Definition of tangent vector to a timelike geodesic

[TimWilliams87]

TL;DR Summary

Clarifying definition of tangent vector to a timelike geodesic

I am considering the definition of a tangent vector field $\psi^{\mu}$ to a timelike geodesic and slightly confused as to how it works for spacetimes.

If a curve is parametrised by some parameter $\lambda$, the tangent to the curve is given by a four-vector $dx^{\mu}/ d \lambda$, as expected for the equation for a tangent to a curve in a manifold, but I am not sure what happens with the ''time'' component of the tangent vector.

As an example, if we have flat Minkowski space and a massive particle in the space which follows a timelike geodesic whose spatial components are given by $x_i = ( f_1 (t), f_2 (t), f_3(t))$ such that the curve is parametrised by time, does that mean when you take the tangent vector $\psi^{\mu}$ that the zero component of the tangent four-vector vanishes, or is it constant?

Or is instead that the curve is timelike and parametrised by the proper time $\tau$ so that the spatial components of the tangent vector to the curve vanish (because it contains $t$ not $\tau$) and the time component is constant?

 

[PeroK]

I am considering the definition of a tangent vector field \psi^{\mu} to a timelike geodesic and slightly confused as to how it works for spacetimes.

Please fix your Latex: $\psi^{\mu}$.

Two hashes either side of inline code. Two dollars either side for this;
$$E_0 = mc^2$$

 

[Ibix]

As an example, if we have flat Minkowski space and a massive particle in the space which follows a timelike geodesic whose spatial components are given by x_i = ( f_1 (t), f_2 (t), f_3(t)) such that the curve is parametrised by time, does that mean when you take the tangent vector $\psi^{\mu}$ that the zero component of the tangent four-vector vanishes, or is it constant?

The parameter $\lambda$ becomes the proper time, $\tau$, for a timelike curve, but the $t$ component of a vector is the component parallel to the direction of increase of the coordinate time (in coordinate systems that have a time coordinate, of course). So the $t$ component of the four velocity would be $dt/d\tau$.

 

[TimWilliams87]

The parameter $\lambda$ becomes the proper time, $\tau$, for a timelike curve, but the $t$ component of a vector is the component parallel to the direction of increase of the coordinate time (in coordinate systems that have a time coordinate, of course). So the $t$ component of the four velocity would be $dt/d\tau$.

So does that mean the $t$ component of the tangent vector is constant (isn't $dt/d\tau = 1$) and that the spatial components of the tangent vector finish (since they contain $t$ but the derivative is with respect to $\tau$?

 

[Ibix]

So does that mean the $t$ component of the tangent vector is constant (isn't $dt/d\tau = 1$)

$dt/d\tau$ is the inverse time dilation factor. No, it is not 1 in general.

the spatial components of the tangent vector finish (since they contain $t$ but the derivative is with respect to $\tau$?

I assume you mean are the spatial components of the four velocity zero? No, clearly not - the position of the object is changing in general. If you want to know $dx/d\tau$ you need to write $x(\tau)$ or use the chain rule to get the derivative from $x(t)$.

 

[Orodruin]

The parameter $\lambda$ becomes the proper time, $\tau$, for a timelike curve, but the $t$ component of a vector is the component parallel to the direction of increase of the coordinate time (in coordinate systems that have a time coordinate, of course). So the $t$ component of the four velocity would be $dt/d\tau$.

It is perfectly possible to use coordinate time $t$ as the curve parameter for a timelike curve. The tangent vector will have time component 1 and the vector therefore has norm < 1 (unless the spatial components all vanish). This means the vector is not the 4-velocity, but it is a tangent vector.

 

[Ibix]

It is perfectly possible to use coordinate time $t$ as the curve parameter for a timelike curve. The tangent vector will have time component 1 and the vector therefore has norm < 1 (unless the spatial components all vanish). This means the vector is not the 4-velocity, but it is a tangent vector.

Right - I got four-velocity in my head instead of the more general tangent vector. Parameterised by coordinate time, the tangent vector has norm $d\tau/dt$ times that of the four velocity.

 

[cianfa72]

Right - I got four-velocity in my head instead of the more general tangent vector. Parameterised by coordinate time, the tangent vector has norm $d\tau/dt$ times that of the four velocity.

Ok, since the four-velocity requires to employ the proper time $\tau$ along the timelike worldline as worldline's parameter.

 

[robphy]

note: The 4-momentum is also a tangent vector. The “time component (according to an observer) of that 4-momentum” (a dot product with the observer 4-velocity) is associated with “energy of the particle according to that observer”.