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Still trying to teach myself some basic differential geometry in relation to general relativity. I've read that, in relation to basis vectors $e_{\nu}=\partial_{\nu}$ and basis one-forms $\omega^{\mu}=dx^{\mu}$, then $$e_{\nu}\omega^{\mu}=\partial_{\nu}dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\nu}}=\delta_{\nu}^{\mu}.$$ Fair enough, but why does the partial derivative operator $\partial_{\nu}$ (which I thought meant $\frac{\partial}{\partial x^{\nu}}$ , where the slot is for an arbitrary function) when applied to $dx^{\mu}$ give $\frac{\partial x^{\mu}}{\partial x^{\nu}}$ and not $\frac{\partial\left(dx^{\mu}\right)}{\partial x^{\nu}}$? Or does $\frac{\partial x^{\mu}}{\partial x^{\nu}}=\frac{\partial\left(dx^{\mu}\right)}{\partial x^{\nu}}$? But if that's the case what does $\partial_{\nu}x^{\mu}$ give? It can't also be correct that

$$\partial_{\nu}x^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\nu}}=\delta_{\nu}^{\mu},$$ but I can't see why not. I'm confused. Thanks.

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  • $\begingroup$ Oh well, I've studied tensor calculus and relativity in the seventies but don't understand this either. Seems that it's sometimes kind of sports to make things incomprehensible with advanced physics. $\endgroup$ – Han de Bruijn Oct 16 '15 at 16:32
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I like Amitai Yuval's answer, but let me try to say something a little more basic. It's necessarily a bit vague, but maybe it will give you the idea of what's going on.

The coordinates $x^1, \dots, x^n$ define functions on your manifold. In a physics context, $x^1$, $x^2$ and $x^3$ may represent position with respect to a particular set of axes whilst $x^4$ labels time. Tangent vectors are little arrows at points in your space, and we have a nice basis at each point given by the arrows pointing along each of the coordinate directions. In our example these would correspond to arrows in the directions of the $x^1$-, $x^2$- and $x^3$-axes, and a fourth arrow which points forwards in time but has no spatial component. We write these vectors as $\partial/\partial x^a$ or $\partial_a$, but this is just notation for now.

A one-form is 'a thing you plug vectors into': you feed it a vector and it spits out a number which depends linearly on the input. For each $a$ we can define a one-form $\mathrm{d}x^a$ to be the one-form which gives $1$ when you input $\partial_a$ and $0$ when you input $\partial_b$ for $b \neq a$. Again this is just notation. In other words, the statement $\mathrm{d}x^a(\partial_b)=\delta^a_b$ is basically a tautology. Back to our example, if you plug a vector $v$ into $\mathrm{d}x^4$ then what you get out is the time component of $v$.

Now, given a vector $v$ at a point $p$ and a function $f$ defined near $p$ you can take the directional derivative of $f$ in the direction $v$, which you could write as $v(f)$ (although personally I don't like this notation). If you take $v$ to be $\partial_a$ then this directional derivative really just is the partial derivative $\partial f/\partial x^a$. This is why the notation $\partial_a$ or $\partial/\partial x^a$ is sensible. So in our example $f$ could be something like density and then $\partial_4(f)$ represents the time-derivative of the density.

Thinking of the vector $v$ in the expression $v(f)$ as variable, we get a thing which takes vectors and spits out numbers (given the vector $v$ it outputs $v(f)$). In other words it defines a one-form: the differential (or derivative) of $f$, which we write as $\mathrm{d}f$. Happily, the one-form we defined to be $\mathrm{d}x^a$ purely as a piece of notation really just is the differential of the function $x^a$.

So in terms of directional derivatives it is true that $\partial_a(x^b)=\delta^b_a$. Whilst in terms of feeding vectors into one-forms it is true that $\mathrm{d}x^a(\partial_b)=\delta^a_b$. By the above comment about differentials, relating directional derivatives to one-forms, these statements are really saying the same thing.

The expression you write as $\partial_\nu \mathrm{d}x^\mu$ is a bit ambiguous. It definitely doesn't mean 'differentiate $\mathrm{d}x^\mu$'. The sensible interpretation (which makes it equal to $\delta^\mu_\nu$) is that you're feeding the one-form into the vector using double-duality. But this is probably a little too abstract at present.

