JVNY Posted November 16, 2013 Share Posted November 16, 2013 Can anyone help explain whether gravity slows light going directly up or down a gravity well over an extended (not infinitesimally small) path, if possible using the two uploaded Minkowski diagrams of linear acceleration to help a student of Special Relativity understand the answer? For figure 1, consider two ships that begin at rest in the lab frame: Rear at x=0.5 and Front at x=1.0. Simultaneously, each flashes a light toward the other and begins to accelerate at a constant proper rate (Rear at a=2, and Front at a=1). These assumptions ensure that the ships maintain their proper distance in their own frame and that lines of simultaneity for the ships in their own frame go through the origin in the lab frame. The first observation is that the flashes arrive at the two ships simultaneously in their own frame. Analogizing to gravity, the speed of light between two directly vertical points is the same regardless of the direction of travel. This seems consistent with gravity not being a force; it does not retard light moving out of the well, nor advance light moving into the well. The second observation is that the light flashes cross to the left of the center point between the ships. Analogizing to gravity, light seems to travel vertically slower when lower in the well (regardless of travel direction), and faster when higher in the well (regardless of travel direction). This seems consistent with gravity being curvature, if there is an invisible dimension in which there is more curvature toward the rear in acceleration, and toward the bottom in gravity. To test this, see figure 2, adding a third ship (Center) that starts at x=0.75. Simultaneously, Center flashes light forward and backward, and the ships begin to accelerate in order to maintain proper distance in their own frame (Rear at a=2, Center at a=1.33, and Front at a=1). When the flashes hit the other ships, they reflect and return to Center. The flashes strike in the following order: (1) the forward flash from Center strikes Front; (2) the rearward flash from Center strikes Rear; (3) the forward flash returns to Center; and (4) the rearward flash returns to Center. Analogizing to gravity, this seems to confirm that light travels slower deeper in the well, and faster higher in the well. The above seems consistent with sources saying that light travels at different speeds depending upon where it is in gravity. But other sources say that complete General Relativity shows that light always travels at the same speed regardless of gravity. These sources refer to the effects of gravity on clocks and measuring rods. The analysis above does not depend on clock rates; it uses only the order of arrival of simultaneous flashes to determine whether a flash is faster, slower, or the same speed as another flash. So, does this mean that General Relativity says that measuring rods at the rear shrink, such that a flash is actually traveling a greater distance when it is toward the rear (or deeper in the well)? Even though the acceleration rates were set such that the ships maintain their proper distance (as all of the discussions of Born rigid motion say can be done)? Has there been an experiment like the second diagram above, sending simultaneous signals up and down, then reflecting and returning them to the same source? One author suggests such an experiment, although he appears to conclude that the downward flash will return to the source first (rather than the upward flash). See Petkov, "Probing the anisotropic velocity of light in a gravitational field: another test of general relativity," http://arxiv.org/abs/gr-qc/9912014. Thanks. Diagrams.pdf Link to comment Share on other sites More sharing options...

Ankit Gupta Posted November 16, 2013 Share Posted November 16, 2013 i dont think so that gravity will affect photon as it has no mass and if mass is zero then gravitational force also becomes 0 (as per my information) but as we can see it can not escape from black hole then i think it is due to gravity (i an not sure about this). Link to comment Share on other sites More sharing options...

