I How Does Special Relativity Affect Perceived Distances Between Planets?

[dwspacetime]

TL;DR Summary

Relative distance

Quite simple question. There are two planets A and B, they are 4 light years apart measured by tom on planet A. Say both A and B remain stationary. Mary flys 0.8c to tom on planet A. When she passes planet B she should see Tom or planet A 2.4 light years away, am I correct? What about tom now? Does he see both planet B and Mary 4 light years away? Thanks

 

[Orodruin]

At the moment Mary flies past planet B, she is 4 ly away in Tom’s rest frame.

 

[Orodruin]

However …

What about tom now?

… you have inadvertently stumbled into the hornet’s nest. What do you mean by “now”? It does not mean the same thing to Tom as it does to Mary because of the relativity of simultaneity. Therefore, more care must usually be taken to specify which “now” one is talking about.

However, Tom will always find Mary to be 4 ly away in his rest frame when she passes B, because B is 4 ly away.

You also should be careful not to interpret “see” as in actually seeing. When Mary passes B it will still be 4 years (in Tom’s rest frame) until Tom actually sees the light from that event. Mary at that time (again in Tom’s rest frame) will only be 0.8 ly away, but he will just see her as passing B 4 ly away.

I find it less confusing for students if instead of “see” I use language as above. For example:
Mary is 4 ly away at that time in Tom’s frame.

 

[dwspacetime]

At the moment Mary flies past planet B, she is 4 ly away in Tom’s rest frame.

That is right when we know Tom is at rest. Let's take out both planets which could make us unconsciously theat them as a reference. Both Tom and Mary can equally think the himself or herself moving towards the other at 0.8c so both could see the other 2.4 light years away.

 

[Orodruin]

That is right when we know Tom is at rest. Let's take out both planets which could make us unconsciously theat them as a reference. Both Tom and Mary can equally think the himself or herself moving towards the other at 0.8c so both could see the other 2.4 light years away.

No. Please see my second post.
You are confusing yourself by forgetting that “now” means different things to Tom and Mary.

 

[Ibix]

so both could see the other 2.4 light years away.

Depends on what time you mean. They're moving relative to each other, and you just removed the landmarks! At some point they'll measure the other to be 2.4 ly away, sure.

 

[dwspacetime]

No. Please see my second post.
You are confusing yourself by forgetting that “now” means different things to Tom and Mary.

But when Mary passes B is tom 2.4 or 4 ly away from her frame?

 

[Sagittarius A-Star]

When she passes planet B she should see Tom or planet A 2.4 light years away, am I correct?

She calculates, that, with reference to her standard inertial rest frame, planet A is 2.4 LY away. Planet A will reach her in 3 years, according to her watch.

 

[Ibix]

But when Mary passes B is tom 2.4 or 4 ly away from her frame?

If Tom is at rest with respect to the planets, Mary will measure him to be 2.4ly away; Tom will measure Mary to be 4ly away.

They are measuring different things and both calling them "the distance", of course.

 

[dwspacetime]

She calculates, that, with reference to her standard inertial rest frame, planet A is 2.4 LY away. Planet A will reach her in 3 years.

That is exactly it does make much sense why u don't want the planets to confuse you. Because they make you think Tom is at rest but Mary is moving. But Mary could think she is at rest but tom is moving to her and he should think exact the same as whatever Mary is thinking. If one person fly's away from the other they both can claim the others time is slower.

 

[Orodruin]

That is exactly it does make much sense why u don't want the planets to confuse you. Because they make you think Tom is at rest but Mary is moving. But Mary could think she is at rest but tom is moving to her and he should think exact the same as whatever Mary is thinking. If one person fly's away from the other they both can claim the others time is slower.

Because both will find the other’s time running slower. This is fine because simultaneity is relative.

However, it is not what is going on in your example.

 

[dwspacetime]

If Tom is at rest with respect to the planets, Mary will measure him to be 2.4ly away; Tom will measure Mary to be 4ly away.

They are measuring different things and both calling them "the distance", of course.

