Mordred Posted July 5, 2016 Share Posted July 5, 2016 (edited) Thanks I forgot a formula. [latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex] The above is defining a uniform static field. [latex]\Omega[/latex] is a dimensionless value for density. As were just defining a static field we don't need the time parameter. We haven't detailed any particle motion yet. Ok so now the Principle of Equovalence. You can google that term for more detail but in the same format as above [latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex] [latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex] Denotes the gravitational field above. Edited July 5, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 When you say "the physical distance of objects in the gravitational field", do you mean the distance between objects based on how long light takes to travel between them? i.e. the space's size expands near massive objects, causing light to have to travel further to get to the other object. what is ds^2 again? What does the term "line element" mean? Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 (edited) Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism) [latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex] Provided ds^2 is invariant [latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation [latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex] With the line element invariance [latex]\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x[/latex] The inverse of the metric tensor transforms as [latex]\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x[/latex] Edited July 5, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 Ok, so does that mean that ds^2 is a distance in meters? Why not just say 'd' or is the s^2 a reference to 2D space? If so it is very confusing having ^2 mean different things in different parts of the equation. Or is it simply the line element's length (distance) squared? Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 Ok, so does that mean that ds^2 is a distance in meters? Why not just say 'd' or is the s^2 a reference to 2D space? If so it is very confusing having ^2 mean different things in different parts of the equation. Or is it simply the line element's length (distance) squared? The s^2 indicates a light like seperation of two events on a 2d sphere. Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 Ok - how confusing, but never mind... So Can you explain (in words) what the full equation is saying? For example: The light distance between two events on a 2D sphere = ??? - ??? - The volume of a spherical shell at radius r Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 (edited) Here this link will explain it better and save me a ton of latex. https://en.m.wikipedia.org/wiki/Line_element Keep in mind I still haven't shown a particles movement. All the steps above just detail the static geometry. GR is all about relating events. So you don't want your coordinates that describe those events to move. (Though in some cases you have no choice ie expansion) Right now I need to take a break and check through my old notes. Make sure I got the covariant and contravariant terms correct. My old handwriting is a bit messy lol I'm placing these into a single post as I may use this again. Though I messed up some of the original posts lol In the presence of matter or when matter is not too distant physical distances between two points change. For example an approximately static distribution of matter in region D. Can be replaced by tve equivalent mass [latex]M=\int_Dd^3x\rho(\overrightarrow{x})[/latex] concentrated at a point [latex]\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})[/latex] Which we can choose to be at the origin [latex]\overrightarrow{x}=\overrightarrow{0}[/latex] Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex] [latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex] Where [latex] G_n=6.673*10^{-11}m^3/KG s^2[/latex] and [latex]r\equiv||\overrightarrow{x}||[/latex] According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element. [latex]ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2[/latex] Where [latex]d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2[/latex] denotes the volume element of a 2d sphere [latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere. The general relativistic form is. [latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex] By comparing the last two equations we can find the static mass distribution in spherical coordinates. [latex](r,\theta\varphi)[/latex] [latex]G_{\mu\nu}=\begin{pmatrix}1+2\phi_N/c^2&0&0&0\\0&-(1+2\phi_N/c^2)^{-1}&0&0\\0&0&-r^2&0\\0&0&0&-r^2sin^2(\theta)\end{pmatrix}[/latex] Now that we have defined our static multi particle field. Our next step is to define the geodesic to include the principle of equivalence. Followed by General Covariance. Ok so now the Principle of Equivalence. You can google that term for more detail but in the same format as above [latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex] [latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex] Denotes the gravitational field above. Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism) [latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex] Provided ds^2 is invariant [latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation [latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex] With the line element invariance [latex]\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x[/latex] The inverse of the metric tensor transforms as [latex]\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x[/latex] In GR one introduces the notion of covariant vectors [latex]A_\mu[/latex] and contravariant [latex]A^\mu[/latex] which is related as [latex]A_\mu=G_{\mu\nu} A^\nu[/latex] conversely the inverse is [latex]A^\mu=G^{\mu\nu} A_\nu[/latex] the metric tensor can be defined as [latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=diag(1,1,1,1) which denotes the Kronecker delta. Finally we can start to look at geodesics. Let us consider a free falling observer. O who erects a special coordinate system such that particles move along trajectories [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex] Specified by a non accelerated motion. Described as [latex]\frac{d^2\xi^\mu}{ds^2}[/latex] Where the line element ds=cdt such that [latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex] Now assunme that the motion of O changes in such a way that it can be described by a coordinate transformation. [latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex] This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation. [latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex] Where the new line element is given by [latex]ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu[/latex] and [latex] g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}[/latex] and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex] Denote the metric tensor and the affine Levi-Civita connection respectively. There done. Edited July 5, 2016 by Mordred 1 Link to comment Share on other sites More sharing options...

