# Spacetime Interval in non-inertial frames.

1. Jan 15, 2012

### The1337gamer

The interval between two events ds^2 = -(cdt)^2 + x^2 + y^2 + z^2 is invariant in inertial frames. I was wondering, if this same interval still applies and is invariant in non-inertial frames?

2. Jan 15, 2012

### Mentz114

The form of the line element will be the same if the coords (t,x,y,z) are transformed by a member of the Poincare group of transformations.

But other transformations can change the form of the line element. For instance if we go to (u,v,y,z)

u = x-t
v= x+t
y=y
z=z

then the line element becomes ds2 = -2dudv + dy2 + dz2

3. Jan 15, 2012

### Matterwave

There is still an invariant interval, but it will no longer take that form in a non-inertial frame.

4. Jan 15, 2012

### pervect

Staff Emeritus
In general the invariant interval will be a quadratic form ,http://en.wikipedia.org/w/index.php?title=Quadratic_form&oldid=467500079.

If your generalized coordinates are t,x,y,z and you have to points, represented by coordinates t1,x1,y1,z1 and t2,x2,y2,z2, then your quadratic form for the interval will have variables dt,dx,dy,dz -- where dt = t2-t1, dx=x2-x1, dy=y2-y1, dz=z2-z1. For example, it might be (just to illustrate the concept clearly)

-0.8 dt^2 + 1.3 dx^2 + 1.2 dy^2 + 1.1 dz^2

The coefficents of the quadratic form can be represented by a matrix - the wiki article talks a bit about this - if you want a more detailed treatment it should be covered in books about linear algebra. This matrix is called the "metric", which hopefully you've at least herad mentioned.

5. Jan 15, 2012

### jfy4

Your question has really been answered, so sorry if some think this is over kill, but I'll write down the interval for you too, since it hasn't been written yet in this thread, just mentioned.
$$c^2 \, d\tau^2=ds^2=g_{\alpha\beta}dx^{\alpha}dx^{\beta}=dx_{\alpha}dx^{\alpha}$$
This is a geometric invariant. Hope this helps a little.

6. Jan 16, 2012

### The1337gamer

Thanks for the feedback. I know a bit about the metric tensor, it is a bilinear form that takes two vectors from the tangent space of our spacetime to a scalar, i think.

So i think im correct in saying the 4x4 matrix representing the metric is the same on all points in flat spacetime (minkowski space).

I haven't read much into GR yet, but in the general spacetime, does the metric tensor change at each point in the space?

7. Jan 16, 2012

### Matterwave

In general, the metric tensor will change as you move from point to point in space-time. If the metric tensor does not change as you move along a certain vector field, then that vector field is a so-called "killing vector field".

8. Jan 16, 2012

### Staff: Mentor

Yes, except that you need to specify that you are talking about an inertial frame or orthonormal basis. The components of the metric expressed in e.g. a rotating frame will not be the same at all points in flat spacetime.

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