Intuition The big pictureA vector lives in a space. To do geometry (lengths, angles, dot products) you need a ruler that tells you how to measure. That ruler is the metric tensor g μ ν g_{\mu\nu} g μν
WHAT it does: converts between two "dialects" for describing the same vector — contravariant components V μ V^\mu V μ covariant components V μ V_\mu V μ WHY we need it: in non-Cartesian / curved / Minkowski spaces, V μ V^\mu V μ V μ V_\mu V μ different lists of numbers . The dot product A ⃗ ⋅ B ⃗ \vec A\cdot\vec B A ⋅ B ∑ A μ B μ \sum A^\mu B^\mu ∑ A μ B μ g μ ν A μ B ν g_{\mu\nu}A^\mu B^\nu g μν A μ B ν HOW : multiply by g μ ν g_{\mu\nu} g μν lower , by its inverse g μ ν g^{\mu\nu} g μν raise .
Definition Contravariant vs covariantPick a basis { e μ } \{\mathbf e_\mu\} { e μ }
Contravariant components V μ V^\mu V μ V = V μ e μ \mathbf V = V^\mu \mathbf e_\mu V = V μ e μ opposite to the basis — shrink the basis, the components grow.Covariant components V μ V_\mu V μ projecting with the dual/reciprocal basis. They transform like the basis.
In plain Cartesian coordinates with orthonormal axes the two coincide, which is why high school never mentions them.
Intuition Why "opposite" naming?Change units from metres to centimetres. A basis vector e x \mathbf e_x e x smaller (1 cm < 1 m). But the number of those units in a fixed physical displacement gets bigger (300 cm vs 3 m). The components vary contra (against) the basis — hence contra variant, upper index.
g μ ν ≡ e μ ⋅ e ν g_{\mu\nu} \equiv \mathbf e_\mu \cdot \mathbf e_\nu g μν ≡ e μ ⋅ e ν g μ ν = g ν μ g_{\mu\nu}=g_{\nu\mu} g μν = g ν μ every pairwise dot product of basis vectors. The inverse metric g μ ν g^{\mu\nu} g μν g μ α g α ν = δ μ ν ( Kronecker delta ) . g^{\mu\alpha}g_{\alpha\nu} = \delta^{\mu}{}_{\nu}\qquad(\text{Kronecker delta}). g μα g α ν = δ μ ν ( Kronecker delta ) .
WHY this is the right object. The squared length of V = V μ e μ \mathbf V = V^\mu \mathbf e_\mu V = V μ e μ ∥ V ∥ 2 = V ⋅ V = ( V μ e μ ) ⋅ ( V ν e ν ) = V μ V ν ( e μ ⋅ e ν ) = g μ ν V μ V ν . \|\mathbf V\|^2 = \mathbf V\cdot\mathbf V = (V^\mu\mathbf e_\mu)\cdot(V^\nu\mathbf e_\nu) = V^\mu V^\nu\,(\mathbf e_\mu\cdot\mathbf e_\nu) = g_{\mu\nu}V^\mu V^\nu. ∥ V ∥ 2 = V ⋅ V = ( V μ e μ ) ⋅ ( V ν e ν ) = V μ V ν ( e μ ⋅ e ν ) = g μν V μ V ν . g μ ν g_{\mu\nu} g μν forced on us the moment we want lengths from components.
Intuition Define the covariant component as a projectionDefine V μ ≡ V ⋅ e μ V_\mu \equiv \mathbf V\cdot\mathbf e_\mu V μ ≡ V ⋅ e μ V \mathbf V V
Derivation.
V μ = V ⋅ e μ = ( V ν e ν ) ⋅ e μ = V ν ( e ν ⋅ e μ ) = g μ ν V ν . V_\mu = \mathbf V\cdot\mathbf e_\mu = (V^\nu \mathbf e_\nu)\cdot \mathbf e_\mu = V^\nu(\mathbf e_\nu\cdot\mathbf e_\mu) = g_{\mu\nu}V^\nu. V μ = V ⋅ e μ = ( V ν e ν ) ⋅ e μ = V ν ( e ν ⋅ e μ ) = g μν V ν .
