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Re: Gravitational Time Dilation on Flat Earth
« Reply #20 on: December 11, 2021, 07:30:33 AM »
It has everything to do with it. The Earth will be moving at a different rate to a ball that is released from a higher altitude versus a ball at a lower altitude. They do not experience the same thing.

No one is releasing a "ball" of anything. I have no idea what you're referring to. Please refer yourself to this previous example - The scenario is: Two people just standing, each staring at a clock - Both clocks sync'd to the same time - observer A is in the basement and observer B is 150 floors above.

Again, does the constant rate mentioned in the wiki actually change the higher up you go? Is observer B's clock running faster than observer A's? And if so, how does UA make that happen?

At some point the photons or signals are moving independently of the clocks, through a medium, to the detector below. The Earth is accelerating into the photons, causing the higher clock to appear to tick faster. See the previous examples from literature and look up what accelerate means.

How does UA make that time difference happen? According to the wiki, Observer A and his clock is accelerating upward at exactly the same as observer B and his clock along with the building itself. There should be no difference, everyone and everything is moving together at the same rate. How does UA cause photons to move independently of everything else that is moving upward together?

Look up the difference between linear movement and accelerated movement.





The Earth is accelerating, not moving at a linear rate.

The Earth will, therefore, accelerate into a long line of photons at a greater rate than a shorter line of photons.

You didn't answer the question. Observers A & B, the clocks, the building they are in, the earth, are all accelerating at exactly the same rate, together. How does UA make the clocks tick differently?

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Offline Tom Bishop

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Re: Gravitational Time Dilation on Flat Earth
« Reply #21 on: December 11, 2021, 07:33:29 AM »
You didn't answer the question. Observers A & B, the clocks, the building they are in, the earth, are all accelerating at exactly the same rate, together. How does UA make the clocks tick differently?

The earth isn't moving towards the released photons or signals at the same rate at all times. Once the signal of the lower clock hits the detector at the bottom the earth is still moving at an increasing pace into the line of photons from the second higher clock.

If a clock at a lower height releases a photon at the same time as a clock from a higher height, the photons will not experience the earth hitting them at the same velocity. The earth is moving into the photon released from the second clock at a greater rate, as it has more time to build up speed.
« Last Edit: December 11, 2021, 07:42:53 AM by Tom Bishop »

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Re: Gravitational Time Dilation on Flat Earth
« Reply #22 on: December 11, 2021, 07:44:14 AM »
You didn't answer the question. Observers A & B, the clocks, the building they are in, the earth, are all accelerating at exactly the same rate, together. How does UA make the clocks tick differently?

The earth isn't moving towards the released photons or signals at the same rate at all times. Once the signal of the lower clock hits the detector at the bottom the earth is still moving at an increasing pace into the line of photons from the second higher clock.

That's not what the wiki says. Your wiki says, "Objects on the earth's surface have weight because all sufficiently massive celestial bodies are accelerating upward at the rate of 9.8 m/s^2 relative to a local observer immediately above said body."

So I guess that's not entirely true. How does UA modulate earth not "moving towards the released photons or signals at the same rate at all times"? How does that work? What are these modulations in time that UA is creating? Are they measurable?

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Offline Tom Bishop

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Re: Gravitational Time Dilation on Flat Earth
« Reply #23 on: December 11, 2021, 08:01:58 AM »
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How does UA modulate earth not "moving towards the released photons or signals at the same rate at all times"? How does that work?

It works by grasping the concept of acceleration and visualizing why it would hurt more when walking off a chair than a skyscraper.

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Re: Gravitational Time Dilation on Flat Earth
« Reply #24 on: December 11, 2021, 08:12:02 AM »
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How does UA modulate earth not "moving towards the released photons or signals at the same rate at all times"? How does that work?

It works by grasping the concept of acceleration and visualizing why it would hurt more when walking off a chair than a skyscraper.

That's not the scenario. No one is stepping off something - They are standing still. All things are at rest but accelerating upward together, at the same rate. Your example does not apply.

How does UA change the speed of the clock when it is at rest?

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Offline Tom Bishop

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Re: Gravitational Time Dilation on Flat Earth
« Reply #25 on: December 11, 2021, 08:38:53 AM »
The photons and electomagnetic signals aren't physically attached to the building or clocks when they are released. Everything isn't moving upwards together at all times.

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Re: Gravitational Time Dilation on Flat Earth
« Reply #26 on: December 11, 2021, 06:43:40 PM »
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According to this time should speed up as light moves towards the surface of the earth, not slow down. Your examples do not make any sense at all

Shortest distance between two points is straight.  The red line is going straight, the blue line is not.  Going the same speed as the red line, it would take the blue line longer to reach the edge, because it is not going straight.  If you couldn’t perceive the curve, it would appear that the blue line is going slower because it takes longer to go the same perceived distance.

