Why are the equations for dt/du and Dt[a]/Du equal in the geodesic equation?

[Apashanka Das]

In the geodesic equation why is dt/du=λ(u)t ,where t is the tangent vector along the curve
and why Dt[a]/Du=λ(u)dx[a]/du equated same,as given in hobson

 

[Orodruin]

There is no relevant coordinate independent meaning with dt/du unless what you mean is ##Dt/Du = \nabla_{t}t##. Also, the relevant interpretation of ##dx/du## is ##dx/du = t## so the equations are the same.

 

[Apashanka Das]

Mr. Orodruin can you please explain the first elaborately ,its very much needed

 

[PeterDonis]

Mr. Orodruin can you please explain the first elaborately ,its very much needed

Please note that you originally marked this thread as "A", which indicates a graduate level knowledge of the subject matter. I have changed it to "I", which indicates undergraduate level, because your question can't really be answered at the "B" level. But even at the undergraduate level, the expressions @Orodruin is using should be fairly common knowledge; and in any case, explaining them "elaborately" as you request would go far beyond the scope of a PF discussion. What background do you have in this area?

 

[Apashanka Das]

Ok please give some precise explanation

 

[PeterDonis]

please give some precise explanation

We need to know your background in this area. What textbooks have you studied?

 

[Apashanka Das]

Nothing to tell just a post graduate student at st xaviers college ,the question strikes me that's why I m asking ,I have read Hobson

 

[PeterDonis]

I have read Hobson

Do you mean this?

https://books.google.com/books/about/General_Relativity.html?id=xma1QuTJphYC

If so, can you give some specific page/section/equation references for what is confusing you? So far you have given no real context, which makes it very difficult to know what you are having trouble understanding.

 

[Apashanka Das]

yes it is this book,
pg no-76,section-3.17,equation -3.44,3.45,3.46,3.47

 

[Orodruin]

Nothing to tell just a post graduate student at st xaviers college ,the question strikes me that's why I m asking ,I have read Hobson

I am sorry, but it is impossible to decipher what you are actually asking about. Can you reformulate the question in such a way that it becomes clear what you want to know? Please explain what it is in the equations you listed that gives you problems. If you are reading these things, I assume that you are at least relatively familiar with calculus on manifolds?

Also, nobody is going to "explain something elaborately" as in writing you a textbook segment. It is up to you to write a question where it is clear what precisely you have problems with. If you do not, you are very unlikely to get a satisfactory answer.

 

[PAllen]

I can see something confusing in the notation used in that book. It appears little d is used to specify the abstract notion of absolute derivative along a curve, while big D is used for the coordinate representation. Also little d is used for ordinary derivative on coordinate basis. Thus, the first equation after 3.41 is just a translation from abstract to coordinate basis. Then, 3.45 follows trivially. The rest is restricting consideration to affine parameters.

[edit: above, substitute 3.44 for 3.41]

 

[PAllen]

Actually, Hobson uses big D for the intrinsic (or "absolute") derivative along a curve. [See eq(3.37) in section 3.14.]

@Apashanka Das : I guess the 1st part of your question is about why the notion of "geodesic" is defined by the requirement that the intrinsic derivative of the tangent vector ##{\mathbf t}## along the curve should be a multiple of itself, i.e., why
$$\frac{d{\mathbf t}}{du} ~=~ \lambda(u) {\mathbf t} ~,$$where ##u## is an arbitrary parameter. Then, since this ordinary derivative is not covariant, Hobson switches to the intrinsic derivative ##D/Du## which involves the connection ##\Gamma##, ending up at eq(3.45).

Since I don't have the book, and some pages (including the one you reference) are not accessible to me via the link, I have to intuit the meaning. I agree with you and state as much for the the interpretation of D. However, several usages of d applied to a vector without a coordinate basis are all consistent (IMO) with this being an abstract absolute derivative, as distinct from d applied to components which is always an ordinary derivative. The discussion from 3.39 to 3.41 seems consistent with my interpretation, but I cannot see pages where certain quantities are formally defined. In any case, I find the notation and ambiguity about what derivative is meant pointlessly confusing. Note also the discussion of covariant derivatives on page 69. To me, this also supports my interpretation that Hobson uses ordinary derivative symbols applied to an abstract vector as equivalent to the covariant (or absolute) derivative. See especially just before 3.32 on this page.

