So, I was trying to prove that "flux across a surface of constant $x$" shouldn't be loosely called "flux in the $x$-direction" by using the inner product approach to compute the angle $\theta$ between the number flux four-vector, $N$ and the one-form $X$ that represents surfaces of constant $x$ which is given by $\arcsin\left(\frac{\langle N,X\rangle}{|N||X|}\right)$. For this to be true, it mustn’t equal to $0$. Then I set up the vector $N$ so that it's timelike, and I encountered a problem: the magnitude of $N$ is imaginary, and thus I get a complex angle. I think something's wrong here. But no matter how I try to compute the angle between the vector $N$ and one-form $X$, it gives me the same result. If my result is correct, I need a physical intuition to make sense of it. If it's wrong, I need a guide to solve this problem (using the same approach if possible). Here's my working:

$$\theta=\sin^{-1}\left(\frac{\langle \textbf{N},\widetilde{dx}\rangle}{|\textbf{N}||\widetilde{dx}|}\right)$$ \begin{align} {\rm where}\, \textbf{N}&= {\rm number}\,{\rm flux}\ 4{\textrm -}{\rm vector}\\ \widetilde{dx} &= {\rm one}\,{\rm form}\,{\rm that}\,{\rm represents}\,{\rm surfaces}\,{\rm at}\,{\rm constant}\,r \end{align}

I demand that $\textbf{N}\cdot\textbf{N}$ is negative (timelike vector).

E.g., let


$$\textbf{N}\cdot\textbf{N}= \begin{bmatrix}3\\2\\0\\0\end{bmatrix}\begin{bmatrix}3\\2\\0\\0\end{bmatrix}=-5$$


$$\widetilde{dx}=\begin{bmatrix}0\ 1\ 0\ 0 \end{bmatrix},\,|\widetilde{dx}|=1$$

$$\langle\textbf{N},\tilde{dx}\rangle=\begin{bmatrix}0\ 1\ 0\ 0 \end{bmatrix}\cdot\begin{bmatrix}3\\2\\0\\0\end{bmatrix}=2$$

Therefore, \begin{align} \theta&=\sin^{-1}\left(\frac{\langle \textbf{N},\widetilde{dx}\rangle}{|\textbf{N}||\widetilde{dx}|}\right)\\ &=\sin^{-1}\left(\frac2{(i\sqrt{5})(1)}\right)\\ &=\sin^{-1}\left(\frac{-2}{5}i\right) \end{align}

It's obvious that $\theta$ is a complex number. Why does this happen, or is there any error in my working?

  • 3
    $\begingroup$ Hi Ray00, and welcome to Physics Stack Exchange! We use MathJax for formulas; could you edit your post to have that instead of the picture? (If you don't know how, eventually someone should be able to do it for you, but it's best if you can do it yourself.) Pictures shouldn't be used for text or formulas, only for things that truly need to be images. $\endgroup$
    – David Z
    Commented Mar 6, 2018 at 6:17
  • $\begingroup$ This made more sense when you used cosine with the inner-product. If you use sine, it would make more sense to use the outer-product. $\endgroup$
    – robphy
    Commented Mar 7, 2018 at 17:31
  • $\begingroup$ @robphy, I used cos earlier when I first asked this question. But that’s because I wrongly assumed that one form is a vector. E.g: Say I have a vector $V$ in the $y$-direction and a one-form $X$ which represents surfaces of constant $x$. Therefore, the vector $V$ is tangent to the one-form $X$, and after applying inner product I get 0 since there’s no intersection. However that doesn’t make sense because you would obtain $\theta=90$, which contradicts the fact that the one-form and vector is tangent to each other. So, I changed it to be sin to correct the problem. $\endgroup$
    – Ray_00
    Commented Mar 8, 2018 at 6:40
  • $\begingroup$ Sir, I’m still new in this, but doesn’t the outer product produces a tensor? So, how do I get an angle from that? And will it be able to produce a real number answer? Or is my approach (the vector product approach) actually wrong in the first place? If possible I need your guidance to solve this problem. Thank you. $\endgroup$
    – Ray_00
    Commented Mar 8, 2018 at 6:47

1 Answer 1


This answer won't address the flux issue, but rather focus on the angle issue.

Given a vector and a one-form, what does "angle" mean geometrically? [I guess one could try to define it algebraically... (e.g. the inverse-sin of a quantity or the inverse-cos of a quantity) but then without a geometric interpretation, it becomes difficult to make sense of results, especially unexpected ones.]

Given two vectors in Euclidean space, the angle could be described as the arc-length of an arc of the unit-circle intercepted by those vectors. Then the inner product could be expressed in terms of the cosine of that angle. (That the cosine is related to the Euclidean metric can be seen by trying to compute the arc-length integral... one finds the ordinary trig functions arise.)

Given two timelike future-directed 4-vectors in Minkowski, the angle [a.k.a. the rapidity] could be described as the Minkowski arc-length of the intercepted arc of the unit-Minkowski circle [a hyperbola] in the plane spanned by those two 4-vectors. Then the Minkowski inner product could be expressed in terms of the hyperbolic cosine of that [rapidity] angle. (That the hyperbolic-cosine is related to the Minkowski metric can be seen by trying to compute the Minkowski arc-length integral... one finds the hyperbolic trig functions arise.) If one tries to use a cosine, then argument of the cosine might not be real valued.

For other combinations of 4-vectors, it might not be so easy to define [where is the arc-of-the-hyperbola?]. In particular, given a timelike 4-vector and spacelike 4-vector, what is the arc-length of the intercepted arc? first, what is the intercepted-arc? then what is its arc length?

Note that these are metrical issues since one needs a metric to map two vectors to a scalar.

However, given a vector and a one-form, no metric is required... so, the notion of angle doesn't make sense [in the way it was defined above]. (If your one-form is really the metric-dual of a vector, then your problem is really that of two vectors and the metric.)

In short, your construction and calculation doesn't make much sense to me.

Toward a resolution of your problem, it might be worth trying to define geometrically what this "angle" you seek is.


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