Calculation of Proper Acceleration

In summary, the conversation discusses the concept of covariant derivatives and their appropriate use in different scenarios. It is argued that covariant derivatives are suitable for dealing with fluids, where the dynamical variables are fields, but not necessarily for calculating the orbit of a single point particle. The correct concept for this is the absolute derivative, which involves Christoffel symbols. However, it is noted that in some cases, using covariant derivatives may still give the correct answer. The conversation also touches on the issue of extending vectors to vector fields, which is necessary for certain calculations in general relativity.
  • #1
pervect
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In another thread, Bill_K writes:

Bill_K said:
I think you meant to write aμ = UααUμ instead, and this is the form in which it is often quoted, but is conceptually incorrect.

The covariant derivative is a four-dimensional gradient operation and as such can only be applied to quantities that are functions of all four dimensions, such as f(x,y,z,t). This formula for aμ only makes sense if you are talking about the motion of a continuous fluid, in which the velocity is a field, Uμ(x,y,z,t), and then ∇αUμ is defined and meaningful, and denotes a rank two tensor field.

But we are talking here about the motion of a single point particle, in which Uμ(τ) is a function only of the proper time along the world line, and is undefined elsewhere. By analogy, in the Newtonian mechanics of projectile motion, it would make no sense at all to talk about the "gradient" of the projectile's velocity vector.

The correct concept is not the covariant derivative but the absolute derivative, D/Dτ, and this is what must be used to differentiate world-line functions. In a manner similar to the covariant derivative, it involves Christoffel symbols, one for each tensor index. E.g. for any vector Vμ(τ) defined along a world line with tangent vector Uμ, the absolute derivative of Vμ is

DVμ/Dτ ≡ dVμ/dτ + Γμνσ UνUσ

Wald, is one of the sources that does it the "wrong" way. But it seems that while the wrong way is rather "sloppy", it seems to work in practice. I' would like to see if this "slopiness" could lead to incorrect answers.

If we imagine that parallel transport was defined by a vector field, rather than a curve, would you agree that we could use aμ = UααUμ ?

Wald's argument is that parallel transport turns out to depend only on the values of the vector field along the curve. If I'm reading this right and Wald is correct, then _any_ vector field will give the right answer, as long as its orbit generates the curve in question.

This would imply that while the formulation is sloppy, it's not so bad as to give wrong answers.

I'd like to be sure I'm not missing something though,.
 
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  • #2
If a the 4-velocity defined along the worldline is extendable to a vector field defined on a neighbourhood of (the image of) the worldline, then aμ = UααUμ makes sense, but when the 4-velocity is not extendable to a neigbourhood, then this expression cannot be used, but the absolute derivative can be used. For (a non-physical) example, consider a closed timelike curve. At the event where the curve intersects itself there are, in general, two different 4-velocities.
 
  • #3
As I said in the earlier thread, the issue is a conceptual one.

Expressions involving covariant derivatives are appropriate when dealing with a fluid, where the dynamical variables are in fact fields. But, in order to calculate the orbit of mercury, or Thomas precession say - anything involving the world line of a particle - am I really required to imagine that the velocity vector has been extended continuously off the world line? And then afterwards say the result is independent of how I extended it! Doing it in terms of a covariant derivative is just a habit, and perhaps a lack of familiarity with the other concepts, we wouldn't dream of doing it that way in Newtonian mechanics. If anything, doing it the "right" way is simpler.

In these days we are so careful about saying exactly what we mean in other contexts - drawing the distinction between a vector and a 1-form, for example - I don't think it's being overly pedagogic to distinguish between a function of one variable and a function of four variables! :wink:
 
  • #4
Bill_K said:
In these days we are so careful about saying exactly what we mean in other contexts - drawing the distinction between a vector and a 1-form, for example - I don't think it's being overly pedagogic to distinguish between a function of one variable and a function of four variables! :wink:

No, it's a very good point, and the fact that a major textbook apparently got it wrong makes it an even better one.

But from the point of damage control, it sounds like as long as I'm not dealing with timelike curves, I won't have a problem with the sloppy formula giving "wrong" answers, rather similar to the way that Feynman's trick of differentiating under the integral sign doesn't usually cause problems (given that most functions in physics are differentiable).
 
