Looking to understand time dilation

ghwellsjr

Yes, I'm not disagreeing with anything you said, I'm merely pointing out that we shouldn't think that it's necessary to say that the two "are at rest in the same inertial frame" because when they are at rest with each other in one inertial frame, they will be at rest with each other in all inertial frames.

matheinste

That's fine.

Matheinste.

Mike_Fontenot

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame. There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary. That preferred inertial frame is the one in which he is stationary.

Of course, other inertial observers (who are moving at some constant velocity with respect to the above inertial observer) will prefer a different inertial frame (the one in which THEY are stationary).

What special relativity says is that there is no single inertial frame that ALL inertial observers prefer. There is no single inertial reference frame that is UNIVERSALLY special.

Mike Fontenot

DaleSpam

Only as a matter of personal and computational convenience, not as a matter of physical necessity. And there may be some situations where a different frame is more convenient, e.g. the center of momentum frame for a collision.

You have never provided any substantiation for this, and it is only even possibly true if you define "elementary calculations" such that it is a tautology. Otherwise the principle of relativity ensures that all inertial reference frames will give the same predictions for the result of any given measurement and your statement is wrong.

ghwellsjr

The term "preferrred frame" has a specific meaning. You are making the same mistake that I am dealing with someone else on in this thread:

http://www.physicsforums.com/showthr...=442132&page=2

It doesn't mean a personal preference like "I prefer blondes". Even if everyone in the world preferred blondes, it wouldn't mean they are "UNIVERSALLY special". I hope you don't think the word "special" in Special Relativity has anything to do with the concept of "preferred". This has nothing to do with anyone's preferences. It has to do with whether we base our physics on the concept of an æther.

Mike_Fontenot

It's not a matter of physical necessity that you refrain from constantly hitting yourself in the head with a hammer, either.

The well-known results of time-dilation and length-contraction are directly available for your use (when you are an inertial observer), provided that you choose to use the usual Lorentz coordinates, in an inertial frame in which you are stationary. And the time coordinate, with that choice, corresponds to the time shown on your OWN watch. The spatial coordinate, with that choice, correponds to the lengths as reported by your OWN rulers and measuring tapes. There is a REASON why Einstein chose those coordinates, when developing his special theory.

In the spirit of general relativity, you are of course free to choose some other set of coordinates, by transforming those Lorentz coordinates in an almost unlimited number of ways, provided that the eigenvalues of the resulting metric are either {1, -1, -1, -1} or {-1, 1, 1, 1}, assuming that spacetime is everywhere flat.

In those alternative coordinate systems, your "clocks" would behave in very odd ways, compared to ordinary clocks. And likewise for your "rulers". Real work-a-day experimental physicists would seldom, if ever, choose those types of "measuring devices" to do their experiments in their laboratories. There is a REASON that the phrase "in the laboratory frame" is used so often.

Not if you're one of the two people involved in the impending collision. I've spent a lot of hours flying airplanes and gliders, constantly needing to avoid mid-air collisions. In all that time, I've NEVER used the center of mass coordinate system to help me avoid mid-air collisions.

Mike Fontenot

Mike_Fontenot

No, the term "special" in "special relativity" refers to the special case of limitless flat spacetime (no gravitational fields).

My other use of the term "special", as in "There is no single inertial reference frame that is UNIVERSALLY special", referred to the fact that in flat spacetime, there is no single inertial reference frame that is preferred by ALL inertial observers.

Mike Fontenot

ghwellsjr

You got the word "special" right, I just don't know why you can't get the word "preferred" right.

ghwellsjr

I thought your CADO formula was for a traveler going at substantial speeds, traveling incredible distances for almost an entire liftetime to "know" at any instant of time how old his daughter was back at home.

