Is the interval the height of the triangle?

[Saw]

A FAQ is why the spacetime interval has a minus sign.

A geometrical answer is that the interval is always the shortest path (the straight line) between two points. If we talk about spatial points, that interval is the hypotenuse of a right angle triangle, whose sides are the projections of such path over the X and Y axes of any coordinate system, no matter its orientation (that is, after rotation). If we talk instead about spacetime points (events), the shortest path turns out to be, instead, the height of the triangle, formed in this case by the projections over the Time (hypotenuse) and the X (basis) axes.

One does not visualize that, however, in a Minkowski diagrams, where you either have:

(i) two independent drawings (one for each frame) or
(ii) two overlapping drawings, where one has perpendicular axes, whilst the other's are rotated in opposite directions.

I have looked for an alternative. Epstein diagrams (see the book "Relativity visualized") are interesting, because he plays with proper time, which is after all what goes in 4-velocity, but they are single diagrams. A step further was this picture, based on which I have drawn this other one:

A little explanation:

v = 0.5 c

The thick blue line connecting (timelike) events P and R is the proper time and the height of the triangle whose hypotenuse is the red coordinate time and whose basis is the red coordinate length separating those events.

The thick red line connecting (timelike) events P and Q is the proper time and also the height of the triangle whose hypotenuse is the blue coordinate time and whose basis is the blue coordinate length between those events.

Interesting, isn´t it? Any comment?

 

[bobc2]

A FAQ is why the spacetime interval has a minus sign.

One does not visualize that, however, in a Minkowski diagrams, where you either have:

(i) two independent drawings (one for each frame) or
(ii) two overlapping drawings, where one has perpendicular axes, whilst the other's are rotated in opposite directions.

Not true. You missed one of the Minkowski diagrams (from Loedel). Here we use a symmetric space-time diagram. You can use the Pythagorean theorem directly. There's no need for hyperbolic calibration curves. The red and blue coordinates have the same scale.

I haven't quite figured out your diagram, because it's not obvious how you would draw a photon world line that would give a ratio of 1:1 in both blue and red coordinates. And what would be the meaning of a pair of orthogonal coordinates added to this diagram?

[edit] On closer inspection of your diagram, it looks like your blue and red coordinates may be scaled the same after all, and your diagram seems to be the equivalent of mine--yours is just rotated so as to place one time axis in a vertical position. I'll look at it a little more. However, in my diagram it is pretty obvious how the hyperbolic calibration curves lay on the sketch. I think the calibration curves are going to look a little strange on your diagram. What are those yellow lines in your diagram? First, I thought they were photon world lines, but they don't seem to bisect the angle between X4 and X1.

 

[Saw]

On closer inspection of your diagram, it looks like your blue and red coordinates may be scaled the same after all, and your diagram seems to be the equivalent of mine--yours is just rotated so as to place one time axis in a vertical position.

Yes, and this different rotation also places the X axis of the other frame in the horizontal position. But for the rest it also looks to me as the same diagram as yours.

I think the calibration curves are going to look a little strange on your diagram.

How do you that?

What are those yellow lines in your diagram? First, I thought they were photon world lines, but they don't seem to bisect the angle between X4 and X1.

They are the photon lines. The original drawing was the typical train example where light is flashed from the mid-point of the train to its back and front and reflects back. If you let the light be trapped in a wagon in the blue frame, you get a similar result, where light pulses hit equidistant places in equal times and reflect back to meet in the origin point.

It is true that if you look at the diagram as a 2D one, in my display the photon lines do not bisect the coordinate systems. In this respect, the orientation you use is clearer. By the way, thanks for those pictures, I had seen them from time to time but never scrutinized them and had not realized they are what I was looking for. Yet the advantage of the display I have used may be to reveal that the yellow lines still bisect all coordinate systems if you look at the diagram as a 3D one.

 

[Saw]

If you let the light be trapped in a wagon in the blue frame, you get a similar result, where light pulses hit equidistant places in equal times and reflect back to meet in the origin point.

