I Clock synchronization for ring-riding observers on rotating disk

[ergospherical]

Yeah that’s fine. I mean, sure, in the abstract index notation I guess $T^aS^b$ is a different tensor than $S^b T^a$ in that the slot order is reversed ($T \otimes S(u, v) = S \otimes T(v, u)$), but the two are in correspondence anyway.

 

[cianfa72]

Yeah that’s fine. I mean, sure, in the abstract index notation I guess $T^aS^b$ is a different tensor than $S^a T^b$ in that the slot order is reversed ($T \otimes S(u, v) = S \otimes T(v, u)$), but the two are in correspondence anyway.

ok, anyway w.r.t. the contraction with $\nabla_b\omega_a$ to get the scalar it should be fine, I believe.

 

[cianfa72]

in the abstract index notation I guess $T^aS^b$ is a different tensor than $S^b T^a$ in that the slot order is reversed ($T \otimes S(u, v) = S \otimes T(v, u)$), but the two are in correspondence anyway.

We had a specific thread some months ago about tensor index notation.

From the tensor perspective do you think $T^aS^b$ and $S^bT^a$ are really two different tensors just because the names of their slots starting from left has been changed (reversed in this case) ?

In both cases if we fill respectively slot $a$ and $b$ with same vectors $u$ and $v$ we get the same answer.

 

[cianfa72]

The tensor product’s generally not commutative, no.

Not sure to grasp it. We're not talking about tensor product commutativity (surely it is not), we are discussing the notation. I think $TS(-,-)$ and $ST(-,-)$ are actually different tensors; however if we assign names to their ordered slots starting from left (using Latin letters such as $a$ and $b$ as required in abstract index notation) why they are not the same ?

For instance since $T^aS^b\omega_a \beta_b = S^bT^a\omega_a\beta_b$ for any dual vector $\omega_a$ and $\beta_b$ then $T^aS^b$ and $S^bT^a$ should be actually the same tensor.

From this perspective $T^aS^b$ and $S^aT^b$ instead are really different tensors (e.g. $T^aS^b\omega_a \beta_b \neq S^aT^b\omega_a\beta_b$)

 

[ergospherical]

In the abstract index notation $T^a S^b = T \otimes S$ and $S^b T^a = S \otimes T$.

 

[cianfa72]

In the abstract index notation $T^a S^b = T \otimes S$ and $S^b T^a = S \otimes T$.

So are they really two different tensors? In that case coming back to post#53 why $-Z^aY^b\nabla_b \omega_a = -Y^bZ^a\nabla_b\omega_a$ ?

Btw, from this perspective $T^{ab}$ should be the same as $T^{ba}$, I guess (same tensor $T(-,-)$ just with ordered slots labelled using different names).

Thanks for your time !

 

[DrGreg]

In the abstract index notation $T^a S^b = T \otimes S$ and $S^b T^a = S \otimes T$.

I'm not an expert in the differences between abstract index notation and (non-abstract) Ricci calculus, but I'd be surprised if that really was true.

$T^a S^b = T \otimes S$ implies a convention that the first "slot" is denoted by $a$ and the second "slot" is denoted by $b$. But $S^b T^a = S \otimes T$ is incompatible with that convention.

To my way of thinking, $T^a S^b = T \otimes S$ is ambiguous, and to avoid ambiguity should really be written $T^a S^b = (T \otimes S)^{ab}$.

And then, if I'm right, you can write $$(T \otimes S)^{ab} = T^a S^b = S^b T^a, $$and you can also write $$(T \otimes S)^{ba} = T^b S^a = S^a T^b. $$

Could someone who is experienced in abstract index notation confirm whether I'm right or not?

 

[cianfa72]

IMO the free indices $a$ and $b$ in the abstract index notation are arbitrary letters and I didn’t pay attention to alphabetical ordering to name the slots.

Hence $T^aS^b$ should be the same (2,0) tensor as $T^bS^a$ ?

 

[cianfa72]

Hence $T^aS^b$ should be the same (2,0) tensor as $T^bS^a$ ?

Btw, Wald in section 2.4 makes clear that $T_{ab} \neq T_{ba}$ except for symmetric tensor of course.

Any thought ?

 

[Ibix]

Any thought ?

I think there's a notational weakness with tensors in general. Take the four index Riemann tensor. It describes the change to a vector if you move it around an infinitesimal loop defined by two infinitesimal displacements. By convention, the "output" is the first index, the "input" is the second, and the loop definitions are the third and fourth. But there is absolutely no way to tell from $R^a{}_{bcd}$ which slot is which - you just have to know which one to contract what tensor with.

So imagine we define some tensor $W^{ab}$ and contract with a one-form $\omega_a$. It's immediately clear that if $W^{ab}=S^aT^b$ then $\omega_aW^{ab}=\omega_aS^aT^b$, but if $W^{ab}=T^aS^b$ then $\omega_aW^{ab}=\omega_aT^aS^b$ and these are different. But are the two $W$s really different? If we keep track of which index relates to $S$ and which to $T$ and make sure we contract with the right one then no. If we just contract with the first index in either case then yes. But I don't know a way to notate which slot means which without referring back to the definition of the tensor or simply memorising a convention.

 

[cianfa72]

At the beginning of this thread we talked about 'stationary' congruence like the Langevin congruence in Minkowski spacetime. Langevin congruence is defined as stationary just because its worldlines are integral orbits of a timelike Killing vector field (KVF).

If the above is correct then, by definition, given a spacetime the existence of at least one stationary congruence suffices to define it as a stationary spacetime, right ?

 

[PeterDonis]

by definition, given a spacetime the existence of at least one stationary congruence suffices to define it as a stationary spacetime, right ?

Yes.

 

[cianfa72]

So in Minkowski spacetime there are some types of timelike KVFs (actually an infinite families of them): inertial KVFs, Rindler KVFs and Langevin KVFs (the corresponding integral orbits define indeed stationary congruences). Both the first two are also static since they are hypersurface orthogonal whilst the third is not.

 

[PeterDonis]

So in Minkowski spacetime there are some types of timelike KVFs (actually an infinite families of them): inertial KVFs, Rindler KVFs and Langevin KVFs (the corresponding integral orbits define indeed stationary congruences). Both the first two are also static since they are hypersurface orthogonal whilst the third is not.

Yes. Note, however, that while the inertial KVFs are timelike everywhere, the others are not; they are only timelike in restricted open regions of the spacetime (in the Rindler case, the appropriate "wedges" where the hyperbolas that are the integral curves of the KVF are timelike; in the Langevin case, an open "tube" where the radius is small enough to make the helical integral curves of the KVF timelike).