tag:blogger.com,1999:blog-40057503068016452342020-11-21T15:23:04.237+01:00Spacetime and GeometryI am reading Spacetime and Geometry : An Introduction to General Relativity – by Sean M Carroll. The blog contains answers to his exercises, commentaries, questions and more.Unknownnoreply@blogger.comBlogger130125tag:blogger.com,1999:blog-4005750306801645234.post-77961905891825534622020-11-17T16:03:00.003+01:002020-11-18T14:28:58.131+01:00Constants and calculations<div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-O4bfXTr0zX8/XHe6e2rjgKI/AAAAAAABE3Y/ThVoNAuKRcsm7JXecOP1-czxMIfB86AMACPcBGAYYCw/s907/Plimpton_322.jpg" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" data-original-height="629" data-original-width="907" height="222" src="https://1.bp.blogspot.com/-O4bfXTr0zX8/XHe6e2rjgKI/AAAAAAABE3Y/ThVoNAuKRcsm7JXecOP1-czxMIfB86AMACPcBGAYYCw/w320-h222/Plimpton_322.jpg" title="Babylonian equations" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Babylonian equations<br /></td></tr></tbody></table>Carroll uses natural units in his book as do I on this web site. In natural units the speed of light, Planck's constant and Boltzmann's constants ##c=\hbar=h/2\pi=k=1##. In our equations these terms are left out. They need to be put in again to do real calculations and it's not always obvious how to do it. When Carroll does calculations he often uses centimeters and grams. I prefer to use SI units: meters and kilograms. </div><div><br /></div><div>It is also common to use geometric units which are units in which ##c=G=1## (so mass and energy have the same units as length and time). You cannot also have ##\hbar=1## in these units.</div><div><br /></div><div>In natural units the Schwarzschild radius is ##R_S=2GM##. I suppose that in geometric units it is ##R_S=2M##. In SI units it is ##2GM/c^2##. The ##c^2## makes a big difference and is easy to forget!</div><div><br /></div><div>The document contains a list in SI units of the values of universal constants and other handy constants like the mass of the sun. The table also gives the conversion factors from natural to SI units, which are derived from their dimensions. Two examples of their use are given at the end followed by a list of small and large prefixes (Terra, peta, pico etc) and their meanings. </div><div><a href="https://drive.google.com/open?id=1AQJtesVCZMp8C2nRhhABUWE7SGrYmSki" target="_blank">Commentary Constants and conversion factors.pdf</a> (4 pages)</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-55236343017851277962020-11-10T16:01:00.001+01:002020-11-11T10:33:20.893+01:00Plotting geodesics of Schwarzschild<div style="text-align: left;">I did some experiments with trying to plot solutions to the geodesics of Schwarzschild. They might give orbits of stars round the black hole Sagittarius A* at the centre of our galaxy. The geodesic equations become three simultaneous second order differential equations which give what I call the chugger equations below:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-9oC-jMQKpkI/X6pxfLqQ0GI/AAAAAAABI1g/fMPrbBbYke86BhXdNtpUJQ-5X0c983l0ACLcBGAsYHQ/s608/pic01.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="235" data-original-width="608" src="https://1.bp.blogspot.com/-9oC-jMQKpkI/X6pxfLqQ0GI/AAAAAAABI1g/fMPrbBbYke86BhXdNtpUJQ-5X0c983l0ACLcBGAsYHQ/s16000/pic01.png" /></a></div><div><a href="https://1.bp.blogspot.com/-SVWDZGo8ijE/X6qklJoKc0I/AAAAAAABI1w/A_flj-iTgoUnwcG5l5aiUtfuCzO9ug6ZgCLcBGAsYHQ/s297/pic01.png" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em; text-align: center;"><img border="0" data-original-height="266" data-original-width="297" src="https://1.bp.blogspot.com/-SVWDZGo8ijE/X6qklJoKc0I/AAAAAAABI1w/A_flj-iTgoUnwcG5l5aiUtfuCzO9ug6ZgCLcBGAsYHQ/s0/pic01.png" /></a>##t## is the coordinate time and ##r,\theta## are the usual plane polar coordinates. We are only considering curves in one plane. A prime indicates a derivative with respect to ##\lambda## which is the affine parameter from the geodesic equation and could be proper time. ##G## is old Newton's gravitational constant, ##M## is the mass of Sagittarius A* and ##c## is the speed of light.</div><div><br /></div><div><a href="https://1.bp.blogspot.com/-Xv3_fvdbZu4/X6qlMjjTWeI/AAAAAAABI18/1iLsymNSWtQhcIHAF4GuIe-nv6m8CDyRgCLcBGAsYHQ/s297/pic01.png" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em; text-align: center;"><img border="0" data-original-height="267" data-original-width="297" src="https://1.bp.blogspot.com/-Xv3_fvdbZu4/X6qlMjjTWeI/AAAAAAABI18/1iLsymNSWtQhcIHAF4GuIe-nv6m8CDyRgCLcBGAsYHQ/s0/pic01.png" /></a>We start with some initial values of ##t,t^\prime,r,r^\prime,\theta,\theta^\prime## and select some ##d\lambda## . Then we use those to get new values of ##t,t^\prime,r,r^\prime,\theta,\theta^\prime## from the chugger equations again and again and again and put them in successive rows of a spreadsheet. The spreadsheet then uses the values of ##r,\theta## to plot the curve. One of the stars orbiting Sagittarius A* is S2 and we use that as our example. It completes an orbit about every 16 years. The first image was my first attempt. The shape is good (it follows the top part of the Newtonian ellipse) until it gets pretty close to the black hole. The second image shows a close up of the same curve near the black hole.</div><div><br /></div><div>The dashed line is the ellipse that S2 would follow according to Newton's equations.</div><div><br /></div><div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-gl62RImF2bI/X6ql99XJy2I/AAAAAAABI2I/TCfvoK-4j5g_C0TGRaLsB3u2jyQrdrBQQCLcBGAsYHQ/s297/pic01.png" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="267" data-original-width="297" src="https://1.bp.blogspot.com/-gl62RImF2bI/X6ql99XJy2I/AAAAAAABI2I/TCfvoK-4j5g_C0TGRaLsB3u2jyQrdrBQQCLcBGAsYHQ/s0/pic01.png" /></a></div><br />If I 'manually' decreased ##d\lambda## when near the black hole and then increased it again as S2 departed I could get the third image which shows the last leg of the approximation. To be able to do that conveniently I had to program the chugger equations in VBA. To do 1,000 iterations takes about three minutes.</div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-iJXAgNkllxU/X6qm6QImX-I/AAAAAAABI2Q/3MWqbfxDrOo79Mc7T8wWSAvKHoa9sLs7QCLcBGAsYHQ/s297/pic01.png" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="267" data-original-width="297" src="https://1.bp.blogspot.com/-iJXAgNkllxU/X6qm6QImX-I/AAAAAAABI2Q/3MWqbfxDrOo79Mc7T8wWSAvKHoa9sLs7QCLcBGAsYHQ/s0/pic01.png" /></a></div><br />Finally I did all the calculations in Excel adjusting ##d\lambda## as it went along and got the fourth image from a 20,000 row spreadsheet. The calculation time is about one second. It was the best so far but still not good enough. The furthest distance (apsis) of S2 from the black hole decreases by 4% - it should be the same. However the furthest distance did advance by 0.0012 radians. I calculated (with help from Carroll) that it should be 0.0035 radians.</div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-_1GfyaJvcEo/X6uvd2Qb76I/AAAAAAABI2o/YVOaN3mE6uc5XtvlNIiDOVZvR68iNCX6gCLcBGAsYHQ/s528/pic01.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="475" data-original-width="528" height="360" src="https://1.bp.blogspot.com/-_1GfyaJvcEo/X6uvd2Qb76I/AAAAAAABI2o/YVOaN3mE6uc5XtvlNIiDOVZvR68iNCX6gCLcBGAsYHQ/w400-h360/pic01.png" width="400" /></a></div><br />For a bit of fun I also made an artist's impression of precession of the perihelion. It's simply done from the exact solution to Newton's equations which is an ellipse.</div><div><br /></div><div>Further material including spreadsheets and VBA at</div><div><a href="https://drive.google.com/open?id=1sv_JhwKQl45CoLeng51z30930CzDDsqP" target="_blank">Commentary 5.4 Geodesics of Schwarzschild plotted.pdf</a> (10 pages)</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-65609097468772115112020-11-03T17:26:00.001+01:002020-11-03T17:26:49.679+01:00Solar orbit plotter<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen="" class="BLOG_video_class" height="266" src="https://www.youtube.com/embed/ey_PMqTv6Kk" width="320" youtube-src-id="ey_PMqTv6Kk"></iframe></div><div style="text-align: left;">Here's how the plotter works</div><div>Newton's second law is$$<br />\vec{F}=m\vec{a}<br />$$Newton's law of gravity is$$<br />\vec{F}=\frac{GMm}{r^2}<br />$$They give you two differential equations of motion in polar coordinates$$<br />\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2=-\frac{GM}{r^2}<br />$$$$<br />2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}=0<br />$$Kepler's laws are</div><div><ol style="text-align: left;"><li>The orbit of a planet is an ellipse with the Sun at one of the two foci.</li><li>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</li><li>The square of a planet's orbital period is proportional to the cube of the length of the semi-major axis of its orbit.</li></ol></div><div>The second differential equation quickly gives you Kepler's second law. It is more difficult to get$$<br />r=\frac{P}{1+e\cos{\left(\theta-\phi\right)}}<br />$$That is a circle when ##e=0##, an ellipse when ##\left|e\right|<1##, a parabola when ##\left|e\right|=1## and a hyperbola when ##\left|e\right|>1##. ##\phi## is the angle of the axis of symmetry of the curve. ##P## is a magic constant. The equation gives you Kepler's first law. </div><div><br /></div><div>Newton must have been very pleased when he did those. Part of the job was inventing calculus! </div><div><br /></div><div>The initial conditions determine ##P,e,\phi##. That requires more work if you only know the initial position and velocity.