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  • $\begingroup$ Thanks. In your para 2, could an example of these $x^{1}$ etc coordinates on the manifold be Cartesian coordinates $x,y,z$ (assuming we're in Euclidean space). If these coordinates were defined in terms of spherical coordinates $r,\theta,\phi$, we could say $z=r\cos\phi$ and thus $\frac{dz}{dr}=\cos\phi$ and $dz=\cos\phi dr$. Is $dz$ a one-form? Also, in para 4 you say vector $v=\frac{\partial}{\partial x^{a}}$. In para 3 you say if we plug $v$ into $dx^{4}$ we get the time component of $v$. But surely $\frac{\partial}{\partial x^{a}}dx^{4}=\frac{\partial}{\partial x^{4}}dx^{4}=1$? $\endgroup$ – Peter4075 Oct 17 '15 at 13:00
  • $\begingroup$ Yes, if you're in 3d Euclidean space then you could take $x^1$, $x^2$ and $x^3$ to be the Cartesian coordinates. You're right that $\mathrm{d}z$ is a one-form, but when you pass to spherical coordinates you should get $\mathrm{d}z=\cos \phi \mathrm{d}r-r \sin \phi \mathrm{d}\phi$. $\endgroup$ – Jez Oct 17 '15 at 15:33
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    $\begingroup$ As I mentioned at the end of my answer, it's a bit ambiguous to write a vector followed by a one-form. The correct statement is $\mathrm{d}x^4(\partial_a)=\delta^4_a$. Why do you think you should replace the index $a$ by $4$ in the last line of your comment? $\endgroup$ – Jez Oct 17 '15 at 15:37
  • $\begingroup$ $\mathrm{d}z/\mathrm{d}r$ is not really a thing, since $z$ is a function of three variables: $r$, $\theta$ and $\phi$ (although it's actually independent of $\theta$). $\endgroup$ – Jez Oct 17 '15 at 15:39
  • $\begingroup$ Sorry, I meant to use the total differential of $z$ $(\mathrm{d}z=\cos\phi\mathrm{d}r-r\sin\phi\mathrm{d}\phi)$. But at least now that I'm using a simple example, the language of coordinate functions is becoming a little easier to visualise. I used $a=4$ in $\mathrm{d}x^{4}(\partial_{a})=\delta_{a}^{4}$ because any other value would give zero. Does that mean time component of $v$ in this case equals 1? $\endgroup$ – Peter4075 Oct 17 '15 at 17:14
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It is only a matter of notation. There are a few different ways to look at things, the following is one of those ways.

Take a coordinate chart $\varphi:U\to\mathbb{R}^n$. Then $\varphi$ consists of $n$ real valued functions which we write as $x^1,\ldots,x^n$. The derivative of the $i$th function is $dx^i$, and this is how we get the local frame of the cotangent bundle $dx^1,\ldots,dx^n$.

On the other hand, the chart $\varphi$ also induces a local frame of the tangent bundle, which we write as $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}.$

Now, there are two standard notations for the derivative of $f$ in the direction $X$ - one of them is $X(f)$ and the other is $df(X)$. Using the former, we can write$$\frac{\partial}{\partial x^i}(x^j)=\delta_i^j,$$whereas the latter yields$$dx^j\left(\frac{\partial}{\partial x^i}\right)=\delta_i^j.$$Both equations mean the same.

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  • $\begingroup$ I'm afraid that's above my head. Any chance of a more high school level type answer? I'm OK with contravariant vector components as tangent vectors, and one-form components as gradients and that these things live at a point on a manifold (4d for spacetime). I understand that $$e_{\nu}\omega^{\mu}=\delta_{\nu}^{\mu}.$$ But "charts", "bundles" and "$n$ real valued functions" lose me. $\endgroup$ – Peter4075 Oct 14 '15 at 14:32
  • $\begingroup$ Thanks to Amitai Yuval, but the second paragraph loses me. What are these $n$ real valued functions? Are they coordinate transformations from one chart to another? The reason for my question was trying to understand how the product of a one-form and a contravariant vector gives a scalar. Hence my trying to see how the product of the basis vectors $e_{\nu}\omega^{\mu}$ gives $\delta_{\nu}^{\mu}$. As is painfully obvious, my knowledge of manifolds is via a very basic introduction to the physics of spacetime. $\endgroup$ – Peter4075 Oct 15 '15 at 17:53
  • $\begingroup$ @Peter4075 I see what you mean. Unfortunately, I don't know how I can help at the moment. It seems to me that you should pick an introductory book on manifolds and find in it the basic terminology. It may take a few days, but if you really want to understand geometry, it is definitely worth the effort. $\endgroup$ – Amitai Yuval Oct 15 '15 at 18:02
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I found this question because I was having the same problem. I had to figure it out and I believe i did. Way too late for the original questioner, but I respond now for people with the same question. I think this is the solution:

If you were reading somebody's book who uses the same notation as Carroll's book on GR, you probably saw three different d's. Carroll uses $d$ for differentials, $\partial{}$ for partial derivatives and the letter d for a one-form. Note the italics $d$ versus regular d. Schutz uses the ~ with the d and writes $\tilde{\text{d}}$ for a one form, and I like that better.

