String Theory: Chapter 2 Problem Set Solutions for Barton Zwiebach's "A First Course in String Theory"


Warning: These are my own solutions for the given problems. If you are a student, use these at your own risk. They have not been subjected to grading in a university setting. Your professor may require more explicit statements, or other details worked out in the calculations.




Quick Calculations

Quick Calculation 2.1:

Problem Statement: Verify that the invariant \( ds^{2} \) is preserved under the \( x^{1} \) boost Lorentz transformation.

Solution Heuristic: Coordinate substitution to illustrate \( ds^{2} \) does not change.
(1) Define the \( x^{1} \) boost Lorentz transformation for \( x^{\mu} \rightarrow x'^{\mu} \)
(2) Substitute the Lorentz transformed coordinates in terms of \( x^{\mu} \) into the interval \( ds'^{2} \)
(3) Prove that \( ds'^{2} = ds^{2} \) by showing that \( ds^{2} \) is not changed by the transformation.

Solution:

Step (1): x-boost Lorentz transform \( x^{\mu '} = \Lambda^{\mu '}_{\phantom{\mu '}\mu} x^{\mu}\)

$$\begin{bmatrix} x'^{0} \\ x'^{1} \\ x'^{2} \\ x'^{3} \end{bmatrix} = \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x^{0} \\ x^{1} \\ x^{2} \\ x^{3} \end{bmatrix} $$

This gives us Equation 2.36 from the textook, where: \( \gamma = (1 - \beta^{2} )^{- \frac{1}{2}} \), \( \beta = \frac{v}{c} \), and \( x^{0} = ct \):

\( x'^{0} = \gamma ( x^{0} - \beta x^{1} ) \)
\( x'^{1} = \gamma ( - \beta x^{0} + x^{1} ) \)
\( x'^{2} = x^{2} \)
\( x'^{3} = x^{3} \)

Step 2: Substitute \( x^{\mu} \) values into \( x'^{\mu} \) components of \( ds'^{2} \)

\( ds'^{2} = (x'^{0})^{2} - (x'^{1})^{2} - (x'^{2})^{2} - (x'^{3})^{2} \)
\( ds'^{2} = \gamma^{2} [ (x^{0} - \beta x^{1})^{2} - (- \beta x^{0} + x^{1})^{2} ] - ( x^{2} )^2 - (x^{3})^{2} \)
\( ds'^{2} = \gamma^{2} [ (x^{0})^{2} - \beta x^{0}x^{1} - \beta x^{1}x^{0} + \beta^{2}(x^{1})^{2} - ( \beta^{2} (x^{0})^{2} - \beta x^{0}x^{1} - \beta x^{1}x^{0} + (x^{1})^{2} ) ] - ( x^{2} )^2 - (x^{3})^{2} \)

The coordinates \(x^{0} \) and \(x^{1} \) are commutative, therefore we combine and rearrange:

\( ds'^{2} = \gamma^{2} [ (x^{0})^{2} + ( - 2\beta x^{0}x^{1} + 2\beta x^{0}x^{1}) + \beta^{2}(x^{1})^{2} - \beta^{2} (x^{0})^{2} - (x^{1})^{2} ) ] - ( x^{2} )^2 - (x^{3})^{2} \)
\( ds'^{2} = \gamma^{2} [ (x^{0})^{2} + \beta^{2}(x^{1})^{2} - \beta^{2} (x^{0})^{2} - (x^{1})^{2} ) ] - ( x^{2} )^2 - (x^{3})^{2} \)

Rearrange to group the \(x^{0} \) and \(x^{1} \) terms, because we need to pull out a factor to cancel \( \gamma^{2} \):

\( ds'^{2} = \gamma^{2} [ ( (x^{0})^{2} - \beta^{2} (x^{0})^{2} ) - ((x^{1})^{2} - \beta^{2}(x^{1})^{2} ) ) ] - ( x^{2} )^2 - (x^{3})^{2} \)
\( ds'^{2} = \gamma^{2} (1 - \beta^{2}) [ (x^{0})^{2} - (x^{1})^{2} ] - ( x^{2} )^2 - (x^{3})^{2} \)
\( ds'^{2} = \frac{(1 - \beta^{2})}{(1 - \beta^{2})} [ (x^{0})^{2} - (x^{1})^{2} ] - ( x^{2} )^2 - (x^{3})^{2} \)

Step (3): Prove that \( ds'^{2} = ds^{2} \) under the x-boost Lorentz transformation

Therefore the \( \gamma^{2} \) term cancels identically, leaving \( ds'^{2} = ds^{2} \):

\( ds'^{2} = (x^{0})^{2} - (x^{1})^{2} - ( x^{2} )^2 - (x^{3})^{2} \)
\( ds^{2} = (x^{0})^{2} - (x^{1})^{2} - ( x^{2} )^2 - (x^{3})^{2} \)
\( \Rightarrow ds'^{2} = ds^{2} \) Q.E.D.

Quick Calculation 2.2:

Problem Statement: Consider two Lorentz vectors \( a^{\mu} \) and \( b^{\mu} \).

(A) Write the Lorentz transformations \( a^{\mu} \rightarrow a'^{\mu} \) and \( b^{\mu} \rightarrow b'^{\mu} \), analogous to Equation 2.36 in Quick Calculation 2.1.
(B) Verify that \( a^{\mu}b_{\mu} \) is invariant under these transformations.

Solution Heuristic:

(A) Apply \( a'^{\mu} = \Lambda^{\mu '}_{\phantom{\mu '}\mu} a^{\mu}\) and \( b'^{\mu} = \Lambda^{\mu '}_{\phantom{\mu '}\mu} b^{\mu}\) for the \( x^{1} \) direction boost.
(B) Calculate scalar product \( a'^{\mu}b'_{\mu} \) and show it equals \( a^{\mu}b_{\mu} \).

