[Pete Svarrior]

Did you miss this? I’m talking about 2 separate jumps 5 minutes apart.

Indeed. However, that's inconsistent with your claim that the relative velocity between the jumper and the Earth has increased. If the jumps are separate, then the velocity at the start of each jump is 0. You know, the lack of motion.

See, the problem is that you keep contradicting yourself at every step. It's impossible for all your statements to be true, so we're left picking out the ones that were closest to plausible.

No, that’s not what I believe but according to numerous statements in the wiki, that is what the FE position is.

I am reliably informing you that it's not.

Or do I misunderstand that the FE position is that an object doesn’t “fall”, but that the earth comes up to meet it?

The two are one and the same from a physics standpoint. You cannot have one and not the other. So, yes, you are misunderstanding not only the FE position, but also the most basic physics behind any physical model, Round Earth or Flat.

What does that object do during the time it takes for the earth to meet it (in a vacuum)?

Laymen would call it "falling".

The scenario I’ve described is conceptually the same as the bowling ball/feathers video that is in the wiki.

It emphatically is not, because regardless of which units you settle on (you still haven't answered that), your numbers are a complete mess.

“At the moment of release the feathers cease being accelerated upwards, are inert in space”. Synonyms for “inert” are “motionless”, “unmoving”, “stationary”

Yes. In other words, their velocity is not 200,000m/s, km/s, mph, or anything. Their velocity, relative to the Earth, is 0m/s.

(is this where I am supposed to ask “motionless”, “unmoving”, “stationary”  relative to what”?)

Indeed, you catch on quickly! It's a shame you didn't think to ask yourself that before you lunged into this diatribe.

I think “hover” would be an applicable synonym as well since it means to be motionless in the air.

For an infitesimally short period of time, this would be correct, but that does not make hovering and falling one and the same. Immediately after release, the distinction between falling and hovering would become rather apparent. In one scenario, you start with a velocity of 0m/s and an acceleration of 9.81m/s^2. In the other, the initial velocity is 0m/s and the acceleration is 0m/s^2. Relative to the Earth, of course.

1) when you perform the experiment the first time on a flat earth with UA, would a clock on the bowling ball and feathers show them meeting the ground in the same amount of time as a clock that is being accelerated on the ground?

No - there would be a marginal, almost unobservable difference. The same is true for RET and gravity. Indeed, there would be no observable difference between the RET and FET scenario. This would become apparent if you used numbers that aren't nonsense.

To give you an idea, in Clyde's scenario that difference would be roughly 1s vs ~1.0000000033s. Just over three billionths of a second of a difference, rather than the three tenths you're proposing. This is why being countless degrees of magnitude is not good for you.

One also has to wonder why the bowling ball and feathers didn’t disintegrate while smashing into the atmolayer.

Because, unlike your assertion, they do not suddenly start yeeting themselves into the air at 200,000 of some unspecified unit. They start at 0m/s relative to Earth, and accelerate until they reach terminal velocity, or until stopped by some other force. It comes down to your numbers being complete nonsense again. Fix those, and you'll start making sense of a lot of things.

2) Second question is if you perform the same experiment a second time, say 5 minutes later, would a clock on the bowling ball and feathers measure the same amount of time to meet the ground as it measured the first time?

Indeed - the two drops would be exactly the same from an earthly observer's frame of reference. You forgot to ask yourself my favourite question - relative to what?.

Again, the answer is no because during the intervening 5 minutes, the accelerating flat earth has increased its velocity

Relative to what?

but the clock on the bowling ball and feathers is just as inert (motionless, unmoving, stationary and still hovering) for the second experiment as it was for the first.

Relative to what? And no, a bowling ball that's been dropped is not hovering, it's falling. You can test this yourself. Try dropping a bowling ball over your foot. Will you see it hover, or is your foot in imminent danger?

That means the relative velocity between the earth and the bowling ball and feathers is different for the first and second experiment

Impossible - as you just stated, in both scenarios the bowling ball starts motionless. In other words, its velocity relative to the Earth is 0m/s.

Nope, its a basic principle of GR

You have no grasp of basic classical mechanics (see above), and are in no position to even get started with GR.

I didn’t answer your question because it has nothing to do with the concept that is being discussed.  How fast the jumper sees the ground approaching him is irrelevant.

Of course it's relevant. Your argument presupposes the impossible. Between that and your fast and loose approach to numbers, it renders everything you say incoherent.

