Understanding Time Dilation: How Fast Do You Have to Go?

[Sherlock]

The problem is that you are still trying to hold on to a Pre-Relativistic notion of "time", and this is incompatable with how Relativity treats time.

What's pre-Relativistic about my notion of time? Time is what you read on a clock. A clock is an oscillator of one sort or another, and an accumulator that indexes (counts) the oscillations. The time of an event (in Relativity) is the reading on a clock next to the event. And from Relativity we know that as an oscillator's state of motion changes, then it's period and rate of timekeeping changes.

Is it the idea that so called 'empty space' isn't empty that you disagree with? This idea is part and parcel of standard modern physics, afaik.

 

[jtbell]

Is it the idea that so called 'empty space' isn't empty that you disagree with? This idea is part and parcel of standard modern physics, afaik.

That's true, but it's not used as an "explanation" for time dilation in mainstream modern physics. If you have some non-mainstream ideas that you'd like to develop, then according to PF's rules you need to submit them to the "Independent Research" forum.

https://www.physicsforums.com/forumdisplay.php?f=146

Advocacy of non-mainstream theories is off-topic in the other forums (such as this one).

 

[Sam Woole]

No, not exactly. (And, let me say that I had the same problem grasping this as you seem to be having. I should also say that I'm not a physicist, so don't take anything I write as being necessarily correct -- of course, I'm sure the mentors will be on top of it. :-))

Thank you, Sherlock. You inspired me on the idea of natural clock. As a result I felt the time dilation idea cannot be properly justified.

 

[Sam Woole]

This is wrong. You can ALWAYS do an experiment to detect that you are accelerating. You cannot do an experiment to detect if you're moving without using another frame as a reference. The accelerating frame can always tell that it is accelerating, and can tell that another frame isn't.

Zz.

ZapperZ, maybe you were talking about experiments in earth, or in a gravitational field. If you were in a space where gravity is zero, can you detect you are accelerating?

Even if I got it wrong with this acceleration, which was considered as absolute as JesseM pointed out, do we know which motion, relative or absolute, will cause time dilation?

 

[JesseM]

ZapperZ, maybe you were talking about experiments in earth, or in a gravitational field. If you were in a space where gravity is zero, can you detect you are accelerating?

Yes, you will experience G-forces when you accelerate in space, just like how when you're in a car that's accelerating you feel yourself pushed back into the seat. From the point of view of an inertial frame, this isn't a true "force" like gravity (it's sometimes called a http://www.hcc.hawaii.edu/~rickb/SciColumns/FictForce.04Feb96.html for this reason), it's just that the car seat is accelerating and it has to overcome the inertia of your body to accelerate it to the same speed. But from the point of view of your own non-inertial frame, it feels just like a force is pulling you backwards.

Even if I got it wrong with this acceleration, which was considered as absolute as JesseM pointed out, do we know which motion, relative or absolute, will cause time dilation?

According to relativity time dilation is a function of velocity rather than acceleration. But if two clocks are moving apart at constant velocity, then in each clock's own rest frame it will be the other clock that is running slower--the only situation where you get an objective answer to which clock is "really" behind is the one where one clock turns around (accelerates) and moves back towards the first clock, so they can meet at a single location in space and see which clock is behind when they meet.

 

[Janus]

What's pre-Relativistic about my notion of time? Time is what you read on a clock. A clock is an oscillator of one sort or another, and an accumulator that indexes (counts) the oscillations. The time of an event (in Relativity) is the reading on a clock next to the event.

okay, so far.

And from Relativity we know that as an oscillator's state of motion changes, then it's period and rate of timekeeping changes.

No, we do not know this from Relativity. What we know from Relativity is that when we compare clocks between relatively moving frames, the clocks' rates differ.
The Pre-Relativitic notion is a tacit one of "absolute time". When you posit that the acceleration wrt space physically affects a clock to change its rate you are establishing a Prefered Frame of Reference to which all motion can be judged. Because the only way that this would give the results like that of Relativity is for the intial acceleration to change the clock's rate, then the clock's rate continues to run slow until the clock decelerates, whereupon its rate speeds back up. The only way this would work is if there was a preferred frame of reference to which you measured acceleration wrt. In Other words, an absolute state of rest. But if you have an absolute state of rest, you also have an absolute or "natural" rate of time (that of objects at a state of absolute rest)
Realtivity denies the existence of both; there is no preferred frame of reference or absolute time rate.

Is it the idea that so called 'empty space' isn't empty that you disagree with? This idea is part and parcel of standard modern physics, afaik.

The fact that empty space is not empty is not the point. The point is that you cannot use it as a preferred frame of referrence by which absolute motion can be measured, nor is it any interaction with this space that is responsible for time dilation.

 

[pervect]

i gotto tell i find these equations funny, for example i can just say;

insert lower V here / insert higher V here, V can not be lower than 1..

if both V is equal than the answer is undefined

And now try going over 1..