Solution:

(A) Write out \( a^{\mu} \rightarrow a'^{\mu} \) and \( b^{\mu} \rightarrow b'^{\mu} \)

$$\begin{bmatrix} a'^{0} \\ a'^{1} \\ a'^{2} \\ a'^{3} \end{bmatrix} = \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a^{0} \\ a^{1} \\ a^{2} \\ a^{3} \end{bmatrix} $$

$$\begin{bmatrix} b'^{0} \\ b'^{1} \\ b'^{2} \\ b'^{3} \end{bmatrix} = \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} b^{0} \\ b^{1} \\ b^{2} \\ b^{3} \end{bmatrix} $$

This gives analogs of Equation 2.36 from the textook, where: \( \gamma = (1 - \beta^{2} )^{- \frac{1}{2}} \), \( \beta = \frac{v}{c} \), and time is ct:

\( a'^{0} = \gamma ( a^{0} - \beta a^{1} ) \)
\( a'^{1} = \gamma ( - \beta a^{0} + a^{1} ) \)
\( a'^{2} = a^{2} \)
\( a'^{3} = a^{3} \)

\( b'^{0} = \gamma ( b^{0} - \beta b^{1} ) \)
\( b'^{1} = \gamma ( - \beta b^{0} + b^{1} ) \)
\( b'^{2} = b^{2} \)
\( b'^{3} = b^{3} \)


(B) Scalar product \( a^{\mu}b_{\mu} \) = \( a'^{\mu}b'_{\mu} \)

Equation 2.29 from the textbook:

\( a \cdot b \equiv a^{\mu}b_{\mu} = \eta_{\mu\nu} a^{\mu}b^{\nu} = -a^{0}b^{0} + a^{1}b^{1} + a^{2}b^{2} + a^{3}b^{3} \)

Where:

$$ \eta_{\mu\nu} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

Therefore for \( a'^{\mu}b'_{\mu} \), we substitute in the \( a^{\mu} \) and \( b^{\mu} \) components under the Lorentz boost:

\( a'^{\mu}b'_{\mu} = -a'^{0}b'^{0} + a'^{1}b'^{1} + a'^{2}b'^{2} + a'^{3}b'^{3} \)
\( a'^{\mu}b'_{\mu} = - \gamma^{2} ( a^{0} - \beta a^{1} )( b^{0} - \beta b^{1} ) + \gamma^{2} ( - \beta a^{0} + a^{1} )( - \beta b^{0} + b^{1} ) + a^{2}b^{2} + a^{3}b^{3} \)

The cross terms of \(a^{0}b^{1}\) and \(a^{1}b^{0}\) cancel out under addition:

\( a'^{\mu}b'_{\mu} = - \gamma^{2} (a^{0}b^{0} - \beta a^{0}b^{1} - \beta a^{1}b^{0} + \beta^{2} a^{1}b^{1}) + \gamma^{2} (\beta^{2}a^{0}b^{0} - \beta a^{0}b^{1} - \beta a^{1}b^{0} + a^{1}b^{1}) + a^{2}b^{2} + a^{3}b^{3} \)
\( a'^{\mu}b'_{\mu} = \gamma^{2} (- a^{0}b^{0} - \beta^{2} a^{1}b^{1}) + \gamma^{2} (a^{1}b^{1} + \beta^{2}a^{0}b^{0}) + a^{2}b^{2} + a^{3}b^{3} \)

Rearrange to pull out the term to cancel \( \gamma^{2} \):

\( a'^{\mu}b'_{\mu} = - \gamma^{2} ( a^{0}b^{0} - \beta^{2}a^{0}b^{0}) ) + \gamma^{2} (a^{1}b^{1} - \beta^{2} a^{1}b^{1}) + a^{2}b^{2} + a^{3}b^{3} \)
\( a'^{\mu}b'_{\mu} = \gamma^{2} (1 - \beta^{2}) (- a^{0}b^{0} + a^{1}b^{1} ) + a^{2}b^{2} + a^{3}b^{3} \)

Then we are left with the componenets for \( a^{\mu}b_{\mu} \), and see that it is invariant under the Lorentz transformation:

\( a'^{\mu}b'_{\mu} = - a^{0}b^{0} + a^{1}b^{1} + a^{2}b^{2} + a^{3}b^{3} \)
\( \Rightarrow a'^{\mu}b'_{\mu} = a^{\mu}b_{\mu} \) Q.E.D.

Quick Calculation 2.3:

Problem Statement: Prove there is no Lorentz transformation taking coordinates \( (x^{0}, x^{1}, x^{2}, x^{3}) \rightarrow ( x'^{0}, x'^{1}, x'^{2}, x'^{3} ) = ( x^{+}, x^{-}, x^{2}, x^{3} ) \)

Solution Heuristic: Use the \( x^{1} \) Lorentz boost as a counter-example showing the light-cone coordinate system is self-contradictory. This shows that coordinate system is a basis for doing calculations, but cannot be transformed into without violating Lorentz invariance. This is not a violation of special relativity, the coordinate axes are on the light cone.

Solution:

Light-cone coordinate system from Equation 2.50 of the textbook:

\( x^{+} \equiv \frac{1}{\sqrt{2}} (x^{0} + x^{1}) \)
\( x^{-} \equiv \frac{1}{\sqrt{2}} (x^{0} - x^{1}) \)

Assume \( x^{1} \) direction Lorentz boost, transforming \( x^{\mu} \rightarrow x'^{\mu} \), setting \( x'^{\mu} \) equal to light-cone coordinates:

\( x^{+} = \frac{1}{\sqrt{2}} (x^{0} + x^{1}) = x'^{0} = \gamma ( x^{0} - \beta x^{1} ) \)
\( x^{-} = \frac{1}{\sqrt{2}} (x^{0} - x^{1}) = x'^{1} = \gamma ( -\beta x^{0} + x^{1} ) \)
\( x^{2} = x'^{2} = x^{2} \)
\( x^{3} = x'^{3} = x^{3} \)

By inspection of \( x^{+} \), \( \beta = -1 \) and \( \gamma = \frac{1}{\sqrt{2}} \). However, \( \gamma = \frac{1}{ \sqrt{1 - \beta^{2}} } = \infty \).

Quick Calculation 2.4:

Problem Statement: Verify that the given energy and momentum equations satisfy Equation 2.67

Solution Heuristic: Substitute into equation and derive right hand side.

(1) Substitute E and \( \overrightarrow{p} \) into the left hand side of Equation 2.67
(2) Pull out a factor to cancel \( \gamma \).

Solution:

Step (1): Substitute in \( E= \gamma mc^{2} \) and \( \overrightarrow{p} = \gamma m\overrightarrow{v} \).

Equation 2.67: \( \frac{E}{c^{2}} - \overrightarrow{p} \cdot \overrightarrow{p} = m^{2}c^{2} \)

The substitution into the left-hand side yields:

\( \frac{E}{c^{2}} - \overrightarrow{p} \cdot \overrightarrow{p} = \frac{ \gamma^{2} m^{2} c^{4} }{c^{2}} - \gamma^{2} m^{2} v^{2} = \gamma^{2} (m^{2}c^{2} - m^{2}v^{2}) \)

Step (2): Pull out a \( \beta^{2} \) factor to cancel \( \gamma^{2} \).

