Partial derivative as a vector

1. Oct 9, 2008

snoopies622

How is $$\frac{\partial}{\partial t}$$ a vector?

The original context of my question is located in post #5 of my most recent and very short-lived thread, “covariant vs. contravariant time component…”, located here

and it had to do with Schwarzschild coordinates and the energy of a particle.

I understand that if one takes the partial derivative of every component of a vector, the result is a new vector. But when the original vector isn’t specified, what does it mean?

2. Oct 9, 2008

Ben Niehoff

MTW is confusing on this aspect.

The point is that all the directional derivatives at a point on a manifold constitute a vector space, and that vector space is isomorphic to the tangent space. So we can make an identification between them.

For simplicity, think of a curved 2-surface S sitting in 3-dimensional Euclidean space. To find a tangent to the surface at a point P, you want to find all the tangents to all curves on S passing through P; these tangents will all lie in a plane, and they will form a vector space: the tangent space. One can identify each tangent vector with a directional derivative operator at P.

That's all it is, really. I think MTW confuses the issue by over-stressing the importance of intrinsic vs. extrinsic curvature. Just because the spacetime manifold doesn't have to sit within a higher-dimensional Euclidean space, doesn't mean you can't imagine it does for the sake of understanding.

Just remember that a vector is not an arrow between two points. A vector is an abstract object that lies within a vector space, which is defined by the axioms of addition and scaling.

3. Oct 9, 2008

atyy

An intuitive way (not rigourous!) to think of it is: why is a vector a directional derivative? Imagine a room filled with air that is not at a uniform temperature. When you make a trajectory through space as a function of time, the temperature will change as a function of time along your trajectory. In this sense, your velocity is defining a rate of change of temperature. So whenever we have a trajectory through a point, we can use the velocity at that point to define a directional derivative operator (on the temperature, or whatever else is in the air).

4. Oct 9, 2008

snoopies622

Let me see if I am following you correctly so far. Suppose T(x,y,z) is temperature as a function of location.

The gradiant of T is $$<\frac {\partial T}{\partial x},\frac {\partial T}{\partial y},\frac {\partial T}{\partial z}>$$.

If I move in the room with velocity $$V= <\frac {\partial x}{\partial t},\frac {\partial y}{\partial t},\frac{\partial z}{\partial t}>$$

and I then take the dot product of the gradiant (co-)vector and the velocity vector, I get scalar $$\frac {dT}{dt}$$-- the rate of temperature change (for me) per unit time. This scalar is the directional derivative, which is not a vector. Am I to identify the velocity vector V as the directional derivative, and to think of it simply as $$\frac{\partial}{\partial t}$$?

5. Oct 9, 2008

Ben Niehoff

The space of directional derivative operators is a vector space. It can be identified with the tangent space. Did you read my post?

6. Oct 9, 2008

snoopies622

Between
and
I thought you were using "directional derivative" and "directional derivative operator" interchangably. Do you mean something that acts on a directional derivative, or something that acts on something else to produce a directional derivative?

7. Oct 9, 2008

Ben Niehoff

No, I mean an operator like

$$\nabla_{\vec x}$$

that differentiates a [scalar, vector, or tensor] in a specific direction. In particular, the partial derivative operators $\partial / \partial x^i$ constitute the basis of a vector space.

For simplicity, think about just two dimensions. Write

$$u = a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y}$$

$$v = c \frac{\partial}{\partial x} + d \frac{\partial}{\partial y}$$

And then show that u and v are elements of a vector space.

8. Oct 10, 2008

snoopies622

Oh, so $$\frac{\partial}{\partial t}$$ is simply another way of representing $$e_{t}$$, the basis vector for the time direction? Well now that I understand.

Does it then follow that the energy of a particle in a static gravitational field is $$e_{t} \cdot \vec{p}$$ where $$\vec{p}$$ is the four-momentum vector, and my original question about whether one should use p0 or p0 becomes unnecessary since $$e_{t} \cdot \vec{p} = \omega^{t} \cdot \tilde{p}$$ no matter what the coordinate system?