Raising. Multiply both sides by g α μ g^{\alpha\mu} g α μ g α μ g μ ν = δ α ν g^{\alpha\mu}g_{\mu\nu}=\delta^\alpha{}_\nu g α μ g μν = δ α ν g α μ V μ = g α μ g μ ν V ν = δ α ν V ν = V α . g^{\alpha\mu}V_\mu = g^{\alpha\mu}g_{\mu\nu}V^\nu = \delta^\alpha{}_\nu V^\nu = V^\alpha. g α μ V μ = g α μ g μν V ν = δ α ν V ν = V α .
Mnemonic "g eats the matching index, hands you the new one"Metric Down, Lower Down — g g g down indices lowers . g g g up indices (g μ ν g^{\mu\nu} g μν raises . The repeated (summed) index always appears once up, once down.
Raise/lower one index at a time , contracting with g g g T μ ν = g ν α T μ α , T μ ν = g ν β T μ β , R μ ν = g α β R μ α ν β . T^{\mu}{}_{\nu} = g_{\nu\alpha}T^{\mu\alpha},\qquad T^{\mu\nu}=g^{\nu\beta}T^\mu{}_\beta,\qquad R_{\mu\nu}=g^{\alpha\beta}R_{\mu\alpha\nu\beta}. T μ ν = g ν α T μα , T μν = g ν β T μ β , R μν = g α β R μα ν β . WHY one at a time: each application of g g g one free index with a dummy, converting exactly that slot.
A neat consistency check: lower then raise must give back the original.
g μ β ( g β ν V ν ) = ( g μ β g β ν ) V ν = δ μ ν V ν = V μ . ✓ g^{\mu\beta}(g_{\beta\nu}V^\nu) = (g^{\mu\beta}g_{\beta\nu})V^\nu = \delta^\mu{}_\nu V^\nu = V^\mu.\ \checkmark g μ β ( g β ν V ν ) = ( g μ β g β ν ) V ν = δ μ ν V ν = V μ . ✓
Worked example (a) Minkowski spacetime, signature( − , + , + , + ) (-,+,+,+) ( − , + , + , + ) g μ ν = d i a g ( − 1 , 1 , 1 , 1 ) g_{\mu\nu}=\mathrm{diag}(-1,1,1,1) g μν = diag ( − 1 , 1 , 1 , 1 ) U μ = ( 2 , 1 , 0 , 0 ) U^\mu=(2,1,0,0) U μ = ( 2 , 1 , 0 , 0 )
U 0 = g 0 ν U ν = g 00 U 0 = ( − 1 ) ( 2 ) = − 2 U_0 = g_{0\nu}U^\nu = g_{00}U^0 = (-1)(2)=-2 U 0 = g 0 ν U ν = g 00 U 0 = ( − 1 ) ( 2 ) = − 2 Why? diagonal metric ⇒ only ν = 0 \nu=0 ν = 0 U 1 = g 11 U 1 = ( 1 ) ( 1 ) = 1 U_1 = g_{11}U^1 = (1)(1)=1 U 1 = g 11 U 1 = ( 1 ) ( 1 ) = 1 So U μ = ( − 2 , 1 , 0 , 0 ) U_\mu=(-2,1,0,0) U μ = ( − 2 , 1 , 0 , 0 ) that is the whole point of an indefinite metric.
Norm: U μ U μ = ( − 2 ) ( 2 ) + ( 1 ) ( 1 ) + 0 + 0 = − 3. U_\mu U^\mu = (-2)(2)+(1)(1)+0+0 = -3. U μ U μ = ( − 2 ) ( 2 ) + ( 1 ) ( 1 ) + 0 + 0 = − 3.
