# Tensor Analysis - Request for opinion

1. Oct 14, 2003

### pmb

[SOLVED] Tensor Analysis - Request for opinion

Seems that a few people refer to things like vectors and tensors as quantities which are invariant. For example

Dr. Bertschinger (Cosmologist at MIT) has online notes at http://arcturus.mit.edu/8.962/notes/gr1.pdf
"Introduction to Tensor Calculus for General Relativity,"

In it he writes

"Scalars and vectors are invariant under coordinate transformations;
vector components are not."

this meaning that the vector is a geometric quantity which has a coordinate independant meaning. Call this Meaning Number 1

This is an unusual use of the term "invariant" since that term usually is synonymous with scalar = tensor of rank zero. Call this Meaning Number 2

My question is this - How many go by #1 and how many go by #2 and how many dirive the meaning from context?

Pete

Last edited by a moderator: Apr 20, 2017
2. Oct 15, 2003

### HallsofIvy

Staff Emeritus
Most people do not "go by" either one. Most people understand the difference between "invariant" and "invariant under coordinate transformations".

3. Oct 15, 2003

### pmb

So I take it that you argree that an invariant is a tensor of rank zero? I also take it that you understand that there is a difference between something being "an" invariant and something having the property of "being" invariant (under a coordinate transformation as you say).

The reason I posted this question was due to a few people who didn't understand that what you just said is true. I was curious as to how many people think that.

Pmb

4. Oct 15, 2003

Staff Emeritus
But tensors are not "invariant under coordinate transformations". Tensor equations are. Which means that constant tensors T_mu,nu = k are invariant, but not general tensors (a vector is a tensor of rank 1). You can "transform tensors away" by changing coordinates. This is how you can eliminate (local) gravity in a free falling frame of reference.

5. Oct 15, 2003

### pmb

There is a sense in which tensors are invariant. See Bertschinger's notes above for an explanation/description.

A tensor is a geometrical object which has a meaning independant of the coordinate system. If I change coodinate systems I don't change the tensor. That is what Bertschinger means when he says that a vector remains invariant under a coordinate transformation.

For example: The vector R points North East and has a magnitude of a. If I change coordinates then R remains unchanged.

Personally I find that usage confusing and I try to avoid it. But some relativists use the term in that way.

However the statement "You can 'transform tensors away' by changing coordinates." needs some clarification.

If the components of a tensor vanish in one coordinate system then they vanish in all coordinate systems. The meaning of Einstein's statement that a gravitational field can be transformed away means something different than going from a non-vanishing tensor to a vanishing one (since that is impossible).

Note that the "gravitational field" tensor is the metric.

If, in Minkowski coordinates (t,x,y,z), the metric is not the Minkowski metric g_uv = diag(1,-1,-1,-1) then it is said that there is a gravitational field at that point. If the metric is the Minkowski metric then it is said that there is no a gravitaional field at that point.

What you can transform away are the components of the gravitational field which Einstein defined to be the Christoffel symbols. If, in Minkowski coordinates, the Christoffel symbols don't vanish at a point then there is a gravitational field at that point. But the Christoffel symbols are not parts of a tensor.

This is all explained in Einstein's 1916 paper which, I believe, is available online. See - http://www.alberteinstein.info/

Pete

6. Oct 15, 2003

### HallsofIvy

Staff Emeritus
Okay, it would be better to say that Tensor EQUATIONS are "invariant under coordinate transformations". That is, if A= B (A and B tensors) is true in one coordiante system then it is true in any coordinate system (which is basically the definition of "tensor").

7. Oct 15, 2003

### pmb

Thank you for your input/opinion. Much appreciated!

Pete