Yes, if someone managed to jump off a chair and found themselves plummeting towards the Earth at near-light-speed, a lot of physics as we know it would break in exciting ways. It still wouldn't make milliseconds a unit of velocity, and it wouldn't make velocity the same as acceleration, but some things would indeed break. However, it's satisfying that very big "if" that's your current problem.

What does matter is that if jumper’s clock objectivelymeasures one second to hit the ground, then a clock on the ground will objectively measure 1.3s to hit the ground.

For this to be the case, the diver would have to be approaching the Earth at ludicrous speeds. He is not. And that's why the relative speed matters. As Clyde rightly pointed out, it is not 200,000ms. So, now you just need to actually calculate it, and then plug the results into the calculator you found online without understanding what it does. This time, with correct units.

By the way, I'm giving you a chance here. This is like your umpteenth troll account. You can drop the act, or you can join your other pricelessalts. We won't have fake RE'ers sully the reputation of what is already the losing side.

[Rog]

For an infitesimally short period of time, this would be correct

,

And in that infinitesimal moment when the jumper/ball and feathers are inert and motionless and hovering, what is the relative velocity between the objects and the earth?

Because that is when the clocks start and they stop when the objects meet the ground. What happens in between the clocks starting and stopping is irrelevant and doesn't effect the time that the clocks measure. Either the clocks measure the same amount of time for the objects to go from hovering to meeting the ground or they don’t.

So for the first and second experiments,  what is the relative velocity between the jumper/feathers and bowling ball and the earth in that infinitesimal moment of hovering?  (remember we are in a vacuum)

That’s really all that matters.  Can you answer that? I don’t expect an exact number, but is it close to c?

You have no grasp of basic classical mechanics (see above), and are in no position to even get started with GR

.

In general relativity, gravity can be regarded as not a force but a consequence of a curved spacetime geometry where the source of curvature is the stress–energy tensor (representing matter, for instance

https://en.wikipedia.org/wiki/Geodesics_in_general_relativity

By the way, I'm giving you a chance here. This is like your umpteenth troll account. You can drop the act, or you can join your other pricelessalts. We won't have fake RE'ers sully the reputation of what is already the losing side

Not trying to pretend to be anything.  Just stating my views, which don't change.  What name I use to do that shouldn't matter.

[Pete Svarrior]

And in that infinitesimal moment when the jumper/ball and feathers are inert and motionless and hovering, what is the relative velocity between the objects and the earth?

0 m/s

So for the first and second experiments,  what is the relative velocity between the jumper/feathers and bowling ball and the earth in that infinitesimal moment of hovering?  (remember we are in a vacuum)

That would be 0 m/s, and whether or not we're in a vacuum (we're not) does not affect this answer. After all, you did just specify motionlessness, so no other answer is possible. You are, quite literally, asking "if the velocity of an object is 0 m/s, then what is its velocity?" Hint: it will not be 200,000ms.

That’s really all that matters.  Can you answer that? I don’t expect an exact number, but is it close to c?

It is not close to c. It is approximately c away from c.

[Clyde Frog]

I'm pretty sure at this point that Rog thinks there's some preferred FoR where clocks have some default rate of ticking. That's the only way I can imagine this terrible argument not coming from a place of trolling.

[Pete Svarrior]

I'm pretty sure at this point that Rog thinks there's some preferred FoR where clocks have some default rate of ticking. That's the only way I can imagine this terrible argument not coming from a place of trolling.

He has a long history of similarly terrible arguments coming from previous accounts, and he's been permabanned over obvious trolling on most of these, because usually he starts digging up countless old threads and spamming them all with the same argument. This account is probably going to join unless he suddenly improves.

He has a weird obsession with UA, but he really doesn't know how to tackle it. Nonsensical numbers with no explanation, meaningless units, and a complete lack of understanding of mechanics are all part of his MO.

My favourite of his arguments was about how spirit levels wouldn't work under UA. He did not know how spirit levels work, and thought that the reason the bubble moves is due to different gravitational forces being applied to each end if the tool isn't level. https://forum.tfes.org/index.php?topic=15343.0

[Rog]

That would be 0 m/s, and whether or not we're in a vacuum (we're not) does not affect this answer. After all, you did just specify motionlessness, so no other answer is possible. You are, quite literally, asking "if the velocity of an object is 0 m/s, then what is its velocity?" Hint: it will not be 200,000ms.

I asked what the relative velocity between the objects and the earth was at the time the jumper was hovering, not the individual velocities of each.  And it just dawned on me what you don’t understand. One of those simple, fundamental things that you assume people know, and are surprised when they don’t.

