Relativity - Oh dear, here we go again!

[BillyJoe]

Paul, please stop.

.8823 c.

Paul, you need a lesson in reading comprehension.
Really.
Or perhaps you just made a mistake.
I will give you the benefit.

Go back to that thread by Ziggurat and see if you can understand how he is correct and you are...wrong....again!

That's a good boy!

BJ

 

[Fredrik]

Acceleration doesn't cause time dilation. The velocity of one frame in the other frame is what matters.

The only thing that acceleration does in the twin paradox problem is to bring a third inertial frame into the discussion. Time dilation occurs both when the rocket is moving away from Earth and when it's coming back. The funny thing that happens when the rocket turns around and comes back is that the "planes" of simultaneity gets tilted in the opposite direction.

Consider two events A and B that simultaneous in the Earth frame. They are not simultaneous in the other two frames. Let's say that A is the earlier event in the rocket's frame when it's moving away from Earth. When the rocket turns around, B becomes the earlier event in the rocket's frame (which is now a different frame than before).

This is all pretty easy to see in a space-time diagram.

 

[Fredrik]

Sorry but what does x=vt mean?

"Step 3

[qimg]http://casa.colorado.edu/~ajsh/sr/construct3.gif[/qimg]
Draw a rectangle, sides 45o from vertical, with the origin at the centre and Cerulean at one corner. The rectangle represents the path of light rays which Cerulean uses to define a hypersurface of simultaneity, as described on the simultaneity page, Spacetime diagram illustrating simultaneity from Cerulean's point of view. Draw the extra diagonal across this rectangle. The diagonal is a hypersurface (reduced to a line) of simultaneity, a `now' line, for Cerulean. The now line lies along t = vx in Vermilion's frame, and along t´ = 0 in Cerulean's frame. "

Earlier they said "x=vt" which is fine; displacement = vel x time.

But Time = vel x displacement ? I don't get that!

The t' axis is just the position of the physical observer who uses the primed coordinates. At any point on that axis, we have x=vt (because he's moving with velocity v). The x' axis is the set of events that the physical observer who uses the primed coordinates considers simultaneous with the event with coordinates (0,0). At any point on that axis, we have t=vx. This is just saying that the slope of the t' axis is 1/v and the slope of the x' axis is v.

 

[Fredrik]

The claim that SR can't handle acceleration is a pet peeve of mine, so I'll respond to that, even though some good answers have already been given.

SR only says that space-time can be mathematically represented by Minkowski space, which is just the set R^4 with a funny metric. A coordinate system on a 4-dimensional manifold such as Minkowski space is just a function from Minkowski space into R^4. And since Minkowski space is R^4 (except for the metric), any "nice enough" function that maps R^4 onto itself is a coordinate system.

The set of all inertial frames is just a subset of the set of all coordinate systems. Among the other systems, there are some that can be interpreted as describing accelerated motion.

Note how absurd it would be if a problem such as the "twin paradox" couldn't be handled by SR. It would mean that there's something fundamentally wrong with the foundations of mathematics. (Because you can use axiomatic set theory to construct the integers, the integers to construct rational numbers, the rational numbers to construct real numbers, and the real numbers to construct Minkowski space).

 

[Fredrik]

So if there were three clocks A B and C involved: Clocks A and B Synchronise as they pass. A continues and passes clock C. C is moving in the same direction but faster than B. C Synchronises to A and continues passes B. All clocks always show the same time. Effectively there is no time dilation.

Yes there is. Three clocks traveling at different speeds in the same direction will tick at different rates.

Let’s say that A, B and C are spaceships. B has two Synchronised clocks and gives one to A as he passes. This clock is rapidly accelerated to the frame of A. Then A passes and gives the clock to C. The clock is now accelerated to the frame of C. Due to the acceleration it has undergone the passed clock is time dilated compared to the clock that B kept.

You're describing instantaneous acceleration, but it doesn't matter if the acceleration takes a very short time or a very long time. The ticking rates depend only on the relative velocities of the clocks.

 

[ynot]

Yes there is. Three clocks traveling at different speeds in the same direction will tick at different rates.

You're describing instantaneous acceleration, but it doesn't matter if the acceleration takes a very short time or a very long time. The ticking rates depend only on the relative velocities of the clocks.

I can start to understand how it might work if time dilation is caused by acceleration because one object is doing something different from the other. With relative velocities I can only see them doing the same thing however, just in different directions. If two objects have different velocities they are essentially either moving together or apart at any given time. Neither can be said to have more velocity than the other, so how can one experience more or less time delay than the other? Any time delay should be equal and opposite.

