While trying to understand the second law of Newton from "An Introduction to Mechanics" by Kleppner and Kolenkow, I came across the following lines that I don't understand:

"It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector. Although this turns out to be the case, it is not obviously true. For instance, if mass were different in different directions, acceleration would not be parallel to force and force and acceleration could not be related by a simple vector equation. Although the concept of mass having different values in different directions might sound absurd, it is not impossible. In fact, physicists have carried out very sensitive tests on this hypothesis, without finding any variation. So, we can treat mass as a scalar, i.e. a simple number, and write $\vec{F} = m\vec{a}$."

The lines above lead me to question:

  1. Why is it not" obviously true" that force behaves like a vector?

  2. Why is it not impossible for mass values to be different in different directions?

  • 1
    $\begingroup$ From what I gather, it is asking you to consider the consequence of mass depending on the direction of its movement. What would the implications be if $\vec{F} = \vec{m}\vec{a}$? This would yield different results for force if mass varied according to a particular direction. But it goes on to say that no variation has been found, so for all intents and purposes, consider mass as a scalar quantity. Thus, force only depends on the direction of the acceleration of the mass. This is the usual way of expressing Newton's second law, $\vec{F} = m\vec{a}$ $\endgroup$
    – bleuofblue
    Nov 20, 2017 at 6:21
  • $\begingroup$ @bleuofblue what confused me the most is the line "Although the concept of mass having different values in different directions might sound absurd, it is not impossible." $\endgroup$
    – R004
    Nov 20, 2017 at 6:51
  • $\begingroup$ @R004 I think I could mess things up even more, but in some models you actually have different masses depending on the direction of motion. For instance, you can consider the effective mass of an electron in a semiconductor. It depends in the direction considered, due to the symmetry of the crystal. However, this is NOT the real mass of the electron, but a model to describe it. $\endgroup$
    – JackI
    Nov 20, 2017 at 6:59
  • 1
    $\begingroup$ Can anyone point me the to a reference for 'the very sensitive tests on this hypothesis'? $\endgroup$
    – lalala
    Nov 20, 2017 at 7:41
  • 4
    $\begingroup$ What don't you understand? What do you think it means? What confuses you? Do not make answerers guess about these questions. It is not clear what information you're looking for. (Perhaps if you had spent the time to try to answer those questions yourself, you would have been able to sort out the passage on your own. This is the mindset required of students of physics, researchers, and members of Stack Exchange; formulate clear questions and then try to answer them before you post.) $\endgroup$
    – jpmc26
    Nov 20, 2017 at 9:12

8 Answers 8


I think that an example will clear out your doubts.

Consider a 3D system in which you have three axis $xyz$. Consider a force that can be written as: $$ \mathbf{F} = F\hat{x} + F\hat{y} +F\hat{z} $$ Therefore, this force is identical in each direction. If we have three different masses depending on the direction we are considering: $$ m_x, m_y, m_z$$ Then we will have different accelerations depending on the direction in which we are studying the motion: $$ a_x = F/m_x, a_y = F/m_y, a_z = F/m_z$$ And therefore, the vector acceleration $\mathbf{a} $ will not be parallel to the vector force $\mathbf{F} $, since in each direction the masses (and therefore, the components of the acceleration) are different.

However, this has shown up to be wrong, according to some experiment. Therefore, the book is saying you that there is an evidence that the mass is identical regardless of the spatial direction you are considering.

  • 4
    $\begingroup$ As a potential example of something with different "masses" in different directions, consider something which is stubborn and does not want to accelerate in a given direction, but will move freely in another. People try to act this all the time. However, we see that at the physics level, particles do not do this. Particles have the same mass in all directions, as you say. We can use this observation to then show how people "act stubborn" by re-directing the forces elsewhere, rather than having to explain it as them having different "masses" in different directions. $\endgroup$
    – Cort Ammon
    Nov 21, 2017 at 1:55

Force could be treated as a vector even if mass behaved different in different directions. In this case, mass would not be a simple scalar, but it would be what is called a second rank tensor. Treating vectors as column $3 \times 1$ matrices, force and acceleration would be

$ \mathcal{F}=\left( \begin{matrix} F_x\\ F_y\\ F_z\\ \end{matrix} \right) \quad \text{and} \quad \mathcal{A}=\left( \begin{matrix} a_x\\ a_y\\ a_z\\ \end{matrix} \right); $

the second rank tensor representing mass would be simply a $3 \times 3$ matrix:

$ \mathcal{M}= \left( \begin{matrix} m_{xx} & m_{xy} & m_{xz}\\ m_{yx} & m_{yy} & m_{yz}\\ m_{zx} & m_{zy} & m_{zz}\\ \end{matrix} \right); $

and Newton's second law would read deceptively like its well known form, $\mathcal{F}=\mathcal{M}\mathcal{A}$, but definitely not with the same content:

$ \left( \begin{matrix} F_x\\ F_y\\ F_z\\ \end{matrix} \right) = \left( \begin{matrix} m_{xx} & m_{xy} & m_{xz}\\ m_{yx} & m_{yy} & m_{yz}\\ m_{zx} & m_{zy} & m_{zz}\\ \end{matrix} \right) \left( \begin{matrix} a_x\\ a_y\\ a_z\\ \end{matrix} \right) \quad \text{or} \quad \begin{matrix} F_x = m_{xx}a_x+m_{xy}a_y+m_{xz}a_z,\\ F_y = m_{yx}a_x+m_{yy}a_y+m_{yz}a_z,\\ F_z = m_{zx}a_x+m_{zy}a_y+m_{zz}a_z. \end{matrix} $

According to this general formulation, the resistance a body has against a change on its state of motion is direction-dependent. In particular, if a force acts in one direction only, the body may accelerate in different directions as well depending on which elements of the mass matrix are nonvanishing.

In some sense, the above relations are similar to the rotational dynamics of a rigid body, where torque relates to angular acceleration by means of a "rotational inertia" matrix. The consequences in such case are less abstract. If a body is set to spin along some direction, its spinning direction usually deviates from the original direction in "unexpected" ways depending on the rotational inertia matrix of the body, initial conditions and applied torque. For instance, a spinning top will precess due to the action of gravity, but this effect is much more impressive as seen here at 35:20; it will precess even if gravitational torque is absent if it is set to spin along a direction other than its symmetry axis, like here at 0:54; for a very dramatic situation of instabilities on the free rotation, see here (starting at 0:28).

Nevertheless, the two cases may be very different in general. The most general rotational inertia matrix has the very important feature of being symmetric (it is the same as its transpose) what guarantees it can be diagonalized so that torque and angular acceleration can be made parallel in some well chosen coordinate system. On the other hand, the most general mass matrix may not have such property (not that I can tell at least). Because of that, it is possible, for some values of the elements of the mass matrix, that a force along a single direction acting on a body's center of mass produces an acceleration in some perpendicular direction only.

In the context of classical mechanics, as far as I can see, experiments dealing only with translations (forces applied at the center of mass of the body) may not be sufficient to determine all nine elements of its mass matrix, requiring experiments dealing with rotations of the body as well.

In the context of quantum mechanics, as far as now (see this Wikipedia page), experiments indicate the mass matrix for a single body is diagonal with all mass elements equal, so mass can be taken as a scalar quantity.

  • $\begingroup$ Nice! For context, you might mention that this is the relationship we have between torque, rotational inertia and angular acceleration. $\endgroup$
    – David Elm
    Nov 20, 2017 at 17:28
  • $\begingroup$ Minor remark: You recover Newton's second law for M = m times identity matrix, not when all elements are m. But what's the meaning of those off-diagonal elements? Does it imply that you can accelerate in the x-direction by pushing in y-direction? $\endgroup$
    – M. Stern
    Nov 20, 2017 at 18:03
  • $\begingroup$ @M.Stern thanks for your remark! I edited my answer to correct this. I will try to complement my answer later to deal with your questions and David Elm's above remark. $\endgroup$ Nov 21, 2017 at 11:37

Consider the motion of rigid body and the role of mass in it:

  1. Translation: if we want to change the velocity of a body we need to apply a force. The difficulty with which the body changes its velocity depends on its mass, i.e. in translation the mass, $m$, is a measure of inertia and it is the same in all directions, i.e. regardless in which direction you will choose to apply the force, which causes a translation, it will lead to the same change in speed - acceleration.

  2. Rotation: if we want to change the angular velocity of a rotating body we need to apply a torque, $\vec{\tau} = \vec{F} \times \vec{r}$. The difficulty with which the body changes its angular velocity depends on its moments of inertia in the specific axis of rotation. Moments of inertia are components of a tensor, which reflects the fact that rotating the body, with the same angular velocity, on different axis of rotation requires different torque, i.e. in rotation the moments of inertia, $I_{ij}$, are the measure of inertia and they are different in different directions.

Observing the second case, you should be able to understand what the authors were probably referring to when they say: "...Although this turns out to be the case, it is not obviously true...".The authors probably try to present a more general perspective by noting that the fact that the amount of substance containing in an object, which is known as mass of that object is scalar, because it has been observed/measured that everywhere the mass of the object is constant, which is not implied by anything. For further reading, check the notion of isotropy.


There could be an additional nuance to this question related to the definition of mass in special and general relativity.

