Mordred Posted July 5, 2016 Share Posted July 5, 2016 (edited) Why would measuring the rotation curve of the Andromeda Galaxy require a round trip for light? Secondly why would the speed of light make any difference to measurement points of a distant galaxy, when the galaxy is far enough away the light paths from each edge of the galaxy essentially follows the same path as it approaches Earth.? Your last post makes little sense 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 Msec^-1 should have been msec^-2 Seeing light from a distant galaxy is not a measure of light's speed. How do we know exactly when it was emitted? In any case if there is Dark Matter the light would take longer to reach us anyhow due to the extra gravitational potential so how could we tell the difference? I'm not quite sure what point you are making in your second point - are you talking about lensing? Link to comment Share on other sites More sharing options...

Mordred Posted July 5, 2016 Share Posted July 5, 2016 (edited) Visualize a galaxy 1 billion light years away. Draw a line from the centre of that galaxy to observer on Earth. Now draw a line from the outer edge of that galaxy the same observation point. How much angle between those two lines will you have. (Assume the galaxy has the same radius as the Milky way) Now your statement about two way signals. Were certainly not going to sendva signal from Earth to that Galaxy and wait for its return to take s measurement. (Assuming no expansion) that would take us 2 billion years to make a single measurement. When we measure a distant stellar object were making a measurement as it was in the past. Through Cosmological redshift we can approximate its distance. With calculation we can calculate its proper distance now and its proper distance when the light source originated. However for that we must work initally in commoving coordinates prior to Proper coordinates. Here is two articles I wrote a few years back. One is on redshift and expansion. The other is universe geometry and how it affects light paths. http://cosmology101.wikidot.com/redshift-and-expansion http://cosmology101.wikidot.com/universe-geometry Thankfully your now familiar with line element for the metrics for page two of the last article. http://cosmology101.wikidot.com/geometry-flrw-metric/ Other good study guides are. http://arxiv.org/pdf/hep-ph/0004188v1.pdf :"ASTROPHYSICS AND COSMOLOGY"- A compilation of cosmology by Juan Garcıa-Bellido http://arxiv.org/abs/astro-ph/0409426 An overview of Cosmology Julien Lesgourgues PS the lightcone calculator in my signature does these calculations Probably one of the better articles is Lineweaver and Davies. I'll have to find a new link to it. Hrmm the full article has moved Ah found it. http://www.google.ca/url?sa=t&source=web&cd=9&ved=0ahUKEwiWw7OLwd3NAhUM_mMKHdc0BqwQFgg1MAg&url=http%3A%2F%2Fwww.dark-cosmology.dk%2F~tamarad%2Fpapers%2Fthesis_complete.pdf&usg=AFQjCNHLzxKUp15sqgaDF2B8NU6i4xnBdg&sig2=GwUOC4EW-TeSeL2l6a2mnA Edited July 5, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 6, 2016 Author Share Posted July 6, 2016 Ok, I have had a quick look at your web pages - I don't have time to read through fully at the moment. What is the point/question you are trying to make? Link to comment Share on other sites More sharing options...

Mordred Posted July 6, 2016 Share Posted July 6, 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. I think have misread what you posted here Link to comment Share on other sites More sharing options...

Declan Posted July 6, 2016 Author Share Posted July 6, 2016 Ok, no problem... Can you point me to a webpage that shows the derivation of the 1+2*phi/c^2 term? I tried to find it in Wikipedia but only found the metric but no derivation of the elements in the matrix... Link to comment Share on other sites More sharing options...

