# How is this deduced? (Differentiation of tensors)

In Schutz's An Introduction to General Relativity, he talked about how to differentiate tensors. Here is a step that I cannot understand.

$$\frac{d\mathbf{T}}{d\tau} = \left( T^{\alpha}_{\beta, \gamma} \tilde{\omega}^\beta \otimes \mathbf{e}_\alpha \right) U^\gamma$$

and from that he deduced

$$\nabla \mathbf{T} = \left( T^{\alpha}_{\beta, \gamma} \tilde{\omega}^\beta \otimes \tilde{\omega}^\gamma \otimes \mathbf{e}_\alpha \right)$$

Can anybody help me understand what is going on? What is the relationship between $$\frac{d\mathbf{T}}{d\tau}$$ and $$\nabla \mathbf{T}$$? Are they the same? Why is $$U^\gamma$$ transformed into $$\tilde{\omega}^\gamma$$?

• The vector field of components $U^{\gamma}$ is tangent to the path in spacetime parameterized by $\tau$ Commented Apr 8 at 15:22

The "abstract" covariant derivative $$\nabla$$ maps tensors of type $$(i, j)$$ to tensors of type $$(i, j+1)$$. More precisely, the covariant derivative adds one dual slot to the tensor. Using the convention of adding this slot to the front of the tensor, we have $$\nabla T (U, \ldots) = \nabla_U T(\ldots)$$ where $$\ldots$$ represents the original slots of $$T$$.
Intuitively, what $$\nabla T$$ is doing is that it is taking the covariant derivative of $$T$$ in each dimension and "assembling" them into an object one rank higher. The "directional" covariant derivative is then obtained by contracting this additional slot with the direction $$U$$.
It's the same with the ordinary directional derivative, in vector calculus, of a function $$f$$ in the direction $$\mathbf{u}$$: $$\text{grad}\, f \cdot \mathbf{u} = g(\text{grad}\, f,\mathbf{u}) = \mathrm{d}f(\mathbf{u})$$ where $$\mathrm{d}f$$ is analogous to $$\nabla T$$ and $$\mathrm{d}f(\mathbf{u})$$ is analogous to $$\nabla_U T$$. The only difference here is that $$\mathrm{d}f$$ (which is a covector) is the covariant derivative of $$f$$ whereas what we normally call the gradient of $$f$$ is a vector which is obtained by raising the index of $$\mathrm{d}f$$. The directional derivative is likewise obtained by contracting with $$\mathbf{u}$$, giving the scalar $$\mathrm{d}f (\mathbf{u})$$.
The motivating reason for all of this is the chain rule. If we have a curve $$x^\mu (\lambda)$$, the rate of change of $$f$$ along this curve is $$\frac{\mathrm{d}f}{\mathrm{d}\lambda} = \frac{\partial f}{\partial x^\nu} \frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda}$$
Thus, we see that it separates into two distinct effects: the first term is precisely $$\mathrm{d}f$$ which represents the level sets of $$f$$ while the second term is the tangent vector and represents the speed and direction at which you are moving through the level sets. This why they combine nicely together to give the rate of change of $$f$$ along the curve.
In your notation, $$U = \mathrm{d}/\mathrm{d}\tau$$ so $$\mathrm{d}T/\mathrm{d}\tau$$ represents $$\nabla_U T$$, although it's not clear what ordering convention is being used. Note that you cannot write upper and lower indices in the same place. The order of the slots matters.