Formulation

Two twins, alice and bob. alice is in one space ship hanging out, and waiting for her brother bob, who is taking a trip to alpha centauri.

according to the theory of relativity, the stationary twin, alice, sees time passing more slowly for the traveling twin, bob, than he himself experiences the passing of time. messages from him arrive more sparsely than he sends them. she looks in the window as he passes by, and sees his clock running slow. indeed, she sees him moving more slowly, even his heartbeat is slowed! the net effect is that bob seems to not age as much as alice does, for the duration of the trip.

ok, great. the time dilation is mostly derived from the fact that the speed of light is measured the same for any observer. but einstein included another postulate in his theory: the relativity principle, which was known even to galileo. namely, that the laws of physics are the same in any intertial frame. that all motion is relative.

so let s look at the problem from bob s frame. he s flying along in his spaceship, but in his frame, he is at rest, and alice is in motion, in the opposite direction. he also sees her clock running slow, for the same reasons. her heartbeat is also slower, and she ages less, according to his measurements in his restframe.

now the paradox is immediate. bob thinks alice aged less, and alice thinks bob aged less. so when he gets home, and they embrace each other, which twin is older? certainly they can t both be right.

this apparent paradox is the source of constant debate on this board, and this thread is my attempt to put the question to bed.

as a side note, let me say what the purpose of this thread is not. it is not to debate whether the postulates of relativity are correct. it is not to debate whether time dilation and length contraction are logically consistent, real, etc. please take that sort of thing to another thread.

Resolution

In fact, with a more complete understanding of relativity theory, one can see that there is no paradox. the problem stems from the word inertial.

the principle of relativity says that all inertial frames are equally valid for doing physics. it makes no such claims about accelerated frames.

this is perfectly understandable. imagine playing ball on a merry-go-round. if you throw the ball into the air, it will appear to curve around due to the coriolis effect. Newton s Law F=ma no longer holds. even with no net external forces acting on it, the ball exhibits an acceleration. The laws of physics are all the same in all inertial frames, but they may need to be modified to be used in a noninertial frame.

and it should be obvious that bob cannot be in an inertial frame if alice is. alice and bob started in the same inertial frame, at some point. after all, they once shared the womb. if bob is to make it to alpha centauri, he must fire his rockets, and gain some speed relative to alice. in order to do so, he must accelerate. eventually, he must decelerate, and eventually turn around.

so the frames are fundamentally different. the basic equations of special relativity (SR) apply only to inertial frames, so we must be very careful about what we apply to bob s frame.

in the language of general relativity (GR), we say that all inertial objects follow geodesics through spacetime. so alice s and bob s paths through spacetime are fundamentally different, and have different lengths, and therefore when the trip is over, both alice and bob can agree that bob is younger.

let me repeat that: once all is said and done, relativity answers unambiguously, bob is younger!

let s look at the details, first from the inertial frame. that is, from alice s perspective.

according to einstein s equivalence principle, a gravitational field is locally indistinguishable from an accelerated frame, so if we can describe the problem with SR using an accelerated frame, we can equally well describe the problem as a stationary observer in a gravitational field, using the machinery in GR. so we can try it from bob s accelerated frame next.


ok, so let s get started!

Solution: Inertial Frame

first, lets setup the problem. i would like to assume the simplest possible accelerated path for bob. that is, constant thrust for the first half of the journey, and at the halfway point, he begins decelerating, reaching alpha centauri, and then accelerating back to the halfway point, then reaching alice at the end of the journey, with no relative motion.

let T denote proper time (that is, time measured on bob s watch), throughout. alice s proper time will be t, which is also the usual relativistic time coordinate.

bob s rockets exert a constant thrust on his rocket, so his acceleration is f(T). (which we will take to be a constant). after a time T, bob s rocket is going du=f(T) dT faster. that is his change of velocity, in his frame.

in alice s inertial frame, we must use the relativistic addition of velocities rule:

v(T+dT)=[v(T)+du]/[1+v(T) du/c²]

therefore

[v(T+dT)-v(T)]/dT = du/dT [1-v(T)v(T+dT)/c²]

dv/[1-v²/c²] = f(T) dT

integrating yields:

v(T) = c tanh {1/c * Int[ f(T) ] } = c tanh h(T)

where h(T) = 1/c Int[ f(T)]

x(T) = c Int[ sinh h(T)]

t(T) = Int[ cosh h(T)]

now, we can simply plug in our value for f, and read off the time that has passed for the accelerated observer.

if f(T) = g, the gravitational acceleration on earth, to make bob comfortable, then h(T) = gT/c, and t(T) = c/g sinh( gt/c )

we re pretty much done now, let s just plug in some numbers. let s assume it s two light years to alpha centauri, so we accelerate for the first light year, and decelerate for the second.

bob s proper time at one light year is 1.29451 Yrs. alice s is 1.71507 Yrs. looks like alice is getting older!

the calculations look much the same for the other legs of the trip. by the time bob get s home, bob s clock read T = 5.178 Yrs, and alice s clock read t(T) = 6.860 Yrs.

assuming bob and alice both understand the physics of spacetime, then they would both have predicted this result.

as we see, alice is older, by almost 2 yrs.

Thanks, lethe, I enjoyed reading this.

Related questions for anyone: Suppose the example involves a trip to a galaxy 2 million light years away. Bob will make a one-way trip in 20 proper years. At the halfway point, then, he must measure the galaxy as < 10 light years away. To him does the galaxy look like it’s < 10 light years away, or 1 million light years away? I mean, is the galaxy spread across his field of vision, or is it still a faint speck? If the former, then when Bob decelerates does the galaxy first shoot away (back towards 1 million light years away) and then approach again as he closes in on it?

Good question zanket. I'm rushing so haven't thought about it much, but I think the distance will appear roughly the same as a stationary observer at the same place. Here's a question for you: If an observer could accelerate at a super-high acceleration (without dying, which he can't), his clock would slow way down. Would it then be possible - in his own Frame Of Reference, not an inertial F.O.R. - for him to see himself as surpassing light speed? Remember his clock is slow, so the time it takes to get to his destination a fixed distance away will be less for him than for someone observing him from an inertial F.O.R. What do you think, zanket? How's about you lethe? Just an idea from left field....

p.s. Thanks for the math, that should shut some people up (though probably not all of them).

Aaron

I’m thinking the opposite, that to Bob the galaxy looks like it’s < 10 light years away. This is because Bob sees objects foreshortened along his axis of motion. Passing stars look like discs, so the space between the stars must be foreshortened as well to make the galaxy appear closer and fill up Bob’s field of vision.

Regarding Bob seeing himself as appearing to surpass light speed, yes he would when he reached (sqrt(0.5) * c) = 0.707c relative to passing stars. At that speed he’d be covering 1 light year—as measured when he was at rest with respect to the stars—every proper year. But he’s still moving at “only” 0.707c. Super-high acceleration isn’t needed for this effect; at 1g it’s attainable within a proper year.

lethe,

Are you saying that the time difference between Alice and Bob only occurs during the parts of the trip in which Bob is accelerating/decelerating (in your example, Bob is acceleration/deceleration during his entire journey)?

If Bob's average acceleration during the entire journey is under 9.8 m/s^2, will Alice be younger then Bob at the reunion?

Tom

Originally posted by Prosoothus
lethe,



If Bob's average acceleration during the entire journey is under 9.8 m/s^2, will Alice be younger then Bob at the reunion?

Tom

I thought we already cleared this up.

The answer is No.

Practical example:

Earth has a mass of 6e24 kg and a radius of 6378,000 m which gives a surface gravity of :

g(e) = GM/r² = 6.673e-11 * 6e24 / 6378,000² = 9.84... m/s²

Uranus has a mass of 8.68e25 kg and a radius of 25,559,000 m which gives a surface gravity of :

g(u) = 6.673e-11 * 8.68e25 / 25,559,000² = 8.86.... m/s²

From this, if time dilation was proportional to merely the value of g-force, you would expect a clock on Earth to run more slowly than one on Uranus.

But....

The time dilation formula for Gravity is:

t = t0/sqrt(1-2GM/Rc²)

Where t0 is the time as measured by a clock far away enough away to be considered "outside" the gravity field for all practical purposes.

For Earth, and assuming t0 = 1sec this works out to:

t(e) = 1/sqrt( 1 - 2(6.673e-11)(6e24)/(6378000(9e16)))

= 1.000000000697... sec

Meaning that the Earth surface clock will tick 1 sec for every

1.000000000697... secs of the reference clock.

for the surface of Uranus:

t(u) = 1/sqrt( 1 - 2(6.673e-11)(8.68e25)/(25559000(9e16)))

= 1.000000002517... sec

Which means a clock on the surface of Uranus runs slower than one on Earth by a factor of 1.00000000182

So simple g-force is not enough.

With the acceleration felt with Bob, the equivalent g-field is uniform and extends all the way from him to the Earth, so you have to take the relative positions into account when determining the relative clock rates.

Originally posted by Janus58
So simple g-force is not enough.

With the acceleration felt with Bob, the equivalent g-field is uniform and extends all the way from him to the Earth, so you have to take the relative positions into account when determining the relative clock rates.

Quick question. I thought that I was following your (very good) explanation all the way up to these two sentences. What has to be taken into account if g-force isn't enough?

Originally posted by Canute
Quick question. I thought that I was following your (very good) explanation all the way up to these two sentences. What has to be taken into account if g-force isn't enough?