Relative velocity is a measurement of velocity between two objects as determined in a single coordinate system. Relative velocity is fundamental in both classical and modern physics, since many systems in physics deal with the relative motion of two or more particles. In Newtonian mechanics, the relative velocity is independent of the chosen inertial reference frame. This is not the case anymore with special relativity in which velocities depend on the choice of reference frame. If an object A is moving with velocity vector v and an object B with velocity vector w, then the velocity of object A relative to object B is defined as the difference of the two velocity vectors .

https://en.wikipedia.org/wiki/Velocity

IOW, the relative velocity between A and B is the difference between their velocities as measured within their own reference frames. The rate of change of the position of the object within its own frame as a function of time within its own frame, and the direction of that change within its own frame.  There is no way to determine the relative velocity without first defining an independent velocity in the individual frames since the relative velocity is the difference between them

This means that it is perfectly appropriate to speak of velocity with respect to a single reference frame and in relativity,  you don’t have to qualify the word “velocity” with the phrase  “with respect to what” because it is understood to be within its own reference frame unless you are specifically talking about relative velocity wrt another frame.

So I’ll ask another way.  What is the velocity of our jumper’s clock within in his own reference frame at the time he is hovering and what is the velocity of the ground clock within its own reference frame at the time the jumper’s clock is hovering?
 
Just subtract whatever you think the velocity of the ground clock in its reference frame (a close approximation is fine) at the time the jumper is hovering and the velocity of the jumper’s clock in its reference frame at the time the jumper is hovering (I assume it is zero) and that should give you the number you believe should be the relative velocity between them.

That’s how you calculate relative velocity in relativity, not Newtonian mechanics and that is the number we can use in the time dilation calculator. It should be simple enough for you to answer so I'll assume anything other than actual numbers as an inability or unwillingness to answer.

It’s also worth noting that If our jumper starts with 0 velocity in his own reference frame, in that infinitesimal moment he is inert and ends with 0 velocity 1s later, he hasn’t moved within his own reference frame and has not “fallen”, in any meaningful sense of the word. 1s*0 velocity is 0 distance.

[Clyde Frog]

You're getting so close! This is exciting!

If only it were possible to define a FoR where, immediately prior to jumping, the disc and the jumper were both at 0m/a. It's a tricky problem, eh?

[RonJ]

Per the question of where the energy for comes from; since it is beneath the earth and inaccessible that is a question easily left as unknown. While we can directly see and experience the mechanical action of the earth's upward movement, we are ignorant of the energy source below. The phenomenon of "gravity" is as equally deficient in its explanation for where all of the energy comes from for matter to pull matter, and that usually gets glossed over.

There isn’t any energy necessary for the perception of gravity to occur.  Nothing has been ‘glossed’ over.  It’s well known that mass influences relative time.  If you step off the chair your clock moves a bit faster, and you traverse thru spacetime just a bit quicker than the earth does and very quickly you close the distance between you and the earth and your mass and the mass of the earth try to occupy the same place at the same time.  This is the point where you feel the force on your feet.  That force is what makes your journey thru space time, with your faster clock, the same as the earth as per the well know formula F=MA.  Therefore, inertial mass, and your measured mass are identical because you are effectively measuring the same thing.  If you want to solve a mystery why not think about how mass can slow down time relative to another clock on a different smaller mass?

[Pete Svarrior]

I asked what the relative velocity between the objects and the earth was at the time the jumper was hovering

Indeed. And, since you once again said "hovering", the answer is 0m/s.
 

IOW, the relative velocity between A and B is the difference between their velocities as measured within their own reference frames. The rate of change of the position of the object within its own frame as a function of time within its own frame, and the direction of that change within its own frame.

That's an extremely and needlessly convoluted way of representing it, keeping in mind that speed is relative. Nonetheless, since the two bodies are stationary relative to each other, there exists no frame of reference (or, indeed, a set of FoR's) in which their relative velocity will be anything other than 0m/s.

This means that it is perfectly appropriate to speak of velocity with respect to a single reference frame and in relativity,  you don’t have to qualify the word “velocity” with the phrase  “with respect to what” because it is understood to be within its own reference frame unless you are specifically talking about relative velocity wrt another frame.

So, the actual reason you normally don't have to be very pedantic about reference frames is that they can be assumed to be obvious. For example, in scenarios that take place on Earth, we would normally assume the frame of reference to be the Earth. For problems set in outer space, we would often assume an external inertial observer.

The fact that you don't know how to work with reference frames illustrates the harm of that approach, but hey, good enough for high school I guess.

