Units of a vector in a velocity vs time graph?

[Joseph M. Zias]

This post parallels a post I made in electrical engineering regarding the S plane. I thought I would post an equivalent in basic physics.
So, given a graph of velocity vs time we have on the vertical axis meters/sec and the hormonal axis just meters. Given a plot of V vs t we know the area under the curve is equal to distance and the slope of the curve is acceleration. What would be the units of the magnitude of a vector (polar form) on the graph. Lets say M = square root (V^2 + t^2). What is the unit of M.

 

[anuttarasammyak]

V and t have different physical dimension so V^2+t^2 is meaningless.

By using dimensionless quantity
V^*=\frac{V}{V_{unit}}
t^*=\frac{t}{t_{unit}}
we can transfer it to pure mathematics which enables V*^2+t*^2 , but I do not think it is of your interest.

 

[A.T.]

What would be the units of the magnitude of a vector (polar form) on the graph.

What is that magnitude supposed to represent physically?

 

[robphy]

Not all vectors have a magnitude.
And for those that do, it doesn't have to be associated with the Pythagorean theorem.

The most important feature of a vector is the parallelogram rule of addition.,
not "something with a magnitude and direction".

 

[Joseph M. Zias]

It seems we can use magnitude for a V vs t vector. Suppose we have V=5m/s and t=5s. We then have M=square root 50. Now if we start with M=square root 50, angle 45, and disassemble it to an x and y component we would, of course, have t=5 seconds and y=5 m/s. We have just spread out the square root of 50 whatever, assigning part to t and part to V in the appropriate amounts. Thus the whatever becomes t for the x component and V for the y component. I have never seen a physics text address this, but I would not doubt is was covered in some advanced mathematics.

 

[Vanadium 50]

Math doesn't work that way.

 

[Joseph M. Zias]

Please explain.

 

[robphy]

https://en.wikipedia.org/wiki/Euclidean_vector#Euclidean_and_affine_vectors

However, it is not always possible or desirable to define the length of a vector. This more general type of spatial vector is the subject of vector spaces (for free vectors) and affine spaces (for bound vectors, as each represented by an ordered pair of "points"). One physical example comes from thermodynamics, where many quantities of interest can be considered vectors in a space with no notion of length or angle.

https://en.wikipedia.org/wiki/Vector_space

In mathematics and physics, a vector space (also called a linear space) is a set whose elements, often called vectors, may be added together and multiplied ("scaled") by numbers called scalars. Scalars are often real numbers, but can be complex numbers or, more generally, elements of any field. The operations of vector addition and scalar multiplication must satisfy certain requirements, called vector axioms.

https://en.wikipedia.org/wiki/Vector_space#Vector_spaces_with_additional_structure

Normed vector spaces and inner product spaces​

"Measuring" vectors is done by specifying a norm, a datum which measures lengths of vectors, or by an inner product, which measures angles between vectors. Norms and inner products are denoted |𝑣|

and ⟨𝑣,𝑤⟩,

respectively. The datum of an inner product entails that lengths of vectors can be defined too, by defining the associated norm |𝑣|:=⟨𝑣,𝑣⟩.

Vector spaces endowed with such data are known as normed vector spaces and inner product spaces, respectively.

 

[Orodruin]

Please explain.

m/s and m are different physical units. There is simply no possible meaning in adding their squares. You cannot add 5 kg to 3 meters. It simply has no meaning.

 

[Joseph M. Zias]

It seems that in the case of Velocity vs time we can draw the graph and calculate distance or acceleration but are we then also saying that is not permissible to draw a vector since the magnitude is is undefined or not permitted?
My similar question in the EE group regarding the S plane seems to have resolved as both axis can be shown to have units of sec^-1 and that can be added. Vectors are routinely used in that case.

 

[Orodruin]

Just because you can draw it in 2D does not mean it is a vector space.

 

[A.T.]

It seems that in the case of Velocity vs time we can draw the graph and calculate distance or acceleration but are we then also saying that is not permissible to draw a vector since the magnitude is is undefined or not permitted?

You can draw it, but is has no physical meaning. Unless you define some transformation, that makes the dimensions of the axes equivalent. Like when space and time where combined into space-time via the factor c.

My similar question in the EE group regarding the S plane seems to have resolved as both axis can be shown to have units of sec^-1 and that can be added. Vectors are routinely used in that case.

No problem If the axes have the same dimension.

 

[robphy]

Below, I draw $\vec w = \vec u + \vec v$,
where $\vec u$ is one horizontal unit and $\vec v$ is one vertical unit.

Q: What is the magnitude of the vector sum $\vec w$?
A: It depends.

• Are these vectors "displacements on a plane in Euclidean geometry"?
  $|\vec w|=\sqrt{|\vec u|^2+|\vec v|^2}= \sqrt{2}$
  
  
• Are these vectors "displacements on a spacetime diagram in special relativity"?
  $|\vec w|=\sqrt{|\vec u|^2-|\vec v|^2}= 0$
  (In this case, $\vec w$ is a "lightlike displacement" along the light-cone.)
  https://mathworld.wolfram.com/Lightlike.html
  https://en.wikipedia.org/wiki/Minkowski_space#Causal_structure
  
  
• Are these vectors "displacements in a city street using https://en.wikipedia.org/wiki/Taxicab_geometry "?
  $|\vec w|=|\vec u|+|\vec v|= 2$
  
  
• Are these vectors "displacements on a PV-diagram in thermodynamics":
  $|\vec w|=\color{red}{\tt undefined}$
  (See my earlier post which quotes Wikipedia.)

 

[Orodruin]

Are these vectors "displacements on a PV-diagram in thermodynamics"

It should be noted that thermodynamic state space is generally not a vector space - it is typically a differentiable manifold. As such, displacements of a thermodynamic state belong to the tangent space of the current state, which is a vector space, but differs from state to state.

It is perfectly possible to have base vectors of different physical dimension as long as the tensorial quantities one constructs take this properly into account. For example, the coordinate base vectors in polar coordinates on the Euclidean plane, i.e., $\partial_r$ and $\partial_\varphi$ have physical dimension 1/L and 1, respectively, but still span the same tangent spaces as the regular Cartesian base vectors.