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The earth isn't moving towards the released photons or signals at the same rate at all times. Once the signal of the lower clock hits the detector at the bottom the earth is still moving at an increasing pace into the line of photons from the second higher clock

I get the concept of what you are trying to explain and you are actually right.  But you don’t understand the implications of what you are saying. So I’ll help you out.

This is the scenario that you are describing.


The first flash travels the distance L1 and the second flash travels the shorter distance L2. It is a shorter distance because the ship is accelerating and has a higher speed at the time of the second flash. So if the two flashes were emitted from clock A one second apart, they would arrive at clock B  at less than one second since the second flash doesn’t spend as much time on the way. The same thing will also happen for all the later flashes   From outside the rocket, it is clear that the distance between the clocks is getting shorter and shorter, but this is what you don’t understand….inside the rocket, the distance between the clocks remains the same.

An observer outside the rocket, who is not accelerating sees the distance the light travels decreasing.  An observer inside the rocket doesn’t. That is the sense in which acceleration warps space time.  Distance (and therefore time) is measured differently depending on if you are in accelerating frame or not.

A couple of other things worth noting.

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Note that Rindler observers with smaller constant x coordinate are accelerating harder to keep up. This may seem surprising because in Newtonian physics, observers who maintain constant relative distance must share the same acceleration. But in relativistic physics, we see that the trailing endpoint of a rod which is accelerated by some external force (parallel to its symmetry axis) must accelerate a bit harder than the leading endpoint, or else it must ultimately break. 
https://en.wikipedia.org/wiki/Rindler_coordinates

IOW, the clock at the bottom and the clock at the top cannot be accelerating at the same rate

It’s also worth noting that the longer the acceleration continues, the distance the light has to travel would get shorter and shorter resulting in the difference between rates of the clock increasing.  The clock at the top would get faster and faster relative to the clock at the bottom.  But that is not what we see on earth.  The difference in the clock rates are static as long as the relative distance between them doesn’t change. The difference in the clock rates at different elevations don’t change over time on earth, so there is no constant acceleration with increasing velocity on earth.
« Last Edit: December 11, 2021, 06:52:05 PM by Rog »

Re: Gravitational Time Dilation on Flat Earth
« Reply #27 on: December 11, 2021, 11:45:42 PM »
The photons and electromagnetic signals aren't physically attached to the building or clocks when they are released. Everything isn't moving upwards together at all times.

I'm not sure that you fully understand how so-called atomic clocks work Tom.     There are no photons involved at all.

An atomic clock uses the resonance frequencies of atoms as its resonator. According to Encyclopedia Britannica, the
resonator is "regulated by the frequency of the microwave electromagnetic radiation emitted or absorbed by the quantum
transition (energy change) of an atom or molecule."  The advantage of this approach is that atoms resonate at extremely
consistent frequencies.

If you take any atom of cesium (for example) and get it to resonate, it will resonate at exactly the same frequency as
any other atom of cesium.   Cesium-133 oscillates at precisely 9,192,631,770 cycles per second.

And this is why each of the ground floor and the top floor clocks of our skyscraper are keeping different times—around
4ns (nanoseconds) for a 2,000 foot building.  The difference is explained by Einstein's theory of relativity, which established
that time is connected to the strength of gravity at the point where it's measured. This phenomenon affects the relative
motion of electrons orbiting the nucleus of an atom.

—I'm not sure what you mean by "Everything isn't moving upwards together at all times".  Could you elaborate please?


Re: Gravitational Time Dilation on Flat Earth
« Reply #28 on: December 19, 2021, 03:32:55 AM »
The photons and electomagnetic signals aren't physically attached to the building or clocks when they are released. Everything isn't moving upwards together at all times.

I've noted that in another FE thread, it was claimed that all satellites move upwards at the
same time as the Earth
, which explains why the Earth never collides with them as it moves
upwards towards them. 

But according to Tom, this is not the case.  How can these conflicting action statements be concomitant?

Re: Gravitational Time Dilation on Flat Earth
« Reply #29 on: December 20, 2021, 10:29:09 PM »
The earth isn't moving upwards at a set velocity. It is accelerating. More specifically, the surface of the Earth is accelerating upwards into the things above it.

If you walk off of the edge of a chair and go into freefall it will hurt a lot less than if you walk off the edge of a skyscraper. In the skyscraper situation you are inert in space and the Earth has more time to build up velocity and smash into you.