Worked example (b) 2D polar coordinatesLine element d s 2 = d r 2 + r 2 d θ 2 ⇒ g μ ν = ( 1 0 0 r 2 ) ds^2 = dr^2 + r^2 d\theta^2 \Rightarrow g_{\mu\nu}=\begin{pmatrix}1&0\\0&r^2\end{pmatrix} d s 2 = d r 2 + r 2 d θ 2 ⇒ g μν = ( 1 0 0 r 2 ) g μ ν = ( 1 0 0 1 / r 2 ) g^{\mu\nu}=\begin{pmatrix}1&0\\0&1/r^2\end{pmatrix} g μν = ( 1 0 0 1/ r 2 ) V μ = ( V r , V θ ) V_\mu=(V_r,V_\theta) V μ = ( V r , V θ )
V r = g r r V r = V r V^r = g^{rr}V_r = V_r V r = g r r V r = V r V θ = g θ θ V θ = V θ / r 2 V^\theta = g^{\theta\theta}V_\theta = V_\theta/r^2 V θ = g θ θ V θ = V θ / r 2 Why the 1 / r 2 1/r^2 1/ r 2 The θ \theta θ e θ \mathbf e_\theta e θ r r r
Worked example (c) Off-diagonal metricg μ ν = ( 1 1 1 2 ) g_{\mu\nu}=\begin{pmatrix}1&1\\1&2\end{pmatrix} g μν = ( 1 1 1 2 ) V μ = ( 3 , 1 ) V^\mu=(3,1) V μ = ( 3 , 1 )
V 1 = g 1 ν V ν = g 11 V 1 + g 12 V 2 = ( 1 ) ( 3 ) + ( 1 ) ( 1 ) = 4. V_1 = g_{1\nu}V^\nu = g_{11}V^1+g_{12}V^2 = (1)(3)+(1)(1)=4. V 1 = g 1 ν V ν = g 11 V 1 + g 12 V 2 = ( 1 ) ( 3 ) + ( 1 ) ( 1 ) = 4. V 2 = g 21 V 1 + g 22 V 2 = ( 1 ) ( 3 ) + ( 2 ) ( 1 ) = 5. V_2 = g_{21}V^1+g_{22}V^2 = (1)(3)+(2)(1)=5. V 2 = g 21 V 1 + g 22 V 2 = ( 1 ) ( 3 ) + ( 2 ) ( 1 ) = 5. Why include both terms? Off-diagonal g 12 ≠ 0 g_{12}\neq0 g 12 = 0 not ignore cross terms.
V μ V_\mu V μ V μ V^\mu V μ Why it feels right: in Cartesian orthonormal coordinates g μ ν = δ μ ν g_{\mu\nu}=\delta_{\mu\nu} g μν = δ μν are equal — that's everyone's first experience.
Fix: they are equal only when g = 1 g=\mathbb 1 g = 1
Common mistake Dot product= ∑ μ A μ B μ = \sum_\mu A^\mu B^\mu = ∑ μ A μ B μ Why it feels right: the familiar Euclidean formula.
Fix: the invariant is A μ B μ = g μ ν A μ B ν A^\mu B_\mu = g_{\mu\nu}A^\mu B^\nu A μ B μ = g μν A μ B ν one up, one down . Two ups (or two downs) is not a valid contraction.
g μ ν g_{\mu\nu} g μν raise .
Why it feels right: "the metric does index gymnastics, so use it for everything."
Fix: down-indexed g μ ν g_{\mu\nu} g μν lowers ; you need the inverse g μ ν g^{\mu\nu} g μν g g g
Recall Feynman: explain to a 12-year-oldImagine a treasure map drawn on stretchy rubber. To say where the treasure is you can either count steps ("3 east, 2 north") or describe how strongly it lines up with each direction. On a flat un-stretched map these two descriptions give the same numbers. But if the rubber is stretched more in one direction, the two descriptions disagree — and you need a little "stretch table" (the metric) to translate one into the other. Lowering an index = use the stretch table; raising = use the undo table.