xyzt Posted November 17, 2013 Share Posted November 17, 2013 (edited) Can anyone help explain whether gravity slows light going directly up or down a gravity well over an extended (not infinitesimally small) path, if possible using the two uploaded Minkowski diagrams of linear acceleration to help a student of Special Relativity understand the answer? For figure 1, consider two ships that begin at rest in the lab frame: Rear at x=0.5 and Front at x=1.0. Simultaneously, each flashes a light toward the other and begins to accelerate at a constant proper rate (Rear at a=2, and Front at a=1). These assumptions ensure that the ships maintain their proper distance in their own frame and that lines of simultaneity for the ships in their own frame go through the origin in the lab frame. The first observation is that the flashes arrive at the two ships simultaneously in their own frame. Analogizing to gravity, the speed of light between two directly vertical points is the same regardless of the direction of travel. This seems consistent with gravity not being a force; it does not retard light moving out of the well, nor advance light moving into the well. The second observation is that the light flashes cross to the left of the center point between the ships. Analogizing to gravity, light seems to travel vertically slower when lower in the well (regardless of travel direction), and faster when higher in the well (regardless of travel direction). This seems consistent with gravity being curvature, if there is an invisible dimension in which there is more curvature toward the rear in acceleration, and toward the bottom in gravity. To test this, see figure 2, adding a third ship (Center) that starts at x=0.75. Simultaneously, Center flashes light forward and backward, and the ships begin to accelerate in order to maintain proper distance in their own frame (Rear at a=2, Center at a=1.33, and Front at a=1). When the flashes hit the other ships, they reflect and return to Center. The flashes strike in the following order: (1) the forward flash from Center strikes Front; (2) the rearward flash from Center strikes Rear; (3) the forward flash returns to Center; and (4) the rearward flash returns to Center. Analogizing to gravity, this seems to confirm that light travels slower deeper in the well, and faster higher in the well. The above seems consistent with sources saying that light travels at different speeds depending upon where it is in gravity. But other sources say that complete General Relativity shows that light always travels at the same speed regardless of gravity. These sources refer to the effects of gravity on clocks and measuring rods. The analysis above does not depend on clock rates; it uses only the order of arrival of simultaneous flashes to determine whether a flash is faster, slower, or the same speed as another flash. So, does this mean that General Relativity says that measuring rods at the rear shrink, such that a flash is actually traveling a greater distance when it is toward the rear (or deeper in the well)? Even though the acceleration rates were set such that the ships maintain their proper distance (as all of the discussions of Born rigid motion say can be done)? Has there been an experiment like the second diagram above, sending simultaneous signals up and down, then reflecting and returning them to the same source? One author suggests such an experiment, although he appears to conclude that the downward flash will return to the source first (rather than the upward flash). See Petkov, "Probing the anisotropic velocity of light in a gravitational field: another test of general relativity," http://arxiv.org/abs/gr-qc/9912014. Thanks. This is a complicated question, the answer has two parts: 1. Vesselin Petkov is a crank, this is known in the physics community. 2. Your question is answered trivially in most textbooks: 2.1 LOCALLY, the speed of light in a gravitational field is CONSTANT, if it weren't the whole GR theory would fall apart. 2.2. Nevertheless, the COORDINATE speed of light is variable. Indeed, start with the reduced Schwarzschild metric: [math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)[/math] Light follows null geodesics, therefore: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)[/math] giving: [math]\frac{dr}{dt}=c(1-r_s/r)[/math] Th LHS represents the COORDINATE speed of light, i.e., the speed of light measured by a distant observer. As one can easily see, it approaches 0 when the light approaches the event horizon , i.e. [math]r \to \infty[/math]. 2.3. When starting from the complete metric, things get even more complicated, resulting into what is taught as "photon orbits" in GR classes. For example: [math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] Since light follows null geodesics: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] 2.3,1. If there is no radial motion (i.e [math]dr=0[/math]) then we obtain the so-called spherical orbits": [math]0=(1-r_s/r)(cdt)^2-r^2 (d\theta)^2[/math] The tangential speed is: [math]r \omega=\pm c \sqrt{1-r_s/r}[/math] 2.3.2 If there is both radial and tangential motion, then: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] so: [math]\frac{dr}{dt}= \sqrt {(1-r_s/r)^2 c^2-(1-r_s/r) (r \omega )^2}[/math] Edited November 17, 2013 by xyzt 3 Link to comment Share on other sites More sharing options...

JVNY Posted November 18, 2013 Author Share Posted November 18, 2013 Thanks. XYZT, is there any way that you can explain why this slowing down occurs in a way that will resonate with someone who is only so far studying special relativity? The assumption is that the light moves directly down the gravity well, reflects off of something there, and then returns directly up the gravity well (I suspect that this is what you call "radial" in both directions). Various sources say that it is a result of gravity contracting rulers and dilating clocks near the mass; or curving of space, or curving of space and time. Yet there should be no curving of the light's path like the famous experiments viewing stars during a solar eclipse, because the path is straight down then straight back up; and still the slowing occurs for the distant observer. I am hoping for an Einstein explanation to a five year old Link to comment Share on other sites More sharing options...

