The measurements are NOT reciprocal.
Right:
With lc of the B-capsule,
A measures the B capsule as length d/g, on his axis of simultaneity Ax.
B measures the A capsule as length d/g, on his axis of simultaneity Bx.
The measurements are reciprocal.

Measure the length of a metal rod.
Heat it to a higher temperature.
Measure the length again and find its length increased.
The length changed, in your own frame, without any relative motion of the rod.
There are no rigid rods.

14. ### Neddy BateValued Senior Member

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2,121
I don't know what you are talking about.

If d is the proper length of each capsule, then:

1. Ann measures the length of her own capsule A to be d (NOT LENGTH CONTRACTED) because the velocity of capsule A in frame A is zero (SO IT IS NOT LENGTH CONTRACTED IN THAT FRAME), and Ann measures the length of capsule B to be d/γ (LENGTH CONTRACTED) because the velocity of capsule B in frame A is non-zero.

2. Bill measures the length of his own capsule B to be d (NOT LENGTH CONTRACTED) because the velocity of capsule B in frame B is zero (SO IT IS NOT LENGTH CONTRACTED IN THAT FRAME), and Bill measures the length of capsule A to be d/γ (LENGTH CONTRACTED) because the velocity of capsule A in frame B is non-zero.

3. Thus the measurements are perfectly reciprocal.

...

Note also that there is no need to say anything ridiculous like, "Ann can't detect that her capsule is length contracted in her own frame, because her measuring rod is also length contracted in her own frame." That is not needed at all, because statement 1 says otherwise: "Ann measures the length of capsule A to be d (NOT LENGTH CONTRACTED) because the velocity of capsule A in frame A is zero (SO IT IS NOT LENGTH CONTRACTED IN THAT FRAME)."

Last edited: Jul 11, 2020
15. ### Confused2Registered Senior Member

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607

Looking at my diagram above - which is (hopefully) the same as Neddy's with a few extras.

I'll admit I had as much difficulty with these drawings as with Phyti's original drawings. Using t from one frame in the other I found very confusing.

Do we a agree -
Clock A and clock B are the same clock viewed from the same frame (A) or viewed from a frame (B) moving relative to the clock at velocity v.
Same clock, different view.
Are Photon Alice and Photon Derek (and even Colin) one and the same?

Clock B seems to include a plot (a quarter circle) of u and v where u²t²+v²t² = c²t² where c²t² is set to 1.
ut and vt are shown as lying on that circle. When the need arises we can read off ut for any given vt without knowing why *. Is that a fair interpretation of the intent?

After time t in the stationary and/or moving frame it is shown that the photon Colin (the same photon as Alice/Derek?) hasn't reached the detector. Looking at the projection onto the y axis of the intersection on u²t²+v²t² = c²t²=1 where v=0.5c we see Colin is only 0.866 the way there.
Given Colin's lack of progress after time t in one/both/the other frame compared to the stationary frame we conclude the moving clock will tick at 0.866 times the rate of the stationary clock. A fair interpretation?

I can't resist saying how much clearer the above seems than
u²t'²+v²t'² = c²t'² where u²t'² = c²t².

* In general u²t²+v²t² = c²t² is going to be true - what took me some time is (hopefully) working out the significance of the circle.

Last edited: Jul 12, 2020
16. ### Confused2Registered Senior Member

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607
phyti - I think we have already have enough trouble with the basics.

17. ### phytiRegistered Senior Member

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492
Neddy Bate;

If the simple 3-sided triangle is accepted as representing the experiment, then there is no acceleration from outbound to inbound since the duration of the change is t=0.
The clock rate would be slow as viewed by the other, for both legs, i.e. reciprocal. The spacetime path, determines the accumulated time for a clock, and these are not symmetrical.
Even classical physics states, t=x/v. The time to move a distance x, is inversely proportional to speed v. The greater the speed, the lesser the time. Many miss this and wonder why time is less for a longer line on the spacetime drawings.

The simplest solution I know is:
the rest clock A emits t signals until the B clock returns,
the B clock emits t' signals until it returns. t'=t/gamma <
What is it?

18. ### phytiRegistered Senior Member

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492
Neddy Bate;
It seems you aren't interpreting the graphic correctly.
SR requires both ends of an object be located simultaneously.
A is black, B is green, each ship is length d, g=gamma.
The near end of both ships are at 0 for all measurements.
left:
A measures location of far end of his ship at d, on his axis of simultaneity Ax.
A measures location of far end of green ship at d, on his axis of simultaneity Ax.
B ship length/A ship length =1.
B measures location of far end of his ship at f, on his axis of simultaneity Bx.
B measures location of far end of black ship at e, on his axis of simultaneity Bx.
A ship length/B ship length =1/g^2.
Measurements are not reciprocal.
right (with lc):
A measures location of far end of his ship at d, on his axis of simultaneity Ax.
A measures location of far end of green ship at d/g, on his axis of simultaneity Ax.
B ship length/A ship length =1/g.
B measures location of far end of his ship at f, on his axis of simultaneity Bx.
B measures location of far end of black ship at e, on his axis of simultaneity Bx.
A ship length/B ship length =1/g.
Measurements are reciprocal

19. ### Neddy BateValued Senior Member

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2,121
Well, it is not possible for the twin to go from positive velocity to negative velocity without having a velocity of zero in between. All you have to do is replace the sharp corner of the Minkowski diagram's triangle with a small radius, and you should see the velocity go through zero at the middle of the voyage.