In the case of many RE'ers here, it becomes crucial to ask the question, because they get confused by the existence of multiple FoR's. They do not undersrand what it means for the Earth to be accelerating with regard to a local inertial observer who's just left the Earth's surface. They're the type that concludes that such an acceleration is impossible because the Earth would approach the speed of light. In short, high school assumptions mean they default to the "outer space" scenario, and they don't know what to do with the fact that a FoR has been specified.

In your case, it becomes crucial to ask the question at every step, because your FoR's are a complete mess. Like, fucked beyond repair. You simultaneously act as if the Earth was accelerating and as if it wasn't, and the same goes for the jumper. You also imply the existence of a universal FoR, which will net you a Nobel prize as soon as you're ready to prove its existence. You're mixing your FoR's up constantly because you neglect to decide on what you're actually considering. It's the same problem as your units. You flip between mph, km/s, m/s, and milliseconds (for some reason) without adjusting the numbers when you do so. You also measure acceleration in units of velocity at times. The problem isn't that your argument is wrong - the problem is that you haven't even approached making a coherent argument. For all I know, you could just be spitting words out of GPT-3 and pasting them here.

There is no way to determine the relative velocity without first defining an independent velocity in the individual frames since the relative velocity is the difference between them

Of course there is. You just need a basic grasp of how mechanics works.

This means that it is perfectly appropriate to speak of velocity with respect to a single reference frame and in relativity,  you don’t have to qualify the word “velocity” with the phrase  “with respect to what” because it is understood to be within its own reference frame unless you are specifically talking about relative velocity wrt another frame.

No, that's complete nonsense. Every body's velocity in its own reference frame is always 0m/s, because no body is moving away from itself. Also, it is relativity that strictly requires you to define your frames of reference.

So I’ll ask another way.  What is the velocity of our jumper’s clock within in his own reference frame

Always, invariably, 0m/s. This is because a body's own frame of reference refers to points on the body, and those will never move away from themselves.

Just subtract whatever you think the velocity of the ground clock in its reference frame (a close approximation is fine) at the time the jumper is hovering and the velocity of the jumper’s clock in its reference frame at the time the jumper is hovering (I assume it is zero) and that should give you the number you believe should be the relative velocity between them.

No - this defies physics, since you failed to transform your measures into a singular frame of reference. Subtracting two velocities without standardising them first will lead you to wildly spurious results.

For example, in Clyde's scenario (which you failed to calculate, but hey ho), the jumper is falling at 10m/s relative to the Earth. However, if we took your approach, we'd be getting 0m/s-0m/s=0m/s. In fact, your approach would always lead to the same calculation, which makes it rather useless for anything that happens to be in motion.

That’s how you calculate relative velocity in relativity

It really, really, really isn't. What you're looking for is Lorentz transformations and the velocity addition formula. In this case, since the bodies aren't moving at relativistic speeds relative to each other, the Galilean approach will yield a very good estimate with much less effort.

I'll assume anything other than actual numbers as an inability or unwillingness to answer.

I already provided you with these numbers while making fun of you before. I am shocked, truly shocked that you haven't read my responses.

It’s also worth noting that If our jumper starts with 0 velocity in his own reference frame, in that infinitesimal moment he is inert and ends with 0 velocity 1s later, he hasn’t moved within his own reference frame and has not “fallen”, in any meaningful sense of the word.

Indeed, he would not fall away from himself. This is why using the body itself as a reference frame for a 1-body system is useless. However, he likely did fall in a meaningful sense of the word, as long as you pick a reference frame that makes a modicum of sense.

1s*0 velocity is 0 distance.

Indeed. You are beginning to understand the problem with your approach.

[Rog]

No - this defies physics, since you failed to transform your measures into a singular frame of reference. Subtracting two velocities without standardising them first will lead you to wildly spurious results

You don’t use the LT when figuring time dilation. You can’t use relative time to calculate time dilation, because relative time is the solution. Read that as many times as necessary for it to sink in. No pun intended, but using relative time to calculate relative time is the very definition of circular logic.  If you tried,  there would be no such thing as time dilation.  Time dilation only becomes relevant at relativistic speeds.  If you use the LT to transform relativistic speeds to non-relativistic speeds, it ceases to be a meaningful concept.

And that would be in all of physics, not just when talking about FE vs. RE. I’m sure you’re familiar with the twin paradox.  The traveling twin comes back younger than the stay at home twin.  If you try and use the LT in that scenario to “standardize” the relative velocity of the twins, they would be the same age when the traveling twin returns. There would be no apparent paradox because there would be no time dilation.