Now, what I think you saw was d$x_i (\frac{\partial{}}{dx^i}) =\delta^j_i\;$ and what it meant was $\;\tilde{\text{d}}x_i (\frac{\partial{}}{dx^i}) =\delta^j_i$ and not $\;dx_i (\frac{\partial{}}{dx^i}) =\delta^j_i$. That is, what was being said is that a one-form basis takes the vector basis and produces $\delta$.

The question now becomes, what is the one form basis? So to be complete, let me say this: The basis denoted $\partial{}_i$ is generated in this way: A one form and a vector produce a invariant scalar. This is bedrock and we can write the invariant scalar as: $d\phi/dS.$ Where $\phi$ is a scalar field and S is a path. The description of the scalar field $\phi$ will change with a change of coordinates as will the description of a path S. But the invariant scalar $d\phi/dS$ will not change with a change of coordinates.

Now, our invariant scalar, the contraction of a vector and a one form, $d\phi/dS$ is: $$\frac{d\phi}{dS} = \frac{\partial{\phi}}{\partial{x}}\frac{dx}{dS} + \frac{\partial{\phi}}{\partial{y}}\frac{dy}{dS}...\tag{EQ 1}$$

In "the modern way" we see a vector as a tangent vector to some curve through some (unnamed) scalar function. That is, in the above equation, from the point of view of a vector, $\phi$ could be any scalar function, so we have: $$a\;vector = \frac{d}{dS} = \frac{\partial{}}{\partial{x}}\frac{dx}{dS} + \frac{\partial{}}{\partial{y}}\frac{dy}{dS}...$$ The right side of the above equation takes the form of a vector and we can see that the vector will have components $\frac{dx^i}{dS}$ and basis $\frac{\partial{}}{\partial{x_i}}$ or $\partial{}_i$ This leaves the definition of a one-form, I always thought of it as: $$d\phi = \frac{\partial{\phi}}{\partial{x}}dx + \frac{\partial{\phi}}{\partial{y}}dy... $$ But that leads back to the same problem stated in the question. If the one-form was written as above, the contraction with the vector written above as $d/dS$ would yield: $$\frac{d\phi}{dS} = \frac{\partial{\phi}}{\partial{x}}\frac{dx}{dS}\frac{\partial{}}{\partial{x}}dx + \frac{d\phi}{dy}\frac{dy}{dS}\frac{\partial{}}{\partial{y}}dy...\;\;(wrong)$$

This cannot be correct. But it was exactly the same result as was found when mixing up the symbols d and $d$ as described in the question. When I didn't know the solution to the first error, this reinforcement of error nearly drove me mad.

So, I believe the correct way to look at the one-form is the same as the modern way of looking at a vector. That is, we see a one-form as the rate of change of a scalar field along some (unnamed) path $$\frac{d\phi}{d} = \frac{\partial{\phi}}{\partial{x}}\frac{dx}{d} + \frac{\partial{\phi}}{\partial{y}}\frac{dy}{d}...$$

The contraction now looks proper as: $$\frac{d\phi}{dS} = \frac{\partial{\phi}}{\partial{x}}\frac{dx}{dS}\frac{\partial{}}{\partial{x}}\frac{dx}{d} + \frac{d\phi}{dy}\frac{dy}{dS}\frac{\partial{}}{\partial{y}}\frac{dy}{d}...$$

and above equation reduces to our definition (EQ 1). Lastly, we can write d$x$ =$dx/d$ or more clearly $\tilde{\text{d}}x$ =$dx/d$

I do still have a problem with Carroll's equation 2.25 which states: $$\text{d}x^u(\partial{}_v)=\frac{\partial{x^u}}{\partial{x^v}} $$ Where we now get $$\text{d}x^u(\partial{}_v)=\frac{d{x^u}}{\partial{x_v}} $$ But this is close enough for me.

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