# Show that getting parallel transported does not change angle between them- Tensors [closed]

I must tell you that I have never seen this kind of question in Tensor Analysis. Our professor had set up this question in our exam, but I don't know whether it belongs to tensors or not. The question goes like thiS:

1. If a vector $u^{i}$ gets parallel transported along a curve S, then $$u^{i}{}_{; j}~ \frac{dX^{j}}{dS}~=~0.$$ 2. If the angle between $u^{i}$ and $v^{j}$ is $\theta$, show that getting parallel transported does not change angle between them.

• Your notation seems a bit non-standard to me, especially the part with the $\mathrm{d}X/\mathrm{d}S$. Is it a vector field? Please elaborate a bit on your question. Commented Jul 18, 2014 at 17:21
• That has really been the problem for me as well,first of all, is there any mistakes in my question...Even I am struggling to understand the question Commented Jul 18, 2014 at 17:27
• Also, with respect to which connection do we parallel transport (Levi-Civita naturally comes to mind)? Commented Jul 18, 2014 at 17:29
• I want to rephrase this question as- Show that the angle between two vectors remain invariant or does not change in parallel displacement parallel transport. Commented Jul 19, 2014 at 0:59

A vector $u^a$ is parallel-transported along the integral curve of tangent vector $V^a \equiv \frac{dX^a}{d s}$ if we have $V^a\nabla_a u^b =0$ (a vector parallel-transported with respect to itself defines a geodesic). The angle between two vectors $v^a$ and $u^a$ is given by $\cos\theta = \frac{v^a u_a}{\sqrt{v^av_a u^bu_b}}$, hence we take: $$V^a \nabla_a (u^b v_b) = v_bV^a \nabla_a u^b+ V^a u^b \nabla v_b = 0,$$ since the vectors $v^a$ and $u^a$ are parallel-transported with respect to $V^a$. Since $V^a \nabla_a (u^b u_b) = 0$ for the same reason, $\theta$ remains constant along the integral curve of $V^a$.
• I'm using $\nabla_a$ instead of ${}_{;a}$ and $s$ is the affine parameter of $X^a$. Commented Jul 21, 2014 at 23:11