I In what formation does this simple block universe exist?

[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.