Einstein Relatively Easy

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In the previous article The Riemann curvature tensor part I: derivation from covariant derivative commutator, we have shown a way to derive the Riemann tensor from the covariant derivative commutator, which physically corresponds to the difference of parallel transporting a vector first in one way and then the other, versus the opposite.

Another interpretation is in terms of relative acceleration of nearby particles in free-fall.

Imagine a cloud of particles in free-fall. Let us suppose that an observer is travelling with one of the particles, and that he looks at a nearby particle and measures its position in local inertial coordinates. In special relativity, it will move in a straight line at constant speed with no acceleration. But what happens in a gravitational field?

As we recall from our article Geodesic equation and Christoffel symbols ,  a geodesic generalizes the notion of a "straight line" to curved spacetime.

Here we will show how the evolution of the separation measured between two adjacent geodesics, also known as geodesic deviation can indeed be related to a non-zero curvature of the spacetime, or to use a Newtonian language, to the presence of tidal force.

So let us pick out two particles following two very close geodesics.

Their respective path could be described by the functions xμ(τ) (reference particle) and yμ(τ)≡xμ(τ) + ξμ(τ) (second particle) where τ (tau) is the proper time along the reference particle's worldline and where ξ refers to the deviation four-vector joining one particle to the other at each given time τ.

The relative acceleration Aα of the two objects is defined, roughly, as the second derivative of the separation vector ξα as the objects advance along their respective geodesics.

Our goal in this article is to show that this relative acceleration is related to the Riemann tensor by the following equation

for which a null Riemann tensor leads to a null relative acceleration between the particles, which is equivalent to say that the spacetime is flat.

So lets try to demonstrate it.

As each particle follows a geodesic, the equation of their respective coordinate is:

In each of these equations, the Christoffel symbol is evaluated at each particle's x and y respective position. As the separation among particles is infinitesimal, we can therefore evaluate Christoffel symbol at yα(Τ ) position by a Taylor series development.

With the assumption that yα(τ) = xα(τ) + ξα(τ) and by replacing this last expression in the y particle's geodesic equation, we get:

where the Christoffel Symbol and its first order derivatives are now evaluated in xα(Τ )

By developing all the terms in parenthesis, and cancelling out those terms in second order with respect to ξ  we get:

By using the geodesic equation of particule x:

we finally can write:


where uμ ≡ dxμ/dτ is the four-velocity vector of the reference particle.

We now have an expression for dξα/dτ, but as usual, this isn’t the total derivative of the four-vector ξ, since its derivative could also get a contribution from the change of the basis vectors  as the object moves along its geodesic. To get the total derivative, we have

By replacing the dummy index α by σ in the second term and from the definition of the Christoffel Symbol or Connection coefficient we get

So that

Since we’re still dealing with the condition that ξ is a four-vector, its derivative with respect to proper time is also a four-vector, so we can find the second absolute derivative by using the same development as for the first order derivative


We can rewrite the second term as the Christoffel symbols depends on τ by  depending on the position of the reference particle


 By using the geodesic equation, we can rewrite the third term, i.e developing out duμ/dτ

Also in order to obtain an epression with μ, ν and ξ only we can rewrite the last term by renaming the dummy indices σ and β


By using the equation of Christoffel symbol Taylor's serie above and replacing ν by σ in the first term, we get


So finally we can write, replacing all the terms

 By cancelling out the first and fifth terms and taking out the common factor uμuνξσ

Since this is still a tensor equation, the quantity in brackets is a tensor and we can define  the Riemann tensor as the opposite of this quantity

Then we can rewrite the above equation in a shorter expression, known as the geodesic deviation equation

Since the only quantity in this equation that depends intrinsically on the metric is the Riemann tensor, we see that if it is identically zero, spacetime is flat, but if only one component of this tensor is non-zero, spacetime is curved.

We’ll explore some more properties and examples of the Riemann tensor in future posts.




"The essence of my theory is precisely that no independent properties are attributed to space on its own. It can be put jokingly this way. If I allow all things to vanish from the world, then following Newton, the Galilean inertial space remains; following my interpretation, however, nothing remains.."
Letter from A.Einstein to Karl Schwarzschild - Berlin, 9 January 1916

"Quantum mechanics is certainly imposing. But an inner voice tells me that it is not yet the real thing. The theory says a lot, but does not really bring us any closer to the secret of the 'old one'. I, at any rate, am convinced that He is not playing at dice."
Einstein to Max Born, letter 52, 4th december 1926

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