xyzt Posted November 18, 2013 Share Posted November 18, 2013 (edited) Yet there should be no curving of the light's path like the famous experiments viewing stars during a solar eclipse, because the path is straight down then straight back up; and still the slowing occurs for the distant observer. I am hoping for an Einstein explanation to a five year old Correct, there is no curving in the case of radial trajectory. Let's look at the formula I derived: [math]\frac{dr}{dt}=c(1-r_s/r)[/math] For the observer at infinity ([math]r=\infty[/math]) [math]\frac{dr}{dt}=c[/math], i.e. such an observer measures light speed to be ...exactly [math]c[/math]. For large values of [math]r[/math], the observer at infinity still measures light pulses to arrive to him/her at close the speed of light. As the light source approaches the event horizon (i.e. [math]r \to r_s[/math]), the light pulses appear to slow more and more, as [math]r[/math] decreases. When the light source reaches the EH, light never appears to arrive to the distant observer. It is as if the BH exerts and ever increasing "attraction" on the photons. Indeed, photons emitted from behind the EH never "escape", they can never be seen by the distant observer. Does this help? Edited November 18, 2013 by xyzt Link to comment Share on other sites More sharing options...

JVNY Posted November 19, 2013 Author Share Posted November 19, 2013 It does mathematically, but I was hoping for something conceptual (even visual). For example, it seems obvious that a light flash will have lower velocity traveling from 12 o'clock on a clock face to 6 o'clock if it gets bent around by gravity and has to travel around the rim of the face. It does not travel at c measured as net distance traveled divided by time, because it travels a greater total distance (the distance along the curve). But how does one conceptualize the light flash going slower than c when it travels radially directly from 12 o'clock to the center of the face, then back again? Or, say, from 12 o'clock straight along the diameter to 6 o'clock? One way to think of it is to turn the clock face 90 degrees so that you look directly at the rim, then position yourself to look directly at 3 o'clock, then observe the end of the second hand as it travels from 12 o'clock to 6 o'clock. You will effectively be looking at the second hand traveling in only one dimension (a rightward line from 12 o'clock on your left to 6 o'clock on your right -- assume that your vision is not good enough to see the depth involved). In that case, the second hand will travel to your right more slowly the closer it is to 12 and 6 o'clock (because it will be traveling mostly toward or away from you, even though you cannot really see that). It will travel faster around 3 o'clock (because there is very little curvature there, so it is mostly traveling to your right, with very little travel toward or away from you). Is that the sort of thing that happens in spacetime with a radial light path? That seems to put everything in space terms, not in time terms, so it is probably not right as a description of spacetime. Elsewhere (in discussions of Shapiro delay), authors say that the effect in the direction toward or away from the gravitational mass is the result of gravitational time dilation, which suggests that the second hand analogy is wrong. Link to comment Share on other sites More sharing options...

xyzt Posted November 19, 2013 Share Posted November 19, 2013 It does mathematically, but I was hoping for something conceptual (even visual). For example, it seems obvious that a light flash will have lower velocity traveling from 12 o'clock on a clock face to 6 o'clock if it gets bent around by gravity and has to travel around the rim of the face. It does not travel at c measured as net distance traveled divided by time, because it travels a greater total distance (the distance along the curve). But how does one conceptualize the light flash going slower than c when it travels radially directly from 12 o'clock to the center of the face, then back again? Or, say, from 12 o'clock straight along the diameter to 6 o'clock? One way to think of it is to turn the clock face 90 degrees so that you look directly at the rim, then position yourself to look directly at 3 o'clock, then observe the end of the second hand as it travels from 12 o'clock to 6 o'clock. You will effectively be looking at the second hand traveling in only one dimension (a rightward line from 12 o'clock on your left to 6 o'clock on your right -- assume that your vision is not good enough to see the depth involved). In that case, the second hand will travel to your right more slowly the closer it is to 12 and 6 o'clock (because it will be traveling mostly toward or away from you, even though you cannot really see that). It will travel faster around 3 o'clock (because there is very little curvature there, so it is mostly traveling to your right, with very little travel toward or away from you). Is that the sort of thing that happens in spacetime with a radial light path? That seems to put everything in space terms, not in time terms, so it is probably not right as a description of spacetime. Elsewhere (in discussions of Shapiro delay), authors say that the effect in the direction toward or away from the gravitational mass is the result of gravitational time dilation, which suggests that the second hand analogy is wrong. Both analogies are wrong. The correct analogy is done in terms of energy not time-distance. The closer the light source is to the EH , the lower the energy of the emitted photons, so they "climb" the gravity well towards the distant observer "slower" (and slower). Link to comment Share on other sites More sharing options...