If in Tom's frame Mary is 4ly away. In Mary's frame Tom is also 4ly away since there is no difference btw the two frames

 

[Ibix]

That is exactly it does make much sense why u don't want the planets to confuse you. Because they make you think Tom is at rest but Mary is moving. But Mary could think she is at rest but tom is moving to her and he should think exact the same as whatever Mary is thinking. If one person fly's away from the other they both can claim the others time is slower.

It's not to do with the planets. It's that point B is defined to be 4ly away from Tom in Tom's rest frame. If you'd defined it to be 4ly away from Mary in her rest frame it'd be the other way around.

This problem is not symmetric between Tom and Mary because that's how you chose to define it.

 

[Orodruin]

If in Tom's frame Mary is 4ly away. In Mary's frame Tom is also 4ly away since there is no difference btw the two frames

This is simply wrong. You are again missing the relativity of simultaneity. You are assuming “now” means the same to Tom and Mary. It doesn’t.

 

[dwspacetime]

This is simply wrong. You are again missing the relativity of simultaneity. You are assuming “now” means the same to Tom and Mary. It doesn’t.

Let's forget my "now", what I mean is that Tom and Mary are equal. If one thinks the other is 4 away the other should think the same. Planets A and B could be Tim and Dawn.

 

[Ibix]

If one thinks the other is 4 away the other should think the same.

You mean they should think the same thing at the same time. And they don't use the same definition of "the same time", as we keep telling you. Failing to understand that is where you are coming unstuck.

 

[Orodruin]

Let’s do the following exercise on a piece of paper:

• Draw two parallel lines (call them A and B) at some distance apart. Call this distance d.
• Draw a line C that is slanted relative to the other two lines.
• When C passes B, call this intersection CB and draw a line D orthogonally to C from CB until it hits A. Call the length of D d’.
• You should find that d’ > d.

If you instead draw a line from CB that was orthogonal to B you would obviously find distance d again.

The above is the exact geometric analogue of your described situation but in Euclidean space. The exact same thing holds in spacetime with the exception that the geometry is different so you will find d’ < d instead.

 

[Orodruin]

Let's forget my "now", what I mean is that Tom and Mary are equal. If one thinks the other is 4 away the other should think the same. Planets A and B could be Tim and Dawn.

As i said in #3 already: When you talk about things like this you are simply not allowed to forget that “now” means different things to different observers. It forms a crucial part and you cannot hope to understand the theory without understanding this.

Edit: As you can perhaps deduce from my signature, failure to understand the relativity of simultaneity is by far the most basic error people make when they are learning special relativity …

 

[dwspacetime]

Let’s do the following exercise on a piece of paper:

• Draw two parallel lines (call them A and B) at some distance apart. Call this distance d.
• Draw a line C that is slanted relative to the other two lines.
• When C passes B, call this intersection CB and draw a line D orthogonally to C from CB until it hits A. Call the length of D d’.
• You should find that d’ > d.

If you instead draw a line from CB that was orthogonal to B you would obviously find distance d again.

The above is the exact geometric analogue of your described situation but in Euclidean space. The exact same thing holds in spacetime with the exception that the geometry is different so you will find d’ < d instead.

Thanks. I'll get back on this. "Now" it's time for something else.

 

[Ibix]

Let's forget my "now", what I mean is that Tom and Mary are equal. If one thinks the other is 4 away the other should think the same. Planets A and B could be Tim and Dawn.

Here's a spacetime diagram showing Tom and Mary. It's drawn using Tom's frame, so Tom is at rest.

If you haven't come across spacetime diagrams before, they are "maps" of spacetime, with time (as defined in this frame) drawn up the page and space (again, as defined in this frame) across it. Tom, therefore, appears as a vertical line (red on the diagram) because he doesn't change his position in this frame. Mary is a slanted line (blue on the diagram) because she is moving. Notice that Tom has been placed at 1ly, and Mary reaches 4ly from him at t=1. Let's add a line showing Tom's idea of "all of space at the moment that Mary is 4ly away from him":

Tom's idea of space is the fine red line, and you can see that Mary is 4ly away. However, Mary does not share this definition of "space" because she does not share the same notion of "at the same time". Her version of "all of space at the same time" is shown as a fine blue line here:

Notice that the fine blue line passes through Mary at the same time as the red line, but it does not pass through Tom at the same time as the red line. This is why they measure different distances - they are measuring along different slices of spacetime, and they are doing that because they don't share a notion of "at the same time". The interval along the fine blue line between the two thick lines is 2.4 ly.