I can't figure out what you are trying to say or ask here, sorry.

No offense;I am simply curious, is the equation formed by the idea that you can't exceed c, or the equation came with test results in favour of the constant c?

And i really would appreciate if i can get a link to the journal that this equation was published and was it Einstein himself? I don't even know who published what yet..I am quite new to this topic, and need to learn a lot before i become convinced that the instruments i am using are reliable..I have quite a problem with information you see, i never trust anything

You can find an online reference to these equations in the sci.physics.faq, which is a reliable source for information on relativity

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

You could also open up just about any textbook on relativity and find these same equations.

Historically I'm not positive if it was Einstein who derived them or not - I very much suspect it was, but to me the history isn't so important. Many other physicists than Einstein have done good physics, he is not the only person capable of doing good physics.

These equations can be derived from the _Lorentz transform_ which is the basis of relativity. The derivation of these equations from the Lorentz transform is done in the sci.physics.faq link I posted above. The Lorentz transform equations were the equations originally published in Einstein's first paper on relativity.

The Lorentz transform was originally inspired by the Michelson Morley experiments which showed the constancy of the speed of light. Being the heart and soul of relativity, the Lorentz Transform has since then been verified by many other experiments.

 

[pmb_phy]

Hi I was wondering about how fast do you have to go before time dilation comes into effect

P.S. I'am no Qauntum physics professor so I don't know a lot about relativity

See

Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63(6), 568 (1995).

The article is online at
http://hussle.harvard.edu/~gabrielse/gabrielse/papers/1995/RelativisticMassAJP.pdf

Relativistic mass differerences of 0.03% of c have been detected.

Pete

 

[werner heisenberg]

Both twins agree that only 3.944 years have elapsed on earth. But what counts as far as aging goes is the time elapsed on the clocks that move along with each twin. .

How can a machine (the watch) be affected by the speed at which is going the guy who wears it? I mean, it seems pretty weird to think in a watch changing the rate at which its hands rotate

 

[ZapperZ]

How can a machine (the watch) be affected by the speed at which is going the guy who wears it? I mean, it seems pretty weird to think in a watch changing the rate at which its hands rotate

Because the mechanism that keeps time in the watch is governed by electromagnetic principles, the SAME principle with which we describe light in special relativity.

Zz.

 

[Xargoth]

I am new to the topic,too..But i can already say, this topic means bussiness;

It's like "Forget all you have learned till this day. And welcome the reality" stuff..

 

[jtbell]

How can a machine (the watch) be affected by the speed at which is going the guy who wears it?

As far as the guy who's wearing it is concerned, the watch always ticks at the same rate, no matter what his speed is. Look at it this way: so long as the guy is in an inertial reference frame, he's completely entitled to consider himself to be at rest and the "other guy" (the observer) to be moving. From his point of view, why should his watch be affected by how other people are moving? Especially when you consider that there might be two people watching him, moving at different speeds!

The way many physicists like to think of length contraction and time dilation is that they're sort of like geometrical perspective effects. Consider a rod that's one meter long. If it's at right angles to your line of sight, you "see" it as being its full length. But if you rotate the rod so that it's partly pointing towards you, at an angle, you "see" it as being shorter. Even though this is "merely" a geometrical perspective effect, it actually makes a physical difference in some situations. After all, if you rotate a one-meter rod to a sufficient angle, you can make it go through a door that's only a half-meter wide!

 

[pervect]

How can a machine (the watch) be affected by the speed at which is going the guy who wears it? I mean, it seems pretty weird to think in a watch changing the rate at which its hands rotate

It's best to think not of watches being "affected" by the speed, which assumes implicitly that there is some sort of "master watch" to which they can all be compared. This "master watch" is the idea of absolute time, which is the source of much of the confusion.

Instead, assume that it's a fact of nature that when you build the best watches possible, synchronize them so that they agree, then move them around via different paths and re-unite them at the destination, that they no longer agree when they are re-united.

 

[Sherlock]

And from Relativity we know that as an oscillator's state of motion changes, then it's period and rate of timekeeping changes.

No, we do not know this from Relativity. What we know from Relativity is that when we compare clocks between relatively moving frames, the clocks' rates differ.

Relativity and the differential timekeeping experiments say that if you change an oscillator's state of motion compared to a previous state then it's rate of timekeeping (hence it's period) changes. The previous state (the reference state) is some specific configuration wrt which the oscillator's state of motion (eg., its velocity) is anomalous (has changed).

As a reference state we start out with clocks A and B next to each other on Earth and we observe them to be keeping time at exactly the same rate in this state. No matter how, or how fast, we move the clocks around, as long as we keep them next to each other their readouts are exactly the same. Then we move only clock A to the moon and back. During the interval of A's round trip to the moon it's state of motion wrt the reference state has, by definition, been altered -- and we note that when the clocks are brought back together, then the time shown on A's readout is different from B's readout by an amount in very close accordance with the predictions of Relativity. We then reset them and note that they are keeping time at exactly the same rate.

