# Four Vectors in Spacetime

• I

## Summary:

Just want to confirm my understanding that the length of four-vectors, and their dot products, have the same value in all frames, not just inertial ones.
I know that the mathematical form of the line element of spacetime is invariant in all inertial reference frames, namely
$$ds^2 = -(cdt^2) + dx^2 + dy^2 + dz^2$$
From what I understand, the actual spacetime distance between two events is the same numerical quantity in all reference frames, inertial or not. It is not however the case that the line element describing distances will have this mathematical form in all frames.

Going off of this, if we have a four-vector connecting two events in spacetime, then its length is the spacetime distance between those events. Its length should be the same as calculated in all frames, not just inertial ones. And similarly, since the vector describes a physically immutable concept, operations such as scaling it, calculating its length, or adding and dotting it with other vectors should have the same result in all frames, right? The reason I ask is that in Hartle's Gravity: An Introduction to Einstein's General Relativity, when he introduces four vectors in chapter 5, he claims that these operations are invariant in inertial reference frames, but does not speak more generally about all reference frames. He claims that the scalar product is the same in all inertial frames, that vectors orthogonal in one inertial frame are in all, etc. I just want to confirm that I'm correct in assuming these statements are true in every frame, inertial or not.

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Ibix
Going off of this, if we have a four-vector connecting two events in spacetime,
That's not a four vector - it's a line. Vectors exist in the tangent space and, although they have directions that correspond to directions in spacetime, they don't connect events. Lines do. The distinction can be (and often is) ignored in flat spacetime, but it is very important in curved spacetime.

That said, yes, the length of a line is the same in any coordinate system. It could not be otherwise if we want to do physics with this theory - for example, there exists a timelike worldline representing the path a rocket takes from Earth to Mars. The interval along it is the flight time. If I switch coordinate systems that mustn't change the flight time.

Operations like scaling and rotation don't generally make sense for lines. They do make sense for vectors, and yes you are correct that they are also coordinate independent. Be wary - the representations can be complicated. As a trivial example, consider the vector (1,1) at the origin in Cartesian coordinates on a 2d Euclidean plane. Scaling it by a factor of 2 gives you (2,2). But in polar coordinates the original vector is ##(\sqrt 2,\pi/4)## and the scaled version is ##(2\sqrt 2,\pi/4)##.

PeroK
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The reason I ask is that in Hartle's Gravity: An Introduction to Einstein's General Relativity, when he introduces four vectors in chapter 5, he claims that these operations are invariant in inertial reference frames, but does not speak more generally about all reference frames. He claims that the scalar product is the same in all inertial frames, that vectors orthogonal in one inertial frame are in all, etc. I just want to confirm that I'm correct in assuming these statements are true in every frame, inertial or not.
Chapter 5 is entitled Special Relativistic Mechanics. In SR, there is the concept of global inertial reference frames and the theory of SR as set out in chapter 5 assumes, unless otherwise stated, that we are dealing with globla inertial reference frames.

Eventually, later in the book, when he considers curved spacetime, the concept of global inertial reference frames must be abandoned, as must the concept of a four-vector as a directed line segment in spacetime. Four-vectors become purely local objects, defined in the local tangent space. But, that's for later chapters. As far as chapter 5 is concerned four-vectors are defined as directed line segments in flat spacetime and global inertila reference frames are assumed.

PeroK
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That's not a four vector - it's a line. Vectors exist in the tangent space ...
Hartle hasn't reached that stage yet. In chapter 5 he defines a four-vector as a directed line segment in flat spacetime.

Ibix
Hartle hasn't reached that stage yet. In chapter 5 he defines a four-vector as a directed line segment in flat spacetime.
Ok - I don't have Hartle.

I must say I don't like this approach. My preference would be to state upfront that lines, coordinate differences, and vectors are distinct things. For a start, a line can be curved even in a flat space and coordinate differences are only (more or less) interchangeable with vector components in Cartesian coordinates.

• Dale
Orodruin
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But in polar coordinates the original vector is (√2,π/4)(2,π/4)(\sqrt 2,\pi/4) and the scaled version is (2√2,π/4)(22,π/4)(2\sqrt 2,\pi/4).
Trying to use polar coordinates at the origin are we?

The fact of the matter is that a vector scaling by a factor of two always means scaling its components by a factor of two regardless of the coordinate system. What you are thinking of here is the coordinates of a point.

• Ibix
Ibix
What you are thinking of here is the coordinates of a point.
You see, this is why I don't like treating coordinates and (position) vectors as interchangeable. I'd been doing it for twenty years before I made a serious stab at learning general relativity and now I have trouble keeping the concepts separate even when I'm trying to think carefully about it. • Dale
Dale
Mentor
I know that the mathematical form of the line element of spacetime is invariant in all inertial reference frames, namely
$$ds^2 = -(cdt^2) + dx^2 + dy^2 + dz^2$$
From what I understand, the actual spacetime distance between two events is the same numerical quantity in all reference frames, inertial or not. It is not however the case that the line element describing distances will have this mathematical form in all frames.

Going off of this, if we have a four-vector connecting two events in spacetime, then its length is the spacetime distance between those events. Its length should be the same as calculated in all frames, not just inertial ones. And similarly, since the vector describes a physically immutable concept, operations such as scaling it, calculating its length, or adding and dotting it with other vectors should have the same result in all frames, right?
When you are dealing with non inertial frames the way that you write the interval is: $$ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}$$ This expression is manifestly covariant so it works in all reference frames (inertial or not) and in all spacetimes (flat or curved). From this expression the one you wrote can be easily obtained from knowing the components of ##g_{\mu\nu}## in an inertial frame.

The following are also manifestly covariant.
Scaling and calculating length (squared): $$g_{\mu\nu}(n \ dx^{\mu})(n \ dx^{\nu})=n^2 g_{\mu\nu}dx^{\mu} dx^{\nu}$$

Adding and dotting with other vectors: $$g_{\mu\nu}(dx^{\mu}+dw^{\mu})dv^{\nu} =g_{\mu\nu}dx^{\mu}dv^{\nu}+ g_{\mu\nu}dw^{\mu} dv^{\nu}$$

• vanhees71