  • 2
    $\begingroup$ The concept of rotation now helps me realize that. Appreciate your answer. $\endgroup$
    – R004
    Nov 20, 2017 at 7:50

There's a different perspective which starts from conservation of momentum: It is experimentally observed that, in a collision of involving particles 1 and 2, one can always find two scalars $m_1$ and $m_2$ so that $$ m_1\vec v_{1b}+m_2\vec v_{2b}=m_1\vec v_{1a}+m_2\vec v_{2a} \tag{1} $$ where $\vec v_{ib}$ are the velocities before collisions, and $\vec v_{ia}$ the velocities after collisions. In fact, this type of experiment only allows the deduction of the ratio $m_1/m_2$. (This is a Newtonian definition with idealized particles from which pieces cannot be chipped off.)

Having established that $m_i$ is a scalar through conservation of momentum, a force $\vec F$ is what is responsible for a change in momentum. Since $m\vec v$ is clearly a vector because $\vec v$ is a vector, the force is related to a change in $\vec v$, i.e. a difference between two vectors, and must therefore be a vector itself.

The absolute values of $m_1$ and $m_2$ (rather than their ratios) can be obtained by taking the ratio of $m_i$ against some standard reference mass (which is what is done in practice since we have a reference kilogram).


In Newton physics, mass is the resistance of an object to linear acceleration. In other words, you define mass by applying a force to a system and by analyzing its liner acceleration.

In this sense, it is not necessary that the linear acceleration be parallel to the applied force. The analog of this is inertia, which is the resistance of an object to angular acceleration. In the case of inertia for example, you may apply a torque only about $x$ axis and find out that the body gets an angular acceleration about $y$ axis as well. This is why inertia is a second rank tensor, not a scalar.

If mass was like inertia (if we ever found out that applying a force in $x$ direction can cause a linear acceleration in $y$ direction), we wound need mass not as a scalar but as a second rank tensor (or a matrix in the more common language). Then, we would have: $$f_x=\vec{m_x}\cdot \vec{a}\\f_y=\vec{m_y}\cdot \vec{a}\\f_z=\vec{m_z}\cdot \vec{a}$$


The accelerating system may be more complex that a single body. Consider an example of a ball in a pipe. If you apply a force to a ball along the pipe, then you would move only the ball inside the pipe. So your effective mass for the equation is the mass of the ball. However, if you apply a force to the ball at 90 degrees, then you would move both the ball and the pipe. So your effective mass is a sum of the mass of the ball and the mass of the pipe. Finally, if you apply a force at any other angle, your effective mass would be the mass of the ball plus the appropriate fraction of the mass of the pipe. Thus in a complex system the effective mass can be different in different directions.


If mass is a vector quantity then how does one find the total mass of two vector masses after they are combined?
Is there really any evidence that two objects of equal magnitude mass $m$ when joined together exhibit variations in the magnitude of their combined mass varying from $0$ to $2m$?

  • 3
    $\begingroup$ This is a strange answer consisting just from questions. I think you dismiss the possibility of a non-scalar too easily. Added mass in fluid mechanics, for example, is not a simple scalar, but a tensor. See the answer by @andrehgomes why you should expect a tensor, not a vector. $\endgroup$ Nov 20, 2017 at 13:25

Consider a solid sphere with mass $m_1$. Now lets say there is a hole through the center that runs horizontally from end to end. Inside this hole is another mass $m_2$ that slides without friction.

If you were to push the sphere horizontally you will have $F_x = m_1 a_x$. But if you push vertically you would get $F_y = (m_1 + m_2) a_y$ since both masses would need to be accelerated equally. $$ \mathbf{F} = \mathcal{M}\, \mathbf{a} $$ $$ \pmatrix{F_x \\ F_y \\ F_z} = \begin{bmatrix} m_1 & & \\ & m_1+m_2 & \\ & & m_1 + m_2 \end{bmatrix} \pmatrix{a_x \\ a_y \\ a_z} $$

So the relationship between forces and accelerations is not always a simple vector scaling operation, but more complex.


The only physical stipulation for $\mathcal{M}$ is that it is symmetric matrix.

In robotics, whenever you have multiple connected rigid bodies you have the concept of Articulated Inertia which is exactly as shown above. This matrix in general is not a scalar multiple of the identity matrix, but has varying diagonal elements and cross terms in the off-diagonal.

  • $\begingroup$ When you write "If you were to push the sphere horizontally you will have $F_x = m_1 a_x$. But if you push vertically you would get $F_y = (m_1 + m_2) a_y$ since both masses would need to be accelerated equally." you are cheating. In the former case you are considering the acceleration of one mass in isolation and in the other you are considering the acceleration of a system consisting of two masses. If you consider the acceleration of the center of mass of the system you get the same magnitude in both cases. $\endgroup$ Dec 9, 2017 at 21:39
  • $\begingroup$ And you get the same ratio $F_{\text{net},1}/a_1$ in both cases if you include the reaction force from the bead. Either way if you examine a consistent physical quantity in both cases you find it to be a vector. $\endgroup$ Dec 9, 2017 at 21:41

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