A small question

[uncool]

Redshift being equal in all directions only points to it being a sphere, not the surface of a hypersphere.

It expanding equally in all directions for all points is just like a circle expanding and all the points on it expanding equally. Just look at it. You'll see what I mean

-Uncool-

[Spyman]

I didn't know that the concept of hypersphere is used to explain redshift. I certaintly don't believe that is necessary.

 

At this point' date=' i would focus upon the definition of 'hypersphere' most importantly, does the idea have direction of motion constant, yet you return to where you were.[/quote']I don't belive the concept is used to explain redshifts but it's one way to explain the uniform measurements of such things in all directions, (3D).

 

If the motion outside the 3D count then so should regression speed also...

 

My point was that I tend to agree with You that there is a contradiction between relativity and a hyperspherical universe.

 

Therefore I don't need to end anywhere else.

 

Redshift being equal in all directions only points to it being a sphere' date=' not the surface of a hypersphere.

It expanding equally in all directions for all points is just like a circle expanding and all the points on it expanding equally. Just look at it. You'll see what I mean[/quote']If we are in the center of the sphere yes, but are we really ?

 

And if we are not then how can it be equal in all directions like in a circle ?

 

Togheter with Big Bang one solution is hyperspherical but it may be a wrong one.

[Severian]

I am a bit confused as to why Tom doesn't regard this problem as simple. It is (as far as I can tell) because there is a perfect symmetry in the problem. (So you don't need to work it our mathematically.)

 

Johnny5 is almost right with his first post. Because there is a perfect symmetry you cannot tell the brothers apart, and therefore each will see the other brother's watch in exactly the same state - with less time on it than their own. But you must remember that they are making frame dependent measurements, so this is not a contradiction since the measurements are not directly corresponding to each other. For example, someone in a third frame travelling at half the speed of the travelling brother (and coincidentally meeting them at the same time that they coincide) would see the same time on both of their watches (which would be both slower than his). Alternatively, if one of the brothers now accelerates to the other's rest frame, he will find himself younger than his brother (it is not the acceleration with makes the time dilation, but the acceleration makes it observable by allowing them to make measurements in the same frame).

[J.C.MacSwell]

OK, what is wrong with this in principle:

 

Two clocks go by the Earth at near c in opposite directions.

 

As they go by they are synchronized by a signal from Earth.

 

Each assumes the other clock to be much much slower.

 

They go "straight" around the Universe/hypersphere.

 

When they meet again Earth is long gone.

 

As they pass by they prepare to resynchronize via an Earth station our descendants have built and positioned for the occasion.

 

How do their watches compare?

[Spyman]

I am a bit confused as to why Tom doesn't regard this problem as simple. It is (as far as I can tell) because there is a perfect symmetry in the problem. (So you don't need to work it our mathematically.)

 

Johnny5 is almost right with his first post. Because there is a perfect symmetry you cannot tell the brothers apart' date=' and therefore each will see the other brother's watch in exactly the same state - with less time on it than their own. But you must remember that they are making frame dependent measurements, so this is not a contradiction since the measurements are not directly corresponding to each other. For example, someone in a third frame travelling at half the speed of the travelling brother (and coincidentally meeting them at the same time that they coincide) would see the same time on both of their watches (which would be both slower than his). Alternatively, if one of the brothers now accelerates to the other's rest frame, he will find himself younger than his brother (it is not the acceleration with makes the time dilation, but the acceleration makes it observable by allowing them to make measurements in the same frame).[/quote']Do you mean: If the brother that stays on Earth gets onboard a spaceship and accelerates to the same speed as the other when he pass - then he will be the younger one ?

 

And if both accelerate to the third frame their clocks will be syncronized again ?

 

Somehow that seems both simple and logic, the frame dependency is the key.

[Johnny5]

OK' date=' what is wrong with this in principle:

 

Two clocks go by the Earth at near c in opposite directions.

 

As they go by they are synchronized by a signal from Earth.

 

Each assumes the other clock to be much much slower.