 

[Apashanka Das]

Ohh,that means here Hobson has defined 'geodesic' along which the the parallel transport of the tangent vector is takes place considering direction ,and 'affinely parametrised curve' along which parallel transport of the tangent vector considering direction and magnitude both,I may be wrong,need correction

 

[strangerep]

Since I don't have the book, and some pages (including the one you reference) are not accessible to me via the link, I have to intuit the meaning. I agree with you and state as much for the the interpretation of D. However, several usages of d applied to a vector without a coordinate basis are all consistent (IMO) with this being an abstract absolute derivative, as distinct from d applied to components which is always an ordinary derivative.

Aaargh! You are right -- this is clear from p72. (I tried to insert corrections in my earlier post but it was too messy, so I deleted the entire post to avoid confusing anyone else.)

When Hobson uses ##d## for a vector it's the absolute derivative. But when switches to components of the vector, he must use ##D## for the absolute derivative. I agree this is a confusing notation -- since Hobson was successful in confusing me.

 

[vanhees71]

##\newcommand{\bvec}[1]{\boldsymbol{#1}}##
Well, the point is that you have to distinguish between vectors and vector components. All physicists (including myself) are very sloppy not doing so all the time, and unfortunately we don't even take the time to explain it to the students in Physics 101, when we should.

Let's write ##\bvec{V}## for a vector (bold italics), which is a function of some scalar parameter (e.g., a world-line parameter for the motion of a particle in spacetime). Then let ##\bvec{b}_{\mu}## be an arbitrary basis. Then of course
$$\bvec{W}(\lambda)=\frac{\mathrm{d}}{\mathrm{d} \lambda} \bvec{V}(\lambda)$$
is still a vector, and no reference has been made to the basis and no components occurred yet. So let's see, how the components of ##\bvec{W}## with respect to the given basis are related to the components of ##\bvec{V}## wrt. the same basis:
$$\bvec{W}=\frac{\mathrm{d}}{\mathrm{d} \lambda} (V^{\mu} \bvec{b}_{\mu})=\bvec{b}_{\mu} \frac{\mathrm{d} V^{\mu}}{\mathrm{d} \lambda} + V^{\mu} \frac{\mathrm{d} \bvec{b}_{\mu}}{\mathrm{d} \lambda}=\dot{V}^{\mu} \bvec{b}_{\mu} + V^{\mu} \dot{\bvec{b}}_{\mu}.$$
Since now ##\dot{\bvec{b}}_{\mu}## is again a vector, I can write it in terms of the basis, defining the connection coefficients
$$\dot{\bvec{b}}_{\mu}={\gamma^{\nu}}_{\mu} \bvec{b}_{\nu},$$
and finally we get
$$\bvec{W}=\dot{\bvec{V}}=\bvec{b}_{\nu} (\dot{V}^{\nu} + {\gamma^{\nu}}_{\mu} V^{\mu}).$$
For convenience, when calculating with components in the Ricci calculus one defines some kind of derivative operator,
$$W^{\nu} = \mathrm{D}_{\lambda} V^{\nu} = \dot{V}^{\nu} + {\gamma^{\nu}}_{\mu} V^{\mu}.$$
Of course you can generalize this principle in many ways, e.g., to introduce the Christoffel symbols in affine bases, as usually used in introductory GR.

Then you have space-time coordinates ##q^{\mu}## (no vector components!), and you can define four basis vectors as tangents to the coordinate lines, i.e., curves defined by only varying ##q^{\mu}## and keeping the other three space-time coordinates fixed. The corresponding basis vectors are intuitively written as
$$\bvec{b}_{\mu}=\partial_{\mu}$$
Now if you have a vector field ##\bvec{V}=V^{\mu} \bvec{b}_{\mu}##, you can define a tensor field in the usual way by
$$\bvec{W}=\mathrm{d} q^{\mu} \otimes \partial_{\mu} \bvec{V}.$$
Here ##\mathrm{d} q^{\mu}## defines the dual basis of dual space to the holonomous basis ##\partial_{\mu}##.