</div><div><br /></div><div>Full details (except inventing calculus) in <a href="https://drive.google.com/open?id=1pbrj6VqLKPnqQ73TQLn7bZAfxz9nPiOm" target="_blank">Commentary 5.4 Orbital toypdf</a>. (9 pages)</div><div>The spreadsheets for plotting the curves are at </div><div><a href="https://drive.google.com/open?id=1AbHcGAzf0rUIav508oOGrAljnWY9nH9W" target="_blank">Commentary 5.4 Orbital toy.xlsx</a> </div><div><a href="https://drive.google.com/open?id=1e4N_LNf39k1_Z-4LVNf55fOmlab2pyfe" target="_blank">Commentary 5.4 Orbital toy.xlsm</a> (with animation macros)</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-17798756571343160102020-10-22T15:41:00.004+02:002020-10-23T16:03:54.297+02:00Maximally symmetric universes<div style="text-align: left;"><div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-CR0jkvqFbLM/X5GKOCYD-3I/AAAAAAABIIA/o5yG2huQ32IJW8nJOOjFE6DUhX9skhVFwCLcBGAsYHQ/s471/junk2.png" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="471" data-original-width="270" height="400" src="https://1.bp.blogspot.com/-CR0jkvqFbLM/X5GKOCYD-3I/AAAAAAABIIA/o5yG2huQ32IJW8nJOOjFE6DUhX9skhVFwCLcBGAsYHQ/w229-h400/junk2.png" width="229" /></a></div>In section 8.1 we meet maximally symmetric universes. That's universes where every point in spacetime is the same. I think the only ones are de Sitter, Minkowski and Anti de Sitter. We've done flat, boring Minkowski. De Sitter and Anti de Sitter are more interesting and we do conformal diagrams for both of them. At the end of the section Carroll does conformal anti de Sitter and casually draws some geodesics on it without saying how he plotted them. I was able to put my newly learnt skills to good use and show the same curves as he did😀. My version is on the left.</div><div><br /></div><div>He also jumps back to section 3.1 and uses the equation there for the Riemann tensor. It is$$<br />R_{\rho\sigma\mu\nu}=\kappa\left(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu}\right)<br />$$where ##\kappa## is a constant. That equation easily becomes$$<br />R_{\ \ \ \sigma\mu\nu}^\rho=\kappa\left(\delta_\mu^\rho g_{\sigma\nu}-\delta_\nu^\rho g_{\sigma\mu}\right)<br />$$so it looks like there's a really way to calculate the fiendish Riemann tensor for these special cases. Unfortunately $$<br />\kappa=\frac{R}{n\left(n-1\right)}<br />$$where ##n## is the number of dimensions and ##R## is the curvature scalar (aka Ricci scalar) which is constant in a maximally symmetric universe. Nevertheless to calculate it you have to calculate the Riemann tensor first! B*gger.</div><div><br /></div><div>I still wanted to check that it was all true. We already did it for a the surface of a unit sphere, S², and have found it was constant at 2 and its Riemann tensor does satisfy the formula above. So S² is maximally symmetric. All points on the surface of a sphere are equal. The metric for conformal anti de Sitter is$$<br />{ds}^2=\frac{\alpha^2}{\cos^2{\chi}}\left(-{dt^\prime}^2+{d\chi}^2+\sin^2{\chi}{d\Omega_2}^2\right)<br />$$where ##\alpha## is some constant. So its Riemann tensor is trickier!</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-fb9MCRgqmgA/X5GKtYuxJRI/AAAAAAABIII/QzuJfIauTpw-fl8ooOdDnkJS1TvOfBc6ACLcBGAsYHQ/s481/junk.png" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="289" data-original-width="481" src="https://1.bp.blogspot.com/-fb9MCRgqmgA/X5GKtYuxJRI/AAAAAAABIII/QzuJfIauTpw-fl8ooOdDnkJS1TvOfBc6ACLcBGAsYHQ/s320/junk.png" width="320" /></a></div>As I had already calculated the Christoffel symbols for that to make the diagram above, I calculated its Riemann tensor and then its curvature scalar which is ##-12\alpha^{-2}##. So it is constant and satisfies the formula as I checked with a cunning spreadsheet.</div><div><br /></div><div>Calculating that Riemann tensor, which is a relatively easy one, took 9 hours 45 minutes over three days and six sittings. I got faster as I went along, notwithstanding long lunches preceded by a refreshing dry martini.</div><div><br /></div><div>Read all about it at</div><div><a href="https://drive.google.com/open?id=1p5-qxrQ6FsuOJ3APTuflONutKefzwP-6" target="_blank">Commentary 8.1 Maximally symmetric universes.pdf</a> (12 pages)</div><div>and</div><div><a href="https://drive.google.com/open?id=1Mo2YPUBo1SmE2KMa3sMTiLgZ3UKcqlRr" target="_blank">Commentary 8.1 Maximally symmetric universes calculations.pdf</a> (23 pages)</div></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-36970809532043410862020-10-09T12:09:00.000+02:002020-10-09T12:09:00.077+02:00Plotting geodesics numerically<div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-6FFVvQ-s_Q4/X4AyBNpBYQI/AAAAAAABHzY/38F70_EtQWoWZtGDN-cnlmuIeZIXBjNIQCLcBGAsYHQ/s290/geodesics.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="288" data-original-width="290" src="https://1.bp.blogspot.com/-6FFVvQ-s_Q4/X4AyBNpBYQI/AAAAAAABHzY/38F70_EtQWoWZtGDN-cnlmuIeZIXBjNIQCLcBGAsYHQ/s0/geodesics.gif" /></a></div><div style="text-align: left;">Geodesic equations are sets of second order differential equations and are usually somewhere between hard and impossible to solve analytically. The following are the geodesic equations for the surface of a sphere (S²):</div><div style="text-align: left;"><div>$$\frac{d^2\theta}{d\lambda^2}-\sin{\theta}\cos{\theta}\left(\frac{d\phi}{d\lambda}\right)^2=0$$</div><div>$$\frac{d^2\phi}{d\lambda^2}+2\cot{\theta}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}=0$$They are about the simplest you can get and I still don't know if it's possible to solve them. When you have the solution you will be able to plot the geodesic curves as on the graph above.</div><div><br /></div></div><div style="text-align: left;"><b>This road block is very annoying </b>and I have finally busted through it. It only took a couple of days! I started with some very simple examples to test that what I was doing was correct and tested the theory on said S² which I had explored in March 2019. It all works and the general procedure for constructing geodesics is quite straightforward really! What a nice surprise.😀</div><div style="text-align: left;"><br /></div><div style="text-align: left;">Read all about it here <a href="https://drive.google.com/open?id=1qzANYOjSLczfSqsdGcoVs9OelxFz8Xr_" target="_blank">Plotting a differential equation.pdf</a> (8 pages with lots of picture). It's a short instruction manual on how to do it yourself.</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-12282498499707577562020-09-30T16:43:00.001+02:002020-09-30T16:43:46.318+02:00Very obscure bug in Google Blogger<p></p><div class="separator" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em; text-align: center;"><img border="0" data-original-height="180" data-original-width="180" src="https://1.bp.blogspot.com/-fjsZhVLEZvA/X3SSY5xO6fI/AAAAAAABHvw/okRdqrA8CFoV7qA-wc1oEI4fUt2hL4jKACLcBGAsYHQ/s0/junk.png" /></div>I use Google Blogger for this blog. I also use MathJax for displaying equations which are in a code called Latex. Google recently introduced major changes to Blogger and it screwed up the equation display. I prepare text for a post in MS-Word so we might have something like this (but replace all £ signs by $ signs)<br /><div><div>££\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0</div><div>££</div></div><div>and it should come out like this$$<br />\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0<br />$$but sadly it now comes out like this</div><div><div>$$</div><div>\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0</div><div>$$</div></div><div>which is not what is wanted!</div><div><br /></div><div>To move the Latex code from MS-Word was simply a matter of copying it and pasting it as plane text (Ctrl+Shift+V).</div><div><br /></div><div>In the old version of Blogger linefeeds produced HTML <br/>, in the new version they produce </div><div>, which screws up Mathjax. The old version and the new version of the HTML are shown below (once again, replace all £ signs by $ signs)</div><div><br /></div><div><div>££ <br/>\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0<br/>££</div><div><br /></div><div>££ </div><div></div><div>\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0 </div><div>££</div></div><h4 style="text-align: left;">Solution</h4><div><ol style="text-align: left;"><li>To begin with I just edited the latex removing linefeeds and reinstating with shift-enter. That replaces </div><div> by <br/>. You also have to make sure that the text style is Normal not Paragraph. The editor sometimes seems to start in the latter mode.</li><li>Then I discovered Search replace in HTML editing view. So now I just replace all </div><div> by <br/>.</li></ol></div><div><b><span style="color: red;">It's still rather tedious</span></b>. I wish the lovely people at Google would fix it.</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-60038943920293906152020-09-17T12:14:00.006+02:002020-09-17T12:20:36.607+02:00Curvature in two dimensions<p>Here we calculate formulas for curvature in two dimensions. Specifically we give coordinates, metrics, Christoffel symbols, Riemann tensors (only twice) and scalar curvature (or Ricci scalar) for ellipsoids, elliptic paraboloids and hyperbolic paraboloids.</p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-6IhaCGNMv3g/X2M0XmmCTzI/AAAAAAABHuA/Qdd-Jx4IeVQEOLo-CAheZXpi3mFsHrKRQCLcBGAsYHQ/s1143/junk.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="357" data-original-width="1143" height="170" src="https://1.bp.blogspot.com/-6IhaCGNMv3g/X2M0XmmCTzI/AAAAAAABHuA/Qdd-Jx4IeVQEOLo-CAheZXpi3mFsHrKRQCLcBGAsYHQ/w545-h170/junk.png" width="545" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Images from Wikipedia: <a href="https://en.wikipedia.org/wiki/Ellipsoid" target="_blank">Ellipsoid</a> , <a href="https://en.wikipedia.org/wiki/Paraboloid" target="_blank">Paraboloid</a><br /></td></tr></tbody></table><p>We also find a general formula for the scalar curvature in two dimensions which only requires one component of the Riemann tensor. On the way we find the formulas to calculate all the other non zero Riemann components from the one. With coordinates ##\left(\theta,\phi\right)## which are naturally used for the ellipsoid, the formula for the scalar curvature is $$<br />R=\frac{2}{g_{\phi\phi}}\left(\partial_\theta\Gamma_{\phi\phi}^\theta-\partial_\phi\Gamma_{\theta\phi}^\theta+\Gamma_{\theta\theta}^\theta\Gamma_{\phi\phi}^\theta+\Gamma_{\theta\phi}^\theta\Gamma_{\phi\phi}^\phi-\Gamma_{\phi\theta}^\theta\Gamma_{\theta\phi}^\theta-\Gamma_{\phi\phi}^\theta\Gamma_{\theta\phi}^\phi\right)<br />$$This is very similar to the formula given at the very end of the Wikipedia article on <a href="https://en.wikipedia.org/wiki/Gaussian_curvature" target="_blank">Gaussian curvature</a> ##K## which is$$<br />K=-\frac{1}{E}\left(\frac{\partial}{\partial u}\Gamma_{12}^2-\frac{\partial}{\partial v}\Gamma_{11}^2+\Gamma_{12}^1\Gamma_{11}^2-\Gamma_{11}^1\Gamma_{12}^2+\Gamma_{12}^2\Gamma_{12}^2-\Gamma_{11}^2\Gamma_{22}^2\right)<br />$$The article also reveals that the scalar curvature is twice the Gaussian curvature. The Wikipedia formula might be better written with all the one indices replaced by ##u## and all the 2 indices replaced by ##v##. Then reverting to ##\theta,\phi## as indices, the