But other than that, YES, the traveling twin DOES say that the stay-home twin's clock rate is slow during both of the inertial legs of the journey. But that is only the clock rate, not the clock reading.

Correct. But for the two astronauts ANN and BILL passing each other in deep space, you never stated anything about a turnaround point, so the situation is completely symmetrical and reciprocal. Meanwhile, none of this has anything to do with your bizarre claim that my office is length contracted as I sit here, but that I cannot detect it because my measuring tape is also length contracted. That is just wrong, or at least unnecessary at best.

The simultaneity lines have different angles on the two different parts of the Minkowski diagram's triangle.

So the traveling twin says these clock readings are simultaneous for the outbound leg (for the case of gamma=2):
STAY-HOME 0, TRAVELER 0
STAY-HOME 1, TRAVELER 2
STAY-HOME 2, TRAVELER 4
Note that the rate of the stay-home twin's clock is half the rate of the traveler's clock, because this is the frame of the traveler. (i.e., For every tick of the stay-home twin's clock there are 2 ticks on the traveler's clock.)

And the traveling twin says these clock readings are simultaneous at middle of the turnaround where the velocity is zero (for the case of gamma=2):
STAY-HOME 8, TRAVELER 4

And the traveling twin says these clock readings are simultaneous for the inbound leg (for the case of gamma=2):
STAY-HOME 14, TRAVELER 4
STAY-HOME 15, TRAVELER 6
STAY-HOME 16, TRAVELER 8
Note that again, the rate of the stay-home twin's clock is half the rate of the traveler's clock, because this is the frame of the traveler. (i.e., For every tick of the stay-home twin's clock there are 2 ticks on the traveler's clock.)

So there is still no need for the traveler to conclude that he ever had time dilation in his own frame. Instead, in the traveler's frame, it was always the stay-home twin who had the time dilation. This, along with the change in the angle of the simultaneity lines, explains how the traveler accumulates less time than the stay-home twin, WITHOUT saying that the traveler ever had time dilation in his own frame.

Last edited: Jul 13, 2020
20. ### Neddy BateValued Senior Member

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2,121
Yes, to measure the length of a moving object, you need to locate both ends simultaneously.

Ann measures Bill's capsule to have length d/gamma, and measures her own capsule to have length d.

Bill measures Ann's capsule to have length d/gamma, and measures his own capsule to have length d.

Length contraction is reciprocal in inertial frames. I don't know what the other diagram is for if it says that length contraction is not reciprocal, because length contraction is reciprocal for inertial frames, and so is time dilation. Full stop.

And none of this helps you get out of your bizarre claim that my office is length contracted as I sit here, but that I cannot detect it because my measuring tape is also length contracted. You need to rethink that.

Last edited: Jul 13, 2020
21. ### Neddy BateValued Senior Member

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2,121
Yes, but in my diagram, I just re-used the graphics that phyti had already made, for my own convenience. I would have had to mirror the image if I wanted it to look like clock B was moving from right-to-left instead of left-to-right.

But the diagram can be interpreted either way -- it is rather like looking out the window of a moving train. If you look out a window on the left side of the train, the ground appears to move from right to left, but if you look out a window on the right side of the train, the ground appears to move from left to right. It makes no difference to the idea behind it.

My purpose in showing the two different reference frames was not about moving any further forward. It was just to show that the time dilation is always in the "other guy's" reference frame, never one's own.

22. ### Confused2Registered Senior Member

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607
My first successful analysis of the muon experiment was the result of seeing that the muon is moving through the Earth frame at v and the Earth is moving through the Muon frame at (-)v .. I do (unfortunately) remember how long it took me to see that .. it was only by constant repetition that I finally saw what I wasn't seeing .. that the Earth is moving through the muon frame at the same (-ve) velocity as the muon is moving through the Earth frame. I'm sure I was trying to make some point when I called you out on the sign .. yes, it was that the Earth is moving through the muon frame at the same (-ve) velocity as the muon is moving through the Earth frame. If there is anyone else out there like me then maybe constant repetition of the same point might help.

23. ### phytiRegistered Senior Member

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492
Confused2;
The vertical axis is p, any direction perpendicular to the motion in x.
All clocks are a standard model, so they function the same in all cases.
a, b, c are three different observations, so consider them independently of the others.
Here 'relative rest' is defined as 'two observers moving with the same velocity'.
This eliminates the black & white 'motion or rest' of Newton.
Now two frames can be moving at v in the same direction and be at rest without contradiction.

view a: Description by a person moving with the clock, to show a half clock cycle
SR states all inertial frames are equivalent, so to specify a speed is redundant.

view b: Description by a person watching a clock move past at .5c. He observes the light moving at an angle relative to the vertical, i.e. in 2 dimensions. He can resolve the signal into horizontal and vertical components, for the purpose of explaining time dilation. The light moves a distance ut in the moving clock while it moves a distance ct in his clock. He concludes the moving clock is simultaneously running slower than his own clock.

view c: Simultaneous description by a person moving with the moving clock from view b.
t is always the time of the person making the description.
The distance ut from view b must be converted to ct' since it is his prediction for a moving observer. In view c, it becomes ct since it is the time for that person in that frame, who expects light speed to be c.