If you use an orthogonal frame, then indices can be raised and lowered freely. Normally if you want to lower an index, you have to multiply by the metric but since the metric is identity matrix, it disappears. This why I often do my calculations in an arbitrary frame. For me this can be less confusing than using an orthogonal one.

Thanks, but I think since the coordinate is not normalised, I still have to keep $l_a$ and its time derivative is not zero. Just to clarify the coordinate is orthogonal but not normalised. @Deane

Sorry. I don’t think both both equation 1 and the next equation can both be correct, as your calculation shows.

But it has to be correct as it is just relaxation to a free energy!

Well, since the metric is time dependent there has to be a time derivative of the metric in one of the equations, depending on which one you start with. Your calculation demonstrates this.

A possible way out of this is to assume the frame is time dependent

The frame is indeed time dependent. Is there any chance this could solve the issue? @Deane

You've assumed a covariant Kronecker delta is metric-compatible. Only a mixed Kronecker delta is. You therefore can't rewrite $l_a^2\delta_{ab}\nabla_k\nabla^kT^b$ as $l_a^2\nabla_k\nabla^k\left(\delta_{ab}T^b\right)$.

@J.G. the non-zero elements of the metric are $g_{11}=l_1^2$, $g_{22}=l_2^2$ and $g_{33}=l_3^2$. The metric is a rank 3 tensor. All the other elements are zero. So the metric is indeed $g_{ab}=l_a^2 \delta_a^b$ and the compatibility has to satisfy

@J.G. is correct.

Sorry, I am confused. How does this solve the issue? How is it related to fram being time depenedent? I can just write $\nabla_k \nabla^k g_{ab} t^b= g_{ab} \nabla_k \nabla^k t^b$ by metric compatibility and then sub the elements of the metric. @Deane

I can just write $\nabla_k \nabla^k g_{ab} t^b= g_{ab} \nabla_k \nabla^k t^b$ by metric compatibility and then sub the elements of the metric. @J.G.

You can put $g_{ab}$ inside or outside the $\Delta^2$. What you can't do is split it into two factors, one inside, one out. The specific way the metric diagonalizes here is a coordinate artefact. A mixed Kronecker delta is just as invariant as a scalar, just as an identity matrix is invariant under rotation. But a covariant Kronecker delta isn't.

So do you agree that for the defined metric, I can just write $\nabla_k \nabla^k g_{ab} T^b = g_{ab} \nabla_k \nabla^k T^b = l_a^2 \nabla_k \nabla^k T^a$? where in the last step I have done a summation on b. @J.G.

No, I don't. There are two issues. The first is that $T^a=\delta^a{}_bT^b$. The second is that we can't contract two of three $a$s.

@J.G. Do you agree that the following is correct? i think it is as it's just writing down different terms of the summation on b: $g_{ab}\nabla_k \nabla^k T^b= g_{a1}\nabla_k \nabla^k T^1+g_{a2}\nabla_k \nabla^k T^2+g_{a3}\nabla_k \nabla^k T^3$

I think your comment was helpful and I proceed a bit more now. I just have a tricky questions for you: is this relation $g^{ca} \partial_t g_{ab} = \frac{1}{2} \partial_t (g^{ca} g_{ab})$ correct? @J.G.

I gotta say that if I were to do this calculation, I would follow the advice in my first comment and substitute the special form of the metric only at the end. That avoids all the subtleties you are running into.

I did what you suggested. Could you please have a look at the modified question? @Deane

If your frame is time dependent, you have to tell us how it moves under the flow

Here's another issue: I don't really know what $F$ is. Could you provide an example of an $F$ that appears in your work or whatever you're reading?

$F = \int dv f$ @Deane where $f$ is free energy density

@Deane the motion of the frame could be given by anything. say with some functionality that does not affect the dynamics of the vector field T

Could you provide a formula for an example of $f$?

@Deane the formula was in the question $f= \nabla_k T^m \nabla^k T_m$

@Deane where T is a vector