To complete the picture, we can also draw the same graph in Mary's frame:

Now Mary is represented by a vertical line and her "all of space" is a horizontal line. You can read off the horizontal axis that the spacing is 2.4 ly, but again, Tom is measuring something different.

Your problem is that you keep hoping you can pretend the different slopes of the two fine lines don't matter. They do.

 

[Sagittarius A-Star]

That is exactly it does make much sense why u don't want the planets to confuse you. Because they make you think Tom is at rest but Mary is moving. But Mary could think she is at rest but tom is moving to her and he should think exact the same as whatever Mary is thinking. If one person fly's away from the other they both can claim the others time is slower.

Time-dilation and length-contraction are frame-dependent. Your original scenario doesn't change only by removing the planets as landmarks.

• In Tom's rest frame, Mary needs only 3 years of her proper time to meet him, counted from when she was 4 LY away, because of her time-dilation.
• In Mary's rest frame, Tom needs only 1.8 years of his proper time to meet her, counted from when he was 2.4 LY away, because of his time-dilation.

If you prefer another (reciprocal) scenario, you could think of 2 other planets, one at rest with Alice and the other "at rest" 4 LY away from Alice (behind planet A).

 

[Orodruin]

The interval along the fine blue line between the two thick lines is 2.4 ly.

Just to interject here: Even though the fine blue line segment looks longer in the drawing that the red, it actually isn’t. The reason it looks longer is that the drawing is made on a piece of screen with Euclidean geometry such that Pythagoras’ theorem $a^2 + b^2 = d^2$ holds. This is the main point where the geometry of spacetime differs. The “Pythagorean theorem of spacetime” instead reads $d^2 = x^2 - t^2$ (note the minus sign!)
(Actually $c^2 t^2$, but we use units where c = 1 ly/y)

The other effect of this different geometry is that the red lines are orthogonal to eachother - as are the blue lines - in all diagrams.

 

[Histspec]

In addition to the diagrams of Ibix, one can choose a frame in which Mary and Tom are moving with same speed in opposite directions, which makes the situation look more symmetric.

On the following image, the thick blue line is the worldline of Tom and the fine blue lines denote simultaneous events in Tom's frame, while the thick red line is the worldline of Mary and the fine red lines denote simultaneous events in Mary's frame.
The markings on the fine blue line connecting Tom's time 0 to Mary's worldline tell you that this distance in Tom's frame is 4 at Mary's time 0.
The markings on the fine red line connecting Mary's time 0 to Tom's worldline tell you that this distance in Mary's frame is contracted to length 2.4 at Tom's time 3.1.
The markings on the fine blue line connecting Tom's time 3.1 to Mary's worldline tell you that this distance in Tom's frame is contracted to 1.44 at Mary's time 1.9.
The markings on the fine red line connecting Mary's time time 1.9 to Tom's worldline tell you that this distance in Mary's frame is contracted to 0,864 at Tom's time 4.4.
The markings on the fine blue line connecting Tom's time 4.4 to Mary's worldline tell you that this distance in Tom's frame is contracted to 0,5184 at Mary's time 2.6.
etc.....

 

[Sagittarius A-Star]

The markings on the fine blue line connecting Tom's time 0 to Mary's worldline tell you that this distance in Tom's frame is 4 at Mary's time 0.
The markings on the fine red line connecting Mary's time 0 to Tom's worldline tell you that this distance in Mary's frame is contracted to length 2.4 at Tom's time 3.1.

Yes.
The (constant) distance between the planets is contracted in Mary's frame.