 

[MortFurd]

I can start to understand how it might work if time dilation is caused by acceleration because one object is doing something different from the other. With relative velocities I can only see them doing the same thing however, just in different directions. If two objects have different velocities they are essentially either moving together or apart at any given time. Neither can be said to have more velocity than the other, so how can one experience more or less time delay than the other? Any time delay should be equal and opposite.

In the case of velocities, it isn't a time delay. The clocks run at different rates, so if they should ever be brought together again they will show different times. The longer they maintain their different velocities, the greater that difference will be.

 

[Paulhoff]

Go back to that thread by Ziggurat and see if you can understand how he is correct and you are...wrong....again!

BJ

Funny, I see no final answer that was a number.

Paul



BJ is very appropriate for you name.

 

[Ziggurat]

I would say “Yes” and I don’t understand your “No” (even followed by “not really“ . A straight line is a direct path between two points. A curved line is an indirect path between two points, and therefore a longer distance. The extra distance of a curve comes from the fact that it’s an indirect path.

Sure, it comes from the fact that it's an indirect path, but not from the bend itself. Consider a right triangle. The hypoteneuse is the shortest path between the two far corners. To calculate the length of the other path, you add the length of the two other sides. What you do NOT do is calculate any length for the right angle itself. That is what I meant. And it's equivalent in relativity: acceleration is like the bend.

So are you saying that a straight line is a shorter distance but a longer time interval, and a curved line is a longer distance but a shorter time interval? Perhaps it would help if you could clarify exactly what you mean by “time interval“.

I mean the time experienced by an object following that trajectory, also refered to as the "proper time".

Here's the Euclidean metric in 2D:
ds² = dx² + dy²This gives the shortest path between two points as being a straight line. Take an easy example: a straight line between (x=0,y=0) and (x=1,y=0), which has a length of 1. For this path, dy=0. For any other path, the dx contribution will remain the same, but dy will not equal zero, so the path is longer.

Now here's the Minkowski metric (using only one spatial dimension for simplicity and dropping the c for simplicity):
ds² = dx² - dt²Note the minus sign. Now consider a straight line from (x=0,t=0) to (x=0,t=1). If we integrate the metric along this path, we get
s² = -1
With this metric, s is imaginary, but that's not important (we could have put the negative on the x instead of the y and everything would work just as easily), all that means is the distance is time-like and not space-like. It's the magnitude of s (in this case, 1) which tells us how much time passes for something following this path. Now here's where it gets interesting: the straight path has an s² of -1, but if we take a non-straight path (so that dx is no longer zero), then the magnitude of s² will be smaller. What does this mean? It means that any curved path has a shorter time interval. So any object traveling along that curved path will experience less time than an object traveling along a straight line between those two end points. And it turns out that this applies regardless of which two endpoints you choose, as long as you're dealing with time-like paths (in other words, no going faster than light). The spatial part of the Minkowski metric is still Euclidean-like, so spatial distances are still shortest with straight lines.

Don’t really understand this. Is this because the travelling twin has undergone acceleration and the other hasn’t? If so, it seems that acceleration is the thing that causes time dilation.

In one sense, yes, but in another sense, no. You need acceleration to get a curved path. But as my previous example with circular orbits was intended to show, the details of the acceleration are not what matter. You can have two curved paths between the same end points with identical proper time (the time interval experienced by traveling along that path) but radically different accelerations.

 

[BillyJoe]

I can start to understand how it might work if time dilation is caused by acceleration because one object is doing something different from the other. With relative velocities I can only see them doing the same thing however, just in different directions. If two objects have different velocities they are essentially either moving together or apart at any given time. Neither can be said to have more velocity than the other, so how can one experience more or less time delay than the other? Any time delay should be equal and opposite.

I thought Ziggurat's explanation was pretty good:
(I've highlighted a couple of important sentences at the end)

Consider what happens if you adopt the reference frame of the outbound twin. The earthbound twin experiences time dilation, and the traveling twin doesn't for part of his journey. But if you STICK with this reference frame, what happens when the traveling twin heads back towards earth? The earthbound twin is still traveling away, and still time-dilated, in this outbound frame. But now the traveling twin is moving faster than the earthbound twin, and is now experiencing even greater time dilation than the earthbound twin. And if you do the calculations, you'll find that this increased time dilation for the return leg more than makes up for the lack of time dilation for the outbound leg, when viewed from the SINGLE reference frame of the outbound journey. So it's not that time dilation doesn't apply to both twins: it most certainly does. It's that you can't just change reference frames in the calculation without accounting for that change.