thing in brackets in the first formula is ##R_{\ \ \ \phi\theta\phi}^\theta## and in the second is ##R_{\ \ \ \theta\theta\phi}^\phi##. The relationship mentioned above between Riemann components is<br />\begin{align}<br />R_{\ \ \ \theta\theta\phi}^\theta&=\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\<br />R_{\ \ \ \theta\phi\theta}^\theta&=-\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\<br />R_{\ \ \ \phi\theta\phi}^\theta&=\partial_\theta\Gamma_{\phi\phi}^\theta-\partial_\phi\Gamma_{\theta\phi}^\theta+\Gamma_{\theta\theta}^\theta\Gamma_{\phi\phi}^\theta+\Gamma_{\theta\phi}^\theta\Gamma_{\phi\phi}^\phi-\Gamma_{\phi\theta}^\theta\Gamma_{\theta\phi}^\theta-\Gamma_{\phi\phi}^\theta\Gamma_{\theta\phi}^\phi&\phantom {10000}\nonumber\\<br />R_{\ \ \ \phi\phi\theta}^\theta&=-R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\<br />R_{\ \ \ \theta\theta\phi}^\phi&=-\frac{g_{\theta\theta}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\<br />R_{\ \ \ \theta\phi\theta}^\phi&=\frac{g_{\theta\theta}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\<br />R_{\ \ \ \phi\theta\phi}^\phi&=-\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\<br />R_{\ \ \ \phi\phi\theta}^\phi&=\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber<br />\end{align}So now it is easy to work out that the mysterious ##E## in the Wikipedia formula should be ##g_{\theta\theta}## which is the same as ##g_{11}##. </p><p>For the record the scalar curvatures for the ellipsoids, elliptic paraboloids and hyperbolic paraboloids are respectively$$<br />R_{El}=\frac{2b^2}{\left(a^2\cos^2{\theta}+b^2\sin^2{\theta}\right)^2}<br />$$$$<br />R_{Ep}=\frac{2a^2}{\left(1+a^2r^2\right)^2}<br />$$$$<br />R_{Hp}=\frac{-8}{\left|g\right|^2a^2b^2}<br />$$These are a bit vague until you know what the coordinate systems are but you can see what the sign of the curvature is which is what I was interested in.<br /></p><h2 style="text-align: left;">Why did I do all this?</h2>This was quite a project. It was sparked off by the following:<br /><i>On <a href="https://www.physicsforums.com/threads/what-is-the-definition-of-a-black-hole.992458/#post-6382574" target="_blank">Physics Forums</a> Ibix said</i>:<br />I don't know where you [JoeyJoystick] are getting numbers for the mass of the universe from - our current understanding is that it's infinite in size and mass.<br /><i>I said</i>:<br />Universe infinite in size and mass? That's extraordinary. Where can I read more about it, please?<br /><i>Ibix said</i>:<br />I would think Carroll covers it - chapter 8 of his lecture notes certainly does. The flat and negative curvature FLRW metrics are infinite in extent and have finite density matter everywhere. Modern cosmological models are a bit more complicated, but retain those features.<p></p><p>Chapter 8 in the book is on Cosmology and about 50 pages long. Part 8 of the lecture notes is on Cosmology and contain 15 pages. I think those are the places to look, but after a quick look I did not find anything specifically about the size of the Universe. <br /><br />If the universe has flat or negative curvature locally and then we presume that applies everywhere, that implies it is infinite in extent. Presumably shortly after the big bang the universe was finite in extent, so at some time it must have changed from finite to infinite! Maybe that's inflation?<br /><br />So I decided to test the flatness idea in two dimensions....<br />Summary in <a href="https://drive.google.com/open?id=16Bu8MTbUUfHem_pMoT-VhYEvBdUAl7Ec" target="_blank">Commentary 8 Curvatures 2D.pdf</a> (6 pages), mostly pictures.<br />Full details in <br /><a href="https://drive.google.com/open?id=1nUe8ah6M1vzYlwISOS-iFbndKR5oA9GI" target="_blank">Commentary 8 Curvatures 2D calculations.docx</a><br />or <a href="https://drive.google.com/open?id=1qAmOlFdRNPM49-71ex8rlUCpWxr5OnQf" target="_blank">Commentary 8 Curvatures 2D calculations.pdf</a> (24 pages)</p>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-42098166742202209862020-08-19T17:02:00.001+02:002020-08-19T17:08:39.594+02:00Einstein-Rosen bridges: Wormholes in Schwarzschild spacetime<p> </p><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen="" class="BLOG_video_class" height="465" src="https://www.youtube.com/embed/1f2Hb7ibyI4" width="560" youtube-src-id="1f2Hb7ibyI4"></iframe></div><div class="separator" style="clear: both; text-align: left;">Nearing the end of section 5.7 Carroll discusses wormholes connecting regions IV and I of the Kruskal diagram. These wormholes are also called Einstein-Rosen bridges.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">It was supposed to be impossible to travel between regions I and IV of the Kruskal diagram and here Carroll shows us how it can almost be done. He is very brief, <span style="color: red;"><b>his diagram is wrong</b></span>, but luckily I found a paper by Peter Collas and David Klein which goes into much more detail and helped me understand. I was even able to plot diagrams of the wormhole which takes you from the depths of region IV to the depths of region I. Sadly there is never enough time and my plot is, admittedly, a bodge. The real calculations would be too complicated.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">There is another way for an intrepid explorer from region I (where we live) to get a glimpse of region IV. After they cross the event horizon (the dashed line ##r=R_s##) they could look 'down and to the left' and they could see light coming in from region IV. They could even meet another explorer from region IV. However they could never tell us back in region I what they learnt and would eventually perish in the singularity.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Read it here <a href="https://drive.google.com/open?id=1zzVzNhxNqeU7WlpMBuFN-ZAt0EswZ0pP" target="_blank">Commentary 5.7a Wormholes.pdf</a> (3 pages).</div><p></p>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-15952669411165817082020-08-13T10:00:00.001+02:002020-08-13T10:00:05.903+02:00Big Bang!<p> Now we want to do a conformal diagram for an expanding universe. The metric equation is$$<br />{ds}^2=-{dt}^2+t^{2q}\left({dr}^2+r^2{d\Omega}^2\right)<br />$$and ##0<q<1\ ,0<t<\infty\ ,\ 0\le r<\infty##. It should be pretty easy because we did most of the heavy lifting when we did the conformal diagram for flat spacetime.<span style="color: red;"> <b>However I think Carroll made another mistake!</b></span></p><p>We introduce the coordinate ##\eta## with ##{dt}^2=t^{2q}{d\eta}^2## and we get a metric$$<br />{ds}^2=\left[\left(1-q\right)\eta\right]^{2q/\left(1-q\right)}\left(-{d\eta}^2+{dr}^2+r^2{d\Omega}^2\right)<br />$$The part on the right is the same as the flat metric with ##t\rightarrow\eta## so we can use all the work we did before to transform that into$$<br />{ds}^2=\omega^{-2}\left[-{dT}^2+{dR}^2+\sin^2{R}{d\Omega}^2\right]<br />$$with$$<br />\omega^{-2}=\left(\frac{\left[\left(1-q\right)\eta\right]^{q/\left(1-q\right)}}{\left(\cos{T}+\cos{R}\right)}\right)^2<br />$$and a bit of work on that gives $$<br />\omega=\left[\left(1-q\right)\sin{T}\right]^{q/\left(q-1\right)}\left(\cos{T}+\cos{R}\right)^{1/\left(1-q\right)}<br />$$But Carroll says that$$<br />\omega=\left(\frac{\cos{T}+\cos{R}}{2\sin{T}}\right)^{2q}\left(\cos{T}+\cos{R}\right)<br />$$<b><span style="color: red;">I'm pretty sure that Carroll is wrong, even though his formula is more attractive.</span></b> I also worked out how he went wrong. Carroll writes "The precise form of the conformal factor is actually not of primary importance" (because you throw it away for the diagram). Perhaps that's why he did not check it very carefully.</p><p>And here's the diagram<br /></p><div style="text-align: left;"><a href="https://1.bp.blogspot.com/-aATS8M3Rdi0/XzFo-eXqcHI/AAAAAAABHqQ/L366MjoI6J0j2U7OfGvmESbJrjay97lsACLcBGAsYHQ/s442/junk.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em; text-align: center;"><img border="0" data-original-height="435" data-original-width="442" src="https://1.bp.blogspot.com/-aATS8M3Rdi0/XzFo-eXqcHI/AAAAAAABHqQ/L366MjoI6J0j2U7OfGvmESbJrjay97lsACLcBGAsYHQ/s0/junk.png" /></a></div><p></p><p>At the singularity very near ##t=0## space can apparently be as big as you like. Never fear: ##r## might be big but ##t^{2q}## will be very small, so distances are very small too.</p><p>Read all the details at <br /><a href="https://drive.google.com/open?id=1WQyGHHvRRwrHXASg9Tvu84tqsWGUileo" target="_blank">Commentary App H Conformal Diagram Expanding Universe.pdf</a> (6 pages including a diversion on values of ##q##)</p><div><br /></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-33715624336198608992020-08-08T12:12:00.006+02:002020-08-08T17:31:29.873+02:00Conformal Diagrams<p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-J25-TNZAnFs/Xy5oZdwyihI/AAAAAAABHp0/qcOuMSgeRCQAuaV28YkV7fn41Yb84PmOwCLcBGAsYHQ/s960/Semay%2BCarroll%2BKeeling2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="546" data-original-width="960" height="328" src="https://1.bp.blogspot.com/-J25-TNZAnFs/Xy5oZdwyihI/AAAAAAABHp0/qcOuMSgeRCQAuaV28YkV7fn41Yb84PmOwCLcBGAsYHQ/w512-h291/Semay%2BCarroll%2BKeeling2.png" width="576" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div>Continuing my studies of conformal transformations and diagrams I move on to appendix H, follow Carroll's logic carefully and attempt to plot his conformal diagram of Minkowski space which he shows in Fig H.4 and I have copied above in the centre. My effort is on the right. The diagrams are similar except that the curves of constant ##t##, the Minkowski coordinate, have gradient 0 nowhere on his diagram and twice on mine. And for lines of constant ##r## the score is 1,3 (gradient ##\infty##). I was distressed. Carroll does not give explicit equations for the curves so there is quite a long chain of calculation to get them plotted. I triple checked it and could find no error so I ransacked the internet and found the short paper from from 2008 by Claude Semay, title "Penrose-Carter diagram for an uniformly accelerated observer". The first part is only about an inertial observer and Semay draws a conformal diagram for her with lines of constant ##t,r## just like mine. I have reproduced half his diagram on the left.<b><span style="color: red;"> So I think Carroll has made a mistake in his Figure H.4 - perhaps he just guessed at the curves!