The markings on the fine blue line connecting Tom's time 3.1 to Mary's worldline tell you that this distance in Tom's frame is contracted to 1.44 at Mary's time 1.9.
The markings on the fine red line connecting Mary's time time 1.9 to Tom's worldline tell you that this distance in Mary's frame is contracted to 0,864 at Tom's time 4.4.
The markings on the fine blue line connecting Tom's time 4.4 to Mary's worldline tell you that this distance in Tom's frame is contracted to 0,5184 at Mary's time 2.6.
etc.....

No.
The (decreasing) distance between Tom and Mary is usually not called contraction. Lorentz contracted can be a distance, that is constant over time, like the length of a ruler or the distance between planets A and B.

 

[Ibix]

In addition to the diagrams of Ibix, one can choose a frame in which Mary and Tom are moving with same speed in opposite directions, which makes the situation look more symmetric.

...which is particularly powerful in this case, because it illustrates where the symmetry is and is not in this scenario. You could easily extend Mary's line downwards another two years and draw left-right reversed versions of the fine lines, and then the diagram would be symmetric. But @dwspacetime chose not to do that (and actually only considered the two fine lines through Mary's zero event), so had an asymmetric setup. That choice is why the scenario (not the physics, just the setup) prefers Tom's rest frame.

 

[Histspec]

No.
The (decreasing) distance between Tom and Mary is usually not called contraction. Lorentz contracted can be a distance, that is constant over time, like the length of a ruler or the distance between planets A and B.

I tried to convey that the asymmetry in dwspacetime's example (which follows from his choice of starting with distance AB=4 corresponding to a specific ruler) does not contradict the fact the length contraction remains fully symmetric in both frames along the entire way until they meet.
That is, Tom determines the proper length between A and B using a ruler of length 4, which is contracted in Mary's frame to 2.4, which momentarily coincides with another ruler of proper length 2.4 at rest in Mary's frame, which is contracted in Tom's frame to 1.44, which momentarily coincides with another ruler of proper length 1.44 at rest in Tom's frame, which is contracted in Mary's frame to 0.86 etc....

 

[Ibix]

which momentarily coincides with another ruler of proper length 2.4 at rest in Mary's frame

Well, the problem here is that "momentarily" is a rather slippy concept. The process that Mary calls "measuring the instantaneous distance between me and Tom" doesn't take place at an instant in any other frame. One end of each ruler momentarily coincide, yes, but frames have different views on what "momentarily" means along the ruler.

Possibly a better illustration of your point is to extend worldlines from the light year markers of Tom's zero ruler. That represents a long ruler (or perhaps more practically a series of bouys) moving with Tom, and you will be able to see that they intersect one Mary-instant of Mary's ruler at less than one light year intervals. That also shows the truth, that taking "now" to mean different planes makes them intersect extended 4d objects at different "angles". So length contraction is fundamentally no more mysterious than slicing a sausage at an angle and getting elliptical sections instead of circular.

 

[Histspec]

Well, the problem here is that "momentarily" is a rather slippy concept. The process that Mary calls "measuring the instantaneous distance between me and Tom" doesn't take place at an instant in any other frame. One end of each ruler momentarily coincide, yes, but frames have different views on what "momentarily" means along the ruler.

Which is precisely why the 2.4 ruler at rest in Mary's frame is contracted to 1.44 in Tom's frame.

Possibly a better illustration of your point is to extend worldlines from the light year markers of Tom's zero ruler. That represents a long ruler (or perhaps more practically a series of bouys) moving with Tom, and you will be able to see that they intersect one Mary-instant of Mary's ruler at less than one light year intervals. That also shows the truth, that taking "now" to mean different planes makes them intersect extended 4d objects at different "angles". So length contraction is fundamentally no more mysterious than slicing a sausage at an angle and getting elliptical sections instead of circular.

 

[Sagittarius A-Star]

fact the length contraction remains fully symmetric in both frames along the entire way until they meet.

Important to consider:

• As I said, length contraction relates to a distance, that is constant over time, like the distance between planets A and B.
• The rest-length relates to 1 frame only, therefore the setup (not the physics, as @Ibix mentioned) becomes asymmetric.
• The complete distance between planets A and B in the OP must be used to calculate the proper travel time of Mary between the start event at planet B and the arrival event at planet A. In Mary's frame, length contraction of this distance must be considered. In Tom's frame, time-dilation of Mary must be considered to come to the same invariant result.