JesseM

But your own watch can only be used to assign coordinates to events on your own worldline, an accelerating observer could come up with multiple non-inertial coordinate systems which all agree that the coordinate time between events on his worldline is equal to his own proper time between those events, but which disagree about the time between events far from him or about simultaneity of distant events. There is no physical reason to think any of these non-inertial coordinate systems more accurately represents his "own measurements" of time than any other.
For an inertial observer "their own" ruler presumably means an inertial ruler at rest relative to themselves. But what does it mean for an accelerating observer? Does it mean that they are using an accelerating ruler, and if so how to decide how parts far from them are accelerating (is the ruler accelerating in a Born rigid way for example?) Or are you supposing that at each moment they define "their own" ruler to be a different inertial ruler which is instantaneously at rest relative to themselves at that moment?
Again this is meaningless if you haven't defined precisely what "your clocks" and "your rulers" means for an accelerating observer, and why you think the accelerating observer doesn't have a choice of how clocks far from his own worldline should move or be synchronized with one another, and likewise why he doesn't have a choice about how points on his own (accelerating?) ruler far from his own worldline should themselves be moving (remember that in SR there is no simple notion of how the back end of a 'rigid ruler' should accelerate if we know how the front end is accelerating, since there is no well-defined notion of 'rigidity' for objects undergoing arbitrary acceleration, physical rulers will behave like silly putty or slinkys when you accelerate one end)

DaleSpam

Very cute. Completely irrelevant to the topic, but cute.

So Mike, are you going to continue to dodge the challenge and hide from the issue? After a dozen or so requests you have had plenty of opportunity but you still can't even define your terms let alone demonstrate your claim. I suspect that you know that your claim is wrong.

ghwellsjr

I think Mike has defined his terms "elementary measurements" and "elementary calculations". Look at this post:
And now this one:
It looks like Mike thinks that Einstein was claiming that you have to use the inertial frame in which you are stationary to make SR work.

Then, somehow, he slips from a perpetually inertial frame to a non-inertial frame when the traveler starts his voyage:
And I'm sure he really believes this is what Einstein was promoting.

Grimble

But yes I agree with what you are saying here! So where does this show that I am going wrong?

Now this is where you lose me, What has SR and transforming measurements with LT possibly got to do with ACCUMULATED time?

LT is used to transform measurements: what a clock currently reads, not a calculation of the difference between two readings on a clock.

LT is a function of the CURRENT velocity. It has absolutely nothing to do with the history of how their relative speed might have varied over the course of their travel relative to one another.

DaleSpam

Which is a complete misunderstanding of SR, particularly the first postulate. I think it is obvious to everyone besides Mike.

matheinste

"when the traveller has returned and come to rest once again they will have identical ages once again."

"It is only if the traveller keeps on travelling past the home twin on his return that the age difference is apparent and it will be apparent to each of them."

These two statements are incorrect.

Perhaps I have missed something from your previous posts that puts these statements in context, if so forgive me and point them out.

Don't forget you cannot use one LT for the whole of the journey for the travellers non inertial frames.

Matheinste

ghwellsjr

The Lorentz Transform assumes that there has only ever been one velocity relative to the frame of reference. In this case, you can use LT to calculate both the difference in clock readings between the stationary one and the moving one, and/or you can calculate the relative tick rate between the two clocks. If you then accelerate the moving clock, you have to be very careful about how you describe what is happening to get a precise unique result. Usually, people just take a short cut and assume that the accelerating clock is gradually changing to its new tick rate as calculated by LT. Then to get an idea of the difference in the clock readings, you have to integrate, or multiply the time interval that a clock was ticking at a particular rate by its tick rate to find the accumulated time at the point of acceleration. And remember, this is all being done from the one frame of reference of the stationary observer (or any other single frame of reference).

In other words, if there were only one velocity involved, you could use LT directly to calculate the difference in clock readings OR you could use the tick rate multiplied by the time interval and you will get the same answer but when you introduce a new velocity, you cannot use the former method to calculate the difference in clock readings, you have to use the later twice, once for each time interval that the clock was moving at each speed and then add them together.

DaleSpam

Careful here. The velocity referred to here is the velocity between two inertial reference frames, not the velocity of some object in a given inertial frame. I am sure that you understand, but some people may misread that and think that you meant that the Lorentz transform cannot be used if an object is accelerating.