It changes direction but not speed and this is proved in experiments.

But is there any more "deep" reason as that space has isotropy and energy must be conserved per example?


Since the acceleration is the derivative of the velocity with respect to time, if they are perpendicular we have that

$$ \mathbf{a}\cdot\mathbf{v} = \dfrac{d\mathbf{v}}{dt}\cdot\mathbf{v} = 0 $$

But this can be expressed as

$$ \dfrac{d\mathbf{v}}{dt}\cdot\mathbf{v} = \dfrac{d}{dt}(\mathbf{v}\cdot\mathbf{v}) - \mathbf{v}\cdot\dfrac{d\mathbf{v}}{dt} = \dfrac{d}{dt}(\mathbf{v}\cdot\mathbf{v}) = 0 $$

This means that the value $\mathbf{v}\cdot\mathbf{v}$ should keep constant over time. This value is, in fact, the squared velocity vector length, therefore:

$$ \mathbf{v}\cdot\mathbf{v} = v^2 = \text{cte} \Rightarrow v = \text{cte} $$

This means also that kinetic energy keeps constant over time since it depends on the mass and on the squared velocity.


Let d$\vec{s}$ be the infinitesimal displacement along the trajectory followed by the body and $\vec{F}$ the force acting on it. We know from the work-energy theorem that the work done by a force is linked to the difference in the kinetic energy through the relation

$$ \Delta W = \int_A^B \vec{F} \cdot d\vec{s} = \Delta K = K_B - K_A $$

where $A$ and $B$ are two points of the trajectory and, for a point-like body of mass $m$, $K = \frac{1}{2}mv^2$. Since the velocity $\vec{v}$ is always tangent to the trajectory, if $\vec{F}$ is orthogonal to $\vec{v}$ it will also be orthogonal to $d\vec{s}$. As a result, $\Delta W = \Delta K = 0$, which means that $v_B = v_A$.

In other words, a force that is orthogonal to the velocity is also orthogonal to the trajectory and hence cannot do work on, that is, cannot change the speed of, a body.


So, I assume you mean a force perpendicular to the current movement direction. Due to the isotropy and homogeneity of space the momentum needs to be conserved, or more exactly all 3 components of momentum need to be conserved (in 3D space). This is a consequence of Noethers theorem, or can be shown from analysing the effects of these symmetries on the Lagrangian of the system. Of course, this conservation law holds for all coordinate systems, so you could position your x,y,z coordinate system any way you like and all 3 components are conserved.

What happens when you have a force in a certain direction, is that this symmetry is broken, and only the momentum perpendicular to the force direction are conserved. That is why in this case it makes sense to align your coordinate system with this force, so you have 2 components that remain constant while 1 component changes (the velocity component in force direction, according to $\mathbf{F} = \frac{d \mathbf{p}}{dt}$). If you choose a non-aligned coordinate system, then all 3 components will change.

This is of course the power of using vectors to describe these phenomena, as the underlying principle is uniquely defined (the momentum in the direction of the force changes) irrespective of your chosen coordinate representation. While specific representation might look different (3 components changing for one specific set of coordinates, or 1 component changing and 2 constant for another), they really do describe the same underlying phenomenon

  • $\begingroup$ I don't think this answer is helpful; a much simpler explanation would be much more appropriate due to the level of the question (see Jaime_mc2's great answer). On the other hand you state "What happens when you have a force in a certain direction, is that this symmetry is broken, and only the momentum perpendicular to the force direction are conserved.". Which is exactly the OP's question; you are just using other words to say "It happens, because that is what happens" $\endgroup$ – Daniel Duque Mar 18 at 14:13
  • $\begingroup$ OP was specifically referring to a "deep" reason including symmetries, so I gave an answer that explains why symmetry/homogeneity of space leads to momentum conservation, and why breaking of this symmetry leads to the momentum in a specific direction not to be conserved. I also have to disagree that the quoted sentence just restates the original question - please note that every explanation includes the questions topic at some point. $\endgroup$ – Brkn Kybrd Mar 19 at 8:37

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