September 21, 2016 9:3  WSPC/Book Trim Size for 9in x 6in  sol03

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Electrodynamics A (PHY 5346) Fall 2016 Solutions Set 2:

(7) Excercise (CW 5 points): Lorentz group Lie matrix generator. The Matrix L is:

L = ( \begin{matrix}l_{0}^{0} & l_{1}^{0} & l_{2}^{0} & l_{3}^{0} \\ l_{0}^{1} & l_{1}^{1} & l_{2}^{1} & l_{3}^{1} \\ l_{0}^{2} & l_{1}^{2} & l_{2}^{2} & l_{3}^{2} \\ l_{0}^{3} & l_{1}^{3} & l_{2}^{3} & l_{3}^{3} \end{matrix} ) .

(1) The matrix g is:

g = ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) .

By matrix multiplication:

−gL = ( \begin{matrix}−l_{0}^{0} & −l_{1}^{0} & −l_{2}^{0} & −l_{3}^{0} \\ l_{0}^{1} & l_{1}^{1} & l_{2}^{1} & l_{3}^{1} \\ l_{0}^{2} & l_{1}^{2} & l_{2}^{2} & l_{3}^{2} \\ l_{0}^{3} & l_{1}^{3} & l_{2}^{3} & l_{3}^{3} \end{matrix} ) .

(2) The transpose of L is:

L̃ = (\begin{matrix}l̃_{0}^{0} & l̃_{0}^{1} & l̃_{0}^{2} & l̃_{0}^{3} \\ l̃_{1}^{0} & l̃_{1}^{1} & l̃_{1}^{2} & l̃_{1}^{3} \\ l̃_{2}^{0} & l̃_{2}^{1} & l̃_{2}^{2} & l̃_{2}^{3} \\ l̃_{3}^{0} & l̃_{3}^{1} & l̃_{3}^{2} & l̃_{3}^{3} \end{matrix} ) = ( \begin{matrix}l_{0}^{0} & l_{0}^{1} & l_{0}^{2} & l_{0}^{3} \\ l_{1}^{0} & l_{1}^{1} & l_{1}^{2} & l_{1}^{3} \\ l_{2}^{0} & l_{2}^{1} & l_{2}^{2} & l_{2}^{3} \\ l_{3}^{0} & l_{3}^{1} & l_{3}^{2} & l_{3}^{3} \end{matrix} ) .

(3) By matrix multiplication:

L̃g = ( \begin{matrix}l_{0}^{0} & −l_{0}^{1} & −l_{0}^{2} & −l_{0}^{3} \\ l_{1}^{0} & −l_{1}^{1} & −l_{1}^{2} & −l_{1}^{3} \\ l_{2}^{0} & −l_{2}^{1} & −l_{2}^{2} & −l_{2}^{3} \\ l_{3}^{0} & −l_{3}^{1} & −l_{3}^{2} & −l_{3}^{3} \end{matrix} ) .

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(4) Set −gL = L̃g and compare the components:

( \begin{matrix}−l_{0}^{0} = l_{0}^{0} & −l_{1}^{0} = −l_{0}^{1} & −l_{2}^{0} = −l_{0}^{2} & −l_{3}^{0} = −l_{0}^{3} \\ l_{0}^{1} = l_{1}^{0} & l_{1}^{1} = −l_{1}^{1} & l_{2}^{1} = −l_{1}^{2} & l_{3}^{1} = −l_{1}^{3} \\ l_{0}^{2} = l_{2}^{0} & l_{1}^{2} = −l_{2}^{1} & l_{2}^{2} = −l_{2}^{2} & l_{3}^{2} = −l_{2}^{3} \\ l_{0}^{3} = l_{3}^{0} & l_{1}^{3} = −l_{3}^{1} & l_{2}^{3} = −l_{3}^{2} & l_{3}^{3} = −l_{3}^{3} \end{matrix} ) .

The resulting matrix is:

( \begin{matrix}0 & l_{1}^{0} & l_{2}^{0} & l_{3}^{0} \\ l_{1}^{0} & 0 & l_{2}^{1} & l_{3}^{1} \\ l_{2}^{0} & −l_{2}^{1} & 0 & l_{3}^{2} \\ l_{3}^{0} & −l_{3}^{1} & −l_{3}^{2} & 0 \end{matrix} ) .