ajb Posted July 5, 2016 Share Posted July 5, 2016 (edited) ...I said it is possible to understand Relativity without getting involved with the Tensor maths. We think not. I can explain exactly how Time Dilation, Length Contraction and Mass Increase works without involving any Tensors. When you calculate anything it seems that you will need tensors (and similar). It is also to understand qualitatively how GR works without knowing how to do the Tensor maths. You maybe able to produce some results for very special space-times and matter without the full machinary, but still you really need differential geometry and tensors are vital. The Tensor maths is a neat mathematical tool for doing exact calculations on the non-linear equations governing the motions of test particles within space-time - agreed. More than that, tensor calculus is needed to write down the equations before you even come to solve them - including the Einstein Field Equations. In my comment I was talking about qualitative understanding (i.e. understanding the principles, rather than the numbers). But the principles are stated mathematically. However, now may be a good time to help me understand the maths that you are familiar with. Edited July 5, 2016 by ajb Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 (edited) Man that took hours to latex lol. Ajb if you would for errors when you get a chance I'd appreciate it. Now what did we learn. First I described a static field of uniform density. This means that the stress tensor is also static. (You cannot change one without the other). Then I described in mathematical detail the equivalence principle. Follows by the principle of covariance. From there I detailed the Kronecker delta. Once I had these established I then detailed how the Geodesic equation is derived. (Which details particle freefall) within that field. As an added bonus I described the Levi-Civita connection. Edited July 5, 2016 by Mordred 1 Link to comment Share on other sites More sharing options...

ajb Posted July 5, 2016 Share Posted July 5, 2016 Man that took hours to latex lol. Ajb if you would for errors when you get a chance I'd appreciate it If I spot anything I will shout. However, I think details are not to be worried about right now ... Declan did not know the term 'line element' nor did he know what the volume element on a sphere is (including writing it in standard angular coordinates.) By default this means he knows almost nothing about general relativity. But that is okay, none of us were born knowing. It takes a lot of hard work. Declan now has some idea of what he has to do - this is exactly what this forum should be about. Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 Thanks I plan on posting that post into the Relativity forum. I might pin it into the What is space thread. Link to comment Share on other sites More sharing options...

swansont Posted July 5, 2016 Share Posted July 5, 2016 Here we go again... To Swansont: How can we test for an Aether? Bradley, and Michelson and Morley, did. They got wildly differing results. Your formulation implies we are moving through this flow towards the galactic center. What measurable effects will this have? Shouldn't light slow down when swimming upstream against it? Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 I just read through the wiki on line elements. Why didn't you just tell me that it is Pythagorous theorem in 4D? Apart from the confusing symbols used, the tensor is effectively just calculating the line length given 4 dimensions of minute length changes. (At least for normal space time). I guess when space is curved then the 4 different minute lengths will be slightly different - giving a different total length. Does this sound about right? Link to comment Share on other sites More sharing options...

ajb Posted July 5, 2016 Share Posted July 5, 2016 The line element ( which is build from the metric) gives you the notion of a distance between two near by points. You are right that you can think of this as a generalisation of Pythagorous theorem, but not just to 4-d, but to more general manifolds than just R^n. You can then use this to understand the length of a path. Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 Ok, sure - up to n dimensions. To Swansont: Yes, but given the extra acceleration is only in the order of 1x10^-11 the flow rate would be quite low. Anyhow if Dark Matter exists it would slow the rate at which light takes to escape the Galaxy too due to the extra gravitational potential (which I guess in GR terms would mean the creation of extra space that the light would have to travel through). The Michelson Morley experiment is not a valid test for an aether due to the length contraction of all the apparatus in the direction of movement through the aether. I show this to be the case in my paper. Link to comment Share on other sites More sharing options...

