I In what formation does this simple block universe exist?

[PeterDonis]

I took what was said to mean something else.

If you mean that you take the map with north at the top and just turn it on its side so west is at the top, then you don't have two different maps, you just have one map. But we were talking about two different maps. So, again, I don't think you have fully thought through the implications of what you are saying.

 

[Ibix]

I think there is ambiguity when you say "But roads are in different places on a map when I rotate it!". This does not seem true by how I am interpreting it.

Here's a crude map of a made up town. I've highlighted one of the roads in red:

Here's a rotated version:

The red highlighted road is in a different place, to the left of the river instead of the right. Clearly things have moved and are in different places.

I suspect what you actually mean by "changing where the roads are on the map" is that the Euclidean relationships between them have changed - that the perpendicular distance between the worldlines as measured with a ruler on the Minkowski diagram has changed in a way more complex than a simple scaling, and that angles between lines have changed. You are correct. They have. This is due to the fact that you are drawing a plane that does not respect the laws of Euclidean geometry, so you cannot represent it on a plane that does respect those laws without distorting something. But that just means your two maps are distorted in different ways - it doesn't mean that they are maps of different things. For example, here's a NASA composite of the Earth as a Mercator projection:

The north and south poles lie at the top and bottom of the map, as usual. Here's another Mercator projection - drawn by exactly the same process - but with the points where the international date line and Greenwich meridian cross the equator at the top and bottom of the map, which puts the north pole in the middle and the south pole at the edge.

South America and Australia point in opposite directions compared to the original, Africa is massively distorted, and the whole of Europe and Asia is much smaller. Do you think this is a map of another planet? Because that seems to be equivalent to your beliefs about Minkowski diagrams. Or is it just that spherical geometry is different from Euclidean geometry and any flat map of a sphere is distorted, and any two maps are differently distorted? Because that's what's really going on, both in the Mercator projection and the Minkowski diagrams.

Original Mercator projection image credit: NASA, via Visible Earth.

 

[student34]

Correct. For example, on the diagram we measure those angles with a protractor, but in spacetime we measure those angles with a speedometer or a radar gun. On the diagram, all distances $ds^2=dx^2+dy^2$ are measured with a ruler, but in spacetime intervals $ds^2=-dt^2+dx^2$ are only measured with rulers if $ds^2>0$ while if $ds^2<0$ then the interval is measured with a clock.

The diagram is an accurate map of spacetime, but like maps of the earth, the geometry is not identical. There is a 1-to-1 map between points in the diagram and events in spacetime, but the metric is different so distances and angles can be different between the map and the physical world.

Here is a question that I think will help me "form" the universe in the OP. For the sake of simplicity, let's say that the universe that we see in the diagram is actually that big in scale.

I see that when t = 0, the formula turns into Pythagorean theorem, and all that would be left are 3 dots in both diagrams. Would it be true to say that we would be looking at an actual slice of the block as it exists?

 

[student34]

Here's a crude map of a made up town. I've highlighted one of the roads in red:
View attachment 296802
Here's a rotated version:
View attachment 296803
The red highlighted road is in a different place, to the left of the river instead of the right. Clearly things have moved and are in different places.

I suspect what you actually mean by "changing where the roads are on the map" is that the Euclidean relationships between them have changed - that the perpendicular distance between the worldlines as measured with a ruler on the Minkowski diagram has changed in a way more complex than a simple scaling, and that angles between lines have changed. You are correct. They have. This is due to the fact that you are drawing a plane that does not respect the laws of Euclidean geometry, so you cannot represent it on a plane that does respect those laws without distorting something. But that just means your two maps are distorted in different ways - it doesn't mean that they are maps of different things. For example, here's a NASA composite of the Earth as a Mercator projection:
View attachment 296804
The north and south poles lie at the top and bottom of the map, as usual. Here's another Mercator projection - drawn by exactly the same process - but with the points where the international date line and Greenwich meridian cross the equator at the top and bottom of the map, which puts the north pole in the middle and the south pole at the edge.
View attachment 296807
South America and Australia point in opposite directions compared to the original, Africa is massively distorted, and the whole of Europe and Asia is much smaller. Do you think this is a map of another planet? Because that seems to be equivalent to your beliefs about Minkowski diagrams. Or is it just that spherical geometry is different from Euclidean geometry and any flat map of a sphere is distorted, and any two maps are differently distorted? Because that's what's really going on, both in the Mercator projection and the Minkowski diagrams.

Original Mercator projection image credit: NASA, via Visible Earth.

Ok thanks I will keep this in mind. My confusion is because I thought that the slices acually existed that way.