In addition to that, time dilation is measured by clocks, not by people. The LT is about what people observe, not about how clocks keep time.  What the clock measures is in no way effected by what one observer perceives the other observer's velocity to be, or what they observe the velocity of the clocks to be. 

Time dilation doesn’t mean that a stationary observer perceives a moving clock to run slower. It means a moving clock physically keeps time at a slower rate than a stationary clock. It is a purely kinematic phenomena. Synchronize two clocks.  Send one on a rocket traveling around space for a period time.  When it comes back it will show a lesser elapsed time than clocks on earth. The rate at which it measures time is effected only by the actual velocity the clock physically experiences,  the rate of its change of position within its own reference frame. The greater the rate of change, the slower it measures time. And that, btw (looking at you, Tom), is why we can’t exceed c.  Time stops when velocity reaches c.

If you want a more technical explanation, time dilation applies when two events occur in the same place in the same reference frame..  IOW, when you are calculating time dilation, you are calculating how long it takes the same clock to click twice.  The LT relates the position and time of one event in two different reference frames..  The time dilation formula can be derived from the time component of the LT,  but they are used for different purposes and in different situations.
 

Indeed, he would not fall away from himself. This is why using the body itself as a reference frame for a 1-body system is useless. However, he likely did fall in a meaningful sense of the word, as long as you pick a reference frame that makes a modicum of sense.

As explained above, it is only in the reference frame of the jumper that you can correctly measure his change in position over time.   The time component of the time dilation formula is proper time and proper time is the time an observer measures in his own frame of reference.

You might find this helpful.  And also note, that at no time is the reader instructed to “standardize” velocities.
https://www.toppr.com/guides/physics-formulas/time-dilation-formula/

[Pete Svarrior]

You don’t use the LT when figuring time dilation.

We're not discussing time dilation yet - you haven't made it that far. We can't do so until you've corrected your nonsensical velocities. Before we can discuss the consequences of your scenario, you need to make your scenario consistent with basic physics. You have a long laundry list of errors to work through, but you haven't made a start yet.

I'd suggest to begin with not using milliseconds as a unit of velocity, and instead deciding on one of the three units you've been using interchangeably.

Then, you need to define your FoR's, this time correctly; and understand that the velocity of someone standing on the Earth relative to the Earth will always be zero, because as long as they're standing there, they are not moving relative to the Earth. Similarly, their velocity relative to themselves will always be zero - if you do not understand that a body can't move away from itself, well, we've got more fundamental issues than relativity or mechanics to work thorugh.

Finally, a few smaller issues, e.g. someone that jumps off a chair will not be falling for 5 minutes, and they will not be accelerating at break-neck ratios long past their terminal velocity.

Once you've done all that, we can plug the corrected numbers into the calculator you found online and which you don't understand to find out that the actual measured difference in time would be imperceptible, and, very importantly, identical to what would be measured under RET assumptions. Best of luck!

In addition to that, time dilation is measured by clocks, not by people. The LT is about what people observe, not about how clocks keep time.  What the clock measures is in no way effected by what one observer perceives the other observer's velocity to be, or what they observe the velocity of the clocks to be.

Find out what an "observer" is in physics. You might notice that your own "helpful" article refers to observers and observer time. It does so for a reason.

And no, Lorentz Transformations aren't "about what people observe". You're just going to have to learn the basics of relativity before you can discuss it in any meaningful way.

As explained above, it is only in the reference frame of the jumper that you can correctly measure his change in position over time.

On the contrary, it is the only frame of reference in which the jumper will never move, and his change in position will always be 0m. This is because the jumper will never move relative to himself. He will not become more distant, nor less distant, from himself. You either don't understand what a frame of reference is, or what it means for something to "move". I assume the former.

You might find this helpful.  And also note, that at no time is the reader instructed to “standardize” velocities.
https://www.toppr.com/guides/physics-formulas/time-dilation-formula/

Of course. After all, it only uses one velocity - there is nothing to standardise, because an apporpriate FoR was already chosen for you. Care to guess what that singular velocity is in your scenario, and how you could define it?

[Rog]

Then, you need to define your FoR's, this time correctly; and understand that the velocity of someone standing on the Earth relative to the Earth will always be zero, because as long as they're standing there, they are not moving relative to the Earth. Similarly, their velocity relative to themselves will always be zero - if you do not understand that a body can't move away from itself, well, we've got more fundamental issues than relativity or mechanics to work thorugh.

