Exercises — Metric tensor — raising - lowering indices

The metric is diagonal. WHAT we do: invert a diagonal matrix by reciprocating each diagonal entry. WHY this works: for a diagonal matrix, $g^{\mu \nu }{g}_{\nu \rho }={\delta }^{\mu }{}_{\rho }$ decouples into independent equations $g^{ii}{g}_{ii}=1$ (no sum), so $g^{ii}=1/g_{ii}$. $g^{\mu \nu }=\left(\begin{array}{cc}1& 0\\ 0& \frac{1}{4}\end{array}\right).$ To raise an index you use the up-indexed inverse metric $g^{\mu \nu }$.

• (i) $A^{\mu }{B}_{\mu }$ — valid. The repeated index $\mu$ is once up, once down. This is the invariant dot product.
• (ii) $A^{\mu }{B}^{\mu }$ — not valid as a tensor expression. The index $\mu$ appears twice up; there is no metric doing the pairing, so this is basis-dependent junk.
• (iii) $g_{\mu \nu }{A}^{\mu }{B}^{\nu }$ — valid. The metric supplies the missing "down" so that $\mu$ and $\nu$ are each paired up–down. In fact (iii) equals (i), because $g_{\mu \nu }{B}^{\nu }=B_{\mu }$.

$V_{\mu }=g_{\mu \nu }{V}^{\nu }$. With the identity metric only the diagonal survives: $V_1=g_{11}{V}^1=(1)(3)=3,V_2=g_{22}{V}^2=(1)(-5)=-5.$ So $V_{\mu }=(3,-5)=V^{\mu }$. Reveal: when $g=1$, lowering changes nothing — this is exactly why school never distinguishes the two.

Lower with the down-indexed metric: $V_r=g_{rr}{V}^r=(1)(1)=1,V_{\theta }=g_{\theta \theta }{V}^{\theta }=(r^2)(3)=(4)(3)=12.$ So $V_{\mu }=(1,12)$. WHY the factor 4: ${\mathbf{e}}_{\theta }$ has length $r=2$, not $1$; the metric entry $g_{\theta \theta }=r^2=4$ builds that stretch into the dot product.

Diagonal ⇒ each component lowered independently: $p_0=g_{00}{p}^0=(-1)(5)=-5,p_1=(1)(3)=3,p_2=0,p_3=(1)(4)=4.$ So $p_{\mu }=(-5,3,0,4)$. Only the time component flips sign — the fingerprint of the indefinite (Lorentzian) metric.

Cross terms are alive, so both metric entries in each row contribute: $V_1=g_{11}{V}^1+g_{12}{V}^2=(2)(1)+(1)(4)=6,$ $V_2=g_{21}{V}^1+g_{22}{V}^2=(1)(1)+(2)(4)=9.$ So $V_{\mu }=(6,9)$. WHY both terms: $g_{12}\ne 0$ means ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ are not perpendicular, so projecting $\mathbf{V}$ onto ${\mathbf{e}}_1$ picks up a piece of the $V^2$-direction too.

Inverse of a 2×2 $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is $\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Determinant $=2\cdot 2-1\cdot 1=3$: $g^{\mu \nu }=\frac{1}{3}\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right).$ Now raise $V^{\mu }=g^{\mu \nu }{V}_{\nu }$ with $V_{\nu }=(6,9)$: $V^1=\frac{1}{3}(2\cdot 6+(-1)\cdot 9)=\frac{1}{3}(12-9)=1,$ $V^2=\frac{1}{3}((-1)\cdot 6+2\cdot 9)=\frac{1}{3}(-6+18)=4.$ So $V^{\mu }=(1,4)$ ✓ — lowering then raising is the identity, as $g^{\mu \alpha }{g}_{\alpha \nu }={\delta }^{\mu }{}_{\nu }$ guarantees.

