# Basic Q about Vector/tensor definition and velocity

• I
HI,
I have read about tensors and generally understand the concept. One core thing about Vector/Tensor as I understand it is that its magnitude and direction should not change with change in coordinate system. I get that when I write vector and also when I use matrix transformation.
However here is my confusion. I see a train moving at say 60km/hr when I am at rest as frame of reference say to North. However from another frame moving north at 40km/hr, the train moves at 20 km/hr north. So here in new frame, the direction did not change but magnitude of vector as seen from new frame is now 20 instead of 60. So it appears that vector does change with respect to frame so is Velocity a tensor?
Can someone explain in easy to understand way?
Thanks

Related Differential Geometry News on Phys.org
fresh_42
Mentor
HI,
I have read about tensors and generally understand the concept. One core thing about Vector/Tensor as I understand it is that its magnitude and direction should not change with change in coordinate system. I get that when I write vector and also when I use matrix transformation.
However here is my confusion. I see a train moving at say 60km/hr when I am at rest as frame of reference say to North. However from another frame moving north at 40km/hr, the train moves at 20 km/hr north. So here in new frame, the direction did not change but magnitude of vector as seen from new frame is now 20 instead of 60. So it appears that vector does change with respect to frame so is Velocity a tensor?
Can someone explain in easy to understand way?
Thanks
The moving frame of reference is a continuously changing basis which adds up to 40 and thus removes the 40 from the 60. The overall vector is still 60, which you see, if you choose comparable frames of reference, namely both at rest at different points, instead of hiding part of the vector in the velocity of coordinates.

The moving frame of reference is a continuously changing basis which adds up to 40 and thus removes the 40 from the 60. The overall vector is still 60, which you see, if you choose comparable frames of reference, namely both at rest at different points, instead of hiding part of the vector in the velocity of coordinates.
Thank a lot for your time and post.
I am almost getting it. Can you elaborate a bit more with example how this happens. As I understand every frame can be moving so is the true magnitude of velocity 20 or 60 (Note direction is not changing in this Ex) If I observe change of magnitude how can I say vector is invariant? I think I understand it when I draw different codt system and get new components and prove that today magnitude of vector is unchanged but hard to visualize a practical example. Really appreciate help!

fresh_42
Mentor
Thank a lot for your time and post.
I am almost getting it. Can you elaborate a bit more with example how this happens. As I understand every frame can be moving so is the true magnitude of velocity 20 or 60 (Note direction is not changing in this Ex)
It's 60. It is invariant for any fixed coordinate system, e.g. whether you measure ##km\cdot h^{-1}## or ##mph##. The representation isn't the same, i.e. the coordinates themselves change, even more if you smoothly change them all the time, as in my example ##v(t)=-22.7178\cdot t +60\,.## That's the difficulty here in general. We can only write them down in numbers, but that requires a scaling which is deliberately chosen. The physical reality remained the same, the velocity vector attached to the locomotive.
If I observe change of magnitude how can I say vector is invariant?
The numbers for the coordinates change, but the train is still running at 60.
I think I understand it when I draw different codt system and get new components and prove that today magnitude of vector is unchanged but hard to visualize a practical example. Really appreciate help!

stevendaryl
Staff Emeritus
HI,
I have read about tensors and generally understand the concept. One core thing about Vector/Tensor as I understand it is that its magnitude and direction should not change with change in coordinate system. I get that when I write vector and also when I use matrix transformation.
However here is my confusion. I see a train moving at say 60km/hr when I am at rest as frame of reference say to North. However from another frame moving north at 40km/hr, the train moves at 20 km/hr north. So here in new frame, the direction did not change but magnitude of vector as seen from new frame is now 20 instead of 60. So it appears that vector does change with respect to frame so is Velocity a tensor?
Can someone explain in easy to understand way?
Thanks
The difficulty that you're having with velocity as a vector is actually an important insight. Velocity is a vector with respect to spatial coordinate transformations, but not with respect to transformations that involve time.

The transformation rule for the components of a vector under a coordinate transformation is this:

Let $x^i$ be your original coordinate system (where $i$ is an index that ranges over the number of dimensions; for example, $x^1 = x$, $x^2 = y$, $x^3 = z$). Let $y^a$ your new coordinate system (where again $a$ is the index). If $V^i$ is the ith component of vector $V$ in coordinate system $x^i$, and $V^a$ is the ath component in the new coordinate system, then:

$V^a = \sum_i V^i \frac{\partial y^a}{\partial x^i}$

Note that it's impossible for all the components $V^a$ to be zero unless all the components $V^i$ are zero, as well.

But that rule doesn't work when your coordinate change is a change of reference frames and $V$ is the velocity. Let's just take one spatial dimension, $x$. Then the coordinate transformation corresponding to a change of reference frames is:

$y = x - u t$

where $u$ is the relative velocity between the frames. If $V^x$ is the x-component of the velocity in the $x$ coordinate system, then in the $y$ coordinate system, the component of velocity is:

$V^y = V^x - u$

This is not of the form $V^a = \sum_i V^i \frac{\partial y^a}{\partial x^i}$.

