Vector components under a translation

[Geekster]

Ok...

I am asked how a vector's components transform under a translation of coordinates.

From mathworld:

Since vectors remain unchanged under translation, it is often convenient to consider the tail A as located at the origin when, for example, defining vector addition and scalar multiplication.

Does that imply that the components used to describe the vector remain unchanged?

If you and I see a car drive east at 50 Km/h and you are standing at what you call the origin, and I am standing let's say 5m along what you are calling the positive y-axis, and I call my point the origin with my basis vectors being parrallel to your basis vectors, then do we both give the same components to describe the vector for the car?

To me it seems like this should not always be the case. After all, one position vector has different components then another, right? If we just shift these around by translation, but keep the same direction, then haven't we changed the magnitude from one coordinate system to the next?

 

[nrqed]

Ok...

I am asked how a vector's components transform under a translation of coordinates.

From mathworld:


Does that imply that the components used to describe the vector remain unchanged?

If you and I see a car drive east at 50 Km/h and you are standing at what you call the origin, and I am standing let's say 5m along what you are calling the positive y-axis, and I call my point the origin with my basis vectors being parrallel to your basis vectors, then we do both give the same components to describe the vector for the car?

To me it seems like this should not always be the case. After all, one position vector has different components then another, right? If we just shift these around by translation, but keep the same direction, then haven't we changed the magnitude from one coordinate system to the next?

It all depends on what you mean by "the vector of the car".
The position vector (usually represented by {\vec r}) does depend on the position of the origin. But it does not actually enter any physical equation (it always appears in the combined form {\vec r_f} - {\vec r_i} \equiv \Delta {\vec r} called the displacement vector). All the other physical vectors (velocity, acceleration, etc) are independent of the position of the origin.

Mathematically, a vector is defined an oriented segment with a certain magnitude and direction independent of any coordinate system. (so in some sense, the "position" vector is not truly a vector. The real vector is the displacement vector which does enter the equations of physics but is independent of the choice of coordinate system. )

 

[Geekster]

I still don't get it...

So let's say I have a vector (1,1,1) using standard basis. Now I want to translate the coordinates so my new origin is at (1,1,1) with the basis vectors being parallel to the standard basis vectors. Then is my vector's coordinates still (1,1,1) relative to the new coordiantes?

Wait...I think I see what is meant here. You can shift the the coordinates (by translation at least) around all you want, but the vector just moves along with the coordinates?

If that is correct then it would mean that the magnitude and direction for which the car is moving (above example) would be the same for either observer, only the car's position vector relative to my coordinates would be different than the position vector relative to the other guys coordinates.

So really position doesn't fit the idea of what a vector is, even if under many of the usual vector operations position vectors act like vectors.

Ok...that's kind of an abstract idea, but I think I get it.

 

[nrqed]

I still don't get it...

So let's say I have a vector (1,1,1) using standard basis. Now I want to translate the coordinates so my new origin is at (1,1,1) with the basis vectors being parallel to the standard basis vectors. Then is my vector's coordinates still (1,1,1) relative to the new coordiantes?

Yes. If you visualize vectors, they are directed line segments ("arrows") with a tail and a tip. If you slide the coordinate system, the arrow won't change. The location of the tip and tail *will* change but not the arrow.

It depends if you want to think in terms of mathematics or in terms of physics. But if you use physics, pick a certain vector, say the velocity vector of an object. That won't change under translation.

But the question can be addressed at different levels. Even under rotations of the coordinate system the vector will not change. However, if you write the vector in unit vector notation {\vec A} = A_x {\hat i} + A_y {\hat j} + A_z {\hat k} you have to be careful. The vector {\vec A} is completely independent of the coordinate system. But the values of the *components* do depend on the coordinate system. Under *translations* they do not change but under a rotation of the coordinate system, the *components* will change.

Wait...I think I see what is meant here. You can shift the the coordinates (by translation at least) around all you want, but the vector just moves along with the coordinates?

No, the vector will move with respect to the coordinates.

If that is correct then it would mean that the magnitude and direction for which the car is moving (above example) would be the same for either observer, only the car's position vector relative to my coordinates would be different than the position vector relative to the other guys coordinates.

Yes.

So really position doesn't fit the idea of what a vector is, even if under many of the usual vector operations position vectors act like vectors.

Ok...that's kind of an abstract idea, but I think I get it.

Yes, position is kind of an exception in that sense.

 

[quasar987]

I'm going to say some things, it may or may not clear up your confusion but hopeful it will!

Say in coordinate system S, the position vector of some object is \vec{r}. If we look at this same object from the coordinate system S', which is obtained by translating the origin of S by a vector \vec{d}. That is to say, in S, the position vector of the origin of S' is \vec{d}. Then the position of the object in S' is given by \vec{r} \ '=-\vec{d}+\vec{r}. Arrange so you see that clearly. Draw the two coordinate systems and all the vectors. This is the transformation equation you are looking for.

If the object is moving, and its path in S is \vec{r}(t), then its velocity in S is d\vec{r}/dt. In S' its velocity is then

\frac{d\vec{r} \ '}{dt}=-\frac{d\vec{d}}{dt}+\frac{d\vec{r}}{dt} = 0+\frac{d\vec{r}}{dt}=\frac{d\vec{r}}{dt}

I.e. the components of the velocity vector are the same wether we look at the object from one coordinate syetem or from another one translated with respect to the first.