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?

Answer 1

You look at the tangent space to be able to not have to worry about the effects of curvature. You can see that fairly intuitively on a sphere. If you think about segments of a path along the equator, from one point of view they are parallel. From another point of view they are clearly not. So in the tangent space, which is flat, this consideration is avoided. There are ways to deal with this, but they get very complicated very quickly. For example, if you wanted to compare vectors that were separated, one possible way is to parallel transport them (meaning, keep their angle with the geodesic constant) until they are at the same point. And then compare them in the tangent space. But that's getting very far afield.

The ordinary way to describe two vectors as parallel is if you can overlap one on the other. That looks like so for two vectors $ \overrightarrow X $ and $\overrightarrow Y $. If we can then find a real number $K$ not equal to zero such that $ \overrightarrow X = K \overrightarrow Y $ then they are parallel.

Being a relativistic 4-vector complicates it a bit. This is because the 4-vector's length-squared is defined as $ x_1^2+x_2^2+x_3^2-x_0^2 $ that is, the the norm is not necessarily positive, even for non-zero vectors. (Let me be lazy about that for the rest of this post and always mean length-squared even if I say length.) A vector representing a particle at rest, for example, only has the $x_0$ component and has negative norm, meaning it is "time-like." If it is zero that is called light-like. And positive is space-like.

So massive particles in a fluid should all have time-like 4-vectors for their motion, even at rest. (If you were talking about a "fluid" consisting of photons, then each would have a light-like 4-vector with zero norm. Let me neglect that one.)

So consider two objects that are moving in the lab frame with regular 3-velocity of $\overrightarrow {V_1}$ and $\overrightarrow {V_2}$. So that looks like so. The 4-vector velocity (with c=1) for each particle is like so.

$\gamma_1 ( 1, \overrightarrow {V_1})$ and $\gamma_2 ( 1, \overrightarrow {V_2})$

Here $\gamma = \frac{1}{\sqrt{1-V^2}} $ is the usual relativistic dialation factor. Note that the 4-vectors for these particles then have magnitude $ \gamma^2 (-1+V^2) = -1$ so they are time-like.

So when are these parallel? In one sense, they are parallel when the 3-velocities are parallel. That is, the usual boring old familiar idea of parallel velocity.

However, that $x_0$ term is in there. So, in another sense, they are not parallel unless $\overrightarrow {V_1} = \overrightarrow {V_2}$. If this is not true, if they have a relative velocity even if it is parallel, then their $x_0$ terms will drift relative to eachother.

What does it all mean?

If you were looking for how the fluid behaved regarding whether it was getting denser or less dense you might want the 3-velocity. Converging 3-velocities will probably mean getting denser. Probably that's where you want to go for star formation, for example. If you needed to compare the clock readings of different particles you would need to look at 4-velocity. That might be what you needed for cosmology.

Answer 2

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.