Note that: \( \beta^{2} = \frac{v^{2}}{c^{2}} \)
\( \Rightarrow m^{2}v^{2} = m^{2}c^{2}\frac{v^{2}}{c^{2}} = \beta m^{2}c^{2} \)

Substitute this expression for \( m^{2}v^{2} \) above:

\( \gamma^{2} (m^{2}c^{2} - m^{2}v^{2}) = \gamma^{2} (m^{2}c^{2} - \beta^{2} m^{2}c^{2}) = \gamma^{2} (1 - \beta^{2}) m^{2}c^{2} \)
\( \Rightarrow \frac{E}{c^{2}} - \overrightarrow{p} \cdot \overrightarrow{p} = m^{2}c^{2} \) Q.E.D.



Problems

Problem 2.1

Problem Statement: Exercises with units.

(a) Find the relation between coulombs (C) and esus.
(b) Explain (1) the meaning of the unit K (degrees [sic] kelvin) used for measuring temperature; and (2) explain its relation to the basic length, mass, and time units.
(c) Construct a dimensionless number using the charge e of the electron (defined in Gaussian units), \(\hbar\), and c.

Solution Heuristic:

(a) Use the S.I. constant in Coulomb's Law and the definition of esu, then rearrange and solve for C.
(b) Set the Boltzmann constant to 1, then [K] will have units of energy.
(c) The Gaussian electron charge is esu. So write out dimensions for esu, h-bar, and c. Arrange a ratio where the dimensions cancel.

Solution:

Problem 2.1 (a): Find the relation between coulombs (C) and esus.

\( esu^{2} = 10^{-9} N \cdot m^{2} \)
\( \frac{1}{4 \pi \epsilon_{0}} = 8.99 x 10^{9} \frac{N \cdot m^{2}}{C^{2}} \)

In Coloumb's Law there is 8.99 x 10^{9} N when two charges are 1 meter apart. Let the charges equal 1 and rearrange the C term:

\( C^{2} = 8.99 x 10^{9} N \cdot m^{2} = 8.99 x 10^{9} x 10^{9} esu^{2} \)
\( C^{2} = 8.99 x 10^{18} esu^{2} \)
\( C = \sqrt{8.99} x 10^{9} esu \)

Therefore 1 Coulomb is approximately \( 3 x 10^{9} esu \). Q.E.D.


Problem 2.1 (b): Meaning of Kelvin

(1) Explain the meaning of the unit K (degrees [sic] kelvin) used for measuring temperature:

The S.I. definition of Kelvin is set in terms of the measured value of the Boltzmann constant \( k_{B} \), which is Joules per Kelvin. Joules have units defined by Planck's constant, the speed of light, and the hyperfine transition duration of Caesium-133. Mass, Length, and Time. This means setting the value \( k_{B} = 1\) rearranges as [Kelvin] = [Joules] dimensionally.

(2) Explain the unit K in terms of mass, length, and time:

\( k_{B} = 1 = J \cdot K^{-1} \)
\( \Rightarrow [K] = [J] = [N \cdot m] \)
\( \Rightarrow [K] = \frac{M L^{2}}{T^{2}} \) Q.E.D.


Problem 2.1 (c): Construct a dimensionless number using the charge e of the electron (defined in Gaussian units), \(\hbar\), and c.

Gaussian [e]: esu = \( M^{\frac{1}{2}} L^{\frac{3}{2}} T^{-1} \)
\( [\hbar] \): \( M L^{2} T^{-1} \)
[c]: \( L T^{-1} \)

Note that the fractions of the esu units go away when squared, yielding \( M L^{3} T^{-2} \). These are the dimensions of multiplying \( \hbar \) and c:

\( e^{2} = M L^{3} T^{-2} \)
\( \hbar c = M L^{3} T^{-2} \)
\( \Rightarrow \frac{e^{2}}{\hbar c} = \) Dimensionless. Q.E.D.

In Heaviside-Lorentz units: \( \frac{e^{2}}{4 \pi \hbar c} \)

This is \( \alpha \) the fine structure constant, and evaluates to approximately \( \frac{1}{137} \):

\( \frac{ (1.602 x 10^{-19} x 2.998 x 10^{9} esu)^{2} x 10^{-9}}{ (1.054 x 10^{-34} N \cdot m \cdot s) (2.998 x 10^{8} \frac{m}{s}) } = \frac{2.307 x 10^{-28}}{3.160 x 10^{-26}} = 0.00730 \)
Note that \( \frac{1}{137} = 0.007299 \)


Problem 2.2

Problem Statement: Lorentz transformations for light-cone coordinates.

Consider the coordinates \( x^{\mu} = (x^{0}, x^{1}, x^{2}, x^{3}) \) and the associated light-cone coordinates \( ( x^{+}, x^{-}, x^{2}, x^{3} ) \). Write the following Lorentz transformations in terms of the light-cone coordinates (which is to say, without reference to \( x^{0}, x^{1} \)).

(a) A boost with velocity parameter \( \beta \) in the \( x^{1} \) direction.
(b) A rotation with angle \( \theta \) in the \( x^{1}, x^{2}\) plane.
(c) A boost with velocity parameter \( \beta \) in the \( x^{3} \) direction.

Solution Heuristic:

(a) \( x^{1} \) Boost
Step 1: Lorentz transform \( x^{\mu} \rightarrow x'^{\mu} \) for \( x^{1} \) direction boost, change \( x'^{\mu} \) to the light-cone coordinate system. Step 2: Re-express the \( x^{\mu} \) components \( x^{0}, x^{1} \) in terms of \( x^{+}, x^{-} \)
Step 3: Simplify with x' light-cone coordinates on the left-hand side and x light-cone coordinates on the right-hand side.

(b) Rotation \( \theta \) in the \( x^{1}, x^{2}\) plane.
Step 1: Write out the \( x^{3} \)-axis rotation matrix in four dimensions.
Step 2: Re-express \( x'^{\mu} \) terms in light-cone coordinates.
Step 3: Re-express \( x^{0}, x^{1} \) components in terms of \( x^{+}, x^{-} \).
Step 4: Substitute \( x^{+}, x^{-} \) into \( x^{0}, x^{1} \) terms.
Step 5: Simplify.

