Jump to content

The twin paradox and other variants.


mistermack
 Share

Recommended Posts

A recent thread touched on the subject, but not in a satisfactory manner and the thread got locked.

I think that the twin paradox goes right to the heart of SR, so should have it's own thread.

I hope this one won't get hijacked with irrelevant stuff.

 

As someone who has struggled to reconcile the case of travelling twins and it's apparent paradox with the interchangeability of inertial frames in Special Relativity, I want to share my progress with others who might want a clearer picture.

 

I just found two pages on the net that give a brilliantly clear explanation for the outline of the paradox, and the explanation for some of it, and I recommend them to anyone: Just click on the links :

 

The spotlight topic The case of the travelling twins.

 

The spotlight topic Twins on the road

 

They are pages from Einstein-online, and I recommend them as thoroughly clear and readable.

What I haven't found there yet, is an analysis of the case where a third twin is introduced, apparently removing the role that acceleration is playing. I'll keep looking at that.

 

 

 

 

 

 

Link to comment
Share on other sites

I thought that Einstein belatedly invoked the moving twin's acceleration-deceleration-(turnabout)-acceleration-deceleration as being the key to explaining why the paradox wasn't a paradox at all.

But those links above dismiss the need to invoke any acceleration (& deceleration).

 

I guess that the above links dismiss acceleration (& deceleration) to counter the more modern example of the paradox (this might be the third twin referred to).

If i remember aright the modern form uses a third clock C (heading towards the stationary twin-clock A) which passes the outgoing twin-clock B, & the 2 clocks C & B instantly synchronise (thusly eliminating the deceleration-acceleration at the turn point).

Perhaps they also eliminate the initial acceleration by having twin B already moving when A & B first meet (at the start), & then eliminate the deceleration at the end by simply comparing clocks C & A as they pass.

 

I would be surprised if the above links satisfactorily explain that the paradox doesn't exist (without invoking acceleration). Einstein had to invoke acceleration.

Edited by madmac
Link to comment
Share on other sites

I would be surprised if the links satisfactorily explain that the paradox doesn't exist.

 

 

Why not try reading them and find out. (Clue: of course they do. Because it doesn't.)

 

 

 

Einstein was struggling even when using acceleration.

 

Nonsense.

Please stop posting your ignorant twaddle in the science forums. Note that the OP requested that this wasn't hijacked by irrelevant stuff (i.e. anything you might say).

Link to comment
Share on other sites

!

Moderator Note

Satisfactory explanation is in the eye of the viewer, and commentary of that sort is not up for discussion here. Stay on-topic. Further distractions will be hidden.

 

(As an aside: in SR clocks moving at different speeds can't be synchronized. A short discussion as to why can take place in another thread if one is interested)

Link to comment
Share on other sites

I offered a analysis in the locked post that addressed this issue. Introducing the third observer as a "third twin" is a bit misleading. As I pointed out in may last analysis, there is no way to arrange things with this set up where all three observers will agree that all three are the "same age" at the same time. The only way that three observers with different relative motions with respect to each other will agree on each other's respective clock readings is if they are co-located. In this scenario, two of the observers are co-located at the start and two at the end, but only one of them in the same observer in both cases. At no point are all three ever co-located.

 

The whole idea of "resolving" the twin paradox to to show how the two twins will agree at both the start and finish; that, while they may disagree as to what is happening during various points of the experiment, they both end up agreeing on the same end result.

 

In this respect (the only one that really matters), the so-called "triplet" paradox is even easier to explain. None of the observers change inertial frames, so all we have to do is examine it from the three different inertial frame.

 

The following diagrams present the three rest frames for the three observers Blue is our "stay at home" observer, Green starts co-located with him and then travels away, and Red starts some distance away, and passes Green on his way to joining the "stay at home" observer.

 

These diagram also show the light carrying visual information so we can note not only what each frame says the respective clock readings are, but what each observer would visually see as happening.

 

We start with Blue.

 

post-222-0-44028800-1491496877_thumb.gif

 

Green and Blue start with the same reading. As Green recedes, its clock ticks at a slower rate. And by the time it meets B Blue's Time reads 4 and Green's 3.46 (up to this point is doesn't matter what Red's time is, as it will be set to match Green's at this point.)

