# What is an invarient observable in GR

1. Dec 4, 2011

### Naty1

https://www.physicsforums.com/showthr...=548148&page=3 [Broken],
post #36,

Peter Donis posted this:

"....But we *can* describe a generic spacetime in GR using a curved geometry (*which* curved geometry depends on the specific spacetime), and we *can* describe any given curved geometry using various coordinate charts, and transform between them.

We can also show that any physical observable in GR (such as the spacetime curvature observed around a given object by an observer traveling on a given worldline) is described by an invariant, something that is the same in all coordinate charts."

Can someone explain the last sentence a little further. I realize gravitational spacetime curvature is an invarient, but I'm not sure about the "....any physical observable..." part. What does that mean?? I'm taking "worldine" as any timelike curve...with a comoving observer....

So, how do I compare the the boldface statement, for example, with the fact that different speed observers would measure a different size,or say, different kinetic energy, for the object, dependent on their relative velocities.

thanks....

Last edited by a moderator: May 5, 2017
2. Dec 4, 2011

### atyy

You can invariantly specify which event and worldline things are taken relative to.

3. Dec 4, 2011

### Staff: Mentor

To expand on atyy's response, suppose I want to measure the energy and momentum of an electron. To keep things simple, suppose all motion is along the x-axis, so I can ignore the y and z coordinates.

The way I describe the electron's energy and momentum is to assign it a 4-momentum vector with components (in my frame) (E, p). Normally that's all we say about it, and that makes it seem like everything we're talking about is frame-dependent, with no "invariants" present. But in fact we have just specified three invariants. The first is the electron's 4-momentum; it has different components in different frames, but in itself it is a 4-vector, i.e., an invariant geometric object whose components transform in a certain way under Lorentz transformations.

The second invariant is my 4-velocity, which is also a 4-vector, and in my frame has components (1, 0). You can see that if I take the inner product of my 4-velocity with the electron's 4-momentum, I will get the electron's energy E: (E, p) * (1, 0) = E * 1 - p * 0 = E.

The third invariant is a unit vector in the x-direction, which I use to define the "x axis" of my frame. This vector has components (0, 1), and if I take minus the inner product of this vector with the electron's 4-momentum, I get the electron's momentum p in the x-direction: - (E, p) * (0, 1) = - E * 0 + p * 1 = p. We could define similar unit vectors in the y and z directions if we needed to use all three spatial coordinates.

(Btw, the technical term for what I have just defined is a "frame field", which is used a lot in General Relativity. There are a number of complications that arise when gravity is present that I won't go into here.)

The key point in all the above is that all three of these 4-vectors can be transformed into someone else's frame, where it looks like all the components are different: but the two inner products I defined above will still give the *same* results. Here's how that goes: first, let's factor out the electron's rest mass from its 4-momentum: (E, p) = m (g, gv), where g is the Lorentz gamma factor for the electron, and v is its velocity in my frame (I'm using units where c = 1). We thus have E = mg, p = mgv.

Now transform to some other frame moving at speed - u in the x-direction relative to mine; this gives a Lorentz gamma factor h, so my 4-velocity in this frame is h (1, u). (Note that we have set this up so that the other frame is moving in the *opposite* direction to the electron, as seen in my frame.) The electron's speed in this new frame will be w, where we can calculate w from u and v using the relativistic velocity addition law. So the electron's 4-momentum will be mk (1, w), where k is the gamma factor corresponding to w.

The first inner product above, in the new frame, looks like this:

E = mg = h (1, u) * mk (1, w) = m hk (1 - uw)

A little algebra should convince you that g = hk (1 - uw), so E is the same in both frames; in other words, I can express the "energy of electron X measured in frame Y" in terms of invariants that will be the same no matter which frame I calculate them in. You can work through a similar analysis for momentum by Lorentz transforming the x-axis unit vector above and showing that p = mgv comes out the same in the new frame.

4. Dec 4, 2011

### PAllen

A further observation is that any book on differential geometry (even ancient ones with pure index gymnastics) derive lots of rules for deriving scalars from tensors, both by contraction and integration. Any scalar is strictly invariant. A full definition of anything that is an actual measurement in GR should state the derivation of one or more scalars, generally involving features of the world line of a measuring instrument. My one sentence summary is: If your description of purported measurement cannot be stated as the construction of one or more scalars, it is not a valid measurement description in GR.

5. Dec 4, 2011

### Staff: Mentor

Yes, in this terminology, what I was describing was expressing measurements of energy and momentum of the electron in terms of scalars derived by contracting 4-vectors (an inner product is a contraction of two vectors using the metric).

6. Dec 5, 2011

### Naty1

As I feared, my question was, ultimately, due to a lack of thinking on my part.....but I
really like the descriptions above and I gained some insights......so it was worthwhile....

Observables in relativity ARE easier to interpret than in quantum mechanics!!!!!