I now tried to draw that and realized it is not true. In fact, it is not true in the red frame, either. The only thing that the diagram truly reflects seems to be, after all, the events.

 

[bobc2]

I now tried to draw that and realized it is not true. In fact, it is not true in the red frame, either. The only thing that the diagram truly reflects seems to be, after all, the events.

Saw, again, your diagram is fully equivalent to mine. Except, you should have drawn the photon world lines so that they bisect the angle between X4 and X1 for both blue and red coordinate systems. I've added in photon world lines (green) in this manner to assure the speed of light is the same in both coordinate systems.

Illustrating the train-lighting strike example, I'm doing the version in which the lightning at two ends of the train are simultaneous in the blue frame of reference. So, the initial strikes are events A and B. The red guy, sitting in the middle of his train, sees the flash from A at the event C. The blue guy, sitting in the middle of his train, sees flashes from both A and B simultaneously at event D. Finally, the red guy sees the flash from B as event E (he first sees the A flash at event C, then much later sees the B flash as event E). So, again we have the interesting aspect of special relativity that events simultaneous for one observer will not be simultaneous for another observer who is in motion relative to the first.

Also, I can see how to put in the hyperbolic calibration curves that keep track of proper times for any observer moving from the origin at any speed relative to your blue and red observers. Maybe I'll put them in (a little too lazy right now).

I really appreciate you sharing this space-time diagram. It is a useful tool for helping with thinking through implications of special relativity and space-time. Thanks.

 

[Saw]

Saw, again, your diagram is fully equivalent to mine. Except, you should have drawn the photon world lines so that they bisect the angle between X4 and X1 for both blue and red coordinate systems. I've added in photon world lines (green) in this manner to assure the speed of light is the same in both coordinate systems.

Great! How stupid I was. I should have realized that, if the two diagrams are equivalent, they just differ in the angle of rotation, mine should also allow for a bisecting photon line, just with a different inclination...

 

[Saw]

I have corrected the picture trying to make the yellow line bisecting the two coordinate systems. It is not very precisely drawn but I think it exemplifies well relativity of simultaneity and time dilation. I hesitate about how to show length contraction.

 

[bobc2]

Good job on that, Saw. You would have a hard time showing the length contraction, because your two train cars are not the same size to begin with. I've prepared another sketch using my original symmetric space-time diagram for red and blue. Red and blue objects have the same sizes in each one's rest frame.

The red object is shorter than the blue object in Blue's instantaneous simultaneous 3-D space. And the blue object is shorter than the red object in Red's instantaneous simultaneous 3-D space.

But you can see why each guy "sees" the other's object as being shorter than his own. It's because they are living in different 3-D cross-sections of the 4-dimensional universe. Their different cross-section vews account for the different lengths.

You can also see why the time dilation happens in special relativity. For this example we will say that 4-dimensional objects in the form of clocks are there in 4-dimensional space along with the blue and red 4-dimensional objects (The tip end of one hand on a clock would be a fiber in the 4-dimensional configuration of a spiral with axis extending along the 4th dimension). The red object (instantaneous 3-D object) distance from the origin is not as great as the blue object distance from the origin as viewed by the blue guy in his (blue's) simultaneous space. And adapting the convention that each observer moves along his own X4 axis at the speed of light, c, then a change in clock time along the X4 dimension is dt = dX4/c. Thus, Blue "sees" and earlier time on the red object's clock. And correspondingly, of course the red guy would see an earlier time on the blue object's clock. So, each one "sees" the other guy's clock as running slow (time dilation).

 

[Saw]

You would have a hard time showing the length contraction, because your two train cars are not the same size to begin with. I've prepared another sketch using my original symmetric space-time diagram for red and blue. Red and blue objects have the same sizes in each one's rest frame.

Yes, definitely your display makes things easier. It is true that my cars are not the same size and I do not know how to make them so. However, at least, what the diagram should show is that a given car's length is longer in its rest frame (= when calculated as per its own simultaneity line) than in the other frame (= when measured as per the latter's simultaneity line). Thus in my drawing it is clear that the blue car is shorter when cut by the red horizontal line than when cut by its blue tilted line. But if we cut the red car with the blue tilted line, it becomes longer than in its own frame... I wonder why.