JVNY Posted November 19, 2013 Author Share Posted November 19, 2013 That works, thanks. Link to comment Share on other sites More sharing options...

xyzt Posted November 19, 2013 Share Posted November 19, 2013 That works, thanks. Check out the Pound-Rebka experiment. Link to comment Share on other sites More sharing options...

JVNY Posted November 19, 2013 Author Share Posted November 19, 2013 On second thought, it works well for light going up the gravity well. But it does not explain why light also goes more slowly when it travels down the gravity well. The Minkowski diagrams attached to the first posting show that light also travels more slowly going rearward (down the well), not just forward (up the well). What explains this (expressed conceptually or visually)? In the first diagram, for example, it takes the same time for a flash to go rearward as forward (by the fact that the two flashes start simultaneously in the ships' frame from each end, and strike the opposite ends simultaneously in the ships' frame). In the second diagram, the flash from the center to the rear takes the same amount of center ship proper time as the flash takes to return from the rear to the center. I did not include these proper times in the diagram, but you can run the numbers and confirm that the rearward flash hits the rear ship when the center ship's clock reads T=0.3041. After reflecting off the rear ship, the flash strikes the center ship when the center's clock reads T=0.6082. It seems that travel in either radial direction is slower. This seems to be borne out by the Shapiro experiments -- when you break the signal's trip down into component segments, there is a segment toward the sun and a segment away from the sun, and light travels slower on both of these legs. Following the approach you suggest, the photos should gain energy on the first leg (going toward the sun), which would counteract the loss of energy going away from the sun, leaving the only net effect being the non-radial travel as the signal curves around the sun. Link to comment Share on other sites More sharing options...

xyzt Posted November 19, 2013 Share Posted November 19, 2013 (edited) On second thought, it works well for light going up the gravity well. But it does not explain why light also goes more slowly when it travels down the gravity well. It doesn't, it goes faster. You need to replace the source going towards the EH with a source going away from the EH. You need to replace the observer at infinity with the one hovering just above the EH. In effect, you need to swap the positions of the observer and the light source and redo the math as I will do for you in the next section. The Minkowski diagrams attached to the first posting show that light also travels more slowly going rearward (down the well), not just forward (up the well). What explains this (expressed conceptually or visually)? Let's fix the ideas, using the Equivalence Principle we replace the gravitational body with an accelerated rocket, flying in an area devoid of any gravitational field. Put the light source on the "floor" and shine it towards the "ceiling". Light appears to traverse the length of the rocket in time [math]ct_{up}=L+0.5at^2_{up}[/math] because, as the light moves "up", the ceiling "runs away" by the distance [math]0.5at^2_{up}[/math]. So, [math]t_{up}=\frac{c-\sqrt{c^2-2aL}}{a}[/math] Now' let's swap the roles of the source and the observer, i.e. put the source on the "ceiling" and the observer on the "floor". The equation becomes: [math]ct_{down}+0.5at^2_{down}=L[/math] because the floor is accelerating towards the light pulse, so the distance L is covered from both ends, giving: [math]t_{down}=\frac{-c+\sqrt{c^2+2aL}}{a}[/math] An astute observer notices that [math]t_{down}<t_{up}[/math]. It is as if the light moves faster "down" the gravity well. Edited November 19, 2013 by xyzt Link to comment Share on other sites More sharing options...

JVNY Posted November 20, 2013 Author Share Posted November 20, 2013 Yes, that is clearly the case in the lab frame. Say the elevator is at rest in the lab frame. Simultaneously (in both elevator and lab frame) there are two flashes, one up form the floor and one down from the ceiling, and the elevator begins to accelerate upward. The two flashes cross below the middle point of the elevator in the lab frame for the reasons you describe. Since they are at the same place when they cross, simultaneity is not relative there: they also cross below the middle in the elevator frame. However, the results are not symmetrical for the ultimate events of the flashes striking the floor and the ceiling. These two events are separated in space, so simultaneity is relative. In the ground frame, the downward flash strikes the floor first. In the elevator frame the two flashes strike the floor and the ceiling simultaneously. This is from the first Minkowski diagram attached to the first post. I am focused on what happens in the elevator frame, because that is the frame that Einstein uses as the one that is equivalent to being in gravity. Does that make sense? Link to comment Share on other sites More sharing options...

