Vectors and the metric tensor

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 5; Problems 5.2, 5.3.

We can define a general vector {\mathbf{A}} in terms of the basis vectors {\mathbf{e}_{i}} in a given coordinate system:

\displaystyle  \mathbf{A}\equiv A^{i}\mathbf{e}_{i} \ \ \ \ \ (1)

This is analogous to the definition of the infinitesimal displacement that we met earlier: {d\mathbf{s}=dx^{i}\mathbf{e}_{i}}. This has a couple of consequences. First, since the basis vectors are not necessarily either unit vectors or orthogonal, this definition may be different from the usual definition of a vector that you’re used to from linear algebra courses.

Second, we require the transformation of a vector’s components between coordinate systems to be the same as the components of {d\mathbf{s}}, which means that

\displaystyle  A^{\prime i}=\frac{\partial x^{\prime i}}{\partial x^{j}}A^{j} \ \ \ \ \ (2)

Finally, the square of a vector follows the same pattern as the square of the increment {ds^{2}}:

\displaystyle  A^{2}=\mathbf{A}\cdot\mathbf{A}=g_{ij}A^{i}A^{j} \ \ \ \ \ (3)

As an example, consider the case of uniform circular motion. From elementary physics, we know that, in polar coordinates, the radial component of the velocity {v^{r}=0} (since the object is always at the same distance from the origin) and the tangential component is {v^{\theta}=v}. Using the metric for polar coordinates, this means that

\displaystyle   v^{2} \displaystyle  = \displaystyle  g_{ij}v^{i}v^{j}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  1\times0\times0+r^{2}\times v^{\theta}\times v^{\theta}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \left(v^{\theta}\right)^{2}r^{2}\ \ \ \ \ (6)
\displaystyle  v^{\theta} \displaystyle  = \displaystyle  \frac{v}{r} \ \ \ \ \ (7)

To transform this vector to rectangular coordinates, we have

\displaystyle  v^{\prime i}=\frac{\partial x^{\prime i}}{\partial x^{j}}v^{j} \ \ \ \ \ (8)

where the primed system is rectangular and the unprimed is polar. So

\displaystyle   v^{x} \displaystyle  = \displaystyle  \frac{\partial x}{\partial r}v^{r}+\frac{\partial x}{\partial\theta}v^{\theta}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -r\sin\theta\frac{v}{r}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  -v\sin\theta\ \ \ \ \ (11)
\displaystyle  v^{y} \displaystyle  = \displaystyle  \frac{\partial y}{\partial r}v^{r}+\frac{\partial y}{\partial\theta}v^{\theta}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  r\cos\theta\frac{v}{r}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  v\cos\theta \ \ \ \ \ (14)

The square is invariant, since using the rectangular metric

\displaystyle   v^{2} \displaystyle  = \displaystyle  g_{ij}v^{i}v^{j}\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \left(-v\sin\theta\right)^{2}+\left(v\cos\theta\right)^{2}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  v^{2} \ \ \ \ \ (17)

Now let’s look at the inverse problem. This time we have an object moving at a constant speed {v} in the {+y} direction, so that {v^{x}=0}, {v^{y}=v}. To convert this to polar coordinates, we need the derivatives

\displaystyle   \frac{\partial r}{\partial x} \displaystyle  = \displaystyle  \frac{x}{\sqrt{x^{2}+y^{2}}}\ \ \ \ \ (18)
\displaystyle  \frac{\partial r}{\partial y} \displaystyle  = \displaystyle  \frac{y}{\sqrt{x^{2}+y^{2}}}\ \ \ \ \ (19)
\displaystyle  \frac{\partial\theta}{\partial x} \displaystyle  = \displaystyle  -\frac{y}{x^{2}+y^{2}}\ \ \ \ \ (20)
\displaystyle  \frac{\partial\theta}{\partial y} \displaystyle  = \displaystyle  \frac{x}{x^{2}+y^{2}} \ \ \ \ \ (21)

Then we get

\displaystyle   v^{r} \displaystyle  = \displaystyle  \frac{y}{\sqrt{x^{2}+y^{2}}}v\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  v\sin\theta\ \ \ \ \ (23)
\displaystyle  v^{\theta} \displaystyle  = \displaystyle  \frac{x}{x^{2}+y^{2}}v\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \frac{\cos\theta}{r}v \ \ \ \ \ (25)

If the object starts at {\left(x,y\right)=\left(b,0\right)} at {t=0}, then {y\left(t\right)=vt} and {x\left(t\right)=b}. In polar coordinates we get

\displaystyle   r\left(t\right) \displaystyle  = \displaystyle  \sqrt{b^{2}+\left(vt\right)^{2}}\ \ \ \ \ (26)
\displaystyle  \theta\left(t\right) \displaystyle  = \displaystyle  \arctan\frac{vt}{b}\ \ \ \ \ (27)
\displaystyle  v^{r} \displaystyle  = \displaystyle  \frac{vt}{\sqrt{b^{2}+\left(vt\right)^{2}}}v\ \ \ \ \ (28)
\displaystyle  v^{\theta} \displaystyle  = \displaystyle  \frac{b}{b^{2}+\left(vt\right)^{2}}v \ \ \ \ \ (29)

At {t=0}, the motion is entirely in the {\theta} direction, since the object is moving tangent to the circle {r=b} at that time. As time increases, the motion gradually transfers over to the radial direction, with {\lim_{t\rightarrow\infty}v^{r}=v}.

12 thoughts on “Vectors and the metric tensor

  1. rbwang

    Hello!
    I am interested in your posts in this site, and, I’d like to do the same thing.
    But I failed at the start, because my latex code didn’t work on my post.
    Could you teach me how do you write your post using latex?

    Sincerely.

    Reply
  2. rbwang

    Hi,
    I converted the tex file into a html file using latex2wp.py, copied and pasted the text of the html file onto my post, however I still failed.
    Could you give me more information on how to use your strategy?

    Regards.

    Reply
    1. growescience

      Sorry, really can’t say much more, since that’s basically what I do and it works. Just to check: you are using wordpress.com (not a user-installed version of wordpress)? If so, can you enter some Latex by hand, such as y=x^2 ?

      Do you paste the HTML into the Text window (not Visual) on the New post page? Is every Latex bit in the HTML enclosed in $latex….$ tags?

      When you say it failed, what do you mean? Do you get the non-Latex text showing up but not the Latex?

      Reply
  3. Serkan

    Can you explain how we do know second requirement because I saw it first time and difficult to understand? Also, I tried geometry, it works very well.

    Reply
    1. growescience

      As
      far as I can tell from Moore’s book, this is just the definition of a vector. That is, a vector is any quantity that transforms in the same way as the infinitesimal displacement ds.

      Reply
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      1. growescience

        The fact that you are getting “Formula does not parse” messages means that wordpress is trying to interpret your code as Latex, so it seems that much works anyway.
        I tried copying the Latex code that appears in the tooltip when you hover the mouse over the “Formula does not parse” and pasting it into one of my own posts. (You can do this in the Chrome browser by right-clicking on the “Formula does not parse” message and selecting Inspect element, then copy the code from the Elements box.) The only thing wrong that I can see is that after the displaystyle there are a couple of nbsp; characters. If you delete those, then the formulas parse OK. These nbsp’s don’t appear in the output I get from latex2wp, so I’m guessing they creep in when you’re writing the Latex in the first place. Have you tried writing Latex directly in the WordPress editor in the browser using the $latex… syntax? If that works, then you should check whatever tool you are using to write the Latex before sending it through latex2wp.

        Reply
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