Relative position in the gravitational field.

Example:

Assume a uniform gravitational field. (one in which acceleration due to gravity does not fall off with distance. )

You have two clocks, clock A above clock B. The field is uniform so each clock experiences exactly the same g-force. But clock A is higher than clock B and has a greater gravitational potential, So it will run faster than clock B(as measured from B). If you you had a third clock C, which was even higher than A, it would still feel the same g-force but would run even faster than B than A does.

Also, If you increase the stength of the G- field you will increase the time differences between the clocks.

So it is both strength of field and relative position in the field that determines the relative clock rates.

Thanks.

I'll risk an even more naive one. If clock speed is dependant on gravatic potential (position in field) and not just on the felt force of gravity how does a clock tell how fast it ought to be running. In other words in a uniform field of gravity how does the clock know where it is in the field?

By the thickness of the ether wich may vary,

Gravity is no pulling force but actually etherpressure pushing you back to earth because massive objects (like the earth) displace a lot of ether.

If you accelerate, it is like trying to swim faster through water, the water (ether) will offer resistance displacing it, so inertia is just another side-effect of etherpressure...

Another question:
Suppose we create a pair of entangled electrons , let's call electron #1, Alice and electron #2, Bob,

Now we accelerate bob in a huge synchronotron, we change the quantum state of alice and measure Bob, would the quantum state of nonlocality Bob change immediately or be time-dilated due to the accelleration?

Originally posted by Canute
Thanks.

I'll risk an even more naive one. If clock speed is dependant on gravatic potential (position in field) and not just on the felt force of gravity how does a clock tell how fast it ought to be running. In other words in a uniform field of gravity how does the clock know where it is in the field?

It doesn't have to. Remember, there is no absolute reference system of time for each clock to judge itself by. Each clock is its own refernce system. All you can say is that from clock A's perspective, clock B is running slow. But from clock B's persepective it is clock A that is moving fast. So how does clock B know how much faster clock A should be running? Because the information of what time is showing on clock A has to pass through the length of the gravitational field to get to clock B.
(It's like asking how an object dropped from the height of clock A "knows" how fast it should be moving when it reaches clock B. )

Ether?! I thought we cleared THAT up 50 years ago. There is no proven ether. Am I crazy, where are your sources?

ANYWAY, my question stems from the original on this thread. Lethe sited that the theory of relativity states that all motion is relative. The problem seemed to be cleared up by the fact that bob is accelerating, and alice is not. But why? who is to say that bob is accelerating and Alice is not?
In a universe where only bob, his spaceship, and alice existed, bob and alice would be gaining distance between the other at a constant accelerated rate. Why could Alice say that she is not accelerating and bob is? Bob ALWAYS finds himself at rest (givin the theory of relativity) and so he SHOULD conclude that Alice is accelerating RELATIVE TO HIM. He could also conclude that he is accelerating RELATIVE TO ALICE.

In this example, both can be said to be accelerating and both moving. Therefore when you say that they have different inertial frames, that is still in itself relative.

In a different example, lets say that Alice and Bob were a fair distance apart and each being pulled by their own equally massive black hole. But Bob is closer to his black hole and therefore "accelerates" faster toward it. Are they in the same or different intertial frames?

Originally posted by Janus58
It doesn't have to. Remember, there is no absolute reference system of time for each clock to judge itself by. Each clock is its own refernce system. All you can say is that from clock A's perspective, clock B is running slow. But from clock B's persepective it is clock A that is moving fast. So how does clock B know how much faster clock A should be running? Because the information of what time is showing on clock A has to pass through the length of the gravitational field to get to clock B.
(It's like asking how an object dropped from the height of clock A "knows" how fast it should be moving when it reaches clock B. )
If I'm derailing the discussion ignore this, I'll ask another time.

I would say that your example at the end is not quite equivalent, since the acceleration of clock A is determined simply by the gravity acting on it. However there is nothing (I presume) that can act on a clock that can 'inform' it of its position in a uniform gravity field.

It seems to follow, since the difference in the various clock speeds (A,B &C in your first example), is a function of the observation of them, that all unaccelerated clocks in a uniform gravity field run at the same speed until they are observed by someone elsewhere in the field. In other words it is not that clock A and B are running at different speeds, it is that the information that passes between them is affected by their relative positions in the field. I think I'm wrong here but it seems to follow from your explanation.

Frencheneesz,

In a universe where only bob, his spaceship, and alice existed, bob and alice would be gaining distance between the other at a constant accelerated rate. Why could Alice say that she is not accelerating and bob is? Bob ALWAYS finds himself at rest (givin the theory of relativity) and so he SHOULD conclude that Alice is accelerating RELATIVE TO HIM. He could also conclude that he is accelerating RELATIVE TO ALICE.

You're right, both of them are accelerating relative to each other. However, only one of them is feeling the force of acceleration. This force appears to be causing the time dilation, and not the acceleration itself.

Originally posted by zanket
I’m thinking the opposite, that to Bob the galaxy looks like it’s < 10 light years away. This is because Bob sees objects foreshortened along his axis of motion. Passing stars look like discs, so the space between the stars must be foreshortened as well to make the galaxy appear closer and fill up Bob’s field of vision.

Regarding Bob seeing himself as appearing to surpass light speed, yes he would when he reached (sqrt(0.5) * c) = 0.707c relative to passing stars. At that speed he’d be covering 1 light year—as measured when he was at rest with respect to the stars—every proper year. But he’s still moving at “only” 0.707c. Super-high acceleration isn’t needed for this effect; at 1g it’s attainable within a proper year.

that sounds about right. bob will see the distance to his goal contracted. and yes, you can have an "effective speed" as fast as you like.

Originally posted by Prosoothus
lethe,
If Bob's average acceleration during the entire journey is under 9.8 m/s^2, will Alice be younger then Bob at the reunion?

Tom

bob will be younger for any value of g you would like to choose (except 0).

Originally posted by Prosoothus
You're right, both of them are accelerating relative to each other. However, only one of them is feeling the force of acceleration. This force appears to be causing the time dilation, and not the acceleration itself.

I’d say that Alice is passively accelerating (relative only to Bob), whereas Bob is actively accelerating (feels it), and due to this active acceleration Bob measures all space along his axis of motion contracted, whereas Alice measures only Bob contracted along his axis of motion. Therein lies the asymmetry in their respective trips. Both are traveling relative to each other. But Bob travels less distance; therefore less time elapses on his clock than on Alice’s.

Originally posted by lethe
that sounds about right. bob will see the distance to his goal contracted.

And if this is true, to show how cool relativity is, Bob could, in principle, without leaving our Solar system, read the billboards on a planet in a distant galaxy.

lethe,

bob will be younger for any value of g you would like to choose (except 0).

Wouldn't time be dilating for Alice as well, since Alice is in the Earth's gravitational field during Bob's entire journey?

Aren't there two time dilations in the twin paradox:

1) Bob's time dilation due to his accelerating/decelerating.

2) Alice's time dilation resulting from her presence in a gravitational field.

Originally posted by Janus58
It doesn't have to. Remember, there is no absolute reference system of time for each clock to judge itself by. Each clock is its own refernce system. All you can say is that from clock A's perspective, clock B is running slow. But from clock B's persepective it is clock A that is moving fast. So how does clock B know how much faster clock A should be running? Because the information of what time is showing on clock A has to pass through the length of the gravitational field to get to clock B.
(It's like asking how an object dropped from the height of clock A "knows" how fast it should be moving when it reaches clock B. )
Thanks, but I still feel that this is not logical. Perhaps the problem is the uniform field, which is presumably an impossibility in reality. What you seem to suggest is that the clocks don't run at any particular speed relative to each other until their speeds are compared by someone. Am I muddled?

Originally posted by Canute
Thanks, but I still feel that this is not logical. Perhaps the problem is the uniform field, which is presumably an impossibility in reality. What you seem to suggest is that the clocks don't run at any particular speed relative to each other until their speeds are compared by someone. Am I muddled?

It might be more correct to say that the relative speed of the clocks depends upon the reference system from which the comparison is made.

For example, Bob carries two clocks with him. One he keeps in the nose of his ship, and one in the tail. During the acceleration phase of his trip he will note that the clock in the nose will run slighty faster than the one in tail (according to GR).

Alice, OTOH, only notices that, at any given moment, each clock has an identical velocity with respect to her, and thus, for her, both clocks run at the same speed.

If bob now decelerates by firing his retro-rockets, for him, the situation reverses and the clock in the tail runs faster. By the time he once again comes to rest with respect to Alice, the clocks will be sychronized again. And both he and Alice will agree as to how much total time ha elasped on them.

(this works even if Bob Flips his ship to brake rather than fires retro-rockets, because from Alice's view now, the two clock did not have equal speeds WRT herself for the whole trip. While it was flipping, the clock in the nose loses speed and the clock in the tail gains speed so they can switch positions. Thus for this period of time, for her, the nose clock will run just that much faster than the tail clock, to make up for the difference in time that Bob sees in them at the end of trip)

I understand that. But these are the effect of acceleration. We were talking about the time dilation that that g-force or acceeration did not explain, the effect of relative position in the uniform gravity field.

Originally posted by Canute
I understand that. But these are the effect of acceleration. We were talking about the time dilation that that g-force or acceeration did not explain, the effect of relative position in the uniform gravity field.

As far as Bob is concerned, the difference in the two clocks can be because of their relative positions in a uniform gravity field.(Due to the Equivalence Principle) Both Clocks experience the the same g-force but one runs faster than the other.