 

[robphy]

Here's a spacetime diagram on rotated graph paper, following the color scheme of @Ibix in #20 .
You can count the diamonds to find
QU is a spatial displacement of length 2.4 in Mary's frame
and
RT is a spatial displacement of length 2.4 in Tom's frame.
(Note that from similar-triangles (\frac{PQ}{PZ}=\frac{4}{5}=\frac{2.4}{3}\frac{RT}{RZ}).)

 

[dwspacetime]

Here's a spacetime diagram showing Tom and Mary. It's drawn using Tom's frame, so Tom is at rest.
View attachment 336789
If you haven't come across spacetime diagrams before, they are "maps" of spacetime, with time (as defined in this frame) drawn up the page and space (again, as defined in this frame) across it. Tom, therefore, appears as a vertical line (red on the diagram) because he doesn't change his position in this frame. Mary is a slanted line (blue on the diagram) because she is moving. Notice that Tom has been placed at 1ly, and Mary reaches 4ly from him at t=1. Let's add a line showing Tom's idea of "all of space at the moment that Mary is 4ly away from him":
View attachment 336790
Tom's idea of space is the fine red line, and you can see that Mary is 4ly away. However, Mary does not share this definition of "space" because she does not share the same notion of "at the same time". Her version of "all of space at the same time" is shown as a fine blue line here:
View attachment 336791
Notice that the fine blue line passes through Mary at the same time as the red line, but it does not pass through Tom at the same time as the red line. This is why they measure different distances - they are measuring along different slices of spacetime, and they are doing that because they don't share a notion of "at the same time". The interval along the fine blue line between the two thick lines is 2.4 ly.

To complete the picture, we can also draw the same graph in Mary's frame:
View attachment 336792
Now Mary is represented by a vertical line and her "all of space" is a horizontal line. You can read off the horizontal axis that the spacing is 2.4 ly, but again, Tom is measuring something different.

Your problem is that you keep hoping you can pretend the different slopes of the two fine lines don't matter. They do.

Sorry I just came back on this. I disagree. All your explanations is based on that you already decided that Tom is stationary which already eliminates the paradox. Let's forget about the two planets and Tom and Mary are moving towards each other or Tom is moving towards Mary. We can do the same as above but think the red line is Mary and the blue line is Tom. What we get "NOW"? I guess the planets just confused you. they can be any other moving things as Jane and Dawn or you can image another two planets C and D are moving relative to A and B. or A and B are moving relative C and D.

 

[PeroK]

A and B are moving relative C and D.

The simplest scenario is that A and B pass each other with a relative speed of $0.8c$.

In the rest frame of A: when A's clock reads 4 seconds, B's clock reads 2.4s.

In the rest frame of B: when B's clock reads 4 seconds, A's clock reads 2.4s. Or, alternatively, when B's clock reads 2.4s, A's clock reads 1.44s.

Are you on board with that?

 

[dwspacetime]

The simplest scenario is that A and B pass each other with a relative speed of $0.8c$.

In the rest frame of A: when A's clock reads 4 seconds, B's clock reads 2.4s.

In the rest frame of B: when B's clock reads 4 seconds, A's clock reads 2.4s. Or, alternatively, when B's clock reads 2.4s, A's clock reads 1.44s.

Are you on board with that?

I do not know what you try to say. yes I get that. So what is wrong with making the red line Mary? if the red line is Mary Tom is 4 lys instead of 2.4 lys away

 

[PeroK]

I do not know what you try to say. yes I get that.

Okay, so it follows that:

In the rest frame of A: when A's clock reads 4s, B is 3.2 light seconds from A (and B's clock reads 2.4s).

In the rest frame of B: when B's clock reads 2.4s, A is 1.92 light seconds from B (and A's clock reads 1.44s).

Are you okay with that?

 

[Ibix]

All your explanations is based on that you already decided that Tom is stationary

No - it's all based on your personal choice to start the experiment when Mary is 4ly away from Tom using Tom's rest frame. That choice that you made is what creates the asymmetry in this set up. There is no paradox.

