What is the most general meaning of the expression covariant, but not manifestly covariant? Suppose I have a general (local) change of coordinates, $x^{\prime} = f(x)$, on an $(n+1)$-dimensional smooth manifold on which some classical fields are defined, say $A_{\alpha}(x_0,x_1,...,x_n)$, which transform into $A_{\alpha}^{\prime}(x^{\prime})$. Suppose the fields $A_{\alpha}(x)$ satisfy some equations of motion, where $x_0 = t$. How should these EOMs look like to be covariant with respect to the given change of coordinates, but not manifestly covariant? Could you explain in plain words the difference between the 2 forms of the EOMs?

If possible, can you write down a practical example of such a situation encountered in physics?


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    $\begingroup$ Maxwell's equations in terms of electric and magnetic fields are covariant but not manifestly covariant. The equation $\partial_\mu F^{\mu\nu} = j^\nu$ is manifestly covariant. $\endgroup$ – knzhou Aug 22 '16 at 0:34
  • $\begingroup$ @knzhou Thx for input, but can you explain in plain words the difference between manifestly covariant vs covariant but not manifestly covariant? I can see that it has to do with the kind of objects one uses for expressing the EOMs, and their transformation properties, but I would like to see it expressed clearly in words. You can write an answer and I will vote it! $\endgroup$ – user22208 Aug 22 '16 at 0:45
  • $\begingroup$ @knzhou If the form of EOMs remains the same (since this is the meaning of covariance), then what makes one form manifest and the other not manifest? If you know please share. $\endgroup$ – user22208 Aug 22 '16 at 0:58
  • $\begingroup$ @Qmechanic I've searched and saw you're the best at symmetry questions. Could you please help me with this one? Maybe it sounds too silly for you, but I really want to know the answer. All the people I've asked refer to Maxwell's equations written in 2 forms like above, but they cannot describe in words the difference. $\endgroup$ – user22208 Aug 22 '16 at 1:20
  • $\begingroup$ @user22208 FYI, you can't ping users who haven't yet commented or otherwise been involved with a post. See meta.stackexchange.com/questions/43019/… $\endgroup$ – jjc385 Aug 22 '16 at 23:25

In my experience, we usually call an expression manifestly covariant under some transformation if all the objects which appear in the transformation transform as tensors (or tensor fields) under the transformation.

For example, Maxwell's equations are not manifestly covariant under Lorentz transformations when written in terms of $\vec E$ and $\vec B$ fields, because these do not transform at tensors. On the other hand, Maxwell's equations written in terms of the electromagnetic field strength tensor $F_{\mu\nu}$, 4-vector current $j^\mu$, and '4-derivative' $\partial_\mu$ are manifestly covariant (under Lorentz transformations), because these objects transform as 2-tensors, vectors, and 1-forms, respectively.

Note that it's important to specify which transformations we're talking about. For example, Maxwell's equations in terms of $\vec E$ and $\vec B$ fields are manifestly covariant under rotations.

I don't think the story changes much for local transformations.


I'll expand on jjc385's answer with an example of how equations of motion can arise in a non-manifestly covariant form. In curved spacetime, the Lagrangian density can be written as $\mathcal{L}=\sqrt{\left| g\right|}\mathcal{L}_0$, with $\mathcal{L}_0$ a scalar called the scalar Lagrangian density. The action is then $S=\int d^4x\sqrt{\left| g\right|}\mathcal{L}_0$, also a scalar. The equation of motion $\frac{\partial\mathcal{L}}{\partial\phi}=\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}$ is not manifestly covariant, because the two sides of the equation aren't tensors in a general spacetime. However, you can derive the equation of motion from first principles in another form, $\frac{\partial\mathcal{L}_0}{\partial\phi}=\nabla_\mu\frac{\partial\mathcal{L}_0}{\partial\nabla_\mu\phi}$. (That's assuming $\phi$ has nothing to do with the metric tensor; if it does, we get a more complicated result, but the point I'm making remains the same.) This only uses tensors, so it is manifestly covariant.

  • $\begingroup$ +1. I should have mentioned partial derivatives aren't diffeomorphism covariant. $\endgroup$ – jjc385 Aug 22 '16 at 6:21

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