 

[Ibix]

Here is a question that I think will help me "form" the universe in the OP. For the sake of simplicity, let's say that the universe that we see in the diagram is actually that big in scale.

I see that when t = 0, the formula turns into Pythagorean theorem, and all that would be left are 3 dots in both diagrams. Would it be true to say that we would be looking at an actual slice of the block as it exists?

The $t=0$ line is what an observer at rest in that frame would call "space at time zero", yes (or one dimension of it, at least). That's why you are getting three dots - they are what you would normally think of as "the objects at that time". The worldlines are the three dots as they appear in the block universe - three dots, extended in time.

And as you note, distances in the $t=0$ (or any other constant value) obey Pythagoras - space is Euclidean. Spacetime is not.

 

[Dale]

I see that when t = 0, the formula turns into Pythagorean theorem, and all that would be left are 3 dots in both diagrams. Would it be true to say that we would be looking at an actual slice of the block as it exists?

Small nitpick. It is $dt=0$ not $t=0$. The $dt=0$ means that time is constant, i.e. one fixed moment in time. But it could be anyone fixed moment you choose, not just $t=0$.

Yes if $dt=0$ then the remaining part of the metric is just the Euclidean metric. You are looking at all of space at one fixed moment of time. Space has the same metric as the paper, so that is undistorted

 

[student34]

The $t=0$ line is what an observer at rest in that frame would call "space at time zero", yes (or one dimension of it, at least). That's why you are getting three dots - they are what you would normally think of as "the objects at that time". The worldlines are the three dots as they appear in the block universe - three dots, extended in time.

And as you note, distances in the $t=0$ (or any other constant value) obey Pythagoras - space is Euclidean. Spacetime is not.

But then when does the diagram, say on the left, start to diverge into the Minkosky geometry? It seems like we are just going to build the diagram by using constants of time.

 

[Dale]

But then when does the diagram, say on the left, start to diverge into the Minkosky geometry?

Whenever the metric has both a + and a - term

 

[PeterDonis]

when does the diagram, say on the left, start to diverge into the Minkosky geometry?

It doesn't "diverge" into Minkowski geometry. The geometry of the diagram as a whole is Minkowski. The geometry of each individual slice of constant time is Euclidean. These are just facts about the geometry. There is no "divergence" from one to the other.

It seems like we are just going to build the diagram by using constants of time.

The full spacetime is a "stack" of slices of constant time, yes. But that in no way requires that the geometry of the full spacetime must be Euclidean, just because the geometry of the individual slices is.

 

[PAllen]

Think of concentric 2-spheres embedded in Euclidean 3-space. Each sphere is non-Euclidean. This does not require the containing space to be non-Euclidean. Similarly, the fact that you can embed Euclidean slices in Minkowski spacetime in no way requires the Minkowski spacetime to be Euclidean.

 

[Ibix]

But then when does the diagram, say on the left, start to diverge into the Minkosky geometry? It seems like we are just going to build the diagram by using constants of time.

Any non-horizontal line through either diagram is distorted in the representation. It's only on surfaces where the time coordinate is the same (i.e., a horizontal line) that the metric @Dale showed becomes the familiar Pytharoras' theorem.

 

[student34]

Whenever the metric has both a + and a - term

But then wouldn't every 0 dt moment turn into the diagram?

 

[Dale]

But then wouldn't every 0 dt moment turn into the diagram?

No, a $dt=0$ slice has a metric with only + terms. So a $dt=0$ slice has ordinary Euclidean geometry

 

[student34]

No, a $dt=0$ slice has a metric with only + terms. So a $dt=0$ slice has ordinary Euclidean geometry

Yes that I understand. Maybe I did not explain what I meant properly.

For example, the diagram on the left, at t = 0 and dt = 0, we seem to get points on the blue worldlines at the x axis. Then if we go to a dt = 0 moment very close to t = 0, it seems that we get another dot on the blue world lines of the diagram. If we keep doing this, don't we get the diagram?

 

[Ibix]

You can build up a diagram that way, yes. But the same is true of the other diagram.

 

[Dale]

If we keep doing this, don't we get the diagram?

Yes, the full 2D spacetime diagram can be assembled from a series of 1D diagrams. This is called a foliation and each subspace is called a leaf. Furthermore, the metric of the 1D sub diagrams (leaves) is Euclidean.

More importantly, spacetime (neglecting gravity) can be foliated as a series of 3D leafs where the metric in each 3D leaf is straight Euclidean.

 

[student34]

You can build up a diagram that way, yes. But the same is true of the other diagram.