So velocity is not a vector when it comes to transformations involving time.

In Special Relativity, however, time is treated as a coordinate like the spatial coordinates. Then there is a notion of "velocity" in Special Relativity that IS a vector, but it's a spacetime vector, not a spatial vector.

WWGD, Orodruin and fresh_42
Orodruin
Staff Emeritus
Homework Helper
Gold Member
The moving frame of reference is a continuously changing basis which adds up to 40 and thus removes the 40 from the 60. The overall vector is still 60, which you see, if you choose comparable frames of reference, namely both at rest at different points, instead of hiding part of the vector in the velocity of coordinates.
I believe that this answer partially obscures a fundamental physics concept, the equivalence of inertial frames. You cannot say that a velocity is 60 m/s without specifically referring to what it is 60 m/s relative to. Usually, this is subtextual and assumed to be relative to the ground rest frame, but it is not generally the case. This is true in classical mechanics as well as in relativity. A frame moving with a constant velocity relative to the ground is also a valid inertial frame and there is a priori no preference for stating that the velocity "is 60 m/s" over the velocity being 40 m/s (or any other value smaller than c). I believe the answer of @stevendaryl is more physically sound. Velocity is a vector under spatial transformations - not under Galilei transformations.

fresh_42
Mentor
I believe the answer of @stevendaryl is more physically sound. Velocity is a vector under spatial transformations - not under Galilei transformations.
I think that, too, so thanks for correction.

WWGD
Gold Member
2019 Award
The difficulty that you're having with velocity as a vector is actually an important insight. Velocity is a vector with respect to spatial coordinate transformations, but not with respect to transformations that involve time.

.
What kind of object is it, if not a vector?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
What kind of object is it, if not a vector?
As @stevendaryl said, it is a vector under spatial transformations. Changes of inertial frames involve Galilei transformations, which are space-time transformations and Galilean space-time does not really work in the same fashion as, e.g., Minkowski space.

WWGD
stevendaryl
Staff Emeritus
What kind of object is it, if not a vector?
Here's the issue: Roughly speaking, a vector (technically, a tangent vector) on a space is like an arrow that points from one location in that space to another. Now, you might think that a velocity vector does that: It points from where an object is located at time $t$ to where it is located at time $t+\delta t$. But the problem is that there is no frame-independent to relate a point at one time to a point at another time. Is it the same point? If so, that means the velocity is zero. Is it a far distant point? If so, the velocity is huge.

So velocity is only a vector if you stick to a specific frame, which allows you to relate points at different times.

WWGD
WWGD
Gold Member
2019 Award
Here's the issue: Roughly speaking, a vector (technically, a tangent vector) on a space is like an arrow that points from one location in that space to another. Now, you might think that a velocity vector does that: It points from where an object is located at time $t$ to where it is located at time $t+\delta t$. But the problem is that there is no frame-independent to relate a point at one time to a point at another time. Is it the same point? If so, that means the velocity is zero. Is it a far distant point? If so, the velocity is huge.

So velocity is only a vector if you stick to a specific frame, which allows you to relate points at different times.
Thanks; somewhat-OT: EDIT Reminds of some bundle objects that keep track of all frames of reference , since there is no preferred frame. There is a group action that takes you from any one frame to another.
https://en.wikipedia.org/wiki/Frame_bundle

Last edited:
stevendaryl
Staff Emeritus
Something that can fix the difficulty is to go to 4-dimensional vectors, rather than 3-dimensional vectors. This is done in SR, but it was never done with Newtonian physics, but in hindsight people could have done it.

Here's the idea: Distinguish between two different notions of "time": There is $t$, which is a coordinate used to locate where you are in spacetime, and another parameter, $s$, which is the elapsed time on your watch. Your trajectory is then defined by 4 different numbers:
1. $\frac{dx}{ds}$
2. $\frac{dy}{ds}$
3. $\frac{dz}{ds}$
4. $\frac{dt}{ds}$
Now, it just happens to be that in Newtonian physics, the last number is a constant, $1$. So the dynamics of the 4th dimension are pretty uninteresting in Newtonian physics. But introducing this 4th dimension has a number of mathematically convenient consequences:
1. The 4-velocity is a true vector (unlike 3-velocity). If you change coordinates from one coordinate system $x,y,z,t$ to another, $x', y', z', t'$, then your 4-velocity changes as follows: $V^{i'} = \sum_j \frac{\partial x^{i'} }{\partial x^j} V^j$
2. The "g-forces" due to an accelerated coordinate system are treated in the same way that fictitious forces such as the Coriolis force and Centifugal force are treated in a rotating coordinate system.