(c) \( x^{3} \) Boost
Step 1: Re-express \( x^{0}, x^{1} \) in terms of \( x^{+}, x^{-} \).
Step 2: Lorentz transform \( x^{\mu} \rightarrow x'^{\mu} \) for \( x^{3} \) direction boost, change \( x'^{\mu} \) to the light-cone coordinate system.
Step 3: Re-express the \( x^{\mu} \) components \( x^{0}, x^{1} \) in terms of \( x^{+}, x^{-} \)
Step 4: Simplify with x' light-cone coordinates on the left-hand side and x light-cone coordinates on the right-hand side.


Solution:

Problem 2.2 (a): \( x^{1} \) Boost

Step 1: Lorentz transform \( x^{\mu} \rightarrow x'^{\mu} \) for \( x^{1} \) direction boost, change \( x'^{\mu} \) be the light-cone coordinate system.

\( x'^{0} = \gamma (x^{0} - \beta x^{1}) \)
\( x'^{1} = \gamma (-\beta x^{0} + x^{1}) \)
\( x'^{2} = x^{2} \)
\( x'^{3} = x^{3} \)

Note the definitions of the light-cone coordinates, expressed in terms of the boosted coordinate system:

\( x'^{+} = \frac{1}{\sqrt{2}} ( x'^{0} + x'^{1} ) \)
\( x'^{-} = \frac{1}{\sqrt{2}} ( x'^{0} + x'^{1} ) \)

Substitute in the \( x^{\mu} \) components:

\( x'^{+} = \frac{1}{\sqrt{2}} ( \gamma ( x^{0} - \beta x^{1} ) + \gamma (-\beta x^{0} + x^{1}) ) \)
\( x'^{-} = \frac{1}{\sqrt{2}} ( \gamma ( x^{0} - \beta x^{1} ) - \gamma (-\beta x^{0} + x^{1}) ) \)


Step 2: Re-express the \( x^{\mu} \) components \( x^{0}, x^{1} \) in terms of \( x^{+}, x^{-} \).

In this particular case you can rearrange \( x^{0}, x^{1} \) terms to simply re-express it into the definition of \( x^{+}, x^{-} \):

\( x'^{+} = \frac{\gamma}{\sqrt{2}} [ ( x^{0} - \beta x^{0} ) + ( x^{1} - \beta x^{1} ) ] \)
\( x'^{+} = \frac{\gamma ( 1 - \beta )}{\sqrt{2}} ( x^{0} + x^{1} ) \)
\( \Rightarrow x'^{+} = \gamma ( 1 - \beta ) x^{+} \)

\( x'^{-} = \frac{\gamma}{\sqrt{2}} [ ( x^{0} + \beta x^{0} ) + ( - x^{1} - \beta x^{1} ) ] \)
\( x'^{-} = \frac{\gamma ( 1 + \beta )}{\sqrt{2}} ( x^{0} - x^{1} ) \)
\( \Rightarrow x'^{-} = \gamma ( 1 + \beta ) x^{-} \)


Step 3: Simplify with x' light-cone coordinates on the left-hand side and x light-cone coordinates on the right-hand side.

Note that \( \gamma = \frac{1}{\sqrt{1 - \beta^{2}}} \), leaving a square root in the denominator:

\( x'^{+} = \frac{1 - \beta}{\sqrt{1 - \beta^{2}}} x^{+} \)
\( x'^{-} = \frac{1 + \beta}{\sqrt{1 - \beta^{2}}} x^{-} \)

This can be fixed with some algebra manipulation. The square-root of the square of the numerators equal the numerators, where we keep the positive term:

\( x'^{+} = \frac{\sqrt{(1 - \beta)^{2}}}{\sqrt{1 - \beta^{2}}} x^{+} \)
\( x'^{-} = \frac{\sqrt{(1 + \beta)^{2}}}{\sqrt{1 - \beta^{2}}} x^{-} \)

Then notice that the denominator can be factored into opposite signed terms, and the whole thing goes under a square root:

\( x'^{+} = \sqrt{\frac{(1 - \beta)(1 - \beta)}{(1 - \beta) (1 + \beta)}} x^{+} = \sqrt{\frac{(1 - \beta)}{(1 + \beta)}} x^{+} \)
\( x'^{-} = \sqrt{\frac{(1 + \beta)(1 + \beta)}{(1 - \beta) (1 + \beta)}} x^{-} = \sqrt{\frac{(1 + \beta)}{(1 - \beta)}} x^{-} \)

Therefore the light-cone coordinates for the \( x^{1} \) boost are:

\( x'^{+} = \sqrt{\frac{(1 - \beta)}{(1 + \beta)}} x^{+} \)
\( x'^{-} = \sqrt{\frac{(1 + \beta)}{(1 - \beta)}} x^{-} \)
\( x'^{2} = x^{2}\)
\( x'^{3} = x^{3} \) Q.E.D.

(Remark: This is similar to the expression for the longitudinal relativistic Doppler effect.)


Problem 2.2 (b): Rotation \( \theta \) in the \( x^{1}, x^{2}\) plane.

Note that this is only a rotation in a spatial plane, but re-expressed in terms of light-cone coordinates. There are no factors of \( \gamma \) or \( \beta \) in the derivation.

Step 1: Write out the \( x^{3} \)-axis rotation matrix in four dimensions.

$$ \begin{bmatrix} x'^{0} \\ x'^{1} \\ x'^{2} \\ x'^{3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & cos(\theta) & -sin(\theta) & 0 \\ 0 & sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x^{0} \\ x^{1} \\ x^{2} \\ x^{3} \end{bmatrix} $$

This matrix multiplication produces the following equations:

\( x'^{0} = x^{0} \)
\( x'^{1} = cos(\theta)x^{1} - sin(\theta)x^{2} \)
\( x'^{2} = sin(\theta)x^{1} + cos(\theta)x^{2} \)
\( x'^{3} = x^{3} \)


Step 2: Re-express \( x'^{\mu} \) terms in light-cone coordinates.

Since we have the light-cone coordinate definition:

\( x'^{+} = \frac{1}{\sqrt{2}} ( x'^{0} + x'^{1} ) \)
\( x'^{+} = \frac{1}{\sqrt{2}} ( x'^{0} - x'^{1} ) \)

We can substitute in the \( x'^{\mu} \) terms from above:

\( x'^{+} = \frac{1}{\sqrt{2}} ( x^{0} + cos(\theta)x^{1} - sin(\theta)x^{2} ) \)
\( x'^{-} = \frac{1}{\sqrt{2}} ( x^{0} - cos(\theta)x^{1} + sin(\theta)x^{2} ) \)
\( x'^{2} = sin(\theta)x^{1} + cos(\theta)x^{2} \)
\( x'^{3} = x^{3} \)


Step 3: Re-express \( x^{0}, x^{1} \) components in terms of \( x^{+}, x^{-} \).

