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Does the physical unit determine the length of the unit vector?

Physics Asked on March 31, 2021

A vector quantity, like position, can be expressed as (using one dimension for simplicity):

$$textbf{r}=(1 text{m})hat{textbf{i}}$$

What determines how "long" the unit vector $hat{textbf{i}}$ is?

Could we pick any of the following vectors as unit vector?

Enter image description here

For example, if an object is $3, text{m}$ away from the origin of a coordinate system (again one dimensional for simplicity) then scaling by $3$ the first vector is not the same as scaling by $3$ the second vector. But both have units of meter so they must have the same length $3,text{m}$. This is what I can’t understand. Does the physical unit determine the length of the unit vector?

3 Answers

A unit vector is a vector of length 1, by definition. Of course, that's somewhat cheating.

What "a length of 1" actually is is defined by the coordinate system you choose. I can make a coordinate system where a vector of length 1 is 1 meter, or I can make one where a vector of length 1 is 1 foot. But once I choose my coordinate system, I can't change it.

What might be confusing is that the idea of a "unit vector" is a purely mathematical concept, with mathematical properties. For use in physics, we always have to do something to map our problem into mathematical terms, such as converting real life concepts to vectors using units and frames. Your concerns for "how long is a unit vector" are actually in the mapping from the physical world into mathematics. Only once we are there, in the language of math, does a "unit vector" have any particular meaning.

Indeed, it is quite common to select a coordinate system with convenient units to make the math easier to manipulate. Physicists have been known to use "natural units", where both the speed of light and plank's constant are unit vectors in their corresponding dimensions (oft written $hbar = c = 1$). For you and I, this is about as unnatural as it can get, but for physicists this choice can be very convenient. Many of the equations they have to work with have conversion factors that one must keep track of. In natural units, many of these become 1, so they can be hand waved away. It didn't change the answer, but it did make it easier to manipulate the symbols.

To your edit, it will actually be easier to speak in 2 dimensions rather than 1. 1 dimension makes it easy to degenerate into non-vector thinking, and that can obscure things easier.

Now usually we don't teach the difference between vectors and coordinate vectors in school. But in this case, I believe you are getting confused because you're conflating the two. I think having two concepts can sometimes offer clarity.

Let's set up a two dimensional test example. In this example, there is an object that I could reach if I walked 3000 meters east, then turned north and walked 4000 meters north. We can speak of a vector from me to that object, and give it a name like $vec v$. We can say things like "the length of $vec v$ is 5000 meters."

We can contrast this with a coordinate vector like $langle3000, 4000rangle$ or $3000hat i + 4000hat j$. Those two things are best thought of as a measurement of the vector. But there's some details there. What if I claimed that same vector was $langle3, 4rangle$ because it was 3 kilometers east and 4 kilometers north? Would I be wrong? Not really, I just measured it differently. What if I measured it as $langle4, 3rangle$ because it was 4 kilometers north and 3 kilometers east? Would I be wrong? The answers to these questions are hidden deep inside $hat i$ and $hat j$. These are vectors that had a length to them. If you know how many meters $hat i$ is, you can figure out how many meters $3hat i$ is.

To this end, we typically notate coordinate vectors with a basis, which is the set of vectors we used to do the measurement. I might say $[v]_B=langle3000, 4000rangle$ where $B$ is a basis consisting of a pair of 1 meter vectors pointing in the east and north directions respectively. It is this concept of a basis which ties the original vector (which had units) to the coordinate vector (which is a purely mathematical vector that has no units).

Thus it would be wrong for me to say $[v]_B = langle3, 4rangle$, because that isn't the measurement of that vector in basis $B$. However, I am welcome to define a new basis $C$, built from a 1km vector pointing east and a 1km vector pointing north, and I am welcome to say $[v]_C = langle3, 4rangle$. In fact, I can even do strange things like make a basis $D$ built from a 2km vector pointing north, a 1km vector pointing east (note I just switched directions AND used different length vectors), and I can say $[v]_D = langle2, 3rangle$. The latter example is probably not all that useful, but it can be done. All I needed to do is keep track of my basis vectors.