xyzt Posted November 20, 2013 Share Posted November 20, 2013 (edited) Yes, that is clearly the case in the lab frame. Say the elevator is at rest in the lab frame. Simultaneously (in both elevator and lab frame) there are two flashes, one up form the floor and one down from the ceiling, and the elevator begins to accelerate upward. The two flashes cross below the middle point of the elevator in the lab frame for the reasons you describe. Since they are at the same place when they cross, simultaneity is not relative there: they also cross below the middle in the elevator frame. However, the results are not symmetrical for the ultimate events of the flashes striking the floor and the ceiling. These two events are separated in space, so simultaneity is relative. The math that I wrote for you in the previous post can be used to calculate exactly what happens in the lab frame. You can use the equations for the times [math]t_{up}, t_{down}[/math] that I derived for you. In the ground frame, the downward flash strikes the floor first. Yes. In the elevator frame the two flashes strike the floor and the ceiling simultaneously. You don't know that, don't be so sure, the elevator frame is accelerated, you would need to do a transformation from the Earth frame (inertial) into the elevator frame (accelerated) in order to find out the strike times. Do you know the equations for hyperbolic motion? I am focused on what happens in the elevator frame, because that is the frame that Einstein uses as the one that is equivalent to being in gravity. Does that make sense? Be careful, the elevator frame is not an inertial frame (it is accelerated wrt the inertial frame of the Earth). I recommend that you do all calculations in the inertial frame, as I did, it is much easier. Edited November 20, 2013 by xyzt 1 Link to comment Share on other sites More sharing options...

JVNY Posted November 20, 2013 Author Share Posted November 20, 2013 OK, thanks. Link to comment Share on other sites More sharing options...

Iggy Posted November 22, 2013 Share Posted November 22, 2013 On second thought, it works well for light going up the gravity well. But it does not explain why light also goes more slowly when it travels down the gravity well. The Minkowski diagrams attached to the first posting show that light also travels more slowly going rearward (down the well), not just forward (up the well). That is my understanding as well It seems that travel in either radial direction is slower. Indeed -- either direction. You probably know, but if not: your diagrams in your pdf are Rindler coordinates. The speed of light in that case is [math]xg[/math] (where x is the x position and g is acceleration). As c and x are proportional in c = xg, the greater the x (i.e. the greater the distance from the mass) the greater the speed of light. For example, the following are g=1 rindler coordinates (I got it [url=http://en.wikipedia.org/wiki/Rindler_coordinates#Relation_to_Cartesian_chart]here[/here]). The small 45° blue and green lines I added are light rays. The blue are 1:1 (run over rise), or 1c. The green is 1:3, or one third c. The speed of light (and the x coordinate) are zero at the dotted line -- called the rindler horizon. As you see, the speed of light is c=xg, so the smaller the x the smaller c: This seems to be borne out by the Shapiro experiments Exactly. Shapiro delay confirms it. Light passing a mass is slowed on approach and slowed likewise on departure. yep yep Link to comment Share on other sites More sharing options...

xyzt Posted November 22, 2013 Share Posted November 22, 2013 (edited) That is my understanding as well Indeed -- either direction. You probably know, but if not: your diagrams in your pdf are Rindler coordinates. The speed of light in that case is [math]xg[/math] (where x is the x position and g is acceleration). As c and x are proportional in c = xg, the greater the x (i.e. the greater the distance from the mass) the greater the speed of light. Couldn't possibly be correct since [math]xg[/math] is measured in [math]\frac{m^2}{s^2}[/math]. What you wrote is total nonsense. Exactly. Shapiro delay confirms it. Light passing a mass is slowed on approach and slowed likewise on departure. yep yep This is incorrect as well, Shapiro delay is due to the fact that the light path is "lengthened" by the amount [math]r_s ln \frac{X_1 X_2}{R^2}[/math] from the "straight line" [math]X_1+X_2[/math]. See Rindler, "Relativity Special , General and Cosmological" page 237, for example: [math]c \Delta t_{12}=X_1+X_2+r_s ln \frac{X_1 X_2}{R^2}[/math] Contrary to your fringe claim, the speed of light is NOT "slowed down", the trajectory follows a geodesic, so the distance traveled is larger than the Euclidian distance. See here , for a nice illustration of the effect. Edited November 22, 2013 by xyzt Link to comment Share on other sites More sharing options...