The only way that clock A's accumulated time for the trip could have been different than B's is if the period of A's oscillator had undergone some change(s) during the trip. So, this is not just a matter of perspective. Clock A physically changed due to the change in it's state of motion. This implies that clock A was interacting with something different during the trip than clock B was, or that clock A was interacting differently with the essentially the same thing as B was, or some combination of both.

The Pre-Relativitic notion is a tacit one of "absolute time". When you posit that the acceleration wrt space physically affects a clock to change its rate you are establishing a Prefered Frame of Reference to which all motion can be judged. Because the only way that this would give the results like that of Relativity is for the intial acceleration to change the clock's rate, then the clock's rate continues to run slow until the clock decelerates, whereupon its rate speeds back up.
The only way this would work is if there was a preferred frame of reference to which you measured acceleration wrt. In Other words, an absolute state of rest. But if you have an absolute state of rest, you also have an absolute or "natural" rate of time (that of objects at a state of absolute rest)
Realtivity denies the existence of both; there is no preferred frame of reference or absolute time rate.
The fact that empty space is not empty is not the point. The point is that you cannot use it as a preferred frame of referrence by which absolute motion can be measured, nor is it any interaction with this space that is responsible for time dilation.

I'm not saying that there's an absolute time rate or that the contents (and 'behavior') of empty space is the same in all regions. Our reference state is always some configuration of objects (eg., clocks and their positions wrt each other and earth), which is compared to different sequences of configurations in which the relative position(s) of one of the objects (eg., one of the clocks) is anomalous wrt the reference state.

It seems to be fairly certain that acceleration is physically affecting the periods of accelerating oscillators. For some situations we can take the average velocity of the anomalous clock wrt the reference state and accurately predict its timekeeping rate for an interval using just the transformation of SR (that is, by calculating the average dilation in the period of its oscillator). For some situations, the field equations of GR are required.

The fact that empty space isn't empty is precisely the point if one wants to give some physical meaning to the prediction and observation that differential acceleration and velocity wrt a previous objective state produces measurable physical effects. An absolute or universal reference state is not only probably non-existent (undetectable anyway) , but absolute values for time and distance are impossible given our measurement conventions.

One point of this discussion is that, for pedagogical purposes, the geometrical interpretations of Relativity can be obfuscating rather than clarifying.

 

[pervect]

Relativity and the differential timekeeping experiments say that if you change an oscillator's state of motion compared to a previous state then it's rate of timekeeping (hence it's period) changes.

Not necessarily.

The previous state (the reference state) is some specific configuration wrt which the oscillator's state of motion (eg., its velocity) is anomalous (has changed).

As a reference state we start out with clocks A and B next to each other on Earth and we observe them to be keeping time at exactly the same rate in this state. No matter how, or how fast, we move the clocks around, as long as we keep them next to each other their readouts are exactly the same. Then we move only clock A to the moon and back. During the interval of A's round trip to the moon it's state of motion wrt the reference state has, by definition, been altered -- and we note that when the clocks are brought back together, then the time shown on A's readout is different from B's readout by an amount in very close accordance with the predictions of Relativity. We then reset them and note that they are keeping time at exactly the same rate.

So far so good.

The only way that clock A's accumulated time for the trip could have been different than B's is if the period of A's oscillator had undergone some change(s) during the trip.

No. This is not the only solution. An alternative and in many ways better solution is to say that clocks cannot be compared at different points in space. The reason that clocks cannot be compared at different poitns in space is that simultaneity is not absolute.

Thus, we can't compare clock A to clock B until after we bring the clocks back together.

Depending on the details of how we bring the clocks back together, either clock A or clock B could read longer. For instance, if we launch B into space, stop it, and bring it back to Earth, A will read longer.

But we could just as well wait a bit, then launch clock A at an even higher speed to "catch up" with clock B. B would then read the longer time when it "caught up" with A.

 

[Intuitive]

If X = heigth and Y = width and Z = length, wouldn't acceleration effect Z as a product of length, It doesn't seem to clash with relativity but seems to fit very well for affecting time as length, Is this part of the solution of why relativity actually ratios an individuals perspective of how time flows in accourdance to SP?

Just a curious

 

[Sherlock]

Thank you, Sherlock. You inspired me on the idea of natural clock. As a result I felt the time dilation idea cannot be properly justified.

I think the idea of time dilation is very much justified. So, you might have drawn some conclusions from what I've written that I wasn't intending.

In the usual twin-clock scenario. The reference state is two identical clocks, A and B, keeping time at exactly the same rate, sitting next to each at some location on earth.