Mordred Posted July 6, 2016 Share Posted July 6, 2016 (edited) I'm not sure I would be able to find that solution on the Internet. Most Textbooks and articles usually skip right to the Scwartzchild metric as an example. Instead of covering the non relativistic Newton potential. However I do have the solution in my notes. It will take me some time to post the latex. (Which I will do once I am free for sufficient time) lol hopefully my old notes are still detailed enough and legible. This isn't precisely the same solution but its close. The symbology used is slightly different for the G_00 element. http://www.google.ca/url?sa=t&source=web&cd=8&ved=0ahUKEwiU9OvO9N3NAhUQHGMKHUqWA1YQFgg1MAc&url=http%3A%2F%2Fwww.math.uchicago.edu%2F~may%2FVIGRE%2FVIGRE2010%2FREUPapers%2FTolish.pdf&usg=AFQjCNF2yRQkUsx2hfa2by5mSWoIOxymlA&sig2=Py9ERsx_AKyp79263uc9UA Its essentially the same as you are deriving the Newton potential limit from the Einstein field equations. So will be variations though roughly equivalent. In the notation I used the N subscript simply denotes Newton as a identifier. I'LL still post the solution I have but having another example solution never hurts. PS how I found this solution is google. "Deriving Newton potential from Einstein field equations" then add pdf to refine your search. Other note... the solution in the link above is far easier to follow than the solution I have. Edited July 6, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 6, 2016 Author Share Posted July 6, 2016 Ok thanks - I'll have a look soon... Link to comment Share on other sites More sharing options...

ajb Posted July 6, 2016 Share Posted July 6, 2016 The derivation of the metric Mordered gives is in Ryder's book on general relativity. I like Ryder, he is usually good at showing enough steps to get you through his calculations. However parts of the derivation are hidden on googlebooks preview. Maybe you could buy this book. Link to comment Share on other sites More sharing options...

Declan Posted July 6, 2016 Author Share Posted July 6, 2016 Hmmm I'll see what I can find online first... I have not found a very satisfactory answer yet, but two things I found that seem to result in a 2*phi result are: (1) a calculation for the escape velocity in a Newtonian gravity field. (2) the comparison of the GR metric to Newtonian gravity in high gravitational environment where the effect is double the classical result. I remember that originally the equation for the precession of mercury was out by a factor of two - is this in any way similar? From what I can see from the tensor for the Newtonian limit given by Mordred, the factors in the matrix describe the gravitational time dilation affect on the time parameter, and the curvature of space on the next diagonal element in the matrix (in curvilinear coordinates). Is that right? The texts on the Metric just seem to say let g00 = 2*phi but don't explain why... Link to comment Share on other sites More sharing options...

ajb Posted July 6, 2016 Share Posted July 6, 2016 Hmmm I'll see what I can find online first... Cheak your personal messages ;-) Link to comment Share on other sites More sharing options...

Mordred Posted July 6, 2016 Share Posted July 6, 2016 The derivation of the metric Mordered gives is in Ryder's book on general relativity. I like Ryder, he is usually good at showing enough steps to get you through his calculations. However parts of the derivation are hidden on googlebooks preview. Maybe you could buy this book. Cool there's one I hadn't read yet. I may just pick up a copy to go with my collection. 40 plus textbooks just isn't enough. Lol that and my collection drives my wife crazy hehe Link to comment Share on other sites More sharing options...

Declan Posted July 6, 2016 Author Share Posted July 6, 2016 So it seems that the gravitational time dilation formula 1+del(phi)/c^2 appears in the Christoffel symbol in the geodesic equation, and when h00 is calculated and then added to the normal flat space time is becomes the 1+2*phi/c^2 form that appears in the metric tensor. Do you agree? Actually the form in the Christoffel symbol is without the 1+ at the start, that gets added when the normal space is added on. It seems a bit odd that the gravitational time dilation equation doesn't appear in its normal form in the matrix, as the time dimension would be affected by that time dilation factor as a result of the metric, wouldn't it, as that is the effect of the gravitational field on the time dimension. Does some other part of the function modify the result, thus forming the normal time dilation equation in the final result? Link to comment Share on other sites More sharing options...

