# Co ordinates in Space Time

• September 21st, 2017, 08:17 AM
geordief
Co ordinates in Space Time
Suppose we have an "origin" in SpaceTime.

It emits regular pulses of light .

From this origin (O) 2 objects(A and B) are sent in orthogonal directions (A traveling 4/3 times faster than B) and at all times they use measuring equipment (light emitters and receivers) to measure their current distance from O and also each other.

After a predetermined length of time (the same length of time measured by both A and B) both A and B measure their distance from each other and also O.

If A measures (a multiple of)4 light seconds to O and B measures (the same multiple of ) 3 light seconds to O is it the case that both A and B will measure (the same multiple of ) 5 light seconds to each other in the absence of any massive bodies causing a local gravitational field?

Is this what is meant by Space Time being flat in the absence of gravity?
And if 5 light seconds is not measured does this show that Space Time has in fact been curved by the presence of a massive body?

If there are no mistakes here ,can I pat myself on the back?:)

EDIT:I now realize that the relative motions of A and B and O complicates(=messes up) the calculation but can I factor them out and make them (unrealistically) stationary wrt each other) at the moment of measurement?

Do my "flat" and "curved" descriptions now hold?
• September 21st, 2017, 04:46 PM
mathman
The measurements will be as you say, unless the speeds are a significant fraction of the speed of light.
• September 21st, 2017, 07:05 PM
Janus
Quote:

Originally Posted by geordief
Suppose we have an "origin" in SpaceTime.

It emits regular pulses of light .

From this origin (O) 2 objects(A and B) are sent in orthogonal directions (A traveling 4/3 times faster than B) and at all times they use measuring equipment (light emitters and receivers) to measure their current distance from O and also each other.

After a predetermined length of time (the same length of time measured by both A and B) both A and B measure their distance from each other and also O.

If A measures (a multiple of)4 light seconds to O and B measures (the same multiple of ) 3 light seconds to O is it the case that both A and B will measure (the same multiple of ) 5 light seconds to each other in the absence of any massive bodies causing a local gravitational field?

Is this what is meant by Space Time being flat in the absence of gravity?
And if 5 light seconds is not measured does this show that Space Time has in fact been curved by the presence of a massive body?

If there are no mistakes here ,can I pat myself on the back?:)

EDIT:I now realize that the relative motions of A and B and O complicates(=messes up) the calculation but can I factor them out and make them (unrealistically) stationary wrt each other) at the moment of measurement?

Do my "flat" and "curved" descriptions now hold?

Here's a table showing in order, A's speed relative to O, B's speed relative to O, A and B's relative speed to each other as measured by A or B, and the ratio of this speed to the speed of A relative to O. It increases by increments of 0.1c for A's speed.

 0.1 0.133 0.23 2.3 0.2 0.266 0.443 2.22 0.3 0.4 0.625 2.08 0.4 0.53 0.769 1.92 0.5 0.667 0.875 1.75 0.6 0.8 0.946 1.58 0.7 0.933 0.988 1.41

Notice that as A's speed increases, even though the ratio between A and B's speeds with respect to O remain a constant, the ratio between the relative speed between A and B compared to the speed of A decreases.

Thus if A is moving at 0.3c relative to O and B is moving at 0.4c relative to O, then after n seconds by its clock, A will measure the distance between it and 0 as being n*0.3 light sec, B will be n*0.4 light sec from O in n seconds by its clock and A and B will measure their relative distances as being n*0.625 light sec. If A is moving at 0.6c, these distances will be n*0.6 light sec, 0.8 light sec and 0.946 light sec.
• September 21st, 2017, 08:20 PM
geordief
Quote:

Originally Posted by Janus
Quote:

Originally Posted by geordief
Suppose we have an "origin" in SpaceTime.

It emits regular pulses of light .

From this origin (O) 2 objects(A and B) are sent in orthogonal directions (A traveling 4/3 times faster than B) and at all times they use measuring equipment (light emitters and receivers) to measure their current distance from O and also each other.

After a predetermined length of time (the same length of time measured by both A and B) both A and B measure their distance from each other and also O.

If A measures (a multiple of)4 light seconds to O and B measures (the same multiple of ) 3 light seconds to O is it the case that both A and B will measure (the same multiple of ) 5 light seconds to each other in the absence of any massive bodies causing a local gravitational field?

Is this what is meant by Space Time being flat in the absence of gravity?
And if 5 light seconds is not measured does this show that Space Time has in fact been curved by the presence of a massive body?

If there are no mistakes here ,can I pat myself on the back?:)

EDIT:I now realize that the relative motions of A and B and O complicates(=messes up) the calculation but can I factor them out and make them (unrealistically) stationary wrt each other) at the moment of measurement?