But then can't we say that the diagrams are slices of the spacetime and exist in the block the way they appear on our screens, as a 2d Euclidean space?

 

[student34]

Yes, the full 2D spacetime diagram can be assembled from a series of 1D diagrams. This is called a foliation and each subspace is called a leaf. Furthermore, the metric of the 1D sub diagrams (leaves) is Euclidean.

More importantly, spacetime (neglecting gravity) can be foliated as a series of 3D leafs where the metric in each 3D leaf is straight Euclidean.

Then why can't the diagrams exist like they do on our screens as 2d Euclidean planes?

 

[PeroK]

But then can't we say that the diagrams are slices of the spacetime and exist in the block the way they appear on our screens, in a 2d Euclidean space?

So, for all $t$, Minkowski spacetime is Euclidean; hence Minkowski spacetime is Euclidean. QED

 

[robphy]

But then can't we say that the diagrams are slices of the spacetime and exist in the block the way they appear on our screens, as a 2d Euclidean space?

Given two such slices, consider a point-event A one slice and another point-event B on the other.
A Euclidean distance you might assign to a segment from A and B is not equal
to the Minkowski (spacetime) interval from A to B.
The Euclidean distance does not correctly capture or encode the physics relating A and B.
You can use a Euclidean ruler on your diagram... but it doesn't tell you about the physics involved.

By the way, what I am saying applying also applies to a position-vs-time graph in physics 101.
The length of an arbitrary line in the graph measured with a ruler has no physical interpretation.

 

[PeterDonis]

But then can't we say that the diagrams are slices of the spacetime and exist in the block the way they appear on our screens, as a 2d Euclidean space?

The diagrams are not spacetime, any more than a map of the Earth is the Earth. You are confusing the diagrams with reality. They're not reality, they're pictures of some aspect of reality, taken from different viewpoints. So your question here doesn't even make sense; it's like asking if two pictures of you taken from different angles "exist the way they appear", as though having two pictures of you meant there were two yous.

 

[student34]

So, for all $t$, Minkowski spacetime is Euclidean; hence Minkowski spacetime is Euclidean. QED

I don't know if you are joking or not.

 

[PeterDonis]

I don't know if you are joking or not.

He is.

 

[PeroK]

I don't know if you are joking or not.

I was trying to interpret what you were saying.

 

[PeroK]

In mathematics, when you have a proof and a counterexample to that proof, in general it's the proof that tends to be wrong.

 

[Dale]

Then why can't the diagrams exist like they do on our screens as 2d Euclidean planes?

Because the metric when $dt\ne0$ is not Euclidean.

The metric of the leaves is not the same as the leaves of the foliated space. Indeed, it cannot possibly be since the foliated space has more dimensions than the leaves.

Note, this property is not peculiar to Minkowski geometry. It is a general fact of foliations. Think of an onion. Each layer (leaf) has a 2D spherical metric. When you put them all together the 3D Euclidean metric describes distances between points in the onion as a whole.

 

[student34]

Given two such slices, consider a point-event A one slice and another point-event B on the other.
A Euclidean distance you might assign to a segment from A and B is not equal
to the Minkowski (spacetime) interval from A to B.
The Euclidean distance does not correctly capture or encode the physics relating A and B.
You can use a Euclidean ruler on your diagram... but it doesn't tell you about the physics involved.

I am really just trying to visualize what I can and use the math that is relevant, for starters. That is why I posted the simplest example that I could think of. I am trying to understand all true statements about the formation of the block in my OP, and most importantly, how all these true statements about the block relate to each other.

This seems to be very hard for me to do. I just really want to understand this anyway I can.

 

[student34]

The diagrams are not spacetime, any more than a map of the Earth is the Earth. You are confusing the diagrams with reality. They're not reality, they're pictures of some aspect of reality, taken from different viewpoints. So your question here doesn't even make sense; it's like asking if two pictures of you taken from different angles "exist the way they appear", as though having two pictures of you meant there were two yous.

Like I told robphy, I am interested in how these true statements about the block in my example relate to each other. For me, I still see contradictions with these statements; this tells me that there is still something quite important that I am not understanding.

For example, you said that we can build the diagram using moments in time, which, I believe, creates an image of the slice as seen on our screen exactly how the slice exists in the block. But then I am told that the image of the slice is not a true reflection of the slice itself.

I am still confused about that.

 

[student34]

In mathematics, when you have a proof and a counterexample to that proof, in general it's the proof that tends to be wrong.

But how do I know which is proving the other wrong? Is Minkowsky geometry proving Euclidean geometry wrong or vice versa?

