Spacetime Interval & Metric: Equivalent?

[jmatt]

This may seem an odd question but it will clear something up for me. Are "The spacetime interval is invariant." and the "The spacetime metric is a tensor." exactly equivalent statements? Does one imply more or less information than the other?

Thanks!

 

[Matterwave]

They aren't exactly equivalent, but they are closely related. The space time interval is $ds^2=g_{ab}dx^a dx^b$. Because the metric $g_{ab}$ is a tensor, contracting it with (infinitesimal) vectors $dx^{a}$ will yield an invariant scalar $ds^2$.

 

[jmatt]

They aren't exactly equivalent, but they are closely related. The space time interval is $ds^2=g_{ab}dx^a dx^b$. Because the metric $g_{ab}$ is a tensor, contracting it with (infinitesimal) vectors $dx^{a}$ will yield an invariant scalar $ds^2$.

Thanks! A little confused why you referred to $ds^2$ as a scalar. Isn't it a four-vector?

 

[bapowell]

Thanks! A little confused why you referred to $ds^2$ as a scalar. Isn't it a four-vector?

Nope. It's the scalar product of (differential) four-vectors.

 

[Matterwave]

Thanks! A little confused why you referred to $ds^2$ as a scalar. Isn't it a four-vector?

Why would it be a 4 vector? What are the 4 components that you are thinking of? It is a scalar because it's just 1 single number, which does not change under an arbitrary Lorentz transformation.

 

[jmatt]

Got it now. Thanks very much.