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  4. How does special relativity explain dimensional components ...
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How does special relativity explain dimensional components ...

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Offline Dimensional (OP)

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How does special relativity explain dimensional components ...
« on: 24/06/2022 23:28:58 »
of space and time of a moving observer and a observer at rest?  So if object A observes object B coming towards it very fast, A is at rest and should only have the time component and no spacelike component.  But B can say the same thing.  How does SR explain the componentry of this kind of scenario?
« Last Edit: 25/06/2022 00:01:09 by Dimensional »
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Re: How does special relativity explain dimensional components ...
« Reply #1 on: 25/06/2022 00:37:22 »
Quote from: Dimensional on 24/06/2022 23:28:58
How does SR explain the componentry of this kind of scenario?
Not sure what you mean by componentry.  Observer A would say B time is dilated and it's length is contracted.  Observer B would say A is time dilated and length contracted.  Is that what you are asking about?
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Re: How does special relativity explain dimensional components ...
« Reply #2 on: 25/06/2022 03:05:02 »
Quote from: Origin on 25/06/2022 00:37:22
Quote from: Dimensional on 24/06/2022 23:28:58
How does SR explain the componentry of this kind of scenario?
Not sure what you mean by componentry.  Observer A would say B time is dilated and it's length is contracted.  Observer B would say A is time dilated and length contracted.  Is that what you are asking about?
Sort of. 

In the light cone, do the timelike trajectories of objects have a time component and a space component?
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Re: How does special relativity explain dimensional components ...
« Reply #3 on: 25/06/2022 04:48:33 »
Hi.

   I'm still not certain what your question was about.

Quote from: Dimensional on 24/06/2022 23:28:58
How does SR explain the componentry of this kind of scenario?
    An objects motion is described by a 4-velocity.   That is frame dependant.   To determine the 4-velocity in one frame from the 4-velocity in another frame,  you apply a Transformation (e.g. a Lorentz transformation).

Best Wishes.
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Re: How does special relativity explain dimensional components ...
« Reply #4 on: 25/06/2022 05:55:33 »
Quote from: Eternal Student on 25/06/2022 04:48:33
Hi.

   I'm still not certain what your question was about.

Quote from: Dimensional on 24/06/2022 23:28:58
How does SR explain the componentry of this kind of scenario?
    An objects motion is described by a 4-velocity.   That is frame dependant.   To determine the 4-velocity in one frame from the 4-velocity in another frame,  you apply a Transformation (e.g. a Lorentz transformation).

Best Wishes.

So just like we would break down a vector in a "normal" 2d space.  For example, an object moves in a certain direction at 4 units in the x direction and 3 units in the y direction.

Does time and space have separate components like that?
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Re: How does special relativity explain dimensional components ...
« Reply #5 on: 25/06/2022 13:02:01 »
Hi.

Quote from: Dimensional on 25/06/2022 05:55:33
Does time and space have separate components like that?
   Yes and No.

The yes bit:
    4-vectors are what are important in spacetime.    These have 4 components,   3 of them are called spatial components and the other component is called the time component.  You can write the spatial components first and the time component last but it's more common to write the time component first.   It's also fairly common to start numbering the components from 0 and not from 1.   The final slightly confusing thing you might see is that if you had a 4-vector  X   then you may see the components written as  X0,  X1,  X2  and  X3.    Superscripts instead of subscripts like  X0,  X1  can be used.
    For example    r =    ( ct,  x,  y,  z)  is the usual way of writing  the position 4-vector of an object.    It has a time component  ct  =    c (the speed of light)  multiplied by the position of the object on the time axis, t.    It has spatial components   x, y, z   which are the position of the object along the x, y and z axis respectively.    Now you could use t as the time component instead of ct but the algebra turns out to be much easier if you use  c (the speed of light) multiplied by t   as the time component.