 

[Saw]

in my drawing it is clear that the blue car is shorter when cut by the red horizontal line than when cut by its blue tilted line. But if we cut the red car with the blue tilted line, it becomes longer than in its own frame... I wonder why.

I have realized the second part was not true. In this case, it is not so apparent but LC is there, the red car as cut by the blue tilted line is also shorter. I have painted it below as a fucsia line, which is actually shorter than the red car in its own frame.

 

[bobc2]

I have realized the second part was not true. In this case, it is not so apparent but LC is there, the red car as cut by the blue tilted line is also shorter. I have painted it below as a fucsia line, which is actually shorter than the red car in its own frame.

I've rotated my diagram by 90 degrees, so it is the same as yours now, except that my blue and red are the same size as 4-dimensional objects.

[edit] I jumped to the conclusion too quickly that red and blue cars must be the same size to demonstrate length contraction. That's true only if you are comparing the length of the blue car to the red for the demonstration of length contraction (as I was doing in my sketch). However, you have demonstrated it quite well in comparing the lengths measured in two different coordinate systems on the same car. You demonstrated this quite effectively for the red car and then the blue car. Good job, Saw.

 

[Saw]

I've rotated my diagram by 90 degrees, so it is the same as yours now, except that my blue and red are the same size as 4-dimensional objects.

Thank you. So I have learned about these Loedel diagrams. I have googled a little and seen a few good explanations on the subject, including some here in PF. It appears that their problem is if one introduces more frames, but I like them for the reason stated at the OP. They are consistent with the math in that they show the interval as the side of the right triangle, with full reciprocity. If I am not mistaken, the standard Minkowski diagram joining two frames lacks that symmetry. If the proper time path is in the worldline of a particle at rest in the non-rotated frame, it is the side of the triangle. But if it is in the skewed frame, it is the hypotenuse. As to length contraction, the car of the rotated frame looks length contracted when cut by the horizontal simultaneity line, but not vice versa. I have made a drawing to check this:

[edit] I jumped to the conclusion too quickly that red and blue cars must be the same size to demonstrate length contraction. That's true only if you are comparing the length of the blue car to the red for the demonstration of length contraction (as I was doing in my sketch). However, you have demonstrated it quite well in comparing the lengths measured in two different coordinate systems on the same car. You demonstrated this quite effectively for the red car and then the blue car. Good job, Saw.

Thanks again. I am an amateur on this, so that is encouraging . I have put all effects together (RS, TD and LC) in the same drawing:

 

[bobc2]

Thanks again. I am an amateur on this, so that is encouraging . I have put all effects together (RS, TD and LC) in the same drawing:

Nice work, Saw. You're no amateur. And I like the way these diagrams seem to suggest the spatial character of the 4th dimension with world lines expressed as parametric equations with time as a parameter:

X4 = ct
X1 = ct ...for the photon world line with velocity, c.

X4 = ct
X1 = vt ...for other world lines associated with objects moving with velocity, v, along the X1 direction.

It's just like parametric equations for objects moving in our familiar X-Y plane:

X = V_xt
Y = V_yt

 

[Saw]

I like the way these diagrams seem to suggest the spatial character of the 4th dimension with world lines expressed as parametric equations with time as a parameter:

X4 = ct
X1 = ct ...for the photon world line with velocity, c.

X4 = ct
X1 = vt ...for other world lines associated with objects moving with velocity, v, along the X1 direction.

It's just like parametric equations for objects moving in our familiar X-Y plane:

X = V_xt
Y = V_yt

Yes, with the difference that here to get the invariant interval, through combination of the two motions, in the T or X4 axis (at c) and in the X or X1 axis (at v), in math you use a minus sign (instead of plus) = geometrically the shortest path is not the hypotenuse but the side of the triangle. By the way, this means that the absolute space-time perspective is the one that sees the interval as the side of the triangle. For example, if we look at the distance between events P and Q, the blue frame would see that path as the hypotenuse. That is the wrong approach, for the purpose of finding invariance. Whose approach is the absolute one? I tend to think that it is the proper frame's one, but sometimes I hesitate. What do you think?

