# Coordinate differentials

• I
There is an ambiguity in certain texts that I want to clarify, atleast it seems ambiguous to me. When describing the differential line element in Relativity by the differential dx's, are they to be infinitesimal vectors or just infinitesimal increments. They are labeled coordinate differentials in some and I've heard others refer to them as vectors also.

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strangerep
When describing the differential line element in Relativity by the differential dx's, are they to be infinitesimal vectors or just infinitesimal increments.
Both. (Infinitesimal) coordinate increments transform like vectors. (Use a multi-variable Taylor series to prove this.)

Both. (Infinitesimal) coordinate increments transform like vectors. (Use a multi-variable Taylor series to prove this.)
What do you mean both? Could you clarify on this please? I understand how they transform but what are they? I suppose they are vetors because technically they have direction and magnitude but where do they come from?

pervect
Staff Emeritus
There is an ambiguity in certain texts that I want to clarify, atleast it seems ambiguous to me. When describing the differential line element in Relativity by the differential dx's, are they to be infinitesimal vectors or just infinitesimal increments. They are labeled coordinate differentials in some and I've heard others refer to them as vectors also.

The general convention that my text (MTW) uses is that dx is just a number, not a vector. dx, can be regarded as an operator, which when applied to a vector, yields a scalar. In this form dx takes an argument. It is also described as linear functional of a vector.

With the second usage, dx is a sort of vector known as a dual vector, also known as a 1-form.

[wiki]

In general, displacments on curved surfaces (an example of which is the sphere) don't add. so they are not vectors. An ambiguous, but useful example : if you start at the equator of a sphere the size of the Earth, and go 500 miles north, then 500 miles east, you don't wind up at the same location as if you went 500 miles east first, then 500 miles north. The example is ambiguous because we haven't defined what we mean by "go north" and "go east".

The mathematically pure way of dealing with this issue of displacements not adding is to define a tangent space to the manifold (in this case the manifold is a sphere, and the tangen space at some point p is just a plane tangent to the sphere at point p). Then the displacements in the tangent space do add. Note that every point on the manifold (the sphere) has it's own tangent space (tangent plane).

The general convention that my text (MTW) uses is that dx is just a number, not a vector. dx, can be regarded as an operator, which when applied to a vector, yields a scalar. In this form dx takes an argument. It is also described as linear functional of a vector.

With the second usage, dx is a sort of vector known as a dual vector, also known as a 1-form.

[wiki]

In general, displacments on curved surfaces (an example of which is the sphere) don't add. so they are not vectors. An ambiguous, but useful example : if you start at the equator of a sphere the size of the Earth, and go 500 miles north, then 500 miles east, you don't wind up at the same location as if you went 500 miles east first, then 500 miles north. The example is ambiguous because we haven't defined what we mean by "go north" and "go east".

The mathematically pure way of dealing with this issue of displacements not adding is to define a tangent space to the manifold (in this case the manifold is a sphere, and the tangen space at some point p is just a plane tangent to the sphere at point p). Then the displacements in the tangent space do add. Note that every point on the manifold (the sphere) has it's own tangent space (tangent plane).
Yes, on curved surfaces you need a connection when dealing with an interval I believe. I am just wondering what would the dx's line element distance formula of relativity be regarded as.

strangerep
What do you mean both?
are [the dx's] to be infinitesimal vectors or just infinitesimal increments?
and I answered "both". I'm not sure what other word I could have used. Anyway...

Could you clarify on this please?
Following is a partial extract from some notes I wrote a while back.

1.1 From Coordinates to Vector Fields:

Let ##x^\mu(s)## be the coordinates of an arbitrary curve with (arbitrary) parameter ##s##. Tangent vectors along the curve are claimed to be given by $$\dot x^\mu ~:=~ \frac{dx^\mu(s)}{ds} ~,$$ but how do we know they are "vectors"?