</span></b><div><p></p><p></p></div><div><div>Carroll lists the important parts of the diagram</div><div>##i^+=## future timelike infinity (##T=\pi,R=0##)</div><div>##i^0=## spatial infinity (##T=0,R=\pi##)</div><div>##i^-=## past timelike infinity (##T=-\pi,R=0##)</div><div>##J^+=## future null infinity (##T=\pi-R,0<R<\pi##)</div><div>##J^-=## past null infinity (##T=-\pi+R,0<R<\pi##)</div><div>Carroll use a symbol like ##\mathcal {J}## not ##J## which he calls "scri". It is hard to reproduce.</div><div><br /></div><div>Conformal diagrams are spacetime diagrams with coordinates such that the whole of spacetime fits on a piece of paper and moreover light cones are at 45° everywhere. The latter makes it easy to visualize causality. Since Minkowski spacetime has 45° light cones, if coordinates can be found which have a metric which is a conformal transformation of the Minkowski metric, the job is done. The first part of appendix H is devoted to finding conformal coordinates for flat Minkowski spacetime, expressed in polar coordinates - presumably to ease our work later in spherically symmetrical manifolds such as Schwarzschild. So we start from that metric:$$<br />{ds}^2=-{dt}^2+{dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)<br />$$</div><div>On the way to finding conformal coordinates we tried coordinates$$<br />\bar{t}=\arctan{t}\ \ ,\ \ \bar{r}=\arctan{r}<br />$$which certainly pack spacetime into the range$$<br />-\frac{\pi}{2}<\bar{t}<\frac{\pi}{2}\ ,\ 0\le\bar{r}<\frac{\pi}{2}<br />$$as you will see below if you press the button. Carroll says it might be fun to draw the light cones on that, so I made a movie:</div><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.blogger.com/video.g?token=AD6v5dxl9Yve0feZvB5wAXOJ2Vc8wXc2-2Fl3O8Z4l1Io0KQjAz0L-lW8U0i-QSnjOTUtjfMUzVld1uZBnoXeXlaSQ' class='b-hbp-video b-uploaded' frameborder='0' /></div><div class="separator" style="clear: both; text-align: center;">Light cone at various ##\bar{r}##</div><div><br /></div><div><span style="color: red;"><b>Carroll's Figure H.2 is also quite confusing.</b></span> It does not show the ##u,v## axes and I naturally assumed that the ##u## axis pointed down and to the right. It does not. It does the opposite.</div><div><br /></div><div>Read all the details at <a href="https://drive.google.com/open?id=1vkSPkdHH9ii5Kzm7tj4XY2XMam4Hbx7j" target="_blank">Commentary App H Conformal Diagrams.pdf</a> (12 pages) </div></div>Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-4005750306801645234.post-58786623162747781482020-08-04T15:58:00.003+02:002020-08-04T15:58:19.051+02:00Exercise Appendix G1 Conformal Null Geodesics<h2 style="text-align: left;">Question</h2><div>Show that conformal transformations leave null geodesics invariant, that is, that the null geodesics of ##g_{\mu\nu}## are the same as those of ##\omega^2g_{\mu\nu}##. (We already know that they leave null curves invariant; you have to show that the transformed curves are still geodesics.) What is the relationship between the affine parameter in the original and conformal metrics?</div><h2 style="text-align: left;">Answer</h2><div>The answer to this is a bit feeble I think - so feeble that I forgot to post it for ten days. It relies on Carroll's assertion in section 3.4 on the properties of geodesics that from some kind of equation like his 3.58 you can always find an equation that satisfies the geodesic equation and he gives us the relationship of the affine parameter to the magic equation. It would have been nice to prove the assertion but I think that was out of scope.</div><div><br /></div><div>It's at: <a href="https://drive.google.com/open?id=1w_31djS4zzOFTohtjNo50yjbqThOuRmr" target="_blank">Ex G1 Conformal Null Geodesics.pdf</a> (a mere two pages).</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-37755787020006308962020-07-23T17:43:00.003+02:002020-07-23T17:47:43.970+02:00Conformal Transformations<div>I want to understand the conformal diagrams in section 5.7 so I must read appendix G then H. Most of this is just checking Carroll's formulas for the conformal 'dynamical variables' - things like the connection coefficients and the Riemann tensor. It is eye bogglingly dense.</div><div><br /></div><div>Conformal transformations all start when you multiply each component of the metric by a scalar ##\omega## which may depend on the coordinates. So we have a conformal metric $$<br />{\widetilde{g}}_{\mu\nu}=\omega^2g_{\mu\nu}<br />$$Then we want find things like the Riemann tensor in the 'conformal frame'. It's quite easy to show that it is$$<br />{\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho+\nabla_\mu C_{\ \ \ \nu\sigma}^\rho-\nabla_\nu C_{\ \ \ \mu\sigma}^\rho+C_{\ \ \ \mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda<br />$$where$$<br />C_{\ \ \ \mu\nu}^\rho=\omega^{-1}\left(\delta_\nu^\rho\nabla_\mu\omega+\delta_\mu^\rho\nabla_\nu\omega-g^{\rho\lambda}g_{\mu\nu}\nabla_\lambda\omega\right)<br />$$</div><div>Carrol then says "it is a matter of simply plugging in and grinding away to get"$$<br />{\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho-2\left(\delta_{[\mu}^\rho\delta_{\nu]}^\alpha\delta_\sigma^\beta-g_{\sigma[\mu}\delta_{\nu]}^\alpha g^{\rho\beta}\right)\omega^{-1}\nabla_\alpha\nabla_\beta\omega<br />$$$$<br />+2\left(2\delta_{[\mu}^\rho\delta_{\nu]}^\alpha\delta_\sigma^\beta-2g_{\sigma[\mu}\delta_{\nu]}^\alpha g^{\rho\beta}+g_{\sigma[\mu}\delta_{\nu]}^\rho g^{\alpha\beta}\right)\ \omega^{-2}\left(\nabla_\alpha\omega\right)\left(\nabla_\beta\omega\right)<br />$$</div><div>I'm glad it wasn't complicated because getting to that took two dense pages part of which is shown below. He's also used the antisymmetrisation operator [], which is very clever but hard work. It also screws up my latex generator which does not like ]'s in indices.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-rhB7qdgvKWk/Xxmu6FZvguI/AAAAAAABHnk/ZaUJl0EA100k034t5nZMzmaG3TKSnPSkwCLcBGAsYHQ/s934/junk2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="934" data-original-width="709" height="625" src="https://1.bp.blogspot.com/-rhB7qdgvKWk/Xxmu6FZvguI/AAAAAAABHnk/ZaUJl0EA100k034t5nZMzmaG3TKSnPSkwCLcBGAsYHQ/w475-h625/junk2.png" width="475" /></a></div><div class="separator" style="clear: both; text-align: left;">See that in searchable form at <a href="https://drive.google.com/open?id=1HzKO2ykUhBqhdgD6-XeBjbtE0vAwzO7q" target="_blank">Commentary App G Conformal Transformations.pdf</a> (10 gruelling pages)</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-27492327530563061522020-07-18T15:45:00.003+02:002020-07-18T15:46:05.220+02:00Coordinates and basis vectors<div>I was still not really sure what is meant by the statement that partials form a coordinate basis. Then if they do, is their character (null, timelike or spacelike) related to the metric? At last I asked on Physics forums and PeterDonis had the final word. So now I do know what is meant by partials forming a coordinate basis and there is some relationship to the metric.</div><div><br /></div><div>Carroll writes$$<br />\frac{d}{d\lambda}=\frac{dx^\mu}{d\lambda}\partial_\mu<br />$$"Thus the partials ##\left\{\partial_\mu\right\}## do indeed represent a good basis for the vector space of the directional derivatives, which we can therefore safely identify with the tangent space."</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-4TBqP2b3tH0/XxL7B6j-z3I/AAAAAAABHmA/mTPYDVCBOzAgMpRAQ8zzAKXrxpfsVO6vACLcBGAsYHQ/s756/junk.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="366" data-original-width="756" height="243" src="https://1.bp.blogspot.com/-4TBqP2b3tH0/XxL7B6j-z3I/AAAAAAABHmA/mTPYDVCBOzAgMpRAQ8zzAKXrxpfsVO6vACLcBGAsYHQ/w500-h243/junk.jpg" width="500" /></a></div><div>We have a parabola parameterized by ##\lambda## given by$$t=\frac{\lambda^2}{10}\ ,\ \ x=\lambda$$so$$\frac{dt}{d\lambda}=\frac{\lambda}{5}\ \ ,\ \ \ \frac{dx}{d\lambda}=1$$and the tangent vector ##V=d/d\lambda## has components ##\left(dt / d\lambda , dx/ d \lambda\right)## at ##\left(t,x\right)##.</div><div><br /></div><div>In plane polar coordinates ##\left(r,\theta\right)## the ##\partial_\theta## basis vector is along lines with parameters ##\left(k_r,\lambda\right)##. So the ##\theta## basis vectors lie on concentric circles around the origin. Similarly ##r## basis vectors are radial.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Vwhid3FMV7M/XxL7G-EVhPI/AAAAAAABHmE/C0yCmYsUQK0ffgaj5JDOGgwP7lAHBQNAwCLcBGAsYHQ/s474/junk2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="474" data-original-width="470" height="400" src="https://1.bp.blogspot.com/-Vwhid3FMV7M/XxL7G-EVhPI/AAAAAAABHmE/C0yCmYsUQK0ffgaj5JDOGgwP7lAHBQNAwCLcBGAsYHQ/w396-h400/junk2.png" width="396" /></a></div><div>All you need for plane polar coordinates is a line segment to measure ##\theta## from and one end to serve as the origin. The line is normally drawn horizontally. That is not essential.</div><div><br /></div><div>See how it all hangs together: <a href="https://drive.google.com/open?id=1qYpubQmANO0Np_ETMf0U7F837LKDv__f" target="_blank">Commentary 2.3 Coordinates and basis vectors.pdf</a> </div><div>And the thread on physics forums <a href="https://www.physicsforums.com/threads/null-basis-vectors-metric-signatures-near-kruskal.991408/ " target="_blank">here</a>.</div><div><br /></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-43636619646927187592020-07-14T16:16:00.011+02:002020-07-16T17:45:27.239+02:00The Metric<div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-53IhoV5iRbY/XwnfXy9UoOI/AAAAAAABHjY/dYc0_Y2wPwwVyiPwihOysrWcTIS2bhwnQCLcBGAsYHQ/s772/junk2.png" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" data-original-height="643" data-original-width="772" src="https://1.bp.blogspot.com/-53IhoV5iRbY/XwnfXy9UoOI/AAAAAAABHjY/dYc0_Y2wPwwVyiPwihOysrWcTIS2bhwnQCLcBGAsYHQ/s320/junk2.png" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Kruskal from <a href="https://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect10_2007.pdf" target="_blank">From Stanford</a></td></tr></tbody></table>I started to write this in May 2019 when I read section 2.5 and it has remained a work in progress until now. I have quite a collection of metrics: 3D Euclidean, Plane polar, Spherical polar, Surface of sphere (S2), Minkowski, Spherical Minkowski, Schwarzschild, Eddington-Finkelstein, Kruskal. I will add more as I find them.