Four Vector Algebra (Part 6 of the series)

/ nihan10

Part 5

Last time, we have seen what four vectors are, and a small introduction to how we can use use them. Here, we will look into more detail into the working of four vectors, and lay the grounds for the interested reader to delve into tensor algebra.

Before moving on though, I recommend you take a look at this article. It is just a recap of all the formulae that we have encountered up until now, and mainly, I have elaborated on how to put in the speed of light ‘c’ into the equations.

Managing units in equations of SR

Four Vector Algebra:

Before we start, lets set a few notations from the beginning. I will use latin alphabets (a,b,c…..i,j,k…..z) as labels for the space coordinates (i.e. 1, 2, 3) only and greek alphabets (\alpha, \beta, ....... \omega ) to denote spacetime coordinates. Also, just as a small reminder on the Einstein summation convention, all the following expressions are equivalent:

x'^{\bar{\alpha}} = \Lambda_{\beta}^{\bar{\alpha}} x^{\beta} = \Lambda_{\gamma}^{\bar{\alpha}} x^{\gamma}

and also, to illustrate the dummy and “non-dummy” indices:

\displaystyle x'^{\bar{\theta}} = \displaystyle \Lambda_{\omega}^{\bar{\theta}} x^{\omega}

where \bar{\theta} is the “non dummy” index and must be maintained carefully, while \omega is the dummy index.

Also, for our case of motion along X axis, it can be illustrated by the following figure:

Now that we are all set, lets move on.

A general four vector is defined by a collection of numbers, i.e. its components in some frame O:

\vec{A} \longrightarrow \ (A^0, A^1, A^2, A^3) = \{A^{\alpha}\}

and by the rule that its components in another frame O’ are given by:

A^{\bar{\alpha}} = \Lambda_{\beta}^{\bar{\alpha}} x^{\beta}

That is, its components transform in the same way as the co-ordinates do. Also, if the components of the vector are determined in one frame of reference, then its components in all other frames of reference are uniquely determined through the Lorentz transformation.

Basically, four vectors are just like three vectors and follow almost all the laws applicable to three vectors. The basic rules obeyed by four vectors are:

\vec{A} + \vec{B} \longrightarrow \ (A^0 + B^0, A^1 + B^1, A^2 + B^2, A^3 + B^3)

\mu \vec{A} \longrightarrow \ (\mu A^0, \mu A^1, \mu A^2, \mu A^3)

where \mu is a scalar.

Thus, four vectors add according to the parallelogram rule. Notice, that, as is in the case of three vectors, if all the components of the four vector are not dimensionless, then they must have the same dimensions, as they cannot be added together otherwise. This was our main motivation to express time in ‘meters’ so that we could create the displacement four vector.

From now on, unless explicitly stated, we will call four vectors simply as vectors.

Basis vectors:

As the more observant of you might have noticed, we are slowly heading towards a formalism that allows us to express the vectors in the form of arrays, and the Lorentz transformations in the form of a matrix which links the vectors from two different frames. That is actually the motive behind introducing such abstraction, i.e. so that we can use the techniques of linear algebra. To do this, we will need to introduce the concept of basis vectors.

Basis vectors, in any coordinate system O have the following form:

\vec{e_0} \longrightarrow \ (1, 0, 0, 0)

\vec{e_1} \longrightarrow \ (0, 1, 0, 0)

\vec{e_2} \longrightarrow \ (0, 0, 1, 0)

\vec{e_3} \longrightarrow \ (0, 0, 0, 1)

Thus, basis vectors are just the generalization of the \hat{i},\ \hat{j},\ \hat{k}  that we have used in ordinary three vector algebra. Each of the basis vector represents one axis of whichever coordinate system we happen to be in.

