37 |
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2.2 Vector algebra |
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In the last line we use the summation convention (remember always to write the index on e as a subscript in order to employ the convention in this manner). The meaning of Eq. (2.11)
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A is the linear sum of four vectors A0e , A1e , etc. |
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is that |
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Transformation of basis vectors
The discussion leading up to Eq. (2.11) could have been applied to any frame, so it is equally true in O¯ :
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α |
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A |
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Aα¯ e |
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¯ |
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A is also the sum of the four vectors A0¯ |
1¯ |
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e1, etc. These are not the same |
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This says that |
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, A |
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O |
four vectors as in Eq. (2.11), since they are parallel to the basis vectors of ¯ and not of |
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A. It is important to understand that the expressions |
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O, but they add up to the same vector |
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are not obtained from one another merely by relabeling dummy indices. |
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Aα eα |
and Aα¯ eα |
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Barred and unbarred indices cannot be interchanged, since they have different meanings.
Thus, |
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is a different set of numbers from |
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, just as the set of vectors |
{ ¯ |
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is different |
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eα |
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from {eα }. But, by definition, the two sums are the same: |
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Aα e |
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Aα¯ e |
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(2.12) |
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α = |
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α¯ |
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and this has an important consequence: from it we deduce the transformation law for the basis vectors, i.e. the relation between {eα } and {eα¯ }. Using Eq. (2.7) for Aα¯ , we write Eq. (2.12) as
α¯ β Aβ eα¯ = Aα eα .
On the left we have two sums. Since they are finite sums their order doesn’t matter. Since the numbers α¯ β and Aβ are just numbers, their order doesn’t matter, and we can write
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Aβ α¯ |
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e |
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Aα e . |
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β α¯ |
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Now we use the fact that β and |
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α are dummy indices: we change β to α and α to |
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α e |
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A |
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eα . |
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This equation must be true for all sets |
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A is an arbitrary vector. Writing it as |
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A ( ¯ α e |
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eα ) 0 |
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we deduce |
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0 for every value of α, |
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α e |
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eα |
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or |
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(2.13) |
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eα |
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38 |
Vector analysis in special relativity |
This gives the law by which basis vectors change. It is not a component transformation: it gives the basis {eα } of O as a linear sum over the basis {eα¯ } of O¯ . Comparing this to the law for components, Eq. (2.7),
A |
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α A , |
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α |
we see that it is different indeed.
The above discussion introduced many new techniques, so study it carefully. Notice that the omission of the summation signs keeps things neat. Notice also that a step of key importance was relabeling the dummy indices: this allowed us to isolated the arbitrary Aα from the rest of the things in the equation.
An example
O |
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O |
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. Then the matrix [ ¯ α ] is |
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Let ¯ move with velocity v in the x direction relative to |
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( ¯ α ) |
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where we use the standard notation |
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γ := 1/√(1 − v2). |
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A |
(5, 0, 0, 2), we find its components in |
O |
by |
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Then, if |
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O
¯ ¯ ¯
A0 = 00A0 + 01A1 + · · ·
=γ · 5 + (−vγ ) · 0 + 0 · 0 + 0 · 2
=5γ .
Similarly,
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A1¯ = −5vγ , |
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A2¯ = 0, |
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A3¯ = 2. |
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5vγ , 0, 2). |
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Therefore, |
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O |
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The basis vectors are expressible as |
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eα |
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α e |
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or
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e0 = 00e¯ + 10e¯ + · · ·
0 1
= γ e¯ − vγ e¯ .
0 1
39 |
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2.2 Vector algebra |
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Similarly, |
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e1 = −vγ e0¯ + γ e1¯ , |
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= e3¯ . |
This gives O’s basis in terms of O¯ ’s, so let us draw the picture (Fig. 2.1) in O¯ ’s frame: This transformation is of course exactly what is needed to keep the basis vectors pointing along the axes of their respective frames. Compare this with Fig. 1.5(b).
Inverse transformations
¯
The only thing the Lorentz transformation β α depends on is the relative velocity of the two frames. Let us for the moment show this explicitly by writing
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α (v). |
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β |
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β |
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Then |
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eα |
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If the basis of O is obtained from that of O¯ by the transformation with velocity v, then the reverse must be true if we use −v. Thus we must have
eμ¯ = ν μ¯ (−v)eν . |
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In this equation I have used μ¯ and ν as indices to avoid confusion with the previous formula. The bars still refer, of course, to the frame O¯ . The matrix [ ν μ¯ ] is exactly the
matrix [ ¯ α ] except with changed to − . The bars on the indices only serve to indicate
β v
v
the names of the observers involved: they affect the entries in the matrix [ ] only in that the matrix is always constructed using the velocity of the upper-index frame relative to the
t |
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e0 |
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e0 |
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e1 |
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Figure 2.1 |
Basis vectors of |
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and |
O¯ |
O¯ |
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as drawn by . |
40 |
Vector analysis in special relativity |
lower-index frame. This is made explicit in Eqs. (2.14) and (2.15). Since v is the velocity of O¯ (the upper-index frame in Eq. (2.14)) relative to O, then −v is the velocity of O (the upper-index frame in Eq. (2.15)) relative to O¯ . Exer. 11, § 2.9, will help you understand this point.
We can rewrite the last expression as
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v)eν . |
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Here we have just changed μ to |
¯. This doesn’t change anything: it is still the same four |
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equations, one for each value of |
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¯. In this form we can put it into the expression for eα |
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Eq. (2.14): |
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eα = β α (v)e ¯
β
¯
= β α (v) ν ¯ (−v)eν . (2.16)
β
In this equation only the basis of O appears. It must therefore be an identity for all v. On
the right there are two sums, one on |
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first, then the right is a sum over the basis {eν } in which each basis vector eν has coefficient
¯ |
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β
Imagine evaluating Eq. (2.16) for some fixed value of the index α. If the right side of Eq. (2.16) is equal to the left, the coefficient of eα on the right must be 1 and all other coefficients must vanish. The mathematical way of saying this is
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β α (v) ν ¯ (−v) = δν α ,
β
where δν α is the Kronecker delta again. This would imply
eα = δν α eν ,
which is an identity.
Let us change the order of multiplication above and write down the key formula
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This expresses the fact that the matrix [ ν |
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sum on ¯ is exactly the operation we perform when we multiply two matrices. The matrix (δν α ) is, of course, the identity matrix.
The expression for the change of a vector’s components, |
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(v)A |
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also has its inverse. Let us multiply both sides by ν |
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v |
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β ( |
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(v)A |
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= δν α Aα
= Aν .