# Thread: Scalar as A One-Dimensional Vector

1. Excuse me ... .

I have a question about scalar quantity ... .

Can a scalar be regarded as a one-dimensional vector ... ?

For example, the continuum electric charge is a scalar quantity which can be regarded to have two directions in real line , namely, the positive direction () and the negative direction () ... .

In Lagrangian Mechanics, an electrical system can be analogized as a mechanical system, with the electric charge as a generalized co-ordinate, and as a generalized velocity ... .

Suppose that here is an electrical circuit which consists of an inductor of inductance constant and a voltage source as function of time ... .

The Lagrangian of such system is ... .

The Euler-Lagrange’s equation of this system is

as its equation of motion ... .

Thus, the Newton’s second law for this electrical circuit is , where can be analogized as constant mass, and can be analogized as time-dependent-force ... .

Thank you for the answer ... .

2.

3. Can a scalar be regarded as a one-dimensional vector ... ?
I'm not so sure about that. I think the proper definition for "scalar" is a quantity which is invariant under rotations or reflections of the coordinate system; this does not necessarily hold for 1-dimensional vectors, does it ?
In your example, the charge should be independent on how you define your coordinate system; however, if you regard it as a 1-dimensional vector, then a reflection of your coordinate system would flip its sign, so they don't seem to be the same to me.

4. Originally Posted by Markus Hanke
In your example, the charge should be independent on how you define your coordinate system; however, if you regard it as a 1-dimensional vector, then a reflection of your coordinate system would flip its sign, so they don't seem to be the same to me.
In my example, the charge was regarded as a "one-dimensional position vector" in a "charge-space" ... . In other word ... , ... . The mirror can be placed at "position" (for example) ... .

Since is a one-dimensional vector space, then each element is a one-dimensional vector ... .

5. Originally Posted by trfrm

In my example, the charge was regarded as a "one-dimensional position vector" in a "charge-space" ... . In other word ... , ... . The mirror can be placed at "position" (for example) ... .

Since is a one-dimensional vector space, then each element is a one-dimensional vector ... .
I see what you are saying, but it still does not feel quite right to me. In the general case, the magnitude of a vector, in my mind, should depend on the coordinate system chosen, even in one dimension. This would make it not the same as a scalar. Perhaps someone more versed in mathematics can clear this up for us; I could only find this :

time - Is a 1D vector also a scalar? - Physics Stack Exchange

6. Originally Posted by Markus Hanke
In the general case, the magnitude of a vector, in my mind, should depend on the coordinate system chosen, even in one dimension.
Thanks ... . In my mind, the charge plays a role as a “vector” in “charge-space” but not in configuration-space ... . The charge plays a role as a scalar in configuration space ... .

In the “charge-space”, the magnitude of “vector charge” depends on choosing "charge co-ordinate system" ... . But, in the configuration space, the value of the charge (as a scalar) does not depend on choosing co-ordinate system ... .

The “charge-space” is different from the configuration space ... , as “momentum-space” which is part of “classical-phase-space” ... .

I’m sorry if my opinion was wrong ... .

7. My gut feeling says no, but I'm no pure mathematician. Here's my reasoning.

Let's say we take some abstract vector,

and multiply it by a scalar ,

If we let be a (one dimensional) vector, then we get an expression that looks like

which isn't defined for vectors. Moreover, can have any dimension, since the scalar product should hold true for any vector by definition. As far as I know there's no definition for multiplying two vectors without the same dimensions.

I'm not sure of the rigor of this proof or if it's even correct, but that's my reasoning.

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