As clock A travels throughout the solar system, then as long as the observational context is just A wrt B or B wrt A, then the time dilation effects seen by A and B will be symmetrical.

But since we've defined the reference state to include the original Earth location from which B never moves, and we know that clock A's state of motion wrt this reference state has changed while B's hasn't; and we know from experiments that they show different accumulated times for the trip interval; then, wrt this reference state and the interval of clock A's round trip, the average period of clock A's oscillator has dilated (by an amount pretty accurately predicted by Relativity) wrt the period of clock B's oscillator.

 

[Sherlock]

Relativity and the differential timekeeping experiments say that if you change an oscillator's state of motion compared to a previous state then it's rate of timekeeping (hence it's period) changes.

Not necessarily.

Ok, you can complicate things (as you do below) so that an experimental demonstration becomes impractical or so that the experiment doesn't tell you anything, but what's the point of that?

We restrict the degrees of freedom so that we're clearly comparing the rate of a clock which moved wrt a specific reference state with the rate of a clock which didn't move wrt that reference state.

The results of such comparisons are fairly clear. The clock whose state of motion has changed wrt the reference state accumulates time at a different rate. So, the average period of it's oscillator is different during the interval in which its average velocity wrt the reference state is different.

The only way that clock A's accumulated time for the trip could have been different than B's is if the period of A's oscillator had undergone some change(s) during the trip.

No. This is not the only solution. An alternative and in many ways better solution is to say that clocks cannot be compared at different points in space. The reason that clocks cannot be compared at different poitns in space is that simultaneity is not absolute.

Thus, we can't compare clock A to clock B until after we bring the clocks back together.

Depending on the details of how we bring the clocks back together, either clock A or clock B could read longer. For instance, if we launch B into space, stop it, and bring it back to Earth, A will read longer.

But we could just as well wait a bit, then launch clock A at an even higher speed to "catch up" with clock B. B would then read the longer time when it "caught up" with A.

Ok, but we specify a reference state, so naturally we need to reassemble that state in order to make any meaningful comparisons between the clocks.

EDIT: I changed my mind about this. Because we specify a reference state involving a frame of reference in addition to the two clocks, then we don't need to bring the clocks back together in order to make meaningful statements about which clock is moving slower.

We of course do need to keep one clock in the reference state, and keep them both running, and continue tracking them both during the separation interval.

 

[JesseM]

Sherlock, when you say "reference state", do you mean the same thing as a "reference frame" or something different? If it's something different, can you explain, maybe with an example?

 

[Sam Woole]

I think the idea of time dilation is very much justified. So, you might have drawn some conclusions from what I've written that I wasn't intending..

Hi, Sherlock, when I said I was inspired by the term natural clock, I did not mean you were a dissident against time dilation.

In the usual twin-clock scenario. The reference state is two identical clocks, A and B, keeping time at exactly the same rate, sitting next to each at some location on earth.

As clock A travels throughout the solar system, then as long as the observational context is just A wrt B or B wrt A, then the time dilation effects seen by A and B will be symmetrical.

But since we've defined the reference state to include the original Earth location from which B never moves, and we know that clock A's state of motion wrt this reference state has changed while B's hasn't; and we know from experiments that they show different accumulated times for the trip interval; then, wrt this reference state and the interval of clock A's round trip, the average period of clock A's oscillator has dilated (by an amount pretty accurately predicted by Relativity) wrt the period of clock B's oscillator.

Here you were repeating the other experts' idea and experiments (by Hafele-Keating). But my difficulty with time dilation was on the symmetrical aspect. As long as it is symmetrical, time dilation cannot be justified because both clocks are producing equal time intervals.

Now supporters of time dilation changed the game rules by introducing acceleration. I said this because when the twin paradox was first introduced to give support to the time dilation idea, acceleration was not specified. Acceleration came into the game only after some dissitents attacked the weird idea by means of uniform motion. (I read about it but I could not quote the source.)

Even if we accept the change of rules, I believe time dilation still cannot be justified, especially when spaceships were meant to return to earth. For the spaceship to return, it must undergo deceleration, such as NASA's spacecraft re-entering the earth. Theoretically the spaceship's acceleration should equal its deceleration. Hence your "oscillator dilated" would be cancelled. Don't you think so? Are the twins of the same age?

 

[JesseM]

Here you were repeating the other experts' idea and experiments (by Hafele-Keating). But my difficulty with time dilation was on the symmetrical aspect. As long as it is symmetrical, time dilation cannot be justified because both clocks are producing equal time intervals.

No, it is "symmetrical" because the second clock runs slow in the first clock's frame, while the first clock runs slow in the second clock's reference frame.

Now supporters of time dilation changed the game rules by introducing acceleration. I said this because when the twin paradox was first introduced to give support to the time dilation idea, acceleration was not specified. Acceleration came into the game only after some dissitents attacked the weird idea by means of uniform motion. (I read about it but I could not quote the source.)

