I understand by expanding out the Riemann tensor, that the Bianchi identities can be derived within a local inertial frame (LIF) by taking the partial derivatives of the Riemann tensor relations in a LIF using the fact that Christoffel symbols are zero in a LIF. This gives

$$\nabla_e R_{abcd} + \nabla_c R_{ab,de} + \nabla_d R_{abec}=0,$$ where $\nabla$ is the covariant derivative and $R$ is the Riemann tensor.

It has been suggested to me that this is a tensor equation and hence holds in all coordinates, but I am not sure I understand this argument. Could someone please suggest an intuitive way to show that this holds in all coordinates?


1 Answer 1


The most powerful behaviour of a tensor equation is precisely this feature that if the tensor components of a tensor equation holds in one frame then in another frame, up to a coordinate transformation this equation is also valid. The simplest way to see that is the following:

Consider the tensor equation:

$$A^{\mu \nu} - B^{\mu \nu} = C^{\mu \nu} \tag{1} $$

If the components of $$A^{\mu \nu} - B^{\mu \nu} = 0$$ in the frame $S$, then in a frame $S'$ this components are also zero.


$$A'^{\mu \nu} - B'^{\mu \nu} = \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}A^{\delta \gamma} - \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}B^{\delta \gamma} = \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}(A^{\delta \gamma} - B^{\delta \gamma}) = \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}(0) \implies $$

$$\implies A'^{\mu \nu} - B'^{\mu \nu} = 0 $$

In the case of Bianchi Identities with a ($LIF = S$) coordinate system (which the Christoffell symbols can be made zero, but not their derivatives) you just need to perform another general coordinate transformation for Riemann tensor and covariant derivatives to see that the equation is valid in the $S' \neq (LIF)$

  • $\begingroup$ Thanks, I can see that now from maths. Any ideas for an intuitive non-mathematical demonstration? $\endgroup$ May 11, 2019 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.