How Can Time Derivatives Be Taken in GR and SR with Orthogonal Basis Vectors?

[davidge]

Since in GR and SR the basis vectors are generally orthogonal, how can we take derivatives of position with respect to time? For example, the current four-vector is $$J^{\alpha} = \sum_n e_{n} \frac{\partial x^{\alpha}}{\partial t} \delta^{3}(x - x_{n})$$ where n labels the n-th particle. In this case the derivative will generally not be zero, so how can time be orthogonal to $x^{i}$?

 

[Ibix]

The basis vectors of Euclidean space are orthogonal. How can we write dy/dx?

You're differentiating the coordinate values, not the basis vectors, along some path with respect to some parameter that indexes events along that path. Time (either proper or coordinate) is a convenient parameter for time-like worldlines.

 

[Dale]

$$J^{\alpha} = \sum_n e_{n} \frac{\partial x^{\alpha}}{\partial t} \delta^{3}(x - x_{n})$$

Hmm, this quantity doesn't seem right. It doesn't look like a four vector at all.

Since in GR and SR the basis vectors are generally orthogonal, how can we take derivatives of position with respect to time?

The orthogonality of the basis vectors doesn't have much to do with taking derivatives of the a world line. Such a derivative is simply a tangent vector to a worldline, which is clearly a valid vector regardless of the basis.

 

[vanhees71]

In special relativity the current is (nearly) correct
$$j^{\mu}(x) = \sum_n e_n \frac{\mathrm{d} x_n^{\mu}}{\mathrm{d} t} \delta^{(3)}(\vec{x}-\vec{x}_n(t)).$$
It can be written in manifestly covariant form by introducing an arbitrary world-line parameter $\lambda$,
$$j_{\mu}(x) = \sum_n e_n \int \mathrm{d} \lambda \frac{\mathrm{d} x_n^{\mu}}{\mathrm{d} \lambda} \delta^{(4)}(x-x_n(\lambda)).$$
In GR you must be a bit more careful since here $\mathrm{d}^4 x$ is of course not invariant, but $\mathrm{d}^4 x \sqrt{-g}$ is, where $g=\det(g_{\mu \nu})$. Thus the covariant $\delta$ distribution is $\frac{1}{\sqrt{-g}} \delta^{(4)}(x-x_n(\lambda))$, and that's why the current density is given by
$$j^{\mu}(x) = \sum_n e_n \int \mathrm{d} \lambda \frac{1}{\sqrt{-g}} \frac{\mathrm{d} x_n^{\mu}}{\mathrm{d} \lambda} \delta^{(4)}(x-x_n(\lambda)).$$
For massive particles you can use each particle's proper time as world-line parameter.

 

[davidge]

The basis vectors of Euclidean space are orthogonal. How can we write dy/dx?

So why in many textbooks we are told that $\partial x^{\mu} / \partial x^{\nu} = \delta^\mu{}_{\nu}$(1)? For example, substituting $x$ and $y$ in (1) we would get $\partial x / \partial y = \delta^x{}_{y}$, meaning that $x$ does not vary with $y$ and vice-versa.

@vanhees71 thank you for explaining it to me.

Such a derivative is simply a tangent vector to a worldline, which is clearly a valid vector regardless of the basis

Please see the example I give above to @Ibix.

 

[Dale]

So why in many textbooks we are told that $\partial x^{\mu} / \partial x^{\nu} = \delta^\mu{}_{\nu}$(1)? For example, substituting $x$ and $y$ in (1) we would get $\partial x / \partial y = \delta^x{}_{y}$, meaning that $x$ does not vary with $y$ and vice-versa

You should not substitute in x and y like that. $\mu$ and $\nu$ are indexes that range over all the coordinates.

So if you are thinking of 2D Cartesian coordinates then $\mu$ is not just x or y but is an index which ranges over both. If you like to think in terms of matrix notation then
$$ \frac{\partial x^{\mu}}{\partial y^{\nu}} = \begin{pmatrix}
\partial x/\partial x & \partial x/\partial y \\
\partial y/\partial x & \partial y/\partial y
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} = \delta^{\mu}_{\nu} $$

 

[davidge]

$\mu$ and $\nu$ are indexes that range over all the coordinates

Indeed. I think you misunderstood my notation.