</div><div><br /></div><div>We start with a bit of general information about a metric, then list each of the metrics. There's also something on null, spacelike and timelike coordinates; converting metrics to coordinate systems with an example on spherical metrics; four velocities and mysterious interchange of ##\mathrm{d}x## and ##dx##.</div><div><br /></div><div>On this page the coordinate infinitesimals (which are really one forms) should be written with an unitalicized ##\rm{d}## for example ##\mathrm{d}x## not ##dx##. That is very fiddly in Latex so I have not bothered except in the very first occurrence! (Most texts don't bother at all).</div><h2 style="text-align: left;">The glory of the metric</h2><div>The metric ##g_{\mu\nu}## is a symmetric (0,2) tensor and Carroll lists the following to appreciate its "glory". He is following Sachs and Wu (1977).</div><div><div><ol style="text-align: left;"><li>The metric supplies a notion of "past" and "future".</li><li>The metric allows the computation of path length and proper time.</li><li>The metric determines the "shortest distance" between two points, and therefore the motion of test particles.</li><li>The metric replaces the Newtonian gravitational field ##\phi##.</li><li>The metric provides a notion of locally inertial frames and therefore a sense of "no rotation".</li><li>The metric determines causality, by defining the speed of light faster than which no signal can travel.</li><li>The metric replaces the traditional Euclidean three-dimensional dot product of Newtonian mechanics.</li><li>The (inverse) metric lowers (raises) indices: ##U_\mu=g_{\mu\nu}U^\nu\ ,\ U^\mu=g^{\mu\nu}U_\nu##</li></ol></div></div><div>I added the last one. It is related to the second last.</div><div><div><h2 style="text-align: left;">Two and three dimensional geometry metrics</h2><div>These are called Euclidean or Riemannian metrics.</div><div><b>3D Euclidean</b>$${ds}^2=\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$$$$g_{ij}=\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right)$$we explain why the equation and matrix formulations are equivalent. </div><div><b>Plane polar coordinates</b></div><div>Coordinates ##(r,\theta)## radial, angle to ##x## axis$$g_{ij}=\left(\begin{matrix}1&0\\0&r^2\\\end{matrix}\right)$$<b>Spherical polar </b></div><div>Coordinates ##(r,\theta,\phi)## radial, polar, azimuthal (= longitude)$$g_{ij}=\left(\begin{matrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2{\theta}\\\end{matrix}\right)$$Surface of sphere (S2)</div><div>Coordinates ##(\theta,\phi)## polar, azimuthal$$g_{ij}=\left(\begin{matrix}1&0\\0&\sin^2{\theta}\\\end{matrix}\right)$$This metric is often written $${d\Omega}^2={d\theta}^2+\sin^2{\theta}{d\phi}^2$$</div><h2 style="text-align: left;">Relativistic metrics</h2><div>These are called Lorentzian or pseudo-Riemannian metrics. </div><div>We use a ##-+++## signature and the speed of light ##c## is conveniently set to 1.</div><div><b>Minkowski</b></div><div>Coordinates ##(t,x,y,z)## $$g_{\mu\nu}=\left(\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{matrix}\right)$$<b>Spherical Minkowski</b></div><div>Coordinates ##(t,r,\theta,\phi)## ##r,\theta,\phi## as in spherical$$g_{\mu\nu}=\left(\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2{\theta}\\\end{matrix}\right)$$We could also write$${ds}^2=-d\tau^2=-{dt}^2+{dr}^2+r^2{d\Omega}^2$$<b>Schwarzschild metric (spherical polar)</b></div><div>Coordinates are ##(t,r,\theta,\phi)## time, radial, polar, azimuthal$${ds}^2=-d\tau^2=-\left(1-\frac{2GM}{r}\right){dt}^2+\left(1-\frac{2GM}{r}\right)^{-1}{dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)$$We often replace ##2GM## (twice Newton's gravitational constant times the central mass) by ##R_s## the Schwarzschild radius which is the radius of event horizon and the last part by ##r^2{d\Omega}^2##. </div><div><b>Eddington-Finkelstein metric</b></div><div>The Eddington-Finkelstein metric is for the same spacetime as Schwarzschild but with coordinates ##\left(v,r,\theta,\phi\right)## $${ds}^2=-\left(1-\frac{R_s}{r}\right)dv^2+dvdr+drdv+r^2{d\Omega}^2$$$$v=t+r+R_s\ln{\left|\frac{r}{R_s}-1\right|}$$This is the first metric we have listed which is not diagonal.</div><div><b>Kruskal predecessor</b></div><div>On the way to the Kruskal metric we get coordinates ##\left(v^\prime,u^\prime,\theta,\phi\right)## with metric equation$${ds}^2=-\frac{2{R_s}^3}{r}e^{-\frac{r}{R_s}}\left(dv^\prime du^\prime+du^\prime dv^\prime\right)+r^2{d\Omega}^2$$where ##r## is implicitly defined in terms of ##u^\prime,v^\prime## as$$u^\prime v^\prime=-\left(\frac{r}{R_s}-1\right)e^{r/R_s}$$<b>Kruskal</b></div><div>Kruskal coordinates are ##\left(T,R,\theta,\phi\right)## where in terms of Schwarzschild ##t,r## $$T=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\sinh{\left(\frac{t}{2R_s}\right)}$$$$R=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\cosh{\left(\frac{t}{2R_s}\right)}$$and they give a metric equation$${ds}^2=\frac{4{R_s}^3}{r}e^{-\frac{r}{R_s}}\left(-dT^2+dR^2\right)+r^2{d\Omega}^2$$with ##r## implicitly defined from$$T^2-R^2=\left(1-\frac{r}{R_s}\right)e^\frac{r}{R_s}$$</div><div><br /></div><div>Read all that and more on at<a href="https://drive.google.com/open?id=1GMA1VfrvDBogsYEutMCxZTgK6h1chWjz" target="_blank"> Commentary 2.5 The Metric.pdf</a> (11 pages)</div><div><br /></div><div><br /></div></div></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-25942207939542790262020-07-09T17:39:00.002+02:002020-08-12T12:08:38.088+02:00Kruskal coordinates and the maximally extended Schwarzschild solution<div><div>Kruskal coordinates ##\left(T,R,\theta,\phi\right)## are called 'maximally extended' because they cover the whole spacetime except the true singularity at ##r=0##. Indeed, they find some remarkable new regions of spacetime! The history of the discoveries spans 45 years from Einstein and Schwarzschild (1915) to Kruskal (1960).</div></div><div></div><div><br /></div><div>The Kruskal coordinates are related to Schwarzschild coordinates ##\left(t,r,\theta,\phi\right)## by $$T=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\sinh{\left(\frac{t}{2R_s}\right)}$$</div><div>$$R=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\cosh{\left(\frac{t}{2R_s}\right)}$$and give a metric equation$${ds}^2=\frac{4{R_s}^3}{r}e^{-\frac{r}{R_s}}\left(-dT^2+dR^2\right)+r^2{d\Omega}^2$$with ##r## implicitly defined from$$T^2-R^2=\left(1-\frac{r}{R_s}\right)e^{r/R_s}$$From these we can draw a Kruskal diagram showing lines of constant ##t## and ##r## and light cones which miraculously are always at 45° just like in flat spacetime.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-LfbbvbBy4ho/Xwc2c6PJ3LI/AAAAAAABHi4/IVe69ZKN29w_pL-EK8tlYq6Q7ztJR233wCLcBGAsYHQ/s407/junk.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="400" data-original-width="407" src="https://1.bp.blogspot.com/-LfbbvbBy4ho/Xwc2c6PJ3LI/AAAAAAABHi4/IVe69ZKN29w_pL-EK8tlYq6Q7ztJR233wCLcBGAsYHQ/s320/junk.jpg" width="320" /></a></div><div><br /></div><div>The regions above the upper ##r=0## line and below the lower ##r=0## line are not part of spacetime. (##r<0## and ##t>+\infty## or ##t<-\infty## in them). The rest of the diagram is divided into four regions.</div><div><br /></div><div><font face="times">I</font> Right quarter. Normal space time outside the event horizon.👍</div><div><font face="times">II</font> Below the upper ##r=0## line and above the upper ##r=R_s## lines. Inside the event horizon.👎</div><div><font face="times">III</font> Above the lower ##r=0## line and below the lower ##r=R_s## lines. The white hole.💣</div><div><font face="times">IV</font> Left quarter. The unreachable mirror image of normal space time.👻</div><div><br /></div><div>The red light cones, which are always at 45°, are informative. In region <font face="times">I</font> you can always maintain a fixed ##r## and you can always move up towards and into region <font face="times">II</font>. You can never get into regions<font face="times"> III</font> or <font face="times">IV</font>. Once in region<font face="times"> II</font> it is impossible to maintain constant ##r## because lines of constant ##r## are always flatter than 45° so you inevitably arrive at ##r=0##. Region<font face="times"> IV</font> is like region<font face="times"> I</font> and you can only get to region <font face="times">II</font> from it. Carroll says region<font face="times"> III</font> is a the time reverse of <font face="times">II</font> and can be thought of as a white hole. "There is a singularity in the past, out of which the universe appears to spring". Things can only come out of it. They can just get directly to <font face="times">II</font> through the origin, more likely they will go into <font face="times">I</font> or <font face="times">IV</font>.</div><div><br /></div><div>Regions <font face="times">II</font> and <font face="times">III</font> are allowed even though ##t>+\infty## in <font face="times">II</font> and ##t<-\infty## in <span style="font-family: times;">III</span>.</div><div><br /></div><div>Although region <span style="font-family: times;">I</span> and region <span style="font-family: times;">IV </span>are mutually unreachable, if an intrepid explorer went from region<span style="font-family: times;"> I</span> into region <span style="font-family: times;">II</span>, they would be able to see some things that had happened in region <span style="font-family: times;">IV</span>. Likewise an explorer from region <span style="font-family: times;">IV </span>could have a brief look at region <span style="font-family: times;">I</span> before perishing.</div><div><br /></div><div>Find out how to get to Kruskal and a few other things in <a href="https://drive.google.com/open?id=1IxnQrwK9K8bCYWkpKzQYX9dgBOQ3HiNJ" target="_blank">Commentary 5.7 Kruskal coordinates.pdf</a> (11 pages).</div><div><br /></div><div><br /></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-49294001727730873372020-06-27T18:00:00.002+02:002020-06-29T15:24:18.096+02:00Time travel inside a black hole<div dir="ltr" style="text-align: left;" trbidi="on"><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;"><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-eiwjXHwkxGQ/XvnrPDc3UxI/AAAAAAABHgQ/p4wJBvDyJSM6F0vy495xULOrNjosVFPXgCK4BGAsYHg/s392/junk2.