What happens 'inside' the event horizon of a black hole, in particular to photon speed - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-21T21:11:10Z https://physics.stackexchange.com/feeds/question/222712 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/222712 1 What happens 'inside' the event horizon of a black hole, in particular to photon speed Harald https://physics.stackexchange.com/users/73067 2015-12-07T12:24:46Z 2015-12-08T07:52:26Z <p>This is a follow-up question to two answers given <a href="https://physics.stackexchange.com/a/77280/73067">here</a> and <a href="https://physics.stackexchange.com/a/24329/73067">here</a>, where the speed in different coordinate systems is discussed when approaching the Schwarzschild radius. To cite, derived directly from the Schwarzschild metric, we get</p> <p>$$\frac{dr}{dt} = c\left(1-\frac{r_s}{r}\right) = c\frac{r-r_s}{r}$$</p> <p>for the speed of a photon as measured by an external observer. The clock ticking the $dt$ is at infinity, the $dr$ is the Schwarzschild metric's radial coordinate. In the <a href="https://physics.stackexchange.com/a/77280/73067">derivation provided</a>, a step of taking the square root is (suspiciouly?) left out that would also allow</p> <p>$$\frac{dr}{dt} = c\frac{r_s-r}{r}$$</p> <p>to be a solution. Now I do have two questions:</p> <ol> <li>Does the reversed sign solution describe the outgoing photon instead of the incoming, or the other way around, depending on the coordinate directional convention?</li> <li>Is the solution valid for $r&lt;r_s$, inside Schwarzschild radius, and if so, how it can be interpreted that the photon's speed seems to pick up again?</li> </ol> https://physics.stackexchange.com/questions/222712/-/222714#222714 1 Answer by John Rennie for What happens 'inside' the event horizon of a black hole, in particular to photon speed John Rennie https://physics.stackexchange.com/users/1325 2015-12-07T12:43:10Z 2015-12-07T12:43:10Z <p>Yes, the expression for the speed of light should be:</p> <p>$$\frac{dr}{dt} = \pm c \left(1-\frac{r_s}{r}\right)$$</p> <p>where the positive root describes outgoing light (moving in the direction of increasing $r$) and the negative root describes ingoing light (moving in the direction of decreasing $r$).</p> <p>To answer your second question, inside the horizon the Schwarzschild coordinates do not describe anything that is physically relevant to an external observer. For an external observer anything falling into the black hole takes an infinite time to even reach the horizon let alone fall through it. You can feed in values of $r \lt r_s$ into the equation if you wish, but there is no physical significance to the resulting value for $dr/dt$.</p> <p>There are coordinate systems that allow you to attach a <em>speed</em> to light as it crosses the event horizon and procedes inwards, but these coordinates do not correspond to anything any observer could see. If you're interested I go through this in some detail in my answer to <a href="https://physics.stackexchange.com/questions/67682/how-does-light-behave-within-a-black-holes-event-horizon">How does light behave within a black hole&#39;s event horizon?</a>.</p> https://physics.stackexchange.com/questions/222712/-/222723#222723 -2 Answer by John Duffield for What happens 'inside' the event horizon of a black hole, in particular to photon speed John Duffield https://physics.stackexchange.com/users/76162 2015-12-07T14:09:38Z 2015-12-07T14:09:38Z <blockquote> <p>This is a follow-up question to two answers given <a href="https://physics.stackexchange.com/a/77280/73067">here</a> and <a href="https://physics.stackexchange.com/a/24329/73067">here</a>, where the speed in different coordinate systems is discussed when approaching the Schwarzschild radius. To cite, derived directly from the Schwarzschild metric, we get $\frac{dr}{dt} = c\left(1-\frac{r_s}{r}\right) = c\frac{r-r_s}{r}$</p> </blockquote> <p>And when r = r<sub>s</sub>, $\frac{dr}{dt}$ equals zero, as per this graph on JR's previous answer:</p> <p><a href="https://i.stack.imgur.com/XlKh0.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XlKh0.gif" alt="enter image description here"></a> </p> <blockquote> <p>In the <a href="https://physics.stackexchange.com/a/77280/73067">derivation provided</a>, a step of taking the square root is (suspiciously?) left out that would also allow $\frac{dr}{dt} = c\frac{r_s-r}{r}$ to be a solution. </p> </blockquote> <p>It makes no difference. When r<sub>s</sub> = r, $\frac{dr}{dt}$ still equals zero. </p> <blockquote> <p>Does the reversed sign solution describe the outgoing photon instead of the incoming, or the other way around, depending on the coordinate directional convention?</p> </blockquote> <p>Neither. It says that according to the external observer, the speed of light at the event horizon is zero. </p> <blockquote> <p>Is the solution valid for $r&lt;r_s$, inside Schwarzschild radius? </p> </blockquote> <p>No it isn't, because the speed of light can't be less than zero. Speed is a scalar, light can't go slower than stopped. In similar vein there are no rulers less than zero inches long. </p> <blockquote> <p>if so, how it can be interpreted that the photon's speed seems to pick up again?