[Edit] to make it less conceptual or philosophical, to make it a purely geometrical question: under which perspective is the interval always the side of the triangle?

 

[Saw]

Whose approach is the absolute one? I tend to think that it is the proper frame's one, but sometimes I hesitate. What do you think?

[Edit] to make it less conceptual or philosophical, to make it a purely geometrical question: under which perspective is the interval always the side of the triangle?

Now I tend to think that the perspective in question (where there is a right triangle whose height is the interval) is the 2D plane where we draw (the 4D world).

 

[bobc2]

Now I tend to think that the perspective in question (where there is a right triangle whose height is the interval) is the 2D plane where we draw (the 4D world).

Not a bad analysis, Saw. So, the answer to your original question is yes--the interval is the height of the triangle. And that explains the negative sign needed to compute the length of world lines.

You might clarify your comment about the 2D plane where we draw the 4D world. I think I know what you are implying here.

 

[Saw]

You might clarify your comment about the 2D plane where we draw the 4D world. I think I know what you are implying here.

Here 2D is sinonimous of 4D. It is just that, as you well know, for simplicity, we merge all 3 spatial dimensions in 1 or we consider motion only in 1 spatial dimension. What "is" that absolute spacetime plane or world, I do not know, but what seems to be finding a place in my mind is that it is simply a way for solving problems (causality problems) by combining the proper time contributed by one frame and the rest length contributed by the other. Note how this way the Loedel diagram we are using here, albeit rotated, becomes an Epstein diagram, albeit completed with the other frame's view.

In any case, the analysis would be incomplete without considering the other two cases, that is to say, where distance berween two events is lightlike or spacelike. I have started to complete the picture with those cases. No idea for the time being about how this affects the former analysis.

 

[Saw]

Two ideas for your comments before I forget them:

- With regard to the interval in the case of light like events, the frame where the interval is still the height of a triangle... Can it be the photon's frame? I know the concept is disputed but that is what has occurred to me.

- Let us talk in 2D, which we know means 4D. The space time perspective is 2D although inertial frames are rotated in a 3D space.

 

[bobc2]

Two ideas for your comments before I forget them:

- With regard to the interval in the case of light like events, the frame where the interval is still the height of a triangle... Can it be the photon's frame?

No.

Let us talk in 2D, which we know means 4D. The space time perspective is 2D although inertial frames are rotated in a 3D space.

Here is a comment from Harrylin from another post:

https://www.physicsforums.com/newreply.php?do=newreply&p=3748210

Addendum: there are competing explanations for SR, just as there are also many competing explanations for quantum mechanics. Some main explanations of SR:
- the existence of a (3D) "ether" (also called physical space or vacuum)
- the existence of a 4D physical "spacetime" (also called block universe or 4D ether)
- shut up and calculate (a non-explanation, but possibly most popular

An interpretation of your 2-D picture was discussed at length in the context of the block universe model of special relativity. The discussion was eventually locked since it was felt that there was zero physics content. But you can look at some of the discussion here:

https://www.physicsforums.com/showthread.php?t=561344

The idea advanced here was that your 2-D picture begins to look more and more like a real spatial 4-dimensional block universe. In keeping with the monitor's assessment, I don't think it is a good idea to pursue this subject.

 

[Saw]

Yes, I agree that is a thorny and probably little productive subject. Here the idea was to make a pure geometric analysis and I suggest returning to that.

The interval is the vector combination of time - space.

We had concluded that, for time-like events, that mathematical expression matches with what can be visualized in the 2D plane (the "screen plane") as hypotenuse - minor side of the right triangle = height = the path is the wordline of a particle present at both events.

The question of light-like and space-like events was pending.

As to space-like events, the right triangle is also there in the screen plane. Space minus time matches with hypotenuse - minor side of the right triangle = height. But our interval was the opposite, i.e. time - space = (in this case) minor side - hypotenuse = minus height and that gives off the square root of a negative number = an imaginary number = no path at all between the events.