Originally posted by Janus58
As far as Bob is concerned, the difference in the two clocks can be because of their relative positions in a uniform gravity field.(Due to the Equivalence Principle) Both Clocks experience the the same g-force but one runs faster than the other.
I'm really sorry to keep coming back to this but your saying 'as far is Bob is concerned' implies that Bob has observed the two clocks. In this case it makes perfect sense to say, as you have, that they will be perceived as running at different speeds. I'm OK with that bit since there is an observer involved, a compsrison of the two clock speeds by someone.

However I'm still unclear about why they run at different speeds in the first place if different g-forces or accelerations are not involved. There is no way of telling one spot in a unifirm gravity field apart from another. They may have different gravity potentials but this is a 'potential, it is not an actual measurement that can be made on the spot. In all senses the two spots are identical from any measurement made on the spot. If you say that this is not so then perhaps you could explain how such a measurement of potential can be made, and what it is about the distance from the source of gravity that can be measured by reference only to the locally perceptible forces of gravity.

That wasn't too clear so - In a uniform gravity field what is it that gives different positions in the field different potentials except simply their distance from the source of that gravity. There must be something beside distance if you are right about this since it cannot be distance itself that causes the clocks to vary in speed, the clock can have no information about its distance from anywhere. It just responds to immediate environmental conditions.

Originally posted by Canute
I'm really sorry to keep coming back to this but your saying 'as far is Bob is concerned' implies that Bob has observed the two clocks. In this case it makes perfect sense to say, as you have, that they will be perceived as running at different speeds. I'm OK with that bit since there is an observer involved, a compsrison of the two clock speeds by someone.

However I'm still unclear about why they run at different speeds in the first place if different g-forces or accelerations are not involved. There is no way of telling one spot in a unifirm gravity field apart from another. They may have different gravity potentials but this is a 'potential, it is not an actual measurement that can be made on the spot. In all senses the two spots are identical from any measurement made on the spot. If you say that this is not so then perhaps you could explain how such a measurement of potential can be made, and what it is about the distance from the source of gravity that can be measured by reference only to the locally perceptible forces of gravity.

That wasn't too clear so - In a uniform gravity field what is it that gives different positions in the field different potentials except simply their distance from the source of that gravity. There must be something beside distance if you are right about this since it cannot be distance itself that causes the clocks to vary in speed, the clock can have no information about its distance from anywhere. It just responds to immediate environmental conditions.

It sounds to me that you are trying to hold on to the idea of some sort of "base time rate", that all clocks adjust themselves relative to, depending on their "local" conditions. This is not the case. no such "base rate exists". Each clock is its own base rate. There is nothing that tells it how fast to run.

Quick analogy: two men are standing 100yds apart. Due to perspective, each man appears smaller to the other. But there is nothing in each man's local enviroment that "tells" him how small he should look to the other man.

Originally posted by Prosoothus
Wouldn't time be dilating for Alice as well, since Alice is in the Earth's gravitational field during Bob's entire journey?

Aren't there two time dilations in the twin paradox:

1) Bob's time dilation due to his accelerating/decelerating.

2) Alice's time dilation resulting from her presence in a gravitational field.

A good question; I’ve thought of this as well. The answer is yes to both questions, but Alice’s time dilation is negligible compared to Bob’s. Recall my comment:

I’d say that Alice is passively accelerating (relative only to Bob), whereas Bob is actively accelerating (feels it), and due to this active acceleration Bob measures all space along his axis of motion contracted, whereas Alice measures only Bob contracted along his axis of motion. Therein lies the asymmetry in their respective trips. Both are traveling relative to each other. But Bob travels less distance; therefore less time elapses on his clock than on Alice’s.

This isn’t entirely accurate as you pointed out; due to gravitational time dilation, Alice measures space contracted too. The contraction of space that she measures is significant only to a limited distance, the distance at which Earth’s gravitational field becomes negligible, a debatable distance but let’s say it’s well within the orbit of Mars. Bob, meanwhile, measures all space along his axis of motion contracted. Bob applies his 1g acceleration to all the distance, whereas Alice applies her 1g to only a few hundred kilometers above her; at higher altitudes her acceleration relative to that altitude (or objects falling from that altitude, if you wish) significantly drops off.

Bob measures the space between himself and Alice contracted on average by, using lethe’s figures, ((6.860 – 5.178) / 6.860) = 24.5% (time dilation always matches spatial contraction percentage-wise), whereas Alice might measure the space between herself and Bob contracted on average by 0.001% (I made that up but the actual figure is likely smaller). These percentages are a discount from the distance each would measure to the other if they were at rest with respect to each other in deep space. Since Alice’s 0.001% discount is negligible compared to Bob’s, he can count on aging less than Alice by the full 24.5%.

Now you might wonder, why does rocket acceleration apply to all space, whereas gravitational acceleration rapidly diminishes with distance? Well, rocket acceleration applies only along your axis of motion, whereas gravitational acceleration applies in all directions. So the differences balance out, so to speak.

alice is in an inertial frame.

You're right! Next time I'll actually read your post :) So let's imagine Alice is on Earth; then Prosoothus's questions are applicable.

lethe,

alice is in an inertial frame.

Are you sure about that? If Alice was falling as a result of the Earth's gravitational field, wouldn't she then be in an inertial frame? Isn't the surface of the Earth, which is stopping her from falling, actually preventing her from being in an inertial frame?

Tom

Originally posted by Prosoothus
lethe,



Are you sure about that? If Alice was falling as a result of the Earth's gravitational field, wouldn't she then be in an inertial frame? Isn't the surface of the Earth, which is stopping her from falling, actually preventing her from being in an inertial frame?

Tom

If you go back and read the initial post, it is said that Alice is "hanging around in a space ship" Not "sitting on the Surface of the Earth". In other words she is floating in space. Lethe did it this way to avoid the whole Dilation by Earth's gravity issue.

If you go back and read the initial post, it is said that Alice is "hanging around in a space ship" Not "sitting on the Surface of the Earth". In other words she is floating in space. Lethe did it this way to avoid the whole Dilation by Earth's gravity issue.

Oh, OK. Disregard my last two posts. :)

Originally posted by Janus58
It sounds to me that you are trying to hold on to the idea of some sort of "base time rate", that all clocks adjust themselves relative to, depending on their "local" conditions. This is not the case. no such "base rate exists". Each clock is its own base rate. There is nothing that tells it how fast to run.

Quick analogy: two men are standing 100yds apart. Due to perspective, each man appears smaller to the other. But there is nothing in each man's local enviroment that "tells" him how small he should look to the other man.
There is no "basetime rate" I agree. But the clocks must have SOME rate. Or do they just intantly adjust to the correct relative value at the moment of Bob's measurement? (A nice QM idea actually).

Originally posted by Canute
There is no "basetime rate" I agree. But the clocks must have SOME rate. Or do they just intantly adjust to the correct relative value at the moment of Bob's measurement? (A nice QM idea actually).

No instant adjustment is needed. Each clock just runs at a "normal" rate, It will always be the "other" clock that is moving fast or slow. If you then bring the clocks together in the same reference system they will agree on the direction and magnitude of the time difference between them, they just won't agree upon [i]how[i] that time difference came about.

Originally posted by Janus58
No instant adjustment is needed. Each clock just runs at a "normal" rate, It will always be the "other" clock that is moving fast or slow. If you then bring the clocks together in the same reference system they will agree on the direction and magnitude of the time difference between them, they just won't agree upon [i]how[i] that time difference came about.
What do you mean by 'normal' rate?

lets say that Alice and Bob were a fair distance apart and each being pulled by their own equally massive black hole. But Bob is closer to his black hole and therefore "accelerates" faster toward it. Are they in the same or different intertial frames?

This definition (http://www.jca.umbc.edu/~george/html/courses/glossary/inertial_frames.html) says “a frame in free-fall under the influence of gravity is an inertial frame.” But these are two frames in free-fall under different influences of gravity. That Bob & Alice are in different inertial frames is provable by showing that they observe the other’s clock running at a different rate than their own while free falling.

Let’s examine their relative clock rates. The other’s light would first be redshifted as it escapes the other black hole. The light would then blueshift as it falls into their black hole, if they weren’t free falling, but they are, so no blueshift is noticed. (When you free fall the blueshift is not observable; you are accelerating at a rate that exactly “soaks up” the blueshift.) The net redshift means a slower clock is observed. Nothing about Bob’s greater rate of acceleration fundamentally changes this. So we’ve proven they are in different inertial frames.

Originally posted by Canute
What do you mean by 'normal' rate?

This means your own clock’s rate; that is, a rate of 100%. Your own clock runs at a constant rate that you can designate as 100%. You could notice another’s clock running at a different rate, say 50% (an hour elapses on the other clock for every 2 hours elapsed on your clock) or 200% (2 hours elapses on the other clock for every hour elapsed on your clock).

Originally posted by zanket
This means your own clock’s rate; that is, a rate of 100%. Your own clock runs at a constant rate that you can designate as 100%. You could notice another’s clock running at a different rate, say 50% (an hour elapses on the other clock for every 2 hours elapsed on your clock) or 200% (2 hours elapses on the other clock for every hour elapsed on your clock).
Yes I understand this. My question related to my question to Janus as to why, if two clocks are stationary (relatively speaking) within a uniform gravity field they would run at different rates.