You don't know your history. Einstein specified each reference frame must be in a state of uniform linear motion (ie no acceleration) in his original 1905 paper, and the twin paradox was certainly not thought up until after that.

Even if we accept the change of rules, I believe time dilation still cannot be justified, especially when spaceships were meant to return to earth. For the spaceship to return, it must undergo deceleration, such as NASA's spacecraft re-entering the earth. Theoretically the spaceship's acceleration should equal its deceleration. Hence your "oscillator dilated" would be cancelled. Don't you think so? Are the twins of the same age?

There is no distinction between "acceleration" and "deceleration" in physics--whether an accelerating object is increasing in velocity or decreasing in velocity depends on your reference frame, this is true in Newtonian physics as well.

Look, whether or not you believe relativity is physically correct (and all the experimental evidence says it is), if you're trying to argue that it doesn't make mathematical sense you're just being foolish. The Lorentz transformation is a perfectly consistent way to transform between different coordinate systems, and if a clock moving at velocity v is ticking at \sqrt{1 - v^2/c^2} the normal rate in a given frame, then if you know some basic calculus it's clear that given the clock's velocity as a function of time v(t), the total time elapsed between two times t_0 and t_1 must be given by the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. You can prove mathematically that the answer to the total time elapsed between two points on the clock's worldline (a 'worldline' is a path through spacetime) in one frame will be equal to the answer in another reference frame if you use the Lorentz transform to describe the worldline in the two different frames. And you can also prove that the path between two points with the greatest elapsed time will always be the straight one, so any non-straight worldline between those same points will have elapsed less time, and straight worldline = no acceleration while non-straight worldline = at least some acceleration. This is analogous to the fact that the shortest distance between two points in 2D space is always a straight line, any non-straight path between the two points will have a greater length.

 

[Sherlock]

Sherlock, when you say "reference state", do you mean the same thing as a "reference frame" or something different? If it's something different, can you explain, maybe with an example?

If I'm understanding what reference frame means, then the reference state includes 3 reference frames -- the earth, clock A, and clock B. But, since clock B doesn't move from it's location on earth, then the reference frame that is clock B is essentially the same as the reference frame that is it's location on earth.

According to relativity time dilation is a function of velocity rather than acceleration. But if two clocks are moving apart at constant velocity, then in each clock's own rest frame it will be the other clock that is running slower--the only situation where you get an objective answer to which clock is "really" behind is the one where one clock turns around (accelerates) and moves back towards the first clock, so they can meet at a single location in space and see which clock is behind when they meet.

As long as B remains in the original location on Earth while A is moving wrt the original configuration (the reference state), and as long as we can continue to track A's motion, then I think we get an objective answer to which clock is really behind, and by how much, whether A and B are reunited.

You mention that time dilation is a function of velocity rather than acceleration. Acceleration is velocity per unit of time.

 

[JesseM]

If I'm understanding what reference frame means, then the reference state includes 3 reference frames -- the earth, clock A, and clock B. But, since clock B doesn't move from it's location on earth, then the reference frame that is clock B is essentially the same as the reference frame that is it's location on earth.

As long as B remains in the original location on Earth while A is moving wrt the original configuration (the reference state), and as long as we can continue to track A's motion, then I think we get an objective answer to which clock is really behind, and by how much, whether A and B are reunited.

Just because A and B start out in a single rest frame, why should that mean that this frame's answer to which clock is later ticking faster is more correct than any other? Why does the "original state" matter, especially since the choice of what moment to pick as the beginning is totally arbitrary? If at an even earlier time A and B had been traveling through space at a high velocity relative to earth, then they came to rest on Earth and stayed there for a while, then B took off from the Earth and returned to its original velocity, which frame would you use to decide which clock is really ticking faster?

You mention that time dilation is a function of velocity rather than acceleration. Acceleration is velocity per unit of time.

Acceleration is change in velocity per unit time--if an object isn't changing velocity, its acceleration is zero. Anyway, what's your point?

 

[pervect]

The results of such comparisons are fairly clear. The clock whose state of motion has changed wrt the reference state accumulates time at a different rate. So, the average period of it's oscillator is different during the interval in which its average velocity wrt the reference state is different.

Ok, but we specify a reference state, so naturally we need to reassemble that state in order to make any meaningful comparisons between the clocks.

EDIT: I changed my mind about this. Because we specify a reference state involving a frame of reference in addition to the two clocks, then we don't need to bring the clocks back together in order to make meaningful statements about which clock is moving slower.

We of course do need to keep one clock in the reference state, and keep them both running, and continue tracking them both during the separation interval.

*IF* you make the additional assumption to use a specific reference frame's definition of simultaneity, your logic is sound. But this additional assumption is required, and leads to a seeming paradox.