Iggy Posted November 22, 2013 Share Posted November 22, 2013 Couldn't possibly be correct since [math]xg[/math] is measured in [math]\frac{m^2}{s^2}[/math] In natural units it is gx. In non-natural units it is gx/c. I'm happy to explain natural units if you are unfamiliar. This is incorrect as well, Shapiro delay is due to the fact that the light path is "lengthened" by the amount [math]r_s ln \frac{X_1 X_2}{R^2}[/math] from the "straight line" [math]X_1+X_2[/math]. See Rindler, "Relativity Special , General and Cosmological" page 237, for example. Sure thing, X. We'll catch up later. I'm going to talk to JVNY now. Link to comment Share on other sites More sharing options...

xyzt Posted November 22, 2013 Share Posted November 22, 2013 (edited) In natural units it is gx. In non-natural units it is gx/c. I'm happy to explain natural units if you are unfamiliar. I am quite familiar, thank you. Trouble is, the errors that you posted have nothing to do with natural units, have to do with the straight nonsense of this caliber: As c and x are proportional in c = xg, the greater the x (i.e. the greater the distance from the mass) the greater the speed of light. Edited November 22, 2013 by xyzt Link to comment Share on other sites More sharing options...

JVNY Posted November 22, 2013 Author Share Posted November 22, 2013 Thanks to all for the answers. I will follow the suggestions to read the various references, because I seem to have gotten in a bit over my head here. Link to comment Share on other sites More sharing options...

Iggy Posted November 23, 2013 Share Posted November 23, 2013 (edited) I am quite familiar, thank you. Trouble is, the errors that you posted have nothing to do with natural units, have to do with the straight nonsense of this caliber: I can't tell if you're honestly confused or if you're trolling. v=gx/c in natural units is v=gx (because in natural units c=1) In both cases v and x are proportional. v=gx is the speed of light in natural units in Rindler coordinates. I'm sorry you didn't know the formula for the speed of light in Rindler coordinates. I'm sorry you didn't know why your dimensional analysis on that formula didn't work -- try it on v=gx/c -- you'll get m/s. I can tell you want to argue. Is there anyone else in the thread that wants to explain this to him in light of his upcoming objections? I'm extremely swamped with work. Edited November 23, 2013 by Iggy Link to comment Share on other sites More sharing options...

xyzt Posted November 23, 2013 Share Posted November 23, 2013 (edited) I can't tell if you're honestly confused or if you're trolling. v=gx/c in natural units is v=gx (because in natural units c=1) In both cases v and x are proportional. v=gx is the speed of light in natural units in Rindler coordinates. I'm sorry you didn't know the formula for the speed of light in Rindler coordinates. I'm sorry you didn't know why your dimensional analysis on that formula didn't work -- try it on v=gx/c -- you'll get m/s. I can tell you want to argue. Is there anyone else in the thread that wants to explain this to him in light of his upcoming objections? I'm extremely swamped with work. But this is not what you wrote, what you wrote is the nonsense [math]c=xg[/math]. Not to mention the wrong things you posted about Shapiro delay. Edited November 23, 2013 by xyzt -2 Link to comment Share on other sites More sharing options...

Iggy Posted November 24, 2013 Share Posted November 24, 2013 But this is not what you wrote, what you wrote is the nonsense [math]c=xg[/math]. People rely on information here to be accurate. [math]c=xg[/math] is indeed the speed of light in Rindler coordinates in natural units. According to you, [math]xg[/math] cannot be a speed because it yields a dimensionality of length squared per time squared. Such ambiguity is the nature of natural units. Wikipedia explains it: Natural Units: Advantages and disadvantages Greater ambiguity: Consider the equation a = 10^{10} in Planck units. If a represents a length, then the equation means a = 1.6×10^{−25} m. If a represents a mass, then the equation means a = 220 kg. Therefore, if the variable a was not clearly defined, then the equation a = 10^{10} might be misinterpreted. By contrast, in SI units, the equation would be (for example) a = 220 kg, and it would be clear that a represents a mass, not a length or anything else Natural Units - Advantages and disadvantages Not only did I define [math]c=xg[/math] as the speed of light (removing any ambiguity), I drew a diagram demonstrating [math]xg[/math] is indeed the speed of light in Rindler coordinates at multiple values of x and t. Even after giving the SI formula (xg/c) so that you could confirm the dimensionality is meters per second you continue provoking. I welcome you to tell me what the formula for the speed of light in Rindler coordinates is. If mine is nonsense then please correct me. What is the formula for the speed of light in Rindler coordinates? 1 Link to comment Share on other sites More sharing options...