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="392" data-original-width="385" height="320" src="https://1.bp.blogspot.com/-eiwjXHwkxGQ/XvnrPDc3UxI/AAAAAAABHgQ/p4wJBvDyJSM6F0vy495xULOrNjosVFPXgCK4BGAsYHg/s320/junk2.png" /></a></div></span></div><div>We can now use the geodesic that we have found and tested to plot a spacetime diagram all the way to the centre of a black hole. The graph shows the result for our famous beacon, or even for a foolish astronaut.</div><div><br /></div><div><div>As we saw before they dawdle at the event horizon for ever (##t\rightarrow\infty## as ##r\rightarrow2GM##). But we know you can cross an event horizon and when that happens in the distant future they hurry backwards in time and soon get to a reasonable ##r,t##! </div><div><br /></div><div>Meanwhile the astronaut looking at a wristwatch sees proper time, ##\tau##, ticking by steadily and reaches the centre in finite time.</div><div><br /></div><div>Luckily the astronauts time travel cannot be observed from the outside. Our red faces are saved.</div></div><div><br /></div><div>Given everything we have done before the calculations are simple. It's a one pager in <a href="https://drive.google.com/open?id=1DeYzQ2viHnLb6tPhw62k3SNwiuT6VKfY" target="_blank">Commentary 5.6 Time Travel.pdf</a>.</div></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-31568060844123505322020-06-27T17:50:00.001+02:002020-06-27T17:50:11.696+02:00Schwarzschild Black Holes - The Geodesic<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-MEQApslXp-U/XvS3aizdLhI/AAAAAAABHa0/SmcocSX89RULfuvbRWD03JWcLstfA30GgCK4BGAsYHg/s917/junk.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="583" data-original-width="917" height="398" src="https://1.bp.blogspot.com/-MEQApslXp-U/XvS3aizdLhI/AAAAAAABHa0/SmcocSX89RULfuvbRWD03JWcLstfA30GgCK4BGAsYHg/w625-h398/junk.png" width="625" /></a></div><div class="separator" style="clear: both; text-align: left;">(1) and (2) are the ##t## and ##r## geodesic equations in the Schwarzschild metric which we found in <a href="https://www.general-relativity.net/2020/04/geodesics-of-schwarzschild.html" target="_blank">section 5.4</a>. (2) becomes a bit simpler on a radial path. That's (3). Geodesic equations are meant to give you trajectories of freely falling particles parametrised by ##\lambda##. (4) and (5) are the equation of a particle (or beacon) falling along a radius into a black hole from a distance ##r_*##. We used them to plot a beacon's path <a href="https://www.general-relativity.net/2020/06/beacon-falling-into-black-hole-revisited.html" target="_blank">here</a>. They come from (6) which we calculated in <a href="https://www.general-relativity.net/2020/06/exercise-55-observer-and-beacon-outside.html" target="_blank">exercise 5.5</a> where we also calculated (7). </div><div class="separator" style="clear: both; text-align: left;">##t## is coordinate time, ##r## is the distance from the centre. They are the coordinates.</div><div class="separator" style="clear: both; text-align: left;">##\lambda## is an affine parameter (it is proportionate to the length along the line).</div><div class="separator" style="clear: both; text-align: left;">##R_s## is the Schwarzschild radius (radius of event horizon).</div><div class="separator" style="clear: both; text-align: left;">##\theta,\phi## are the other spherical polar coordinates (polar and azimuth), which we can ignore.</div><div class="separator" style="clear: both; text-align: left;">##r_*## is the radial distance from which the test particle, or beacon, is dropped.</div><div class="separator" style="clear: both; text-align: left;">##\tau## is the proper time, which can be an affine parameter for a massive particle.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><b>The question is: Is the path given by (4) and (5) a real geodesic? That is, does it satisfy (1) and (3)?</b></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><font size="5">And the answer is <b>YES.</b></font></div><div class="separator" style="clear: both; text-align: left;"><b><br /></b></div><div class="separator" style="clear: both; text-align: left;">I studied these geodesics three months ago and they have never yet been useful. This is the first time that they have even clicked with anything!</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Here's how they click: <a href="https://drive.google.com/open?id=10VEJzUOvVSDucsBasye3f0ORfp3IyGnL" target="_blank">Commentary 5.4#1 Geodesics of Schwarzschild.pdf</a> (only 2 pages really)</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-44967878897135919432020-06-24T16:41:00.001+02:002020-06-25T16:01:02.888+02:00Proper acceleration, Spaghettification and G2<div dir="ltr" style="text-align: left;" trbidi="on"><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-JZ5L4JLP-X8/XvNho9ngm5I/AAAAAAABHZ0/ai5FbTWlSJ08zW_SF0lPI_BRqscTN3D8wCLcBGAsYHQ/s1600/junk.jpg" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="168" data-original-width="300" src="https://1.bp.blogspot.com/-JZ5L4JLP-X8/XvNho9ngm5I/AAAAAAABHZ0/ai5FbTWlSJ08zW_SF0lPI_BRqscTN3D8wCLcBGAsYHQ/s1600/junk.jpg" /></a></div>After that Exercise 5.5 I thought I was an expert on falling into a black hole and and that I could calculate the 'proper acceleration' and spaghettification which is the term used for what happens when acceleration differs so much in different parts of your body that you get stretched out like the doomed lady on the right.<br /><br />Then there is there is the matter of the gas cloud known as G2, which was discovered heading towards Sagittarius A*, the black hole at the centre of our galaxy, in 2011 by some folk at the Max Planck Institute. G2 was destined to come closest to Sagittarius A* in Spring 2014 "with a predicted closest approach of only 3000 times the radius of the event horizon". There was great excitement because spaghettification and great fireworks were expected. However nothing much happened and G2 continues on its way, orbiting Sagittarius A*.<br /><br />I attempted to do some calculations and tested them on <a href="https://www.physicsforums.com/threads/questions-about-spaghettification-and-gas-cloud-g2.990493/" target="_blank">Physics Forums</a> and got adverse comments from PeterDonis. Ibix was more positive "I think your maths is correct ...". PeterDonis showed me the 'correct' way of calculating proper acceleration and then dragged me back to the geodesic deviation equation which is the right way to calculate spaghettification. But I still don't fully understand said equation and how to use it😭. I also learnt a bit more about units: 'natural' and 'geometric'. PeterDonis is a hard task master.</div><div dir="ltr" style="text-align: left;" trbidi="on"><br /></div><div dir="ltr" style="text-align: left;" trbidi="on">The correct way to calculate proper acceleration gives infinite acceleration at the event horizon. One benefit of that is that it tells you that you cannot escape falling through it, once close enough. That is true. The drawback is that it makes the radial change in acceleration also infinite. So you will get ripped up at the event horizon. That is not true given a big enough black hole.<br /><br />Here are my calculations (about four pages) and what I learnt (another three).<br /><a href="https://drive.google.com/open?id=1hecx4Toize7XDGt3ZW-8JV-plL-Iugrv" target="_blank">Commentary 5.6#4 Proper acceleration.pdf</a> </div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-41176675673341369902020-06-15T15:48:00.002+02:002020-06-17T16:50:39.584+02:00Beacon falling into a black hole, revisited<div dir="ltr" style="text-align: left;" trbidi="on"><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div>Having done <a href="https://www.general-relativity.net/2020/06/exercise-55-observer-and-beacon-outside.html" target="_blank">Exercise 5.5</a> and discovered lots of things about a beacon being dropped into a black hole we can now revisit <a href="https://www.general-relativity.net/2020/04/schwarzschild-black-holes.html" target="_blank">Commentary 5.6</a> where we struggled with Carroll's figure 5.8 and Carroll's claims about increasing intervals observed at a safe distance ##r_\ast## from the centre of a black hole. Back then I ran into several problems: 1) I could not calculate the equation for the radial geodesic, 2) When I used some invented curve, that was more or less the right shape, I had to estimate ##\Delta\tau_1##'s along the beacon path, 3) Having done that the intervals measured by the observer did often not increase.<br /><br />Now we have an equation of motion for the beacon, that is the radial geodesic. I thought it would never be useful. We now find that it is useful it, and it is the intimidating$$<br />t=\frac{\sqrt{r_\ast-R_S}}{\sqrt{R_S}}\left[\sqrt{r_\ast-r}\sqrt r-\left(r_\ast+2R_S\right)\sin^{-1}{\left(\frac{\sqrt r}{\sqrt{r_\ast}}\right)}\right]<br />$$$$<br />+R_S\ln{\left|\frac{\left(r_\ast-R_S\right)\sqrt r+\sqrt{R_S}\sqrt{r_\ast-R_S}\sqrt{r_\ast-r}}{\left(r_\ast-R_S\right)\sqrt r-\sqrt{R_S}\sqrt{r_\ast-R_S}\sqrt{r_\ast-r}}\right|}<br />$$$$<br />+\frac{\pi\left(r_\ast+2R_S\right)\sqrt{r_\ast-R_S}}{2\sqrt{R_S}}<br />$$(##R_s=2GM## is the Schwarzschild radius) so we can plot that on a graph and the first problem is resolved. Moreover we found the proper speed of the beacon$$<br />\frac{dr}{d\tau}=-\sqrt{\frac{R_s\left(r_\ast-r\right)}{rr_\ast}}<br />$$By inverting and integrating that we have an expression for ##\tau## along the path and we can calculate ##\Delta\tau_1##'s which fixes the second problem. We already knew how to plot the return flight of the photon and when we do so, we find that the intervals measured by the observer do <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-56-lmztpa1A/Xud4jRIvXJI/AAAAAAABHYc/qABmoQntJZsGuKXX_SiRDWxabihXUV6agCLcBGAsYHQ/s1600/junk3.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="197" data-original-width="192" src="https://1.bp.blogspot.com/-56-lmztpa1A/Xud4jRIvXJI/AAAAAAABHYc/qABmoQntJZsGuKXX_SiRDWxabihXUV6agCLcBGAsYHQ/s1600/junk3.png" /></a></div>increase for successive signals. So problem 3 was fixed! However, we are still not out of the woods. My second attempt is shown on the right with amounts in natural units. The observer is at ##r_\ast=15## and the event horizon at ##2GM=10##. We consider three photons emitted at ##a,b,c## separated by ##\Delta\tau_1=6##. We can measure the intervals seen by the observer and they do increase but the photon world line is disappointingly flat, unlike Carroll's. Spacetime is almost flat up to event ##b##. So we have to start very close to the event horizon with ##r_\ast=12## and zoom in to the area marked by the red rectangle. Eventually we achieve something like Carroll's (with a bonus photon from ##d##) as shown below. The diagram on the left is fairly bare like Carroll's the same one on the right has numbers added in natural units. ##\Delta\tau_1## was a very small ##0.5## and the intervals observed at ##r_\ast## were an order of magnitude larger and increasing as Carroll predicted. The observer might have wanted to subtract out the, easily calculated, flight times of the returning photons. The intervals between emissions still increase as can be seen by the increasing vertical distances of ##a,b,c,d## on the diagram.