Mordred Posted July 6, 2016 Share Posted July 6, 2016 (edited) Yes thats essentially correct. I don't know the model procedure Ryder used. The procedure I have was done via the geodesic equation by modelling the parallel transport of two photons null geodesics. Then calculating calculating the slight deviation of those two null geodesics. Which is why the procedure I have is lengthy and complex. Newton limit is modelled as a scalar field. The [latex]H_{\mu\nu}[/latex] is used to model permutations of that scalar field. Its one of those hidden tensors in the EFE. For example I can model a vector or scalar field. Then model a gravity wave of that field with the [latex]H_{\mu\nu}[/latex] tensor. Newton gravity didn't include time dilation. This one of the key aspects that seperate Newton gravity from SR. Think of the Newton limit as pre Minkowskii. To model the Newton limit you need to describe a system where time dilation is neglibible. Which is why a uniform matter distribution is often used. If all observers is in the same mass density there is no time dilation. ( this makes the Newton limit handy to model gravity without the added complexity of time dilation) Edited July 6, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 6, 2016 Author Share Posted July 6, 2016 So if no time dilation in the Newtonian limit example, what do those two elements in the metric concerning 2*phi etc do? How do they affect the time and r dimensions? Link to comment Share on other sites More sharing options...

Mordred Posted July 6, 2016 Share Posted July 6, 2016 (edited) So from that above. Take a cloud of plasma with uniform distribution. The centre of mass of that cloud is where your gravitational potential vectors will be pointing to. There isn't enough anistropy mass density in that cloud to have time dilation. Newton limit describes this system beautifully. So if no time dilation in the Newtonian limit example, what do those two elements in the metric concerning 2*phi etc do? How do they affect the time and r dimensions? Let's ask a counter question. What is the in the function of the metric tensor and the Ricci tensor/ Ricci scalar? https://en.m.wikipedia.org/wiki/Ricci_curvature Hint on the metric tensor. Think back to the time like geodesics. Read that section here. https://en.m.wikipedia.org/wiki/Metric_tensor Remember geodesics also occur in Euclidean flat space. Looking at that section then think about this earlier statement The procedure I have was done via the geodesic equation by modelling the parallel transport of two photons null geodesics. Then calculating calculating the slight deviation of those two null geodesics. Which is why the procedure I have is lengthy and complex. Its a bit tricky but if you think about the quoted statement above you should be able to answer the question below So if no time dilation in the Newtonian limit example, what do those two elements in the metric concerning 2*phi etc do? How do they affect the time and r dimensions? If not let me know and I will post the parallel transport Kronecker delta relations I have ! Moderator Note Strong Hint. Uniform Density+ Centre of mass. If it helps plot this on a graph paper. Place a particle of equal mass at each x,y coordinate. Then place centre of mass at coordinate 0,0. What type of path will particles free fall. What is the deviation between two of those geodesic paths Edited July 6, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 6, 2016 Author Share Posted July 6, 2016 I read those wiki pages already yesterday. Is the cloud of plasma extending into space around the origin such that Shell theorem is applied as r changes? Or is the mass being considered as all being at the origin? Either way though, without time dilation shouldn't the element in the matrix for the time dimension just be -1 as it is in normal flat space? This element determines the length of the line element in the time dimension right? So the distance in time between two points would stay the same wouldn't it, as the rate of time is the same everywhere in the space? Or am I misunderstanding the length of the line element for the time dimension: is it the time taken for light to travel along the geodesic between the two points? Link to comment Share on other sites More sharing options...