If you like to think in terms of matrix notation then

Yea. But the problem persists, namely that $\partial x^\mu / \partial x^\nu = \delta^\mu {}_{\nu}$ indicates that the basis vectors are orthogonal. Ibix says in post #2 that this derivative is in fact a derivative of the coordinate values and not the basis vectors itself, but still we have the problem that e.g. $\partial x^\mu / \partial \tau = \delta^\mu{}_{\tau}$ and this will be zero for $\mu \neq \tau$, where $\tau$ can be an arbitrary parameter or the proper time.

I hope you understand that I'm using the same symbol to label indices and variables as well.

 

[PeterDonis]

we have the problem that e.g. $\partial x^\mu / \partial \tau = \delta^\mu{}_{\tau}$

No. $\tau$ is not a coordinate and you can't substitute it into coordinate equations as an index.

 

[davidge]

$\tau$ is not a coordinate

Oh, I thought this was the reason for the derivative not to vanish. Thanks for confirming it to me.

Now, in a rest reference frame, the time coordinate is the proper time, right? Does it mean that we cannot evaluate a calculation using the time as measured in a rest reference frame (e.g. the frame moving along with a particle)?

 

[PeterDonis]

in a rest reference frame, the time coordinate is the proper time, right?

Proper time for an object at rest in the frame, if the object at rest is in free fall and we are talking about an inertial frame in SR or a local inertial frame in GR, yes. Otherwise this will not be true in general.

 

[Dale]

that ∂xμ/∂xν=δμν∂xμ/∂xν=δμν\partial x^\mu / \partial x^\nu = \delta^\mu {}_{\nu} indicates that the basis vectors are orthogonal

No. That does not indicate orthogonality. Two vectors $a^{\mu}$ and $b^{\mu}$ are orthogonal if $g_{\mu\nu}a^{\mu}b^{\nu}=0$ which is not the same as the condition you wrote.

Ibix says in post #2 that this derivative is in fact a derivative of the coordinate values and not the basis vectors itself

Sure, but that is more a matter of notational laziness than anything else. You can always write the basis vectors explicitly if you wish.

still we have the problem that...

Hmm. That expression doesn't make sense to me. $\delta^{\mu}_{\nu}$ is a tensor with two indices. In your expression $\mu$ is an index, but $\tau$ is a scalar.

Edit: oh, I see @PeterDonis pointed that out already

 

[Dale]

Now, in a rest reference frame, the time coordinate is the proper time, right?

I disagree with this statement. Proper time is an invariant quantity (a scalar field defined only on the worldline of the particle). The time coordinate is a frame variant quantity defined on the entire chart. They are not the same, even in the particle's rest frame.

For one thing, proper time does not define any notion of simultaneity, but coordinate time does.

That said, on the worldline of the particle they are equal. Sometimes this is expressed by saying that the coordinate time is "adapted" to the proper time of the particle.

 

[PeterDonis]

on the worldline of the particle they are equal.

Only for the special case I stated before, of an inertial frame that is the rest frame (globally in SR, locally in GR) of an object in free fall.

For a simple counterexample in the more general case, Rindler coordinate time is not equal to proper time along the worldlines of objects at rest in Rindler coordinates (except for one particular object at a particular Rindler $x$ coordinate).

 

[Dale]

Only for the special case I stated before, of an inertial frame that is the rest frame (globally in SR, locally in GR) of an object in free fall

Well, you can have other coordinates which are adapted to other objects. Rindler coordinates are an example, and Schwarzschild coordinates can be adapted to some non inertial objects.

 

[haushofer]

So why in many textbooks we are told that $\partial x^{\mu} / \partial x^{\nu} = \delta^\mu{}_{\nu}$(1)?

My 2 cents: because the 4 coordinates x^{\nu} describe directions which are, per definition (!), linear independent.

However, if you switch from one coordinate system x^{\nu} to another one, y^{\mu}( x^{\nu}), y and x are per definition dependent on each other (even more; the transformation should be invertible)! The x^{\nu} are still lineair independent, and the y^{\mu} too. This means that
<br /> \frac{\partial x^{\mu}}{ \partial x^{\nu}} = \frac{\partial y^{\mu} }{ \partial y^{\nu}} = \delta^\mu{}_{\nu} \,.<br />

Your \tau is a coordinate on a wordline, not in spacetime (also called the target space). That's why you only need one; a line is 1-dimensional.

 

[PeterDonis]

you can have other coordinates which are adapted to other objects.

Yes, but coordinate time will not, in general, be the same as proper time for those objects. See the second paragraph of the post of mine that you quoted. Schwarzschild spacetime is an even better example, since in that case coordinate time is not equal to proper time for any observer that is static (at rest in the coordinates) at finite $r$.