Originally posted by Canute
Yes I understand this. My question related to my question to Janus as to why, if two clocks are stationary (relatively speaking) within a uniform gravity field they would run at different rates.

It seems like we are just going around in circles.

There is nothing about local conditions that tells a clock how fast or slow to run. There are iglobal conditions that tell it how fast other clocks run wrt it. The global condition in a uniform field is the other clock's position in that field relative to to the first clock.

Originally posted by Janus58
It seems like we are just going around in circles.
Agree. Perhaps we are doomed never to find the exit.

There is nothing about local conditions that tells a clock how fast or slow to run. There are iglobal conditions that tell it how fast other clocks run wrt it. The global condition in a uniform field is the other clock's position in that field relative to to the first clock. [/B]

There is something very odd about this discussion. I will try to be clear about what I find odd.

Thing one is the question of how fast a clock runs if there are no other clocks. How is it detirmined (in the sense of caused). I would presume that it is determined by the local force of gravity. Why would its speed then differ from any other realtively stationary clock subject to the same force of gravity?

You say differences in clock speeds will be caused by position in field. Now if local gravity and/or accelleration determine the speed of the clock then I am happy, for that is what I has always thought. But position in field does not seem to be capable of having the effect that you suggest. How can gravity potential, which cannot be measured or sensed locally, have any effect on clock speed?

I have tried to put the question as clearly as I can. If you think you have already answered it then say so and I'll break the circle by retiring gracefully, albeit unenlightened.

Let’s examine the relative rates of 2 clocks in the top and base of an accelerating rocket. An observer stands next to each clock. Their own clock runs at 100%, naturally. Imagine that the top clock continuously emits photons forming a readable image of the clock’s display, and these photons move towards the base of the rocket at c, the speed of light. Meanwhile the observer at the base accelerates towards the photons. If the photons were instead basketballs, then the base observer could expect to hit them at a greater speed than they were released from the top. But, as Einstein postulated (and experiments have confirmed), everyone directly measures the speed of light as c. This is made possible because the clocks run at precisely the relative rates required to offset any otherwise expected change in the light’s directly measured speed. For example, if the base observer would have expected to accelerate to hit the photons at 2c, then instead the top clock runs at 200% of the rate of the base clock. This allows the base observer to directly measure the speed of the incoming photons at c despite the acceleration towards them. (The photons hit the base at 2c/2 = c.) When the base observer looks at the images of the top clock formed by the incoming photons, sure enough the top clock runs twice as fast as the base clock, and the top observer ages twice as fast as the base observer.

Einstein’s principle of equivalence, roughly stated, says that if all you know is that you’re in a rocket accelerating at 1g, you can’t tell if that rocket is on the launch pad with its engines off, or accelerating in space with its engines on; gravity is equivalent to acceleration. So the top and base could just as well be in a building; those observers would also notice relatively different clock rates.

The uniformity of gravity isn’t a determinant as to the clocks running at different rates. Suppose hypothetically that the gravity is exactly 1g at both the top and base of a building. Photons emitted from the top would still accelerate to hit the base at > c if not for the difference in clock rates. Likewise photons emitted from the base would still decelerate to hit the top at < c if not for the difference in clock rates.

If this doesn’t answer your questions, rephrase. Here’s a book that painlessly explains this: Relativity Visualized (http://www.amazon.com/exec/obidos/tg/detail/-/093521805X/qid=1045997115/sr=1-1/ref=sr_1_1/002-0252450-6232802?v=glance&s=books).

Thanks but I'm not sure how this realates to my question. Which was not about the effects of time dilation due to either local gravity or acceleration. I am told that position in the field, even a uniform field, also plays a role. It is this that I find illogical.

Canute,

It is this that I find illogical.

Welcome to the club. :)

If you want to see something really illogical, then check out the Lorentz transformations for calculating the one-way speed of light relative to a moving observer.

Tom

Originally posted by Canute
I am told that position in the field, even a uniform field, also plays a role.

Hmm. I covered how the position in the field, even a uniform field, plays its role.

Originally posted by zanket
Hmm. I covered how the position in the field, even a uniform field, plays its role.
In that case I apologise. I do not remember you doing this. I thought you just covered the equivalence of gravity and acceleration. I have yet to understand why, in the absence of any gravity, acceleration or velocity differential, two clocks should vary in speed. I am missing something somewhere.

Originally posted by Prosoothus
Canute, Welcome to the club. :)

If you want to see something really illogical, then check out the Lorentz transformations for calculating the one-way speed of light relative to a moving observer.Tom
Hmm. Thanks but I suspect they would be beyond my mathematics. It is very odd. I'm no phycisist but thought I had a decent layman's understanding of relativity. Now this 'position in the field' thing turns up. I haven't come across it before and am having trouble getting anyone to explain it. Is it a spurious addition to Einstein by someone here?

Originally posted by Canute
I thought you just covered the equivalence of gravity and acceleration. I have yet to understand why, in the absence of any gravity, acceleration or velocity differential, two clocks should vary in speed.

Do you mean gravity differential and acceleration differential? If yes, I covered that in the paragraph starting with “The uniformity of gravity...” Can you be more specific about what you mean by “position in the field?”

Originally posted by zanket
Do you mean gravity differential and acceleration differential? If yes, I covered that in the paragraph starting with “The uniformity of gravity...” Can you be more specific about what you mean by “position in the field?”
My problems started with this post from Janus58

"Assume a uniform gravitational field. (one in which acceleration due to gravity does not fall off with distance. )

You have two clocks, clock A above clock B. The field is uniform so each clock experiences exactly the same g-force. But clock A is higher than clock B and has a greater gravitational potential, So it will run faster than clock B(as measured from B). If you you had a third clock C, which was even higher than A, it would still feel the same g-force but would run even faster than B than A does.

Also, If you increase the stength of the G- field you will increase the time differences between the clocks.

So it is both strength of field and relative position in the field that determines the relative clock rates."

I am trying to clear up how relative position in a uniform gravity field can effect clock speed in this way, and how increasing a uniform field uniformly can have any effect on their relative speeds.

Originally posted by Canute
Hmm. Thanks but I suspect they would be beyond my mathematics. It is very odd. I'm no phycisist but thought I had a decent layman's understanding of relativity. Now this 'position in the field' thing turns up. I haven't come across it before and am having trouble getting anyone to explain it. Is it a spurious addition to Einstein by someone here?

Check out this site, and go down to section (c) "Displacement of Spectral Line Towards the Red" it is taken from a book By Einstein himself.

http://bartleby.com/173/a3.html


He decsirbes what happens to clocks on a rotating system and equates it to a gravtiational field.

Note that the time rate difference between the center of the disk and the rim is determined by the speed at the rim. Also, an object standing at the rim would feel a g_force porportional to v^2/r .

One could use a number of different combinations of v and r to get the same g-force at the rim. For instance, you could use a radius of 1m and a rim speed of 3.13 m/s to get 1g (9.8m/s^2) or 2m and 4.427m/s . Since the time factor depends on rim speed, you could have two clocks on two disks, each experiencing the same value of g, yet running at different rates.

Also, you will note the he says that it is the Gravitational potential differential that accounts for the time rate difference. (IOW, the amount of work it would take to move a unit mass from the rim to the center.)

If you place two clocks in different points of a uniform gravity field, along a line that lies in the direction of the field, They will be be at different gravitational potentials. (you would have to perform work to move the lower clock to the position of the higher clock.)

While nothing locally can tell each clock how fast to run, each can can know the nature of the field it is in, and the vertical direction and distance of the other clock, thus it has all the info it needs to tell how much faster or slower the other clock must run compared to itself.

It isn't a matter of each clock deciding how fast to run, it is how each clock views other clocks as running depending on their relative reference frames.

It is the same as two clocks moving with a constant velocity relative to each other. Each clock sees the other as moving slower than itself, and each clock has an equally valid right to its preception of events.

Originally posted by Canute
I am trying to clear up how relative position in a uniform gravity field can effect clock speed in this way, and how increasing a uniform field uniformly can have any effect on their relative speeds.

OK. What Janus58 said is true. I suggest not focusing on a uniform gravity field; whether uniform or not the clocks still run at different rates.

Because gravity is equivalent to acceleration, you can ignore gravity altogether and just focus on an accelerating rocket, where it may be simpler to see what’s causing the clock rate difference. That’s what I did in the first paragraph of my explanation above.

It boils down to this: In an accelerating rocket, you’d hit incoming photons at a relative velocity faster than c if your clock didn’t run at a different rate than a clock at the source of those photons. The clocks run at different rates as a natural consequence of everyone directly measuring light’s speed as c.

If you are at the base of the rocket, the incoming photons could be sourced from the top of the rocket. So clocks at the top and base of an accelerating rocket run at different rates. Likewise, because gravity is equivalent to acceleration, clocks at the top and base of a building run at different rates.

If these explanations still aren’t working for you, you might try Google. Here’s a search on gravitational time dilation (http://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=%22gravitational+time+dilation%22). Here’s a simple Non-Mathematical Proof of Gravitational Time Dilation (http://www.upscale.utoronto.ca/GeneralInterest/Harrison/GenRel/TimeDilation.html). Also the book I recommended above is excellent.