The "no" bit
1.   There's an unusual way of determining the magnitude of a 4-vector.  You might see it called the "norm" or "Minkowski norm" of a 4-vector.    For simple vectors used in Euclidean space or Newtonian mechanics,  whenever you increase the size of one component the overall magnitude of the vector would increase.   For 4-vectors that's not always the situation,  you can increase the size of one component and end up reducing the overall magnitude of the 4-vector.    The Minkowski metric is described in various places  [For example,   https://phys.libretexts.org/Bookshelves/Modern_Physics/Book%3A_Spiral_Modern_Physics_(D%27Alessandris)/3%3A_Spacetime_and_General_Relativity/3.1%3A_Minkowski_Metric  ].

2.   You are mainly discussing velocities and motion in your posts, rather than just positions.   In ordinary Newtonian mechanics, velocities are just a rate of change of position with respect to the time co-ordinate.   For 4-velocities we can't determine rates of change with respect to a fixed time co-ordinate,  instead we must determine rates of change with respect to what is called the proper time for that object undergoing the motion.
   What this boils down to is that spatial components of the 4-velocity are NOT exactly the spatial components of the ordinary Newtonian or 3-velocity that the object might have.   Instead the spatial components are a multiple of the spatial velocity you would have assigned the object in Newtonian mechanics.  Also it's not a fixed multiple,  the multiple changes according to the Newtonian 3-velocity of the object.    Specifically, the  spatial components of the 4-velocity are given by   γ  (the gamma factor) multiplied by the spatial components of the 3-velocity.

- - - - - - - -
     That might be more detail than you were after.   Overall there is a lot of similarity between  4-velocities used in Special relativity and more conventional velocity vectors you might have seen in Newtonian mechanics.    I've mentioned the differences because, in my limited experience, if we don't then it's human nature to run away with the idea that it's all exactly like Newtonian mechanics and ordinary 3-velocity vectors.   You'll soon hit problems if you do that.
     For example,  it can be useful to consider the magnitude of a 4-velocity vector.  An ordinary object with some positive rest mass always has a 4-velocity vector of magnitude c (the speed of light).   That magnitude can be shared out between the time component and the spatial component of the objects 4-velocity.   An object at rest (in a given frame which we will use to assign the velocity vector) has all of its velocity in the time component while the spatial components would have the value 0.     Meanwhile, an object in motion (in the given frame) has a non-zero value in the spatial components of the 4-velocity and a correspondingly lower value* in the time component.

Best Wishes.

* LATE EDITING:  I don't like this on a second reading.   It's precisely one of those examples where you could have a larger numerical value for the time component but that is actually reducing the overall magnitude of the 4-velocity and not increasing it (because the Minkowski metric subtracts the time component instead of adding it).  It's fair to say the object has less velocity through time and many Pop Sci articles will do this - but it's not correct to imply that the numerical value you find written in the time component of the 4-velocity has to be smaller.
« Last Edit: 26/06/2022 00:45:43 by Eternal Student »
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Offline Dimensional (OP)

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Re: How does special relativity explain dimensional components ...
« Reply #6 on: 25/06/2022 20:52:05 »
Quote from: Eternal Student on 25/06/2022 13:02:01
Hi.

Quote from: Dimensional on 25/06/2022 05:55:33
Does time and space have separate components like that?
   Yes and No.

The yes bit:
    4-vectors are what are important in spacetime.    These have 4 components,   3 of them are called spatial components and the other component is called the time component.  You can write the spatial components first and the time component last but it's more common to write the time component first.   It's also fairly common to start numbering the components from 0 and not from 1.   The final slightly confusing thing you might see is that if you had a 4-vector  X   then you may see the components written as  X0,  X1,  X2  and  X3.    Superscripts instead of subscripts like  X0,  X1  can be used.
    For example    r =    ( ct,  x,  y,  z)  is the usual way of writing  the position 4-vector of an object.    It has a time component  ct  =    c (the speed of light)  multiplied by the position of the object on the time axis, t.    It has spatial components   x, y, z   which are the position of the object along the x, y and z axis respectively.    Now you could use t as the time component instead of ct but the algebra turns out to be much easier if you use  c (the speed of light) multiplied by t   as the time component.

