# Why is kinetic energy a scalar, if we require additional information to represent all it's intrinsic properties?

The usual rule of thumb decides whether the quantity has a direction or not with a sign ($$-$$ or $$+$$) in front.

However, $$KE$$ has also another intrinsic attribute - whether the energy is lost or gained (better phrased to be whether that energy is converted to some other form leading to a net positive or negative sum).

As I understand, vectors are usually required for representing multi-faceted quantities like $$\vec{v}$$ m/s, which require additional information to be fully explained. Thus, wouldn't $$KE$$ be classified as a vector in that sense?

If I give a value of $$KE$$ without any other information (i.e the sign); physically its wrong since someone might assume an accumulation of energy whereas it could actually be the opposite.

tl;dr vectors are required to represent multiple aspects of a physical quantity; I argue kinetic energy thus should also be considered as a vector.

• Would you say that a sum of money should be a vector because it should carry information as to whether it is lost or gained etc.? Commented Nov 22, 2021 at 18:27
• @PhilipWood No, because that's not an intrinsic property of money; it's of the wallet which has lost a certain amount. This is an intrinsic property of Kinetic Energy, or any energy actually. Commented Nov 22, 2021 at 18:30
• It is not the kinetic energy but the change in kinetic energy that can be positive or negative. The KE energy describes a state of the system and there is no meaning to say that is gained or lost.
– nasu
Commented Nov 22, 2021 at 18:55
• Something might lose KE, but that's work and not KE itself. KE = $1/2mv^2$, which is never negative (you'd require negative mass). You can subtract KE from an object, but KE is always positive. Commented Nov 22, 2021 at 19:53
• I'm not sure kinetic energy can be represented with a sign. Only a change in kinetic energy can. Kinetic energy is always a positive quantity. A minus sign can be written as an aritmetic operator to indicate subtraction, but not as a sign. Commented Nov 22, 2021 at 19:53

Why do you assume that a scalar can't have a sign? Scalars can take all reals numbers as values (of course there are scalars, that may only take positive values, e.g. the magnitude of a vector). But e.g. the electric charge is a scalar and take values like $$e$$ or $$-e$$ (with the elementary charge $$e$$).

Also, you seem to mix up energy and work. Work is the change of energy of a system when external forces are acting on it. Energy in contrast is just a number that is assigned to the state of a system (and that is conserved for closed systems).

Whether something is a vector or a scalar (in the sense the terms are used in physics) is a well defined mathematical concept, that depends on the transformation behaviour under coordinate changes. For the elementary classification we only consider rotations of Cartesian coordinate systems: $$\vec r' = D\vec r = \sum_{ij} \vec e_i D_{ij} r_j$$, with an orthogonal matrix $$D$$, $$D^TD = 1$$.

The components of the velocity will then transform just like the components of the position: $$\vec v' = D\vec v$$, and this is the requirement for a quantity to be a vector.

The kinetic energy $$T = \frac 1 2 mv^2 = \frac 1 2 \vec v \cdot \vec v$$, however, stays the same under such rotations: $$T = \frac 1 2 m \vec v \cdot \vec v = \frac 1 2 m \vec v \cdot (1 \vec v) = \frac 1 2 m \vec v \cdot (1 \vec v) = \frac 1 2 m (\vec v \cdot D^T) (D \vec v) = \frac 1 2 m \vec v' \cdot \vec v' = T'$$ (This computation was done for a single particle, but it can be extended to $$n$$-particle systems trivially.)

And this is the definition of a scalar: A quantity that does not change under rotations. While a vector is a quantity whose components transform like the coordinates under rotations.

However, this is not the entire truth when moving beyond Newtonian mechanics. Actually, kinetic energy is a component of a generalized notion of a vector in general relativity. But this has nothing to do with the objections to the scalar nature of the kinetic energy in the question, but with the transformation behaviour under Lorentz transformations. Namely, the 4-vector for momentum is $$(p^\mu) = (T/c, \vec p)$$.

I like this question. So first off

The usual rule of thumb decides whether the quantity has a direction or not with a sign (− or +) in front.

If a quantity has a negative sign it means it has a direction, right? It means it's a vector, right?

Well, not exactly. Don't get me wrong, all vectors do have a sign ($$+/-$$), but that doesn't mean quantities that have a negative sign are vectors.

Want an example of a scalar with a negative magnitude? Try $$-50^o C$$.

We know temperatures are of course not vectors. They are scalars with no directional dependency. Instead, The negative sign here is due to our frame of reference, which we have decided to be at the freezing point of pure water, ie $$0^o C$$. What $$-50^0C$$ tells us is that our current temperature is $$50^oC$$ below/lesser than our frame of reference, not the direction of any sort. (interestingly, using the phrase "below $$0^oC$$" does indeed make it sound like it's directional) Ultimately temperature is a scalar quantity with a negative value but without any directional dependency.

Put differently "All vectors may have a negative sign, but the converse is not true. Having a negative sign is not the absolute test for vector-ness". Check if they satisfy triangular/parallelogram vector addition. (see my related answer)

There are other quantities too with negative values but no directional dependencies, most notably Potential Energy. Here we are considering our frame of reference at infinity, ie the point where the magnitude is 0 is at infinity.