Unlike in Problem 2.2(a), it is not simple to rearrange terms to get \( x^{+}, x^{-} \), so we will rearrange them for substitution:

\( x^{+} = \frac{1}{\sqrt{2}} ( x^{0} + x^{1} ) \)
\( x^{+} = \frac{1}{\sqrt{2}} ( x^{0} - x^{1} ) \)

\( \Rightarrow x^{-} - \frac{1}{\sqrt{2}} x^{0} = - \frac{1}{\sqrt{2}}x^{1} \)

So we substitute this factor of \( x^{1} \) into \( x^{+} \):
\( x^{+} = \frac{1}{\sqrt{2}} x^{0} - ( - \frac{1}{\sqrt{2}} x^{1} ) \)
\( \Rightarrow x^{+} = \frac{1}{\sqrt{2}} x^{0} - ( x^{-} - \frac{1}{\sqrt{2}} x^{0} ) \)
\( \Rightarrow x^{+} = \frac{2}{\sqrt{2}} x^{0} - x^{-} \)

So we re-arrange to find \( x^{0} \):

\( x^{+} + x^{-} = \frac{2}{\sqrt{2}} = \sqrt(2) x^{0} \)
\( \Rightarrow x^{0} = \frac{1}{\sqrt{2}} (x^{+} + x^{-}) \)

We repeat the same procedure to isolate \( x^{1} \) instead:

\( \Rightarrow x^{-} + \frac{1}{\sqrt{2}} x^{1} = \frac{1}{\sqrt{2}}x^{0} \)

\( x^{+} = ( \frac{1}{\sqrt{2}} x^{0} ) + ( \frac{1}{\sqrt{2}} x^{1} ) \)
\( x^{+} = ( x^{-} + \frac{1}{\sqrt{2}} x^{1} ) + ( \frac{1}{\sqrt{2}} x^{1} ) \)
\( \Rightarrow x^{+} = \frac{2}{\sqrt{2}} x^{1} + x^{-} \)

The rearrange to find \( x^{1} \):

\( x^{+} - x^{-} = \sqrt{2} x^{1} \)
\( \Rightarrow x^{1} = \frac{1}{\sqrt{2}} (x^{+} - x^{-}) \)

So the conversion of \( x^{0},x^{1} \) into \( x^{+},x^{-} \) is:

\( x^{0} = \frac{1}{\sqrt{2}} (x^{+} + x^{-}) \)
\( x^{1} = \frac{1}{\sqrt{2}} (x^{+} - x^{-}) \)


Step 4: Substitute \( x^{+}, x^{-} \) into \( x^{0}, x^{1} \) terms.

\( x'^{+} = \frac{1}{\sqrt{2}} ( x^{0} + cos(\theta)x^{1} - sin(\theta)x^{2} ) \)
\( x'^{-} = \frac{1}{\sqrt{2}} ( x^{0} - cos(\theta)x^{1} + sin(\theta)x^{2} ) \)
\( x'^{2} = sin(\theta)x^{1} + cos(\theta)x^{2} \)
\( x'^{3} = x^{3} \)

\( \Rightarrow \)

\( x'^{+} = \frac{1}{\sqrt{2}} ( \frac{1}{\sqrt{2}} (x^{+} + x^{-}) + cos(\theta)\frac{1}{\sqrt{2}} (x^{+} - x^{-}) - sin(\theta)x^{2} ) \)
\( x'^{-} = \frac{1}{\sqrt{2}} ( \frac{1}{\sqrt{2}} (x^{+} + x^{-}) - cos(\theta)\frac{1}{\sqrt{2}} (x^{+} - x^{-}) + sin(\theta)x^{2} ) \)
\( x'^{2} = sin(\theta)\frac{1}{\sqrt{2}} (x^{+} - x^{-}) + cos(\theta)x^{2} \)
\( x'^{3} = x^{3} \)


Step 5: Group the \( x^{+}, x^{-} \) terms and simplify.

\( x'^{+} = \frac{1}{2} [ (1 + cos(\theta) )x^{+} + (1 - cos(\theta) )x^{-} - \frac{sin(\theta)}{\sqrt{2}}x^{2} ] \)
\( x'^{-} = \frac{1}{2} [ (1 - cos(\theta) )x^{+} + (1 + cos(\theta) )x^{-} + \frac{sin(\theta)}{\sqrt{2}}x^{2} ] \)
\( x'^{2} = \frac{sin(\theta)}{\sqrt{2}} (x^{+} - x^{-}) + cos(\theta)x^{2} \)
\( x'^{3} = x^{3} \) Q.E.D.


Problem 2.2 (c): \( x^{3} \) Boost.

Step 1: Re-express \( x^{0}, x^{1} \) in terms of \( x^{+}, x^{-} \)

Taking the identical result from Problem 2.2 (b), Step 3:

\( x^{0} = \frac{1}{\sqrt{2}} (x^{+} + x^{-}) \)
\( x^{1} = \frac{1}{\sqrt{2}} (x^{+} - x^{-}) \)


Step 2: Lorentz transform \( x^{\mu} \rightarrow x'^{\mu} \) for \( x^{3} \) direction boost, change \( x'^{\mu} \) to the light-cone coordinate system.

The Lorentz transformation matrix \( x^{\mu} \rightarrow x'^{\mu} \) for the \( x^{3} \) boost:

$$\begin{bmatrix} x'^{0} \\ x'^{1} \\ x'^{2} \\ x'^{3} \end{bmatrix} = \begin{bmatrix} \gamma & 0 & 0 & -\beta\gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta\gamma & 0 & 0 & \gamma \end{bmatrix} \begin{bmatrix} x^{0} \\ x^{1} \\ x^{2} \\ x^{3} \end{bmatrix} $$

The matrix multiplication yields the transformation equations:

\( x'^{0} = \gamma (x^{0} - \beta x^{3} ) \)
\( x'^{1} = x^{1} \)
\( x'^{2} = x^{2} \)
\( x'^{3} = \gamma (-\beta x^{0} + x^{3}) \)

Then expressing the boosted coordinate system in terms of light-cone coordinates:

\( x'^{+} = \frac{1}{\sqrt{2}} (x'^{0} + x'^{1}) = \frac{1}{\sqrt{2}} ( \gamma (x^{0} - \beta x^{3}) + x^{1} ) ) \)
\( x'^{-} = \frac{1}{\sqrt{2}} (x'^{0} - x'^{1}) = \frac{1}{\sqrt{2}} ( \gamma (x^{0} - \beta x^{3}) - x^{1} ) \)
\( x'^{2} = x^{2} \)
\( x'^{3} = \gamma (-\beta x^{0} + x^{3} ) \)

Notice that the \( x^{+}, x^{-}\) terms are still defined in terms of \( x^{1} \), so this makes a different situation than Problem 2.2 (a). The \( x'^{3} \) term does not appear in the \( x^{+}, x^{-}\) terms, unlike the case of the \( x^{1} \) boost. This will not be a simple rearrangement to yield the \( x^{+}, x^{-}\) form. We will substitute them into \( x^{\mu} \) components.