Do my "flat" and "curved" descriptions now hold?

Here's a table showing in order, A's speed relative to O, B's speed relative to O, A and B's relative speed to each other as measured by A or B, and the ratio of this speed to the speed of A relative to O. It increases by increments of 0.1c for A's speed.

 0.1 0.133 0.23 2.3 0.2 0.266 0.443 2.22 0.3 0.4 0.625 2.08 0.4 0.53 0.769 1.92 0.5 0.667 0.875 1.75 0.6 0.8 0.946 1.58 0.7 0.933 0.988 1.41

Notice that as A's speed increases, even though the ratio between A and B's speeds with respect to O remain a constant, the ratio between the relative speed between A and B compared to the speed of A decreases.

Thus if A is moving at 0.3c relative to O and B is moving at 0.4c relative to O, then after n seconds by its clock, A will measure the distance between it and 0 as being n*0.3 light sec, B will be n*0.4 light sec from O in n seconds by its clock and A and B will measure their relative distances as being n*0.625 light sec. If A is moving at 0.6c, these distances will be n*0.6 light sec, 0.8 light sec and 0.946 light sec.

Thanks,yes I understand that.

Is it possible to introduce two additional objects ,a and b?

Both a and b also have light emitters and receivers.

If a meets A at n seconds on A's clock and b meets B at n seconds on B's clock but both a and b are stationary wrt O at that point do O ,a and b now measure out a perfect Pythagorean triangle?

In other words if a ,b and O exchange light signals at n seconds on A's and B's clock (a being in the immediate vicinity of A and b being in the immediate vicinity of B) will the square of the distance measured between a and b be exactly equal to the sum of the squares of the distances between A and O plus B and O?

In a region removed from gravity fields ,of course.
• September 22nd, 2017, 11:22 AM
Janus
Quote:

Originally Posted by geordief
Quote:

Originally Posted by Janus
Quote:

Originally Posted by geordief
Suppose we have an "origin" in SpaceTime.

It emits regular pulses of light .

From this origin (O) 2 objects(A and B) are sent in orthogonal directions (A traveling 4/3 times faster than B) and at all times they use measuring equipment (light emitters and receivers) to measure their current distance from O and also each other.

After a predetermined length of time (the same length of time measured by both A and B) both A and B measure their distance from each other and also O.

If A measures (a multiple of)4 light seconds to O and B measures (the same multiple of ) 3 light seconds to O is it the case that both A and B will measure (the same multiple of ) 5 light seconds to each other in the absence of any massive bodies causing a local gravitational field?

Is this what is meant by Space Time being flat in the absence of gravity?
And if 5 light seconds is not measured does this show that Space Time has in fact been curved by the presence of a massive body?

If there are no mistakes here ,can I pat myself on the back?:)

EDIT:I now realize that the relative motions of A and B and O complicates(=messes up) the calculation but can I factor them out and make them (unrealistically) stationary wrt each other) at the moment of measurement?

Do my "flat" and "curved" descriptions now hold?

Here's a table showing in order, A's speed relative to O, B's speed relative to O, A and B's relative speed to each other as measured by A or B, and the ratio of this speed to the speed of A relative to O. It increases by increments of 0.1c for A's speed.

 0.1 0.133 0.23 2.3 0.2 0.266 0.443 2.22 0.3 0.4 0.625 2.08 0.4 0.53 0.769 1.92 0.5 0.667 0.875 1.75 0.6 0.8 0.946 1.58 0.7 0.933 0.988 1.41

Notice that as A's speed increases, even though the ratio between A and B's speeds with respect to O remain a constant, the ratio between the relative speed between A and B compared to the speed of A decreases.

Thus if A is moving at 0.3c relative to O and B is moving at 0.4c relative to O, then after n seconds by its clock, A will measure the distance between it and 0 as being n*0.3 light sec, B will be n*0.4 light sec from O in n seconds by its clock and A and B will measure their relative distances as being n*0.625 light sec. If A is moving at 0.6c, these distances will be n*0.6 light sec, 0.8 light sec and 0.946 light sec.

Thanks,yes I understand that.

Is it possible to introduce two additional objects ,a and b?

Both a and b also have light emitters and receivers.

If a meets A at n seconds on A's clock and b meets B at n seconds on B's clock but both a and b are stationary wrt O at that point do O ,a and b now measure out a perfect Pythagorean triangle?

In other words if a ,b and O exchange light signals at n seconds on A's and B's clock (a being in the immediate vicinity of A and b being in the immediate vicinity of B) will the square of the distance measured between a and b be exactly equal to the sum of the squares of the distances between A and O plus B and O?