When Red meets Green, Blue visually sees Green's time as being somewhere between 2-2.5, and won't actually see Red and Green meet until his own clock reads ~6. Thus while Green's time goes from 0 to 3.46 while his clock goes from 0 to 4(time dilation) he sees Green's time tick from 0 to 3.46 while his clock goes from 0 to 6(Doppler shift).

 

From this point on We can ignore Green and concentrate on Red. Red goes from 3.46 to 6.92 while Blue goes from 4 to 8(time dilation), and Blue sees Red go from 3.46 to 6.92 while he goes from 6 to 8(Doppler shift).

No matter which way you examine it, Green leaves reading zero, transfers a clock reading of 3.46 to Red and then arrives at Blue reading 6.92.

 

Now we look at Red's rest frame.

post-222-0-18068100-1491496892_thumb.gif

 

Both Blue and Green start a equal distance away and move towards him, with Green approaching faster, Both clocks tick slow, with Green's ticking the slowest.

He will visually see Green ticking the fastest, and only a short period of time passes between his seeing Green and Blue separate and his meeting up With Green. Green reads 3.46 at this moment and Red sets his clock accordingly.

From this moment on, he no longer cares what happens to Green.

At this moment Blue reads 5, but he visually sees a reading of ~2. and while Blue goes from 5 to 8 while he goes from 3.46 to 6.92(time dilation), in this same period, he sees blue tick from 2 to 8(Doppler shift). In either case, he meets up with Blue as his own clock reads 6.92 and Blue's reads 8

Once again, Blue and Green separate reading the same time, Red meets Green with Green transferring a clock reading of 3.46 to Red and Red and Blue meet up with blue reading 8 and Red 6.92

 

Lastly, Green's rest frame.

post-222-0-30748600-1491500761_thumb.gif

 

Blue separates from Green with its Clock ticking slower. Red starts some distance from Green and approaches at a faster speed than Blue is receding.

When Red reaches Green, Greens clock reads 3.46, which is transferred to Red. Blue reads ~3 at this moment (time dilation), and Green sees Blue reading ~2.

Sometime after Greens clock reads 9, Red catches up to Blue. Red will have ticked from 3.46 to 6.92 while Blue goes from 3 to 8, both due to time dilation. Green does not See red catch Blue until his own clock reads ~14. Thus while his clock goes from 3.46 to 14, he sees Blue's clock go from 2 to 8 and Red's clock go from 3.46 to 6.92, both due to Doppler shift.

 

We start and end with the same results as in the last two frames.

 

All three observers agree upon the outcome and there is no "paradox".

Link to comment
Share on other sites

I offered a analysis in the locked post that addressed this issue. Introducing the third observer as a "third twin" is a bit misleading. As I pointed out in may last analysis, there is no way to arrange things with this set up where all three observers will agree that all three are the "same age" at the same time. The only way that three observers with different relative motions with respect to each other will agree on each other's respective clock readings is if they are co-located. In this scenario, two of the observers are co-located at the start and two at the end, but only one of them in the same observer in both cases. At no point are all three ever co-located.

 

The whole idea of "resolving" the twin paradox to to show how the two twins will agree at both the start and finish; that, while they may disagree as to what is happening during various points of the experiment, they both end up agreeing on the same end result.

 

In this respect (the only one that really matters), the so-called "triplet" paradox is even easier to explain. None of the observers change inertial frames, so all we have to do is examine it from the three different inertial frame.

 

The following diagrams present the three rest frames for the three observers Blue is our "stay at home" observer, Green starts co-located with him and then travels away, and Red starts some distance away, and passes Green on his way to joining the "stay at home" observer.

 

These diagram also show the light carrying visual information so we can note not only what each frame says the respective clock readings are, but what each observer would visually see as happening.

 

We start with Blue.

 

attachicon.giftrip.gif

 

Green and Blue start with the same reading. As Green recedes, its clock ticks at a slower rate. And by the time it meets B Blue's Time reads 4 and Green's 3.46 (up to this point is doesn't matter what Red's time is, as it will be set to match Green's at this point.)