The "no" bit
1.   There's an unusual way of determining the magnitude of a 4-vector.  You might see it called the "norm" or "Minkowski norm" of a 4-vector.    For simple vectors used in Euclidean space or Newtonian mechanics,  whenever you increase the size of one component the overall magnitude of the vector would increase.   For 4-vectors that's not always the situation,  you can increase the size of one component and end up reducing the overall magnitude of the 4-vector.    The Minkowski metric is described in various places  [For example,   https://phys.libretexts.org/Bookshelves/Modern_Physics/Book%3A_Spiral_Modern_Physics_(D%27Alessandris)/3%3A_Spacetime_and_General_Relativity/3.1%3A_Minkowski_Metric  ].

2.   You are mainly discussing velocities and motion in your posts, rather than just positions.   In ordinary Newtonian mechanics, velocities are just a rate of change of position with respect to the time co-ordinate.   For 4-velocities we can't determine rates of change with respect to a fixed time co-ordinate,  instead we must determine rates of change with respect to what is called the proper time for that object undergoing the motion.
   What this boils down to is that spatial components of the 4-velocity are NOT exactly the spatial components of the ordinary Newtonian or 3-velocity that the object might have.   Instead the spatial components are a multiple of the spatial velocity you would have assigned the object in Newtonian mechanics.  Also it's not a fixed multiple,  the multiple changes according to the Newtonian 3-velocity of the object.    Specifically, the  spatial components of the 4-velocity are given by   γ  (the gamma factor) multiplied by the spatial components of the 3-velocity.

- - - - - - - -
     That might be more detail than you were after.   Overall there is a lot of similarity between  4-velocities used in Special relativity and more conventional velocity vectors you might have seen in Newtonian mechanics.    I've mentioned the differences because, in my limited experience, if we don't then it's human nature to run away with the idea that it's all exactly like Newtonian mechanics and ordinary 3-velocity vectors.   You'll soon hit problems if you do that.
     For example,  it can be useful to consider the magnitude of a 4-velocity vector.  An ordinary object with some positive rest mass always has a 4-velocity vector of magnitude c (the speed of light).   That magnitude can be shared out between the time component and the spatial component of the objects 4-velocity.   An object at rest (in a given frame which we will use to assign the velocity vector) has all of its velocity in the time component while the spatial components would have the value 0.     Meanwhile, an object in motion (in the given frame) has a non-zero value in the spatial components of the 4-velocity and a correspondingly lower value in the time component.

Best Wishes.
Thanks, this is very helpful.  But I was afraid of that answer because it leads to my issue in the OP. 

For example, imagine just 2 dimensions of spacetime for simplicity sake.  There is an x axis and a time axis.  Object A thinks it is at rest on the x axis, and object B observes itself as being at rest on the x axis too.  However, they are on a collision course. 