 

[Dale]

Yes, but coordinate time will not, in general, be the same as proper time for those objects

I agree. I think people should never say that coordinate time is the same as proper time.

 

[davidge]

@haushofer I think you misunderstood my notation. In my post #5, $x$ and $y$ are two of the coordinates of a point in a same chart, say point $P$. Then $P$ has coordinates $(x,y,...)$. Now, what you pointed out is for the case where ${x} = (x^0,x^1,...)$ are the coordinates of the point $P$ in some chart and ${y} = (y^0,y^1,...)$ are its coordinates in another chart. Btw, is it correct to say that what we call two reference frames in physics, is in math language roughly just two different coordinate charts?

Proper time for an object at rest in the frame, if the object at rest is in free fall and we are talking about an inertial frame in SR or a local inertial frame in GR, yes. Otherwise this will not be true in general

Ah, ok.

Two vectors $a^{\mu}$ and $b^{\mu}$ are orthogonal if $g_{\mu\nu}a^{\mu}b^{\nu}=0$ which is not the same as the condition you wrote

Oh, I see. Thanks.

$\delta^\mu{}_{\nu}$ is a tensor with two indices

Why is $\delta^\mu{}_{\nu}$ a tensor? Does it transforms like a tensor?

 

[davidge]

I was using $x$ and $y$ for the coordinates of a point, following the suggestion of @PeterDonis on a previous thread. He told me something like: "Don't make things so hard for yourself".
In that thread, I was using abstract notation, $x^0,x^1$ instead of $x,y$.

 

[PeterDonis]

is it correct to say that what we call two reference frames in physics, is in math language roughly just two different coordinate charts?

It depends on how precise you want to be.

The terms "reference frame" and "coordinate chart" are commonly used as synonyms, yes. But "reference frame", strictly speaking, means something different: it means what is also called a "frame field", which is an assignment of four mutually orthogonal unit vectors, one timelike and three spacelike, to every event in some open region of spacetime. The set of vectors at each event represent, heuristically, a clock and three mutually orthogonal rulers used by a (hypothetical) observer at that event to make measurements. This is basically the GR version of the set of measuring rods and clocks that was used by Einstein and others in the early days of SR to describe what they meant by an "inertial frame".

The reason the two terms are often used synonymously is basically a holdover from SR, where, because there are global inertial frames and global inertial coordinate charts in which the frame vectors at every event are also the coordinate basis vectors at that event, there is an exact correspondence between the two concepts. But I think it's a good idea to unlearn this habit, because in GR there is in general no such correspondence: for many, if not most, frame fields of interest in GR, there is no coordinate chart in which the frame vectors at every event are also the coordinate basis vectors at that event. So I think it's a good idea to keep the concepts separate in your mind, and to use the proper term for whichever one you actually mean in a given case.

 

[Dale]

Why is $\delta^\mu{}_{\nu}$ a tensor? Does it transforms like a tensor?

Yes, it is a legitimate tensor and transforms like a tensor. It also has the same numerical value in every coordinate system.

 

[vanhees71]

@haushofer
Why is $\delta^\mu{}_{\nu}$ a tensor? Does it transforms like a tensor?

Physicists are sometimes sloppy in their language. ${\delta^{\mu}}_{\nu}$ are tensor components with respect to a basis. In GR tensors themselves are invariant under general coordinate transformations.

In GR one often uses holonomous coordinate bases, and then components with upper indices transform like the differentials of the coordinates ("contravariant"), i.e.,
$$\mathrm{d} \tilde{x}^{\mu} = \mathrm{d} x^{\nu} \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}},$$
while those with lower indices like $\partial_{\mu}$ (covariant), i.e.,
$$\partial_{\mu}'=\frac{\partial}{\partial \tilde{x}^{\mu}}=\frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}} \partial_{\nu}.$$
From this you have
$${\tilde{\delta}^{\mu}}_{\nu} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\rho}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\nu}} {\delta^{\rho}}_{\sigma} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\rho}} \frac{\partial x^{\rho}}{\partial \tilde{x}^{\nu}} = \frac{\partial \tilde{x}^{\mu}}{\partial \tilde{x}^{\nu}}={\delta^{\mu}}_{\nu},$$
i.e., the tensor components don't change. The Kronecker-$\delta$ thus are invariant tensorcomponents under general coordinate transformations.

 

[stevendaryl]

There is a more abstract way to see that \delta_\mu^\nu is a tensor (or rather, components of a tensor):

A (1,1) tensor is by definition a linear function that takes a vector and returns a vector (see the note below^*). The very simplest such function is the identity function: I(v) = v. If you work out the components of the tensor I, you find that I^\mu_\nu = \delta^\mu_\nu.

