# Invariant spacetime and time dilation

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

1. ### Neddy BateValued Senior Member

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In view C, the speed of light is still the same finite constant, c=29992458 m/s, so the clock is not actually running slow at all in that reference frame. The time is just 0.87 at that moment, in that frame. You describe it as if the clock is running slow, and the person's biological clock also running slow by the same factor, but that is basically what the reference frame of view B would say.

But if you are the person in view C, you would say the clock is not moving at all, but rather the "other" clock is moving at v=-0.5c, and that the "other" clock is the one running at a slower rate, (with "other" meaning the clock shown in view A). It is a minor difference, but I think it is important, because you almost make it sound like there is an absolute rest frame, and clocks moving relative to that frame run slow, but people co-moving with those clocks are just unable to realise that they are running slow. On the contrary, these effects are all due to relative motion, not absolute motion.

3. ### Confused2Registered Senior Member

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607
Looking at Phyti's post:

One of the problems is that that the distance from source to mirror (fig a) is given as ct where t is the the time in a frame moving with respect to the mirrors. Given that ct > ct' we see (correctly) in fig c that the pulse doesn't actually reach the mirror in time ct' (actually shown as ct but this is clearly ct') . The circle of radius ct in fig b would have had physical significance if the distance between the source and mirror had been correctly drawn (as ct') in Fig a but it wasn't.

As with Neddy Bates' comment above:
It seems to be implied that the photon that reaches the mirror in one frame is not the same photon as the one that reaches the mirror in the other frame. There is only one event and it must be the same photon in both frames. As with birds they move their wingtips (pretty much) up and down | but when seen flying past the wingtips go /\/\ - birds don't have special joints which change to suit the observer.

Hopefully the attached image works better.
In fig 2
u²t²=c²t² - v²t²
Comparing with fig 1
u²t²= c²t'²
c²t'²=c²t² - v²t²
t'²=t²(1- v²/c²)

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5. ### Neddy BateValued Senior Member

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My interpretation of Phyti's diagram is that there are two different light clocks. In figure A there is one light clock, stationary with respect to frame A. In figure B there is a second (different) light clock moving with respect to frame B which is the same frame as A only without the stationary clock shown. And finally, in figure C that same second light clock is shown stationary with respect to frame C.

In frame A & B, the stationary clock ticks 1 unit of time, while the moving clock ticks 0.866025 units of time because the moving clock ticks at a slower rate. It is not the same photon, so it does not have to reach the mirror at the same time as the other clock's photon.

But in frame C, the when the clock has only reached 0.866025, the clock is not interpreted as ticking at a slow rate, because it is not running slow in that frame. 0.866025 is just how much time has elapsed since the photon started from the bottom. Frame C would also say that the clock stationary with respect to frame A is moving at velocity v=-0.5c and as such would only have reached 0.866025 * 0.866025 = 0.75 at that time, and so that would be the clock considered to be running slow in frame C.

7. ### Confused2Registered Senior Member

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And if you were seeking clarity...?

8. ### Neddy BateValued Senior Member

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I'm not sure what you are asking. Does this video help at all?

Two different light clocks, two different photons, two different pairs of mirrors.

9. ### Confused2Registered Senior Member

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607
Is an analysis of (possibly) two light clocks clearer to another person than an analysis of one? As I am not 'another person' I have no alternative but to seek out the view(s) of 'another person'.

10. ### Neddy BateValued Senior Member

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Okay, let's make it one clock. Then phyti's frame A is the clock at rest, and phyti's frame B is the same clock moving, and we can just leave frame C out of it. We still get 1.0 tick for the clock at rest, and 0.866025 ticks for the moving clock. But since at time 0.866025 the photon has not reached the mirror yet, we just have to wait until it reaches the mirror, and when it does, then it will have ticked 1.0 tick. Is that better?

11. ### Confused2Registered Senior Member

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607

While I appreciate your clarification of the points made by Phyti would

In fig 2
u²t²=c²t² - v²t²
Comparing with fig 1
u²t²= c²t'²
so
c²t'²=c²t² - v²t²
t'²=t²(1- v²/c²)
require any clarification?

Edit.. at some point I might want to poll another 'another person' (humour intended)

Last edited: Jun 23, 2020
12. ### Neddy BateValued Senior Member

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2,108
Okay, I think your method is clearer than phyti's.

Picking up from where you left off...
t'² = t²(1 - v²/c²)
t'²/t² = (1 - v²/c²)
t'/t = √(1 - v²/c²)

and...
gamma = γ = t/t'
gamma = γ = 1 / √(1 - v²/c²)

13. ### Confused2Registered Senior Member

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607
In fairness I should make clear it was Phyti that first introduced me to the light clock. While our methods may have diverged over the years mine is traceable back to a post by Phyti in 'The Joint' several years ago.
There's a likewise kind'a post by Neddy Bate on this thread that might produce similar fruit if I can just get to grips with the darn thing.