When Red meets Green, Blue visually sees Green's time as being somewhere between 2-2.5, and won't actually see Red and Green meet until his own clock reads ~6. Thus while Green's time goes from 0 to 3.46 while his clock goes from 0 to 4(time dilation) he sees Green's time tick from 0 to 3.46 while his clock goes from 0 to 6(Doppler shift).

 

From this point on We can ignore Green and concentrate on Red. Red goes from 3.46 to 6.92 while Blue goes from 4 to 8(time dilation), and Blue sees Red go from 3.46 to 6.92 while he goes from 6 to 8(Doppler shift).

No matter which way you examine it, Green leaves reading zero, transfers a clock reading of 3.46 to Red and then arrives at Blue reading 6.92.

 

Now we look at Red's rest frame.

attachicon.giftrip2.gif

 

Both Blue and Green start a equal distance away and move towards him, with Green approaching faster, Both clocks tick slow, with Green's ticking the slowest.

He will visually see Green ticking the fastest, and only a short period of time passes between his seeing Green and Blue separate and his meeting up With Green. Green reads 3.46 at this moment and Red sets his clock accordingly.

From this moment on, he no longer cares what happens to Green.

At this moment Blue reads 5, but he visually sees a reading of ~2. and while Blue goes from 5 to 8 while he goes from 3.46 to 6.92(time dilation), in this same period, he sees blue tick from 2 to 8(Doppler shift). In either case, he meets up with Blue as his own clock reads 6.92 and Blue's reads 8

Once again, Blue and Green separate reading the same time, Red meets Green with Green transferring a clock reading of 3.46 to Red and Red and Blue meet up with blue reading 8 and Red 6.92

 

Lastly, Green's rest frame.

attachicon.giftrip3.gif

 

Blue separates from Green with its Clock ticking slower. Red starts some distance from Green and approaches at a faster speed than Blue is receding.

When Red reaches Green, Greens clock reads 3.46, which is transferred to Red. Blue reads ~3 at this moment (time dilation), and Green sees Blue reading ~2.

Sometime after Greens clock reads 9, Red catches up to Blue. Red will have ticked from 3.46 to 6.92 while Blue goes from 3 to 8, both due to time dilation. Green does not See red catch Blue until his own clock reads ~14. Thus while his clock goes from 3.46 to 14, he sees Blue's clock go from 2 to 8 and Red's clock go from 3.46 to 6.92, both due to Doppler shift.

 

We start and end with the same results as in the last two frames.

 

All three observers agree upon the outcome and there is no "paradox".

I wish I could give you +10. The way it stands, I could only give you +1.

A few notes:

1. It is sufficient to do the analysis for half trip only. At the half trip, the travelling twin has clocked 3.46 and the stay at home twin has clocked 4. After that, symmetry takes over.

2. If one wants to take acceleration into account, for a more realistic analysis, I wrote these two sections for wiki long ago. The perspective of the travelling twin is most interesting.

Link to comment
Share on other sites

I wish I could give you +10. The way it stands, I could only give you +1.

A few notes:

1. It is sufficient to do the analysis for half trip only. At the half trip, the travelling twin has clocked 3.46 and the stay at home twin has clocked 4. After that, symmetry takes over.

2. If one wants to take acceleration into account, for a more realistic analysis, I wrote these two sections for wiki long ago. The perspective of the travelling twin is most interesting.

 

I wrote these two sections for wiki long ago.

!

Moderator Note

 

Oh Dear - That means you are xyzt and I claim my five pounds

 

And I am pretty sure we banned xyzt for being an insufferable oik who was rude and abusive (check), downvoted those who dared to disagree (check), whined about being neg-repped which we though was hypocrictical then too (check) had good maths and great latex (check), a great ability with relativity (check), but big holes in his understanding about which he would rabidly attack anyone when they were pointed out (check), and the complete absence of humility when he misspoke and was called on it (final check).

 

Account suspended pending mod-review

 

 

Link to comment
Share on other sites

Even though I like the explanations that I linked to in the OP, from Einstein-online, I'm beginning to have problems with them, having read them a few times.

