# Non Inertial Frames

Hi all, think this might be a silly or trivial question but I've got myself in a bother so thought I'd get some help.

Best to illustrate my question with an example. Take a vector expressed in some chart, then we can find the components of that vector in another chart in the standard way by using partial derivatives. However, it got my thinking why is it the case that one set of coordinates has some dependence of another? ie. why can't all the partial derivatives be zero? Geometrically speaking (I think) we view all frames as charts on our manifold, so of course any chart that is in the atlas corresponds to a frame and by compatibility the partial derivatives are non zero. But whats to stop us say, having some weird accelerated reference frame that isn't compatible with all charts and therefore not in the maximal atlas? How could we get a relation then? (I think theres an easy math answer to this that I'm missing, perhaps by playing around with the two homeomorphisms and using some sort of inverse function theorem or something). A good example would be the standard atlas for the sphere, and stereo graphic projection. As it turns out, these give rise to the same differentiable structure, but I imagine this isn't always the case?

Also...can anyone recommend any good books on GR? Ideally one suitable as reading for an undergrad course, a post grad course and then a bit more, but that's set up as more than just reading for lectures.

WannabeNewton
A frame is not a coordinate system. You should drill that into your head first. A frame is a basis for the tangent space to a differentiable manifold at some point on said manifold.

oh yes you're right sorry, a frame is just any set of linearly independent vectors. Okay wait so now I see. to find my vector component in terms of another frame, I apply my co frame. But what is to stop each element of my co frame, being applied to all vectors in another frame, all being zero? which would imply all vector components are zero

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WannabeNewton
Take a frame ##\{e_{\alpha}\}## and the associated coframe ##\{\omega^{\alpha}\}## for ##T_p M## and ##T^*_p M## respectively. Can you use them to rephrase your question mathematically? It's hard to parse what you're saying when you just use words.

Yes sorry.

Take a new frame \X_n and coframe μ_n. Spose we have a vector V in the first frame, so V = Ʃ[(V)^i]e_i. This vector also has a representation in the new frame, V = Ʃ[(V')^i]X_i. To find the components in terms of one another, we apply the co frame, so that V^j = [(V')^i] Ʃ w^j (X_i). My question is, what is stopping each w^j (X_i) being zero for every i and j? obviously it cannot be as it would imply that V is the zero vector...unless this of course only happens when V is the zero vector, and thats the implication. That's why I was getting confused, is my question just silly/trivial?

I first started thinking about this in coordinates. Say I use the tensor transformation law to transfer from one coordinate system to another, it seems pretty strong to me to assume that every coordinate function has some sort of dependence on the previous system. But maybe I have got myself mixed up, is this the case?

thanks for the input btw!

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WannabeNewton
##\omega^{\alpha}(e_{\beta}) = \delta^{\alpha}_{\beta}## by definition of a coframe.

Two different frames for the same tangent space are related to one another by a linear map whose matrix representation is just the usual change of basis matrix.

yes thats what separates the component on the LHS, but my question is what stops all elements of one coframe being zero when applied to all elements of a separate frame (unless this only occurs when V is the zero vector). But then I am still unsure about the coordinate party.

remember in my defintion above X_i are the elements of a new frame, so w^i (X_j) can by any real number

WannabeNewton
Again, if ##\{e_{\alpha}\}## is a frame for ##T_p M## and ##\{e_{\beta'}\}## is another frame for ##T_p M## then ##e_{\beta'} = \Lambda^{\alpha}{}{}_{\beta'}e_{\alpha}##. If ##\{\omega^{\alpha}\}## is the coframe associated with ##\{e_{\alpha}\}## then ##\omega^{\alpha}(e_{\beta'} ) = \Lambda^{\gamma}{}{}_{\beta'}\omega^{\alpha}(e_{\gamma}) = \Lambda^{\alpha}{}{}_{\beta'}##. This cannot be identically zero because a change of basis matrix cannot be identically zero (review the definition of the change of basis matrix from your introductory linear algebra class).

Oh my days. Thats basic algebra, how have I done that. I guess working on a building site has taken more out of me than I thought. thanks for (apart from to all the people who read this thread) sparing my blushes!

WannabeNewton
Also...can anyone recommend any good books on GR? Ideally one suitable as reading for an undergrad course, a post grad course and then a bit more, but that's set up as more than just reading for lectures.

To address this question, my most favorite book on GR is Wald "General Relativity". I've also been working through Straumann "General Relativity" and have been loving it. Check them out if you can.

just ordered the first one, sounds great, thanks!

WannabeNewton
just ordered the first one, sounds great, thanks!

Sounds good! Yeah it's a really awesome book, I don't think I've seen a more elegant book on GR.

They complement Wald perfectly.

Hi, sorry to bother you but I was wondering if you might help me with one of walds problems. Im pretty sure its not too hard, but I've been out of maths for a while and its making me feel very stupid. The question is question 3 on here www.maths.ox.ac.uk/system/files/coursematerial/2013/2654/2/sheet2.pdf [Broken] but only the bits to do with alpha and beta. so far I used the fact that alpha is harmonic to get some expression involving the christoffel symbols, but it didnt really help. Also you can find a form for the epsilon tensor due to anti symmetry. but again this just leads me to a messy sum which doesnt seem to get the result. Actually, if you could just show me how to show beta is harmonic, that would be great as I'm not sure if I'm manipulating everything right. I should be able to deduce the rest from that. Any help would be greatly appreciated! Tom

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great, excellent thanks for that! very nice solutions. thanks for recommending that book btw, my maths was a bit rusty at first but now I'm getting back into the swing of things. Starting chapter 4 officially tomorrow, I had a brief read through it last night and it look wonderful, am excited!

WannabeNewton