Here’s some math. Suppose you have a 22.6-meter tower. You drop a photon from the top. Its final velocity at the base is given by the standard equation v_final = sqrt(v_initial² + 2ad). If not for everyone directly measuring light’s velocity as c, the photon’s final velocity would be sqrt((299,792,458 m/s)² + (2 * (9.8 m/s²) * (22.6 m))) = 299,792,458.000001 m/s. That’s a ((299,792,458.000001 / 299,792,458) – 100%) = 2.44E-15% difference. Instead, the rates of clocks at the base and top of this tower differ by 2.44E-15%. And that difference was confirmed by experiment using a 22.6-meter tower in 1960.

What's all this stuff about rockets and acceleration!? They are NOT accelerating. They are stationary relative to everything. The whole point of my question is that this has nothing to do with acceleration, as was the point of Jason's original point.

Who’s they? If they are observers at the top and base of a building, it still has to do with acceleration. You are accelerating as you read this. That’s why you feel your chair pressing against you.

Yes, that's what I believe. I take it you agree that relative position in the gravity field where the field is uniform has nothing to do with time dilation.

I agree that whether the field is uniform or not the clocks still run at different rates, as I said before. (Keep in mind that a uniform gravity field exists only in an accelerating rocket or such. On Earth, say, the rate of gravity slightly varies between your head and feet.)

You didn’t answer my question “Who’s they?” But I gather that “they” are observers at the top and base of an accelerating rocket, which is a uniform field. And you say, “They are NOT accelerating. They are stationary relative to everything.” If so, yes, they are stationary relative to each other, but they accelerate relative to the photons traversing between them. That’s why time dilation applies in an accelerating rocket. If not for the difference in the rates of their clocks, they’d directly measure the speed of photons arriving from the other at < or > c.

Originally posted by Canute
Yes, that's what I believe. I take it you agree that relative position in the gravity field where the field is uniform has nothing to do with time dilation.

If you go back and look at zanket's second-to-last post, the last paragragh, you will see that he does not agree with that conclusion.

The math he used to compare the difference between the top and bottom of the tower assumed a constant value of g from top to bottom. IOW, the difference in g between the top and bottom of the tower is so small( about 7e-05 m/s^2) that the tower can be considered as being in a uniform field for all practical purposes.

And using calculations that assume a uniform field you get an answer that agrees with experiment.

Sorry guys but I must be being particularly think here.

Zanket - By 'they' I meant the clocks.

Janus - thanks I now understand (well...) the tower experiment a bit better.

However, wait for it, I remain puzzled. I can see that the maths gives the right answers, but cannot see how this can be. I can see that in the example the transmission of a photon from one end to the other of the tower can be used to fix the relative rates of the clocks at the other end of the tower. What I cannot see is how the clocks know that they should have these relative speeds before the photon is transmitted and measured if the force of gravity acting on each of them is identical and neither is accelerating. In other words what is that gives them these realtive speeds, or don't they have them until they are compared?

If this is a stupid question I'm sorry. However I can't see the answer. What am I missing?

Remember that the clocks are accelerating, relative to photons traversing between them. Even in a uniform field, the lower clock would pass the photons at a greater velocity than would the upper clock, to make the speed of light variable, if not for the time dilation that keeps the speed of light constant. This is described in the Non-Mathematical Proof of Gravitational Time Dilation (http://www.upscale.utoronto.ca/GeneralInterest/Harrison/GenRel/TimeDilation.html).

Since the constancy of the speed of light gives the clocks their relative speeds (which they have whether or not compared), I think the essence of what you’re asking is, how is light’s speed constant despite movement toward or away from the light’s source? Relativity Visualized (http://www.amazon.com/exec/obidos/tg/detail/-/093521805X/ref=pd_bxgy_img_2/104-0289417-6043949?v=glance&s=books) puts it this way:

“Why can’t the speed of light be increased or decreased a little by moving toward or away from the light’s source? In the year 1900 that was the most important question in the world. The speed of light’s immunity from the rules governing the speed of everything else was shocking, embarrassing, and to some philosophers unbelievable. ... Why the speed of light is an absolute is an independent problem, indeed a paradox in and of itself. ... Einstein said to himself, ‘The problem is not with light; the problem is with speed.’ Speed is a measure of space divided by a measure of time. So if the idea of speed is in trouble, it is because the underlying ideas of space and time need alteration. ... The price of making the speed of light absolute, rather than relative, is to make time relative, rather than absolute.”

Many others have asked your question. Einstein answered it, and predicted that the universe works differently than initially expected by our intuition.

I really hate to do this but I can't pass. You seem to be basing your time dilation on the basis of inertial or non-inertial systems.

If so then if I should accelerate at tremendous speeds for a short duration - i.e. reach maximum velocity in "A" clocks click of 1 second, coast for 10 years, decelerate in the clock of 1 sec and run the same scenario in reverse to end up with bob and alice embracing after his long voyage. Bob would only be 4 seconds younger.

Originally posted by MacM
I really hate to do this but I can't pass. You seem to be basing your time dilation on the basis of inertial or non-inertial systems.

If so then if I should accelerate at tremendous speeds for a short duration - i.e. reach maximum velocity in "A" clocks click of 1 second, coast for 10 years, decelerate in the clock of 1 sec and run the same scenario in reverse to end up with bob and alice embracing after his long voyage. Bob would only be 4 seconds younger.

1. Click of 1 second - When u accelerate to get maximum speed ur click of 1 second is actually a click of dilated 1 second.. it could be hours, months and years (as u are increasing the speed, accelereation) for the clock in rest.

2. coast for 10 years - Time dilated "10 years" for the moving clock - that dilation depends on the velosity during the coasting.

So it is not just 4 seconds (for both bob and alice when finally at rest). Difference will be in years.

everneo,

The clicks were set as clock "A" clicks but I was watching to see if the relative velocity would come in again since alice is in space assumed inertial and bob becomes inertial after acceleration so we are back to "Who's in motion and who's at rest and that unfortunately leads back to the long arguement of the 3 clocks.

And that was the time dilation between clocks "A" and "B" since each considers the other going slower and how that gets resolved in actual clocks when returned for comparison.

There still seems to be no basis for physical dilation difference between the clocks where only linear velocity is involved.

MacM,

Thats why i split the answer into 2 parts. 1. accelerating period. 2. Coasting period.

I know 2nd part is going to be an enigma for U and Tom for quite sometime.!

1st part would alone be sufficient to show that the time difference is not just 4 seconds. Since u did not make any reference to that part i assume u agreed..!

everneo,

1st part would alone be sufficient to show that the time difference is not just 4 seconds. Since u did not make any reference to that part i assume u agreed..!

Reply: On this we do not agree. Everything I have ever read and that includes prior posting on this board refers to the slowing of clock "B" not the speeding up of clock "A" . The given clock clicks were stipulated as clock "A" clicks. Which seems to me to keep them to 4 seconds.

As far as the balance of the issue. I posted here for the simple reason that "lethe" solved my problem in my clock example.

1 - I had stipulated a trip where acceleration was not in the mix and had tried to get an evaluation of nothing but linear velocity to affect the clocks so that a dilation calculation of that affect alone could be made.

Numerous objections were made to not including acceleration, turn-around, synchronization of the clock timing, gravity, etc. Frankly I view them as attempts to avoid trying to complete the test and provide an ultimate answer.

2 - In lethe's proposal I realized that if you restrict that amount of time and go primarily linear velocity and make the round trip, since he cleverly put alice in space in absence of gravity and declared her as "inertial", that bob too becomes inertial for most of the trip.

That provides my desired effect of returing the clocks for comparision to show that since Relativity dictates that during such inertial periods each clock appears to run slower than the other, no actual time dilation occurs.

Please show me where that is not the case.

MacM,

First, i've not named any clock as "A" or "B" or "C". ( lethe too did not name the clocks. he took moving clock's proper time as T. ) The "clicks" u r refering to is pertaining to moving clock whose time dilates continuously as it accelerates to get maximum speed. in short, as it getting closer to ur "maximum" velocity the time dialtion also increases. [starting from being equal rate to the clock at rest, the moving clock's time gets dilated as the velocity increases comparing to the clock at rest by minutes then hours and then days and so on within its (moving clock's) click of dilated 1 second.]

Second, lethe did not involve any linear velocity. if u read the third post of this thread there he set up the problem, cleverly and clearly ofcourse , as bob accelerates up to half-way point to alpha centauri and starts decelerating during the other half. his frame of reference is not inertial at any time during his journey. (bob reaches alpha centauri but lethe did not mention whether bob stops and turns back or turns around on the fly without stopping)

I don't know how u got the idea that lethe solved ur three clocks problem here when no 'linear velocity' involved. really i would not like to get into that in this thread..! i think ur 3 clocks problem is closed for its incompleteness and non-viablity to compare instantaneously 3 proper times.

everneo,

I did not mean he solved the problem. He solved the counter arguement that you couldn't perform the test because they would not agree on the colco test timeing using Relativty to stop the clocks.

His round trip with both participants being in space, absent graity allows for a brief period of acceleration and then linear velocity where both participants are inertial.

By havign the round trip the clockls can be returned for a dierect comparision after a long period of relative velocity and only brief periods of acceleration.

With the "No preferred Frame" rule "A" Alice and "B" Bob see each other as in motion and each other as losing time relative to themselves.

At the end of the trip there is no basis for differential of the clocks except for that period of acceleration. That goes to my original post of 3 Clocks wheich was attempting to show the absence or "Real " time dilation due to relative velocity alone.

Originally posted by MacM
At the end of the trip there is no basis for differential of the clocks except for that period of acceleration.

