Spacetime Interval: Is it Invariant Under Rotations?

[Kalidor]

I know that the spacetime interval is the same in coordinate system moving wrt each other at constant speed. But is it true that the spacetime interval is invariant under rotations? If so can you suggest a proof or post a link to one?

 

[Mentz114]

The proper interval is

ds² = -c²dt² - (dx² + dy² + dz²)

and the spatial part is invariant under rotations ( as in 3D Euclidean) , so the whole thing is also.

dx' = cos(a)dx + sin(a)dy

dy' = -sin(a)dx + cos(a)dy

(dx')² + (dy')² = dx² (cos(a)²+sin(a)²) + dy²(cos(a)²+sin(a)²) = dx² + dy²

 

[Meir Achuz]

It is not invariant for a rotation with an angular velocity.

 

[JesseM]

It is not invariant for a rotation with an angular velocity.

I thought the original question was just about inertial frames with spatial axes rotated relative to one another, not about non-inertial rotating frames. In a non-inertial frame the metric line element would have to be something different from ds² = c²dt² - (dx² + dy² + dz²)

 

[Dale]

I know that the spacetime interval is the same in coordinate system moving wrt each other at constant speed. But is it true that the spacetime interval is invariant under rotations? If so can you suggest a proof or post a link to one?

In general, to determine if it is invariant under some transformation simply apply the transformation and see if it simplifies back to the original form as shown by Mentz114.

 

[ApplePion]

I know that the spacetime interval is the same in coordinate system moving wrt each other at constant speed. But is it true that the spacetime interval is invariant under rotations? If so can you suggest a proof or post a link to one?

The proof is almost by definition.

The spacetime intervasl contains the metric which is a covariant 2-tensor, and two spacetime coordinate differentials, each of which are 1-contravariant tensors. So under the rules of tensor transformations, the spacetime interval is forced to be a scalar underr ANY coordinate transformation.