# Getting an intuition of a “Vector Field” in General relativity

In GR the way we define a vector field is :

$$v^a=\sum_\mu v^\mu (\partial_\mu)^a$$

The a in the superscript is the abstract index notation.

I understand this attachment of a vector field if there is function at every point to act upon or along a curve.

Many times in GR there is a vector field attached to the space time without the specification of any functions or curve. What does it mean by this intuitively?

An example: For a time orientable spacetime (M,$g_{ab})$ we always have a timelike vector field.
- Okay I get the part that we have a vector field which is timelike. But what are these vectors at each point, rather what is the physical dimension of these vectors ?
- I am not sure if this is a right thinking or not but I am having trouble understanding the idea of a vector field when we talk about vectors as directional derivatives attached to every point.

How can this vector field be understood intuitively. For example when we talk about Vector fields in Classical field theory we attach a vector to every point in the space like an electric field vector (Having a physical dimension [Newtons/Coulomb] ) for a Electric vector field.

Edit: What is the r vector here in the example? I got everything else!

REPHRASED QUESTION: We define vectors by directional derivatives in Diff.Geo. Then at every point on the manifold we have a unique tangent space where these vectors live. When we have a curve going through these points we can take the directional derivative at every point and have a vector field in coordinate basis at every point along the curve using the parameter of the curve.

But many times in GR we define a vector field without talking about any curve, i.e. we associate a vector to every point on the manifold without talking about any curve. So how do we make the choice of a specific vector on every point on the manifold and called it a vector field if we have no curve and a curve parameter.

• i believe that dimesion is Hz. A vector, id est, a tangent vector on some manifold, and space time is viewed as a manifold, is defined as a map from some space of smooth functions to a real numbers. vector field then is somethin like that but you define at each point of the manifold a vector so you get a field. – Žarko Tomičić Jan 22 '18 at 18:44
• I recomend these: youtube.com/watch?v=V49i_LM8B0E&t=2374s – Žarko Tomičić Jan 22 '18 at 18:45
• so, a vector field is a map from some space of smooth functions to some space of smooth functions actually. – Žarko Tomičić Jan 22 '18 at 18:47
• Did you sort out your issues with tangent vectors and vector fields? – Futurologist Nov 17 '18 at 4:56

One need not specify a function to talk about an object which eats functions and spits out real numbers. We can talk about how such an object would act on any function which we choose to feed it, but we need not specify one function in particular.

It turns out that given any choice of vector, there exists a curve such that the action of the vector on a function can be regarded as the directional derivative of the function along that curve. I expand on this more later in my answer.

The dimensions of a vector are usually Hz (assuming that we parameterize our curves by something with dimensions of time). However, the components of a vector have dimensions of length over time (assuming that the coordinates of the manifold have dimensions of length). This changes if you use different coordinate charts.

Here's a more complete explanation.

Motivation

In terms of raw intuition, the picture that you have in your head is not actually a bad one. Given some manifold $M$ (a sphere, say), at each point $p\in M$ you attach a tangent plane $T_p M$. Each tangent plane is simply a copy of $\mathbb{R}^d$, where $d$ is the topological dimension of $M$ (for a sphere, $d=2$), and vectors can be thought of as the familiar "arrows" which live in $\mathbb{R}^d$.

Picture that in your mind. What you are imagining is called an embedding of the sphere into a 3D space. The notion of gluing a plane to a point on a sphere makes real sense only if you're picturing the sphere as a subset of a larger space.

We would like to avoid this. In particular, if $M$ is supposed to represent the entire universe, it's a bit odd to try to model dynamics on it by appealing to a "top down" view from outside $M$. Instead, we would much prefer to take an intrinsic approach - to define vectors in a space without leaving the space.

The Rough Idea

Imagine that you're walking through your house, and you're measuring the temperature at each point. Let $p$ be a point in your house (notice that I haven't defined the point by some coordinate system or anything - $p$ genuinely represents a point, not a pair of numbers). The function

$$T : \text{House} \rightarrow \mathbb{R}$$ $$p \mapsto \text{Temperature at } p$$

maps points in your house to the temperature at that point.

