Imagine we have $p$-contravariant and $q$-covariant tensor, such that

$$ t~=~t^{i_1, \dots, i_p} _{j_1, \dots, j_q} e_{i_1}\otimes \dots \otimes e_{i_p} \otimes e^{j_1} \otimes \dots \otimes e^{j_q}~\in ~T \left( p,q, L\right),$$

where $e_{i_l}$ are basis vectors of $L$ (arbitrary linear space), $e^{j_k}$ its dual basis in $L^{*}$. My question is following: what is the physical meaning of vector and its dual basises? I mean, for example, how do I interpret 2 times contravariant and 1 time covariant tensor in physics.


2 Answers 2


One way of seeing the physical difference between vectors and covectors is the way they scale when you rescale your coordinates. For example, if I change my distance units from meters to centimeters, then a $\Delta x$, expressed numerically in terms of its components in a chosen basis, goes up by a factor of 100, while a wavelength goes down by a factor of 100. This tells us that displacement is a vector, while frequency is a covector.

This is why gradients are covectors. For example, the gradient of a temperature tells you how much the temperature changes per meter.

The physical distinction between vectors and covectors is often obscured by the fact that if we have a metric, we can freely raise and lower indices. That could give the impression that the distinction is optional or unimportant. But there are practical situations where you don't have a metric, and then the distinction is mandatory and important. Here is an example. If you really want a thorough and careful discussion of this sort of thing, the book to look at is Spacetime, Geometry, Cosmology by Burke.

  • $\begingroup$ Thank you, that is the way to go. I was afraid there will be no such an interpretation. Very thankful for further reading recommendation, also. $\endgroup$
    – Bohdan
    Oct 22, 2013 at 19:24
  • $\begingroup$ frequency is a covector sounds strange to me; never fear: wave_vector_ [sic] to the rescue ;) $\endgroup$
    – Christoph
    Oct 22, 2013 at 19:30

In physics tensors are machines for generating scalars by devouring the appropriate combination of vectors and one-forms. In your example we would expect your tensor to be used in some expression of the form:

$$ \text{scalar} = T^{\alpha\beta}_\gamma a_\alpha b_\beta c^\gamma $$

i.e. given the one-forms $a$ and $b$ and the vector $c$, the tensor represents some physical law for calculating our scalar. Obviously what the law is depends on the context.

  • 1
    $\begingroup$ I would say that this is a mathematical interpretation, not a physical one. $\endgroup$
    – user4552
    Oct 22, 2013 at 18:50

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