Particularly, the second, The spotlight topic Twins on the road

 

The essence of that is that the acceleration changes everything, and after the acceleration, the inertial motion of the second twin is somehow different, for the rest of the trip. (as in the case of the hiker striking off at an angle).

With the hiker, you can clearly see what the difference is. He is straying into another of the spatial dimensions, now travelling in two dimensions, instead of one straight line.

So what is the equivalent difference in the case of the twins paradox? What clear difference is there in the second twin, apart from the speed difference?

Two non-accelerating twins with just a speed difference, we are told, have exchangeable situations. Each experiences time slowed for the other, and both are right.

But according to this, if the speed difference is the result of acceleration, then all bets are off. One twin is ACTUALLY ageing faster than the other, and the two situations are no longer symmetrical. So when they meet up, one is substantially older than the other.

 

I can't see that this is any different, from two twins who HAVEN'T undergone acceleration, but are just in motion relative to each other.

For instance, twin B leaves twin A, being accelerated to speed S.

What if that acceleration takes him right alongside a third twin, C, who was already travelling at a constant speed S ?

You then have two twins, travelling side-by-side at the same speed. Their clocks MUST run at the same speed.

 

So according to that, a clock that has been accelerated, runs no differently to a similar clock that is just in constant inertial motion.

 

If you are to claim that twin B is ACTUALLY ageing slower than A, then that also applies to twin C.

 

So the consequence is that for two twins A and C, in constant motion relative to each other, one is ACTUALLY aging faster than the other.

Link to comment
Share on other sites

I posted a comments on the twins in this recent thread, now closed.

 

No one seems to have taken up on as a way of avoiding the acceleration question.

 

I don't know if I may refer directly to the thread but the post was #109.

Link to comment
Share on other sites

Even though I like the explanations that I linked to in the OP, from Einstein-online, I'm beginning to have problems with them, having read them a few times.

Particularly, the second, The spotlight topic Twins on the road

 

The essence of that is that the acceleration changes everything, and after the acceleration, the inertial motion of the second twin is somehow different, for the rest of the trip. (as in the case of the hiker striking off at an angle).

With the hiker, you can clearly see what the difference is. He is straying into another of the spatial dimensions, now travelling in two dimensions, instead of one straight line.

So what is the equivalent difference in the case of the twins paradox? What clear difference is there in the second twin, apart from the speed difference?The equivalence is that while in this example we are dealing with two spacial dimensions, With the two hikers taking two different routes in that two dimensional space, With the twin paradox, we are dealing with 3 spatial and 1 temporal dimensions wrapped up into space-time, and the twins take different routes through space-time. What complicates this a bit further is that the directions of these dimensions are not fixed but are frame dependent.

To illustrate this, we will look at the diagram from that link again:

post-222-0-98029200-1491588539.png

The image shows Emma walking to the right and up a bit, while Albert first walks to the right and up and then to the right and down. The problem is that Albert and Emma don't see it this way, all they see is that at all times they are walking "forward" with forward defined as the direction they are facing. Thus when Albert first gets to point C, Emma, assuming she is walking at the same pace, is at the position of the Black dot. In order to be abreast of Albert, she would have to be where the red arrow intersects her path. Thus from Albert's perspective, she is "behind" him.

Albert makes his turn. Now the yellow arrow represents being abreast of him, and Emma is now "ahead" of him. Emma did not rush to pass him, neither did he slow down to allow her to. Her position with respect to him has changed relative to the direction he is facing. Note that if we draw a line from Emma's black dot indicating being abreast of her, Albert will be behind her. This is the equivalent of Time dilation where each observer measures the other clock as running slow.

Go back and look at the diagrams for Red and Greens frames I provided in my last post. Note that according to Green when he meets up with Red, the Blue's clock reads 3. According to Red, at that same moment, Blue's clock reads 5. If Green, instead of continuing on, were to accelerate and match speeds with Red, then after the acceleration, his diagram will match Red's and Blue's clock now reads 5. Just like Emma went from being behind of to ahead of Albert by virtue of his changing direction, Blue's clock goes from 3 to 5 by virtue of Green's change of inertial frame.

