Under a coordinate transformation x^μ → x'^μ, a rank-(1,1) tensor T^μ_ν transforms as:
AT'^μ_ν = T^μ_ν (tensors are coordinate-invariant)
BT'^μ_ν = (∂x'^μ/∂x^α)(∂x^β/∂x'^ν) T^α_β
CT'^μ_ν = (∂x^α/∂x'^μ)(∂x'^ν/∂x^β) T^α_β
DT'^μ_ν = (∂x'^μ/∂x^α)(∂x'^ν/∂x^β) T^α_β
Each upper (contravariant) index transforms with ∂x'^μ/∂x^α, and each lower (covariant) index transforms with ∂x^β/∂x'^ν. For a (1,1) tensor, this gives T'^μ_ν = (∂x'^μ/∂x^α)(∂x^β/∂x'^ν) T^α_β. Option A confuses invariance of tensor equations (which hold in all frames) with invariance of components (which change). Option C and D incorrectly assign the Jacobian factors.
Question 2 True / False
The partial derivative of a vector field ∂_μ V^ν is itself a tensor.
TTrue
FFalse
Answer: False
Under a general coordinate transformation, ∂_μ V^ν picks up an extra term involving the second derivative of the coordinate transformation, ∂²x'^ν/∂x^α∂x^β, which spoils the tensor transformation law. This is why the covariant derivative ∇_μ V^ν is needed — the Christoffel symbol connection term cancels the unwanted second-derivative piece, producing an object that transforms as a proper (1,1) tensor. In flat spacetime with Cartesian coordinates the Christoffel symbols vanish and partial derivatives coincide with covariant derivatives, but this is a special case.
Question 3 Short Answer
Explain what it means to 'raise an index' on a covariant vector V_μ, and why the resulting object V^μ is physically the same vector expressed differently.
Think about your answer, then reveal below.
Model answer: Raising an index means contracting with the inverse metric: V^μ = g^{μν} V_ν. The covariant components V_μ and the contravariant components V^μ are two different representations of the same geometric object — the vector V living in the tangent space. The metric provides the isomorphism between the tangent space and its dual (cotangent space). In flat spacetime with Cartesian coordinates, g^{μν} = η^{μν} and raising/lowering only flips the sign of the time component, but in curved spacetime or non-Cartesian coordinates the relationship is nontrivial.
Index raising and lowering is fundamental to GR computation. The metric tensor serves as the bridge between vectors and one-forms, and this duality is built into every tensor equation. Physically, the distinction between V^μ and V_μ reflects the difference between a displacement direction and a gradient direction, which coincide only in orthonormal coordinates.
Question 4 Short Answer
Why does general relativity require that all physical laws be expressed as tensor equations?
Think about your answer, then reveal below.
Model answer: The principle of general covariance states that the laws of physics must take the same form in all coordinate systems, since coordinates are arbitrary labels with no physical content. Tensor equations automatically satisfy this requirement: if a tensor equation holds in one coordinate system, the transformation law guarantees it holds in every coordinate system. Non-tensorial equations, by contrast, can be true in one coordinate system and false in another, making them coordinate-dependent rather than physical. Expressing laws as tensor equations ensures that physics does not depend on the observer's choice of coordinates.
General covariance is the mathematical implementation of the equivalence principle's lesson: no coordinate system is preferred. This does not mean all coordinate systems are equally convenient — Schwarzschild coordinates may be more practical than Kruskal for some problems — but the physics must be the same regardless.