Suppose there is a quantity written as $\sum\limits_\mu \nabla_\mu V^\mu$ which is invariant under a coordinate transformation, i.e. scalar, where $V^\mu=(V^0,V^1,V^2,V^3)$ and $\nabla_\mu$ is a covariant derivative. Must $V^\mu$ be necessarily a vector or something else?

  • 1
    $\begingroup$ I'm not sure what you mean. $V^\mu$ is a vector with components $V^0$, $V^1$, $V^2$, $V^3$. $\endgroup$
    – Jold
    Commented Oct 13, 2014 at 0:26
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    $\begingroup$ $\nabla_{\mu} V^{\mu}$ is always a scalar, or more precisely, a scalar field. Do you perhaps mean "constant" when you are saying "scalar", in both places? $\endgroup$
    – Red Act
    Commented Oct 13, 2014 at 0:54
  • $\begingroup$ You may be interested in this Math.SE post. $\endgroup$
    – Kyle Kanos
    Commented Oct 13, 2014 at 1:43
  • $\begingroup$ I am sorry that I did not clarify it. "scalar" here means an invariant quantity under coordinate transformation. $\endgroup$
    – Traveler
    Commented Oct 13, 2014 at 6:05
  • $\begingroup$ Sorry that I said $V^\mu$ is a vector. Suppose we do not know what the properties of $V^\mu$ is but know $\sum\limits_\mu \nabla_\mu V^\mu$ is a scalar. Do you mean $V^\mu$ must be a vector? $\endgroup$
    – Traveler
    Commented Oct 13, 2014 at 6:36

2 Answers 2


In general, the statement that $\nabla_\mu V^\mu$ transforms as a scalar does not quite fix the transformation properties of $V^\mu$. Rather, the most general such transformation would be

$$V^\mu \mapsto V'^\mu + C^\mu,$$

where $V'^\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu$ is the ordinary vector transformation law, and $C^\mu$ is any quantity satisfying $\nabla_\mu C^\mu = 0$. You can get a large class of such transformations by taking $$C^\mu = \epsilon^{\alpha \beta\gamma\mu}\nabla_\alpha B_{\beta\gamma},$$ for any two form $B_{\beta\gamma}$, since such a quantity will have identically vanishing divergence. Thus, $V^\mu$ is a vector up to the "gauge ambiguity" parameterized by $B_{\beta\gamma}$, or simply another object entirely that doesn't transform covariantly.

  • $\begingroup$ :Thank you for your valuable explanation. Can we construct such an object like $V^\mu$ with non-trivial $C^\mu$ in terms of metrics and derivatives? By the way, can you clarify how to get $\nabla_\mu C^\mu = 0$ from $C^\mu = \epsilon^{\alpha \beta\gamma\mu}\nabla_\alpha B_{\beta\gamma}$? I do not see how it vanishes. $\nabla_\mu \nabla_\alpha-\nabla_\alpha \nabla_\mu$ leads to the Riemann tensor, which is antisymmetric in $(\alpha,\mu)$ and $\epsilon^{\alpha \beta\gamma\mu}$ is so. $\endgroup$
    – Traveler
    Commented Oct 14, 2014 at 10:35
  • $\begingroup$ @Traveler I don't know of any object that would naturally transform in this way. The closest thing I've seen to this kind of modified transformation law is for massless representations of the Lorentz group, i.e. the gauge field $A_\mu$ of electromagnetism, which from a representation viewpoint has a transformation that looks like an ordinary Lorentz transform, plus a gauge transform. It seems that the general $V^\mu$ here might arise in a similar fashion, but I don't know exactly how. $\endgroup$
    – asperanz
    Commented Oct 14, 2014 at 16:59
  • $\begingroup$ @Traveler for the vanishing divergence of $C^\mu$, I just thought of $\nabla_{[\alpha}B_{\beta\gamma]}$ as the exterior derivative of a two form. Then taking the divergence of $C^\mu$ is equivalent to $d d B$, which vanishes identically. If you did it with Riemann tensors, you would get two terms, one for each index of $B_{\beta\gamma}$, and the two contributions should cancel. $\endgroup$
    – asperanz
    Commented Oct 14, 2014 at 17:02
  • $\begingroup$ :Please, look over my calculation as follows and let me know if there are some mistakes. $\nabla_\mu C^\mu = \epsilon^{\alpha \beta\gamma\mu}\nabla_\mu\nabla_\alpha B_{\beta\gamma}=\frac{1}{2}\epsilon^{\alpha \beta\gamma\mu}(\nabla_\mu\nabla_\alpha-\nabla_\alpha\nabla_\mu)B_{\beta\gamma}$ $\endgroup$
    – Traveler
    Commented Oct 15, 2014 at 15:35
  • $\begingroup$ :Using $(\nabla_\mu\nabla_\alpha-\nabla_\alpha\nabla_\mu)B_{\beta\gamma}=R^\delta_{ \beta \alpha \mu}B_{\delta\gamma}+R^\delta_{ \gamma\alpha\mu} B_{\beta\delta}$, $\nabla_\mu C^\mu=\frac{1}{2}\epsilon^{\alpha\beta\gamma\mu}(R^\delta_{ \beta \alpha \mu}B_{\delta\gamma}-R^\delta_{\beta\alpha\mu} B_{\gamma\delta})=\epsilon^{\alpha\beta\gamma\mu}R^\delta_{ \beta \alpha \mu}B_{\delta\gamma}\neq0$. $\endgroup$
    – Traveler
    Commented Oct 15, 2014 at 15:43

When you write $V^\mu$ you mean that $V$ is a vector. Next $\nabla_\mu V^\mu$ is called divergence of a vector. Finally answering your question, a vector field with constant zero divergence is called incompressible or solenoidal – in this case, no net flow can occur across any closed surface (according to the Gauss law). If it is a constant, but not zero, then there is a constant flow across any closed surface.


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