In a region removed from gravity fields ,of course.

Let's use A moving at 0.6c and B moving at 0.8 c for a text case. ( these numbers not only fit your A to B speed ratio, but also lead to nice neat numbers to work with in the rest of the problem because at 0.6c the time dilation factor is 0.8 and at 0.8c it is 0.6)

First consider the frame of a,O, and b. Clock A is traveling from O to a at 0.6c. A will travel 0.6 light sec for every sec that passes in this frame. However, A's clock is time dilated in this frame by a factor of 0.8 Which means that in order for A to pass a when A's clock reads n seconds, a has to be n*0.6/0.8 = n*3/4 light sec from 0 as measured in the rest frame of O, a, and b. B is traveling from 0 to b, and by the same reasoning above, b has to be n* 0.8/06 = n* 4/3 light sec from O for B to pass b when B's clock reads n sec. This means that in the a, O, b rest frame the ratio between the a-O b_O distances is 9:16 not 3:4 The a-b distance will be n* sqrt(337/144) or n* sqrt(137)/12

Now consider the frame of A. for it to reach a when its own clock reads n sec, the a_O distance as measured by A needs to be n*0.6 light sec. The O-b distance will be the same as measured in the a-O-b frame or n* 4/3 light sec. The ratio of these distances is 9:20, neither 3:4 or the 9:16 measured in the a-0-b frame.

In the rest frame of B, the O-b distance is n*0.8 light sec and the O-a distance is n* 3/4 light sec, which leads to a ratio of 15:16 (according to B, the distances O-a and O-b are nearly the same.)

So no, the relationship between the O-a and O-b distances are not the same in all three frames.
• September 24th, 2017, 03:34 AM
geordief
Quote:

Originally Posted by Janus

Let's use A moving at 0.6c and B moving at 0.8 c for a text case. ( these numbers not only fit your A to B speed ratio, but also lead to nice neat numbers to work with in the rest of the problem because at 0.6c the time dilation factor is 0.8 and at 0.8c it is 0.6)

First consider the frame of a,O, and b. Clock A is traveling from O to a at 0.6c. A will travel 0.6 light sec for every sec that passes in this frame. However, A's clock is time dilated in this frame by a factor of 0.8 Which means that in order for A to pass a when A's clock reads n seconds, a has to be n*0.6/0.8 = n*3/4 light sec from 0 as measured in the rest frame of O, a, and b. B is traveling from 0 to b, and by the same reasoning above, b has to be n* 0.8/06 = n* 4/3 light sec from O for B to pass b when B's clock reads n sec. This means that in the a, O, b rest frame the ratio between the a-O b_O distances is 9:16 not 3:4 The a-b distance will be n* sqrt(337/144) or n* sqrt(137)/12

Now consider the frame of A. for it to reach a when its own clock reads n sec, the a_O distance as measured by A needs to be n*0.6 light sec. The O-b distance will be the same as measured in the a-O-b frame or n* 4/3 light sec. The ratio of these distances is 9:20, neither 3:4 or the 9:16 measured in the a-0-b frame.

In the rest frame of B, the O-b distance is n*0.8 light sec and the O-a distance is n* 3/4 light sec, which leads to a ratio of 15:16 (according to B, the distances O-a and O-b are nearly the same.)

So no, the relationship between the O-a and O-b distances are not the same in all three frames.

Thanks again. I have been able to follow your maths and explanation (with a small difficulty that I will mention at the end)

My main point though is that (which may seem obvious to you) is that in all 3 reference frames the a-b distance is calculated using the familiar Pythagorean formula. Is this because spacetime is indeed flat in the absence of local gravitational fields?

To repeat , although this may seem obvious to you (and anyone else) it has taken me perhaps 5 or more years since I began trying to understand the subject before this simple result has become plain to me .(fact can be simpler than fiction to mangle a well known expression)

I have been in all that time labouring under the unconscious half assumption that even in a rest frame (and in the absence of a local gravitational field) the angles of triangles drawn in spacetime might not add up to 180- on account of its 4 dimensional character. Sometimes (often) an illusion lost can be as or more valuable than knowledge gained.:)

To come back to the minor stumble I still have in your explanation it is this s following section:

"The O-b distance will be the same as measured in the a-O-b frame or n* 4/3 light sec"

I don't dispute that (again it must be obvious to most ) but I don't actually "get it" ** . It is a minor point though and maybe it is just down to my unfamiliarity with working through similar examples ....

perhaps it will sink in eventually;-)
• September 24th, 2017, 09:58 AM
Janus
Quote:

Originally Posted by geordief

Thanks again. I have been able to follow your maths and explanation (with a small difficulty that I will mention at the end)

My main point though is that (which may seem obvious to you) is that in all 3 reference frames the a-b distance is calculated using the familiar Pythagorean formula. Is this because spacetime is indeed flat in the absence of local gravitational fields?