ajb Posted July 5, 2016 Share Posted July 5, 2016 The Michelson Morley experiment is not a valid test for an aether due to the length contraction of all the apparatus in the direction of movement through the aether. Not that we can really trust you analysis here, but there are of course many other modern experiments that show no effects of an aether to some large degree of accuracy. Also, we have all the other evidence that GR and SR are good models - all of which suggest the aether is not realised in nature. Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 To ajb: Why can't you trust it - it is simple maths to follow? To Mordred: Getting back to the line element for the gravitational field: where does the 1+2*phi/c^2 term come from. I know it is to do with the gravitational potential, but can you please explain it? Thanks... Link to comment Share on other sites More sharing options...

ajb Posted July 5, 2016 Share Posted July 5, 2016 To ajb: Why can't you trust it - it is simple maths to follow? You did not know what a line element is... thats enough for me not to trust your calculations. Sorry. Anyway, it seems you are making some progress, so don't spoil your hard work in this way. Keep going reading and learning physics and mathematics. Getting back to the line element for the gravitational field: where does the 1+2*phi/c^2 term come from. I know it is to do with the gravitational potential, but can you please explain it? Thanks... I think he has written a 'weak field metric'- so using the Newtonian approximation we see that phi is the gravitational potential as we think of it in Newtonian gravity. Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 Yes I understood that, but why 1+2*phi/c^2 rather than the GR time dilation formula 1 + delta(phi)/c^2 ? Link to comment Share on other sites More sharing options...

ajb Posted July 5, 2016 Share Posted July 5, 2016 There are textbook derivations of this... You use the fact that for the static distribution of matter [math] T^{00} = T_{00} = \rho[/math] and all the other components of the energy-momentum tensor vanish in the Einstein field equations (You can make use of the weak field approximation here also) I am not sure if this is covered in Carroll's notes. It is in Ryder's book on GR, which I don't know very well. (But his QFT book I do like) Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 You did not know what a line element is... thats enough for me not to trust your calculations. Sorry. Anyway, it seems you are making some progress, so don't spoil your hard work in this way. Keep going reading and learning physics and mathematics. I think he has written a 'weak field metric'- so using the Newtonian approximation we see that phi is the gravitational potential as we think of it in Newtonian gravity. Yes thats correct it is the non relativistic Newton weak field limit. The number of steps to derive that, is lengthier than the post itself. 1 Link to comment Share on other sites More sharing options...

ajb Posted July 5, 2016 Share Posted July 5, 2016 (edited) See I do know some GR Edited July 5, 2016 by ajb Link to comment Share on other sites More sharing options...

Bignose Posted July 5, 2016 Share Posted July 5, 2016 (edited) Yes, but given the extra acceleration is only in the order of 1x10^-11 the flow rate would be quite low. Hmmmm. Herrmann, S.; Senger, A.; Möhle, K.; Nagel, M.; Kovalchuk, E. V.; Peters, A. (2009). "Rotating optical cavity experiment testing Lorentz invariance at the 10−17 level". Physical Review D 80 (100): 105011 This is a modern test of aether theory. They found null result to within 10^-17. Your 10^-11 should have very, very, very easily shown up. Explanation? Edited July 5, 2016 by Bignose 3 Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 See I do know some GR 😉 Link to comment Share on other sites More sharing options...

Declan Posted July 5, 2016 Author Share Posted July 5, 2016 To Bignose: The 10^-11 msec^-1 I was referring to was the extra acceleration in spiral galaxies required to explain the orbital velocities that are observed. I was not suggesting the speed of light would be different as measured by any observer in any reference frame. As the speed of light and the rate of time change at the sane time due to the same reason, it will always be measured as c. Any measurements done of the speed of light whilst moving through the space-time medium necessarily involve the signal completing a round-trip so that a timing measurement can be made against the same clock. So any difference in the upstream time as made up for in the downstream time - thus the total time always gives the speed c for light's total travel time. I have shown the maths if this in my paper. Link to comment Share on other sites More sharing options...

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