^*Note: A (1,1) tensor can be characterized in a number of different, equivalent ways:

1. A linear function that takes one vector and returns a vector: A(v) = A^\mu_\nu v^\nu e_\mu
   
2. A linear function that takes one covector and returns a covector: A(B) = A^\mu_\nu B_\mu e^\nu
   
3. A function, linear in each argument, that takes one vector and one covector and returns a scalar. A(v,B) = A^\mu_\nu v^\nu B_\mu

 

[davidge]

"reference frame", strictly speaking, means something different: it means what is also called a "frame field", which is an assignment of four mutually orthogonal unit vectors

The reason the two terms are often used synonymously is basically a holdover from SR, where, because there are global inertial frames and global inertial coordinate charts in which the frame vectors at every event are also the coordinate basis vectors at that event

So... we often use the notation $\partial / \partial x^\mu$ for a coordinate basis vector defined at a point $x$. What would be a notation for the frame vectors? Or would it be meaningless to use a notation for frame vectors?

Thanks @vanhees71 for showing how it transforms.
@stevendaryl Interesting way to see it

 

[PeterDonis]

What would be a notation for the frame vectors?

If you have chosen a coordinate chart, you can write the frame vectors in terms of the coordinate basis vectors (i.e., by giving their components in the coordinate basis), just as for any vectors.

In coordinate-free terms, frame vectors are sometimes written as letters with indices, for example $e_\mu$ (note the lower index), where $\mu$ runs over the four spacetime indices. But this notation can get confusing since the index $\mu$ in this case does not refer to a component of a vector, but to a vector from the set of four vectors.

 

[davidge]

Humm, ok. These frame vectors are the so called non-coordinate basis vectors?

 

[PeterDonis]

These frame vectors are the so called non-coordinate basis vectors?

That term is sometimes used to refer to frame vectors, yes. But it is actually more restricted, since it means taking vectors that point in the same directions as the coordinate basis vectors, and normalizing them to be unit vectors, one timelike and three spacelike. But you can't always do that; there are many coordinate charts in which it is not possible, either because the basis vectors are not orthogonal or because they're not one timelike and three spacelike. But you can define a frame field in any region of spacetime, and express one in terms of any coordinate chart.

 

[davidge]

That term is sometimes used to refer to frame vectors, yes.

Ok, many thanks.

 

[pervect]

is one.)

So... we often use the notation $\partial / \partial x^\mu$ for a coordinate basis vector defined at a point $x$. What would be a notation for the frame vectors? Or would it be meaningless to use a notation for frame vectors?

Thanks @vanhees71 for showing how it transforms.
@stevendaryl Interesting way to see it

The wiki article on "frame fields in general relativity" <<link>> has a fairly standard notation, similar to what Peter described. In the Wiki article, though, instead of writing the basis vectors as $e_0, e_1, e_2, e_3$ , with $e_0$ being the time basis vector and the others the space basis vectors. In the Wiki article they add an arrow over them to emphasis their vecor nature, i.e. $\vec{e_0}, \vec{e_1}, \vec{e_2}, \vec{e_3}$. The arrows are definitely optional, though.

Note that these vectors in the Wiki notation are written in what's called index-free notation, the vector as a whole is given a symbol, the components are not given explicitly. The subscripts here tell you which vector to select from the set of the four vectors in the tetrad, they aren't the usual tensor component indices. The notation for expressing the components of these vectors gets a bit awkward -- I've seen various options. My favorite is to enclose the vector in parenthesis, i.e. to write $(\vec{e_0})^0, (\vec{e_0})^1, (\vec{e_0})^2, (\vec{e_0})^3$ for the four components of $\vec{e_0}$, much as one might write the components of u - or to be consistent $\vec{u}$ - (a vector written in index free notation) as $u^0, u^1, u^2, u^3$.

Usually the vectors in a frame field are orthonormal, i.e. of unit length and orthogonal to each other, though this isn't required. Coordinate basis vectors are usually not unit length. For instance if one has polar coordinates $r, \theta$, the unit vectors are $\partial/\partial r$ (often written as just $\partial_r$), and $(1/r) \, \partial / \partial_\theta$.

 

[PeterDonis]

Usually the vectors in a frame field are orthonormal, i.e. of unit length and orthogonal to each other, though this isn't required.

Are you sure? My understanding is that a frame field requires orthonormal vectors. A coordinate chart does not.