 

They go "straight" around the Universe/hypersphere.

 

When they meet again Earth is long gone.

 

As they pass by they prepare to resynchronize via an Earth station our descendants have built and positioned for the occasion.

 

How do their watches compare?[/quote']

 

There will be a single inertial frame, in which the following statement is true:

 

watch A reads less than watch B AND watch B reads less than watch A simultaneously.

 

if the time dilation formula is true.

 

Regards

 

PS: That they even get to compare watches twice, is a consequence which stems from the built in assumption in the problem, that the universe is 'hyperspherical' whatever that means.

[Severian]

Do you mean: If the brother that stays on Earth gets onboard a spaceship and accelerates to the same speed as the other when he pass - then he will be the younger one ?

 

And if both accelerate to the third frame their clocks will be syncronized again ?

 

Somehow that seems both simple and logic' date=' the frame dependency is the key.[/quote']

 

Yes, that's right. Its just the same as in the twin paradox: the paradox is absent because the travelling brother has undergone frame shifts, which the stationary brother has not.

 

Incidentally, you don't need a closed universe for this. You could just have the travelling brother travel in a very big circle. As long as the acceleration from the circular motion is very small (hence 'very big circle') it should have no noticable effect compared to the time dilation.

[Spyman]

Yes' date=' that's right. Its just the same as in the twin paradox: the paradox is absent because the travelling brother has undergone frame shifts, which the stationary brother has not.

 

Incidentally, you don't need a closed universe for this. You could just have the travelling brother travel in a very big circle. As long as the acceleration from the circular motion is very small (hence 'very big circle') it should have no noticable effect compared to the time dilation.[/quote']Then I think I understand, Thank You !

[J.C.MacSwell]

There will be a single inertial frame' date=' in which the following statement is true:

 

watch A reads less than watch B AND watch B reads less than watch A simultaneously.

 

if the time dilation formula is true.

 

Regards

 

PS: That they even get to compare watches twice, is a consequence which stems from the built in assumption in the problem, that the universe is 'hyperspherical' whatever that means.[/quote']

 

Nicely (and concisely) stated!

[swansont]

There will be a single inertial frame' date=' in which the following statement is true:

 

watch A reads less than watch B AND watch B reads less than watch A simultaneously.

 

if the time dilation formula is true.

[/quote']

 

There are two inertial frames, not one.

[Johnny5]

There are two inertial frames, not one.

 

In my response, i was thinking of the center of mass frame, when they come back to meet.

 

But each is moving at a constant speed in that frame, and hence both of those are inertial frames as well.

 

But I am assuming that as they approach each other, there is a point midway between them, which represents where they will eventually both be simultaneously, and that this is the CM of the system.

 

So, we could say that there are three inertial frames one could refer to, in this problem. I meant the CM frame, I should have said that.

 

Regards

[J.C.MacSwell]

I assumed Johnny meant the Earth station frame, but I think his statement is true (contradictory but true given my assumptions) for any choice of inertial frame.

[Johnny5]

I assumed Johnny meant the Earth station frame, but I think his statement is true (contradictory but true given my assumptions) for any choice of inertial frame.

 

Yes, I meant the earth station frame, which was built.

[swansont]

In my response' date=' i was thinking of the center of mass frame, when they come back to meet.

 

But each is moving at a constant speed in that frame, and hence both of those are inertial frames as well.

 

But I am assuming that as they approach each other, there is a point midway between them, which represents where they will eventually both be simultaneously, and that this is the CM of the system.

 

So, we could say that there are three inertial frames one could refer to, in this problem. I meant the CM frame, I should have said that.

 

Regards[/quote']

 

To an observer in the CM frame, both clocks should read the same.

[Johnny5]

To an observer in the CM frame, both clocks should read the same.

 

How do figure that?

[Guest volumeIII]

There will be a single inertial frame' date=' in which the following statement is true:

 

watch A reads less than watch B AND watch B reads less than watch A simultaneously.

 

if the time dilation formula is true.