Similarly, in the frame O’, the basis vectors will have the same form, with only the indices expressed with bars (like \bar{0} \ ) instead of unbarred indices to indicate that the frames are different. Also, generally, the two basis vectors of different frames will not be equal to each other. We can thus succinctly state that the definition of basis vectors is equivalent to:
(\vec{e_{\alpha}})^{\beta} = \delta_{\alpha}^{\beta}              where  \delta_{\alpha}^{\beta} \longrightarrow \ \textrm{Kronecker delta}

i.e. the\beta component of \vec{e_{\alpha}} is 1 iff \alpha = \beta , otherwise it is 0.

The main use of these basis vectors is that we can express any vector in a particular frame of reference in terms of the basis vectors for that particular frame. Thus,

\vec{A} = A^{\alpha} \vec{e_{\alpha}} = A^{0}\vec{e_0} + A^{1}\vec{e_{1}} + A^{2}\vec{e_{2}} + A^{3}\vec{e_{3}}

(always remember to write the index on \vec{e} as a subscript, allowing us to use the summation convention)

From hereon in, I will drop the arrow sign over the basis vectors due to some problem with the LaTeX rendering of WordPress.

Transforming basis vectors:

The above equation is applicable in other frames also. Thus,

\vec{A} = A^{\bar{\alpha}} e_{\bar{\alpha}}

Notice one important thing though. The above equation might appear deceptively easy to obtain by just substituting the unbarred indices with the barred ones. However, it is not advisable to do that, as both the components and the basis vectors differ in two different frames. However, by definition, the resultant vector obtained from the entire sum will be the same. Thus,

\vec{A} = A^{\alpha} e_{\alpha} = A^{\bar{\alpha}} e_{\bar{\alpha}}

From this equation, we can determine how the basis vectors transform under a Lorentz transformation. Lets look at the entire derivation so that it will give you an idea of how to work with the index notation. So here we go:

first of all, lets convert the components A^{\bar{\alpha}} back into the unbarred components using the Lorentz transformation that we saw in the last part:

A^{\alpha} e_{\alpha} = \Lambda_{\beta}^{\bar{\alpha}} A^{\beta} e_{\bar{\alpha}}

Next, (and pay attention to this step because this is used many times in such manipulations) we interchange the dummy indices on the RHS:

A^{\alpha} e_{\alpha} = \Lambda_{\alpha}^{\bar{\beta}} A^{\alpha} e_{\bar{\beta}}

Next, as all the components on the RHS are just numbers, their ordering does not matter:

A^{\alpha} e_{\alpha} = A^{\alpha} \Lambda_{\alpha}^{\bar{\beta}} e_{\bar{\beta}}

Taking everything on to the LHS and simplifying a bit:

A^{\alpha} (e_{\alpha} - \Lambda_{\alpha}^{\bar{\beta}} e_{\bar{\beta}}) = 0

[now you see why we interchanged the dummy indices earlier]

Now, this equation must hold for all \vec{A} as we assumed it as arbitrary. That gives us:

e_{\alpha} = \Lambda_{\alpha}^{\bar{\beta}} e_{\bar{\beta}}                 for every value of \alpha

This gives us how the basis vectors transform from one frame to another under a Lorentz transformation.

Note, that this NOT a component transformation. We are writing the basis \{ e_{\alpha} \} of O as a linear sum over the basis \{ e_{\bar{\alpha}} \} of O’.

This transformation is just what is required to keep the vector \vec{A} pointing in the same direction in both the frames. The component transformation gives the components of the vector on these new basis vectors.

We are halfway through the four vector algebra that we will require for our computations of momentum and energy later on. In the next part, we will look at the inverse Lorentz transformations and how to represent them in the above formalism. We will also look at the scalar product, and add some finishing touches to four vector algebra.

Part 7

3 thoughts on “Four Vector Algebra (Part 6 of the series)”

  1. Pingback: A no nonsense introduction to Special Relativity Part 5 | Proper Physics

  2. Pingback: A no nonsense introduction to Special Relativity Part 7 | Proper Physics

  3. Pingback: Basis vectors and inner product | Physics Forums

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