# Vectors, their representation and dipole moment

I know that the length of an arrow representing a vector gives represents the magnitude of that vector, now consider a force vector.

$$\vec{F}$$=m$$\vec{a}$$ , therefore in space a force vector would always be larger in length then the acceleration vector right?

Now in physics dipole moment is defined as $$\vec{p}$$=$$\vec{d}$$q

now a displacement vector of a electric dipole points from the negative charge to the positive charge and in space the length of the displacement vector is basically the length of the dipole.

as I am multiplying $$\vec{d}$$ and q therefore I get an even larger value and therefore shouldn't the arrow representing dipole moment be larger then the arrow representing the displacement vector?

But usually in textbooks length of the arrow representing dipole moment is smaller then the length of the arrow representing the displacement vector $$\vec{d}$$, why is it so?

• The dipole moment and the length of displacement (as well as the force and the acceleration) do not have the same units; therefore it makes no sens to compare their length. – EigenDavid Apr 30 '19 at 10:30
• (By the way, the word "then" is different from "than". In the above two instances, "than" should be used instead of "then".) – Digiproc Apr 30 '19 at 10:34

There are many things here.

$$\vec{F}=m\vec{a}$$ , therefore in space a force vector would always be larger in length then the acceleration vector right?

No! Not neccesarily. That would only happen if $$m>1$$. It is well known that multiplying by something smaller than 1 (between 0 and 1) is actually redicing the value. In other words:

$$0.1\times a < a$$

So the "arrow" will be actually smaller if $$m$$ is smaller than one.

## Secondly.

What I wrote above applies to pure math. Numbers smaller than 1 make vectors shorter.

But, we are not scaling by a real number. Mass has units. You are actually doing $$0.1 kg \times \vec{a}$$, for example.

So, the above reasoning applies if and only if we have stablished a scale for each magnitude. If you are using the international system, then a mass of $$0.1 kg$$ would make the force vector in newtons be smaller than the acceleartoin vector in meters.

However, you must be aware that this strongly depends on units. If you use different units for each magnitude, the lenghts can vary quite much.

Lastly.

But usually in textbooks length of the arrow representing dipole moment is smaller then the length of the arrow representing the displacement vector $$\vec{d}$$ , why is it so?

Apart from the fact that the electron charge is really small... Check that textbooks usually illustrate the direction of the dipole moments. We are not interested in a rigurous measurement of their lenght; in fact, the scale is not usually even stablished, so the lenght on the arrow is meaningless. That is because we only care about directions.

The lenght of the arrow is often set to be "big enough for visibility", regardless of their modulus accuracy.

There is no reason to compare vector magnitude values of vectors of different units. Units are (usually) arbitrarily created by humans. We might have decided that the base unit for force is the "glerf", which is the force required to accelerate the Sun to half light speed in 1 femtosecond. In that case the force vector magnitude on a baseball over one second to accelerate it to 10 meters per second would be a very short vector (only a tiny fraction of a "glerf"). There are units that are more "natural" (see https://en.wikipedia.org/wiki/Planck_units). Comparing those might have more meaning, but there isn't consensus on that.