Does the concept of parallel and perpendicular vectors to a given four vector exists?

Question

In relativistic fluid dynamics I encountered a way of writing $partial_mu $ as follows:
$partial_mu = mu_mu D + nabla_mu $. Here $mu_mu $ is the four-velocity. $D=mu^alpha partial_alpha=(partial_t,mathbf 0 ) $ and $nabla_mu= (0,partial_i)$. Can anyone justify how is this relation holding. They provided a reason that it is taking components parallel and perpendicular to the derivatives. What is the concept of parallel vectors in four-vector notation?

Mozibur Ullah · Answer

Well, I'm not well versed in relativistic fluid dynamics ;) - which is why I was talking in general terms!

Basically, tensors live in the tangent space of a manifold and you don't need a metric to define them; however, without a metric you can't define what it means to be orthogonal (in the tangent space) and also, with a metric you get what's called a connection, and this allows you to parallel transport vectors from one point of the manifold to another.

I've glossed over here that tensors actually live not just in the tangent space (this would give you a tensor with just one upper index) but other spaces That you can build from the tangent space, like the cotangent space (which would give you a tensor with one lower index). If you want higher tensors then you 'multiply' appropriate numbers of these types of tensors together.

It looks to me that the equation you've written above is using what's called the covariant derivative. Essentially, when we're in the usual ordinary Euclidean space we use the partial derivative:

$partial_mu$

But when we're in a curved context we change this to what's called the covariant derivative and this is of the form:

$partial_mu + nabla_mu$

The $nabla$ is what's called the connection and this is derived from the metric.

Does the concept of parallel and perpendicular vectors to a given four vector exists?

One Answer

Add your own answers!

Ask a Question