Vectors & Frames Qs: Galilean & Lorentz Transforms

[Mathematicsresear]

Homework Statement

Imagine there exists reference two frames, a frame O which is stationary and another, O' moving relative to O. If there is a four vector A--> (A⁰,A¹,A²,A³). Then why is A^ae^->_a as measured by the observer O equal to A^a'e^->_a' as measured by the observer O'?

The components are different, I know that the lengths of the vectors are the same, regardless of rotation or translation but assuming Galilean relativity, shouldn't velocity vectors not be equal in both frames? Does that mean velocity is not a four component vector? If not, why not? if so, why?

so if I applied the Lorentz transformation, A^a'e^->_a' = Λ_b^a'A^be_a', why is that important?

Are the vectors equal if and only if I applied the Lorentz transformation? If so why? if not? why not?

 

[Gene Naden]

Velocity is not a four-vector because it doesn't take into account time dilation. The four-velocity is
$\frac{d\vec{x}}{d\tau}$ where $\tau$ is the proper time.

Are the vectors equal if and only if I applied the Lorentz transformation? If so why? if not? why not?

I don't understand this part of your question. In some cases a four-vector corresponds to an event in space-time. That event is the same in all frames of reference (in all coordinate systems) although the components with respect to different coordinate systems are different.

 

[vela]

You have to transform the basis vectors as well.

 

[Dale]

The components are different, I know that the lengths of the vectors are the same, regardless of rotation or translation but assuming Galilean relativity, shouldn't velocity vectors not be equal in both frames?

As @vela said, you need to include the basis vectors. The basis vectors in one frame can each be written as a linear combination of the basis vectors in another frame. The coordinate transformation, by design, takes the coordinates in one frame and returns the coordinates of the exact same vector in the other frame. The vector itself is the same vector in both frames.

 

[Ibix]

On a 2d plane, you can draw a vector as an arrow, and write it as $v=v^x\mathbf i+v^y\mathbf j$. That's interpreted as "move $v^x$ times the length of $\mathbf i$ in the direction of $\mathbf i$ and $v^y$ times the length of $\mathbf j$ in the direction of $\mathbf j$". This is simply a rather long-winded version of $v=v^ie_i$.

But there's no obligation to use any particular set of basis vectors. You could decide that instead of $\mathbf i$ and $\mathbf j$ you would prefer to use $\mathbf k$ and $\mathbf l$. Then your vector can be written as $v=v^{x'}\mathbf k+v^{y'}\mathbf l$. This is a long-winded way of writing $v=v^{a'}e_{a'}$. It's still the same vector, though. All you did was change your description of it.

As others have noted, you can transform your basis vectors too - write $\mathbf k$ and $\mathbf l$ in terms of $\mathbf i$ and $\mathbf j$. Your expression is correct, but it's a bit odd in this context to express components as an explicit transform from O to O' but leave the basis vectors in terms of O'.

You are correct that Galilean relativity does not preserve the lengths of vectors. I think this is because they feature time derivatives, and time is not a part of the vector space (like the time-like basis vector in relativity) nor is it an affine parameter (like proper time). But in Einsteinian relativity, everything of interest is a geometric object of some kind in a 4d vector space. So everything behaves like (a more complicated version of) the vectors on my 2d plane.

 

[Orodruin]

I think this is because they feature time derivatives, and time is not a part of the vector space (like the time-like basis vector in relativity) nor is it an affine parameter (like proper time).

Time certainly is a part of Galilean spacetime and you need the transformation of the time basis vector in order to get the transformation properties correctly. For example, the Galilei transformation of the basis vectors are $\vec e_t' = \vec e_t + \vec v$ and $\vec e_i' = \vec e_i$. The 4-velocity of an object is given by $\vec U = \vec e_t + \vec u$, where $\vec u$ is the 3-velocity in the unprimed frame. Expressing the unprimed basis vectors in the primed ones leads to
$$
\vec U = \vec e_t' - \vec v + \vec u
$$
and hence the 3-velocity in the primed frame is $\vec u - \vec v$ (as it should). The spatial metric that you use to determine the length of 3-vectors and the 3-vector corresponding to a 4-vector only has its magnitude conserved if the 4-vector has a time-component that is zero, e.g., the 4-acceleration is
$$
\vec A = \frac{d\vec U}{dt} = 0\vec e_t + \vec a = 0 \vec e_t' + \vec a.
$$
I guess my main point is that Galilean spacetime is also an affine space, just as Minkowski space or the Euclidean plane, and you can deal with basis transformations in precisely the same way. What differs is the additional structure you put on the different spaces.

 

[haushofer]

Maybe page 37,38 of my thesis can help regarding the precise form of the Galilean transformations:

https://www.rug.nl/research/portal/files/34926446/Complete_thesis.pdf

Maybe I'm missing the point, but Galilean boosts are fundamentally different from Lorentz boosts. Galilean boosts only transform spatial coordinates to time coordinates, but not the other way around! So you have to be very careful if you want to calculate how the inner product of two vectors transform under a Galilean boost.

 

[Orodruin]

So you have to be very careful if you want to calculate how the inner product of two vectors transform under a Galilean boost.

To me the point is that Galilean spacetime does not have an inner product on its tangent space. The (spatial) inner product only cares about the spatial subspace.

Galilean boosts only transform spatial coordinates to time coordinates, but not the other way around!

The fun part of this is that the exact opposite is true for the basis vectors. The spatial basis vectors remain spatial whereas the time basis of different inertial frames differ by a spatial vector.