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"Gradient of a tensor field of rank $n$ is a tensor of rank $n + 1 $, In cartesian coordinates".

I came across this statement in the Mathematical physics by Arfken. How does this statement relate to the definition of a tensor?

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  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Dec 9 '20 at 20:57
  • $\begingroup$ Which definition of a tensor do you mean? This is easiest to understand if you define tensors based on the transformations of their components under changes of coordinates, but there are other definitions. $\endgroup$
    – G. Smith
    Dec 9 '20 at 21:07
  • $\begingroup$ The only definition in our textbook is based on "An mathematical object that its components transform in a specified manner under coordinate transformations". I do understand the transformation, but how does that transformation relate to my question? $\endgroup$ Dec 9 '20 at 21:19
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It's common for physicists to think about tensors in terms of the transformation rules for their components when you change the coordinates, such as rotating the coordinate axes.

The two "prototypical" tensors are simple geometric notions: the coordinate differential $dx^\mu$ and the partial derivative operator $\partial_\mu$. They have the transformation rules

$$dx^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^\alpha}dx^\alpha$$

$$\partial_{\mu'}=\frac{\partial x^\alpha}{\partial x^{\mu'}}\partial_\alpha$$

from unprimed to primed coordinates. (I'm using the Einstein summation convention, and also the less-common convention that the primes go on the indices.) These transformation rules are nothing more than an application of the chain rule of calculus.

Tensors of rank 1 ("vectors") are defined as quantities that have contravariant and covariant components that transform in the same way as these simple geometrical quantities:

$$V^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^\alpha}V^\alpha$$

$$V_{\mu'}=\frac{\partial x^\alpha}{\partial x^{\mu'}}V_\alpha\,.$$

Tensors of higher rank are defined by performing the same transformations on each index. For example, a rank-2 tensor has two indices and obeys

$$T^{\mu'\nu'}=\frac{\partial x^{\mu'}}{\partial x^\alpha}\frac{\partial x^{\nu'}}{\partial x^\beta}T^{\alpha\beta}$$

$$T_{\mu'\nu'}=\frac{\partial x^\alpha}{\partial x^{\mu'}}\frac{\partial x^\beta}{\partial x^{\nu'}}T_{\alpha\beta}\,.$$

(Note that these transformations are identical to that of the "tensor product" of two vectors, with components $V^\mu V^\nu$ and $V_\mu V_\nu$.)

Now, consider the case when the coordinate transformations are linear, such as between rotated Cartesian coordinates. Then the partial derivatives $\partial x^{\mu'}/\partial x^\alpha$ and $\partial x^{\alpha}/\partial x^{\mu'}$ are just constants, not functions, and you don’t have to worry about a gradient operator operating on them.

If you combine the above transformation rules, you'll find that the gradient $\partial_\lambda V_{\mu}$ (often written $V_{\mu,\lambda}$) transforms as a tensor of rank 2 and $\partial_\lambda T_{\mu\nu}$ (or $T_{\mu\nu,\lambda}$) transforms as a tensor of rank 3.

So taking the gradient just produces something that transforms as a tensor of one-higher rank.

You can also take the gradient of $V^\mu$ to get a tensor of rank 2 with "mixed" indices, $V^\mu{}_{,\lambda}$, etc.

For nonlinear coordinate transformations, it is possible to define a "covariant derivative" $D_\mu$, more complicated than just $\partial_\mu$, that makes these simple transformation rules continue to work.

This way of thinking about tensors, emphasizing their components, is practical but old-fashioned. Mathematicians tend to hate it. But it was good enough for Weinberg, who wrote a famous textbook on General Relativity this way.

A more modern approach emphasizes the tensor itself and the fact that although its components change when the coordinates change, the tensor as a geometrical object remains unchanged. For example, think of a vector (which, remember, is a rank-1 tensor) such as velocity. It has a magnitude and direction independent of any coordinate system. Similarly, any tensor is actually a geometrically invariant object, even though its coordinates transform.

This is why we write the laws of physics using tensors. We are expressing relationships between geometrical objects which are entirely independent of the coordinate systems that we use to describe them. This is still true even if we use the components-only notation; it’s just less obvious.

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Very roughly, the idea is simply that the partial derivative transforms as a rank 1 covariant tensor under coordinate transformations. Explicitly, if I change coordinates from $x^i$ to $y^i$, then the derivative transforms as $$ \frac{\partial}{\partial y^i} = \frac{\partial x^j}{\partial y^i} \frac{\partial}{\partial x^j} . $$

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