As you walk along some path through $p$, the temperature you measure will be constantly changing. You can therefore define the rate of change of the temperature as you walk through $p$. If you follow the same path at twice the speed, then the rate of change will double; if you follow a different path through $p$, then the rate of change will be different as well.

This is the central idea. If $f$ is a function on the manifold (i.e. your house), then the rate of change of $f$ at a particular point $p$ contains information about the direction at which you move through $p$, as well as the rate at which you walk - which amount to "direction" and "magnitude".

Our job now is to make this notion mathematically precise.

Manifold Charts

Let $M$ be a smooth manifold. A chart $(U,x)$ is an open subset $U\subset M$ and a continuous, invertible map $$x : M \rightarrow \mathbb{R}^d$$ $$p \mapsto x(p)$$

which takes a point $p\in U$ to $x(p)=(x^1,x^2,\ldots,x^d) \in \mathbb{R}^d$. We call $x(p)$ the coordinates of the point $p$ in the chart.

We should pause here to appreciate what this means. Up until now, we've typically defined a point by its coordinates, which means we have been implicitly choosing particular charts. Now, we recognize that a point exists as a point rather than as a list of numbers, and that the coordinates of a point completely depend on what labels we decide to apply to it.

Curves on Manifolds

We define a curve $\gamma:\mathbb{R} \rightarrow M$ to be a map which takes a real number and maps it to a point in the manifold $M$. If you'd like, the real number might represent time, but we aren't talking about physics yet, so for now it's just a parameter.

Derivatives along a Curve

Let $f$ be a smooth function $f : M \rightarrow \mathbb{R}$ (note: the set of smooth functions on $M$ is denoted $C^\infty(M)$), and let $\gamma$ be a curve which passes through the point $p\in M$ such that $\gamma(0) = p$ (this is just for convenience).

Now consider the function $$f \circ \gamma : \mathbb{R} \rightarrow \mathbb{R}$$ $$t \mapsto f\big( \gamma(t) \big)$$

This function takes a real number $t$ and maps it to the value of $f$ at the point in the manifold given by $\gamma(t)$. In other words, it is the value of the function along the curve $\gamma$.

As a function from $\mathbb{R}$ to $\mathbb{R}$, it makes perfect sense to differentiate this object - we define $(f \circ \gamma)'(0)$ to be the directional derivative of $f$ along $\gamma$ at the point $p$.

Such a thing deserves a new notation - we will call it $V_{\gamma,p}$.

$$V_{\gamma,p} : C^\infty(M) \rightarrow \mathbb{R}$$ $$f \mapsto (f\circ \gamma)'(0)$$

It also deserves a name - we say that $V_{\gamma,p}$ is the tangent vector to the curve $\gamma$ at the point $p$. The space of all such objects defined at a point $p$ is denoted $T_pM$, and called the tangent space to $M$ at $p$.

Tangent Spaces

Having defined one tangent vector, we might now embark on a quest for others. Given some curve $\gamma$, is it possible to find another curve $\sigma$ such that $$V_{\sigma,p}f = c V_{\gamma,p}f$$ for any choice of smooth function $f$ and any real number $c$? The answer is yes - therefore, we can multiply these tangent vectors by real numbers to get new tangent vectors.

Can we add them? That is, given two curves $\gamma$ and $\sigma$, does there always exist a third curve $\delta$ such that

$$V_{\gamma,p}f+V_{\sigma,p}f = V_{\delta,p}f$$

for any smooth function $f$? The answer again is yes. I won't prove these two facts (the first is very simple, the latter requires a bit of though) - you can think about them yourself, or simply look them up.

In any case, this implies that $T_pM$ is indeed a vector space - we can add vectors and multiply them by real numbers.

Coordinate Representation of Tangent Vectors

This is the meat and potatoes of your question. Up to this point, hopefully things have been reasonable - but they have been very abstract. It's time to drop to the level of a chart and to see what these things look like when we define a coordinate system.

