How Does Special Relativity Affect Perceived Distances Between Planets?

[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.

 

[dwspacetime]

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.

i guess the reason I was puzzled is because I thought when tom at A and Mary at B the distance is between them is Absolut

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.

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.

I kind think it has nothing to do with "when". the distance btw A and B is 4 lys for Tom nonmatter when since they are stationary to Tom. but to Mary both A and B are moving towards her, it is always 2.4 lys btw them whenever.
According to your equation, to use either Tom or mary as reference, Tom or Mary would have a v=0, the other one is always younger.

.

 

[Ibix]

According to your equation, to use either Tom or mary as reference, Tom or Mary would have a v=0, the other one is always younger.

They don't meet twice in the scenario we are discussing, so one of the limits of the integral isn't defined. If they do meet twice, at least one of them must accelerate so at least one of them can't always have speed zero in the arbitratily chosen inertial frame.

 

[PeroK]

i guess the reason I was puzzled is because I thought when tom at A and Mary at B the distance is between them is Absolute

It's the "when" that is not absolute. That's what I was trying to emphasise.

Although I see now that you have moved on to a different question altogether.

 

[Orodruin]

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

Misconceptions regarding special relativity in general and Minkowski space in particular can often be illuminated by considering the Euclidean space equivalent. What you essentially need to know are what the statements correspond to in Euclidean space. Here, in particular, the following is relevant:

The Euclidean statement corresponding to ”When A finds B to be x away” is ”for a straight line A, the line orthogonal to A crossing the line B at a distance of x”. For brevity, let’s just call x this ”the orthogonal distance from A to B”. Note that only the direction of A is important for this definition.

Let’s consider the equivalent of your scenario.

- Draw two parallel lines A and B. (These correspond to your planets)
- Draw a line T on top of A. (This corresponds to Tom)
- Draw a line M that is not parallel to A and B. (This corresonds to Mary) This line necessarily crosses both A and B.
- Draw the line orthogonal to T which crosses M at a distance of 4. Call this line C and its intersection with M you call MC.
- Determine the orthogonal distance from M to T at the point MC. (Ie, draw a line from MC that is orthogonal to M, find how long it must be to intersect T)

You should find that the result in the last step is not 4. This is completely analogous to your setup, just in a different geometry. Whether the result is smaller or larger than 4 will differ between the geometries because of a sign in the geometry definition, but the geometries are similar enough for this to be a direct analogy.

Now remove lines A and B. Does this change the result?

edit: here is a diagram

Your claim is essentially the Minkowski space equivalent of saying that C has the same length as D.

 

[dwspacetime]

They don't meet twice in the scenario we are discussing, so one of the limits of the integral isn't defined. If they do meet twice, at least one of them must accelerate so at least one of them can't always have speed zero in the arbitratily chosen inertial frame.

I am trying to simply it by spreading the acceleration to speed. to meet again some of the speeds have to be negative. but since is v square, so it doesnt affect the result.

my original question is still a paradox. I was assuming A and B stay stationary to Tom. in the meantime we can also assume planet C and D are stationary with Mary 4 lys apart. Then to Tom C and D are 2.4 years apart.... so "NOW" we have 4 planets.

 

[PeroK]

my original question is still a paradox.

It's not a paradox. There's only your failure to understand the basics of SR.

 

[Orodruin]

to meet again the some of the speeds have to be negative

Pet peeve: Speed is never negative. It is the absolute value of velocity (which may be negative depending on direction and chosen coordinates).

 

[dwspacetime]

Pet peeve: Speed is never negative. It is the absolute value of velocity (which may be negative depending on direction and chosen coordinates).

that is fine. I am trying to tell you they will meet after all kinds of speed. since somebody says they dont meet again assuming no one turning back. if that is true, Tom travels from A to B and wait for mary from A to B also. they do meet again without turning back. if we consider either one of them as a reference they other one will have a speed which will result a less number in time. so the other one is always younger

 

[dwspacetime]

It's not a paradox. There's only your failure to understand the basics of SR.

my failure of understanding your whatever. regardless "when", or whatever "when" you applies to Tom it shoudl applies to Mary. without a reference independent from them whatever assumption you made to one of them you can just switch btw them

 

[dwspacetime]

It's not a paradox. There's only your failure to understand the basics of SR.

at whatever "when" Mary is 4 lys away to Tom there is another "when" the totally opposite is also true

 

[Orodruin]

that is fine. I am trying to tell you they will meet after all kinds of speed. since somebody says they dont meet again assuming no one turning back. if that is true, Tom travels from A to B and wait for mary from A to B also. they do meet again without turning back. if we consider either one of them as a reference they other one will have a speed which will result a less number in time. so the other one is always younger

No, if they both travelled from A to B their frames were not always inertial and therefore could not be used* with the quoted formula.

* You can use noninertial frames as well, but it typically is more mathematically involved and gives the same result

 

[dwspacetime]

It's not a paradox. There's only your failure to understand the basics of SR.

let us put it this way. I agree the "when" of Tom is red is not the same as the "when" Mary is red. but for the same original question I asked there must be a "when" Mary is red and all the charts you drew still apply which is why it is a paradox.

 

[Orodruin]

let us put it this way. I agree the "when" of Tom is red is not the same as the "when" Mary is red. but for the same original question I asked there must be a "when" Mary is red and all the chart you drew still apply which is why it is a paradox.

It is not a paradox. It is geometry. Try the Euclidean space equivalent!

 

[dwspacetime]

let us put it this way. I agree the "when" of Tom is red is not the same as the "when" Mary is red. but for the same original question I asked there must be a "when" Mary is red and all the chart you drew still apply which is why it is a paradox.

It is not a paradox. It is geometry. Try the Euclidean space equivalent!

let's do this. just go back to your original charts. just switch Tom and Mary and forget about the planets. do those charts still apply? why not? I know when you switch them the "when" is not the same as the "when" you did them the first time.

It is not a paradox. It is geometry. Try the Euclidean space equivalent

 

[Orodruin]

You messed up your quotes, but yes, you can do exactly the same statements from Mary’s frame. It is unclear why you think this is a paradox. Again I refer you to try the same in the Euclidean case. It is just a fact of geometry. Nothing strange is going on.

 

[dwspacetime]

It is not a paradox. It is geometry. Try the Euclidean space equivalent!

let's make it very simple. let me modify my original question to this. Tom and Mary in spacetime. in Tom's frame Mary is moving towards him in 0.8C and in Mary's frame Tom is moving towards her in 0.8C. when Mary is 4 lys away from Tom in Tom's frame how far is Mary from Tom? why

 

[Orodruin]

let's make it very simple. let me modify my original question to this. Tom and Mary in spacetime. in Tom's frame Mary is moving towards him in 0.8C and in Mary's frame Tom is moving towards her in 0.8C. when Mary is 4 lys away from Tom in Tom's frame how far is Mary from Tom? why

Define what you mean by ”when”. This is what seems to confuse you.

 

[dwspacetime]

let's make it very simple. let me modify my original question to this. Tom and Mary in spacetime. in Tom's frame Mary is moving towards him in 0.8C and in Mary's frame Tom is moving towards her in 0.8C. when Mary is 4 lys away from Tom in Tom's frame how far is Mary from Tom? why?

 

[Orodruin]

Also specify what distance you want to know and according to what convention.

 

[dwspacetime]

Also specify what distance you want to know and according to what convention.

use whatever convention you want to use. then ask the same question with Tom and Mary switched. you should get the same answer for the two questions by using whatever convention if you are consistent or you are wrong.