I understand that relativity says that both A and B are correct in thinking that they are at rest.  But in reality, because they are on a collision course, only one of them can actually be at rest where r = ( ct,  0,  n/a,  n/a), right?  Or am I missing something here?
« Last Edit: 25/06/2022 21:11:44 by Dimensional »
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Re: How does special relativity explain dimensional components ...
« Reply #7 on: 25/06/2022 21:17:39 »
Quote from: Dimensional on 25/06/2022 20:52:05
I understand that relativity says that both A and B are correct in thinking that they are at rest.  But in reality, because they are on a collision course, only one of them can actually be at rest where r = ( ct,  0,  0,  0), right?  Or am I missing something here?
When you say only one of them is actually at rest, that sets off alarm bells.  I don't see the difference if A thinks they are at rest or if B thinks they are at rest, even though they collide.  From A's point of view he is stationary on the X axis and moving vertically parallel to the Y (time axis).  A sees B moving at an angle to the X and y axis until he intersects A at some point on the Y axis.  Replace the B with A in the above scenario and that is B's viewpoint as stationary on the X axis.
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Re: How does special relativity explain dimensional components ...
« Reply #8 on: 25/06/2022 22:42:26 »
Quote from: Origin on 25/06/2022 21:17:39
Quote from: Dimensional on 25/06/2022 20:52:05
I understand that relativity says that both A and B are correct in thinking that they are at rest.  But in reality, because they are on a collision course, only one of them can actually be at rest where r = ( ct,  0,  0,  0), right?  Or am I missing something here?
When you say only one of them is actually at rest, that sets off alarm bells.  I don't see the difference if A thinks they are at rest or if B thinks they are at rest, even though they collide.  From A's point of view he is stationary on the X axis and moving vertically parallel to the Y (time axis).  A sees B moving at an angle to the X and y axis until he intersects A at some point on the Y axis.  Replace the B with A in the above scenario and that is B's viewpoint as stationary on the X axis.
So then which one is the vector r = ( ct,  0,  n/a,  n/a)?  Is it A, B, neither?
« Last Edit: 25/06/2022 22:45:12 by Dimensional »
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Re: How does special relativity explain dimensional components ...
« Reply #9 on: 25/06/2022 23:48:04 »
Hi.

    You're suggesting that A and B can't both be stationary  -  but they can.    The only restriction is that they can't both be stationary in the same frame of reference.

    Without worrying too much about special relativity,  just in ordinary Newtonian mechanics,  you can see that if A is stationary but B is moving towards it then there will be a collision.    Also if B was stationary but A was moving toward it then there will be a collision.   Either one can be declared to be stationary,  you just can't declare both of them to be stationary.   
    If you examine that restriction carefully, it is just saying that there isn't a frame of reference (a simple co-ordinate system) where the velocity of A would be 0   AND ALSO  the velocity of B would be 0 at the same time.   

   Exactly the same can happen in special relativity -  A  or  B  can be stationary - it's completely your choice.    However, having made that choice you've effectively decided what frame of reference you will be using.   If the two objects were on a collision course then the other object won't be stationary in that frame.

    I don't know if it will help but I'll try and phrase this another way:   The object A does not have a velocity that is absolute and everyone agrees on.   The moment you decide or declare that it is stationary you have made a choice.   You have picked out a frame of reference that you will use from among many that could have been used.   That assigns a velocity of 0 to the object A.   However, that velocity does not become permanently imprinted on the object like some mark that everyone can see.   It is NOT an intrinsic property of that object, it is a frame-dependant property.     If you now declare that object B was actually stationary then you have (unavoidably) abandoned the old reference frame and decided to use a different one.   The velocity of object A in the old frame was 0 but that is not important, it wasn't an absolute or intrinsic property that object A had - and it does not remain as the velocity of object A now you're using a different frame of reference.

Best Wishes.
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Re: How does special relativity explain dimensional components ...
« Reply #10 on: 26/06/2022 01:11:39 »
Quote from: Eternal Student on 25/06/2022 23:48:04
Hi.

    You're suggesting that A and B can't both be stationary  -  but they can.    The only restriction is that they can't both be stationary in the same frame of reference.

    Without worrying too much about special relativity,  just in ordinary Newtonian mechanics,  you can see that if A is stationary but B is moving towards it then there will be a collision.    Also if B was stationary but A was moving toward it then there will be a collision.   Either one can be declared to be stationary,  you just can't declare both of them to be stationary.   
    If you examine that restriction carefully, it is just saying that there isn't a frame of reference (a simple co-ordinate system) where the velocity of A would be 0   AND ALSO  the velocity of B would be 0 at the same time.   

   Exactly the same can happen in special relativity -  A  or  B  can be stationary - it's completely your choice.    However, having made that choice you've effectively decided what frame of reference you will be using.   If the two objects were on a collision course then the other object won't be stationary in that frame.