 

[Dale]

But how do I know which is proving the other wrong? Is Minkowsky geometry proving Euclidean geometry wrong or vice versa?

Neither. What is proven wrong is the assumption that a foliation of Euclidean leaves gives a Euclidean space.

Again, this is a general fact about foliations. Please think about the onion example. Focus on that until you understand it

 

[PeterDonis]

we can build the diagram using moments in time

Yes, but which particular events (points in spacetime) are part of a given "moment of time" is different for different frames. That is why a horizontal line across one of your diagrams (which represents a moment of time in that diagram's frame) passes through different points on the red and blue lines than a horizontal line across the other diagram (which represents a moment of time in that diagram's frame). You can build a full diagram using either frame, but the "slices" you use to build it will "cut" the actual spacetime at different angles.

 

[student34]

Because the metric when $dt\ne0$ is not Euclidean.

I just want to clear this up once and for all. I seem to be getting two conflicting answers to the question, "why can't the diagrams exist like they do on our screens as 2d Euclidean planes?". I apologize if I am misinterpreting your following answers.

Answer #1 (from post #116): "Yes, the full 2D spacetime diagram can be assembled from a series of 1D diagrams."

Answer #2 (from post #126): "Because the metric when $dt\ne0$ is not Euclidean."

 

[Dale]

I just want to clear this up once and for all. I seem to be getting two conflicting answers to the question, "why can't the diagrams exist like they do on our screens as 2d Euclidean planes?". I apologize if I am misinterpreting your following answers.

Answer #1 (from post #116): "Yes, the full 2D spacetime diagram can be assembled from a series of 1D diagrams."

Answer #2 (from post #126): "Because the metric when $dt\ne0$ is not Euclidean."

It is the full diagram but the full metric is not Euclidean.

 

[student34]

It is the full diagram but the full metric is not Euclidean.

I do not know what you mean. In what way is the metric different?

 

[Dale]

I do not know what you mean. In what way is the metric different?

It is $ds^2=-dt^2+dx^2$. The way it is different is that it has a negative term. We have been over this several times.

 

[student34]

It is $ds^2=-dt^2+dx^2$. The way it is different is that it has a negative term. We have been over this several times.

But I thought that if we built the diagram using only moments in time that it would be Euclidean.

 

[PeterDonis]

I thought that if we built the diagram using only moments in time that it would be Euclidean.

You have been told repeatedly that this is not the case. There is no point in continuing this discussion if you are not paying attention.

 

[Dale]

But I thought that if we built the diagram using only moments in time that it would be Euclidean.

I have told you multiple times that is not so, as have others. Think about the onion. Foliation does not work the way you thought.

 

[student34]

Yes, but which particular events (points in spacetime) are part of a given "moment of time" is different for different frames. That is why a horizontal line across one of your diagrams (which represents a moment of time in that diagram's frame) passes through different points on the red and blue lines than a horizontal line across the other diagram (which represents a moment of time in that diagram's frame). You can build a full diagram using either frame, but the "slices" you use to build it will "cut" the actual spacetime at different angles.

By answering yes, do you mean that we can build the diagram exactly how we see it as the image on our computers?

 

[student34]

I have told you multiple times that is not so, as have others. Think about the onion. Foliation does not work the way you thought.

I do not understand the onion analogy. I understand that we can integrate many 2d onion bulbs to create a full onion. But I do not understand how that relates to this. What are we integrating in the example?

 

[Dale]

I do not understand the onion analogy. I understand that we can integrate many 2d onion bulbs to create a full onion. But I do not understand how that relates to this. What are we integrating in the example?

If you didn’t understand it, then why have you waited this long to ask for clarification?

Each leaf of an onion is a sphere. So the metric of each leaf is $ds^2=R^2 d\phi^2 + R^2 \sin^2(\phi) d\theta^2$ which is curved. But when you add all of the leaves together the metric is the flat Euclidean $ds^2=dx^2+dy^2+dz^2$.

The full space does not inherit the metric from the leaves. An onion has spherical leaves that foliate a Euclidean space. Spacetime has Euclidean leaves that foliate a Minkowski space.

 

[PeterDonis]

By answering yes, do you mean that we can build the diagram exactly how we see it as the image on our computers?

That depends on what you mean by "exactly how you see it". The geometry of the actual spacetime that the diagram is a picture of is not Euclidean. We have said this multiple times, but it doesn't seem to get through to you, and you using the phrase "exactly how you see it" indicates to me that you still are thinking you can somehow make spacetime Euclidean by drawing the diagram the right way. You can't.