9. Oct 10, 2008

atyy

Yes, along those lines. I think of the gradient covector as a later concept than the velocity as a directional derivative operator. I think of rate of change of temperature with time dT/dt, and the velocity [dx/dt,dy/dt].

The temperature T is a number at every point T(x,y). The trajectory assigns a time t, also a number, to points along a path [x(t),y(t)]. On the trajectory, T(x(t),y(t))=T(t) is just a normal function of one variable, so we can define dT/dt as the rate of change of temperature with time along the trajectory.

Then using the chain rule:
dT(x(t),y(t))/dt
=(dT/dx)(dx/dt)+(dT/dy)(dy/dt)
=[(dx/dt)(d/dx)+(dy/dt)(d/dy)]T

This will work not only for temperature, but for any function of space like humidity or density, so we can make this operator which maps any function to a number:
d/dt=[(dx/dt)(d/dx)+(dy/dt)(d/dy)]

Comparing the term in the square brackets with the velocity, it makes sense to call this operator the directional derivative operator in the direction of the velocity.

Many different trajectories can have the same velocity at a point, so instead of depending on one particular trajectory, we can use v to represent the velocity of all trajectories having that velocity at that point:
v=[vx(d/dx)+vy(d/dy)]

Amazingly or not, I haven't decided, the velocity operators at a point form a vector space like Ben Niehoff says. You can add different velocity operators at a point, but you cannot add velocity operators at different points, unless the space is special.

Last edited: Oct 10, 2008
10. Oct 10, 2008

atyy

OK, I am going to hazard a guess from George Jones's clues in the thread you referred to.

E=g(d/dt,u)=gab<d/dt,dxa><u,dxb>

In coordinates by definition of basis vectors and dual basis vectors:
<d/dt,dt>=1 for xa=t
<d/dt,dxa>=0 for xa!=t.

So: E=gtb<u,dxb>

For the Schwarzschild metric, gtb=0 except gtt

So: E=gtt<u,dxt>=gttut=ut ... ?

Edit: E=gtt<u,dxt>=gttut=ut ... ?

Last edited: Oct 10, 2008
11. Oct 11, 2008

snoopies622

Interesting. This is forcing me to (finally) work with basis vectors and covectors in a context where there is no background higher-dimensional Euclidean space. For those who might be interested, I first described my problem here https://www.physicsforums.com/showthread.php?t=236974

It seems part of my confusion has been the notation I have been using -- in which Ar represents a component of a vector while er represents an entire vector. I'm not quite finished working it out yet, but so far the results are promising.

Thank you both for your help.

Last edited: Oct 11, 2008
12. Oct 11, 2008

George Jones

Staff Emeritus
Right, but $\partial / \partial t$ is not normalized, i.e., does not have unit length.
Right! This is because $\partial / \partial t$ is a timelike Killing vector, so this works in the more general case of stationary spacetimes. Usually, $E$ is defined as energy per unit mass, so 4-velocity $u$ is used instead of 4-momentum $p$. Note that $E$ is a scalar that has the same value in all coordinate systems, and that happens in the particular case of Schwarzschild coordinates to be equal to $u_0$.
Right! Typically, physicists wouldn't use letters for particular values of indices. Let me reproduce what you wrote this using slightly different notation.

Set

$$\left\{ e_0 = \frac{\partial}{\partial t} , e_1 = \frac{\partial}{\partial r} , e_2 = \frac{\partial}{\partial \theta} , e_3 = \frac{\partial}{\partial \phi} \right\}.$$

Slightly non-standard notation has been used in that $\left\{ e_\mu \right\}$ is not an orthonormal basis. Now,

$$\begin{equation*} \begin{split} E &= g \left( e_0 , u \right) \\ &= g \left( e_0 , u^\mu e_\mu \right) \\ &= u^\mu g \left( e_0 ,e_\mu \right) \\ &= u^\mu g_{0 \mu} \\ &= u_0. \end{equation*} \end{split}$$

Last edited: Oct 11, 2008
13. Oct 11, 2008

snoopies622

I thought the unit-length basis vectors were represented by putting a little circumflex over the coordinate symbol, as in $$e_{\hat{\theta}}$$ instead of $$e_{\theta}$$. Is this not the convention, or do I misunderstand you?