Histspec's diagram in #28 illustrates it nicely, using a frame where Tom and Mary are moving at equal and opposite speeds. There is asymmetry because you chose to key "time zero" off an event on Mary's worldline, but Tom's "time zero" is only the same as Mary's at that one event.

Again: the asymmetry is introduced by you in your experimental setup. It's got nothing to do with me "deciding Tom is at rest". If that is not clear to you then study the spacetime diagrams until it becomes clear. Probably the most important thing to grasp is the difference between Tom's "time zero" and Mary's "time zero". That's the bit that evades a lot of people, and that failing is where most paradoxes spring from.

 

[dwspacetime]

The simplest scenario is that A and B pass each other with a relative speed of $0.8c$.

In the rest frame of A: when A's clock reads 4 seconds, B's clock reads 2.4s.

In the rest frame of B: when B's clock reads 4 seconds, A's clock reads 2.4s. Or, alternatively, when B's clock reads 2.4s, A's clock reads 1.44s.

Are you on board with that?

the reason it is a paradox is because we can not decide which is stationary unless using a reference frame independent of Tom and Mary which are the planets you unconsciously take which already eliminates the paradox and all the rest of spacetime conversion is just a waste of time.

 

[PeroK]

the reason it is a paradox is because we can not decide which is stationary unless using a reference frame independent of Tom and Mary which are the planets you unconsciously take which already eliminates the paradox and all the rest of spacetime conversion is just a waste of time.

I take it from that you are not okay with my simple analysis. Can you say what you think is wrong with my analysis? There are no planets in my analysis. Only A and B moving relative to each other. No one is "stationary".

 

[Ibix]

the reason it is a paradox is

There is no paradox.

we can not decide which is stationary unless using a reference frame independent of Tom and Mary

None of them is stationary in any absolute sense; that's one of the postulates on which the whole theory is based! So you can choose to work in a frame where either Tom or Mary or neither to be stationary. All three choices have been illustrated in this thread, and PeroK is trying to walk you through it with numbers.

 

[dwspacetime]

Okay, so it follows that:

In the rest frame of A: when A's clock reads 4s, B is 3.2 light seconds from A (and B's clock reads 2.4s).

In the rest frame of B: when B's clock reads 2.4s, A is 1.92 light seconds from B (and A's clock reads 1.44s).

Are you okay with that?

no. I disagree. what is wrong with saying

In the rest frame of B: when B's clock reads 4s, A is 3.2 light seconds from B (and A's clock reads 2.4s).

or simply put, just simply replacing A with B, all you said still stands? why not. When Tom measures Mary should X lys to reach him. Mary should measure Tom should take X lys to reach her too.

 

[PeroK]

no. I disagree. what is wrong with saying

In the rest frame of B: when B's clock reads 4s, A is 3.2 light seconds from B (and A's clock reads 2.4s).

That's true also.

or simply put, just simply replacing A with B, all you said still stands?

Yes, you can swap A and B and all statements are also true. The scenario is symmetrical between A and B.

 

[dwspacetime]

There is no paradox.

None of them is stationary in any absolute sense; that's one of the postulates on which the whole theory is based! So you can choose to work in a frame where either Tom or Mary or neither to be stationary. All three choices have been illustrated in this thread, and PeroK is trying to walk you through it with

There is no paradox.

None of them is stationary in any absolute sense; that's one of the postulates on which the whole theory is based! So you can choose to work in a frame where either Tom or Mary or neither to be stationary. All three choices have been illustrated in this thread, and PeroK is trying to walk you through it with numbers.

if I choose Mary as stationary as red line as you drew. Tom should be 4 liys away?

 

[Ibix]

why not.

The relativity of simultaneity: Tom and Mary do not have a shared global notion of what "all of space, now" means. This is what all of the fine red and blue lines on the diagrams show - each one is the slice of spacetime they call "space, now". And, as the diagram in #28 illustrates, those different definitions of "space, now" mean that "distance, now" is different.

This point seems to be the hardest part of relativity for people to grasp, and failing to understand it is how almost all "paradoxes" arise.