And this demonstration that time dilation is completely independant of acceleration:

Here's a nice little problem involving time dilation which might help out a bit. We set up a "rest" reference frame, stick an observer at the origin and let him stay there (along with a clock) for the duration of our experiment. Now we take several volunteers with clocks, sync all the clocks together, and have them start moving in circular trajectories which all start at the origin (and hence return there periodically), which they will travel along at constant speed (with respect to our rest frame). Now one of the nice features of using circular orbits is that the acceleration and the speed are now independent parameters. We can have high-speed trajectories with small accelerations by using large circles, we can have small speed trajectories with large accelerations by using small circles, or any other combination. Now we ask, when these traveling clocks swing by the origin, how do their times compare with our stationary clock?

The answer is quite simply that the travelling clocks are slowed down compared to the stationary clock because of time dilation, by a factor given by their speed. Now, just like the twin paradox, there's an asymmetry in the problem because the moving clocks experience acceleration and the stationary clock doesn't. But here's the interesting bit which becomes clear because of the setup of the problem: the acceleration never actually enters into the calculation! Remember, we can make trajectories which have the exact same speed but different accelerations, or different speeds but the exact same acceleration, but the ONLY thing we need to know to figure out how much time dilation a trajectory experiences is the speed. The answer is totally independent of the value of the acceleration, and so acceleration is, in one sense, completely irrelevant to that observed time dilation.

 

[Gertrude]

Why do you want that?

I mean, in the absence of gravity, you can always just pick an arbitrary inertial reference frame and figure everything out relative to it. What's wrong with doing things that way? Does it merely feel like "cheating" to you, or is there some question you think really can't be answered by this method?

I have been told that SR can handle problems of accelerating frames. I want to see it for the sake of seeing it. If I were to handle relativity in accelerating frames, I honestly don't know where I'd start. I'm just curious what others have done. You are right that practically, it is not very useful. But that doesn't bother me the least...

 

[Fredrik]

I can start to understand how it might work if time dilation is caused by acceleration because one object is doing something different from the other. With relative velocities I can only see them doing the same thing however, just in different directions. If two objects have different velocities they are essentially either moving together or apart at any given time. Neither can be said to have more velocity than the other, so how can one experience more or less time delay than the other? Any time delay should be equal and opposite.

Lorentz contraction and time dilation are properties of space-time, not properties of objects. That's why they don't need to "do" different things. Again, this stuff is almost trivial when you draw space-time diagrams, and I don't see how anyone could possibly understand this without the aid of space-time diagrams.

 

[Ziggurat]

I have been told that SR can handle problems of accelerating frames. I want to see it for the sake of seeing it. If I were to handle relativity in accelerating frames, I honestly don't know where I'd start. I'm just curious what others have done. You are right that practically, it is not very useful. But that doesn't bother me the least...

To start with, are you familiar with how to handle non-inertial reference frames in Newtonian mechanics? That's a pretty well-treated topic. If you think you've got a handle on that, then you can move on to trying it in SR, but if that stuff still confuses you, it's probably a good idea to spend some time figuring it out first.

 

[The Grave]

Galaxies.....

An astro physics problem then?

If we 'observe' a star in a far-off galaxy.....14,000 lya, moving at say 0.6c relative to us, then shouldn't its stars be younger?

If it all started from a single point, the BB, then after so long of being accelerated + high vel., shouldn't we look at far off galaxies and see all young stars?

I mean if their clocks are running so slowly...

But I have to say that I find it hard to accept that "stationary" A, is older than "travelling" B, when really I would think that "both" A + B are travelling.....

So why should the time dilation effect be noticeable to one & not the other?

It seems to be a fix to account for the MM experiment and nothing more.

Griff...

 

[The Grave]

See my thread...

I have been told that SR can handle problems of accelerating frames. I want to see it for the sake of seeing it. If I were to handle relativity in accelerating frames, I honestly don't know where I'd start. I'm just curious what others have done. You are right that practically, it is not very useful. But that doesn't bother me the least...

I believe Yllanes deals with this in my thread "Am I really older than my twin"....

Griff.

 

[ynot]

But I have to say that I find it hard to accept that "stationary" A, is older than "travelling" B, when really I would think that "both" A + B are travelling.....

So why should the time dilation effect be noticeable to one & not the other?

It seems to be a fix to account for the MM experiment and nothing more.

Griff...