Now let's move to Kinetic Energy!

If I give a value of KE without any other information (i.e the sign); physically its wrong since someone might assume an accumulation of energy whereas it could actually be the opposite.

$$KE$$ is always a positive quantity, as $$KE=mv^2$$ and both mass and v^2 are positive, the latter due to maths and the former because we are not considering anitmatter. What you may be referring to must be $$\Delta KE$$ which yes can be negative as it $$\Delta KE=KE_2-KE_1$$

A negative value of $$\Delta KE$$ implies that $$=KE_2 which, yes, means that the Kinetic Energy is lost or given out. Conversely, a positive value of $$\Delta KE$$ indicates that $$=KE_2>KE_1$$ or that the Kinetic energy of magnitude $$\Delta KE$$ is accumulated.

Once again the negative sign appears only due to our frame of reference, which is $$KE_1$$ here, and nothing else.

Sorry if anything sounds incomprehensible, in my defence it was already midnight when I started typing.

• Antimatter also has positive mass (it's the charge, roughly speaking, whose sign changes). Negative mass is an entirely theoretical concept that hasn't been observed (AFAIK) to date. Commented Nov 23, 2021 at 15:09

Can kinetic energy be thought of as a vector, where the direction represents inflowing or outgoing energy? The answer is no, for two reasons:

• First of all, the OP is conflating kinetic energy and work. Really, the OP is thinking about the change $$\Delta KE$$ in kinetic energy during some process.
• There's not really a meaningful way to "add" or "multiply by scalars".

Details to follow:

The kinetic energy of an object is a property of the object itself at a particular moment in time and does not represent the property of some sort of process. Instead, the loss or gain of energy during a process is represented by the work done on that object. The work-energy principle for a single particle reads $$W_{\textrm{done on particle}} = \Delta KE_{\textrm{particle}}\,.$$ Thus, the change in kinetic energy of a particle has a "direction" in the sense that the change in kinetic energy is either negative or positive, representing the fact that the particle is either gaining or losing energy.

We can think of this as a vector if we like, except: can we ascribe any meaning to the sum of such vectors or the multiplication of the vector by a number (this is necessary if we're going to think of $$\Delta K$$ as a vector)? Perhaps, but I don't think it's particularly useful. The sum would work as follows. Take two processes one after the other. The changes in kinetic energy during the two processes are $$\Delta KE_1$$ and $$\Delta KE_2$$, and so the net change in kinetic energy over both processes is $$\Delta KE_1+\Delta KE_2$$. I guess that's fine, but what would multiplying the change in kinetic energy really mean here? That would mean changing the process to one where the work is also multiplied by that number. This doesn't seem like a useful notion.

As I understand, vectors are usually required for representing multi-faceted quantities like $$\vec{v}$$ m/s, which require additional information to be fully explained.

Vectors are required for quantities that have both magnitude and direction.

Thus, wouldn't $$KE$$ be classified as a vector in that sense?

No, because KE only has magnitude and no direction. The only component of the velocity $$v$$ that we need for kinetic energy is its magnitude, or speed, and not its direction.

If I give a value of $$KE$$ without any other information (i.e the sign); physically its wrong since someone might assume an accumulation of energy whereas it could actually be the opposite.

Any sign (+ or -) assigned to KE simply means there is an increase or decrease in KE. It has nothing to do with any "direction" of KE.

Identical objects moving in different directions with the same speed $$v$$ will have the same KE even though the velocity $$\vec v$$ of each object is different. KE is therefore a scalar not vector quantity. Moreover the total KE of a collection of such particles is simply the sum of the KEs of the individual particles. It doesn't matter if all particles are going in the same direction, in which case they will all have the same velocity, or if they are going in completely random directions, in which case the average velocity of the collection will be zero.

Hope this helps.

• Sorry if I'm not firm in these elementary mathematical concepts, but I think $v$ in $\frac{1}{2}mv^2$ is a vector alright; the thing is that $v^2$, or $v \cdot v$, is a dot product yielding a scalar value. Is that correct? Commented Nov 23, 2021 at 17:07
• @Peter-ReinstateMonica I don't disagree that KE is derived from the dot product of vectors, starting with force and displacement. I'm only saying we only care about the magnitude of the velocity, not its direction. I've edited my answer to make it clear. Thanks for your input. Commented Nov 23, 2021 at 17:56

The concept of KE in classical mechanics comes from the Newton's second law, if we make the dot product of each side of the equation with a elementary vector displacement

$$\mathbf {F.dr} = m\mathbf {a.dr} = m \frac{\mathbf {dv}}{dt}\mathbf{.dr} = m\mathbf {dv.}\frac{\mathbf{dr}}{dt} = m(\mathbf {dv).}\mathbf v = \frac{1}{2}m\mathbf {d(v.v)} = d(\frac{1}{2}mv^2)$$

The dot product of the LHS can be positive or negative, depending on the relative direction of the force and the vector displacement. This corresponds to an increase or decrease of the quantity $$\frac{1}{2}mv^2$$.

Both quantities are scalars, as they result from dot products.