I do not understand the geometry of spacetime

[Mates]

I am having a tough time understanding how there are only 4 dimensions in spacetime and not more.

To make my issue easier to explain, and easier for me to understand your feedback, I will use a 2d light cone to explain my issue. I will explain how I am getting 3 dimensions instead of 2. And, this will explain how I am only getting 2 spatial dimensions instead of 1 spatial and 1 temporal. Of course either case is problematic.

A reminder of the spacetime interval:[math]\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2[/math]
In our case we simply have :[math]\Delta s^2 = -(c\Delta t)^2 + \Delta x^2[/math]
In image A below, we see an object (in red) in its light cone moving up through time. As the spacetime interval implies, the object travels through a nonspatial interval. The horizontal black lines that intersect the path of the object are spatial segments. I put them in there to remind us that we can fill the nonspatial interval with spatial points that the object travels through.

My issue arises because every spatial point that the object travels through, it also travels through a point in the spacetime interval.

So as for now, I either see the need for an extra dimension or no need for the negative sign in the spacetime interval, which would just turn it into another spatial dimension.

 

[topsquark]

I am having a tough time understanding how there are only 4 dimensions in spacetime and not more.

To make my issue easier to explain, and easier for me to understand your feedback, I will use a 2d light cone to explain my issue. I will explain how I am getting 3 dimensions instead of 2. And, this will explain how I am only getting 2 spatial dimensions instead of 1 spatial and 1 temporal. Of course either case is problematic.

A reminder of the spacetime interval:[math]\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2[/math]
In our case we simply have :[math]\Delta s^2 = -(c\Delta t)^2 + \Delta x^2[/math]
In image A below, we see an object (in red) in its light cone moving up through time. As the spacetime interval implies, the object travels through a nonspatial interval. The horizontal black lines that intersect the path of the object are spatial segments. I put them in there to remind us that we can fill the nonspatial interval with spatial points that the object travels through.

My issue arises because every spatial point that the object travels through, it also travels through a point in the spacetime interval.

So as for now, I either see the need for an extra dimension or no need for the negative sign in the spacetime interval, which would just turn it into another spatial dimension.

View attachment 34455

When the object travels through the space interval it is traveling through a space-time interval. Any path on that plane is a space-time interval. I don't understand what you are trying to get at?

-Dan

 

[Mates]

When the object travels through the space interval it is traveling through a space-time interval. Any path on that plane is a space-time interval. I don't understand what you are trying to get at?

-Dan

But when we rotate the light cone like in image B, it seems to have a lot of the same properties as a Cartesian Plane. I can get to any point on the plane by using the 2 spatial dimensions that the light travels on. Doesn't there have to be a 3rd dimension if we want a time dimension too?

 

[topsquark]

But when we rotate the light cone like in image B, it seems to have a lot of the same properties as a Cartesian Plane. I can get to any point on the plane by using the 2 spatial dimensions that the light travels on. Doesn't there have to be a 3rd dimension if we want a time dimension too?

Just because you are rotating something, why would you need to include another dimension? (Beyond the one that you need to rotate into, I mean.) If you rotate something in a 2D space (or higher) you don't need to leave the space. Now, you can, Mathematically, rotate out of the space, but unless we have Physical evidence that this extra space exists there is no reason to. The extra dimensions in String Theory are not equipped like this so they don't count.

Now, there are what are called "Kaluza-Klein" theories that add an extra dimension of space to Einstein's space-time that manage to include electromagnetism. (This is actually one of the inspirations for String Theory.) But we've shown that the Kaluza-Klein theories are not correct.

And it doesn't really have a lot of properties of the Cartesian plane. One of the defining features of Minkowski space-time is that the interval (the SR version of the dot product) is not positive definite. It is in Euclidean space. It might look Euclidean but it is anything but.

-Dan

 

[Mates]

Just because you are rotating something, why would you need to include another dimension?

I should not have used that wording. I just meant to show that it starts to look like a Cartesian Plane.

And it doesn't really have a lot of properties of the Cartesian plane. One of the defining features of Minkowski space-time is that the interval (the SR version of the dot product) is not positive definite. It is in Euclidean space.

I am not sure what positive definite means.

Looking at image A, I understand that everything outside of the light cone is essentially a Euclidean 2D space for the red object. Doesn't that mean that inside the light cone there must be another dimension where [math]\Delta s < 0[/math]

 

[topsquark]

I should not have used that wording. I just meant to show that it starts to look like a Cartesian Plane.


I am not sure what positive definite means.

Looking at image A, I understand that everything outside of the light cone is essentially a Euclidean 2D space for the red object. Doesn't that mean that inside the light cone there must be another dimension where [math]\Delta s < 0[/math]

By "positive definite" I mean that, in the Euclidean plane [imath]\Delta s^2 = c^2 \Delta t^2 + \Delta x^2 > 0[/imath] for all intervals [imath](x_1, t_1) \to (x_2, t_2)[/imath]. In Minkowski space it's defined as [imath]\Delta s^2 = -c^2 \Delta t^2 + \Delta x^2[/imath] so it could be positive, negative, or even equal.

Looking at image A, I understand that everything outside of the light cone is essentially a Euclidean 2D space for the red object.

And here's where the problem lies. Why would you say this? Just because it's outside the light cone doesn't mean the space becomes Euclidean. I don't know why you would think that. Inside the light cone [imath]\Delta s^2[/imath] will be negative. Outside it will be positive, but that does not mean it acts Euclidean. We still have that minus sign on the first term.

-Dan