As to light-like events, time - space = 0. I wondered where the right triangle was in this case. A side and a hypotenuse with the same length cannot be… unless one considers that this is a special case of triangle, a triangle with height = 0, that is to say, a straight line. This means that in the end the line connecting the two events is the hypotenuse of that triangle without a height, the photon's path (not the photon's frame as I had initially misinterpreted).

How does it sound now?

 

[bobc2]

As to light-like events, time - space = 0. I wondered where the right triangle was in this case. A side and a hypotenuse with the same length cannot be… unless one considers that this is a special case of triangle, a triangle with height = 0, that is to say, a straight line. This means that in the end the line connecting the two events is the hypotenuse of that triangle without a height, the photon's path (not the photon's frame as I had initially misinterpreted).

How does it sound now?

Your world line as the leg of the triangle still holds up here in the limit. I've sketched the situation using a sequence of the Loedel space-time diagrams, starting with sketch a) and proceding to step e) in the limit. The leg gets smaller and smaller as the red and blue observers approach the speed of light. In the limit the world lines of both red and blue are colinear with the photon world lines. The observers of course would not be able to attain the speed of light; nevertheless the implication is clear that your view of the leg of the triangle is consistent with special relativity in the contexts of time-like, space-like and light-like.

 

[Saw]

To tell you the truth, now I am not sure of anything. See this picture:

First question: should we choose as reference fixed spots or fixed instants, as per the black frame?

Second: in any case, after each step, the relative v is higher and the black frame (with its fixed spots or instants) remains the same, but is it really the same frame?

Third: let us choose anyhow fixed spots as per the black frame. This is what I think you have done. That would correspond to the right-hand side of my drawing. As the messenger between the two spots is faster, logically its proper time is shorter. Consequently, the segment that represents such proper time is also shorter. But in the limit, as v becomes c, a funny thing happens. The segment itself does not disappear. It overlaps the photon's path, so it has a length. It just happens that it has ceased to be the height of a right triangle.

 

[Saw]

I am trying to recapitulate and reformulate. The original question was ultimately a translation issue: given that the interval (the invariant space-time distance between two events) is "such and such" in algebraic language (in particular, a vector subtraction, time minus space), how do you express that in geometric language? I tend to think that post 20 was close to the answer, but would need better wording.

 

[bobc2]

I am trying to recapitulate and reformulate. The original question was ultimately a translation issue: given that the interval (the invariant space-time distance between two events) is "such and such" in algebraic language (in particular, a vector subtraction, time minus space), how do you express that in geometric language? I tend to think that post 20 was close to the answer, but would need better wording.

I don't see where you came up with an imaginary number in post #20. Where does "minus time" come from? What triangle were you using? Sides of triangles aren't imaginary. Also, I didn't understand the arrows going in opposite directions on the coordinates.

 

[Saw]

I don't see where you came up with an imaginary number in post #20.

As you know, the interval S (with c=1) = square root of (Δt^2 - Δx^2).

This is the length of the vector resulting from the vector subtraction "time - space".

When Δx > Δt --> Δx^2 > Δt^2 --> (Δt^2 - Δx^2) < 0 = negative number --> S = square root of a negative number = imaginary number. This means that the events are space-like = no causal influence between them.

Where does "minus time" come from?

S, which is the interval, is not the proper distance between space-like events, which is the opposite, that is to say, the vectorial subtraction "space - minus time" = square root of (Δx^2 - Δt^2), which for the same reason (Δx > Δt) is a positive number.

Also, I didn't understand the arrows going in opposite directions on the coordinates.

In which picture? Where?

 

[bobc2]

As you know, the interval S (with c=1) = square root of (Δt^2 - Δx^2).

This is the length of the vector resulting from the vector subtraction "time - space".

When Δx > Δt --> Δx^2 > Δt^2 --> (Δt^2 - Δx^2) < 0 = negative number --> S = square root of a negative number = imaginary number. This means that the events are space-like = no causal influence between them.