This leads you to define a "covariant derivative" as components of this tensor:
$$\bvec{W}=\mathrm{d} q^{\mu} \otimes \partial_{\mu} (V^{\nu} \bvec{b}_{\nu}).$$
Then you define the affine connection
$$\mathrm{d} q^{\mu} \otimes \partial_{\mu} \bvec{b}_{\nu}=\mathrm{d} q^{\mu} \otimes {\Gamma^{\rho}}_{\mu \nu} \bvec{b}_{\rho}={\Gamma^{\rho}}_{\mu \nu} \partial_{\rho}\mathrm{d} q^{\mu} \otimes \partial_{\rho} $$
to get
$$\bvec{W}=:(\nabla_{\mu} V^{\nu}) \mathrm{d} q^{\mu} \otimes \partial_{\nu} = (\partial_{\mu} V^{\nu} + {\Gamma^{\nu}}_{\mu \rho} V^{\rho})\mathrm{d} q^{\mu} \otimes \partial_{\nu}.$$
This defines the "covariant derivative" of vector components, leading to a 2nd-rank mixed tensor components
$${W_{\mu}}^{\nu} = \nabla_{\mu} V^{\nu}=\partial_{\mu} V^{\nu} + {\Gamma^{\nu}}_{\mu \rho} V^{\rho}.$$
The "connection terms" always occur due to taking derivatives of the basis vectors and writing them again in terms of the basis.

 

[PAllen]

Ok, @vanhees71, that justifies his notation, however, it still seems confusing and uncommon to me.

 

[vanhees71]

Hm, but why is the (very common) notation of covariant derivatives for components confusing? It's used for decades in differential geometry, general relativity, and gauge-field theory...

 

[PAllen]

Hm, but why is the (very common) notation of covariant derivatives for components confusing? It's used for decades in differential geometry, general relativity, and gauge-field theory...

No, what is confusing is the use of ordinary derivative symbols for covariant or absolute derivatives when components are not used. I have literally never seen this before. Your derivation provides some motivation, however all books I've seen use some symbol other than ordinary derivative symbols for abstract coordinate free equations.

 

[martinbn]

I don't see the motivation in vanhees71's post. For me it is equally unusual to see the ordinary derivative symbol there in ##\frac{d}{d\lambda}V(\lambda)##.

 

[vanhees71]

But the point is that if ##\boldsymbol{V}(\lambda)## is a vector and ##\lambda## is a scalar, then ##\mathrm{d} \boldsymbol{V}/\mathrm{d} \lambda## is a vector. So why should you introduce another symbol for such a derivative than the usual one? My motivation was to make clear, why you don't need a special symbol for the vectors themselves and why you need one for the components.

 

[martinbn]

But the point is that if ##\boldsymbol{V}(\lambda)## is a vector and ##\lambda## is a scalar, then ##\mathrm{d} \boldsymbol{V}/\mathrm{d} \lambda## is a vector. So why should you introduce another symbol for such a derivative than the usual one? My motivation was to make clear, why you don't need a special symbol for the vectors themselves and why you need one for the components.

But to define ##\mathrm{d} \boldsymbol{V}/\mathrm{d} \lambda## you need additional structure. Just the differentiable manifold is not enough you need a connection. That's why it is more common to denote it ##\mathrm{D} \boldsymbol{V}/\mathrm{d} \lambda##.

 

[PAllen]

But the point is that if ##\boldsymbol{V}(\lambda)## is a vector and ##\lambda## is a scalar, then ##\mathrm{d} \boldsymbol{V}/\mathrm{d} \lambda## is a vector. So why should you introduce another symbol for such a derivative than the usual one? My motivation was to make clear, why you don't need a special symbol for the vectors themselves and why you need one for the components.

And now, with more thought, I question your whole justification. Independent of any coordinate basis, there is no notion of vector derivative without a connection. Otherwise there is no way to compare or subract vectors at different points, even along a given curve, because the vectors are in different tangent spaces - and the connection is what maps between tangent spaces of nearby points. Given this need for connection to even define a vector derivative, it is confusing to use an ordinary derivative symbol for this.

[edit: I see I cross posted with @martinbn ]

 

[strangerep]

[...] Independent of any coordinate basis, there is no notion of vector derivative without a connection. Otherwise there is no way to compare or subtract vectors at different points, even along a given curve, because the vectors are in different tangent spaces - and the connection is what maps between tangent spaces of nearby points. [...]

That makes me wonder: why then are we allowed to subtract scalars at different points?

The vector tangent space is (isomorphic to) ##{\mathbb R}^4## whereas the "scalar tangent space" is just ##{\mathbb R}##.

I'm guessing the answer has something to do with whether a "natural" mapping between tangent spaces exists. But I don't see a "natural" mapping between 2 copies of ##{\mathbb R}## since arbitrary translations and scale changes are possible.