So I'm going to answer "no". In fact, at this point pretty much any question I think you will ask will have the answer "no", because you are refusing to let go of your flawed conceptual scheme no matter how many times we tell you it's flawed, and you're refusing to change how you think about this problem no matter how many times we tell you that the way you are thinking about it is not working.

 

[PeterDonis]

do you mean that we can build the diagram exactly how we see it as the image on our computers?

Perhaps this will help: maybe what you meant by the title question of this thread is that you have these two different diagrams that you drew in your OP, and you know they're different, and you're wondering which one of those diagrams is the "true" one, the one that shows spacetime the way it "really is".

If that's the case, the answer is that neither diagram shows spacetime the way it "really is". Both of them are diagrams of spacetime, not spacetime itself. Neither of them is a "truer" picture than the other. It's like asking which of two pictures taken of you at the same time from different angles is the "true" picture.

 

[student34]

If you didn’t understand it, then why have you waited this long to ask for clarification?

Each leaf of an onion is a sphere. So the metric of each leaf is $ds^2=R^2 d\phi^2 + R^2 \sin^2(\phi) d\theta^2$ which is curved. But when you add all of the leaves together the metric is the flat Euclidean $ds^2=dx^2+dy^2+dz^2$.

The full space does not inherit the metric from the leaves. An onion has spherical leaves that foliate a Euclidean space. Spacetime has Euclidean leaves that foliate a Minkowski space.

Okay, I understand the analogy now. That is quite interesting.

 

[PeroK]

But how do I know which is proving the other wrong? Is Minkowsky geometry proving Euclidean geometry wrong or vice versa?

It means that you're wrong! Not Minkowsky, nor Euclid. student34 is wrong!

 

[student34]

It is the full diagram but the full metric is not Euclidean.

Does this mean that every horizontal relation between the 3 lines on either graph exist exactly this way in the block, but any diagonal relationship is different than what we see on the graphs?

 

[PeterDonis]

what we see on the graphs?

What do you mean by "what we see on the graphs"? If you mean that "what we see on the graphs" implies Euclidean geometry everywhere, you have already been told, multiple times, by multiple people, that the geometry of spacetime is not Euclidean.

So again, I'm going to answer "no" to your questions here. And rather than repeat the rest of my last paragraph in post #142, and all of my post #143, I will just refer you to them again.

 

[student34]

What do you mean by "what we see on the graphs"? If you mean that "what we see on the graphs" implies Euclidean geometry everywhere, ...

No, only horizontal distances.

 

[PeterDonis]

Does this mean that every horizontal relation between the 3 lines on either graph exist exactly this way in the block, but any diagonal relationship is different than what we see on the graphs?

This question as you ask it is not well posed, for the reasons I have already given. However, there are some valid statements that can be made in this context that might be helpful, although they will most likely just illustrate how much more work you have to do to discard your current intuitions. All of these statements apply to any diagram of Minkowski spacetime that is drawn the way your diagrams in the OP are drawn, i.e., that is a diagram from the viewpoint of some inertial frame.

(1) Any straight line that has a slope (relative to horizontal) of less than 45 degrees is called a "spacelike" line. Any such line represents a spacelike 3-space that has Euclidean geometry. However, there will only be one inertial frame (which will only be the frame the diagram is drawn in if the line is exactly horizontal) in which the Euclidean geometry of this spacelike 3-space is obvious from the metric (by setting $dt = 0$).

(2) Any straight line that has a slope of exactly 45 degrees is called a "null" or "lightlike" line. Any such line represents a portion of a light cone, which is the set of all possible light rays to or from a given event. The "geometry" of a light cone has no simple analogue in ordinary geometry.

(3) Any straight line that has a slope of more than 45 degrees is called a "timelike" line. Any such line represents the worldline of a timelike observer who is always inertial, i.e., always moving in free fall with zero proper acceleration. Every such observer is at rest in some inertial frame; lines that are exactly vertical represent the worldlines of observers who are at rest in the specific inertial frame in which the diagram is drawn. The "geometry" of any worldline is simply a straight line, but this is not very helpful since it just means the points on the line each represent individual events at which the observer's clock reads a particular time, and those times can be treated as real numbers ordered in the usual way from past to future.

(4) To have any region of spacetime that has Minkowski geometry, you must have a region that is represented by an area on the diagram, not a line. No single line, no matter what its slope, will represent any region of spacetime that has Minkowski geometry.

 

[Dale]

Does this mean that every horizontal relation between the 3 lines on either graph exist exactly this way in the block, but any diagonal relationship is different than what we see on the graphs?

Yes, for one thing the diagonal angles on the graph are measured with a protractor, but in spacetime the diagonal angles are measured with a speedometer.