 

Regards

 

PS: That they even get to compare watches twice, is a consequence which stems from the built in assumption in the problem, that the universe is 'hyperspherical' whatever that means.[/quote']

Johnny5,

Help me out here.

Assuming time dilation and your statement are true that after each complete the round trip of the universe, moving in straight line, A and B will see the other's watch slower than his own. As the ships pass each other at the point where earth was located when A and B started their voyage, each releases a piece if paper on which is printed the time of the A and B watches (each assumed slower than the other). What are the times printed on each piece of paper?

[swansont]

How do figure that?

 

They would both be travelling at the same v wrt the observer, and have the same dilation.

[Johnny5]

Johnny5' date='

Help me out here.

Assuming time dilation and your statement are true that after each complete the round trip of the universe, moving in straight line, A and B will see the other's watch slower than his own. As the ships pass each other at the point where earth was located when A and B started their voyage, each releases a piece if paper on which is printed the time of the A and B watches (each assumed slower than the other). What are the times printed on each piece of paper?[/quote']

 

The point of my response, was meant to imply that it cannot be the case that both SR is correct, and the universe is hyperspherical.

 

Here is the logically complete answer for you:

 

1. SR false and not (universe is hyperspherical)

2. SR true and not(universe is hyperspherical)

3. SR false and universe is hyperspherical

 

The possible choices are reduced by one, since both cannot be true.

 

So you can now do a case by case logical analysis.

 

Without knowing the exact meaning of "universe is hyperspherical" its difficult to carry out the logical analysis, but i think it has something to do with your direction isn't changing, yet you can return to your starting point.

 

This kind of space is impossible, but was the subject of a star trek episode.

 

The space ship entered some kind of highly warped region of space, and they couldn't get out, and had no way to know what direction they were moving in, because they couldn't see any stars no matter what direction they looked.

 

They dropped a sound buoy, that pinged the ship, and they began to travel away from the buoy in a straight line at a constant speed.

 

Sound pulses or whatever, were shot from the buoy in all directions, and obviously some hit the hull of the spaceship, and when that happened, there was a 'ping' recorded that everyone could hear.

 

As the ship moved further and further away from the buoy, the loudness of the 'ping's decreased, until they could no longer hear the buoy.

 

Now, they knew for sure that they were traveling in a straight line away from the buoy, and here's how.

 

On the bridge of the enterprise, they had a gyroscope.

 

The spin axis of the gyroscope was aligned with the ship's motion.

 

Had the enterprise strayed in any direction, the gimbals would have reoriented, but left the direction spin axis alone.

 

That's the special feature of a gyroscope, so they had one, and made sure that the gimbals never moved.

 

In other words, if the spin axis, started pointing at different objects on the bridge, they would know they were veering off a straight line path for certain, even though they couldn't see any stars.

 

They could then quickly compensate, so that the spin axis again, pointed at the front of the bridge. So the point is, they could control their speed, and their direction, and keep both constant in time.

 

 

In this episode they did this.

 

And yet, after a few minutes of moving away from the buoy, in a straight line at a constant speed, they began to hear a faint pinging sound, which got louder and louder, until they were back where they started, right next to the buoy.

 

Obviously, our real universe cannot be like this, it is truly impossible.

 

So if that kind of thing can happen in a hyperspherical universe, then the universe for sure isn't hyperspherical.

 

So that would eliminate #3 above.

 

But, if the meaning of hyperspherical universe isn't quite what i said, then some other analysis would be needed.

 

For example, suppose that as one of the twins moved along at a constant speed, his direction slowly changed, so that he really did travel in a circle.

 

And suppose his brother didn't move, i.e. his brother was the sound buoy.

 

So we are imagining the radius of the universe to be enormous, but circumnavigable.

 

So over any short amount of time, the tangential speed v of one of the twins is constant in time, but the direction of motion is slowly varying, and the centripetal acceleration of this particular twin is given by:

 

v^2/R

 

where R is the radius of the universe.

 

I don't think centripetal acceleration affects SR formulas, but I'm not sure.