Therefore, in a situation where the Earth is accelerating upwards if you have a broadcasting photon clock light source at the altitude of a chair and a photon clock at the altitude of a skyscraper, from the perspective of a detector on the floor, those photons would be perceived to be hitting it at different rates of reception. The time for the light source on the skyscraper will appear faster than the light source on the chair.

It is also what would happen between the floor and ceiling inside of a rocket ship accelerating upwards through space.

From p.8 of Cosmological Physics by John A. Peacock, PhD. we read the following:

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  “ Many of the important features of general relativity can be obtained via rather simple arguments that use the equivalence principle. The most famous of these is the thought experiment that leads to gravitational time dilation, illustrated in figure 1.1. Consider an accelerating frame. which is conventionally a rocket of height h, with a clock mounted on the roof that regularly disgorges photons towards the floor. If the rocket accelerates upwards at g, the floor acquires a speed v = gh / c in the time taken for a photon to travel from roof to floor. There will thus be a blueshift in the frequency of received photons, given by Δv / v = gh / c^2, and it is easy to see that the rate of reception of photons will increase by the same factor.

Now, since the rocket can be kept accelerating for as long as we like, and since photons cannot be stockpiled anywhere, the conclusion of an observer on the floor of the rocket is that in a real sense the clock on the roof is running fast. When the rocket stops accelerating, the clock on the roof will have gained a time Δt by comparison with an identical clock kept on the floor. Finally, the equivalence principle can be brought in to conclude that gravity must cause the same effect. Noting that ΔΦ = gh is the difference in potential between roof and floor, it is simple to generalize this to Δt / t = ΔΦ / c^2 ”



“ Figure 1.1. Imagine you are in a box in free space far from any source of gravitation. If the box is made to accelerate ‘upwards’ and has a clock that emits a photon every second mounted on its roof, it is easy to see that you will receive photons more rapidly once the box accelerates (imagine yourself running into the line of oncoming photons). Now, according to the equivalence principle, the situation is exactly equivalent to the second picture in which the box sits at rest on the surface of the Earth. Since there is nowhere for the excess photons to accumulate, the conclusion has to be that clocks above us in a gravitational field run fast. ”

See the bolded. If you imagine yourself running into the line of photons it is apparent that the clock above you would run fast, because you are running into them. This is a physical explanation for how this works under the concept of upwards acceleration.

In contrast, the Round Earth Theory adopts a non-physical explanation for this which occurs in a hidden layer of reality, in which space is bending to cause the apparent speedup of time at different altitudes. In my opinion this is completely ad-hoc. Physics behaves as if the surface of the Earth is accelerating upwards, but that can't work in RE, so they created this space-bending explanation in an untestable layer of reality which seeks to emulate the physics of upwards acceleration.

In one situation, with upwards acceleration, we can describe what is happening on a physical level for why the rate of reception speeds up when the photons are coming from higher altitudes. In the case of space bending, we cannot. We just call it "space bending" like a magic wand and say that it's physically equivalent to upwards acceleration to explain the otherwise unexplainable. Clearly, there is a difference between the two views.

Why should this space bending mechanism cause photons to travel faster from higher altitudes rather than slower from higher altitudes or no difference at all? What physical reason is there other than to claim that it must be the case because that is what is experienced? One quickly finds a lack of answers.
First, kudos for a good citation.  However, I think the argument must be mistaken, because while the bottom of the rocket has accelerated upwards, so has the clock -- meaning, the photons would arrive sooner than expected, but not with a blue shift.  Can you find another citation with a similar argument?  (I can't.)

IMHO, in this case there would be no time dilation because the top of the rocket gets exactly the same acceleration as the bottom.  If there are two clocks, the only time dilation they'd see would be from the trips between them.  (That is, if both started at the bottom of the rocket and one went up and down, we'd see dilation from that trip.  If both started at the bottom and one went up first and the other later, there would be no dilation. And this would be true regardless of what the rocket does.)

So, if this is indeed a case of a bad lesson in print, then there would be no time dilation for a clock atop a mountain on an accelerating Earth.

If it is a good lesson, then your point stands and I must misunderstand relativity.  (I think we can take it for granted that I misunderstand GR.  But I think I have this bit right.)

Edit to add: Note that the wikipedia article (https://en.wikipedia.org/wiki/Gravitational_time_dilation) says "Consider a family of observers along a straight 'vertical' line, each of whom experiences a distinct constant g-force directed along this line ..." and leads to a similar formula.  Note carefully the word "distinct," meaning "different".  In the textbook's example, the same g-force is exerted on the top and bottom.
« Last Edit: December 20, 2021, 10:46:27 PM by drand48 »