  • #5
Bill_K said:
I don't think it's being overly pedagogic to distinguish between a function of one variable and a function of four variables! :wink:

pervect said:
No, it's a very good point

I could agree more! Students often get the multivariable chain rule wrong, because they don`t understand domains of functions.
 
  • #6
Bill_K said:
As I said in the earlier thread, the issue is a conceptual one.

Expressions involving covariant derivatives are appropriate when dealing with a fluid, where the dynamical variables are in fact fields. But, in order to calculate the orbit of mercury, or Thomas precession say - anything involving the world line of a particle - am I really required to imagine that the velocity vector has been extended continuously off the world line?

I agree that it is kind of weird to do that, but it seems to happen in more than one place in the GR literature. For example, the definition of the Riemann curvature tensor is:

[itex]R(u,v,w) = \nabla_u \nabla_v w - \nabla_v \nabla_u w - \nabla_[u,v] w[/itex]

Conceptually, that doesn't make any obvious sense, because [itex]u[/itex], [itex]v[/itex] and [itex]w[/itex] are vectors, not vector fields, so it doesn't make sense to use covariant derivatives. However, one can imagine the process of:
  1. Extend [itex]u[/itex], [itex]v[/itex] and [itex]w[/itex] to vector fields.
  2. Compute the above expression for [itex]R(u,v,w)[/itex]
  3. Note that the value of [itex]R(u,v,w)[/itex] doesn't depend on the vector fields except their value at one point (so the value is actually a function of three vectors, rather than three vector fields).
 
  • #7
stevendaryl said:
I agree that it is kind of weird to do that, but it seems to happen in more than one place in the GR literature. For example, the definition of the Riemann curvature tensor is:

[itex]R(u,v,w) = \nabla_u \nabla_v w - \nabla_v \nabla_u w - \nabla_[u,v] w[/itex]

Conceptually, that doesn't make any obvious sense, because [itex]u[/itex], [itex]v[/itex] and [itex]w[/itex] are vectors, not vector fields, so it doesn't make sense to use covariant derivatives. However, one can imagine the process of:
  1. Extend [itex]u[/itex], [itex]v[/itex] and [itex]w[/itex] to vector fields.
  2. Compute the above expression for [itex]R(u,v,w)[/itex]
  3. Note that the value of [itex]R(u,v,w)[/itex] doesn't depend on the vector fields except their value at one point (so the value is actually a function of three vectors, rather than three vector fields).
This is not correct. In that formula w IS a vector field, and that is all is needed to use the covariant derivative. While tangent vectors u and v are the linear arguments of the tensor.
 
  • #8
TrickyDicky said:
This is not correct. In that formula w IS a vector field

No, it's not. In the expression [itex]R(u,v,w)[/itex] all three are vectors, not vector fields.

--
Daryl McCullough
Ithaca, NY
 
  • #9
In math books, u,v, and w are vector fields. In many references, the word "field" is implicit. Meaning is then given by context. For example, "consider vector v", could mean that v is an element of a tangent space (just a vector), or that v lives in the tangent bundle (a vector field), depending on context.
 
  • #10
George Jones said:
In math books, u,v, and w are vector fields. In many references, the word "field" is implicit. Meaning is then given by context. For example, "consider vector v", could mean that v is an element of a tangent space (just a vector), or that v lives in the tangent bundle (a vector field), depending on context.

But the Riemann curvature tensor acts on vectors, not vector fields.
 
  • #11
stevendaryl said:
But the Riemann curvature tensor acts on vectors, not vector fields.

Actually, I suppose it's a little bit of a moot point. Any operator on vectors immediately can be considered an operator on vector fields. But [itex]R(u,v,w)[/itex] has a meaning for any three vectors, there is no need to think of them as vector fields.
 
  • #12
stevendaryl said:
No, it's not. In the expression [itex]R(u,v,w)[/itex] all three are vectors, not vector fields.

--
Daryl McCullough
Ithaca, NY

First of all, the expression is R(u,v)w, and for Riemann manifolds this equals the covariant derivative for the vector field w in the direction of the tangent vectors u and v minus the same thing in the direction of v and u, the failure of the commutativity of the covariant derivatives leads to nonvanishing curvature.
 
  • #13
stevendaryl said:
But the Riemann curvature tensor acts on vectors, not vector fields.

Acts on both.
 
