I Understand 4-Vectors & Spacetime: Hartle Gravity Chapter 5

[Kashmir]

Hartle, gravity. Chapter 5

"A four-vector is defined as a directed line segment in four-dimensional flat spacetime in the same way as a three-dimensional vector (to be called a three-vector in this chapter) can be defined as a direcied line segment in three-dimensional Euclidean Space"For vectors in 3D, we know that vectors exist independently of the coordinate system used to measure them. Like a position vector. It exists there, it's tip at a point in the 3D space.
We may choose different coordinates to write the vector but all of them say the same thing. They all are representing a vector whose tip ends at the same point.

So similarly should I as the book suggests, think of spacetime as being out there independently of coordinate system, and for example a displacement vector in spacetime is just the line segment from one point to another.

Now once I choose a coordinate system I get components of the vector ( which is independent of the coordinate system) in that coordinate system.
If I apply a Lorenzt boost, I'm actually choosing another coordinate system which thus changes the coordinates of the vector. However they actually represent the same vector whose tip ends at the same point, although that point has different coordinates in different frames.

Is that a correct understanding?

 

[PeterDonis]

"A four-vector is defined as a directed line segment in four-dimensional flat spacetime in the same way as a three-dimensional vector (to be called a three-vector in this chapter) can be defined as a direcied line segment in three-dimensional Euclidean Space"

It is worth noting that this definition actually does not generalize to curved spacetime. I don't have Hartle's book so I don't know if he discusses this later on, but other books on GR certainly do.

So similarly should I as the book suggests, think of spacetime as being out there independently of coordinate system

Yes, definitely.

and for example a displacement vector in spacetime is just the line segment from one point to another.

In flat spacetime, yes, this works. But, as noted above, it does not generalize to curved spacetime. The reason is that in curved spacetime, displacements do not form a vector space. So a different definition of a vector has to be adopted, which is initially less intuitive but which allows vectors (and more generally tensors) to be used in curved spacetime.

Now once I choose a coordinate system I get components of the vector ( which is independent of the coordinate system) in that coordinate system.

Yes.

If I apply a Lorenzt boost, I'm actually choosing another coordinate system which thus changes the coordinates of the vector. However they actually represent the same vector whose tip ends at the same point, although that point has different coordinates in different frames.

Yes.

 

[Ibix]

Is that a correct understanding?

Yes, except I expect you mean "changes the components of the vector" in the penultimate paragraph.

Just to warn that in curved spacetime vectors are slightly more abstract, and are part of a flat vector space called a "tangent space" associated with each event. That means that, although directions can be associated with vectors, the notion of a vector as the straight line linking two points (or events) doesn't work. So your definition is OK for SR, but doesn't generalise to GR.

Edit: Peter beat me to it, I see.

 

[Orodruin]

Since the cat is out of the box I think it is worth underlining the underlying reason that viewing vectors as (finite) directed line segments works in Euclidean and Minkowski space: they are both examples of affine spaces, which means that for any point there exists a translation operation from the set of (tangent) vectors to the space itself. Because of this the space itself is isomorphic to the set of tangent vectors. This identification is lost in a general space(time). Instead, it is possible to view the tangent vector space at a point as the set of infinitesimal displacements from that point (although there are many equivalent ways of viewing this), which do form a vector space.

 

[Kashmir]

Thank you all.