14. ### Neddy BateValued Senior Member

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Which post? I might be able to help.

With the Lorentz transformation equations, one only needs to know the coordinates of the events in question. So, for example with a muon, there is an event when the muon passes a hypothetical earth clock high in the atmosphere, and then there is another event when the muon passes another hypothetical earth clock at ground/sea level. Those two earth clocks are synchronised in the earth frame, but not in the muon frame. The muon frame can have its own hypothetical synchronised clocks, though.

15. ### phytiRegistered Senior Member

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458
Neddy Bate;

My statement: "In view-c you are moving with the clock at .5c."
There is no absolute rest frame, since none is needed. Einstein's SR does what Lorentz's LET does, but without the ether. The reason it works, because 'events don't move'. When light is emitted, it moves outward AS IF its source/origin is a fixed location, and independently of the speed of the source.
The 'outside observer' in view-b , and all observers NOT moving with the clock, see a light path with two components as the explanation for the time dilation of the moving clock.

In view-c, The observer moving with the clock is only aware of any motion relative to himself, i.e. the vertical component, which is moving at <c relative to the clock. The observer is allowed to assume a pseudo rest frame per the equivalence of inertial motion. By SR definition, the light within the clock moves at c, thus the substitution of ct' for ut. His altered biological clock, is included to explain why he is not aware of a change in the local clock rate, and why all observers experience local/proper time.

16. ### phytiRegistered Senior Member

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458
Confused2;

Thanks for staying with me. Your questions motivate me to improve explanations.
The observer in view-b is at rest, the same as in view-a, so the same time axis applies. It shows ct = 1, same distance, different angle, and short of reaching the mirror M.
If all directions are included, photons would expand outward in a sphere, which is not altered by motion of the clock, thus the independency of light speed.
This is the significant difference between matter in motion and light in motion. If an object was projected at M, it would have acquired the speed of the ref. frame and simultaneously moved up and forward, reaching the mirror.

If it's the same experiment viewed by two different observers, each will perceive different trajectories. Just as the train scenario, the passenger sees the object fall from his hand vertically to the floor, while (simultaneously) the platform person sees it fall in a curve to the floor. Trajectories, orbits, world lines, etc. are perceptions, reality in the mind, not physical entities out there, in the world.

17. ### Neddy BateValued Senior Member

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Yes, I agree with all of that.

This is the part that I don't agree with. While it is true that the observer co-moving with the clock only detects the vertical component of the light's motion, it is not moving at <c relative to the clock. It is moving at exactly c relative to the clock. Thus the clock is not running slow in that frame, and neither is the observer's biological clock running slow in that frame. The observer in view B can say those things about the moving clock & Moving observer, but the observer in view C does not say that about himself. He says that about other clocks and observers which are moving relative to himself, but he does not say that his own biological clock is running slow in his own frame, thus making it undetectable that his clock is running slow. That is simply not the case in his own reference frame.

18. ### phytiRegistered Senior Member

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458
NeddyBate;

Looking at the triangle in view-b,
vt< ct, and ut<ct, thus u<c.
If only the clock rate was slow, and not the biological clock, the user would be aware of it.
I never said the observer in view-c was aware of his time dilation, but stated why he was NOT aware. Without an explanation in terms of physics, it's magic.
If there is no biological clock dilation, then the 'twin' scenario is false.
Light speed in any inertial ref. frame is c, only if it's measured.

Almost!
Light speed propagation is absolute in space, thus time dilation and length contraction are real effects as a result of motion relative to light. Examining the gamma factor, you find the only significant factor in SR, 'v/c'.

Perception is what the observer thinks is happening relative to the ref. frame he occupies. His motion cannot alter the physics of the universe, but it can alter his perception of the universe. The effects are measured with the gamma factor, resulting from a fixed, independent speed of light. We can't measure the speed of our ref. frame in space, only the difference between frames. If measurements require material references, that's the best we can do.

As in any round trip reflection of light, the transit time out doesn't always equal the transit time back (as observed by someone outside the frame). Einstein therefore defined them as equal for consistency, since that would be the expectation of an occupant in an inertial ref. frame.
From the 1905 paper or the 1920 book (below):
"That light requires the same time to traverse the same path A to M as for the path B to M is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity."

19. ### Neddy BateValued Senior Member

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Yes, u<c in view B. But it does not follow that in view C the speed of light is therefore actually u. This is SR, so in view C, the speed of light is c, which is not <c.