The seeming paradox is commonly called the "twin paradox", where A concludes that B's clocks are running slow, and B concludes that A's clocks are also running slow.

The key to resolving this paradox is to note specifically that one *does* have to make the extra assumption about what definition of simultaneity one choses to use. There are as many different defintions of "simultaneous" as there are reference frames (which you call reference states). A's definition of simultaneity is not the same as B's definition. This is equivalent to my remark that the results of any clock comparision will depend on the details of the path by which the clock is transported. Until a reference frame is chosen (or a clock transport method is specified), the results of a direct comparison between the spatially separated clocks is not defined.

The recognition of the fact that there *are* different defintions of "simultaneous" is what prevents the twin paradox from actually being a paradox, i.e. it is necessary to recognize the relativity of simultaneity to understand how special relativity is self-consistent.

 

[Sam Woole]

No, it is "symmetrical" because the second clock runs slow in the first clock's frame, while the first clock runs slow in the second clock's reference frame.

You have been saying this a few times. Let me show you what was my understanding. The second clock shows 3 seconds to the observer in the first frame whose clock shows 4 seconds, meaning the second clock is 1 second slower than the first clock. Symmetrically, the first clock shows 3 seconds to the observer in the second frame, whose clock shows 4 second, meaning the first clock is 1 second slower than the second clock. In short, there is always this sequence: 3-4, 4-5, 5-6 ...in each frame. This sequence means that the two clocks constantly have 1 second difference between them. It further means to me, both clocks are measuring identical time intervals, constantly synchronized. One second in one frame is always equal to the one second in another frame. To me, this means there is no time dilation.

You don't know your history. Einstein specified each reference frame must be in a state of uniform linear motion (ie no acceleration) in his original 1905 paper, and the twin paradox was certainly not thought up until after that. There is no distinction between "acceleration" and "deceleration" in physics--whether an accelerating object is increasing in velocity or decreasing in velocity depends on your reference frame, this is true in Newtonian physics as well..

Your words above mean to me, both acceleration and deceleration will cause differential aging, uniform relative motion included. But you also said earlier that the space traveling twin must accelerate "significantly"? Why significantly?

Look, whether or not you believe relativity is physically correct (and all the experimental evidence says it is), if you're trying to argue that it doesn't make mathematical sense you're just being foolish. The Lorentz transformation is a perfectly consistent way to transform between different coordinate systems, and if a clock moving at velocity v is ticking at \sqrt{1 - v^2/c^2} the normal rate in a given frame, then if you know some basic calculus it's clear that given the clock's velocity as a function of time v(t), the total time elapsed between two times t_0 and t_1 must be given by the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. You can prove mathematically that the answer to the total time elapsed between two points on the clock's worldline (a 'worldline' is a path through spacetime) in one frame will be equal to the answer in another reference frame if you use the Lorentz transform to describe the worldline in the two different frames. And you can also prove that the path between two points with the greatest elapsed time will always be the straight one, so any non-straight worldline between those same points will have elapsed less time, and straight worldline = no acceleration while non-straight worldline = at least some acceleration. This is analogous to the fact that the shortest distance between two points in 2D space is always a straight line, any non-straight path between the two points will have a greater length.

I myself cannot prove that Einstein's theory is correct or not. But I know there are plenty of challenges to his math. One was a Canadian Dr. Paul Marmet,
http://www.Newtonphysics.on.ca/, who said:
"Conventional Wisdom, Conventional Logic, Newton's Physics and Galilean coordinates, classical physics can explain all the observed phenomena attributed to relativity. Einstein's Relativity is completely useless." Marmet has a book to support his words.

And there were others who showed, mathematically, that Einstein's math is false as it results in 2=0, etc.

 

[JesseM]

You have been saying this a few times. Let me show you what was my understanding. The second clock shows 3 seconds to the observer in the first frame whose clock shows 4 seconds, meaning the second clock is 1 second slower than the first clock. Symmetrically, the first clock shows 3 seconds to the observer in the second frame, whose clock shows 4 second, meaning the first clock is 1 second slower than the second clock. In short, there is always this sequence: 3-4, 4-5, 5-6 ...in each frame.

No, the difference is not constant. Each frame observes the other clock as ticking slower than its own, not just being behind its own by a constant amount but ticking at the same rate. For example, suppose two clocks are moving at 0.8c relative to each other, and they start out at the same position at which point they both read "0 seconds". Then each clock will observe the other run at 0.6 times its own rate--so in the first clock's frame, when the first clock reads 4 seconds the second clock reads 2.4 seconds, when the first clock reads 5 seconds the second reads 3 seconds, when the first clock reads 6 seconds the second clock reads 3.6 seconds, and so forth. Likewise, in the second clock's frame, when the second clock reads 4 seconds the first reads 2.4 seconds, when the second clock reads 5 seconds the first reads 3 seconds, and so forth. That is what is meant by "symmetrical" here. Again, if you're having trouble understanding how this is possible, I suggest you take a careful look at the diagrams from the example I provided in An illustration of relativity with rulers and clocks.