 

[PeroK]

Okay, so it follows that:

In the rest frame of A: when A's clock reads 4s, B is 3.2 light seconds from A (and B's clock reads 2.4s).

In the rest frame of B: when B's clock reads 2.4s, A is 1.92 light seconds from B (and A's clock reads 1.44s).

Are you okay with that?

To emphasise the point. It also follows that:

In the rest frame of B: when B's clock reads 4s, A is 3.2 light seconds from B (and A's clock reads 2.4s).

In the rest frame of A: when A's clock reads 2.4s, B is 1.92 light seconds from A (and B's clock reads 1.44s).

 

[dwspacetime]

The relativity of simultaneity: Tom and Mary do not have a shared global notion of what "all of space, now" means. This is what all of the fine red and blue lines on the diagrams show - each one is the slice of spacetime they call "space, now". And, as the diagram in #28 illustrates, those different definitions of "space, now" mean that "distance, now" is different.

This point seems to be the hardest part of relativity for people to grasp, and failing to understand it is how almost all "paradoxes" arise.

OK if Mary is stationary as red. does the chart you drew above showing Tom as blue is 4 lys away? why not

 

[Ibix]

View attachment 337735
OK if Mary is stationary as red. does the chart you drew above showing Tom as blue is 4 lys away? why not

Yes it would. But Tom would not say Mary was 4ly away at the same time in that setup, because "same time" means different things to the two of them.

 

[dwspacetime]

Yes it would. But Tom would not say Mary was 4ly away at the same time in that setup, because "same time" means different things to the two of them.

when exacty Mary can say that? I think that is what puzzles me if you are correct

 

[dwspacetime]

Yes it would. But Tom would not say Mary was 4ly away at the same time in that setup, because "same time" means different things to the two of them.

I havent gone through all the replies. I just realized that the distance btw planet A and B are not the same to Tom and Mary. distance is also relative. I guess I consider d btw A and B are the same or absolute as 4 lys to both of them. A and B are stationary with Tom but is not with Mary.

 

[dwspacetime]

Yes it would. But Tom would not say Mary was 4ly away at the same time in that setup, because "same time" means different things to the two of them.

let me ask another question. if both Tom and Mary leave planet A or the same spot at the same time and do all kinds of travels through spacetime with all kinds of speeds (including 0) and accelerations and finally meet again. who is younger?

 

[PeroK]

I havent gone through all the replies. I just realized that the distance btw planet A and B are not the same to Tom and Mary. distance is also relative. I guess I consider d btw A and B are the same or absolute as 4 lys to both of them. A and B are stationary with Tom but is not with Mary.

You don't need planets in your scenario. Planets suggest a natural frame of reference in which the planets are at rest. You don't want that.

Likewise, a spacetime diagram picks out a frame of reference - determined by the specific axes you draw.

That's why I was trying to get you towards a scenario with just A and B moving relative to each other. That scenario has two equally naturals frames of reference - the rest frames of A and B. That allows you to examine the symmetry between A and B.

 

[Ibix]

when exacty Mary can say that? I think that is what puzzles me if you are correct

That depends what you mean by "when" - there are several definitions available. Your confusion comes from the fact that Tom and Mary are using different definitions when they measure different separations "at the same time" and you are implicitly assuming they're using the same one. If they were using the same one then there would be a paradox.

Tom and Mary are going to meet at some point. Both will say that five years before that (by their own clocks) the other one was four light years away. But both will also say that, at the same time they are making this claim the other's clock would show only three years to the meeting. This is how there's no paradox - Tom knows that Mary isn't measuring the same distance at the same time.

if both Tom and Mary leave planet A or the same spot at the same time and do all kinds of travels through spacetime with all kinds of speeds (including 0) and accelerations and finally meet again. who is younger?

You don't need a planet, just for them to meet twice. Pick any inertial frame and write down their velocity as a function of time. Then compute each of their ages using $\int\sqrt{1-v^2/c^2}dt$, where the limits of the integrals are the meeting times specified using your arbitrarily chosen frame. The lower value is the younger one.