Mordred Posted July 6, 2016 Share Posted July 6, 2016 (edited) I read those wiki pages already yesterday. Is the cloud of plasma extending into space around the origin such that Shell theorem is applied as r changes? Or is the mass being considered as all being at the origin? Either way though, without time dilation shouldn't the element in the matrix for the time dimension just be -1 as it is in normal flat space? This element determines the length of the line element in the time dimension right? So the distance in time between two points would stay the same wouldn't it, as the rate of time is the same everywhere in the space? Or am I misunderstanding the length of the line element for the time dimension: is it the time taken for light to travel along the geodesic between the two points? Yes on the distribution. Yes on the geodesic path. The other thing to watch for in regards to the 1+ 1- is your sign convention. So if you add position Hoo to Goo keep track of your sign notation. The other trick is normalized units within the tensors. My advise is to study matrix math Then work into GR àtensors Because of centre of mass two parallel geodesics will deviate from a straight line. This is what's plotted in the metric tensor. The Ricci tensor reflects the curvature changes due time dilation. The Ryder reference you have may be in the (-+++) convention. The workup I did is in the (+---) convention. I don't know if your familiar with tensor calculus. So this link will help. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.ita.uni-heidelberg.de/~dullemond/lectures/tensor/tensor.pdf&ved=0ahUKEwiI26G0_d_NAhVS-mMKHZPzBVYQFgg4MAc&usg=AFQjCNH28SMShwZP6np1MNQZ56WEtHGUHg&sig2=L0nCitet3oBLAmv9x-kGJg Edited July 6, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 I studied matrix maths in high school (up to 3x3 matrices) and was good at it, though I have forgotten some of the details such as which elements to calculate the determinant etc. I am aware of the -+++ +--- convention difference though not the reason why one has to be different from the other three - is it because it is time? Can you explain this easily? When you said 'yes on the distribution' did you mean yes apply shell theorem or yes treat all the mass at the origin? Incidentally the Newtonian limit section says that the gravitational redshift can be derived from the matrix elements due to the equivalence principle, yet this due to time dilation of one gravitational potential relative to another gravitational potential- but you said there is no time dilation in this example. Can you clarify? Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 Yes apply the shell theorem in the Newton limit. Two terms describe uniform. Homogeneous-essentailly no preferred location. Isotropic- no preferred direction. In the plasma case you model system as an ideal gas or fluid that homogeneous and isotropic in distribution. For gravitational redshift you want to map this in the Scwartzchild metric. Which has a preferred direction. Planets and BHs etc. In a uniform mass distribution there is no time dilation. Newton limit. In an increasing or decreasing mass distribution there is Schwartzchild metric. Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 Hmm the Ryder section on Newtonian limit says the redshift can be calculated from the time-time element of the matrix - this seems to be misleading? Anyhow it seems that one needs to calculate the geodesic equation first in order to know what the matrix elements for the metric should be - is that the only way to do it? What would the matrix elements be for the space around a mass (which is all at the origin)? I guess for a weak field such that phi=Gm/r? In that case no shell theorem required. Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 (edited) Actually no your right. I misread the conditions. Which is the matter must be approximately static. Lol wonder how I got off tangent there lol. Got into my head inside a uniform distribution. Probably because I'm more often modelling that in the FLRW metric. Lol gotta hate handwritten personal notes. Ah its specified better under weak field. Pressure contribution is neglibible Newton limit is [latex]p=0[/latex] [Latex]\Lambda=0[/latex] So pressureless matter no relativistic radiation. No cosmological constant. What does Ryder set the system as for Newton limit? Does he give the same equations of state? Edit: Newton limit is still non-relativistic meaning your infalling particle. The geodesic equation we derived is space-time component. Not the time-time needed for redshift. Time-time is a null geodesic. Edited July 7, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 I don't see any reference to relativistic radiation or cosmological constant, just seems to be based on phi=-Gm/r What about the question I asked about the matrix elements? Is the calculation of the geodesic the only way to derive them? Apart from trying to understand the maths, it seems clear that the line element calculation and the geodesic would be the same for GR based on an energy field with variable density, setting the speed of light and time dilation. A point on your edit: surely if there is *any* difference in gravitational potential between the start and end points there will be a time dilation (however small) otherwise there would be no acceleration. Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 (edited) May not denote it in terms of equations of state. Another format to describe Newton limit is [latex]G_{\mu\nu}=\eta_{\mu\nu}+H_{\mu\nu}[/latex] [latex]\eta_{\mu\nu}[/latex] denotes Minkowskii note there is no stress term. Either way the geodesic equation we have is time-like interval. Null geodesic for requires time-like intervals. Space-like intervals are tachyonic. So this doesn't include redshift. For sure there is always some time dilation unless your in the same reference frame. The term non relativistic simply means approximately Euclidean flat. For the [latex]G_{\mu\nu}[/latex] essentailly describes the geodesic line element. The stress tensor and the Ricci/Riehmann modify the metric tensor and vise versa Edited July 7, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 Yes that is what is used. What about my other questions? Link to comment Share on other sites More sharing options...

## Recommended Posts