That’s all that is required. If in 1 of his seconds (proper seconds) Bob accelerates close enough to c relative to Alice, he can contract the distance of his whole trip to a few meters, traverse that at a constant velocity within some proper nanoseconds, and likewise return, in principle of course. If his trip were to the nearest star and back then Alice would have aged some 8 years in Bob’s few proper seconds. If his trip were to the nearest galaxy, Earth would have aged millions of years.

zanket,

That’s all that is required. If in 1 of his seconds (proper seconds) Bob accelerates close enough to c relative to Alice, he can contract the distance of his whole trip to a few meters, traverse that at a constant velocity within some proper nanoseconds, and likewise return, in principle of course. If his trip were to the nearest star and back then Alice would have aged some 8 years in Bob’s few proper seconds. If his trip were to the nearest galaxy, Earth would have aged millions of years.

From Bob's frame of reference his distance is contracted, but from Alice's frame of reference her distance is contracted. That's why relativity states that the speed of light will be c for Alice AND for Bob. The assumption that length contracts for one but not for the other goes against the principle of relativity.

Tom

So is this why my watch never matches other clocks on campus? I accelerate around campus all day, with very different accelerations than the library clock, and so the library clock runs slow. Could I go in a different direction and make it run fast?

Originally posted by Prosoothus
zanket,



From Bob's frame of reference his distance is contracted, but from Alice's frame of reference her distance is contracted. That's why relativity states that the speed of light will be c for Alice AND for Bob. The assumption that length contracts for one but not for the other goes against the principle of relativity.

Tom

From Bob's frame of reference he sees the frame of reference including the Earth, Alpha Centauri, and all the space between as contracted in length, as that is what he sees as moving relative to him.

Alice, only sees Bob and his ship contracted, because that is what is moving relative to Her.

So for instance, Alice see's Bob's ship take 8.6 years, (at .5c) to travel the 4.3 lightyears to Alpha C.

Bob sees himself as taking 7.45 yrs to travel 3.72 ly at the same .5c

No violation of the Principle of Relativity.

Okay, one more attempt to explain this.

Let's look at what both Alice and Bob see.

What Alice sees:

By her clock:

Bob accelerates up to some fraction of c (we'll assume that does this at a high accel and gets up to speed very quickly, though it doesn't make any difference if he were to accel at a lower rate)
This takes some time (t1) on Alice's clock. She then sees him coast for a distance close to 4.3 ly, taking t2 to so. She then sees him decel for time t3, coming back to rest with respect to her.

Total trip time by her clock : t1+t2+t3

By Bob's clock (As seen by Alice):

Since Bob is moving relative to her, she will see Bob's clock run slower than her's during the whole trip. Since she feels no forces during the trip, she only has to take Bob's relative motion into account.

During the acceleration period, She will see his clock go from a rate matching hers to one slower, the final rate determined by what value of c Bob attains. Thus the total elasped time that she sees passing on his clock will be somthing less than t1 (-t1)

He then coasts. She will see his clock run slow by the factor given by the Lorentz transformation. total elapsed time for this phase by his clock will be less than t2(-t2)

Finally during the decel, his time rate will speed back up to match her own as he comes to rest. Again, the time she would see as having passed on his clock during this period will be less than waht passed on her own. (-t3)

Total time elapsed on Bob's clock (as seen by Alice) :
-t1 + -t2 +-t3

Or a total less than what passes on her own.

What Bob sees:

By his clock:

As he acclerates or (conversely sees Alice and Alpha C accelerate with respect to him). He will note that a certain time will pass to accel up to speed. This time will be less than that as noted by Alice because of the effect of length contraction. (he will feel as if he was accelerating as a faster rate Than Alice sees him Accelerating, thus he will take less time to reach final speed.) time for accel, -t1 (with T1 again being the time Alice see pass during this phase.

He coasts. Again becuase of length contraction this phase is shorter in distance for him that the 4.3 ly Alice measures. Thus, at the fracton of c he is traveling this protion of the trip again takes less time by his clock than Alice measured on her's (-t2)

Decel is just the reverse of accel and he measures -t3 passing on his clock.

Total time elapsed on the trip By Bob's clock as seen by Bob:

-t1 + -t2 + -t3

Which agrees with what Alice sees for his clock.

So far, so good.

By Alice's clock (As seen by Bob).

Now this is where people seem to get lost.

As Bob accelerates away from Alice, He can assume (by the Principle of Relativity) that it is actually Alice and Alpha C that are accelerating. But then he has to determine what causes this accel. (Alice didn't have this problem, as she saw Bob's engines firing and Bob being pushed to the rear of his ship by his acceleration)

Alice is obviously increasing speed relative to him, but she does not show any of the signs of someone undergoing high accel. The only expalinantion for this is that Alice is falling in a gravity field, and that the only reason that he, Bob, isn't is that his engines are holding him still with respect to this field. In fact, he can feel this field pulling him down. Upon inspection, he notes that the field is uniform is strength and extends an infinite distance above and below him. Alice is below him and is also increasing speed relative to him, so he will see her clock moving more slowly than his for two reasons, increasing SR dialation and gravitational dilation. But since gravitational dilation depends on diferential of gravitatioanl potential, and during this part o fthe trip Alice isn't very far below him, the GR effect will be relatively small. The time he sees passing on Alice's clock will be less than his, which was less than what Alice see's passing on her clock (call it --t1)

Coasting. The field is gone so now Bob only sees SR dilation during this part on Alice's clock. He will see less time pass on hers than his(which was -t2) so we'll call --t2.

Decelerating. The field is back, but now its reversed direction. Alice is now above him and losing speed. the SR Time dilation will decrease as the velocties match, and the GR dilation has Alice's clock as moving faster. But Alice is much Higher than Bob, now than she was lower than him during the Acceleration phase (By a few lightyears), Thus the gravitational potential difference is much greater. Because of this Bob will se her clock speed up considerably during this phase, and the total time that he sees elaspe on it will be ++++T3

total elasped time on Alice's Clock(as seen by Bob)

--t1 + --t2 + ++++t3
which added together will equal t1+t2+t3.

Bob will see a total elapsed time On Alice's Clock that agrees with What Alice saw as the Total elapsed time on her Clock.

They both agree upon the final result, even though they won't agree upon just how those results came to be.

No matter what combinations of Speed and accelerations you want to use, it will always work out that they will agree as to on whose clock less time has elapsed and by how much.

P.S. I tried using less-than and greater-than signs, but the forum mistook them as HTML brackets.

Janus58,

Do u mean to say,

i) only when Bob accelerates away from Alice, she is falling below Bob in 'gravitional field' and that is the only period her net time dilation is more than Bob's during the entire journey (of Bob) and ;

ii) in other periods ([except coasting], decelerating, accelerating after turn around, coasting and decelerating) she is above Bob in that 'gravitational field' ( hence faster/equal time rate) so net time dilation in these periods is less than or equal to Bob's..?

(just to know whether i followed ur gravitational potential proposition properly )

everneo,

I can only hope he can answer "Yes" to your question.

) only when Bob accelerates away from Alice, she is falling below Bob in 'gravitional field' and that is the only period her net time dilation is more than Bob's during the entire journey (of Bob) and ;


That has been my arguement from day one on this site and my attempt to prove it with the three clock scenario. I did not refer to arbitrary gravity fields but only to the aspect of each seeing the other moving equally slower during linear relative velocity; hence no net time dilation.

It becomes observational and not physical.

MacM,

if u r telling the same what janus58 is telling then u should agree with him when he showed Bob's time is dilated more than Alice's.

Alice's time dilation as observed by Bob, (for Alice there is no change in her time rate) as per janus58's argument, is attributed to free falling of Alice's frame of reference in the 'gravity filed' ( + SR time dilation) as percieved by Bob. Alice & Bob observe Bob's time dilation is due to SR alone. Hope janus58 would make it clear (i think he has already done that going along with ur argument that 'relative velocity causes Alice & Bob to observe the time dilation for both of them equally as required by SR'. in addition he showed, in his own way, why Alice's time dilation is less by GR). better u await his reply and get clarrified and positively drop twin paradox ( & 3 clocks too) for ever..!

everneo,

I am at a bit of a loss.

Hope janus58 would make it clear (i think he has already done that going along with ur argument that 'relative velocity causes Alice & Bob to observe the time dilation for both of them equally as required by SR'. better u await his reply and get clarrified and positively drop twin paradox ( & 3 clocks too) for ever..!

You seem angry that I ever tried to get people to say exactly what is now being said. Why should the agruement be dropped if it is true.

I have never once denied time dilation during acceleration and/or gravity. I openly stated that in the very beginning. My whole effort was infact to show the lack of "Net" time dilation during linear velocity. That makes it "Perception", not "Reality".

But the response's I got were to the contrary. That it was I that didn't understand physics or Relativity. (Plus much more unrelated responses).

I'll not replay that long series here but I am only glad to start to see some correlation here and unless this position is justly retracted I can only hope others begin to see what I am saying about it all.

It all comes down to the presentation and not the issue. I have to take responsibility for the presentation.

Originally posted by MacM
everneo,


I have never once denied time dilation during acceleration and/or gravity. I openly stated that in the very beginning. My whole effort was infact to show the lack of "Net" time dilation during linear velocity. That makes it "Perception", not "Reality".



No, the SR Time Dilation is "Reality". From Alice's frame, Bob only undergoes "SR" time diation, and it is the only way to explain why his clock ends up showing less elapsed time than hers.

