Spacetime Interval: Is it Invariant Under Rotations?

In summary: However, because it is a covariant tensor, the covariant derivative preserves the interval. So for a rotation with an angular velocity, the Spacetime interval is invariant.
  • #1
Kalidor
68
0
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?
 
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  • #2
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
 
Last edited:
  • #3
It is not invariant for a rotation with an angular velocity.
 
  • #4
Meir Achuz said:
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)
 
  • #5
Kalidor said:
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.
 
  • #6
Kalidor said:
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.
 

1. What is the concept of Spacetime Interval?

The Spacetime Interval is a fundamental concept in physics that measures the distance between two events in a four-dimensional spacetime. It combines both the spatial and temporal components of an event and is used to describe the relationship between events in the universe.

2. Is the Spacetime Interval invariant under rotations?

Yes, the Spacetime Interval is invariant under rotations. This means that the measurement of the interval between two events in spacetime will be the same regardless of the observer's frame of reference or the orientation of their coordinate system.

3. How does the Spacetime Interval differ from the Euclidean Distance?

The Spacetime Interval differs from the Euclidean Distance in that it takes into account the relativistic effects of time dilation and length contraction. This allows for the measurement of distances between events to remain consistent in different reference frames, unlike the Euclidean Distance which changes with the observer's perspective.

4. Can the Spacetime Interval be negative?

Yes, the Spacetime Interval can be negative. This occurs when the interval between two events is imaginary, meaning that the events are not causally connected. In this case, the interval is referred to as "spacelike" and represents a distance that cannot be traveled in a finite amount of time.

5. How is the Spacetime Interval used in General Relativity?

The Spacetime Interval is a key component in Einstein's theory of General Relativity. It is used to describe the curvature of spacetime caused by the presence of mass and energy. By calculating the spacetime interval between different events, the curvature of spacetime can be determined and the effects of gravity can be understood.

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