When does the Minkowski metric get non-zero off-diagonals?

[DuckAmuck]

I have only seen scenarios so far where the elements are all along the diagonal, but what are some known cases where there are off-diagonal elements?

Thank you.

 

[Orodruin]

When you go to a non-orthogonal coordinate system. This is the same issue as with non-orthogonal coordinate systems in regular Riemannian geometry.

 

[robphy]

For a simple (and useful) example for Minkowski spacetime, use "light-cone coordinates".
In my conventions, $$u\equiv t+x\qquad v\equiv t-x$$
and $$\hat u\equiv \frac{1}{2}\left( \hat t+\hat x \right)\qquad \hat v\equiv \frac{1}{2}\left( \hat t-\hat x \right).$$
where $\hat{}$ indicates a basis vector (not necessarily a "unit-magnitude" vector).
(Other conventions use factors of $\sqrt{2}$.)
In my signature conventions, using the Minkowski-dot-product, $\quad$ $\hat t \cdot \hat t=1$, $\quad$ $\hat x \cdot \hat x= -1$, and $\quad$ $\hat t \cdot \hat x=0$.

postscript:
Although the basis vectors $\hat u$ and $\hat v$ point along the light cone, and they may look to have a Euclidean-angle of "90-degrees" between them, these basis vectors are not Minkowski-orthogonal (which you can check by computing the Minkowski-dot-product).

 

[kent davidge]

You are correct that the inner product of a fixed pair of vectors does not change with a change in coordinates

As long as one is accepting the equivalence principle, correct? If one is not following the equivalence principle, there would be no need for the inner product to be invariant.

 

[martinbn]

As long as one is accepting the equivalence principle, correct? If one is not following the equivalence principle, there would be no need for the inner product to be invariant.

No, this is just geometry. The equivalence principle is not relevant.

 

[weirdoguy]

As long as one is accepting the equivalence principle, correct?

This has nothing to do with physics, it's a basic fact about tensors.

Edit. Ups, to late.

 

[PAllen]

For a simple (and useful) example for Minkowski spacetime, use "light-cone coordinates".
In my conventions, $$u\equiv t+x\qquad v\equiv t-x$$
and $$\hat u\equiv \frac{1}{2}\left( \hat t+\hat x \right)\qquad \hat v\equiv \frac{1}{2}\left( \hat t-\hat x \right).$$
where $\hat{}$ indicates a basis vector (not necessarily a "unit-magnitude" vector).
(Other conventions use factors of $\sqrt{2}$.)
In my signature conventions, using the Minkowski-dot-product, $\quad$ $\hat t \cdot \hat t=1$, $\quad$ $\hat x \cdot \hat x= -1$, and $\quad$ $\hat t \cdot \hat x=0$.

postscript:
Although the basis vectors $\hat u$ and $\hat v$ point along the light cone, and they may look to have a Euclidean-angle of "90-degrees" between them, these basis vectors are not Minkowski-orthogonal (which you can check by computing the Minkowski-dot-product).

and, of course, the obvious statement that lightlike basis vectors cannot possibly be unit magnitude...since they are zero magnitude

 

[Ibix]

As long as one is accepting the equivalence principle, correct? If one is not following the equivalence principle, there would be no need for the inner product to be invariant.

Are you thinking of the principle of relativity? That's the one that says observables must be frame (or coordinate) independent. I suspect that if you did that, though, you'd have to throw out the whole geometric basis of relativity, and I'm not sure that "inner product" or "vector" would necessarily even have any physical significance.

 

[DrGreg]

Besides robphy's example in post #6, another simple example is to take Minkowski coordinates $(t, x, y, z)$ and define new coordinates$$\begin{align*}
T &= t - \alpha x\\
X &= x\\
Y &= y\\
Z &= z
\end{align*}$$where $\alpha$ is any constant with $|\alpha| < 1/c$. $T$ is still timelike and $X$, $Y$ and $Z$ are still spacelike. This uses a non-standard definition of simultaneity and the one-way coordinate speed of light is not the same in all directions.

 

[Mentz114]

I have only seen scenarios so far where the elements are all along the diagonal, but what are some known cases where there are off-diagonal elements?

Thank you.

The Born chart for a rotating frame has off diagonal elements.

$\pmatrix{{r}^{2}\,{w}^{2}-1 & 0 & 0 & {r}^{2}\,w\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr {r}^{2}\,w & 0 & 0 & {r}^{2}}$

This is coordinate transformation of the Langevin chart

 

[kent davidge]

I suspect that if you did that, though, you'd have to throw out the whole geometric basis of relativity

Yes, I meant if one does not accept the equivalence principle one need not require the inner products to be coordinate independent, but then one is not working with general relativity anymore.

 

[PAllen]

Yes, I meant if one does not accept the equivalence principle one need not require the inner products to be coordinate independent, but then one is not working with general relativity anymore.

An inner product is mathematical construct that is inherently coordinate independent. If some quantity you somehow compute is not coordinate independent, it is not an inner product. It’s like proposing a fractional integer.

 

[kent davidge]

An inner product is mathematical construct that is inherently coordinate independent

What about being inherently basis independent? I thought basis independent was the only general condition.

 

[PAllen]

What about being inherently basis independent? I thought basis independent was the only general condition.

Coordinates provide a basis at every point, so the two are equivalent for this purpose. That is, basis independence implies coordinate independence for inner products.

 

[sweet springs]

Schwartzshild solution is still diagonal, Kerr solution of spinning BH has non-diagonal components in metric.

 

[Orodruin]

Schwartzshild solution is still diagonal, Kerr solution of spinning BH has non-diagonal components in metric.

This thread is about Minkowski space. The metric being diagonal or not depends on the chosen local basis. This is true regardless of what solution you are looking at, but here the title explicitly mentions Minkowski space.