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-5cHBD2gBcV0/Xud4z9qejTI/AAAAAAABHYk/w8hXgplYBycV-J9CKPD44EtEc4KbzkzcACLcBGAsYHQ/s1600/junk.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="378" data-original-width="560" src="https://1.bp.blogspot.com/-5cHBD2gBcV0/Xud4z9qejTI/AAAAAAABHYk/w8hXgplYBycV-J9CKPD44EtEc4KbzkzcACLcBGAsYHQ/s1600/junk.png" /></a></div>As an added bonus, from the equation for the proper speed of the beacon, we can calculate it's finite proper time to the event horizon and to the centre of the black hole. The increasing length along the geodesic for fixed proper time helps us intuit a resolution to the apparent paradox that the beacon 'never seems to get into the black hole'<br /><br />Moreover if we get inside the event horizon, the proper time from there to ##r=0## is ##\pi GM## which is the maximum possible value that we calculated in <a href="https://www.general-relativity.net/2020/04/exercise-53-inside-event-horizon.html" target="_blank">exercise 5.3</a>.<br /><br />Get the details here: <a href="https://drive.google.com/open?id=1sRV2NdWu-JwnG2y6ir8iCwaYa37DJBJs" target="_blank">Commentary 5.6#3 Schwarzschild Black Holes.pdf</a> (5 pages)<br /><br />More intriguing puzzles remain:<br /><br /><ul style="text-align: left;"><li>To relate the monster formula for ##t## to the geodesic equation from section 5.4; </li><li>To think about the beacon's (free falling) inertial coordinate system, its forward and backward light cones and its relationship to the Schwarzschild coordinate system </li><li>Spaghettification and the mysterious G2 gas cloud</li></ul></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-85506034455468119452020-06-06T11:39:00.000+02:002020-06-06T11:39:28.700+02:00Exercise 5.5 Observer and beacon outside black hole<div dir="ltr" style="text-align: left;" trbidi="on"><h3 style="text-align: left;">Question</h3>Consider a comoving observer sitting at constant spatial coordinates ##\left(r_\ast\ ,\theta_\ast\ ,\ \phi_\ast\right)## around a Schwarzschild black hole of mass ##M##. The observer drops a beacon onto the black hole (straight down along a radial trajectory). The beacon emits radiation at a constant wavelength ##\lambda_{em}## (in the beacon rest frame).<br /><br />a) Calculate the coordinate speed of the beacon as a function of ##r##.<br />b) Calculate the proper speed of the beacon. That is, imagine there is a comoving observer at fixed ##r##, with a locally inertial coordinate system set up as the beacon passes by, and calculate the speed as measured by the comoving observer. What is it at ##r=2GM##?<br />c) Calculate the wavelength ##\lambda_{obs}##, measured by the observer at ##r_\ast##, as a function of the radius ##r_{em}## at which the radiation was emitted.<br />d) Calculate the time ##t_{obs}## at which a beam emitted by the beacon at radius ##r_{em}## will be observed at ##r_\ast##.<br />e) Show that at late times, the redshift grows exponentially: ##\lambda_{obs}/\lambda_{em}\propto e^{t_{obs}/T}##. Give an expression for the time constant ##T## in terms of the black hole mass ##M##.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-FEizDo3_L-k/XttivUdAPsI/AAAAAAABHVc/Vha6FkYUzlkuAHoLHWYRhDylrSZl5TgvgCLcBGAsYHQ/s1600/junk2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="595" src="https://1.bp.blogspot.com/-FEizDo3_L-k/XttivUdAPsI/AAAAAAABHVc/Vha6FkYUzlkuAHoLHWYRhDylrSZl5TgvgCLcBGAsYHQ/s1600/junk2.png" /></a></div><h3 style="text-align: left;">Answers</h3>This question is a fascinating can of worms. I needed the help of an unwitting mentor Jeriek Van den Abeele from the University of Oslo. The first crucial help was to use the timelike killing vector constant $$<br />E=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau}<br />$$This came in useful not only for the beacon velocities but also for the travel time of the photon back to the observer at ##\ r_\ast## in question d.<br /><br />I also needed help from Physics Forums on comoving coordinates and I think I understand them now, although there was some dispute about how the (second) comoving observer in question (b) would actually achieve that state. When people talk about comoving coordinates for the Universe, they are talking about something quite different from here: Coordinates comoving with the Universe are growing with the Universe.<br /><br />Part of the worminess arises from confusion about what coordinates are referring to what. There was a profusion of subscripts: ##r_{em},r_\ast,\ R_s,t_\gamma,t_{em},t_{obs},t_b##. The last of those was introduced gratuitously by my mentor who I refer to as Oslo. In my opinion ##t_b\equiv t_{em},\ \ r_b\equiv r_{em}##. I try to avoid these subscripts as much as possible.<br /><br />I got two answers to question (b). One was the proper speed calculated in the strange comoving inertial coordinate system, as Carroll asked, and the other was the proper speed in Schwarzschild coordinates. I called them ##dr^\prime/d\tau^\prime## and ##dr/d\tau##. The first was what Carroll asked for and can be used (but is not essential) in question (c). Apart from that I am not sure how useful it is. ##dr/d\tau## is what would be experienced by an astronaut falling with the beacon and it does not reach the speed of light at the event horizon.<br /><br />Interestingly I calculated the answer to (c) using Doppler redshift + gravitational redshift and Oslo did it in one leap - which was more complex and contained a small error which had no effect. The formulas in the two answers looked quite different but when plotted gave the same lines. Eventually I proved that the formulas were in fact the same.<br /><br />As usual the actual geodesic equations are not used to find out about all these geodesics. Nevertheless I will have a try to see if I can do better with that beacon in section 5.6.<br /><br />Read it all at <a href="https://drive.google.com/open?id=1vRdg4txwxjb1uIQJ3RHwDCiZGvdu6thK" target="_blank">Ex 5.5 Observer and beacon.pdf</a> (11 pages not including other documents)</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-58285579627574220362020-05-13T15:20:00.000+02:002020-05-13T15:20:21.838+02:00Reciprocals of prime numbers<div dir="ltr" style="text-align: left;" trbidi="on">On a program about Gauss (<a href="https://www.bbc.co.uk/sounds/play/b09gbnfj" target="_blank">BBC In Our Time</a>), near 48:00, they said that the reciprocal of a prime number can be a recurring decimal and the maximum length of the recurring part is one less than the number. My curiosity was aroused.<br /><br />Examples are<br />1/7=0.142857142857142857142857142857142857...<br />1/59=0.01694915254237288135593220338983050847457627118644067796610169491525423728813559322033898305084745762711864406779661...<br />the latter was found by a VBA program which could theoretically go to 2 billion digits because that is the maximum size of an array and length of a string. However it would run out of time or memory well before it got there. Calculation the reciprocals of the numbers from 80,000-89,999 and putting them into an Excel spreadsheet found 316 reciprocals which had recurring digits one less than the number you first thought of in 7 hours 50 minutes. 89,989 is prime 1/89,989 has 89,988 recurring digits and the VBA program took 44 seconds just to calculate that. A different programming language would be much faster.<br /><br />I now have a list (in several spreadsheets) of all the reciprocals from 2 to 999,999. It is fascinating.<br /><br />A 'number whose reciprocal has recurring digits one less than the number' is a bit of a mouthful, so I abbreviate it to an ##R_-## ('R minus') number. It is also useful to define a function ##R\left(n\right)## which gives the number of Recurring digits in ##n##. The VBA program calculates ##R\left(n\right)##. For ##R_-## numbers, ##R\left(n\right)=n-1## . A related function is ##U\left(n\right)## which is the number of non-recurring digits, or Unique digits, in the reciprocal. So for example we have<br />##R\left(7\right)=6,\ U\left(7\right)=0## and<br />##\frac{1}{22}=0.0454545\ldots\ \ ,\ \ R\left(22\right)=2,\ U\left(22\right)=1##<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-MGObNo02uqc/Xrvx-vqiqII/AAAAAAABHQw/z0-rDwkNvnIyGevTmaAX10OwFjv56ZC3wCLcBGAsYHQ/s1600/junk.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="802" data-original-width="861" height="372" src="https://1.bp.blogspot.com/-MGObNo02uqc/Xrvx-vqiqII/AAAAAAABHQw/z0-rDwkNvnIyGevTmaAX10OwFjv56ZC3wCLcBGAsYHQ/s400/junk.png" width="400" /></a></div><br />The scatter chart contains a point at ##n,R\left(n\right)## for every number from 2 to 9,999.<br /><br />Given positive integers ##m,n,p## greater than 0 or 1, I have proved that<br />1) ##R\left(n\right)<n## for all ##n##. This is shown by the graph. There are no data points above the main diagonal line which has gradient ##~1##.<br /><br />2) ##R\left(n\right)=n-1## only for prime numbers. Or only prime numbers are ##R_-## numbers.<br /><br />3) ##R_-## numbers occur about 2.6 times less often than prime numbers as shown by the density of the main diagonal line.<br /><br />4) For multiples of ##R_-## numbers ##n## , ##R\left(m\times n\right)## is likely to be ##R\left(n\right)##. These are shown by the less dense diagonal straight lines which have gradients 1/2, 1/3, 1/5 ....<br /><br />5) ##U\left(p\times2^m\times5^n\right)=m\ or\ n## whichever is larger of ##m,n##. (We can have ##m,n=0## in this case and ##p## must not have 2 or 5 as a factor.)<br /><br />6) A consequence of 5 is that two fifths of consecutive numbers have ##Q\left(n\right)=0##.<br /><br />7) Another consequence of 5 is that all prime numbers, except 2 and 5, and therefore all ##R_-## numbers have no non-recurring digits.