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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?

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TimWilliams87 said:

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;
$$E_0 = mc^2$$

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Ibix said:

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 said:

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.

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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”.