So why go through this rigamarole? Well, there are some things you can do once you have component vectors that cannot be done on vectors. Both $vec v$ and $[v]_B$ have a length. The length $|vec v|$ is 5000 meters, and $|[v]_B| = 5000$. But the former required that I think entirely in vectors and do vector logic to figure out the length. Sometimes this is easy, sometimes it's hard. But once I have a coordinate vector, I have something which is nothing more than a vector right out of linear algebra, and we can say that for any $[v]_B = langle x, yrangle$, we know $|[v]_B|=sqrt{x^2+y^2}$. We can "shut up and calculate."

(As an aside, this can go deeper. When you get into angular velocities and accelerations, you will find that angular velocity is a real vector, and behaves like one, but angular acceleration does not. Angular acceleration can only be thought of as a coordinate vector or something more exotic like a bivector... and this is still just weird for me)

So in the end, your example with the object has only one vector, but it may be measured as different coordinate vectors simply by assuming different a different basis (such as choosing a basis vector of "1 meter" rather than a basis vector of "3 meters"). All that matters is that you keep your measurement and your basis consistent with each other. Just because $langle 3000, 4000rangle_B = langle 3, 4rangle_C$ does not mean $langle 3000, 4000rangle_B = langle 3, 4rangle_B$. The former just points out that $B$ and $C$ are different basis, one using vectors that are a meter and one using vectors using a kilometer. The latter is simply not correct arithmetic, and most often arises from us forgetting that the basis vectors were part of the measurement process.

As such, it is most formally correct to move the units out of the coordinate vector. Instead of writing $(3000m)hat i + (4000m)hat j$, its more correct to write $(3000hat i + 4000hat j) [m]$, using the notation from Units of Measurement. This points out that the units are not actually part of the coordinate vector. They're bundled into the definitions of $hat i$ and $hat j$, and all you're doing is noting those units to help you keep them consistent.

Can you write it both ways? Sure. It's just notation. But if you're getting confused, the more strict notations do a better job of keeping things separate.

Answered by Cort Ammon on March 31, 2021

Does the physical unit determine the length of the unit vector?

No.

I can use the same unit vectors $hat{mathbf i}$, $hat{mathbf j}$, and $hat{mathbf k}$ to write vector quantities with any dimensions because these unit vectors have a dimensionless length of $1$.

When I write a position as

$$mathbf r=xhat{mathbf i}+yhat{mathbf j}+zhat{mathbf k}$$

it is the components $x$, $y$ and $z$ that have the dimensions of length, and thus units of meters, feet, lightyears, etc.

When I write a velocity as

$$mathbf v=v_xhat{mathbf i}+v_yhat{mathbf j}+v_zhat{mathbf k}$$

it is the components $v_x$, $v_y$ and $v_z$ that have the dimensions of speed, and thus units of meters per second, feet per minute, lightyears per year, etc.

The same applies to acceleration, momentum, force, electric field, magnetic field, etc. The components of the vector have the dimensions and the units.

When I compute the magnitude (“length”) of a vector with dimensions, such as

$$|mathbf r|^2=sqrt{x^2+y^2+z^2}$$

or

$$|mathbf v|^2=sqrt{v_x^2+v_y^2+v_z^2},$$

its dimensions and units come from the components.

The unit vectors have no dimensions and thus no units. Their magnitude is the dimensionless number $1$. The unit vectors exist as mathematical abstractions completely independent of any physical quantity.

Answered by G. Smith on March 31, 2021

the definition of unit vector is: ${displaystyle mathbf {hat {u}} ={frac {mathbf {u} }{|mathbf {u} |}}}$. So it follows that the unit vector is a dimensionless size and it does not depend on the physical unit.

The tricky point is that the unit vector is a mathematical concept and you tried to think about it in the "real world". In this world every physicals size is measured relatively to some physical unit.

Note that some teacher or books can refer to the unit vector simply as a vector with magnitude 1 in some conventional physics unit (for example vector with magnitude of 1 meter), although it is not rigorously true. If you are taking any physics course it is always better to check with the teacher how he defines it.

Answered by ziv on March 31, 2021

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