Is the covariant basis covariant?
$begingroup$
Not sure where my reasoning is going wrong in the following (and also not sure if this is common notation/phrasing):
Since $e_i=frac{partial vec{R}(z')}{partial z^i}$ if the coordinate system z doubles for instance then the components of the covariant basis halves. However, I imagine the covariant basis is named as such because it varies with/co to the coordinate system. Is there a motivation for defining this otherwise?
I read the wiki article on contra and covariance of vectors but am still a bit misty about how it all works here.
EDIT: I think I get it now. $z^i$ transforms contravariant to the basis of the coordinate system (ie the coordinate system). The "doubling" the coordinate system halves the component $z^i$ doubling the covariant basis and justifying the name covariant. Geometrically the covariant basis is tangent to the coordinate system, and thus changes with it/like it. It then also forms a good basis for describing vectors or tensors on a manifold.
The alternative visual intuition behind the mechanics is that if a coordinate curve triples in total size, (or more generally in any component), ie stretches at each point, the distance travelled along the curve to describe a point is one thirded. The new tangent as you move along the new coordinate curve in components/units a third of the size means that you have to move three times faster (in old coordinates) along the curve to keep up. Therefore the derivative involved in the covariant basis which is this speed, triples. Equivalently, since the covariant basis is defined as tangent to the coordinate curve, if the units are divided by three the components of the tangent must be three times larger, and therefore the covariant/tangent basis too. Again, always varying with the coordinate system.
Is all this correct?
tensors
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
$endgroup$
add a comment |
$begingroup$
Not sure where my reasoning is going wrong in the following (and also not sure if this is common notation/phrasing):
Since $e_i=frac{partial vec{R}(z')}{partial z^i}$ if the coordinate system z doubles for instance then the components of the covariant basis halves. However, I imagine the covariant basis is named as such because it varies with/co to the coordinate system. Is there a motivation for defining this otherwise?
I read the wiki article on contra and covariance of vectors but am still a bit misty about how it all works here.
EDIT: I think I get it now. $z^i$ transforms contravariant to the basis of the coordinate system (ie the coordinate system). The "doubling" the coordinate system halves the component $z^i$ doubling the covariant basis and justifying the name covariant. Geometrically the covariant basis is tangent to the coordinate system, and thus changes with it/like it. It then also forms a good basis for describing vectors or tensors on a manifold.
The alternative visual intuition behind the mechanics is that if a coordinate curve triples in total size, (or more generally in any component), ie stretches at each point, the distance travelled along the curve to describe a point is one thirded. The new tangent as you move along the new coordinate curve in components/units a third of the size means that you have to move three times faster (in old coordinates) along the curve to keep up. Therefore the derivative involved in the covariant basis which is this speed, triples. Equivalently, since the covariant basis is defined as tangent to the coordinate curve, if the units are divided by three the components of the tangent must be three times larger, and therefore the covariant/tangent basis too. Again, always varying with the coordinate system.
Is all this correct?
tensors
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
$endgroup$
add a comment |
$begingroup$
Not sure where my reasoning is going wrong in the following (and also not sure if this is common notation/phrasing):
Since $e_i=frac{partial vec{R}(z')}{partial z^i}$ if the coordinate system z doubles for instance then the components of the covariant basis halves. However, I imagine the covariant basis is named as such because it varies with/co to the coordinate system. Is there a motivation for defining this otherwise?
I read the wiki article on contra and covariance of vectors but am still a bit misty about how it all works here.
EDIT: I think I get it now. $z^i$ transforms contravariant to the basis of the coordinate system (ie the coordinate system). The "doubling" the coordinate system halves the component $z^i$ doubling the covariant basis and justifying the name covariant. Geometrically the covariant basis is tangent to the coordinate system, and thus changes with it/like it. It then also forms a good basis for describing vectors or tensors on a manifold.
The alternative visual intuition behind the mechanics is that if a coordinate curve triples in total size, (or more generally in any component), ie stretches at each point, the distance travelled along the curve to describe a point is one thirded. The new tangent as you move along the new coordinate curve in components/units a third of the size means that you have to move three times faster (in old coordinates) along the curve to keep up. Therefore the derivative involved in the covariant basis which is this speed, triples. Equivalently, since the covariant basis is defined as tangent to the coordinate curve, if the units are divided by three the components of the tangent must be three times larger, and therefore the covariant/tangent basis too. Again, always varying with the coordinate system.
Is all this correct?
tensors
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
$endgroup$
Not sure where my reasoning is going wrong in the following (and also not sure if this is common notation/phrasing):
Since $e_i=frac{partial vec{R}(z')}{partial z^i}$ if the coordinate system z doubles for instance then the components of the covariant basis halves. However, I imagine the covariant basis is named as such because it varies with/co to the coordinate system. Is there a motivation for defining this otherwise?
I read the wiki article on contra and covariance of vectors but am still a bit misty about how it all works here.
EDIT: I think I get it now. $z^i$ transforms contravariant to the basis of the coordinate system (ie the coordinate system). The "doubling" the coordinate system halves the component $z^i$ doubling the covariant basis and justifying the name covariant. Geometrically the covariant basis is tangent to the coordinate system, and thus changes with it/like it. It then also forms a good basis for describing vectors or tensors on a manifold.
The alternative visual intuition behind the mechanics is that if a coordinate curve triples in total size, (or more generally in any component), ie stretches at each point, the distance travelled along the curve to describe a point is one thirded. The new tangent as you move along the new coordinate curve in components/units a third of the size means that you have to move three times faster (in old coordinates) along the curve to keep up. Therefore the derivative involved in the covariant basis which is this speed, triples. Equivalently, since the covariant basis is defined as tangent to the coordinate curve, if the units are divided by three the components of the tangent must be three times larger, and therefore the covariant/tangent basis too. Again, always varying with the coordinate system.
Is all this correct?
tensors
tensors
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
edited Dec 7 '18 at 15:46
Benjamin Thoburn
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
asked Dec 6 '18 at 14:41
Benjamin ThoburnBenjamin Thoburn
352213
352213
add a comment |
add a comment |
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