Fr0m Bob's Frame the SR time Dilation he sees In Alice's clock has to be "Reality" because that is the only way the Total Elapsed time will come out right. If the SR Dilation was just illusion, Then the GR effect would have her clock gain too much time, and he would conclude that more time passed on her clock than she does. The total elapsed time on Alice's clock from his frame is due to the combined effect of SR and GR time dilation.

everneo,

I am not sure if we are saying the same thing but in a different way.

When I say it is "Perception" not "Reality" I am referring to the above statement that only the acceleration portion of the trip produces net physical clock changes.

If you are saying that the "Net Zero" result of linear relative velocity time dilation must also be physical, I suppose it becomes symantics.

Do we agree that the result of linear relative velocity is "Net Zero" physical dilation differential of the clocks?

IF so then I believe there is no actual disagreement other than interpretation of the process netting the agreed to end result.

MacM,

I don't know whether u read janus58's post before ur last post. porbably u were not aware of that as u must be typing. He meant net time dilation by combined SR, GR effects.

Why should i get angry man? or rather why shouldn't i?! i just stated that u had to await janus's reply as me too asked for a clarification. being so, i cannot answer on his behalf isn't it? but before that u started drawing conclusions..! just wished u to get it solved..!

You can also add a third person (Charlie) and his clock, floating somewhere in space near Alpha C, and at rest with respect to Alice.

He will agree with Alice that Bob's shorter elapsed time is due to his relative speed alone. Also Both Charlie and Alice will agree that their clocks remained sychronized during the whole trip.

Bob, on the other hand will say that while Alice's clock followed a pattern of Run slow, run slow, Run very fast, Charlie's clock followed the pattern of Run very fast, run slow, run slow, and that their( Charlie's & Alices's) clocks did not remained sychronized during the whole trip. Though they would record the same total elapsed time at the end on the trip.

Originally posted by zanket
Remember that the clocks are accelerating, relative to photons traversing between them. Even in a uniform field, the lower clock would pass the photons at a greater velocity than would the upper clock, to make the speed of light variable, if not for the time dilation that keeps the speed of light constant.
This is a tautology. It saying that we must assume that the two ends of the physically rigid tower are must be accelerating because otherwise the speed of light wouldn't be constant. I can see that this has to be true. However it is entirely self-referential. You seem to be saying that position in the gravity field MUST be a factor in determing the relative speeds of the clock because otherwise the maths wouldn't work. This is what I suspected.

Many others have asked your question. Einstein answered it, and predicted that the universe works differently than initially expected by our intuition. [/B]
Hmmm. Many others have asked the question you thought I was asking, which is not quite the same thing.

Janus,

Now you really lost me. What happened to the arguement that bob comes back younger? My arguement all along was that Bob doesn't come back younger and I was dubed a crackpot.

I recognize that my arguement wasn't couched in the same manner but no age difference was the result I was trying to show. Everyone disagreed with me?

Originally posted by MacM
Janus,

Now you really lost me. What happened to the arguement that bob comes back younger? My arguement all along was that Bob doesn't come back younger and I was dubed a crackpot.

I recognize that my arguement wasn't couched in the same manner but no age difference was the result I was trying to show. Everyone disagreed with me?

No, Bob does age less on both the outbound and return trips (I just didn't include the return trip in my scenerio). I never said he didn't. Just that while Alice and Charlie's clocks are always sychronized as far as they are concerned, According to Bob, Alice's and Charlie's clocks aren't sychronized for the whole trip.

Janus58,

SO the following was Charlie and Alice?

Though they would record the same total elapsed time at the end on the trip.

Originally posted by Canute
This is a tautology. It saying that we must assume that the two ends of the physically rigid tower are must be accelerating because otherwise the speed of light wouldn't be constant. I can see that this has to be true. However it is entirely self-referential. You seem to be saying that position in the gravity field MUST be a factor in determing the relative speeds of the clock because otherwise the maths wouldn't work. This is what I suspected.

I’d put it differently: That the whole tower is accelerating there can be no doubt; otherwise we’d be floating now. It should accelerate relative to photons in free fall (such as those emitted from either end toward the other) just as it does relative to material objects in free fall. Acceleration is a change in velocity, but no change in light’s velocity is directly measured, therefore clocks at either end must run at different rates. That seems a deduction to me, not a tautology or self-reference.

Originally posted by zanket
I’d put it differently: That the whole tower is accelerating there can be no doubt; otherwise we’d be floating now. It should accelerate relative to photons in free fall (such as those emitted from either end toward the other) just as it does relative to material objects in free fall. Acceleration is a change in velocity, but no change in light’s velocity is directly measured, therefore clocks at either end must run at different rates. That seems a deduction to me, not a tautology or self-reference.
I think I see this, and I am not trying to argue aginst the fact that the clocks will appear to run differently in real life. But... the two ends of the tower are not accelerating relative to each other, since we have assumed that they are each subject to a force of 1g.

I also don't see why the tower should be accelerating relative to the photon. It is certainly going at c in the opposite direction, but not accelerating surely.

If the two ends of the tower are ageing at different rates, as this seems to suggest, then why is this? The acceleration acting on each is the same, so before and after the moment at which we compare their speeds what is going on? In other words if they are not going at the same speed then why not? (I'm talking physical causation here rather than mathematics).

Originally posted by MacM
Janus58,

SO the following was Charlie and Alice?

Correct. I probably should have said "Charlie and Alice" rather than "they" to prevent confusion.

Okay, I think I see this. Assume c was variable. Then it leaves the front of the rocket at c-v where v is the speed of the rocket. But by the time it arrives at the rear of the rocket, the rocket is going faster due to acceleration, so someone at the rear will measure it going at c+(v+) where v+ is the new velocity of the rocket. But since c is not variable, the clock speeds must compensate.

regarding time dilation, even though the time dilation during the constant velocity part of the trip can be added into the dilation at the other parts of the trip that is due to acceleration, you have stated that it is then subtracted out by the clocks speeding up. This is certainly one way to look at it. Another way would be to say that there is only an apparent time dilation during the constant velocity part of the trip, and that the apparent dilation disappears once the two rockets are stationary wrt each other again. Either way, the only NET time dilation occurs during acceleration. Someday, we may have an experiment that can decide which perception reflects reality, or if both are wrong (or right!). But for now, both approaches seem equally valid (though I admit that I prefer thinking of constant velocity time dilation as being only apparent, nothing seems to be gained by thinking otherwise. Occam's razor.)
Aaron

Originally posted by synergy

regarding time dilation, even though the time dilation during the constant velocity part of the trip can be added into the dilation at the other parts of the trip that is due to acceleration, you have stated that it is then subtracted out by the clocks speeding up. This is certainly one way to look at it. Another way would be to say that there is only an apparent time dilation during the constant velocity part of the trip, and that the apparent dilation disappears once the two rockets are stationary wrt each other again. Either way, the only NET time dilation occurs during acceleration. Someday, we may have an experiment that can decide which perception reflects reality, or if both are wrong (or right!). But for now, both approaches seem equally valid (though I admit that I prefer thinking of constant velocity time dilation as being only apparent, nothing seems to be gained by thinking otherwise. Occam's razor.)
Aaron

Two Problems with considering the SR dilation as only apparent.

It doesn't explain Why Alice and Charlie agree that Bob's clock showed less elasped time, As all they see is SR dilation.

Even with Bob, it doesn't work. Look at it this way. Consider the GR dilation which makes Alice's clock speed up as walking up a set of Stairs, and The SR dilation as walking down.

The pattern for Alice's clock then would go some thing like this.

Go down 3 steps, go down 7 steps, climb up 14 steps. Net difference at end, up 4 steps.

If the going down steps were only "apparent" then the steps taken down would only be apparent, and the "real" net difference at the end would be up 14 steps not 4.

Originally posted by Canute
But... the two ends of the tower are not accelerating relative to each other, since we have assumed that they are each subject to a force of 1g.

Yes, we can assume that to keep the example simpler.

I also don't see why the tower should be accelerating relative to the photon. It is certainly going at c in the opposite direction, but not accelerating surely.

If you drop a ball off the top, it accelerates relative to both ends of the tower, right? And both ends accelerate relative to the ball. Well a photon has mass when it’s moving, which it always is. So it should fall like the ball does, and both ends should accelerate relative to the photon. Except the rate of clocks at either end differ, so no change in the photon’s velocity is directly measured.

Synergy has it correct here:

Originally posted by synergy
Okay, I think I see this. Assume c was variable. Then it leaves the front of the rocket at c-v where v is the speed of the rocket. But by the time it arrives at the rear of the rocket, the rocket is going faster due to acceleration, so someone at the rear will measure it going at c+(v+) where v+ is the new velocity of the rocket. But since c is not variable, the clock speeds must compensate.

If you’re in an accelerating rocket, it’s easier to see how you accelerate relative to photons outside the rocket, like those emitted from a star in front of you. After all, you’re accelerating relative to everything that is not being pushed like you are. You could say that you are accelerating relative to space itself. And that space includes the space within your rocket, including the photons emitted within your rocket. What synergy said applies to all photons.

Einstein realized you can’t tell the difference between a rocket accelerating at 1g and a rocket sitting on a launch pad with its engines off; in your seat in the rocket the same force is felt. So as you read this, space is accelerating past you at 1g, toward the center of the Earth. Or, if you prefer, you are accelerating through space at 1g. Anything floating within the space is likewise accelerating relative to you, like photons. Only we can’t directly measure their change in velocity due to the change in the rate of clocks along the radius of a gravitational field, or along the axis of motion of an accelerating rocket.