<br /><br />The first two have mathematical proofs, the rest are really conjectures based on experiments with lots of numbers. The mathematical proofs are algorithmic: One thinks of how to do the calculation and that proves the result.<br /><br />Here's why <a href="https://drive.google.com/open?id=1uC8aFt1rAzXzXynENRooorNknpfOaooC" target="_blank">Reciprocals of prime numbers.pdf</a> (11 pages)<br /><h2 style="text-align: left;">On Gauss and me</h2><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-DXKbHYyPuBo/Xrvzr7EGKOI/AAAAAAABHQ8/Dp_cEykTJJQ4YfHpYi7J4trNM4gjiNMPgCLcBGAsYHQ/s1600/junk.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" data-original-height="561" data-original-width="440" height="320" src="https://1.bp.blogspot.com/-DXKbHYyPuBo/Xrvzr7EGKOI/AAAAAAABHQ8/Dp_cEykTJJQ4YfHpYi7J4trNM4gjiNMPgCLcBGAsYHQ/s320/junk.jpg" width="250" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><a href="https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss" target="_blank">Carl Friedrich Gauss</a></td></tr></tbody></table>On the program they mention the famous story about Gauss when he was age 10 at elementary school adding up all the numbers from 1 to 100. His class was asked to do this by the teacher who, no doubt, thought she could have some quiet time while the class was busy. Gauss added the numbers up in a minute, foiling the teacher's plan. Apparently Gauss used to love telling this story.<br /><br />When I was about 10 or 11 and day dreaming at school assembly, I worked out a formula for the sum of the first ##n## numbers. It is of course ##\Sigma=n(n+1)/2##. I thought I had made an fantastic mathematical and nervously went to tell Mr Fillingham, head maths teacher. He gave me a strange look and showed me the general formulas for arithmetic and geometric progressions. I was crestfallen and never repeated the story.<br /><br />Gauss was given a book of logarithm tables when he was 15 and it had a list of prime numbers in the back. He said that there was poetry in the tables.<br /><br />He spent so much time day dreaming about mathematics that he memorised his times tables up to very large numbers. He would have loved computers. They give you the power to play with very large sets of numbers.</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-69984282766428901702020-04-22T14:56:00.000+02:002020-04-22T14:59:17.138+02:00Exercise 5.3 Inside the event horizon<div dir="ltr" style="text-align: left;" trbidi="on"><div style="text-align: left;"><b>Question</b></div>Consider a particle (not necessarily on a geodesic) that has fallen inside the event horizon, ##r<2GM##. Use the ordinary Schwarzschild coordinates ##\left\{t,r,\theta,\phi\right\}##.<br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><span style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" data-original-height="600" data-original-width="600" height="320" src="https://1.bp.blogspot.com/-aiajGQern6g/XqA7olpzVhI/AAAAAAABHHU/8QYHcosBGVQdCzPeoma_9uMckM-_Z92qQCLcBGAsYHQ/s320/junk.png" width="320" /></span></td></tr><tr><td class="tr-caption" style="text-align: center;"><a href="https://en.wikipedia.org/wiki/Observable_universe" target="_blank">Observable Universe</a></td></tr></tbody></table><br />a) Show that the radial coordinate must decrease at a minimum rate given by $$<br />\left|\frac{dr}{d\tau}\right|\geq\sqrt{\frac{2GM}{r}-1}<br />$$b) Calculate the maximum lifetime for a particle along a trajectory from ##r=2GM## to ##r=0##.<br />c) Express this in seconds for a black hole with mass measured in solar masses.<br />d) Show that this maximum proper time is achieved by falling freely with ##E\rightarrow0##.<br /><div style="text-align: left;"><b>Answers</b></div>The most interesting exercise so far, especially parts b and c. If the sun were a black hole then the maximum lifetime for our particle would be $$<br />{\Delta\tau}_\bigodot=1.55\times\ {10}^{-5}\ \rm{s}<br />$$That's a pretty short time, but if the Sun were a black hole it would have ##2GM=3\ \text{km}## so our test particle would have an average speed of about ##2\ \times\ {10}^8\ \text{m s}^{-1}## which is just below the speed of light and a hundred times faster than the Parker Solar Probe launched in 2018 which should only reach 0.064% the speed of light.<br /><br />However big the black hole is, the average minimum speed for the fall for the centre is constant at$$<br />v_\rm{AvMin}=\frac{2c}{\pi}<br />$$M87* the black hole at the centre of our galaxy is about ##6.5\times\ {10}^9## solar masses so we get$$<br />{\Delta\tau}_\rm{M87\ast}=1.55\times\ {10}^{-5}\times6.5\times\ {10}^9={10}^5\ s=28\ \rm{hours}<br />$$There is a short time to prepare in M87*.<br /><br />We can do the same for a black hole with the mass of the Universe: The observable Universe contains ordinary matter equivalent to ##{10}^{23}## solar masses. So$$<br />{\Delta\tau}_{Universe}=1.55\times\ {10}^{-5}\times{10}^{23}\approx{10}^{18}s=300\ \text{billion years}<br />$$The estimated 'age' of the Universe is 14 billion years, so there is plenty of time inside the big black hole - we have hardly started the journey.<br /><br />See proof and calculations at <a href="https://drive.google.com/open?id=1OOa2kMhiAzCu3CfHJQYCRjSblXQwBgYb" target="_blank">Ex 5.3 Inside the event horizon.pdf</a> (4 pages). Also contains speculations on what happens to a photon and links to other answers.<br />My document on <a href="https://drive.google.com/open?id=1AQJtesVCZMp8C2nRhhABUWE7SGrYmSki" target="_blank">Constants and conversion factors</a> also came in very handy.</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-55762028436572212162020-04-16T16:32:00.000+02:002020-04-19T17:15:18.507+02:00Eddington and Finkelstein take us into a black hole<div dir="ltr" style="text-align: left;" trbidi="on"><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-POMzDwYcwVE/Xpxq8jy6IKI/AAAAAAABHGs/XTe0-B0ZT8YV3yLzg1vz1OEMnsShFC0VwCLcBGAsYHQ/s1600/light%2Bcone.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="195" data-original-width="454" src="https://1.bp.blogspot.com/-POMzDwYcwVE/Xpxq8jy6IKI/AAAAAAABHGs/XTe0-B0ZT8YV3yLzg1vz1OEMnsShFC0VwCLcBGAsYHQ/s1600/light%2Bcone.gif" /></a></div><br /><div style="text-align: center;"></div><div class="separator" style="clear: both; text-align: left;">In the first part of section 5.2 (<a href="https://www.general-relativity.net/2020/04/schwarzschild-black-holes.html" target="_blank">two posts ago</a>) it seemed to be impossible to get inside the Schwarzschild radius. In the second part we look at other coordinate systems and find the Eddington-Finkelstein coordinates which show us how. The metric then takes a different form (which says something about Birkhoff's theorem) and it does not have an infinity at the Schwarzschild radius (##r=2GM##). </div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">The properties of the function ##1-2GM/r## (which causes the trouble) frequently amaze. It keeps eating itself up which is very satisfactory.</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">Eddington-Finkelstein coordinates eliminate the time coordinate ##t## and introduce</div><div class="separator" style="clear: both;">$$v=t+r+2GM\ln{\left(\frac{r}{2GM}-1\right)}$$</div><div class="separator" style="clear: both;">so the relationship between ##v## and ##t## is complicated.</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">The image shows a light cone at various distances out from the centre at ##r=0## . The ingoing light beam always heads for the centre the outgoing beam can get away when ##r>2GM## but flips towards the centre once it originates at a distance less than the Schwarzschild radius (aka the event horizon). The closer the starting point is to the centre, the less room there is for manoeuvre. </div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">I am slightly dubious about the direction of the ingoing side of a light cone inside the Schwarzschild radius.</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">See why and check my maths: <a href="https://drive.google.com/open?id=1nqZAM73DeMazWCVyyXw2Kq9Nri1bsurO" target="_blank">Commentary 5.6#2 Schwarzschild Black Holes.pdf</a> (6 pages)</div></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-26281006776544135932020-04-11T17:24:00.002+02:002020-06-25T17:37:23.272+02:00Geodesics of Schwarzschild In section 5.4 Carroll explores the geodesics of Schwarzschild. These turn out to be almost useless as far as I can see. What are very useful are the Killing vector fields in Schwarzschild. In particular the energy Killing vector field which will eventually enable us to find an equation fro the path of a radially free-falling test particle. I learnt why Killing vectors are so important!<div><br /></div><div>In this section we also write out the useless geodesic equations and work out some potentials. Most of it is pretty dry stuff. So dry that I forgot to post it until I actually used the equations with some success in late June.</div><div><br /></div><div>All at <a href="https://drive.google.com/open?id=1o-EKrykvUFOgacMDT7CD3Of4Zqpj7HHL" target="_blank">Commentary 5.4 Geodesics of Schwarzschild.pdf</a> </div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4005750306801645234.post-20562619235778089152020-04-10T14:54:00.003+02:002020-04-10T14:54:46.939+02:002 years on<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-EBGAl2bXgJI/XpBp3EwuvMI/AAAAAAABHCg/HtsAzDik750UMSRpc37xfEWtq7qewsQ2wCK4BGAsYHg/IMG_20200410_140104.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="518" data-original-width="625" height="331" src="https://1.bp.blogspot.com/-EBGAl2bXgJI/XpBp3EwuvMI/AAAAAAABHCg/HtsAzDik750UMSRpc37xfEWtq7qewsQ2wCK4BGAsYHg/w400-h331/IMG_20200410_140104.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: left;">Two years on and I am about half way through Spacetime and Geometry : An Introduction to General Relativity – by Sean M Carroll. It's probably the best value for money book I have read (er... studied) in my life. When I started I could barely remember how to differentiate. Now I can use the chain rule almost without thinking and the tensor things which I had never met before are a doddle. I am now on the Eddington-Finkelstein metric and discovering how to get into a black hole (which seems to be impossible in the obvious ##t,r,\theta,\phi## coordinates) and why you never get out. I can't praise the book and Sean Carroll highly enough!</div>Unknownnoreply@blogger.com0