S, which is the interval, is not the proper distance between space-like events, which is the opposite, that is to say, the vectorial subtraction "space - minus time" = square root of (Δx^2 - Δt^2), which for the same reason (Δx > Δt) is a positive number.

My only concern is why you are doing vector addition? I thought we were talking about lengths of legs and hypotenuses of triangles in 4-dimensional space. Legs of triangles don't have imaginary lengths. Did you ever encounter imaginary lengths in plane geometry class?

But, that's O.K. We can do the vector additions. But, draw me a picture of the triangle you are talking about, because I don't see where your negative value comes from. When I do the vector addition, I get

DX4 +DX1' = DX1 (where these are 4-D displacement vectors and DX1 is the hypotenuse)

DX1' = DX1 - DX4 (The hypotenuse, DX1, is larger than DX4)

In which picture? Where?

In post #22 all of your coordinates have arrows at the positive end of the coordinate and also at the negative end of the coordinate.

 

[Saw]

I don't see where your negative value comes from. When I do the vector addition, I get

DX4 +DX1' = DX1 (where these are 4-D displacement vectors and DX1 is the hypotenuse)

DX1' = DX1 - DX4 (The hypotenuse, DX1, is larger than DX4)

What you say is true:

DX1' (red leg) = DX1 (blue hypotenuse) - DX4 (blue leg)

But that is the "proper distance" AFAIK, not "the interval", which is:

DX4 (blue leg) - DX1 (blue hypotensue) or (in my notation) DT - DX

In post #22 all of your coordinates have arrows at the positive end of the coordinate and also at the negative end of the coordinate.

I thought it is the usual way to paint the axes of a coordinate system. You also do, it is only that you do not continue the line along the negative axis and so do not paint the second arrow. Does this answer your question or did I misunderstand it?

 

[bobc2]

But that is the "proper distance" AFAIK, not "the interval", which is:

DX4 (blue leg) - DX1 (blue hypotensue) or (in my notation) DT - DX

Again, I thought we were analyzing legs of triangles. But, anyway I get a positive interval (see sketch and interval below). I've also presented two different triangles to represent the same vector addition.

I thought it is the usual way to paint the axes of a coordinate system. You also do, it is only that you do not continue the line along the negative axis and so do not paint the second arrow.

No. You just put the arrow on the positive end of the coordinates. I extended a couple of the coordinates in the sketch below to illustrate this.

 

[Saw]

I have looked at the Wikipedia entry on the ST interval for orientation: http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals

Unfortunately, they use the (-+++) convention and so they get S^2 > 0 (positive number) in spacelike events. But, as they say, if you use the opposite convention, you get the reverse sign. In any case, what is important is that, whatever convention you use, there will be one case where S^2 is negative and hence S is imaginary.

I think that the (+---) convention is preferrable, because with it you get (i) S^2 and S positive in timelike events and (ii) S^2 negative and S imaginary in spacelike events, which can be translated physically as, respectively, (i) there is a causal path between the two events through a massive messenger (whose proper time is S) and (ii) there is no causal path. With both conventions, in lightlike events, S=0, which translates as there is a causal path through a massless messenger, whose proper time is 0.

My concern was only the expression of the same truths in geometric language.

I understand that an imaginary leg is an awkward concept in geometry but you may agree with me that -in the case of spacelike events- that is what we have: no path, no way to communicate one event with the other. A solution may be simply that in those cases the math breaks down and the line connecting the two events in ST does not exist: there is a spatial path but not a ST path. This is a difference with X-Y diagrams because in them both X and Y are identical and hence the fact that in a given frame there is no Y path does not mean that there is no "spatial" path (the X path still qualifies as spatial link), whilst in ST diagrams you need both things and hence if there is X path but no sufficient T, then there is no ST path.

I have updated my drawing. As to arrows, maybe we were looking at different versions. I have taken them away and will think on how they should be placed. And thanks for your help with these reflections!

 

[bobc2]

I have looked at the Wikipedia entry on the ST interval for orientation: http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals

Unfortunately, they use the (-+++) convention and so they get S^2 > 0 (positive number) in spacelike events.