The ##x^\mu## coordinates are not vectors (in general), since they do not transform like vectors under arbitrary changes of coordinates ##x \to z = z(x)##. To see this, expand ##z^\mu## in a Taylor series as follows:
$$z^\mu(x) ~\approx~ z^\mu(0) ~+~ x^\alpha \left. \frac{\partial z^\mu}{\partial x^\alpha} \right|_{x=0} +~ \frac12 x^\alpha x^\beta \left. \frac{\partial^2 z^\mu}{\partial x^\alpha \partial x^\beta} \right|_{x=0} +~ \dots$$ Hence, if ##z^\mu(x)## is anything other than a homogeneous-linear transformation, then
$$z^\mu ~\ne~ \frac{\partial z^\mu}{\partial x^\alpha} \, x^\alpha ~,$$as would be required for ##x^\mu## to be a vector.

On the other hand,
$$z^\mu(s+h) ~\approx~ z^\mu(s) ~+~ h \dot z^\mu(s) ~+~ \frac12 h^2 \ddot z^\mu(s) ~+~ \dots ~\approx~ z^\mu(s) ~+~ h \frac{\partial z^\mu}{\partial x^\alpha} \dot x^\alpha(s) ~,$$(where, in the last step, we used the chain rule). Therefore,$$\lim_{h=0}\, \frac{z^\mu(s+h) - z^\mu(s)}{h} ~=~ \frac{\partial z^\mu}{\partial x^\alpha}\, \dot x^\alpha(s) ~, ~~~~~ \mbox{i.e.,}~~ \dot z^\mu(s) = \frac{\partial z^\mu}{\partial x^\alpha}\, \dot x^\alpha(s) ~,$$ which shows that ##\dot x^\mu(s)## is indeed a vector (anchored at the point on the curve corresponding to ##s##). (Even though the derivative formula is subtracting two indexed quantities, the result is only a vector in the limit sense as ##h\to 0##. For finite ##h,## that formula is adding coordinates, not vectors.)

Hence the differentials ##dx^\mu## are themselves vectors, since ##ds## is a scalar.

[Strictly speaking, I should be saying "vector components" rather than just "vector".]

HTH.

• WWGD, haushofer, dsaun777 and 1 other person
vanhees71
Gold Member
[Strictly speaking, I should be saying "vector components" rather than just "vector".]
That's very important! A tensor is an invariant object, i.e., it does not depend on any basis for co-basis, but the tensor components to!

Well,... you asked: and I answered "both". I'm not sure what other word I could have used. Anyway...

Following is a partial extract from some notes I wrote a while back.

1.1 From Coordinates to Vector Fields:

Let ##x^\mu(s)## be the coordinates of an arbitrary curve with (arbitrary) parameter ##s##. Tangent vectors along the curve are claimed to be given by $$\dot x^\mu ~:=~ \frac{dx^\mu(s)}{ds} ~,$$ but how do we know they are "vectors"?

The ##x^\mu## coordinates are not vectors (in general), since they do not transform like vectors under arbitrary changes of coordinates ##x \to z = z(x)##. To see this, expand ##z^\mu## in a Taylor series as follows:
$$z^\mu(x) ~\approx~ z^\mu(0) ~+~ x^\alpha \left. \frac{\partial z^\mu}{\partial x^\alpha} \right|_{x=0} +~ \frac12 x^\alpha x^\beta \left. \frac{\partial^2 z^\mu}{\partial x^\alpha \partial x^\beta} \right|_{x=0} +~ \dots$$ Hence, if ##z^\mu(x)## is anything other than a homogeneous-linear transformation, then
$$z^\mu ~\ne~ \frac{\partial z^\mu}{\partial x^\alpha} \, x^\alpha ~,$$as would be required for ##x^\mu## to be a vector.