In space-time terms, Blue, Red and Green have different time axis, and when Green accelerates his time axis "Rotates" with him until it matches Red's time axis.

Two non-accelerating twins with just a speed difference, we are told, have exchangeable situations. Each experiences time slowed for the other, and both are right.

But according to this, if the speed difference is the result of acceleration, then all bets are off. One twin is ACTUALLY ageing faster than the other, and the two situations are no longer symmetrical. So when they meet up, one is substantially older than the other.

No, it doesn't. Acceleration is the equivalent to a rotation in space-time, it changes your perspective of space-time compared to your original frame of reference, but it does not "physically" affect your clock rate.

I can't see that this is any different, from two twins who HAVEN'T undergone acceleration, but are just in motion relative to each other.

For instance, twin B leaves twin A, being accelerated to speed S.

What if that acceleration takes him right alongside a third twin, C, who was already travelling at a constant speed S ?

You then have two twins, travelling side-by-side at the same speed. Their clocks MUST run at the same speed.

 

So according to that, a clock that has been accelerated, runs no differently to a similar clock that is just in constant inertial motion.

 

If you are to claim that twin B is ACTUALLY ageing slower than A, then that also applies to twin C.

No one is saying that acceleration has a physical effect on the operation of a clock; that idea comes from a misinterpretation on your part of the example given.

So the consequence is that for two twins A and C, in constant motion relative to each other, one is ACTUALLY aging faster than the other.

No it isn't. Your are bringing preconceived notions into your attempts to understand Relativity that are getting in your way. Until you can shed yourself of them, you will continue to be frustrated.
Link to comment
Share on other sites

Janus, try as I might, I can't follow your method of argument. (I'm not claiming that anything's wrong with it, I just find it hard to follow)

 

I know that in SR, nothing can be regarded as simultaneous, unless it's in the same place at the same velocity.

That doesn't mean that things AREN'T happening simultaneously. It just means we can't observe it at the time.

 

But looking at the past, it's obvious that things WERE happening simultaneously.

 

It's a similar case with these twins. When they get back together, and one is ten times older than the other, it's obvious that one's clock has been running slower than the other's. You can't end up younger, unless your clock was running slower.

 

You could question on which leg of the trip that happened. But there's no question that on one, or both legs, clocks were actually running slower for the travelling twin. Not just observed to be slower, but actually slower.

For the duration of that leg, neither twin was undergoing acceleration.

Why are the two twins's experiences not interchangeable?

Link to comment
Share on other sites

Janus, try as I might, I can't follow your method of argument. (I'm not claiming that anything's wrong with it, I just find it hard to follow)

 

I know that in SR, nothing can be regarded as simultaneous, unless it's in the same place at the same velocity.

That doesn't mean that things AREN'T happening simultaneously. It just means we can't observe it at the time.

That's not quite right, but it'd derail this thread to get into it.

 

But looking at the past, it's obvious that things WERE happening simultaneously.

 

It's a similar case with these twins. When they get back together, and one is ten times older than the other, it's obvious that one's clock has been running slower than the other's. You can't end up younger, unless your clock was running slower.

 

You could question on which leg of the trip that happened. But there's no question that on one, or both legs, clocks were actually running slower for the travelling twin. Not just observed to be slower, but actually slower.

For the duration of that leg, neither twin was undergoing acceleration.

Why are the two twins's experiences not interchangeable?

Take a step back and look at post #5 again.

 

In picture 1, the journey of the "travelling twin" is shown by the green line 'till half way, then the red line. Janus does it this way to show that it's not specifically the acceleration that directly causes the asymmetry, but the multiple inertial frames *. (Instead of one clock that accelerates at half way to go back to blue, green passes their clock reading to red at 3.46).

 

Note that in all of the pictures, i.e. from blue's rest frame, green's rest frame and red's rest frame, there's a single line for blue - they stay in one rest frame. This is different than the travelling twin, their journey is made up of green and red - two rest frames.

 

Also, at all times, each observer (blue, red, and green) considers the other two observers to have slow clocks. But when red gets back to blue, red's clock actually does read less than blues. That first picture shows why. The situation is not symmetrical.

