I am studying Special Theory of Relativity from the book "Special Relativity And Classical Field Theory" by Leonard Susskind. I am not being able to understand the following:

The three space components of a 4-vector may equal zero in your reference frame. You, in your frame, would say that this displacement is purely timelike. But this is not an invariant statement. In my frame, the space components would not all equal zero, and I would say that the object does move in space. However, if all four components of a displacement 4-vector are zero in your frame, they will also be zero in my frame and in every other frame. A statement that all four components of a 4-vector are zero is an invariant statement.

Firstly, if a displacement in my reference frame is purely timelike then why isn't an invariant statement? The author said in another chapter that

The property of being timelike is invariant: If an event is timelike in any frame, it is timelike in all frames.

Secondly, I do not understand the line:

In my frame, the space components would not all equal zero, and I would say that the object does move in space.

The specification of the space components denotes a particular point in space. How can I say whether an object is moving in space from just the information about the space coordinates? After all, from two different reference frames, an object can have two different sets of spatial coordinates and yet be at rest w.r.t. both of them.

Thirdly, I do not understand why

A statement that all four components of a 4-vector are zero is an invariant statement.

Please help me with these doubts.

  • $\begingroup$ Dr Susskind isn't being inconsistent. Being timelike is invariant, but being purely timelike isn't. $\endgroup$
    – PM 2Ring
    May 26, 2020 at 13:57
  • $\begingroup$ can you please explain the difference between the two? $\endgroup$
    – Alice
    May 26, 2020 at 13:57
  • $\begingroup$ Do the answers here help? physics.stackexchange.com/q/169631/123208 Also see the various linked questions on that page. Basically, in a spacetime diagram a purely timelike interval is perfectly vertical, a purely spacelike interval is perfectly horizontal. And a lightlike interval is at 45°. $\endgroup$
    – PM 2Ring
    May 26, 2020 at 14:15
  • 1
    $\begingroup$ Susskind declares a potentially-confusing shortcut: "events are timelike" for "events are timelike relative to the origin". Using standard definitions, "tImelike" describes a vector, e.g. a displacement, a relation involving two events. Single events are not timelike. [His shortcut should be included in the quotes, or else rewritten the more standard terminology.] $\endgroup$
    – robphy
    May 26, 2020 at 20:01

1 Answer 1


In Special Relativity, we use something known as a spacetime diagram. It looks like this:

enter image description here

Now, here we are only looking at 1 dimension of space and dimension of time. The green dot represents and event, that takes place at some point in space, and at some point in time. In this diagram, every object can be viewed as a worldline (see the orange line), and the slope of the worldline gives the speed of the object. The blue, diagonal line you see represents the speed of light. Now any line left to the blue line is called timelike, and any line to the right is called spacelike. The blue line itself is sometimes called 'null'.

Timelike worldlines represent objects with velocity less than $c$, while spacelike ones represent objects with velocity higher than $c$. As nothing can travel faster than light, spacelike worldlines are impossible.

Now if you have a pure timelike worldline, that means your worldline lies on the time axis. Mathematically, that corresponds to $$ds^2 = dt^2 - 0 = dt^2$$ (where $ds$ is the spacetime interval). But when you have a moving object, travelling relative to you, they can argue that they are at rest and you are moving, so your spacetime interval (according to them) would be: $$ds^2 = dt^2 - dx^2$$ which means in their reference frame, your worldline is not purely timelike. So, being purely timelike is not an invariant.

Now, when you generally say timelike, you mean any worldline to the left of the blue line. Which corresponds to you having velocity less than that of light.

Now, if there is an moving observer, they will agree that your worldline is timelike, because your worldline has to lie somewhere in that region (otherwise they would see you moving faster than light, which leads to all sorts of problems). So the fact that your worldline is timelike is invariant, but the fact that the particular worldline you observe is the one also observed by everybody else is not. So in the first case, the moving observer will agree that your worldline is timelike, but will not agree that it is completely (purely) timelike (that is basically lies on the time axis).

As for the statement

A statement that all four components of a 4-vector are zero is an invariant statement.

If you have any four vector, and if you want to know what it's components are in another frame of reference, you basically apply the Lorentz transformations on each of the components, which corresponds to multiplying or dividing the component by some factors.. But in the case where all four components are $0$, when you apply the Lorentz transformations, you basically get $0$ again. So, because the components remain the same in every frame of reference, you get an invariant statement.

Further Info:

Spacetime diagrams: https://en.wikipedia.org/wiki/Minkowski_diagram

Lorentz transformations: https://en.wikipedia.org/wiki/Lorentz_transformation

And another one of my answers, describing invariants and four vectors in more details (in this context you just need part 1): Why is mass an invariant in Special Relativity?


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