This is the question I"ve been asking as well. If we ignore acceleration, A and B are both "travelling" and "stationary" in equal and opposite measure. What applies to one should apply equally to the other. Observing space time diagrams haven't made things any clearer to me as yet. Seems to be too much that you have to accept on "say so" with no "plain english" description to back it up. I will keep observing.

ETA - How can A and B have different velocites when they are either coming together or moving apart at a single, common velocity?

 

[Ziggurat]

An astro physics problem then?

If we 'observe' a star in a far-off galaxy.....14,000 lya, moving at say 0.6c relative to us, then shouldn't its stars be younger?

Sure they are. The tricky thing is that when you talk about time for spatially separated objects, you need to consider simultaneity. Intuitively we think of simultaneity as something which is fixed and equivalent for everyone (that is, everyone agrees about whether or not two events happen at the same time or not). But it isn't. So in our reference frame, our galaxy has, say, its 10 billionth birthday simultaneously with the distant galaxy's, say, 7 billionth birthday. But their notion of simultaneity is different because they're in a different reference frame, and to them, their 10 billionth birthday will be simultaneous with our 7 billionth birthday. It's not that either observation is wrong, it's that the observation depends upon the reference frame, and we're not in the same reference frame.

If it all started from a single point, the BB, then after so long of being accelerated + high vel., shouldn't we look at far off galaxies and see all young stars?

I mean if their clocks are running so slowly...

Sure. But the dominant effect in terms of what we see (not what we observe) is the time delay of the signal (which is a classical effect). And we do indeed see distant galaxies being much younger than our own.

But I have to say that I find it hard to accept that "stationary" A, is older than "travelling" B, when really I would think that "both" A + B are travelling.....

They are. But their travels are not equivalent.

So why should the time dilation effect be noticeable to one & not the other?

It's noticeable on either twin from any reference frame in which the twin you're observing is moving. The trick is that 1) for the traveling twin, you CANNOT pick a reference frame in which he doesn't move for the whole journey, and 2) you cannot pick a reference frame for which the traveling twin doesn't move faster than the earthbound twin for at least one leg of his journey. So you can pick a frame where the traveling twin is motionless for, say, the outbound journey, and the earthbound twin is moving, so the traveling twin won't experience time dilation but the earthbound twin will. But you've got to STICK with that reference frame, then: when the traveling twin returns to earth, he's moving faster than the earthbound twin, and that faster speed means greater time dilation, and in the end he's STILL younger than the earthbound twin, even though from this reference frame the earthbound twin experienced time dilation for his whole journey.

It seems to be a fix to account for the MM experiment and nothing more.

That's because you still don't understand it. It's a beautifully elegant and completely self-consistent theory which explains a whole lot more than just MM, and it's been tested quite rigorously.

 

[Ziggurat]

This is the question I"ve been asking as well. If we ignore acceleration, A and B are both "travelling" and "stationary" in equal and opposite measure.

If you pick a reference frame (ANY reference frame) and stick with it the whole time, then this is manifestly NOT true. I've already pointed this out. You cannot change reference frames without consequence.

What applies to one should apply equally to the other. Observing space time diagrams haven't made things any clearer to me as yet.

I'll try to draw some up over the weekend, but if you're looking at correct diagrams, it should be pretty clear.

ETA - How can A and B have different velocites when they are either coming together or moving apart at a single, common velocity?

That will be made clear from the diagrams I intend to draw up.

 

[ynot]

If you pick a reference frame (ANY reference

If you pick a reference frame (ANY reference frame) and stick with it the whole time, then this is manifestly NOT true. I've already pointed this out. You cannot change reference frames without consequence.

Yes, I can understand it from just one frame, but two frames always exist. In the twins paradox we start with one frame (twins together) that changes to two frames (twins moving apart then together), then back to one frame (twins back together). The comparison of ages is made by both twins when they are back together in a single frame. If we just consider velocity and not acceleration, I still can’t see why the velocity is not equal and opposite for both twins. If the ”travelling” twin ages less than the “stationary” twin, they should both age less as they are equally “stationary” and “travelling”. It seems you are saying that both twins would say to each other “You have aged less than me”.

I'll try to draw some up over the weekend, but if you're looking at correct diagrams, it should be pretty clear.

That will be made clear from the diagrams I intend to draw up.

I await your diagram with eager anticipation. I really appreciate the time and effort you are giving me.

 

[Paulhoff]

This may help you.

http://nobelprize.org/educational_games/physics/relativity/paradox-1.html

Paul