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GarageDweller
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In SR, Four velocity was the proper time derivative of a world line, do we have to take the covariant derivative instead in GR?
An event in spacetime is described by X = (ct, x, y, z). A displacement in spacetime isGarageDweller said:In SR, Four velocity was the proper time derivative of a world line, do we have to take the covariant derivative instead in GR?
I don't understand. That's what I just posted.stevendaryl said:I think nobody has really answered what I thought was the original question.
A point in spacetime not a 4-vector but a spacetime displacemen is a 4-vector.stevendaryl said:But why not? The real fact is that position is not a 4-vector in curved spacetime. It's really just a 4-tuple of scalars.
Boston_Guy said:I don't understand. That's what I just posted.
stevendaryl said:Well, I don't see how your post explained why
[itex]\dfrac{DQ^{\mu}}{d \tau} = \dfrac{dQ^{\mu}}{d \tau} + \Gamma^{\mu}_{\nu \lambda} Q^{\nu} U^{\lambda}[/itex]
doesn't apply in the case [itex]Q^{\mu} = X^{\mu}[/itex]. It doesn't apply because [itex]X^{\mu}[/itex] is not a 4-vector.
No. If you were right then you would be saying that the 4-velocity U is defined by the equationChestermiller said:It certainly applies when using curvilinear coordinates in flat space.
Chestermiller said:It certainly applies when using curvilinear coordinates in flat space. Try it with cylindrical coordinates and see what you get.
stevendaryl said:What do you mean "it certainly applies"? The definition of [itex]U^\mu[/itex] is just
[itex]U^\mu = \dfrac{dX^\mu}{d \tau}[/itex],
not
[itex]U^\mu = \dfrac{dX^\mu}{d \tau} + \Gamma^{\mu}_{\nu}{\lambda} U^\nu X^\lambda[/itex]
The latter is not a correct equation, and it's a circular definition of [itex]U^{\mu}[/itex], since that appears on both sides.
But since you suggested working with cylindrical coordinates, let me work out the latter. Let's have coordinates [itex]r[/itex] and [itex]\theta[/itex], which are defined in terms of cartesian coordinates [itex]x[/itex] and [itex]y[/itex] via
[itex]x = r cos(\theta)[/itex]
[itex]y = r sin(\theta)[/itex]
The connection coefficients [itex]\Gamma^{\mu}_{\nu \lambda}[/itex] turn out to be (derivation skipped)
[itex]\Gamma^{r}_{r r} = \Gamma^{r}_{r \theta} = \Gamma^{r}_{\theta r} = 0[/itex]
[itex]\Gamma^{r}_{\theta \theta} = -r[/itex]
[itex]\Gamma^{\theta}_{r r} = \Gamma^{\theta}_{\theta \theta} = 0[/itex]
[itex]\Gamma^{\theta}_{r \theta} = \Gamma^{\theta}_{\theta r} = \dfrac{1}{r}[/itex]
So the path-derivative, or whatever it is called is:
[itex]\dfrac{DQ^r}{dt} = \dfrac{dQ^r}{dt} - r Q^{\theta} U^{\theta}[/itex]
[itex]\dfrac{DQ^\theta}{dt} = \dfrac{dQ^\theta}{dt} + \dfrac{1}{r} Q^{\theta} U^{r} + \dfrac{1}{r} Q^{r} U^{\theta} [/itex]
In the particular case [itex]Q^\mu = X^{\mu}[/itex], with [itex]X^r = r[/itex] and [itex]X^\theta = \theta[/itex], this becomes:
[itex]\dfrac{Dr}{dt} = \dfrac{dr}{dt} - r \theta U^{\theta}[/itex]
[itex]\dfrac{D \theta}{dt} = \dfrac{d \theta}{dt} + \dfrac{1}{r} \theta U^{r} + \dfrac{1}{r} r U^{\theta} [/itex]
[itex]= \dfrac{d \theta}{dt} + \dfrac{1}{r} \theta U^{r} + U^{\theta} [/itex]
I don't know what those quantities are supposed to mean. Typically, the quantity [itex]X^{\mu}[/itex] is not a vector, and it doesn't make sense to take its covariant derivative, or to parallel-transport it.
Chestermiller said:In cylindrical coordinates, the component of the position vector in the θ direction is zero, not Xθ.
stevendaryl said:This might be a terminology issue about what is the "position" of an object. You are interpreting it to mean the displacement vector from the origin to the location of the object. A displacement vector certainly is a vector (well, in curved spacetime, that's only true in the limit as the displacement is small).
However, the whole point of a coordinate system is to be able to identify points in space by an n-tuple of coordinate values. In order to specify the location of an object in polar coordinates, you have to give two numbers: the distance [itex]r[/itex] from the origin, and the angle [itex]\theta[/itex] that one must pass through on a circle of radius [itex]r[/itex] centered at the origin until one gets to the object. So the location of the object is specified by the pair [itex](r,\theta)[/itex]. This pair is NOT a vector. However, the velocity vector [itex]U^{\mu}[/itex] created by differentiating the pair with respect to [itex]t[/itex] IS a vector: [itex]U^r = \dfrac{dr}{dt}[/itex], [itex]U^{\theta} = \dfrac{d \theta}{dt}[/itex]
As I said, it's a matter of terminology, I suppose, but I think it's pretty useless to describe the "position" of an object by saying "It's at a distance r from the origin along the [itex]\widehat{r}[/itex] direction. It's much more useful to say that the location is the n-tuple of coordinates needed to locate the object.
Four velocity in GR, also known as 4-velocity, is a 4-dimensional vector that describes the motion of an object in curved spacetime. It is a fundamental concept in general relativity that takes into account both the object's spatial and temporal components.
The four velocity is calculated by taking the derivative of an object's position vector with respect to its proper time. This means that it takes into account the effects of both time dilation and length contraction due to the object's motion in curved spacetime.
In general relativity, the concept of parallel transport is used to compare vectors at different points in curved spacetime. However, since spacetime is curved, the direction of a vector may change when it is transported along a curved path. A covariant derivative is needed to account for this change in direction and properly calculate the four velocity.
The four velocity is a crucial concept in general relativity as it allows us to properly describe the motion of objects in curved spacetime. It is used to calculate important quantities such as acceleration and energy in the theory of relativity.
Geodesics are the paths that objects follow in curved spacetime, which are determined by the four velocity. The four velocity vector is tangent to the geodesic, meaning it is parallel to the direction of motion along the path. This relationship is important for understanding the motion of objects in general relativity.