Step 3: Re-express the \( x^{\mu} \) components \( x^{0}, x^{1} \) in terms of \( x^{+}, x^{-} \)

Substitute the \( x^{+}, x^{-}\) terms:

\( x'^{+} = \frac{1}{\sqrt{2}} ( \gamma (\frac{1}{\sqrt{2}} (x^{+} + x^{-}) - \beta x^{3}) + \frac{1}{\sqrt{2}} (x^{+} - x^{-}) ) ) \)
\( x'^{-} = \frac{1}{\sqrt{2}} ( \gamma (\frac{1}{\sqrt{2}} (x^{+} + x^{-}) - \beta x^{3}) - (\frac{1}{\sqrt{2}} (x^{+} - x^{-}) ) ) \)
\( x'^{2} = x^{2} \)
\( x'^{3} = \gamma (-\beta (\frac{1}{\sqrt{2}} (x^{+} + x^{-})) + x^{3} ) \)


Step 4: Simplify

Re-group the \( x^{+}, x^{-} \) terms. Notice that the \( x^{1} \) substitution has no \( \gamma \) term because that direction is not boosted. This will create a \( \frac{1}{\gamma}\) coefficient when pulling out \( \gamma \).

\( x'^{+} = \frac{\gamma}{\sqrt{2}}[ \frac{1}{\sqrt{2}} ( x^{+} + x^{-} ) + \frac{1}{\gamma} ( \frac{1}{\sqrt{2}} (x^{+} - x^{-} ) )] - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)
\( x'^{+} = \frac{\gamma}{2} [ (1 + \frac{1}{\gamma} )x^{+} + (1 - \frac{1}{\gamma})x^{-} ] - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)
\( x'^{+} = \frac{\gamma}{2} [ (\frac{\gamma + 1}{\gamma} )x^{+} + (\frac{\gamma - 1}{\gamma})x^{-} ] - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)
\( \Rightarrow x'^{+} = \frac{\gamma + 1}{2} x^{+} + \frac{\gamma - 1}{2} x^{-} - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)

Then the \( x'^{-} \) term differs from this derivation by only a minus sign in the middle:

\( x'^{-} = \frac{\gamma}{\sqrt{2}}[ \frac{1}{\sqrt{2}} ( x^{+} + x^{-} ) - \frac{1}{\gamma} ( \frac{1}{\sqrt{2}} (x^{+} - x^{-} ) )] - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)
\( x'^{-} = \frac{\gamma}{2} [ (1 + \frac{1}{\gamma} )x^{+} - (1 - \frac{1}{\gamma})x^{-} ] - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)
\( x'^{-} = \frac{\gamma}{2} [ (\frac{\gamma - 1}{\gamma} )x^{+} + (\frac{\gamma + 1}{\gamma})x^{-} ] - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)
\( \Rightarrow x'^{-} = \frac{\gamma - 1}{2} x^{+} + \frac{\gamma + 1}{2} x^{-} - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)

\( x'^{2} = x^{2} \) trivially, and the \( x'^{3} \) term is a simple substitution:

\( x'^{3} = \gamma ( - \frac{\beta}{\sqrt{2}}( x^{+} + x^{-} ) + x^{3} ) \)
\( \Rightarrow x'^{3} = - \frac{\gamma\beta}{\sqrt{2}}( x^{+} + x^{-} ) + \gamma x^{3} \)

Therefore the light-cone coordinate equations for the \( x^{3} \) boost are:

\( x'^{+} = \frac{\gamma + 1}{2} x^{+} + \frac{\gamma - 1}{2} x^{-} - \frac{\gamma\beta}{\sqrt{2}}x^{3}\)
\( x'^{-} = \frac{\gamma - 1}{2} x^{+} + \frac{\gamma + 1}{2} x^{-} - \frac{\gamma\beta}{\sqrt{2}}x^{3} \)
\( x'^{2} = x^{2} \)
\( x'^{3} = - \frac{\gamma\beta}{\sqrt{2}}( x^{+} + x^{-} ) + \gamma x^{3} \) Q.E.D.

Notice how different these equations look from Problem 2.2(a), where the boost is contained within the \( x'^{+},x'^{-} \) coordinates. The choice of the \( x^{1} \) coordinate in their definition is only a convention.


Problem 2.3

Problem Statement: Lorentz transformations, derivatives, and quantum operators

(a) Give the Lorentz transformations for the components \(a_{\mu}\) of a vector under an \(x^{1}\) boost.
(b) Show that the objects \( \frac{\partial}{\partial x^{\mu}} \) transform under an \(x^{1}\) boost the same way as \(a_{\mu}\).
(c) Show that in quantum mechanics the expressions of energy and momentum, in terms of derivatives, can be written compactly as \(p_{\mu} = \frac{\hbar}{i} \frac{\partial}{\partial x^{\mu}} \).

Solution Heuristic:

(a) Lower index \( x^{1} \) boost.

Step 1: Write the \(x^{1}\) boost Lorentz transformation for the four vector \(a^{\mu}\), from Quick Calculation 2.2.
Step 2: Define the relationships between \(a^{\mu}\) and \(a_{\mu}\) using the textbook's sign convention.
Step 3: Re-express \(a'^{\mu}\) as \(a'_{\mu}\) with \(a_{\mu}\) components.

Solution:

Problem 2.3 (a): Give the Lorentz transformations for the components \(a_{\mu}\) of a vector under an \(x^{1}\) boost.

