Let $(r_1,r_2,r_3)$ be the coordinates of a particle $r$ in the coordinate system $\phi$. Let $\{\hat{e_1},\hat{e_2},\hat{e_3}\}$ be the coordinate basis of $\phi$. Why do we define the velocity $v$ of $r$ in $\phi$ as

$$v:= \frac{d}{dt}\sum_{i=1}^3r_i\hat{e_i} \tag{1}$$

instead of just

$$v: = \big(\frac{d}{dt}r_1,\frac{d}{dt}r_2,\frac{d}{dt}r_3\big)~?\tag{2} $$

What special characterizations or advantages does the former definition have over the latter?

  • 2
    $\begingroup$ $\hat{e}_{i}$ could be time dependent. Try polar coordinate :) $\endgroup$ – K_inverse Oct 10 '18 at 4:30
  • $\begingroup$ While that is true, it doesn't really answer my question. $\endgroup$ – J_Psi Oct 10 '18 at 4:32
  • $\begingroup$ It may not be the answer you were looking for but it still should have been posted as an answer, not a comment (@K_inverse please keep this in mind for the future). I'm going to come back and remove these comments after a little while. $\endgroup$ – David Z Oct 10 '18 at 5:30
  1. First of all, if OP's basis doesn't depend on time then the two definitions (1) & (2) coincide.

  2. Q: What does the time-dependence of OP's basis refer to? A: It's the time-dependence of OP's basis relative to another fiducial reference frame, which in Newtonian mechanics usually is assumed to be an inertial frame. If this is the case, then definition (1) is the velocity relative to an inertial frame, while definition (2) is the velocity relative to OP's basis. Note that an accelerated frame of reference leads to fictitious forces.

  3. Example. If OP's basis is Earth-fixed then definition (2) is used in practice to measure velocities on/near the surface of the Earth, say, velocity of cars, velocity of wind, etc.

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Your equations turn out to be the same thing when we are using Cartesian coordinates. This is because each coordinate is independent of location in space. For example at the Cartesian point $(1,1,1)$, the $\hat x$ direction is the same as it is at Cartesian point $(-1,-1,-1)$. However in spherical coordinates, the direction of $\hat r$ is different at each of those spatial coordinates (and in fact in that example they are anti-parallel).

So with that being said, if we have a general vector with spatial and temporal depdence, it is not true in spherical coordinates that it's time derivative is $\left(\frac{dr_1}{dt},\frac{dr_2}{dt},\frac{dr_3}{dt}\right)$, since this assumes a constant direction of the unit vectors that form the basis of our coordinate system.

More explicitly, a vector in any 3D basis is given by $\sum_{i=1}^3r_i\hat {e_i}$, so the time derivative of a general vector$^*$: $$\frac{d}{dt}\sum_{i=1}^3r_i\hat {e_i}=\sum_{i=1}^3\left(\frac{dr_i}{dt}\hat{e_i}+r_i\frac{d\hat{e_i}}{dt}\right)$$

In Cartesian coordinates, $\frac{d\hat{e_i}}{dt}=0$, so this shows the equivalence of your two methods in your question for Cartesian coordinates. But $\frac{d\hat{e_i}}{dt}\neq0$ in spherical coordinates, because as we have seen, the unit vectors are changing direction as we move around in space.

$^*$Another thing to keep in mind is that the position vector itself is not always $\vec r=\sum_{i=1}^3r_i\hat {e_i}$ where all $\hat{e_i}\neq 0$. For example in spherical coordinates $\vec r=r\hat r$, since the position vector always points from the origin to the position of the particle. So the velocity in spherical coordinates ends up being $\vec v=\dot r\hat r+r\dot{\hat r}$

This might be the issue that you discuss in the comments. In Cartesian coordinates we can say that if we are at point $(x,y,z)$ then our position vector is $x\hat x+y\hat y+z\hat z$ But if we use something like spherical coordinates, being at point $(r,\phi,\theta)$ means that the position vector is $r\hat r$, not $r\hat r+\phi\hat\phi+\theta\hat\theta$.

In general terms, being at spatial coordinate $(r_1,r_2,r_3)$ does not translate to having position vector $r_1\hat{r_1}+r_2\hat{r_2}+r_3\hat{r_3}$. You could even argue that how you specify the vectors doesn't depend on how you specify your spatial coordinates. For example, I could choose to express my spatial coordinates in terms of Cartesian coordinates, but use a spherical basis: $\vec r=r(x,y,z)\hat r(x,y,z)$

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  • $\begingroup$ I understand that this is the typical argument given in favor of the latter definition; that because the coordinate basis possibly depends on the coordinates of the particle that they ought to be differentiated as well. What I am wondering is why we even brought the coordinate basis into the definition in the first place? Why not just talk only about the coordinates and then differentiate each coordinate? Certainly, there must be a reason why this is not the case. $\endgroup$ – J_Psi Oct 10 '18 at 7:45
  • $\begingroup$ @J_Psi Do you know that if you are at spatial spherical coordinate $(r,\phi,\theta)$ that your position vector is not $r\hat r+\phi\hat\phi+\theta\hat\theta$? $\endgroup$ – BioPhysicist Oct 10 '18 at 9:50
  • $\begingroup$ @J_psi I have edited my answer to try to address the concern I think you are having here. Let me know if I did not understand your previous comment. $\endgroup$ – BioPhysicist Oct 10 '18 at 10:36
  • $\begingroup$ Yes, I am aware that the coordinate vector cannot, in general, be expressed as the linear combination of coordinate vectors with the respective coordinates as you have pointed out. $\endgroup$ – J_Psi Oct 10 '18 at 18:00
  • $\begingroup$ Let me highlight a statement in your post to attempt to drive home my point if it's still not clear. You say: "So with that being said, if we have a vector with general spatial and temporal dependence, it's not true in [general coordinates] that it's time derivative is [the definition I posited]." But that argument is circular. Of course that is not true for velocity in general, because that is not how we have defined velocity in general. My question is more fundamental: /Why/ is that not how we have chosen to define velocity? $\endgroup$ – J_Psi Oct 10 '18 at 18:04

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