You may want to think through some of that.
 
The time dilation formula is just the Lorentz factor which is part of the formula for transforming velocities

Time dilation effects are already baked into the transformed velocities. Using the transformed velocities as the relative velocity for the time dilation formula on top of that is circular. You are literally using time dilation to calculate time dilation.  You can use the full LT to find time dilation or you can use the time dilation formula, but you can’t use them together without compounding the amount of time dilation.

Of course the proof is in the pudding. According to this calculator, which uses the complete transformation, the time dilation factor at .8c should be 1.666*the observed time. Which happens to be the exact same number the calculator that uses the dilation formula with “untransformed” relative velocity comes up with. Note the formula represented by gamma is the exact same formula used in the link I previously provided and in the time dilation calculator.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html






Interestingly,  if you transform the velocities to achieve a non-relativistic relative velocity, and use that for the time dilation calculator, there is virtually no time dilation, even at nearly the very limit of c.  But according to the LT, there should be about 22s of time dilation. Transforming the velocities makes the whole concept of time dilation moot.  Shouldn’t exist in any significant way. It’s almost as if the time dilation effect was somehow compounded.

[Pete Svarrior]

You're still mixing up your FoR's. Yes, you desperately want to consider a scenario in which something is moving at 0.8c. As before, you forgot to define the observer, or the frame of reference in relation which said something is moving at 0.8c.

Clyde Frog hit the nail on the head a whole ago - you keep acting as if some objective FoR existed. It doesn't. That's the "relativity" part of "relativity".

But according to the LT, there should be about 22s of time dilation.

Which Lorentz Transformation would that be? You're mixing up terms again.

Transforming the velocities makes the whole concept of time dilation moot.  Shouldn’t exist in any significant way.

You've cracked the case. In the scenario you specified, assuming either an observer standing on the Earth (jumper falling as in RET) or the observer being your jumper (Earth accelerating upwards), time dilation is a moot point! The two clocks would need to move at relativistic speeds relative to one another, but, in your scenario, they simply don't. You defined the scenario. Only you can fix it.

[Rog]

You've cracked the case. In the scenario you specified, assuming either an observer standing on the Earth (jumper falling as in RET) or the observer being your jumper (Earth accelerating upwards), time dilation is a moot point! The two clocks would need to move at relativistic speeds relative to one another, but, in your scenario, they simply don't. You defined the scenario. Only you can fix it.

You don't understand.  Its not just "in the scenario I've specified".  Time dilation wouldn't exist at all even if two clocks are moving at relativistic speeds.



If both clocks are moving at .999, then their "transformed relative velocity" is -.999.  If you use that number for time dilation formula, there is still no time dilation.



At the very limit of c, according to you, TD doesn't exist.

If that is what you believe, somebody needs to get busy editing the wiki.  It also contradicts the argument that the earth could never reach c if it were accelerating constantly.

Because the wiki bases the whole argument on the Lorentz factor

[drand48]

I'm trying to understand the objections to Tom's OP.

Are the objections to his experiments?  If so, any such objections would contradict the Equivalence principle of gravity and acceleration, a principle that is generally accepted in modern physics.  Are people objecting to that?

Or are the objections to the interpretation?  Because the interpretation doesn't seem to have anything to do with these arguments and could be objected to on philosophical grounds (such as whether "We can see that the earth moves upwards" or whether we see ourselves or the ball dropping.)  But not on the grounds of what actually happens.  Same thing in both cases.  If there was no gravity and the surface of the Earth was accelerating upwards at 9.8 m/s^2, the experience would be exactly the same as Earth with gravity.

Or are those who are objecting here disagree with the equivalence principle?  It seems to me that people are expecting different outcomes from the experiments.

[Pete Svarrior]

Time dilation wouldn't exist at all even if two clocks are moving at relativistic speeds.

Relative to what? I said nothing about two clocks moving. I specified the velocity of one relative to the other.

If both clocks are moving at .999, then their "transformed relative velocity" is -.999.

Velocity is a vector, not a scalar, so your "transformation" doesn't make an ounce of sense.

Also, what unit is that in? 0.999 what? Not that I'm foreshadowing anything hilarious here. Oh, wait...

If you use that number for time dilation formula, there is still no time dilation.

Yes, if something is moving at just under 1km/s, as you showed in your screenshot, time dilation will indeed be imperceptible. This is because 0.999km/s is very small compared to relativistic speeds.

At the very limit of c, according to you, TD doesn't exist.