</p> </blockquote> <p>It can't. See the history section of the Wikipedia <a href="https://en.wikipedia.org/wiki/Schwarzschild_metric#History" rel="nofollow noreferrer">Schwarzschild metric</a> article, and note this:</p> <p><em>"In 1939 Howard Robertson showed that a free falling observer descending in the Schwarzschild metric would cross the r = r<sub>s</sub> singularity in a finite amount of proper time even though this would take an infinite amount of time in terms of coordinate time t."</em></p> <p>That infinite coordinate time means <em>it hasn't happened yet, and it never ever will</em>. This is what <a href="http://www.co-bw.com/NewsTheelephantan%20theeventhorizon.htm" rel="nofollow noreferrer">The Elephant and the Event Horizon</a> is all about. Look at the Schwarzschild-coordinate picture from MTW. Imagine the light-cone is the elephant:</p> <p><img src="https://i.stack.imgur.com/dk2m6.jpg" alt="enter image description here"></p> <p>Note how the picture is truncated vertically? What that "suspiciously leaves out" is that at r=r<sub>s</sub> the elephant goes to the end of time and back, and is in two places at once. Draw a horizontal line at about t/M=45 and trace across from right to left. The elephant is there at τ = 33.3M, and it's there again at τ = 34.2M. IMHO this is nonsense, along with the idea that you can switch to say KS coordinates to get past this. When light stops an optical clock stops, and the stopped observer doesn't see it ticking normally "in his frame". Light is stopped, so he sees nothing, ever. See <a href="http://mathpages.com/rr/s7-02/7-02.htm" rel="nofollow noreferrer">The Formation and Growth of Black Holes</a> where Kevin Brown refers to the frozen-star interpretation. Most people don't know about it, but IMHO it's the only interpretation that makes sense and fits in with <a href="http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22" rel="nofollow noreferrer">what Einstein said</a>. </p> https://physics.stackexchange.com/questions/222712/-/222770#222770 1 Answer by Timaeus for What happens 'inside' the event horizon of a black hole, in particular to photon speed Timaeus https://physics.stackexchange.com/users/57780 2015-12-07T17:41:02Z 2015-12-07T17:41:02Z <ol> <li><p>Outside the event horizon, the opposite sign does describe decreasing $r$ as opposed to increasing $r.$ However, $dr/dt$ is not a speed, it is a ratio of two coordinates. And you shouldn't call $r$ a radius, it isn't a radial distance and $r-r_s$ is <strong>not</strong> the radial distance from the event horizon. The coordinate $r$ is the <strong>area</strong>l coordinate (the circumference divided by $2\pi$, or $\sqrt{S/4\pi}$ where $S$ is the surface area) which in a curved space is different than the radius. <em>The actual physical speed is $c$.</em></p></li> <li><p>Outside the event horizon, the "solution" wasn't a speed (it was just a limit of a ratio of a spacelike coordinate difference over a timelike coordinate difference) and inside the event horizon the coordinates $r$ and $t$ are no longer spacelike and timelike coordinates respectively. So it becomes worse than not a speed it becomes a ratio of a timelike coordinate over a spacelike coordinate. <em>And when I say becomes, I mean the person calculating the math faces a new challenge when they switch from doing math for the coordinate patch outside to doing math for the coordinate patch inside (and the Schwarzschild coordinates don't cover the horizon so you have to just do math for the outside or just do math for the inside as if they were two different universes, unless you are willing to use other coordinate systems). I don't mean it in any causal way.</em> In particular the Schwarzschild coordinates don't cover the event horizon, so they don't describe a light crossing from the outside to the inside, at best you can describe the portions of spacetime where it is outside (with $r\gt r_s$) and the portions where it is inside (with $r\lt r_s$) and those coordinates never <em>ever</em> <strong>ever</strong> describe a crossing or any connection between the two regions at all. It would be like if you had a universe that consisted of the outside of a ball and another universe that consisted of the inside of a ball, if you don't have the surface of the ball then you just have two universes. The Schwarzschild coordinates do not describe a connected universe with an inside and an outside. If you wanted to use Schwarzschild coordinates for the inside you can, and the physical speed of light is still $c$ and you can still talk about limits of ratios of coordinate differences and they still won't be speeds, and now decreasing $r$ is a timelike direction in the same way increasing $t$ was a timelike direction outside.</p></li> </ol>