Recall that a coordinate chart $x$ is continuous and invertible, so $x^{-1}:\mathbb{R}^d \rightarrow U \subset M$ is perfectly well-defined. Therefore, let's choose a chart $(U,x)$. Then we can write $$V_{\gamma,p}f = (f \circ \gamma)'(0) = (f \circ x^{-1} \circ x \circ \gamma)'(0)$$

We can regard this as the composition of two maps:

1. $(x \circ \gamma):\mathbb{R} \rightarrow \mathbb{R}^d$ takes a real number $t$ and maps it to the coordinates of the corresponding point $\gamma(t)$ in the chart.
2. $(f \circ x^{-1}) : \mathbb{R}^d \rightarrow \mathbb{R}$ takes the coordinates of the point $\gamma(t)$ in the chart and maps them to the value of $f$ at that point.

When we differentiate, we do so using the chain rule:

$$(f \circ x^{-1} \circ x \circ \gamma)'(0) = \partial_a \big(f \circ x^{-1}\big)(x(p)) \cdot \big(x^a \circ \gamma\big)'(0)$$

Note: The symbol $\partial_a$ applies to functions $g:\mathbb{R}^d\rightarrow \mathbb{R}$, and means "the partial derivative of $g$ with respect to the $a^{th}$ slot."

From here, we define a new symbol: $$\left(\frac{\partial}{\partial x^a}\right)_p$$

Here's the idea: The symbol $\partial_a$ is used to differentiate functions on $\mathbb{R}^d$, which take $d$ real numbers as inputs. However, $f$ is a function on the manifold. Its only input is a point $p$, so how do we differentiate it?

The answer is as follows: We take a point $p$, we map it to its coordinates by choosing a chart $x$. We then differentiate the function $f \circ x^{-1}$ (which we know how to do) and evaluate at $x(p)$.

In other words,

$$\left(\frac{\partial}{\partial x^a}\right)_p f := \partial_a\big(f \circ x^{-1}\big) (x(p))$$

and so we find that in the chart $(U,x)$,

$$V_{\gamma,p}f = V^a \left(\frac{\partial }{\partial x^a}\right)_p f$$ where we have defined the real numbers $$V^a \equiv (x^a \circ \gamma)'(0)$$

to be the components of $V$ in the chart.

That's most of the major technical machinery. In the old days, our definition of a vector was the list of numbers $(V^1,V^2,\ldots,V^d)$. Through the lens of differential geometry, these are just a particular choice of label for a more abstract object which exists independent of any particular choice of coordinate system.

You can also calculate what must happen to the components of a vector under a change of chart - this is a useful exercise, so not one that I will reproduce here.

• Thanks a lot for the comment! I actually do know all this but it was really nice to see it all together! But it was really nice of you to explain it so intricately !It is still somehow difficult for me to understand this idea. What I understand is that the tangent vectors are defined by curves and then there is a line those vectors as directional derivatives along the curve. But what is the idea when there are no curves. Contd in next commend – Feynstein Jan 23 '18 at 11:59
• So I can imagine a manifold with every point on it having a unique tangent space with the directional derivative operators (Coordinate basis). From there how do I attach a vector to every point ? Dont I need a function to act upon? Or a curve along which I have to take these derivatives? – Feynstein Jan 23 '18 at 11:59
• Is there a way to think about this using a space time diagram? – Feynstein Jan 23 '18 at 12:00
• What is the idea when there are no curves? I'm not sure what you mean - you can always draw curves on a smooth manifold. – J. Murray Jan 23 '18 at 15:04
• A vector field is intuitively just a choice of tangent vector from the tangent space at each point. It's actually a bit more than that - it's a smooth section of the tangent bundle. This provides us with a notion of a smooth vector field - i.e. one where the "arrows" flow together in some sense, rather than simply pointing in random directions. – J. Murray Jan 23 '18 at 15:06