# Is the spacetime interval a tensor?

Tensors are objects that are invarient under a change of basis representation and whose coordinates change predictably. The spacetime interval is invarient under a change of coordinate representation caused by a change in frame of reference and the coordinates describing it change predictably with the Lorentz transformations.

My question is: Do we consider the spacetime interval to be a tensor, or a tensor-like object in any way mathematically? Or is this similarity something different and I am connecting the wrong dots?

• The space-time interval $ds^2=g_{ij} du^{i}du^{j}$ is defined by the metric tensor - the metric tensor defines the inner product on the tangent space. And $ds^{2}$ is not a tensor - it's a $2-$form - it's ordinary multiplication between $du^{i}$ and $du^{j}$. The length of an infinitesimal tangent vector is $ds$, and the arc length of a curve would be $s=\int^{t_1}_{t_0} \sqrt { g_{ij} \frac {du^{i}}{dt} \frac{du^{j}}{dt}} dt$ - which would be a scalar. – Cinaed Simson Dec 7 '19 at 6:02

Edit: I no longer agree with my answer, and I think it would be appropriate to accept one of the other answers. Depending on what you mean by the spacetime interval, it's either a non-local, non-tensorial thing (if we're talking about $$\Delta s$$), or a 2-form that happens to be preserved by Lorentz transformations (if we're talking about $$ds^2$$).

The spacetime interval is Lorentz invariant. That is, it is a scalar. A scalar is a particular case of a tensor. One could call it a (0,0)-tensor (as in zero covariant and zero contravariant indices).

• Thanks for the answer, does it's invarience under Lorentz transformations make it a tensor or is it a tensor simply because it is a scalar and all scalars are tensors by definition?. And as a follow up if you don't mind, when we transform basis vectors describing tensors by operating on them with a transformation matrix, is this mathematically what we are doing when we change coordinate representations of events in spacetime (with the Lorentz transformations) during frame of reference changes? – Charlie Dec 6 '19 at 0:46
• Tensors are things that transform under Lorentz transformations (or more general coordinate transformations, depending on the context) in a definite way. Specifically, one transforms a tensor by contracting its indices with those of Lorentz transformation tensors. A scalar is a tensor because it transforms in this definite way. But a scalar has no indices so nothing happens to it when you Lorentz transform it. – d_b Dec 6 '19 at 0:56
• As for your second question, if I am reading it correctly the answer is yes. – d_b Dec 6 '19 at 0:57
• @mmeent I guess I am confused then. The definitions of the spacetime interval I have seen make it manifestly a scalar. To be clear, I am talking about the thing one usually writes as $ds$. It seems like what you are talking about is the metric tensor or something like it. – d_b Dec 6 '19 at 15:28
• @d_b The thing people write as $ds^2$ is the metric tensor! – mmeent Dec 9 '19 at 7:15

The answer is technically yes, as has been lined out, but in regards to the second part of the question the answer is an emphatic NO. It should never be treated as a tensor because that only works in flat space and will only lead to future misunderstandings and lots of wasted time.

To give a longer answer: The great thing about most tensors in special relativity is that they turn into similar tensors in general relativity because they are local and locally space-time still looks like Minkowski space. The reason this does not work with the space-time interval is that it is actually non-local. Here local means that it should only depend on what happens in a small neighborhood around the single point you are considering. If you consider the space-time distance between two points it kind of depends on two different points as well as on everything that happens in between.

What you should think about the space-time interval instead is as the length of a curve taken in a specific invariant way. It depends on the curve taken and thus on the end-points and everything inbetween. It just so happens in flat Minkowski-space that for any pair of points there is an obvious curve between two points in the form of a straight line.

The spacetime interval is a bilinear map that takes two (relative position) 4-vectors and produces a scalar. This means that is a rank 2 tensor (more specifically type $$(0,2)$$). In most basic special relativity context, the two 4-vectors are taken to be two copies of the same position 4-vector, but this does not change the mathematical nature of the underlying object.

• I agree with @d_b when he says that you are talking about the metric tensor. The spacetime interval is a single real number and any inertial observer agrees on its value. This makes it a scalar. – MannyC Dec 7 '19 at 1:09

Strictly speaking, yes, because the spacetime interval is a Lorentz scalar. But be aware that many people use the word "tensor" to refer to what is more properly called a tensor field defined on a manifold. These people would not consider the spacetime interval between two points in Minkowski spacetime to be a tensor (although they would consider the local infinitesimal version - the metric - to be one).