 

[PeterDonis]

why then are we allowed to subtract scalars at different points?

Are we? Can you give an example?

 

[strangerep]

Can you give an example?

Well, I was thinking of the case of a scalar-valued function ##f(\lambda)## defined on a curve ##\gamma^i(\lambda)##. Calculating its derivative involves ##f(\lambda+\epsilon) - f(\lambda)##.

 

[martinbn]

The scalars have values real numbers. We can subtract real numbers. The vectors belong to different vector spaces. They may be isomorphic, but there are many different isomorphisms. How do you pick an isomorphism? How do you subtract ##v\in T_pM## and ##w\in T_qM##? On the other hand ##\alpha\in\mathbb R## and ##\beta\in\mathbb R## can be subtracted.

 

[strangerep]

The scalars have values real numbers. We can subtract real numbers. The vectors belong to different vector spaces. They may be isomorphic, but there are many different isomorphisms. How do you pick an isomorphism? How do you subtract ##v\in T_pM## and ##w\in T_qM##? On the other hand ##\alpha\in\mathbb R## and ##\beta\in\mathbb R## can be subtracted.

Yes, that's similar to what I was thinking. But... there's also more than 1 possible isomorphism between different copies of ##\mathbb R##, unless you invoke some extra criteria. E.g., 0 matches 0, and the mapping is isometric (hence no scale changes allowed).

 

[martinbn]

Scalars are defined as maps from the manifold to the real numbers i.e. ##\alpha, \beta : M\rightarrow \mathbb R##. The target set is the same, so we can subtract their values. Vectors are not maps to ##\mathbb R^n##. If they were we could subtract them. But they are derivations of functions at the point (for example). They form a vector space, which is isomorphic to ##\mathbb R^n##.

 

[stevendaryl]

Are we? Can you give an example?

Well, one way of defining tangent vectors is via differences of scalar fields at different locations:

If you have a parametrized path [itex]\gamma(s)[/itex], the corresponding tangent vector [itex]\mathbf{V}(s)[/itex] is that operator which given a scalar field [itex]\Phi(\mathcal{P})[/itex] returns [itex]\frac{d}{ds} \Phi(\gamma(s))[/itex].

 

[stevendaryl]

And now, with more thought, I question your whole justification. Independent of any coordinate basis, there is no notion of vector derivative without a connection. Otherwise there is no way to compare or subract vectors at different points, even along a given curve, because the vectors are in different tangent spaces - and the connection is what maps between tangent spaces of nearby points. Given this need for connection to even define a vector derivative, it is confusing to use an ordinary derivative symbol for this.

[edit: I see I cross posted with @martinbn ]

Okay, but in the context of talking about a smooth manifold with connection (which is the case for GR), the notation [itex]\frac{d\mathbf{V(s)}}{ds}[/itex] is unambiguous.

 

[PAllen]

Okay, but in the context of talking about a smooth manifold with connection (which is the case for GR), the notation [itex]\frac{d\mathbf{V(s)}}{ds}[/itex] is unambiguous.

Unambiguous, yes , if defined by the author. Confusing and nonstandard, also yes. Also, my argument was in response to an argument that connections were only relevant in a coordinate basis - which is not correct.

 

[vanhees71]

The entire vector analysis on differentiable manifolds do not need additional structures a priori. As long as you restrict yourself to alternating differential forms you don't need covariant derivatives.

And now, with more thought, I question your whole justification. Independent of any coordinate basis, there is no notion of vector derivative without a connection. Otherwise there is no way to compare or subract vectors at different points, even along a given curve, because the vectors are in different tangent spaces - and the connection is what maps between tangent spaces of nearby points. Given this need for connection to even define a vector derivative, it is confusing to use an ordinary derivative symbol for this.

[edit: I see I cross posted with @martinbn ]

Why then don't you need a connection when dealing with plain differentiable manifolds without any additional structures? You can go pretty far, introducing tangent and cotangent spaces and introducing differential forms. For this I don't need a connection, let alone a fundamental form (i.e., metric of pseudometric).

Also see the remark on Wikipedia:

https://en.wikipedia.org/wiki/Covariant_derivative#Remarks

 

[martinbn]

If you want to differentiate objects other than functions in the direction of a vector you need a connection.

 

[vanhees71]

Then, why don't I need a connection to define (alternating) differential forms?

 

[martinbn]

Then, why don't I need a connection to define (alternating) differential forms?

Why would you?