To repeat , although this may seem obvious to you (and anyone else) it has taken me perhaps 5 or more years since I began trying to understand the subject before this simple result has become plain to me .(fact can be simpler than fiction to mangle a well known expression)

In all three frames, the O-a,0-b, and a-b distances form a right triangle that follow the rules of the Pythagorean Theorem, they just won't agree as to the exact ratios of the distance lengths. Basically, in flat space-time, the rules of Euclidean geometry apply, while in curved space-time, they do not.
Quote:

I have been in all that time labouring under the unconscious half assumption that even in a rest frame (and in the absence of a local gravitational field) the angles of triangles drawn in spacetime might not add up to 180- on account of its 4 dimensional character. Sometimes (often) an illusion lost can be as or more valuable than knowledge gained.:)

To come back to the minor stumble I still have in your explanation it is this s following section:

"The O-b distance will be the same as measured in the a-O-b frame or n* 4/3 light sec"

I don't dispute that (again it must be obvious to most ) but I don't actually "get it" ** . It is a minor point though and maybe it is just down to my unfamiliarity with working through similar examples ....

perhaps it will sink in eventually;-)
A has a relative velocity to the a-O-b triangle which it parallel to the a-O side. This one would expect the a-O distance to be length contracted(shorter than that as measured in the a-O-b frame) as measured by A.
The b_O side of the triangle is perpendicular to the relative motion and thus would not be measured by A as being length contracted. Therefore A will measure this distance as being the same as that measured in the a-O-b frame.
• September 24th, 2017, 11:21 AM
geordief
Quote:

Originally Posted by Janus

A has a relative velocity to the a-O-b triangle which it parallel to the a-O side. This one would expect the a-O distance to be length contracted(shorter than that as measured in the a-O-b frame) as measured by A.
The b_O side of the triangle is perpendicular to the relative motion and thus would not be measured by A as being length contracted. Therefore A will measure this distance as being the same as that measured in the a-O-b frame.

Yes, I see that now (forgot about/didn't see the significance of the perpendicularity of b_O to A's FoR**.)

** is it correct to say "the perpendicularity of b_O to A's FoR" or should I have said "the perpendicularity of b_O to A's motion"?

Is it possible to have frames of reference perpendicular to each other?
• September 25th, 2017, 03:31 PM
Guitarist
Quote:

Originally Posted by geordief
Is it possible to have frames of reference perpendicular to each other?

No. A so-called frame of reference refers to an arbitrary choice of coordinates - in relativity there are 4 of these coordinates - which for convenience are each taken to be mutually perpendicular. No pair of choices of these 4 coordinate (taken together) can be perpendicular.

Of course in non-flat spacetime this is a bit more complicated
• September 25th, 2017, 04:58 PM
geordief
Quote:

Originally Posted by Guitarist
No. A so-called frame of reference refers to an arbitrary choice of coordinates - in relativity there are 4 of these coordinates - which for convenience are each taken to be mutually perpendicular. No pair of choices of these 4 coordinate (taken together) can be perpendicular.

Of course in non-flat spacetime this is a bit more complicated

Thanks. If there are two FoR s moving wrt each other ,is it possible for one or more of its axes to be perpendicular to an axis in the second FoR?

Maybe that is what Janus was describing in post#7? (If I am not hopelessly confused:confused:)
• September 27th, 2017, 10:52 AM
Guitarist
[QUOTE=geordief;607529]
Quote:

Originally Posted by Guitarist
Thanks. If there are two FoR s moving wrt each other ,is it possible for one or more of its axes to be perpendicular to an axis in the second FoR?

I repeat - the term "frame of reference" is a silly term that physicists use to describe a particular set of coordinates. These are usually taken to be ( because you want all coordinate axes to "measure" the same sort of thing - distance)

Relativistic theories say that you are free to choose a coordinate set relative to which an object is stationary (the physicist's "rest frame") or a set relative to which the same object is moving (the physicist's "moving frame"). This does not - repeat NOT - imply the the coordinates themselves are moving. They cannot.

There is a well-known way to relate these 2 sets of coordinates - the so-called Lorentz transformation.

And given the freedom of choice of coordinate set, then for any object in constant motion relative one coordinate set with an arbitrary direction to it, it is always possible to choose another coordinate set - frame of reference - in which the same object is moving parallel to a single axis.

In this case, the so-called Lorentz transformation will affect only 2 of the 4 possible coordinate axes (the is always one of them)