  • #14
stevendaryl said:
But the Riemann curvature tensor acts on vectors, not vector fields.

From "Riemannian Maniforlds: an Introduction to Curvature" by Lee:

"If [itex]M[/itex] is any Reimannian manifold, the (Riemann) curvature endomorphism is the map [itex]R: T\left(M\right) \times T\left(M\right) \times T\left(M\right) \rightarrow T\left(M\right)[/itex] defined by ..."

Here, [itex]T\left(M\right)[/itex] is the tangent bundle, so, vector fields.
 
  • #15
TrickyDicky said:
First of all, the expression is R(u,v)w,

Are you arguing about the parentheses? There is really no difference in differential geometry between the three types of tensors
(1) A linear function of two vector arguments that returns a function from vectors to vectors, (2) A linear function of three vector arguments that returns a vector, and (3) a linear function of 3 vectors and 1 covector that returns a scalar. In all three cases, you write the components of the tensor [itex]R^i_{jkl}[/itex]

and for Riemann manifolds this equals the covariant derivative for the vector field w in the direction of the tangent vectors u and v minus the same thing in the direction of v and u, the failure of the commutativity of the covariant derivatives leads to nonvanishing curvature.

I don't understand the point of that paragraph. It seems to be just expressing in words exactly what I wrote in symbols:

[itex]R(u,v,w) = \nabla_u \nabla_v w - \nabla_v \nabla_u w - \nabla_{[u,v]} w[/itex]

My point was the weirdness that the left-hand side is a function of three vectors, while the right-hand side depends on three vector fields.
 
  • #16
George Jones said:
From "Riemannian Maniforlds: an Introduction to Curvature" by Lee:

"If [itex]M[/itex] is any Reimannian manifold, the (Riemann) curvature endomorphism is the map [itex]R: T\left(M\right) \times T\left(M\right) \times T\left(M\right) \rightarrow T\left(M\right)[/itex] defined by ..."

Here, [itex]T\left(M\right)[/itex] is the tangent bundle, so, vector fields.

This might be a quibble of terminology, but at every point on the manifold, there is a corresponding tangent space, and a corresponding Riemann tensor that is an operator on vectors of that tangent space. It's ALSO true that the Riemann tensor can be thought of as an operator on vector fields, as well.
 
  • #17
George Jones said:
From "Riemannian Maniforlds: an Introduction to Curvature" by Lee:

"If [itex]M[/itex] is any Reimannian manifold, the (Riemann) curvature endomorphism is the map [itex]R: T\left(M\right) \times T\left(M\right) \times T\left(M\right) \rightarrow T\left(M\right)[/itex] defined by ..."

Here, [itex]T\left(M\right)[/itex] is the tangent bundle, so, vector fields.

My point is that if at some point on the manifold you have three tangent vectors u,v,w, you can compute R(u,v,w). You don't need a vector field.
 
  • #18
stevendaryl said:
My point was the weirdness that the left-hand side is a function of three vectors, while the right-hand side depends on three vector fields.

Why do you keep saying that?
 
  • #19
TrickyDicky said:
Why do you keep saying that?

Because it's true. Let me quote Misner, Thorne and Wheeler on this point:

E. Is Modified Definition Acceptable?

I.e., is Riemann(...,C,A,B) ... a linear machine with output independent of how A, B, and C vary near the point of evaluation? YES! (See exercise 11.2)

...

(1) Riemann is a tensor; despite the appearance of [itex]\nabla[/itex] in its definition, no derivatives actually act on the input vectors A, B, and C.

I'm really just repeating their point; although the mathematical definition of Riemann involves extending the vectors A, B and C to vector fields and using covariant derivatives, Riemann itself is a tensor, a function of vectors at a point.
 
  • #20
TrickyDicky said:
Acts on both.

Sigh. Every operator on vectors can be considered trivially an operator on vector fields. My point is that the other way around doesn't hold; not every operator on vector fields is a legitimate operator on vectors. An example is the covariant derivative. It doesn't make any sense to take a covariant derivative of a vector, only of a vector field. An operator on vector fields is only considered a tensor if it acts as an operator on vectors at each point. The Riemann curvature tensor is a tensor (hence the name "curvature tensor").
 
  • #21
stevendaryl said:
I'm really just repeating their point; although the mathematical definition of Riemann involves extending the vectors A, B and C to vector fields and using covariant derivatives, Riemann itself is a tensor, a function of vectors at a point.