Hopefully this answers your other questions.

Originally posted by Janus58
Two Problems with considering the SR dilation as only apparent.

It doesn't explain Why Alice and Charlie agree that Bob's clock showed less elasped time, As all they see is SR dilation.

Here is my explanation for this within SR. It has to do with how Bob measures space as contracted compared to what Alice & Charlie measure. I’ll focus on Alice & Bob only, but you can apply Alice’s observations below to Charlie as well.

To use an extreme example, suppose Bob had, by when his initial acceleration stopped, contracted his measured remaining distance to the star from 1 light year (as Alice measures) to 1 meter. Let him continue to the star without decelerating. He’s obviously traveling close to c relative to the star, so he’ll cover the remaining distance at his constant velocity within say 1 nanosecond by his clock (I won’t bother to calculate the exact figure).

Let Bob be traveling in a long caravan of rockets at rest with respect to each other, so there’s always a rocket passing Alice. These helper rockets directly measure (to eliminate the Doppler effect) Alice’s elapsed time and email Bob the information.

Now let’s examine relative rates of time. During Bob’s remaining travel to the star, the helper rockets will note that Alice’s clock elapsed just over 1 year (because from Alice’s perspective, at just less than c it took Bob just over 1 year to cross the remaining light year to the star). Meanwhile Bob’s clock elapsed 1 nanosecond. Bob can conclude that Alice’s clock runs faster than his own clock.

Let Alice have a string of observation posts between her and Charlie. These helper observers directly measure (to eliminate the Doppler effect) Bob’s elapsed time and email Alice the information. During Bob’s remaining travel to the star, the helper observers will note that Bob’s clock elapsed 1 nanosecond (because from Bob’s perspective, at just less than c it took him 1 nanosecond to cross the remaining 1 meter to the star). Meanwhile Alice’s clock elapsed just over 1 year. Alice can conclude that her clock runs faster than Bob’s clock. Alice and Bob agree that Bob’s clock shows less elapsed time.

SR teaches that two rockets passing each other at constant velocity measure the other’s clock as running slow. But that presumes that the observers within the rockets see distance equally contracted; that is, not contracted relative to the other. If two observers were once united, then the only way they can subsequently have a relative velocity is if acceleration intervened. In the case of Bob & Alice, Bob’s initial acceleration causes him to measure space along his axis of motion as contracted relative to what Alice measures, which in turn causes him to measure Alice’s clock running faster than his own when his engines are off.

Originally posted by zanket
If you drop a ball off the top, it accelerates relative to both ends of the tower, right? And both ends accelerate relative to the ball. Well a photon has mass when it’s moving, which it always is. So it should fall like the ball does, and both ends should accelerate relative to the photon. Except the rate of clocks at either end differ, so no change in the photon’s velocity is directly measured..
I don't think you see my real questions. WHY are the clocks at the ends of the tower running at different speeds? WHAT determines their speed. They are stationary relative to each other in a uniform field. You're explanation is very good but as far as I can see it does not address these questions.

I appreciate that the maths works, for a photon being communicated along the axis of acceleration, because readings of the clocks will show that light speed is always constant. But is the speed of the clocks always different, even when there is no communication between the clocks. How do thier speeds relate then, and why?

And does asking this make any sense?

Why, when the janitor wanders upstairs at an amble, do he find that the clock at the top of the tower always running differently to the clock in the basement?

Originally posted by Canute
I don't think you see my real questions. WHY are the clocks at the ends of the tower running at different speeds? WHAT determines their speed.

Ultimately these questions are unanswerable. The only deeper you can go toward answering them is to say that the constancy of the speed of light demands that clocks run at different rates, and the speed of light determines their relative rates. The question then becomes, “How is light’s speed constant despite movement toward or away from the light’s source?” which I addressed here (http://www.sciforums.com/f33/s/showthread.php?postid=289391#post289391). One of my books says “This is a key aspect of relativity, indeed of much of modern physics: we focus only on observable, operationally defined quantities, and avoid unanswerable questions.”

I appreciate that the maths works, for a photon being communicated along the axis of acceleration, because readings of the clocks will show that light speed is always constant. But is the speed of the clocks always different, even when there is no communication between the clocks.

There is always such communication. When you flinch you propagate a minute gravitational wave throughout the universe at the speed of light. The wave passes through the Earth, and rather than the Earth disrupting the wave, the wave distorts the Earth, slightly changing its shape. Nothing can stop the communication. Since you’d have to have communication between the clocks just to compare their rates, your question at its essence is “Does a tree falling in the woods make a sound if no one is around to hear it?” And that question is meaningful only if there is nobody in the universe, for the falling tree’s sound waves will themselves propagate gravitational waves throughout the universe.

Since there is always communication between the clocks, and since the speed of light is thought to be constant, then the rates of the clocks always differ.

To be a tad philosophical, every clock in the universe is interconnected through gravity. If their rates did not differ as relativity theory supposes they do, then the speed of light would not be constant and that in turn would lead to problems; for example, you’d then expect two clocks side by side on a table to run at different rates, if one is run by a battery within and the other by a battery affixed to the ceiling. If you kept expanding on the example I think you’d eventually have to explain how we could survive if the speed of light were not constant.

Originally posted by zanket
Ultimately these questions are unanswerable. The only deeper you can go toward answering them is to say that the constancy of the speed of light demands that clocks run at different rates, and the speed of light determines their relative rates. The question then becomes, “How is light’s speed constant despite movement toward or away from the light’s source?” which I addressed here (http://www.sciforums.com/f33/s/showthread.php?postid=289391#post289391). One of my books says “This is a key aspect of relativity, indeed of much of modern physics: we focus only on observable, operationally defined quantities, and avoid unanswerable questions.”
Thanks - that is very clear. I wasn't trying to prove a point but I did supsect that we didn't know why. That's why I queried it whenever anyone posted a circular explanation.

There is always such communication. When you flinch you propagate a minute gravitational wave throughout the universe at the speed of light. The wave passes through the Earth, and rather than the Earth disrupting the wave, the wave distorts the Earth, slightly changing its shape. Nothing can stop the communication. Since you’d have to have communication between the clocks just to compare their rates, your question at its essence is “Does a tree falling in the woods make a sound if no one is around to hear it?” And that question is meaningful only if there is nobody in the universe, for the falling tree’s sound waves will themselves propagate gravitational waves throughout the universe.[/B]
Ok about 'gravity'. But trees in the wood are not quite equivalent, although I see your point, since clocks might have to tick accurately and relatively for always and always tell the right time, they have to synchronise in other words. Or is that just a more complicated equivalence? Perhaps it depends how you look at it.

To be a tad philosophical, every clock in the universe is interconnected through gravity. If their rates did not differ as relativity theory supposes they do, then the speed of light would not be constant and that in turn would lead to problems; for example, you’d then expect two clocks side by side on a table to run at different rates, if one is run by a battery within and the other by a battery affixed to the ceiling. If you kept expanding on the example I think you’d eventually have to explain how we could survive if the speed of light were not constant. [/B]
Yes - but we cannot say that clocks all vary because if they didn't it would be a bit awkward for us. Ultimately there must be a reason that light speed appears to us to be a constant, even if for the moment we just have to accept it.

Thanks for your answers.

Originally posted by Canute
Ultimately there must be a reason that light speed appears to us to be a constant, even if for the moment we just have to accept it.

Thanks for your answers.

Agreed, and you’re welcome.

Canute,

Ultimately there must be a reason that light speed appears to us to be a constant, even if for the moment we just have to accept it.

A few weeks ago, I posted an alternate explanation of why I believe that the speed of light remains constant for an observer (most of the time).

I believe that light uses gravitational fields for propulsion, so the omnidirectional speed of light is equal to c only in the local gravitational field. If an observer is stationairy in the local gravitational field he/she will measure the omnidirectional speed of light to be c. If, however, the observer is moving through the local gravitational field, the observer will witness a change in the speed of light. This "change" in the speed of light will cause chemical and physical reactions to slow down for the observer making it appear that time is dilating. If you want to get more info about my theory, check out this link:

http://www.sciforums.com/showthread.php?s=&threadid=16733

Tom

Originally posted by Prosoothus
Canute,A few weeks ago, I posted an alternate explanation of why I believe that the speed of light remains constant for an observer (most of the time)... If you want to get more info about my theory, check out this link:

http://www.sciforums.com/showthread.php?s=&threadid=16733
Tom
I went through some of it. You're a braver man than I Gungha Din. Who is this Chroot guy who always prevents the discussion from being good natured? I may have a go at rattling his paradigm myself when I spot the moment.

I have no understanding of half of what was being said. However it strikes me that if photons use gravity fields to propel themselves then it will make a difference which way they're going. Thus one might measure photons going in different directions at different speeds from a single observation point. Or does you rtheory say that the velocity of the observer relative to the field cancels these (perceived) differences out? To me it does not seem that it would, although I'm teetering on the edge of ignorance here.

Canute,

Who is this Chroot guy who always prevents the discussion from being good natured?

He was an insecure individual who hung around here for a few months. Fortunately, he left sciforums a week ago and now he's irritating the members of physicsforums.com.

As for your other questions, I will answer them in the "Alternative to Special Relativity" thread so that I don't hijack this thread.

Tom