 

 

* of course, for the travelling twin in the usual thought experiment, it is acceleration that causes them to have been in those multiple rest frames. But that's not the point in post #5.

 

It may be worth pointing out that acceleration is absolute. Imagine stay at home twin and travelling twin both holding very full cups of coffee. Travelling twins' rocket blasts off, travels for a while, then turns and blasts again to go back home. Which twin spilled their coffee? Which twin didn't? This just illustrates that the situation isn't symmetrical, and the stay-at-home twin has stayed in a single rest frame the whole time.

Edited by pzkpfw
Link to comment
Share on other sites

Janus does it this way to show that it's not specifically the acceleration that directly causes the asymmetry, but the multiple inertial frames *.

My problem with that is that if you just take the case of the outward leg, then you have symmetry.

Does the outward clock run slower?

 

And if you just take the case of the return leg, then you have symmetry.

Does the return clock run slower?

Link to comment
Share on other sites

The outward and return journeys cannot be symmetrical.

 

Yes the travelling clock appears to run more slowly to the stay-at-home clock on both legs, but,

 

On the outward leg travel is away from the base so light (or any other) signals take longer and longer to traverse the expanding distance between the two.

 

On the return leg it is the other way round the separating distance is diminishing so signals arrive more and more quickly.

 

This fact has nothing to do with Einstinian relativity, but must be factored in to any analysis.

Link to comment
Share on other sites

The point I'm making isn't about how the clocks appear. It's about how they ran.

 

If the moving twin returns and is one tenth as old as his stationary twin, then his clock was at some point running slower.

It could be on the outward trip, the inward trip, or both.

 

If you just focus on the interval in which his clock is running slower than his twin's, then he is moving at a steady rate relative to his twin.

That's ALL that's happening.

Two twins, separated by distance, moving relative to each other. And one's clock is going faster, the other is going slower. In actual terms, as proven by the different ages when they meet up.

Link to comment
Share on other sites

Try one small step at a time: see Post #5 ...

 

(Ignore the yellow lines for this, don't worry about when they see each others clocks.)

 

A: As seen in pic 1, according to the stay at home twin (blue) what does their (blue) clock read when the travelling twin (green) reaches the halfway point (meets red), and what does the travelling twin clock (green) read?

 

B: As seen in pic 3, according to the travelling twin (green) what does their (green) clock read when they reach the halfway point (meet red), and what does the stay at home (blue) clock read?

 

C: As seen in pic 2, according to the travelling twin (red) what does their (red) clock read * when they reach the halfway point (meet green), and what does the stay at home (blue) clock read?

 

* remember that red got their reading from green here.

 

Something important happens when you compare B: with C: - take note of that (especially the blue clock as seen by green and red) then read the "Relativity of simultaneity" section here: https://en.wikipedia.org/wiki/Twin_paradox#Relativity_of_simultaneity

Edited by pzkpfw
Link to comment
Share on other sites

pz and Janus, I would prefer it if someone pointed out what is actually wrong with what I said, rather than posting the diagrams etc.

 

As far as simultaneity goes, Wikipedia says : "one must understand that in special relativity there is no concept of absolute present."

 

I'm fully prepared to accept that there is no way to work with an absolute present, that doesn't mean that it doesn't exist.

In fact, if the past doesn't exist, and the future doesn't exist, then the absolute present is all that exists.

 

Of course special relativity doesn't work with it, but that's not the same as "it doesn't exist".

 

If I look at two supernovas, both 1000 light years from me, in opposite directions, I can say with confidence that they were happening simultaneously 1000 years ago. Even though they couldn't see each other or interact.

Link to comment
Share on other sites

Of course special relativity doesn't work with it, but that's not the same as "it doesn't exist".

 

 

!

Moderator Note

For the purposes of a scientific argument, yes it does. SR is the accepted science, and we're discussing SR. If anyone wants to delve into the quagmire of "the present" and simultaneity, do so in another thread.

Link to comment
Share on other sites

The point I'm making isn't about how the clocks appear. It's about how they ran.

 

 

 

There is no such thing as absolute time.

 

So there is no such thing as absolute 'running' of a clock.

 

This is another way of saying or a consequence of the fact that every observer will see any clock running differently.