(8) Excercise (5 points): Obtain the result of the previous excercise discussing the elements of g^{αβ} l̃_{β}^{γ} g_{γδ} = −l_{δ}^{α}. Performing the contractions we get l̃_{δ}^{α} = −l_{δ}^{α}. Using now the definition l̃_{δ}^{α} = l_{δ}^{α}, the following equations are obtained (no summations, a = 0, 1, 2, 3, i = 1, 2, 3 and j = 1, 2, 3):

l̃_{a}^{a} = l_{a}^{a} = +l_{a}^{a} ⇒ l_{a}^{a} = −l_{a}^{a} = 0 ,

l̃_{0}^{i} = l_{0}^{i} = −l_{i}^{0} ⇒ l_{i}^{0} = +l_{0}^{i} ,

l̃_{j}^{i} = l_{j}^{i} = +l_{i}^{j} ⇒ l_{i}^{j} = −l_{j}^{i}

and the resulting matrix is the same as before.

(9) Excercise: A rotation and a boost generator.

For S_{3} we have

S_{3} = (\begin{matrix}0 & 0 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) and (S_{3})^{2} = −1_{2,3} with 1_{2,3} = (\begin{matrix}0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) .

Therefore, for n = 1, 2, 3, …

(S_{3})^{2n} = (−1)^{n} 1_{2,3} and (S_{3})^{2n+1} = (−1)^{n} S_{3}

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and the result follows:

e^{−ωS_{3}} = 1 + ∑_{n=1}^{∞}(−1)^{n} \frac{ω^{2n}}{(2n)!} 1_{2,3} − ∑_{n=0}^{∞}(−1)^{n} \frac{ω^{2n+1}}{(2n + 1)!} S_{3}

= (\begin{matrix}1 & 0 & 0 & 0 \\ 0 & cos ω & sin ω & 0 \\ 0 & − sin ω & cos ω & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) .

Similarly we deal with K_{1}:

K_{1} = (\begin{matrix}0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) and (K_{1})^{2} = 1_{1,2} with 1_{1,2} = (\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) .

Therefore, for n = 1, 2, 3, …

(K_{1})^{2n} = 1_{1,2} and (K_{1})^{2n+1} = K_{1}

and

e^{−ζK_{1}} = 1+∑_{n=1}^{∞} \frac{ω^{2n}}{(2n)!} 1_{1,2}−∑_{n=0}^{∞} \frac{ω^{2n+1}}{(2n + 1)!} K_{1} = (\begin{matrix}cosh ζ & − sinh ζ & 0 & 0 \\ − sinh ζ & cosh ζ & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) .

(10) Excercise: Four-dimensional Levi-Civita tensor. (1) Using the Einstein summation convention of the left-hand side, but not on the right-hand side, of the following equation we have

g_{α_{1}α_{2}}g_{β_{1}β_{2}}g_{γ_{1}γ_{2}}g_{δ_{1}δ_{2}}ϵ^{α_{2}β_{2}γ_{2}δ_{2}} = g_{α_{1}α_{1}}g_{β_{1}β_{1}}g_{γ_{1}γ_{1}}g_{δ_{1}δ_{1}}ϵ^{α_{1}β_{1}γ_{1}δ_{1}} .

The right-hand side is zero unless α_{1}β_{1}γ_{1}δ_{1} are a permutation π_{0}π_{1}π_{2}π_{3} the numbers 0, 1, 2, 3 and in that case

ϵ_{π_{0}π_{1}π_{2}π_{3}} = g_{π_{0}π_{0}}g_{π_{1}π_{1}}g_{π_{1}π_{2}}g_{π_{1}π_{3}}ϵ^{π_{0}π_{1}π_{2}π_{3}} = (−1)^{3} ϵ^{π_{0}π_{1}π_{2}π_{3}} ,

which proves ϵ_{αβγδ} = −ϵ^{αβγδ}.