(i) $g_{\mu \nu }{V}^{\mu }{V}^{\nu }=2(1{)}^2+2g_{12}{V}^1{V}^2+2(4{)}^2$. Carefully: $=g_{11}(1)(1)+g_{12}(1)(4)+g_{21}(4)(1)+g_{22}(4)(4)=2+4+4+32=42.$ (ii) Using $V_{\mu }=(6,9)$: $V_{\mu }{V}^{\mu }=V_1{V}^1+V_2{V}^2=(6)(1)+(9)(4)=6+36=42$ ✓. The norm $42$ is the same object however you slice it — that invariance is the whole reason the metric exists.

Lowering the first slot means the first index of $T$ gets contracted with $g$. Since $g$ is diagonal, $g_{\mu \alpha }{T}^{\alpha \nu }=g_{\mu \mu }{T}^{\mu \nu }$ (no sum), i.e. multiply row $\mu$ by $g_{\mu \mu }$:

• Row 1 ($\mu =1$): $\times g_{11}=1$ → $(0,2)$.
• Row 2 ($\mu =2$): $\times g_{22}=4$ → $(12,4)$. $T_{\mu }{}^{\nu }=\left(\begin{array}{cc}0& 2\\ 12& 4\end{array}\right).$

Lowering the second slot multiplies column $\nu$ by $g_{\nu \nu }$:

• Column 1 ($\nu =1$): $\times g_{11}=1$ → $\left(\begin{array}{c}0\\ 12\end{array}\right)$.
• Column 2 ($\nu =2$): $\times g_{22}=4$ → $\left(\begin{array}{c}8\\ 16\end{array}\right)$. $T_{\mu \nu }=\left(\begin{array}{cc}0& 8\\ 12& 16\end{array}\right).$ WHY one index at a time: each $g$ pairs exactly one free slot with a dummy; doing them sequentially guarantees you convert only the slot you intend.

First form $T^{\mu }{}_{\nu }=g_{\nu \beta }{T}^{\mu \beta }$ (lower the second index of $T^{\mu \nu }$): multiply column $\nu$ by $g_{\nu \nu }$:

• Column 1 $\times 1$: $(0,3{)}^{\top }$. Column 2 $\times 4$: $(8,4{)}^{\top }$. $T^{\mu }{}_{\nu }=\left(\begin{array}{cc}0& 8\\ 3& 4\end{array}\right).$ Now contract $\mu =\nu$ (sum the diagonal): $T^{\mu }{}_{\mu }=0+4=4$. This is the invariant trace.

Start from the left and regroup (associativity of contraction): $g^{\mu \beta }\left(g_{\beta \nu }{V}^{\nu }\right)=\left(g^{\mu \beta }{g}_{\beta \nu }\right)V^{\nu }.$ WHAT/WHY: we bracketed the two metrics because their product is the defining relation of the inverse metric: $g^{\mu \beta }{g}_{\beta \nu }={\delta }^{\mu }{}_{\nu }.$ The Kronecker delta ${\delta }^{\mu }{}_{\nu }$ is $1$ when $\mu =\nu$, else $0$; contracting it with $V^{\nu }$ simply renames the index: ${\delta }^{\mu }{}_{\nu }{V}^{\nu }=V^{\mu }.■$ No component values were used — this holds for every metric, diagonal or not, curved or flat.

Lower: $p_{\mu }=g_{\mu \nu }{p}^{\nu }=(-E,E,0,0)$ (time component flips). $p_{\mu }{p}^{\mu }=(-E)(E)+(E)(E)+0+0=-E^2+E^2=0.$ Meaning: the invariant norm is $0$ — the momentum is null (lightlike). This encodes $m^2=-p_{\mu }{p}^{\mu }=0$: the photon is massless. The metric's minus sign is exactly what makes a nonzero-energy object have zero invariant length.

Lower $A$: $A_{\mu }=(-3,1,0,0)$. Then $A_{\mu }{B}^{\mu }=(-3)(2)+(1)(2)+0+0=-6+2=-4.$ Cross-check directly with the metric: $g_{\mu \nu }{A}^{\mu }{B}^{\nu }=-A^0{B}^0+A^1{B}^1=-(3)(2)+(1)(2)=-6+2=-4.✓$ Both routes give $-4$; the negative sign says the vectors' "time overlap" dominates their "space overlap."