Your words above mean to me, both acceleration and deceleration will cause differential aging, uniform relative motion included.

Relative to a particular frame, differential aging depends only on the velocity. But the point is that different frames disagree about who's aging slower when two clocks are moving apart at constant velocity, in order for there to be a single objective answer at least one of the clocks has to turn around and return to meet the other clock, and "turning around" means accelerating. As I said, it can be shown that all frames will agree on their prediction of what the clocks will read when they meet up again (each frame makes this prediction by doing the integral I posted earlier, which depends only on the velocity function v(t) in that frame), and they will all agree that the clock that moved at constant velocity elapsed more time than the clock that accelerated.

But you also said earlier that the space traveling twin must accelerate "significantly"? Why significantly?

I don't remember the context, but I'd guess I probably just said that because if the two clocks are moving apart at relativistic speeds, one will have to accelerate significantly in order for its path to change enough so that it's now moving towards the other clock at relativistic speeds (here 'significantly' can mean either a large burst of acceleration over a short time or a lower rate of acceleration but extended over a long time period). If you just had two clocks moving apart at speeds which were very small compared to light speed, then the acceleration to get one clock to turn around needn't be large. As long as two clocks depart from each other at a single point in space and then meet again later at a single point in space, it will always be the clock that accelerated that has elapsed less time, regardless of the size of the acceleration.

I myself cannot prove that Einstein's theory is correct or not. But I know there are plenty of challenges to his math. One was a Canadian Dr. Paul Marmet,
http://www.Newtonphysics.on.ca/, who said:
"Conventional Wisdom, Conventional Logic, Newton's Physics and Galilean coordinates, classical physics can explain all the observed phenomena attributed to relativity. Einstein's Relativity is completely useless." Marmet has a book to support his words.

It doesn't sound like he's challenging the math in that quote, just the theory. Can you provide a quote from his webpage or his book saying that there is some self-contradiction in the math itself?

And there were others who showed, mathematically, that Einstein's math is false as it results in 2=0, etc.

No there weren't. And if you think there were, why do you think the mathematics community wasn't convinced? Do you think they're all idiots who can't follow a simple proof showing a contradiction? Remember, once something is proved in mathematics then its completely cut-and-dried, there's no room for differences of opinion--if it's been proved, any mathematician competent to evaluate the proof should agree.

 

[Sherlock]

Just because A and B start out in a single rest frame, why should that mean that this frame's answer to which clock is later ticking faster is more correct than any other?

The selection of reference (or rest) frame is, as you note, arbitrary. But, for practical purposes we pick a location on Earth and position the clocks right next to each other. This is a state of motion of the earth-clock A-clock B system. Wrt this state, as clock A travels around the solar system, we know that clock A is the moving clock. Clock A's motion is the anomalous motion wrt the reference state. So, whether we reunite the clocks (reassemble the reference state) or not, we can make some statements about which clock's timekeeping rate has 'really' slowed wrt the rate (which clock B is still keeping time at) observed when both clocks were next to each other on earth. (We're assuming of course that the Earth's regular motion won't be appreciably changed due to A's acceleration away from it, so that if we expanded the observational context,the reference state, to include, say, the entire solar system, then we would see that clock A's motion is the anomalous motion wrt the regular motion of the solar system, and so on.) We observe that clock A's rate has changed due to changing it's state of motion.

Why does the "original state" matter, especially since the choice of what moment to pick as the beginning is totally arbitrary?

The reference state defines a specific state of motion of the clocks. We change clock A's state of motion wrt this state, and note the differences between the rate at which it's oscillator cycles and B's. They're different, so we conclude that clock A's oscillator has undergone some physical changes that clock B's oscillator hasn't -- and, therefore, that clock A's oscillator has undergone some physical changes that it would not have undergone had it remained on Earth next to clock B.

Acceleration is change in velocity per unit time--if an object isn't changing velocity, its acceleration is zero. Anyway, what's your point?

Just that when an object is accelerating, then the period of its oscillator is dilating. During intervals of uniform velocity, the dilation (the oscillator's period) is constant.

 

[Sherlock]

... my difficulty with time dilation was on the symmetrical aspect. As long as it is symmetrical, time dilation cannot be justified because both clocks are producing equal time intervals.

