Four-vectors: basis vectors

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

\displaystyle   \vec{e}_{0} \displaystyle  \stackrel{\mathcal{O}}{\rightarrow} \displaystyle  (1,0,0,0)
\displaystyle  \vec{e}_{1} \displaystyle  \stackrel{\mathcal{O}}{\rightarrow} \displaystyle  (0,1,0,0)
\displaystyle  \vec{e}_{2} \displaystyle  \stackrel{\mathcal{O}}{\rightarrow} \displaystyle  (0,0,1,0)
\displaystyle  \vec{e}_{3} \displaystyle  \stackrel{\mathcal{O}}{\rightarrow} \displaystyle  (0,0,0,1)

In another inertial frame {\bar{\mathcal{O}}}, we can define another set of basis vectors:

\displaystyle   \vec{e}_{\bar{0}} \displaystyle  \stackrel{\mathcal{\bar{\mathcal{O}}}}{\rightarrow} \displaystyle  (1,0,0,0)
\displaystyle  \vec{e}_{\bar{1}} \displaystyle  \stackrel{\mathcal{\bar{\mathcal{O}}}}{\rightarrow} \displaystyle  (0,1,0,0)
\displaystyle  \vec{e}_{\bar{2}} \displaystyle  \stackrel{\mathcal{\bar{\mathcal{O}}}}{\rightarrow} \displaystyle  (0,0,1,0)
\displaystyle  \vec{e}_{\bar{3}} \displaystyle  \stackrel{\mathcal{\bar{\mathcal{O}}}}{\rightarrow} \displaystyle  (0,0,0,1)

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 {A^{\alpha}} we are talking about component {\alpha} in frame {\mathcal{O}}, but when we write {A^{\bar{\alpha}}} we are talking about component {\bar{\alpha}} in frame {\bar{\mathcal{O}}}, but both components refer to the same vector {\vec{A}}. 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 {\bar{\mathcal{O}}} is a different set of vectors from that referring to frame {\mathcal{O}}. Also, each {\vec{e}_{\alpha}} 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 {\vec{e}_{\alpha}} and {\vec{e}_{\bar{\alpha}}} 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

\displaystyle  \vec{A}\stackrel{\mathcal{O}}{\rightarrow}A^{\alpha}\vec{e}_{\alpha}

Note that we are using the summation convention, and that this formula is specific to frame {\mathcal{O}}. In frame {\bar{\mathcal{O}}} we would write

\displaystyle  \vec{A}\stackrel{\mathcal{\bar{\mathcal{O}}}}{\rightarrow}A^{\bar{\alpha}}\vec{e}_{\bar{\alpha}}

This emphasizes again that {\vec{A}} 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 {\vec{A}} above must give the same vector in both frames, we can say

\displaystyle  A^{\alpha}\vec{e}_{\alpha}=A^{\bar{\alpha}}\vec{e}_{\bar{\alpha}}\ \ \ \ \ (1)
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:

\displaystyle  A^{\bar{\alpha}}=\Lambda_{\;\beta}^{\bar{\alpha}}A^{\beta}\ \ \ \ \ (2)
where {\Lambda} is the matrix of Lorentz transformations:

\displaystyle  \Lambda=\left(\begin{array}{cccc} \gamma & -v\gamma & 0 & 0\\ -v\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{array}\right)

Plugging this into 1, we get

\displaystyle  A^{\alpha}\vec{e}_{\alpha}=\Lambda_{\;\beta}^{\bar{\alpha}}A^{\beta}\vec{e}_{\bar{\alpha}}

Note that the summation convention implies that there is a double sum on the right side of this equation: one over {\bar{\alpha}} and one over {\beta}. 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 {\beta} to {\alpha} and {\bar{\alpha}} to {\bar{\beta}}:

\displaystyle   A^{\alpha}\vec{e}_{\alpha} \displaystyle  = \displaystyle  \Lambda_{\;\alpha}^{\bar{\beta}}A^{\alpha}\vec{e}_{\bar{\beta}}
\displaystyle  \displaystyle  = \displaystyle  A^{\alpha}\Lambda_{\;\alpha}^{\bar{\beta}}\vec{e}_{\bar{\beta}}

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

Since this formula must be true for any vector {\vec{A}}, it must be true for each component of {\vec{A}} separately, so we get the transformation formula from {\bar{\mathcal{O}}} to {\mathcal{O}}:

\displaystyle  \vec{e}_{\alpha}=\Lambda_{\;\alpha}^{\bar{\beta}}\vec{e}_{\bar{\beta}}\ \ \ \ \ (3)

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 {\mathcal{O}} to {\bar{\mathcal{O}}}, where the second frame is moving at velocity {+v} with respect to the first. To go the other way, we replace {v} by {-v} in the transformations, which gives the inverse matrix:

\displaystyle  \Lambda^{-1}=\left(\begin{array}{cccc} \gamma & v\gamma & 0 & 0\\ v\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{array}\right)
Remember that {\gamma=1/\sqrt{1-v^{2}}} so has the same value with either of {v} or {-v}. It is readily verified by direct matrix multiplication that {\Lambda^{-1}\Lambda=\Lambda\Lambda^{-1}=I}, the identity matrix. In terms of components, we get

\displaystyle   \Lambda_{\;\alpha}^{\bar{\beta}}(v)\Lambda_{\;\bar{\gamma}}^{\alpha}(-v) \displaystyle  = \displaystyle  \delta_{\;\bar{\gamma}}^{\bar{\beta}}
\displaystyle  \Lambda_{\;\bar{\alpha}}^{\beta}(-v)\Lambda_{\;\gamma}^{\bar{\alpha}}(v) \displaystyle  = \displaystyle  \delta_{\;\gamma}^{\beta}

where {\delta_{\;\bar{\gamma}}^{\bar{\beta}}} 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:

\displaystyle   \Lambda_{\;\bar{\alpha}}^{\gamma}(-v)A^{\bar{\alpha}} \displaystyle  = \displaystyle  \Lambda_{\;\bar{\alpha}}^{\gamma}(-v)\Lambda_{\;\beta}^{\bar{\alpha}}(v)A^{\beta}
\displaystyle  \displaystyle  = \displaystyle  \delta_{\;\beta}^{\gamma}A^{\beta}
\displaystyle  \displaystyle  = \displaystyle  A^{\gamma}

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

\displaystyle   \Lambda_{\;\bar{\gamma}}^{\alpha}(-v)\vec{e}_{\alpha} \displaystyle  = \displaystyle  \Lambda_{\;\bar{\gamma}}^{\alpha}(-v)\Lambda_{\;\alpha}^{\bar{\beta}}(v)\vec{e}_{\bar{\beta}}
\displaystyle  \displaystyle  = \displaystyle  \delta_{\;\bar{\gamma}}^{\bar{\beta}}\vec{e}_{\bar{\beta}}
\displaystyle  \displaystyle  = \displaystyle  \vec{e}_{\bar{\gamma}}

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 {\mathcal{O}} to {\bar{\mathcal{O}}}, we have, for the components of a vector

\displaystyle  A^{\bar{\alpha}}=\Lambda_{\;\beta}^{\bar{\alpha}}(v)A^{\beta}
and for the basis vectors

\displaystyle  \vec{e}_{\bar{\alpha}}=\Lambda_{\;\bar{\alpha}}^{\beta}(-v)\vec{e}_{\beta}

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.

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Comments

  • Santosh Kumar  On Wednesday, 9 January 2013 at 09:44

    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 ?

    • Carlos Jorge  On Thursday, 14 February 2013 at 15:53

      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.

Trackbacks

  • By Tensors: one-forms « Physics tutorials on Monday, 25 April 2011 at 13:38

    [...] are its values when it operates on the four basis vectors in a particular frame. Recall that these basis vectors were defined by the symbol and their transformations were worked out using the Lorentz [...]

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