    I don't know if it will help but I'll try and phrase this another way:   The object A does not have a velocity that is absolute and everyone agrees on.   The moment you decide or declare that it is stationary you have made a choice.   You have picked out a frame of reference that you will use from among many that could have been used.   That assigns a velocity of 0 to the object A.   However, that velocity does not become permanently imprinted on the object like some mark that everyone can see.   It is NOT an intrinsic property of that object, it is a frame-dependant property.     If you now declare that object B was actually stationary then you have (unavoidably) abandoned the old reference frame and decided to use a different one.   The velocity of object A in the old frame was 0 but that is not important, it wasn't an absolute or intrinsic property that object A had - and it does not remain as the velocity of object A now you're using a different frame of reference.

Best Wishes.
I understand what you are saying.  However, there does seem to be an absolute difference between *only* the ct vector (the direction of time) and having a component of ct *with* a component in the x direction.  From an objective point of view, it should be noticeable which object has more of which component. 

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Re: How does special relativity explain dimensional components ...
« Reply #11 on: 26/06/2022 13:53:23 »
Quote from: Dimensional on 26/06/2022 01:11:39
From an objective point of view, it should be noticeable which object has more of which component.
That is implying a preferred frame.  In other words it seems you are saying 'objectively' one of them is 'really' moving in the x direction.  The fact of the matter is the movement can only be relative to another object so there can be no absolute velocity. 
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Re: How does special relativity explain dimensional components ...
« Reply #12 on: 26/06/2022 19:08:23 »
Quote from: Origin on 26/06/2022 13:53:23
Quote from: Dimensional on 26/06/2022 01:11:39
From an objective point of view, it should be noticeable which object has more of which component.
That is implying a preferred frame.  In other words it seems you are saying 'objectively' one of them is 'really' moving in the x direction.  The fact of the matter is the movement can only be relative to another object so there can be no absolute velocity.
But since they are moving towards each other, (keeping this in special relativity) doesn't that mean that at least one of them must be moving in the x direction?
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Re: How does special relativity explain dimensional components ...
« Reply #13 on: 26/06/2022 20:24:27 »
Quote from: Dimensional on 26/06/2022 19:08:23
But since they are moving towards each other, (keeping this in special relativity) doesn't that mean that at least one of them must be moving in the x direction?
Yes, there's clearly movement on the x-axis.  A could say only B is moving and B could say that only A is moving and a 3rd observer may say both A and B are moving and they would all be correct.  No one however can say which is moving in an absolute way.
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Re: How does special relativity explain dimensional components ...
« Reply #14 on: 26/06/2022 22:52:34 »
Quote from: Origin on 26/06/2022 20:24:27
Quote from: Dimensional on 26/06/2022 19:08:23
But since they are moving towards each other, (keeping this in special relativity) doesn't that mean that at least one of them must be moving in the x direction?
Yes, there's clearly movement on the x-axis.  A could say only B is moving and B could say that only A is moving and a 3rd observer may say both A and B are moving and they would all be correct.  No one however can say which is moving in an absolute way.
Since we know for certain that at least one of A or B is moving (or both), then how can either claim to not be moving?
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Re: How does special relativity explain dimensional components ...
« Reply #15 on: 27/06/2022 02:52:07 »
Quote from: Dimensional on 26/06/2022 22:52:34
Since we know for certain that at least one of A or B is moving (or both), then how can either claim to not be moving?
Because velocity is relative.  Let me say it again, velocity is relative.  There is no way to tell If A is moving towards B or B is moving towards A.  I can say A is moving and B is stationary relative to some arbitrary reference point or I could say the opposite based on difference arbitrary reference point.  So either can claim to be stationary because it is all relative.
Quote
How can either claim to be not moving?
I can give a sleeping pill and after you are asleep, I could load you on a spaceship with no windows and send you into space at 1,000,000 kph.  When you awake traveling at 1,000,000 kph it would not feel like you were moving.  As a matter of fact there is no test that you can perform that would tell you if you were traveling at 1,000,000 kph or 0 kph.
« Last Edit: 27/06/2022 03:35:25 by Origin »
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Re: How does special relativity explain dimensional components ...
« Reply #16 on: 27/06/2022 06:10:54 »
Quote from: Origin on 27/06/2022 02:52:07
Quote from: Dimensional on 26/06/2022 22:52:34
Since we know for certain that at least one of A or B is moving (or both), then how can either claim to not be moving?
Because velocity is relative.  Let me say it again, velocity is relative.  There is no way to tell If A is moving towards B or B is moving towards A.  I can say A is moving and B is stationary relative to some arbitrary reference point or I could say the opposite based on difference arbitrary reference point.  So either can claim to be stationary because it is all relative.
Quote
How can either claim to be not moving?
I can give a sleeping pill and after you are asleep, I could load you on a spaceship with no windows and send you into space at 1,000,000 kph.  When you awake traveling at 1,000,000 kph it would not feel like you were moving.  As a matter of fact there is no test that you can perform that would tell you if you were traveling at 1,000,000 kph or 0 kph.
I don't think my point has come across properly. 