(2) We have ϵ_{αβγ_{1}δ_{1}}ϵ^{αβγ_{2}δ_{2}} = 0 unless either γ_{1} = γ_{2}, δ_{1} = δ_{2} or γ_{1} = δ_{2}, δ_{1} = γ_{2} or (otherwise one of the already used numbers will be repeated by α or β of the sums). Therefore,

ϵ_{αβγ_{1}δ_{1}}ϵ^{αβγ_{2}δ_{2}} = a δ_{γ_{1}}^{γ_{2}} δ_{δ_{1}}^{δ_{2}} + b δ_{γ_{1}}^{δ_{2}} δ_{δ_{1}}^{γ_{2}}

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holds. With no summation in the permutation indices π_{0} to π_{3} the constants follow from

ϵ_{αβπ_{2}π_{3}}ϵ^{αβπ_{2}π_{3}} = ϵ_{π_{0}π_{1}π_{2}π_{3}}ϵ^{π_{0}π_{1}π_{2}π_{3}} + ϵ_{π_{1}π_{0}π_{2}π_{3}}ϵ^{π_{1}π_{0}π_{2}π_{3}} = −2 = a , ϵ_{αβπ_{2}π_{3}}ϵ^{αβπ_{3}π_{2}} = ϵ_{π_{0}π_{1}π_{2}π_{3}}ϵ^{π_{0}π_{1}π_{3}π_{2}} + ϵ_{π_{1}π_{0}π_{2}π_{3}}ϵ^{π_{1}π_{0}π_{3}π_{2}} = +2 = b .

(3) We have ϵ_{αβ_{1}γ_{1}δ_{1}}ϵ^{αβ_{2}γ_{2}δ_{2}} = 0 unless α_{2}β_{2}γ_{2} is a permutation of α_{1}β_{1}γ_{1}. There are six such permutations, so that the results is a sum of the form

ϵ_{αβ_{1}γ_{1}δ_{1}}ϵ^{αβ_{2}γ_{2}δ_{2}} = a_{1} δ_{β_{1}}^{β_{2}}δ_{γ_{1}}^{γ_{2}}δ_{δ_{1}}^{δ_{2}} + a_{2} δ_{β_{1}}^{γ_{2}}δ_{γ1}^{δ_{2}}δ_{δ_{1}}^{β_{2}} + a_{3} δ_{β_{1}}^{δ_{2}}δ_{γ_{1}}^{β_{2}}δ_{δ_{1}}^{γ_{2}}
+ b_{1} δ_{β_{1}}^{β_{2}}δ_{γ_{1}}^{δ_{2}}δ_{δ_{1}}^{γ_{2}} + b_{2} δ_{β_{1}}^{δ_{2}}δ_{γ_{1}}^{γ_{2}}δ_{δ_{1}}^{β_{2}} + b_{3} δ_{β_{1}}^{γ_{2}}δ_{γ_{1}}^{β_{2}}δ_{δ_{1}}^{δ_{2}} ,

where it follows from (no summation) ϵ_{π_{0}π_{1}π_{2}π_{3}}ϵ^{π_{0}π_{1}π_{2}π_{3}} = −1 that a_{i} = −1 and b_{i} = 1 holds for i = 1, 2, 3.

(11) Excercise: Addition theorem for transverse velocity components. In K′ the motion with velocity u⃗′ is

x^{′i} = c^{−1} u^{′i} x^{′0} .

With respect to frame K the origin of K′ frame moves with speed v along the x^{1} axis of K. For i = 2, 3 the Lorentz transformations give then

x^{i} = x^{′i} = c^{−1} u^{′i} γ (x^{0} − β x^{1}) .

Dividing by x^{0} gives the velocity components in K:

c^{−1} u^{i} = c^{−1} u^{′i} γ (1 − β c^{−1} u^{1}) = c^{−1} u^{′i} γ (1 − β c^{−1} \frac{u^{′1} + v}{1 + u^{′1}v/c^{2}} ) ,

where in the second step the already calculated equation for u^{1} has been inserted. Bringing everything to the common denominator gives

u^{i} = u^{′i} γ \frac{(1 + β u^{′1}/c) − β c^{−1} (u^{′1} + v)}{1 + u^{′1}v/c^{2}}
= u^{′i} γ \frac{1 − β^{2}}{1 + u^{′1}v/c^{2}} = \frac{u^{′i}}{γ (1 + u^{′1}v/c^{2})} ,

where 1 − β^{2} = 1/γ^{2} was used in the last step.