xyzt Posted November 24, 2013 Share Posted November 24, 2013 (edited) People rely on information here to be accurate. [math]c=xg[/math] is indeed the speed of light in Rindler coordinates in natural units. According to you, [math]xg[/math] cannot be a speed because it yields a dimensionality of length squared per time squared. Such ambiguity is the nature of natural units. Wikipedia explains it: Natural Units: Advantages and disadvantages Greater ambiguity: Consider the equation a = 10^{10} in Planck units. If a represents a length, then the equation means a = 1.6×10^{−25} m. If a represents a mass, then the equation means a = 220 kg. Therefore, if the variable a was not clearly defined, then the equation a = 10^{10} might be misinterpreted. By contrast, in SI units, the equation would be (for example) a = 220 kg, and it would be clear that a represents a mass, not a length or anything else Natural Units - Advantages and disadvantages Not only did I define [math]c=xg[/math] as the speed of light (removing any ambiguity), I drew a diagram demonstrating [math]xg[/math] is indeed the speed of light in Rindler coordinates at multiple values of x and t. Even after giving the SI formula (xg/c) so that you could confirm the dimensionality is meters per second you continue provoking. I welcome you to tell me what the formula for the speed of light in Rindler coordinates is. If mine is nonsense then please correct me. What is the formula for the speed of light in Rindler coordinates? Not only that the units were wrong but also the relationship [math]c=xg[/math] is wrong as well. The speed of light is constant, c. [math]xg[/math] is not constant for the mere reason that x is variable. How about the other gross, error, the one about the ever increasing speed of light in the Shapiro delay? You never addressed your other error on the subject. The crackpottery posted insistently by Iggy is getting a little much. There are two natural frames in which to describe a uniformly accelerating system. They are the accelerating coordinates (i.e. the proper (rest frame) time [math]\tau[/math] and related spatial coordinate [math]X[/math] ) and the coordinates of one inertial frame that matches the accelerating frame at one instant (as represented by [math](x,t)[/math]). The transformation between the coordinate systems is given by: [math]x=-\frac{c^2}{g}+(X+\frac{c^2}{g})cosh(\frac{g \tau}{c})[/math] [math]ct=(X+\frac{c^2}{g})sinh(\frac{g \tau}{c})[/math] where g is the uniform 3-acceleration magnitude. At the pivot point [math]X=- \frac{c^2}{g}[/math] time must be frozen to zero, [math]ct=0[/math] and [math]x=-\frac{c^2}{g}=X[/math], (thus the name pivot point). This doesn't mean in any shape or form that [math]c=xg[/math] as Iggy keeps claiming. Light speed, contrary to Iggy's crank claims, does NOT VARY in relativity, no matter if the motion is accelerated or not, it is a constant. If it did, the whole theory would fall apart. Edited November 24, 2013 by xyzt Link to comment Share on other sites More sharing options...