You've "mistaken" 1km/s for c. You're off by a factor of 300,000, in true keeping with your previous gargantuan errors.

Do remember that I'm still giving you a chance to drop the transparent troll. Will you give it a go, or should we move on?

[Rog]

You've "mistaken" 1km/s for c. You're off by a factor of 300,000, in true keeping with your previous gargantuan errors.

You’re right.  I miscalculated that.  But you are still not getting it.  I am beginning to suspect that misunderstand how relative velocity is calculated.  To be fair, it’s not just you.  There seems to be some misunderstanding for a lot people on this site, even those on the RE side.  So maybe we need to go back to basics.

I think the explanation given on this site is as clear as I’ve seen. https://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

In non-relativistic mechanics the velocities are simply added and the answer is that A is moving with a velocity w = u+v relative to C. But in special relativity the velocities must be combined using the formula
               u + v
         w =  ---------
              1 + uv/c2

This change in the velocity addition formula from the non-relativistic to the relativistic theory is not due to making measurements without taking into account light-travel times, or the Doppler effect. Rather, it is what is observed after such effects have been accounted for. It is an effect of special relativity which cannot be accounted for using newtonian mechanics

.
In other words, this formula takes the effects of time time dilation into account.  How that is done is explained here.

To go from the reference frame of A to the reference frame of B, we must apply a Lorentz transformation on co-ordinates in the following way (taking the x-axis parallel to the direction of travel and the spacetime origins to coincide):
   xB = γ(v)( xA - v tA )
   tB = γ(v)( tA - v/c2 xA )

   γ(v) = 1/sqrt(1-v2/c2)
To go from the frame of B to the frame of C you must apply a similar transformation
   xC = γ(u)( xB - u tB )
   tC = γ(u)( tB - u/c2 xB )
These two transformations can be combined to give a transformation which simplifies to
   xC = γ(w)( xA - w tA )
   tC = γ(w)( tA - w/c2 xA)

             
       u + v
         w = ---------
               1 + uv/c2
This gives the correct formula for combining parallel velocities in special relativity.

Note that the required Lorentz transformations are already baked into the formula. No further transformations are necessary.  This formula is for parallel velocities, but can be rearranged to calculate the relative velocity of objects moving in opposite directions.

The formula can also be applied to velocities in opposite directions by simply changing signs of velocity values, or by rearranging the formula and solving for v. In other words, If B is moving with velocity u relative to C and A is moving with velocity w relative to C then the velocity of A relative to B is given by
            w-u
      v =  ---------
           1–wu/c2

So assigning A as the jumper, B as the earth and C as a stationary observer on the ground, the formula looks like this. (using the 200,000 km/s we started with)
V_ab = Jumper velocity wrt Observer – Earth velocity wrt Observer / 1-wu/c2
V_ab = 2000,000km/s – 0  / 1-0
V_ab = 200,000km/s

The relative velocity of the jumper and the earth is 200,000km/s. I am left to wonder why the jumper doesn't vaporize when he meets the earth.


I have no idea how you think the relative velocity should be calculated because you have never explicitly stated, but if you are doing it any other way, you are doing it wrong.  I gather you think there is some other step somewhere in the process where LT needs to be calculated, but shown above, the LT is already accounted for in the relative velocity that is inputted into the time dilation calculator. But it doesn't really matter how you think the relative velocity should be calculated or what you think it should be.

Figure it however you want.  As long as it is consistent with relativistic speeds, there will always be some observed time dilation.  If, however, your method never results in a relative velocity consistent with relativistic speeds, you've made time dilation a meaningless concept.  That is the point I was trying to make before. If your calculation of relative velocity is never more than say .4c, time dilation doesn't exist. If it can be more than .4, then you still are stuck having to explain  why an observer on the ground doesn't measure a fall time differently than a "falling observer" or why that difference doesn't increase over time if the earth is accelerating.

[Pete Svarrior]

You’re right.  I miscalculated that.

Yes. It's just one of many such "miscalculations", where you just completely ignored units and assumed they'd work out. The moment you fix all of these, your results will start making more sense.

I am beginning to suspect that misunderstand how relative velocity is calculated.  To be fair, it’s not just you.  There seems to be some misunderstanding for a lot people on this site, even those on the RE side.  So maybe we need to go back to basics.

Ah, another close brush with self-awareness! Let's just recall that you thoroughly documented your understanding of velocities in this thread, and that it contradicts your new findings. Don't get me wrong, it's good that you're learning, but you could be honest about it.