On the other hand,
$$z^\mu(s+h) ~\approx~ z^\mu(s) ~+~ h \dot z^\mu(s) ~+~ \frac12 h^2 \ddot z^\mu(s) ~+~ \dots ~\approx~ z^\mu(s) ~+~ h \frac{\partial z^\mu}{\partial x^\alpha} \dot x^\alpha(s) ~,$$(where, in the last step, we used the chain rule). Therefore,$$\lim_{h=0}\, \frac{z^\mu(s+h) - z^\mu(s)}{h} ~=~ \frac{\partial z^\mu}{\partial x^\alpha}\, \dot x^\alpha(s) ~, ~~~~~ \mbox{i.e.,}~~ \dot z^\mu(s) = \frac{\partial z^\mu}{\partial x^\alpha}\, \dot x^\alpha(s) ~,$$ which shows that ##\dot x^\mu(s)## is indeed a vector (anchored at the point on the curve corresponding to ##s##). (Even though the derivative formula is subtracting two indexed quantities, the result is only a vector in the limit sense as ##h\to 0##. For finite ##h,## that formula is adding coordinates, not vectors.)

Hence the differentials ##dx^\mu## are themselves vectors, since ##ds## is a scalar.

[Strictly speaking, I should be saying "vector components" rather than just "vector".]

HTH.
Is it only when parameterized by a curve s that the dx can be vector components of the the tangent vectors dx/ds? The dx's are meaningless unless parameterizing a scalar curve?

strangerep
Is it only when parameterized by a curve s that the dx can be vector components of the the tangent vectors dx/ds?
What we're really working with here is the set of all possible directions on a manifold, from at an arbitrary point ##P##. To express a direction, one needs at least an infinitesimal neighbourhood, (i.e., open set of the manifold), containing ##P##. Thus, the set of all possible directions at ##P## is isomorphic to the tangent vectors of all possible curves on the manifold that pass through ##P##, evaluated at ##P##. [Strictly speaking, I should mention something about "equivalence classes" here, but I'll skip over that.]

So, you can think of the dx's as infinitesimal directions on the manifold (provided you also specify the point ##P##). The imagery of a totality of curves on the manifold just helps us to visualize what's going on, and what might be involved if we wanted to transport a vector anchored at ##P## to another point ##P'## (which is an essential framework needed for analyzing motion of nontrivial bodies and fields in space and time).

The dx's are meaningless unless parameterizing a scalar curve?
As directions at any point on the spacetime manifold, the dx's are never meaningless.

• dsaun777
vanhees71
Gold Member
In a differentiable manifold you don't have vectors from the beginning as in the special case of an affine manifold, but you have to define them as tangent vectors at any point of the manifold, and the most intuitive definition, borrowed from physics, is the definition via smooth curves through a point. Maybe that's the resolution of the confusion in the OP?

What we're really working with here is the set of all possible directions on a manifold, from at an arbitrary point ##P##. To express a direction, one needs at least an infinitesimal neighbourhood, (i.e., open set of the manifold), containing ##P##. Thus, the set of all possible directions at ##P## is isomorphic to the tangent vectors of all possible curves on the manifold that pass through ##P##, evaluated at ##P##. [Strictly speaking, I should mention something about "equivalence classes" here, but I'll skip over that.]

So, you can think of the dx's as infinitesimal directions on the manifold (provided you also specify the point ##P##). The imagery of a totality of curves on the manifold just helps us to visualize what's going on, and what might be involved if we wanted to transport a vector anchored at ##P## to another point ##P'## (which is an essential framework needed for analyzing motion of nontrivial bodies and fields in space and time).

As directions at any point on the spacetime manifold, the dx's are never meaningless.
In this context can you view a manifold as just an integration of tangent vectors along all possible curves? How do you define mathematically your initial point P on the manifold to begin with?

WWGD
Gold Member
In this context can you view a manifold as just an integration of tangent vectors along all possible curves? How do you define mathematically your initial point P on the manifold to begin with?
No, usually the manifold comes first and the tangent vectors are used to model Euclidean space locally. The union of all tangent spaces is a manifold called the tangent bundle which is not the same as the original manifold. The manifold is just a way of smoothly patching pieces of Euclidean space , all of the same dimension. You lose some traits of ( global) Euclidean space in the patching ( like having zero curvature) but you get back some of it through tangent spaces.

• PAllen
strangerep