Iggy Posted November 25, 2013 Share Posted November 25, 2013 (edited) Not only that the units were wrong but also the relationship [math]c=xg[/math] is wrong as well. The speed of light is constant, c. [math]xg[/math] is not constant for the mere reason that x is variable. I recommend reading the following site: Rindler Spacetime One of the fascinating things about general relativity is how it can be brought smoothly from special relativity when considering accelerating observers. In order to describe gravity, general relativity uses the concept of curved spacetime... The result is called Rindler spacetime, described by the so-called Rindler metric. <snip> ...Long story short: when Alice looks at points at her left (remember, gravity points leftwards), she sees a lower speed of light. Is that even possible? That is against the principle of relativity, isn’t it? No! The principle of relativity talks about inertial observers. Alice is not. So, again: points at her left have lower speeds of light. Therefore, relativistic effects are “more notorious”. Even worse: as you move leftwards, this “local speed of light” decreases more and more... until it reaches zero! Exactly at the “special point”, where Alice coordinates behaved badly. What happens there? It’s an horizon! Where time stood still. http://physicsnapkins.wordpress.com/2012/12/13/rindler/ It simplifies the issue greatly. The crackpottery posted insistently by Iggy is getting a little much. There are two natural frames in which to describe a uniformly accelerating system. They are the accelerating coordinates (i.e. the proper (rest frame) time [math]\tau[/math] and related spatial coordinate [math]X[/math] ) and the coordinates of one inertial frame that matches the accelerating frame at one instant (as represented by [math](x,t)[/math]). The transformation between the coordinate systems is given by: [math]x=-\frac{c^2}{g}+(X+\frac{c^2}{g})cosh(\frac{g \tau}{c})[/math] [math]ct=(X+\frac{c^2}{g})sinh(\frac{g \tau}{c})[/math] where g is the uniform 3-acceleration magnitude. At the pivot point [math]X=- \frac{c^2}{g}[/math] time must be frozen to zero, [math]ct=0[/math] and [math]x=-\frac{c^2}{g}=X[/math], (thus the name pivot point). This doesn't mean in any shape or form that [math]c=xg[/math] as Iggy keeps claiming. You plagiarized what you just wrote from section 1.2.1 of this paper. I doubt you know what it means, because it doesn't address what we're talking about at all. In fact, given the transforms from cartesian (inertial) to Rindler (accelerated) coordinates that you plagiarized, one can readily prove c = xg. The transforms are expressed better by wikipedia (mind you they are using natural units): for cartesian [math](t,x)[/math], and Rindler [math](t_A, x_A)[/math] we have: [math]t_A = \frac{1}{g} \mbox{arctanh}\left(\frac{t}{x}\right),\; x_A= \sqrt{x^2-t^2}[/math] [math]t = x_A \, \sinh(gt_A), \; x = x_A \, \cosh(gt_A)[/math] A light ray emitted at [math]x_I[/math] follows the cartesian path [math]x = x_I + vt[/math] (where v is +1 or -1 depending on direction) making its path in Rindler coordinates rather trivial: [math]{x_A}^2 = x^2 - t^2[/math] [math]{x_A}^2 = (x_I + vt)^2 - t^2[/math] [math]{x_A}^2 = {x_I}^2 + 2x_Ivt[/math] [math]{x_A}^2 = {x_I}^2 + 2x_Iv(x_A \sinh(gt_A))[/math] and, using wolfram to solve for [math]x_A[/math] where v = +/- 1: [math]x_A =x_I \exp{(\pm g t_A)}[/math] light emitted at [math](t=0, x = x_I)[/math] therefore has a Rindler speed: [math]\frac{d x_A}{d t_A} = \pm gx_I \exp{(\pm g \cdot 0)}[/math] [math]\frac{d x_A}{d t_A} = \pm gx_I[/math] which is to say that the speed of a ray of light in natural units in Rindler coordinates is c(x) = gx Edited November 25, 2013 by Iggy Link to comment Share on other sites More sharing options...

xyzt Posted November 25, 2013 Share Posted November 25, 2013 (edited) which is to say that the speed of a ray of light in natural units in Rindler coordinates is c(x) = gx I do not know why you insist in posting your crank stuff, over and over. You need to understand the light speed is an invariant, it definitely isn't gx as you keep maintaining. You are utterly confused in your attempt at pushing fringe ideas. What you derived is nothing but the coordinate speed of light in an accelerated frame. As I explained earlier, the coordinate speed of light can be ANYTHING, it is a MEANINGLESS expression. BTW, your muddled derivation can be done CORRECTLY in just a few lines: -start with the Rindler metric [math]ds^2=(\frac{gx}{c^2})^2 (cdt)^2 -dx^2[/math] -light follows null geodesics, [math]ds=0[/math] so [math]\frac{dx}{dt}=\pm \frac{gx}{c}[/math]. -if you use the alternate form of the metric : [math]ds^2=(1+\frac{gx}{c^2})^2 (cdt)^2 -dx^2[/math] you will get a different value for the coordinate speed of light. [math]\frac{dx}{dt}=\pm c (1+ \frac{gx}{c^2})[/math]. Done. It shows clearly that what you get is the coordinate speed of light, totally useless. Mathematically much more rigorous and more general than what you did, physically, totally useless. Light speed is an invariant, c, you need to learn that. Edited November 25, 2013 by xyzt Link to comment Share on other sites More sharing options...

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