Vectors in curvilinear coordinate systems

[ShayanJ]

To specify a vector in cartesian coordinate systems,we assume its tail to be at the origin and give the cartesian coordinates of its head.What about other coordinate systems?
For example,in spherical coordinates,is the following correct?
<br /> a \hat{x}+b \hat{y}+c \hat{z}=\sqrt{a^2+b^2+c^2} \hat{r}+\cos^{-1}{\frac{c}{\sqrt{a^2+b^2+c^2}}}\hat{\theta}+\tan^{-1}{\frac{b}{a}}\hat{\varphi}<br />

I know,it may seem so easy but the point that's making me doubt it,is the space dependent of the basis vectors in spherical coordinates.I just don't know can I just specify the point that the spherical coordinates of the vector are indicating and connect the origin to that point to show the vector or not!Another reason for doubting the process I explained,is the break down of spherical coordinates at the origin.

Thanks

 

[WannabeNewton]

No it isn't. The most obvious way to see why it is false is to look at the units. The left hand side is a sum of 3 meter valued quantities whereas the right hand side is a meter valued quantity plus two dimensionless quantities. Anyways, in spherical coordinates one simply writes position vectors from the origin of the frame as \vec{r} = r\hat{r} where r = \sqrt{x^2 + y^2 + z^2} as before.

 

[ShayanJ]

Yes,you're right.
But that doesn't indicate the direction of the vector so it seems it gives less information than the cartesian form of the vector

 

[WannabeNewton]

Sure it does. \hat{r} encodes the direction.

 

[ShayanJ]

I know,\hat{r} does encode the direction,but it does not give us that direction.

 

[WannabeNewton]

I know,\hat{r} does encode the direction,but it does not give us that direction.

what is "that" direction?

 

[ShayanJ]

I mean when we have a vector in cartesian coordinates,we have its magnitude and direction.But when we have a vector in spherical coordinates,we only have its magnitude and we can't have its direction.

 

[Jorriss]

,we only have its magnitude and we can't have its direction.

Can you explain why you think that?

 

[ShayanJ]

Just consider the vector a \hat{r}.All it tells us is that the tail is at the origin and the head is on a spherical shell of radius a.I see nothing in that which can indicate a special direction.

 

[Jorriss]

Just consider the vector a \hat{r}.All it tells us is that the tail is at the origin and the head is on a spherical shell of radius a.I see nothing in that which can indicate a special direction.

I don't get your point.

If you only indicate the x direction in cartesian coordinates, its not the full picture either?

 

[WannabeNewton]

Just consider the vector a \hat{r}.All it tells us is that the tail is at the origin and the head is on a spherical shell of radius a.I see nothing in that which can indicate a special direction.

\hat{r} = \sin\theta \cos\varphi \hat{x} + \sin\theta \sin\varphi \hat{y} + \cos\theta \hat{z}. Try to derive this: it is just a simple exercise in geometry. This should make it intuitive to you how the unit radial vector encodes direction.

 

[ShayanJ]

I don't get your point.

If you only indicate the x direction in cartesian coordinates, its not the full picture either?

If you read previous posts,you see we concluded that vectors in spherical coordinates are specified by only a radial component.

\hat{r} = \sin\theta \cos\varphi \hat{x} + \sin\theta \sin\varphi \hat{y} + \cos\theta \hat{z}. Try to derive this: it is just a simple exercise in geometry. This should make it intuitive to you how the unit radial vector encodes direction.

I know what you mean,but there are still 2 points:
1-Consider when you have only the spherical coordinates of a vector,not its cartesian ones.
With just a \hat{r} at hand,how can you use that formula?
2-An assumption that we're always making about vectors in such discussions,is that their tail is at the origin.But spherical coordinates break down there and you can't give unique coordinates to the origin.So there is no unique direction for \hat{r} at the origin.

 

[mikeph]

If you read previous posts,you see we concluded that vectors in spherical coordinates are specified by only a radial component.

No! You concluded that vectors are specified by r\hat{r}(\varphi,\vartheta), where r, \varphi, and \vartheta are all components!

The radial component is one of three. The other two determine the direction of \hat{r}, and you need them! If you only know that r = a, then all you know is your point lies on that surface. You don't know \hat{r} until someone tells you the missing two (angular) components.