Okay, now I see your point. Riemann is a tensor field, so Riemann evaluated at a point is is a tensor, and this isn't immediately obvious from the definition of the tensor field Riemann.
 
  • #22
steven, your point only makes sense if you don't consider the Riemann tensor a tensor field, but every book I've consulted defines it as a tensor field. Of course if you evaluate it locally at a point it is a tensor and acts on vectors. But that is trivial isn't it?
 
  • #23
George Jones said:
Okay, now I see your point. Riemann is a tensor field, so Riemann evaluated at a point is is a tensor, and this isn't immediately obvious from the definition of the tensor field Riemann.

I phrased this poorly. I should have written

"Okay, now I see your point. It isn't immediately obvious that the right side of Riemann defines a tensor field, i.e., that Riemann evaluated at a point p is is a tensor, i.e., that that Riemann evaluated at p only depends on tangent vectors at p, and not on the extensions of the tangent vectors to vector fields."

Most math books give a proof of this, i.e., there really is something here that needs proving.
 
  • #24
George Jones said:
I phrased this poorly. I should have written

"Okay, now I see your point. It isn't immediately obvious that the right side of Riemann defines a tensor field, i.e., that Riemann evaluated at a point p is is a tensor, i.e., that that Riemann evaluated at p only depends on tangent vectors at p, and not on the extensions of the tangent vectors to vector fields."

Most math books give a proof of this, i.e., there really is something here that needs proving.
Ok, so maybe is not trivial. Or it is only once you've read the proof ;-)
 
  • #25
Are Killing vectors vectors or vector fields ?

I ask because their covariant derivative is zero. So I suppose they must be vector fields.
 
  • #26
Mentz114 said:
Are Killing vectors vectors or vector fields ?

Vector fields.
 
  • #27
George Jones said:
Vector fields.

That was quick ! Thanks.

Of course I meant the anticommutator of the covariant diff. is zero.
 
Last edited:
  • #28
Mentz114 said:
Are Killing vectors vectors or vector fields ?

I ask because their covariant derivative is zero. So I suppose they must be vector fields.
Vector fields. Their covariant derivative wrt the metric is zero, yes.
Edit: that was quick indeed, George.
 
  • #29
TrickyDicky said:
Vector fields. Their covariant derivative wrt the metric is zero, yes.
Edit: that was quick indeed, George.

I've edited my post ! It's the symmetrized covar. diff. that is zero, isn't it ?
 
  • #31
Mentz114 said:
I've edited my post ! It's the symmetrized covar. diff. that is zero, isn't it ?

Yes, that is actually equivalent to the Lie derivative of the metric wrt the Killing field being zero.
 
  • #32
George Jones said:
George, as academic jokes go that's pretty funny.

TrickDicky said:
Yes, that is actually equivalent to the Lie derivative of the metric wrt the Killing field being zero.
Argh ! Information overload.

Seriously, is it the case taht using the covariant derivative to test for geodesics will be OK if they are geodesic congruences - which I take to mean that the vector is extendable to a field.
 

1. What is proper acceleration?

Proper acceleration is the rate at which an object's velocity changes in its own reference frame. It is a measure of how quickly an object's speed or direction is changing, without taking into account the effects of gravity or other external forces.

2. How is proper acceleration calculated?

Proper acceleration is calculated using the equation a = dv/dt, where a is the proper acceleration, dv is the change in velocity, and dt is the change in time. This equation can be used to find the proper acceleration at any given moment in an object's motion.

3. What are the units of proper acceleration?

The units of proper acceleration are meters per second squared (m/s²) or, equivalently, gravitational acceleration (g). This unit represents the change in velocity over time and is commonly used in physics and engineering calculations.

4. How does proper acceleration differ from coordinate acceleration?

Proper acceleration is a measure of an object's acceleration in its own reference frame, while coordinate acceleration is a measure of an object's acceleration in a specific coordinate system. Proper acceleration takes into account the effects of relativity, while coordinate acceleration does not.

5. What is the significance of proper acceleration in space travel?

Proper acceleration is an important concept in space travel because it affects the perceived gravity and motion of objects in space. It is also a factor in the effects of relativity on space travel and can impact the trajectory and speed of spacecraft.

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