I think the explanation given on this site is as clear as I’ve seen. https://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

Right, but you're once again assuming relativistic speeds, because of your previous collossal errors. None of your examples to date necessitate any of this, and a classical mechanics approach will yield perfectly fine results. Sure, you can use the velocity addition formula (which I already referred you to - good job you've looked at it) if you really want. It'll make an immaterial difference to your calculations.

using the 200,000 km/s we started with

Unfortunately, we can't use that. 200,000km/s is an absolutely ludicrous velocity for a jumper, measured relative to the Earth. Until you've corrected your previous errors - either by correcting the calculations that got you here or correctly defining your FoR's, we're stuck in the Nonsense Zone.

The relative velocity of the jumper and the earth is 200,000km/s. I am left to wonder why the jumper doesn't vaporize when he meets the earth.

Because you defined an impossible scenario and decided to roll with it. This is exactly why you need to fix it.

I have no idea how you think the relative velocity should be calculated because you have never explicitly stated

Of course I did! You're just not a particularly attentive reader. Observe!

That’s how you calculate relative velocity in relativity

It really, really, really isn't. What you're looking for is Lorentz transformations and the velocity addition formula. In this case, since the bodies aren't moving at relativistic speeds relative to each other, the Galilean approach will yield a very good estimate with much less effort.

You already agreed with me that you're happy to use the velocity addition formula by linking to an (oversimplified) explanation of how to perform these calculations. As you can see, if you had simply paid attention, you'd have saved yourself a lot of time.

I gather you think there is some other step somewhere in the process where LT needs to be calculated, but shown above, the LT is already accounted for in the relative velocity that is inputted into the time dilation calculator.

You don't know what "LT" is. Find out.

But it doesn't really matter how you think the relative velocity should be calculated or what you think it should be.

Ahhh, another brush with self-awareness! So blissful!

Figure it however you want.  As long as it is consistent with relativistic speeds, there will always be some observed time dilation.

Indeed. As soon as you find a scenario in which two bodies are moving at relativistic speeds relative to one another, you'll be able to meaningfully consider time dilation. It's just that, so far, you haven't. Furthermore, I propose you will not be able to come up with an earthly scenario in which the speeds would even remotely approach relativistic speeds, but I'm happy for you to try.

why an observer on the ground doesn't measure a fall time differently than a "falling observer"

Because the falling observer's velocity relative to the Earth will generally not exceed their terminal velocity. For a human, that would be something to the tune of 200km/h, or roughly 0.0000002c

or why that difference doesn't increase over time if the earth is accelerating.

It does, to a point. To answer the question of why they won't accelerate to the ridiculous speeds you really want, the answer is twofold:

• Drag - as the falling observer gains momentum relative to Earth, the air resistance increases in magnitude. Eventually, the two reach an equilibrium, and the falling observer ceases to accelerate.
• Time - if you somehow eliminated drag from the equation, you'd still be stuck with the fact that "falling" is usually not sustained. In your scenario of someone jumping off a chair, we can safely assume this will not take longer than a few seconds. Let's say 1 second (wowee, that's a tall chair!) and let's assume an acceleration of 10m/s^2 for simplicity of calculations. Can you, the reader, figure out what the velocity of the falling observer will be just before they touch the ground? Can you compare that number to c?

[AFlat]

I must be missing something here because all of this seems needlessly complicated. Take a simple disc and add basic gravity. Unless you're near the edge then all the lateral forces cancel each other and you're left with a net downward force. If you were near the edge then you'd experience a radial inward force. The closer you got to the edge the stronger this unchecked radial pull would become, so "falling off the edge"  would be no easier than climbing a mountain. This agrees well with the circumferential ice caps and mountain ranges that we see bounding the antarctic oceans. It's no coincidence that all the circumference is free of liquid water.

UA seems loaded with problems, including the need for an accelerating force and the rapid increase in velocity. At a constant acceleration of 9.81 m/s² over a single year we'd approach a significant fraction of the speed of light. This would cause some pretty spectacular distortions where all the stars near the zenith would end up blue-shifted and moved a lot closer to the horizon. Anything that the Earth struck would arrive at relativistic velocities, and even pebbles would pack the punch of an atomic weapon. Happily, this isn't happening. 

[Pete Svarrior]

At a constant acceleration of 9.81 m/s² over a single year we'd approach a significant fraction of the speed of light.

Right. So you've jumped into the middle of an in-depth discussion on relativity just to tell us you haven't read about UA, and that you don't understand the differences between classical mechanics and special relativity.

Don't do that.