Let's imagine that the light clock shown in view A is also present in view B as a stationary clock which can be compared to the moving clock shown in view B. We find that when the stationary clock (view A) has reached 1.000000 tick, the moving clock (view B) has only reached 0.866025 ticks. You can calculate that from view B alone, so you do not have to use view C.

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If you do choose to show view C, you must be careful not to assume that in that frame, when the clock (view C) reaches 0.866025 ticks, that the the light clock in view A must have reached 1.000000 tick. Even though that is true in the reference frame in which view B is stationary, it is not true in the reference frame in which view C is stationary. In that frame, when the clock (view C) reaches 0.866025 ticks, the other clock (view A) has only reached 0.750000 ticks, because it is moving the opposite direction at v=-0.5c. So there we have a similar triangle demonstrating that it is clock A which is ticking slower, at least in frame C. Thus there is no reason to try to explain why the observer traveling with the clock (view C) is unaware that his clock is ticking slow. In that frame, that is not even the clock which is ticking slow!

If you insist on saying that the observer traveling with the clock (C) is unaware that his biological clock is ticking slow, and therefore he is unable to detect that his clock is ticking slow, then you must also say that the observer traveling with the clock (A) is unaware that his biological clock is also ticking slow, and therefore he is unable to detect that his clock is also ticking slow. The situation is completely symmetrical and reciprocal, because time dilation is caused by relative motion, and there is no way to say whether (A) or (C) is the one that is really moving. That is why I object to your explanation regarding the biological clock. It does not hold up to reason in a perfectly symmetrical and reciprocal situation like this.

Last edited: Jun 28, 2020
20. ### phytiRegistered Senior Member

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458
Neddy Bate;

1. I agree and typically use the spacetime graphic for both perceptions (the ref.frame and the moving frame) to show reciprocity. Postulate 1 states each should see the same physical processes.

2. The cause of time dilation is the constant independent speed of light. The gamma factor is a function of 'v/c', the speed of a system of material objects relative to the speed of light. It's part of the transformation from t to t'. The relative speed is only an indication that two frames have different velocities. Will include a pdf on this.

3. The mind/brain functions via chemistry, which is the physics of the electron cloud surrounding each atom, which is based on em interactions. How can the mind be exempt from time dilation effects while the clock is not?

In the case of length contraction, it is not detected because the measuring device is also contracted to the same degree.
If no biological change, why doesn't the observer in the moving frame sense a change in clock rate, especially when moving near light speed?
It's the same factor in both cases, light moving a greater distance to complete a process.
Without the biological clock change, how do you explain the difference in aging for the 'twin scenario'?

21. ### phytiRegistered Senior Member

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458
Here is why observations depend appox. on relative speed, and why the composition of velocities is more than the difference in speed.

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22. ### Neddy BateValued Senior Member

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You are misunderstanding me. I agree with you that a moving clock runs slow, and that a person moving with that clock also biologically ages slow by the same factor. You and I agree on that. But whether something is moving or not (at constant velocity) is relative, not absolute. So when I say "moving clock" and "person moving with that clock" I am specifically not talking about a reference frame in which neither are moving. And in their own reference frame, neither one is moving, so I am not talking about that reference frame.

So, one thing I am trying to point out is that you seem to be focusing on the wrong reference frame when you say that a moving clock is running slow. You seem to be focusing on the clock's own reference frame, and saying the clock is running slow in that frame, which is incorrect. The clock is not moving in that frame, so it is not running slow in that frame.

Another thing I am trying to point out is that you seem to be focusing on the wrong reference frame when you say that a person's biological aging is slow. You seem to be focusing on the person's own reference frame, and saying the person is aging slow in that frame, which is incorrect. The person is not moving in that frame, so he is not aging slow in that frame.

Then you say the clock running slow in its own frame is undetectable because the observer's biological processes are running slow by the same factor. So you are missing the key idea that neither the clock nor the biological aging are running slow in their own reference frame.

Once again you seem to be saying that length contraction happens in the reference frame of the object itself. Likewise, you seem to be saying that the measuring device is length contracted in the reference frame of the object itself, and that therefore the length contraction in that frame is undetectable. But there is no length contraction in the reference frame of the object itself. If there were, it would have to be contracted by an infinite number of different factors, based on the infinite number of different velocities that object has relative to the infinite number of different inertial frames in the universe.

Because in his own reference frame, he is not moving, so his biological aging rate is not running slow in that reference frame. And because in his own reference frame, his clock is not moving, so its tick rate is not running slow in that reference frame.

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'.

Last edited: Jul 2, 2020