Good for them. I used (+++-). And we had no trouble getting the sides of the triangle--our original goal.

But, as they say, if you use the opposite convention, you get the reverse sign. In any case, what is important is that, whatever convention you use, there will be one case where S^2 is negative and hence S is imaginary.

No. I don't think that is what is appropriate. I thought you were onto something interesting when you began investigating lengths of the hypotenuse and sides of triangles. Now. you've cast the problem back into the well worn presentation of the application of the Minkowski metric to get invariant intervals. But, I had originally sensed that you were on the track of something beyond the arbitrary definitions of interval and metric (can you show us a derivation of the metric?).

I think that the (+---) convention is preferrable, because with it you get (i) S^2 and S positive in timelike events and (ii) S^2 negative and S imaginary in spacelike events, which can be translated physically as, respectively, (i) there is a causal path between the two events through a massive messenger (whose proper time is S) and (ii) there is no causal path. With both conventions, in lightlike events, S=0, which translates as there is a causal path through a massless messenger, whose proper time is 0.

Of course there are different preferences. Different metrics are used in different applications. Nuclear physicists may like one convention and other specialties may like the other. That you are free to choose your convention should tell you something.

My concern was only the expression of the same truths in geometric language.

Good. Then don't worry too much about one Minkowski metric. Just focus on how every observer can come up with the same lengths for the sides of a triangle.

I understand that an imaginary leg is an awkward concept in geometry but you may agree with me that -in the case of spacelike events- that is what we have.

No I don't. I showed you it was not imaginary.

No path, no way to communicate one event with the other.

That is totally irrevalent to our problem of finding the length of sides of a triangle. It matters not at all whether we have causal connections, etc. There is an object out there. It exists whether we see it or not, and we can compute the lengths of its sides.

A solution may be simply that in those cases the math breaks down and the line connecting the two events in ST does not exist: there is a spatial path but not a ST path. This is a difference with X-Y diagrams because in them both X and Y are identical and hence the fact that in a given frame there is no Y path does not mean that there is no "spatial" path (the X path still qualifies as spatial link), whilst in ST diagrams you need both things and hence if there is X path but no sufficient T, then there is no ST path.

See the mess you get into when you start misapplying an equation to a very simple problem?

I appologize for being been too argumentative in the recent posts. You've been a good trooper here and brought in the logical points that should be made. I thought some of the other forum members would jump in and team up with you.

I probably don't have anything else to contribute that would not get us into a monitor's lock down.

 

[Saw]

I probably don't have anything else to contribute that would not get us into a monitor's lock down.

Well, yes, I see that we are lately running into circles in our respective comments. I understand you drop from the thread for that reason. I may post some developments in case someone else wants to comment, but in any case I thank you again for your very helpful intervention. You have told me about Loedel diagrams, which is a very interesting tool. It is curious that by adapting an Epstein diagram I had hit on the idea of this sort of diagrams, just with a different inclination, but I was not placing photon lines where they should and might have never realized this without your indication. We have agreed that at least in timelike events the interval is the height of a right triangle, so math and geometry are consistent. We have some discrepancies as to other types of events, but the discussion was fun. Thanks bobc2.

 

[bobc2]

Well, yes, I see that we are lately running into circles in our respective comments. I understand you drop from the thread for that reason. I may post some developments in case someone else wants to comment, but in any case I thank you again for your very helpful intervention. You have told me about Loedel diagrams, which is a very interesting tool. It is curious that by adapting an Epstein diagram I had hit on the idea of this sort of diagrams, just with a different inclination, but I was not placing photon lines where they should and might have never realized this without your indication. We have agreed that at least in timelike events the interval is the height of a right triangle, so math and geometry are consistent. We have some discrepancies as to other types of events, but the discussion was fun. Thanks bobc2.

And thanks to you, Saw. You brought some goot stuff in here. My feelings about the Minkowski metric not really being a metric (true metrics are positive definite) and therefore not applicable to yielding lengths of sides of triangles in general gets out of bounds for this forum.