 

The reason I'm not sure, is simply because i've never thought about it before, not because I don't know how to carry out the analysis.

 

So, in the case where the one twin remained at rest in some frame F, and the other twin entered a rocket, was given a brief impulse, and then just coasted for a very long time...

 

What SR says is undeniable...

 

Lets say that Gonzo remained at rest in inertial reference frame F, and just to help us visualize, let it be the case that the center of mass of Gonzo is also at rest relative to the center of the universe. Just to help you imagine.

 

Then identical twin Bonzo, boards a spacecraft, and the engines are turned on for just a few seconds.

 

Then the fuel runs out, and Bonzo and the spaceship just coast, at a constant speed v.

 

Suppose that the impulse given to Bonzo+spaceship was as follows:

 

Initial speed of Gonzo relative to Bonzo =0, then force is applied for three seconds, then all fuel used up, and no force is ever applied to ship again.

 

Final speed of Gonzo relative to Bonzo = .9999 c

 

Thus, Bonzo is smushed, but presume his wristwatch is still intact, and can circumnavigate the universe. Or if you want Bonzo to live, pretend his ship was superconductive, and that inertial forces are really due to external magnetic fields, and that advanced technology prevented him from being smushed, even under such extreme but brief acceleration. in either case the watch was undamaged.

 

 

Regardless of what you chose, what SR says about time dilation is very very easy to understand, and i will now explain what follows if time dilation formula is true:

 

Let Dt denote an amount of time as measured by Gonzo's watch.

 

Let Dt` denote an amount of time as measured by Bonzo's watch.

 

If there is no such thing as time dilation then

 

Dt = Dt`

 

Which means that if their watches were synchronous at the start of Bonzo's voyage, then upon his circumnavigation of the universe, what he writes on his paper will match what his brother writes as well.

 

At most, any discrepancy would be due to some effect which happened during the acceleration phase of Bonzo's trip, but that only lasted for a few seconds, and would be small.

 

So thats what would be the case if SR is false, if SR is true, then you have the following formula, which Gonzo has to use...

 

Dt = Dt` (1-vv/cc)^-1/2

 

keep in mind, the LHS is what is measured by Gonzo's watch, a 24th century digital timex.

 

Now, the delta t` on the RHS is what Gonzo is supposed to compute to be the amount of time that passes aboard his brother's spaceship. In other words, spaceship time. v denotes the relative speed of Gonzo to Bonzo, or equivalently the speed of the ship in reference frame F.

 

c is the speed 299792458 meters per second.

 

You can now infer that time passes slower aboard the ship, relative to external reference frame F's time.

 

Now, recall that the impulse only lasted for three seconds, but that afterwards, the relative speed of Gonzo to Bonzo was .9999 c.

 

Hence we have this here:

 

Dt = Dt` (1-.99980001)^-1/2

 

Since (.9999)(.9999)=.99980001

 

Now, 1-.99980001 = .00019999

 

so that we have:

 

Dt = Dt` (.00019999)^-1/2

 

So that we have:

 

 

Dt = Dt` (5000.25)^1/2

 

 

since

 

1/.00019999 = 5000.25 and 1/x = x^-1 for any x

 

Now we just take the square root to obtain the following statement, which must be true in the Gonzellian frame F:

 

Dt = Dt` 70.7

 

So, for the sake of using numbers, and not letters, let us suppose that after 20 billion years have passed according to Gonzo's watch, his brother finally is coming into view.

 

What does special relativity theory say that Bonzo's watch will read?

 

Well before trying to answer this, let's figure out the radius R of the universe first. I think i've introduced enough information for that to be computable now.

 

We know the path traveled by Bonzo was a circle, we are stipulating that there were no perturbations to the uniform circular motion of his ship for any reason.

 

Letting R denote the radius of the universe, the distance traveled by Bonzo, in Gonzo's frame is given by the Archimedian formula :

 

C = 2pR

 

We have stipulated that the amount of time of the Bonzo's whole trip, as measured by Gonzo's 24th century digital timex, was 20 billion years.