Your statement here suggests that you're using time dilation to mean the same thing as differential timekeeping. They're not the same thing. The dilation in timekeeping rates of two clocks, A and B, is symmetrical as long as we're only considering the two clocks and nothing else -- A wrt B, and B wrt A is the same dilation as they move wrt each other. But if we bring in another reference frame wrt which one, B, is stationary, then wrt this frame we can say that A's time is dilated while B's isn't. Also, when A and B start out next to each other on, say, a laboratory table, and then we take A on an airplane ride, then we know which clock moved and which clock didn't.
As long as we can find at least one frame of reference wrt which A and B are moving differently, then we have a basis for saying that they are keeping time differently, and, as long as we can continue to track them both, what that difference is for a given interval.

Even if we accept the change of rules, I believe time dilation still cannot be justified, especially when spaceships were meant to return to earth. For the spaceship to return, it must undergo deceleration, such as NASA's spacecraft re-entering the earth. Theoretically the spaceship's acceleration should equal its deceleration. Hence your "oscillator dilated" would be cancelled. Don't you think so? Are the twins of the same age?

Any change in speed or direction is an acceleration. As the traveling twin slows down to land back on Earth his time is still dilated wrt the earthtwin's time, but the dilation decreases as he approaches Earth to land.
As long as the traveling twin is moving wrt the earthbound twin, whether this motion is away from or toward the earthbound twin, then his timekeeping rate will be slower (the period of his oscillators dilated) wrt the earthbound twin. This is because the traveling twin's motion is the anomalous motion. We know that the traveling twin is the one who accelerated.

The twins age differently.

It took me a while to understand this. But, I think I have an intuitive grasp of it now. I know the feeling of being stuck on the idea that the twins should age the same -- even though the equations say differently. Then it sunk in that every measurement depends on time and distance units, and that these units are physically different wrt different states of motion.
I just focused on the idea that the twins are moving differently wrt the natural order of things. So their oscillators are experiencing things, interacting with things a bit differently, and as a result the periods of their oscillators are different, and the values of time and distance units (and even the biological processes) that depend on these periods are also different.

 

[Sherlock]

*IF* you make the additional assumption to use a specific reference frame's definition of simultaneity, your logic is sound. But this additional assumption is required, and leads to a seeming paradox.

The seeming paradox is commonly called the "twin paradox", where A concludes that B's clocks are running slow, and B concludes that A's clocks are also running slow.

The key to resolving this paradox is to note specifically that one *does* have to make the extra assumption about what definition of simultaneity one choses to use. There are as many different defintions of "simultaneous" as there are reference frames (which you call reference states).

A's definition of simultaneity is not the same as B's definition. This is equivalent to my remark that the results of any clock comparision will depend on the details of the path by which the clock is transported. Until a reference frame is chosen (or a clock transport method is specified), the results of a direct comparison between the spatially separated clocks is not defined.

The recognition of the fact that there *are* different defintions of "simultaneous" is what prevents the twin paradox from actually being a paradox, i.e. it is necessary to recognize the relativity of simultaneity to understand how special relativity is self-consistent.

I'd put it this way:
The resolution of paradoxes due to symmetrical time dilation is based on the observation of anomalous motion wrt some defined state or natural process or order. It's based on the observation that clock A's motion is the anomalous motion (comparing A and B) wrt the Earth reference frame. The alternative is that the Earth and clock B accelerated away from clock A -- but I think we can rule that out (anyway, we can determine what is really happening by observing the rest of the solar system).

So, yes, we make assumptions about which states to use as reference states. It's the recognition that some alternatives make more sense than others wrt our observations of the natural order of things that prevents the twin paradox from actually being a paradox.

I don't think this contradicts anything you said.

 

[JesseM]

I'd put it this way:
The resolution of paradoxes due to symmetrical time dilation is based on the observation of anomalous motion wrt some defined state or natural process or order. It's based on the observation that clock A's motion is the anomalous motion (comparing A and B) wrt the Earth reference frame. The alternative is that the Earth and clock B accelerated away from clock A -- but I think we can rule that out (anyway, we can determine what is really happening by observing the rest of the solar system).

It doesn't make any difference who accelerated away from who initially, it's only acceleration after the two clocks depart that determines which one will be behind when they meet again. The perspective of a frame where the Earth is initially moving at 0.99c, and the clock that accelerates away from it is actually slowing down initially, is every bit as valid as the perspective of the Earth's frame.

So, yes, we make assumptions about which states to use as reference states.

No, the way you are using "reference state", basically to mean using the initial conditions to determine which frame's answer is more "correct" than the others, is not an idea that any physicists would find useful, and you haven't given any justification for why the initial conditions should cause us to prefer one frame over another. It's totally arbitrary, as far as I can see (and note my comments earlier about the choice of which state to label the 'initial' one as being equally arbitrary).

I don't think this contradicts anything you said.

I don't think pervect understood what you meant by "reference state", since he seemed to think it just meant the same thing as "reference frame", when actually you are using it to mean an initial state which you use to determine a preferred frame. I'm pretty sure he wouldn't agree with this notion of yours that the initial state tells us that one frame's perspective is more valid than others.