Recall Active recall checkpoint
Define g μ ν g_{\mu\nu} g μν
Derive V μ = g μ ν V ν V_\mu=g_{\mu\nu}V^\nu V μ = g μν V ν
Why must the inverse metric raise indices?
For d i a g ( − 1 , 1 , 1 , 1 ) \mathrm{diag}(-1,1,1,1) diag ( − 1 , 1 , 1 , 1 ) U μ = ( 2 , 1 , 0 , 0 ) U^\mu=(2,1,0,0) U μ = ( 2 , 1 , 0 , 0 ) U μ U_\mu U μ
What is the metric tensor g μ ν g_{\mu\nu} g μν The dot product of basis vectors,
g μ ν = e μ ⋅ e ν g_{\mu\nu}=\mathbf e_\mu\cdot\mathbf e_\nu g μν = e μ ⋅ e ν ; a symmetric rank-2 tensor.
Formula to lower an index? V μ = g μ ν V ν V_\mu = g_{\mu\nu}V^\nu V μ = g μν V ν .
Formula to raise an index? V μ = g μ ν V ν V^\mu = g^{\mu\nu}V_\nu V μ = g μν V ν , where
g μ ν g^{\mu\nu} g μν is the inverse metric.
Defining relation between g μ ν g^{\mu\nu} g μν g μ ν g_{\mu\nu} g μν g μ α g α ν = δ μ ν g^{\mu\alpha}g_{\alpha\nu}=\delta^\mu{}_\nu g μα g α ν = δ μ ν (they are matrix inverses).
Why are V μ V^\mu V μ V μ V_\mu V μ They are equal only when
g μ ν = δ μ ν g_{\mu\nu}=\delta_{\mu\nu} g μν = δ μν (orthonormal Cartesian); otherwise the metric is non-trivial and rescales/mixes components.
Invariant dot product of A A A B B B A μ B μ = g μ ν A μ B ν A^\mu B_\mu = g_{\mu\nu}A^\mu B^\nu A μ B μ = g μν A μ B ν (one up, one down index summed).
In Minkowski ( − , + , + , + ) (-,+,+,+) ( − , + , + , + ) Flips its sign:
V 0 = − V 0 V_0 = -V^0 V 0 = − V 0 .
Polar metric and V θ V^\theta V θ V θ V_\theta V θ g θ θ = r 2 g_{\theta\theta}=r^2 g θ θ = r 2 ,
g θ θ = 1 / r 2 g^{\theta\theta}=1/r^2 g θ θ = 1/ r 2 , so
V θ = V θ / r 2 V^\theta = V_\theta/r^2 V θ = V θ / r 2 .
Dot product A.B = g_mu_nu A^mu B^nu
Intuition Hinglish mein samjhoDekho, ek vector ko describe karne ke do tarike hote hain: contravariant components V μ V^\mu V μ covariant components V μ V_\mu V μ metric tensor g μ ν g_{\mu\nu} g μν g μ ν = e μ ⋅ e ν g_{\mu\nu}=\mathbf e_\mu\cdot\mathbf e_\nu g μν = e μ ⋅ e ν
Index lower karna ho (upar se neeche laana) to g μ ν g_{\mu\nu} g μν V μ = g μ ν V ν V_\mu=g_{\mu\nu}V^\nu V μ = g μν V ν raise karna ho to uska inverse g μ ν g^{\mu\nu} g μν V μ = g μ ν V ν V^\mu=g^{\mu\nu}V_\nu V μ = g μν V ν g g g g g g
Yeh important kyun hai? Kyunki real dot product / length ∑ V μ V μ \sum V^\mu V^\mu ∑ V μ V μ g μ ν V μ V ν g_{\mu\nu}V^\mu V^\nu g μν V μ V ν g = d i a g ( − 1 , 1 , 1 , 1 ) g=\mathrm{diag}(-1,1,1,1) g = diag ( − 1 , 1 , 1 , 1 )