Imagine the spatial (x), temporal (y) plane.  Only one of A and B can have 0 velocity in the x direction, and this is true before you choose the reference point. 
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Re: How does special relativity explain dimensional components ...
« Reply #17 on: 27/06/2022 10:54:24 »
Hi.

Quote from: Dimensional on 27/06/2022 06:10:54
I don't think my point has come across properly. 
    I think your point has been understood.   It's just not agreed and the difficulty now is only trying to explain why.

Quote from: Dimensional on 27/06/2022 06:10:54
Imagine the spatial (x), temporal (y) plane.  Only one of A and B can have 0 velocity in the x direction, and this is true before you choose the reference point.
     Let's just start by saying no:   It's not so much that only one of them can 0 velocity which is important,   instead it's that either one of them could have 0 velocity.   Before you choose a reference frame there's nothing at all that restricts or informs you about the velocity of A or B.   You have completely free choice to assign any velocity you want to A or B,   restrictions for the velocity of the other object don't appear until after you have decided what the velocity of the first object will be.

 
Quote from: Deecart on 26/06/2022 20:58:56
Can someone explain me why we talk about "light cone"  ?
    The original Poster (OP) has requested that you put this question in a new thread.   That may help to avoid confusing the OP.
Mod edit: New topic here https://www.thenakedscientists.com/forum/index.php?topic=85049.0

Best Wishes.
« Last Edit: 27/06/2022 14:14:27 by Halc »
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Re: How does special relativity explain dimensional components ...
« Reply #18 on: 27/06/2022 21:01:07 »
Quote from: Eternal Student on 27/06/2022 10:54:24

     Let's just start by saying no:   It's not so much that only one of them can 0 velocity which is important,   instead it's that either one of them could have 0 velocity.   Before you choose a reference frame there's nothing at all that restricts or informs you about the velocity of A or B.   You have completely free choice to assign any velocity you want to A or B,   restrictions for the velocity of the other object don't appear until after you have decided what the velocity of the first object will be.
I definitely think that we both understand each other's argument. 

Here may be a clearer way to see my issue.  Imagine a very simple universe where there only exists object A and an object B.  They are on a collision course.  Scientist A, (from another dimension) uses object A as a point of reference.  But scientist B (from yet a different dimension than scientist A) uses object B as a reference.  Each scientist is going to end up with a different description of this universe from using special relativity.

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Re: How does special relativity explain dimensional components ...
« Reply #19 on: 27/06/2022 22:10:12 »
Quote from: Dimensional on 27/06/2022 21:01:07
Here may be a clearer way to see my issue.  Imagine a very simple universe where there only exists object A and an object B.  They are on a collision course.  Scientist A, (from another dimension) uses object A as a point of reference.  But scientist B (from yet a different dimension than scientist A) uses object B as a reference.  Each scientist is going to end up with a different description of this universe from using special relativity.
So what's your issue?
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