Step 1: Write the \(x^{1}\) boost Lorentz transformation for the four vector \(a^{\mu}\), from Quick Calculation 2.2.

$$\begin{bmatrix} a'^{0} \\ a'^{1} \\ a'^{2} \\ a'^{3} \end{bmatrix} = \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a^{0} \\ a^{1} \\ a^{2} \\ a^{3} \end{bmatrix} $$

Which yields the Lorentz transformations:

\( a'^{0} = \gamma ( a^{0} - \beta a^{1} ) \)
\( a'^{1} = \gamma ( - \beta a^{0} + a^{1} ) \)
\( a'^{2} = a^{2} \)
\( a'^{3} = a^{3} \)


Step 2: Define the relationships between \(a^{\mu}\) and \(a_{\mu}\) using the textbook's sign convention.

These are upper-index terms. Following convention from Equation 2.16 where a minus sign is added to time, so the minus sign is buried in the invariant interval equation, we have:

\( a_{0} = - a^{0}\)
\( a_{1} = a^{1}\)
\( a_{1} = a^{2}\)
\( a_{1} = a^{3}\)


Step 3: Re-express \(a'^{\mu}\) as \(a'_{\mu}\) with \(a_{\mu}\) components.

Therefore substituting \( a'^{0} \) and \(a^{0}\) with lower index terms and negative signs, along with the other lower indices:

\( - a'_{0} = \gamma ( (-a_{0}) - \beta a_{1} ) \)
\( a'_{1} = \gamma ( - \beta (-a_{0}) + a_{1} ) \)
\( a'_{2} = a_{2} \)
\( a'_{3} = a_{3} \)

Which becomes:

\( a'_{0} = \gamma ( a_{0} + \beta a_{1} ) \)
\( a'_{1} = \gamma ( \beta a_{0} + a_{1} ) \)
\( a'_{2} = a_{2} \)
\( a'_{3} = a_{3} \) Q.E.D.


Problem 2.3 (b): Show that the objects \( \frac{\partial}{\partial x^{\mu}} \) transform under an \(x^{1}\) boost the same way as \(a_{\mu}\).

This checks that the partial derivatives with respect to upper-index coordinates \( x^{\mu} \) behave as a four-vector with lower indices \( x'_{\mu} \). This justifies the gradient symbol \( \partial_{\mu} \).

Consider the chain rule for the boost objects \( \frac{\partial}{\partial x'^{\mu}} \) are a change of coordinates and inverse Jacobian:

\( \frac{\partial}{\partial x'^{0}} = \frac{\partial}{\partial x^{0}}\frac{\partial x^{0}}{\partial x'^{0}} + \frac{\partial}{\partial x^{1}}\frac{\partial x^{1}}{\partial x'^{0}} \)
\( \frac{\partial}{\partial x'^{1}} = \frac{\partial}{\partial x^{0}}\frac{\partial x^{0}}{\partial x'^{1}} + \frac{\partial}{\partial x^{1}}\frac{\partial x^{1}}{\partial x'^{1}} \)

Since this case is an \( x^{1} \) Lorentz boost, we can see that this is the inverse Lorentz transform \( \Lambda_{\phantom{\nu}\mu}^{\nu} = (\frac{\partial {x}^{\nu}}{\partial x'^{\mu}}) \):

$$\begin{bmatrix} x^{0} \\ x^{1} \\ x^{2} \\ x^{3} \end{bmatrix} = \begin{bmatrix} \gamma & \beta\gamma & 0 & 0 \\ \beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x'^{0} \\ x'^{1} \\ x'^{2} \\ x'^{3} \end{bmatrix} $$

Thus, the inverse Lorentz boost in the \( x^{1} \) direction:

\( x^{0} = ( x'^{0} + \beta x'^{1}) \)
\( x^{1} = (\beta x'^{0} + x'^{1}) \)
\( x^{2} = x'^{2} \)
\( x^{3} = x'^{3} \)

Substituting the equations for \(x^{0},x^{1}\) into the inverse Lorentz transform, which operates upon it, gives:

\( \frac{\partial}{\partial x'^{0}} = \frac{\partial}{\partial x^{0}}[\frac{\partial}{\partial x'^{0}}( x'^{0} + \beta x'^{1} )] + \frac{\partial}{\partial x^{1}}[\frac{\partial}{\partial x'^{0}}(\beta x'^{0} + x'^{1})] \)
\( \frac{\partial}{\partial x'^{1}} = \frac{\partial}{\partial x^{0}}[\frac{\partial}{\partial x'^{1}}( x'^{0} + \beta x'^{1} )] + \frac{\partial}{\partial x^{1}}[\frac{\partial}{\partial x'^{1}}(\beta x'^{0} + x'^{1})] \)

Applying the partial derivatives for \( x'^{0}, x'^{1} \) to those equations makes the components \( x'^{0}, x'^{1} \) into 1 or 0:

\( \frac{\partial}{\partial x'^{0}} = \gamma ( \frac{\partial}{\partial x^{0}} + \beta \frac{\partial}{\partial x^{1}} ) \)
\( \frac{\partial}{\partial x'^{1}} = \gamma ( \beta \frac{\partial}{\partial x^{0}} + \frac{\partial}{\partial x^{1}} ) \)
\( \frac{\partial}{\partial x'^{2}} = \frac{\partial}{\partial x^{2}} \)
\( \frac{\partial}{\partial x'^{3}} = \frac{\partial}{\partial x^{3}} \)

By inspection, this is the same mathematical form as the four vector \( a_{\mu} \) under an \( x^{1} \) boost, the result from Problem 2.3 (a):

\( a'_{0} = \gamma ( a_{0} + \beta a_{1} ) \)
\( a'_{1} = \gamma ( \beta a_{0} + a_{1} ) \)
\( a'_{2} = a_{2} \)
\( a'_{3} = a_{3} \)

Therefore we can conclude that the gradient partial derivative of an upper-index four-vector transforms the same way as a lower-index four-vector.


Problem 2.3 (c): Show that in quantum mechanics the expressions of energy and momentum, in terms of derivatives, can be written compactly as \(p_{\mu} = \frac{\hbar}{i} \frac{\partial}{\partial x^{\mu}} \).