 

Now lets convert that to seconds.

 

20 billion years = 6.3 x 10¹⁷ seconds

 

Now, Bonzo's tangential speed v, was stipulated to be .9999c throughout the duration of his ship (well all but three seconds of it anyways).

 

So, we do have enough information to compute the radius R of the universe.

 

v=.9999c = distance traveled/time of travel

 

hence:

 

.9999c = 2pR/(6.3 x 10¹⁷ seconds)

 

and c = 299792458 meters per second hence:

 

.9999(299792458 m/s) = 2pR/(6.3 x 10¹⁷ s)

 

This is one equation in one unknown, the unknown is R, hence we can solve for R, as an explicit number, doing so we find that:

 

R = 3.0005 x 10²⁵ meters

 

So, Gonzo has assumed that the time dilation formula is true, hence Gonzo must use the following formula:

 

Dt = Dt` 70.7

 

And the time, as measured by Gonzo's wristwatch was:

 

6.3 x 10¹⁷ seconds

 

Hence, Gonzo reaches the conclusion that:

 

6.3 x 10¹⁷ = Dt` 70.7

 

From which it follows that:

 

8.9 x 10^15 seconds = Dt`

 

Which we can transform to years.

[Johnny5]

They would both be travelling at the same v wrt the observer, and have the same dilation.

 

Hmm i need a moment. let me answer the previous thing, and I will come back and think about this.

[swansont]

each complete the round trip of the universe, moving in straight line

 

How does one manage a round trip, travelling in a straight line? Either one or both of them have to accelerate, or you have to have curved space, which AFAIK is going to act like an acceleration (and was the point of the OP)

[swansont]

The point of my response' date=' was meant to imply that it cannot be the case that both SR is correct, and the universe is hyperspherical.

[/quote']

 

 

You've neglected the case where a hyperspherical universe violates one of the assumptions of SR.

[Johnny5]

You've neglected the case where a hyperspherical universe violates one of the assumptions of SR.

 

Can you be more precise?

[J.C.MacSwell]

How does one manage a round trip, travelling in a straight line[/b']? Either one or both of them have to accelerate, or you have to have curved space, which AFAIK is going to act like an acceleration (and was the point of the OP)

 

The assumption would be straight lines are geodesics on the surface (read volume) of the hypersphere. If this acts like acceleration for a round trip then everything works out nicely as the symmetry balances everything out.

 

However:

 

If this was the case for a round trip, it would equivalently be the case for any portion of a round trip.

 

Simultaneity would be alive and well.

 

(note to casual readers: this is based on a specific set of hypothetical assumptions stated or implied in post 29, not the real world)

[swansont]

Can you be more precise?

 

 

As is mentioned in a few other posts, the hypersphere does not constitute an inertial frame, which means you can't blindly apply SR to it.

 

As Tom Mattson suggested, one should work out the case of two travellers who, in flat space, travel away from each other, turn around, and return. It seems to me that this shows the dynamics of the issue more clearly. They hypersphere scenario is similar, but has a small acceleration over the whole trip, and it makes it harder to see what's going on. Though once you've solved the simpler case, it will make the hypersphere case a little easier to understand. Walk before you run, and all that.

[Johnny5]

As is mentioned in a few other posts' date=' the hypersphere does not constitute an inertial frame, which means you can't blindly apply SR to it.

[/quote']

 

I was aware of this Dr. Swanson, because, Bonzo is traveling in a space ship which isn't subjected to any external forces, and hence should move in a straight line at a constant speed, in Gonzo's frame F, which was stipulated to be inertial. Yet because the universe was assumed hyperspherical, Bonzo did in fact travel a curved path in Gonzo's inertial frame.

 

That pushes the confusion to definition of inertial frame vs definition of hyperspherical universe.

 

I'm not sure of the definition of 'hyperspherical universe' but then again, the universe isn't hyperspherical, in the sense that the enterprise, guided using a gimbal to keep it moving straight, will not return to the sound buoy.

 

Regards