 

You will not progress until you accept this basic fact.

Edited by studiot
Link to comment
Share on other sites

pz and Janus, I would prefer it if someone pointed out what is actually wrong with what I said, rather than posting the diagrams etc.

 

As far as simultaneity goes, Wikipedia says : "one must understand that in special relativity there is no concept of absolute present."

 

I'm fully prepared to accept that there is no way to work with an absolute present, that doesn't mean that it doesn't exist.

In fact, if the past doesn't exist, and the future doesn't exist, then the absolute present is all that exists.

 

Of course special relativity doesn't work with it, but that's not the same as "it doesn't exist".

 

If I look at two supernovas, both 1000 light years from me, in opposite directions, I can say with confidence that they were happening simultaneously 1000 years ago. Even though they couldn't see each other or interact.

For you, and anybody at rest with respect to you, they will have happened simultaneously. For someone moving at some velocity relative to you (As long as one component of that velocity is parallel to the line joining the two) they will not be simultaneous, even if he were passing you at the same instant as you saw the two supernovae and also saw them explode at the same time. the nonexistence of an absolute simultaneity does not mean that no frame can determine what distant events are simultaneous in that frame, but that frame with different relative motion will not agree as to which events are simultaneous.

This idea that absolute simultaneity must exist is one of those preconceived notions that you are going to have to shed.

 

It is the kind of shift in thinking like that that had to be made to accept a spherical Earth vs a flat one. With a flat Earth, "down" was an absolute and universal direction throughout reality. People stood upright on the World because they were on the "topside". A person trying to stand on the "underside" would fall off.

With a spherical Earth this has to be abandoned. "Down" now pointed to the center of the world. A person standing at the North Pole and one standing at the South pole can both say that they are standing upright even though their heads are pointing in opposite directions. To argue that one is "really" standing upright, and the other just "thinks" he is, is an attempt to hold on to the old "Absolute down" model.

In the same way that the spherical Earth model does away with absolute down, Relativity requires one to do away with absolute time. To claim that one clock has to be "really" running faster than the other is equivalent to sating that the person at the North pole is "really" the one on the Top side of the world.

 

So in my diagrams, the Horizontal direction is "Now" for that frame. For Green's frame when he meet's Red, "right now" Blue's clock reads 3. In Red's frame when he meets Green, "right now" Blues clock reads 5. Red and Green agree on the "right now" of when they meet, but they don't agree as to what time it is for Blue "right now". For Blue, when his clock reads 4 Green and Red are meeting "right now".

 

You need to grasp that the idea of "right now" as an absolute cannot be maintained.

 

Changing deep held ideas can be challenging, but well worth it once the "light goes on".

Edited by Janus
Link to comment
Share on other sites

To claim that one clock has to be "really" running faster than the other is equivalent to sating that the person at the North pole is "really" the one on the Top side of the world.

Except that the evidence of meeting the two twins, one aged twenty, the other aged two, would prove that one's clock "really" ran faster than the other.

One is holding a clock reading twenty years, the other is holding a clock reading two years.

I think the difference in the two twins would also be pretty obvious.

 

I don't know what more proof you would need than that.

 

If the proof of clocks "really" running faster and slower than each other in the past is so strong and obvious, I can't see the problem with accepting that it can be happening in the present.

Link to comment
Share on other sites

Emphasis mine:

[..]

Two non-accelerating twins with just a speed difference, we are told, have exchangeable situations. Each experiences time slowed for the other, and both are right.

 

Obviously that's a contradiction: at best one can be right (except of course that they can be both right about their experience; but I doubt that you mean that).

 

 

[..]

So the consequence is that for two twins A and C, in constant motion relative to each other, one is ACTUALLY aging faster than the other.

 

I agree with your reasoning (and so do many others); however the point of SR is that we cannot determine which one.

Link to comment
Share on other sites

pz and Janus, I would prefer it if someone pointed out what is actually wrong with what I said, rather than posting the diagrams etc.

...

You're wanting a simple intuitive understanding to be handed to you on a platter. With relativity that's pretty hard to do.

 

Please, just do the little exercise in post #18.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.