**Required math: algebra, vectors**

**Required physics: basics of relativity**

We’ve seen some of the basics of four-vectors in relativity. A key point is that although the components of a four-vector vary from one inertial frame to another, the vector itself is an entity independent of the frame in which it is described. The Lorentz transformations are used to transform the components of a vector from one frame to another.

However, in each inertial frame there are four special vectors known as the *basis vectors* for that frame. The vectors are defined as follows

In another inertial frame , we can define another set of basis vectors:

We have to be wary of the notation used here. When we were talking about ordinary (that is, non-basis) four-vectors, putting a bar over the superscript index of a vector’s component means that we are talking about that component in the barred reference frame; the vector itself is still the same vector. That is, when we write we are talking about component in frame , but when we write we are talking about component in frame , but both components refer to the same vector . The components of a given vector will vary from one frame to another.

In the definitions of the basis vectors above, however, the set referring to frame is a *different set* of vectors from that referring to frame . Also, each is an entire vector, and not just a component of a vector. The difference is made explicit in two ways: first, there is an arrow over the symbol, and second, the index is given as a subscript rather than a superscript. One of the problems with learning relativity is in getting used to the various subscripts and superscripts, so it is important to realize that and are completely different vectors; one is *not* the transformation of the other into a different reference frame.

We can write any other four-vector in terms of the basis vectors by saying

Note that we are using the summation convention, and that this formula is specific to frame . In frame we would write

This emphasizes again that is the same vector in both cases, but we are expressing it in terms of different sets of basis vectors in the two reference frames. Remember that barred and unbarred indexes are *not* equivalent, so the second formula above is a different relation from the first one.

Although the basis vectors in two frames are different, we can still use the Lorentz transformations to find one set of basis vectors, given the set in another frame. Since the expansions of the vector above must give the same vector in both frames, we can say

This formula says that two different vector sums are equal, not that both sides of the equation are identical.

We can now use the transformation we worked out earlier for the components of a four-vector:

where is the matrix of Lorentz transformations:

Plugging this into 1, we get

Note that the summation convention implies that there is a double sum on the right side of this equation: one over and one over . Since we can relabel pairs of dummy indexes (as long as we don’t swap barred with unbarred indexes), on the right side, we can relabel to and to :

where we have just reordered the terms in the last line.

Since this formula must be true for any vector , it must be true for each component of separately, so we get the transformation formula from to :

It is worth re-emphasizing that this is a transformation which converts an entire vector into another, different, vector, *not* a transformation formula for components of one vector from one frame to another.

The inverse transformations are fairly easy to derive. The Lorentz transformation matrix above is for a transformation from frame to , where the second frame is moving at velocity with respect to the first. To go the other way, we replace by in the transformations, which gives the inverse matrix:

Remember that so has the same value with either of or . It is readily verified by direct matrix multiplication that , the identity matrix. In terms of components, we get

where is the Kronecker delta symbol, and we’ve made explicit the dependence of each matrix element on the direction of the velocity..

We can use this fact to derive the inverse transformations for both the components of a vector and for the basis vectors.

Starting with the formula 2 the inverse transformation of components of a single vector is found by multiplying both sides by the inverse matrix:

For the basis vectors, we start with 3 and multiply by the inverse matrix in the same way

It’s important to notice that the transformations for components of a vector and of the basis vectors are not the same. For example, if we look at the transformations in each case from to , we have, for the components of a vector

and for the basis vectors

The matrix used in one case is the inverse of that used in the other. This shouldn’t be surprising, since the type of transformation being done is different in the two cases.

## Comments

how does one come to an conclusion that a vector itself remains unchanged in different frames ? As from the basic vector notion i had (common length vector) changes its value from one frame to another.

is it a specialty of a four base vector not to change or is it the definition or does it follow from 4 base vector definition ?

It’s by definition of what a vector is. A vector is an entity in itself, the coordinates associated with a given frame+basis are just a particular representation that might differ from frame to frame. On the logical development first you define what a vector is, and it’s quite abstract from particular representations. Only then you figure out that you can represent it in a given frame by the manner we’re used to.

What changes is the way in wich you represent the SAME vector.

Whenever in physics you say something is a vector, your making a statement about reality that exists apart from your choice of frame, or where you put the origin etc. Ofcourse picking different frames would lead to different representations, but of the same reality. The picture of an “arrow” isn’t the vector, it’s just a representation in a particular frame and basis… Just change the basis from having lenght 1 to another lenght, now your visual representation has a dif “scale”, the arrow would look different when drawn.. but it’s still the same object.

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