Vector Calculus in 1D: ± to Show Magnitude?

[MatinSAR]

[mentor's note - moved from one of the homework help forums]

Homework Statement:: It's a question.
Relevant Equations:: Vector calculus.

Is it true to say that in one dimension I can show vector quantities using ±number instead of a vector?
± can show possible directions in one dimension and that number shows magnitude of quantity.

Thanks.

 

[vanhees71]

Yes!

 

[MatinSAR]

Yes!

Thank you for your time.

@malawi_glenn Thank you for your help.

 

[kuruman]

Is it true to say that in one dimension I can show vector quantities using ±number instead of a vector?
± can show possible directions in one dimension and that number shows magnitude of quantity.

Consider where this is coming from. A 3D vector is written as $$\mathbf{A}=A_x~\mathbf{\hat x}+A_y~\mathbf{\hat y}+A_z~\mathbf{\hat z}$$ where the components ##A_x##, ##A_y## and ##A_z## can be positive or negative. Of course, to write down a vector in this manner, you must have a coordinate system with unit vectors already defined.

In the special case ##A_y=A_z=0##, you have a 1D vector which is formally written as $$\mathbf{A}=A_x~\mathbf{\hat x}.$$ Note that ##A_x## can still be positive or negative.

Informally, the vector notation and the single unit vector are dropped, but the choice of which way the single axis is positive must remain. Ambiguity may arise when one sees in 1D something like ##v = -2~##m/s. Is this a vector equation or not? There is no ambiguity in the vector equation ##\mathbf{v}=(-2)~\mathbf{\hat x}~##m/s or in scalar equation ##v_x=-2~##m/s.

Dropping the vector notation and the subscript (a widespread practice in 1D) conflates the two and could be a source of confusion to people who don't know what's "under the hood." It looks like you were confused about this yourself and that is why you asked your question.

 

[malawi_glenn]

There is also an inbuilt ambiguity due to speed and velocity having the same symbol ##v## in that case

 

[MatinSAR]

There is also an inbuilt ambiguity due to speed and velocity having the same symbol ##v## in that case

Thank you for your help.

Consider where this is coming from. A 3D vector is written as $$\mathbf{A}=A_x~\mathbf{\hat x}+A_y~\mathbf{\hat y}+A_z~\mathbf{\hat z}$$ where the components ##A_x##, ##A_y## and ##A_z## can be positive or negative. Of course, to write down a vector in this manner, you must have a coordinate system with unit vectors already defined.

In the special case ##A_y=A_z=0##, you have a 1D vector which is formally written as $$\mathbf{A}=A_x~\mathbf{\hat x}.$$ Note that ##A_x## can still be positive or negative.

Informally, the vector notation and the single unit vector are dropped, but the choice of which way the single axis is positive must remain. Ambiguity may arise when one sees in 1D something like ##v = -2~##m/s. Is this a vector equation or not? There is no ambiguity in the vector equation ##\mathbf{v}=(-2)~\mathbf{\hat x}~##m/s or in scalar equation ##v_x=-2~##m/s.

Dropping the vector notation and the subscript (a widespread practice in 1D) conflates the two and could be a source of confusion to people who don't know what's "under the hood." It looks like you were confused about this yourself and that is why you asked your question.

Thank you for your detailed answer. I was confused when I start to reread about motion along a straight line in "Fundamentals of Physics (Textbook by David Halliday)".

 

[Mark44]

There is also an inbuilt ambiguity due to speed and velocity having the same symbol ##v## in that case

Textbooks usually distinguish typographically between speed (v) and velocity (v or ##\vec v##), and define speed as the magnitude of the velocity. I.e., ##v = |\vec v|## = |v|.

 

[kuruman]

A related problem with writing 1D vector equations appears in the accelerating hanging mass. Students correctly write down the equation ##~T-mg=ma.## When they are asked to find the tension in the case of a hanging mass accelerating down with acceleration 2 m/s², they correctly substitute -2 m/s² for ##a## in the equation but are puzzled to find out that the substitution of -9.8 m/s² for ##g## gives the wrong answer.