In what formation does this simple block universe exist?

[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.

 

[student34]

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.

Thanks a lot for this. It is a lot to take in all at once, but I am sure I will refer to at to help guide my questions and thought process.

 

[robphy]

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.

... or, maybe better, a rapidity-meter.

 

[Dale]

... or, maybe better, a rapidity-meter.

Yes, which is a speedometer with the gauge calibrated to display rapidity.

 

[PeterDonis]

... or, maybe better, a rapidity-meter.

That's easy, just recalibrate the readout on the speedometer. (Of course, you might have some difficulty for highly ultra-relavistic values...)

 

[student34]

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.

If there were an actual block universe of the same size that these images describe, could these images be actual slices from that block?

 

[PeterDonis]

If there were an actual block universe of the same size that these images describe, could these images be actual slices from that block?

They are both drawings of the same actual slice from the block universe, from different viewpoints.

 

[Dale]

If there were an actual block universe of the same size that these images describe, could these images be actual slices from that block?

Does this matter? It sounds like you are asking if an image of an apple slice is an actual apple slice. This seems like a pointless and silly tangent that I don't particularly want to pursue.

 

[student34]

Does this matter? It sounds like you are asking if an image of an apple slice is an actual apple slice. This seems like a pointless and silly tangent that I don't particularly want to pursue.

No, that's not what I am wondering. I am wondering if the images of these slices could look exactly like actual slices from the block.

 

[student34]

They are both drawings of the same actual slice from the block universe, from different viewpoints.

Since we can't put an image of the block on the screen, are there any parts or points of the block that would actually align with the 2 images that I made?

 

[PeterDonis]

I am wondering if the images of these slices could look exactly like actual slices from the block.

Once again you are failing to grasp what you have been told repeatedly in this thread: there is no such thing as "look exactly like actual slices from the block." Any more than there is any such thing as one particular picture of you, out of a set of many of them all taken at the same time from different viewpoints, looking "exactly like the actual you" whereas the others don't.

The diagrams are diagrams. A diagram is not the same as the thing it's a diagram of. Read that over and over again until it sinks in.

No diagram "looks exactly like the actual thing" it's a diagram of. They're diagrams. They're not the same as the thing they're a diagram of. Read that over and over again until it sinks in.

All diagrams of the same spacetime are on an equal footing: none looks any more "like" the spacetime they're a diagram of than any others. None of them are "real", and all of them give an equally valid representation of "reality". Read that over and over again until it sinks in.

At this point any further question from you that's asking a question you've already asked and had answered multiple times, and which indicates that you still have not grasped the statements above that I told you to read over and over again until they sink in, will cause your question to be deleted and this thread to be closed. It is pointless to keep going around in circles. Either you are able and willing to listen to what we tell you and change what you think accordingly, or you're not.

 

[student34]

I think that there is ambiguity in my wording that is causing frustration. My goal is to geometrically visualize what I can about the actual block, not just its Euclidean transformations. I want to understand it visually when possible.

For simplicity sake, let's assume that for some quantum reason all three objects/particles came into existence and out of existence at the ends of their lines in the images in the OP.

In my attempt to "form" this block, can we determine the block's size in say meters? For example maybe it is close to one second old so maybe it would have a length of 300,000,000 meters and a width of, say, 100 meters. Can we at least make these boundaries? (I understand that my angles in the diagrams will probably not match these measurements)

If the answer is yes, can we fill in anything else within these boundaries. In other words, what other part/s of the block in the OP are Euclidean, if any?

 

[PAllen]

If the answer is yes, can we fill in anything else within these boundaries. In other words, what other part/s of the block in the OP are Euclidean, if any?

This is not a meaningful way to look at things. Going back to an analogy of mine, consider Euclidean 3-space filled with concentric spheres (with an unaccounted for point in the center). Since every sphere is non-Euclidean, can you then say that all of Euclidean 3-space is non-Euclidean except for one point??! In fact every connected 3-d subset of Euclidean 3-space is Euclidean. And there are also families of 2-d subsets all which are Euclidean.

Similarly, for Minkowski spacetime, the whole and every 4-d connected subset are non-Euclidean. On the other hand, you can find families of 3-d subsets all of which are Euclidean, and other families of 3-d subsets none of which are Euclidean.

 

[PeterDonis]

My goal is to geometrically visualize what I can about the actual block, not just its Euclidean transformations.

The transformation that takes one of your diagrams in the OP to the other is not a Euclidean transformation. You have already been told this multiple times. It is a Lorentz transformation. You have just gotten your thread closed since you are failing to follow the instructions I gave you in post #160 about questions already asked and answered.

As for visualizing the actual block as opposed to the diagrams themselves, you have just gotten this thread closed again since you are still failing to grasp one of the statements I told you in post #160 to read again and again until it sunk in.

what other part/s of the block in the OP are Euclidean, if any?

I already answered this for you, in the post of mine that you quoted in your post #151. You said then that you would refer to that post of mine to guide your thought processes. Evidently you have not done that. Go do it.

Thread closed.