\( p_{\mu} = \eta_{\mu \nu} p^{\nu} = (- \frac{E}{c}, p_{x}, p_{y}, p_{z} ) \)

Note the first quantization of the observables on the right-hand side of this equation:

\( E = i\hbar \frac{\partial}{\partial t} \)
\( p_{x} = - i\hbar \frac{\partial}{\partial x} \)
\( p_{y} = - i\hbar \frac{\partial}{\partial y} \)
\( p_{z} = - i\hbar \frac{\partial}{\partial z} \)

Therefore if we define \( p_{\mu} \) as a four-vector, where E is the time component:

\( p_{\mu} = ( - \frac{1}{c} i\hbar \frac{\partial}{\partial t}, - i\hbar \frac{\partial}{\partial x}, - i\hbar \frac{\partial}{\partial y}, - i\hbar \frac{\partial}{\partial z} ) \)

If we express this in terms of \( \frac{\partial}{\partial x^{\mu}} \), the factor of \( \frac{1}{c} \) folds into the time component \( x^{0} \) because of ct. This gives the expression:

\( p_{\mu} = - i\hbar \frac{\partial}{\partial x^{\mu}} \)

Since \( -i = \frac{1}{i} \), this becomes:

\( p_{\mu} = \frac{\hbar}{i} \frac{\partial}{\partial x^{\mu}} \) Q.E.D.


Problem 2.4

Problem Statement: A matrix L that satifies \( L^{T} \eta L = \eta \) is a Lorentz transformation. Show the following.

(a) If \( L_{1} \) and \( L_{2} \) are Lorentz transformations, so is their product \( L_{1} L_{2} \).
(b) If L is a Lorentz transformation then so is the inverse matrix \( L^{-1} \).
(c) If L is a Lorentz transformation then so is the transpose matrix \( L^{T} \).

Solution Heuristic: Matrix multiplication rules and algebra manipulations.

(a) Substitute \( L_{1} L_{2} \) for "L" in the definition. Apply the product transform rule and simplify.
(b) Apply inverses of the terms in the rule, which are identity, then use the matrix rule for inverses of transposes to find the expression \( (L^{-1})^T \eta L^{-1} = \eta \).
(c) Apply inverse multiplications of the result from (b), and note \( \eta = \eta^{-1}\), and show it yields the expression \( L \eta L^{T} = \eta \).

Solution:

Problem 2.4 (a): If \( L_{1} \) and \( L_{2} \) are Lorentz transformations, so is their product \( L_{1} L_{2} \).

Let \( L_{1}, L_{2} \) be Lorentz transformations. Therefore each separately obeys the rule \( L^{T} \eta L = \eta \).

Substitute the product \( L_{1} L_{2} \) into this rule:

\( (L_{1} L_{2})^{T} \eta ( L_{1} L_{2} ) = \eta \)

The matrix transpose rule for products, assuming the scalars are commutative, is \( (A B)^{T} = B^{T} A^{T} \):

\( \Rightarrow (L_{1} L_{2})^{T} \eta ( L_{1} L_{2} ) = (L_{2}^{T} L_{1}^{T}) \eta ( L_{1} L_{2} ) \)
\( \Rightarrow L_{2}^{T} (L_{1}^{T} \eta L_{1}) L_{2} = \eta \)

The associativity property lets us recognize the rule for \( L_{1} \), which allows the middle term to become \( \eta \):

\( L_{2}^{T} \eta L_{2} = \eta \)

Which is just the same definition for \( L_{2} \). Therefore the product \( L_{1} L_{2} \) is itself a Lorentz transformation. Q.E.D.


Problem 2.4 (b): If L is a Lorentz transformation then so is the inverse matrix \( L^{-1} \).

The argument in part (a) cannot be used here, because using \( L^{-1} \) in it would be assuming the conclusion. We start with the definition that \( L^{T} \eta L = \eta \) if L is a Lorentz transform. Multiplying the left hand side by inverse matrices by definition also yields \( \eta \) because they become the identity matrix:

\( [( L^{T})^{-1} L^{T} ] \eta [ L L^{-1} ] = \eta \)

This is justified because \( det(L^{T})det(\eta)det(L) = det(\eta) \) implies \( det(L) = \pm 1 \) in general, which means \(L^{-1}\) as a matrix always exists if L exists. Notice the matrix transpose rule that \( (L^{T})^{-1} = (L^{-1})^{T} \):

\( (L^{-1})^{T} L^{T} \eta L L^{-1} = \eta \)
\( \Rightarrow (L^{-1})^{T} ( L^{T} \eta L ) L^{-1} = (L^{-1})^{T} \eta L^{-1} = \eta \)

This proves that if L is a Lorentz transformation, then its inverse \( L^{-1} \) also obeys the rule for Lorentz transformations, and therefore \( L^{-1} \) must also be a Lorentz transformation. Q.E.D.


Problem 2.4 (c): If L is a Lorentz transformation then so is the transpose matrix \( L^{T} \).

Beginning with the previous result of \( (L^{-1})^{T} \eta L^{-1} = \eta \), and multiplying it through with inverse matrices until it is all inverted:

\( \Rightarrow (L^{-1})^{T} \eta [ L^{-1} L ] = \eta L \)
\( \Rightarrow (L^{-1})^{T} \eta = \eta L \)

\( \Rightarrow \eta^{-1} (L^{-1})^{T} \eta = [\eta^{-1} \eta] L \)
\( \Rightarrow \eta^{-1} (L^{-1})^{T} \eta = L \)

\( \Rightarrow \eta^{-1} (L^{-1})^{T} [\eta \eta^{-1}] = L \eta^{-1} \)
\( \Rightarrow \eta^{-1} (L^{-1})^{T} = L \eta^{-1} \)

\( \Rightarrow L^{-1} \eta^{-1} (L^{-1})^{T} = [ L^{-1} L ] \eta^{-1} \)
\( \Rightarrow L^{-1} \eta^{-1} (L^{-1})^{T} = \eta^{-1} \)

In the Minkowski spacetime, the metric is its own inverse, so \( \eta = \eta^{-1} \), and recall that \( (L^{T})^{-1} = (L^{-1})^{T} \):

\( \Rightarrow L^{-1} \eta (L^{-1})^{T} = \eta \)
\( \Rightarrow L^{-1} \eta (L^{T})^{-1} = \eta \)

Multiplying \( L \) to the left of both sides and \( L^{T} \) to the right of both sides flips the expression out of inverse terms:

\( [L L^{-1}] \eta (L^{T})^{-1} = L \eta \)
\( \Rightarrow \eta [(L^{T})^{-1} L^{T}] = L \eta L^{T}\)
\( \Rightarrow \eta = L \eta L^{T} \)

These are just the transposes of the terms in the rule \( L^{T} \eta L = \eta\). Therefore if L is a Lorentz transformation, \( L^{T} \) is a Lorentz transformation as well. Q.E.D.



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Last updated: 8/8/2020