Does Gravity Slow Light Moving Vertically?

[CaptainPanic]

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Moderator Note

 

Iggy, xyzt, you are both violating rule 1 of our forum (flaming, uncivilized behavior). You have both posted remarks which were aimed only to insult the other person, but which were irrelevant for a third person who is interested in the science only.

 

It is very common to disagree in science. And it is not against the rules to be wrong on this forum. However, if you cannot discuss your differences in a civilized manner, this thread will be closed (the OP has indicated to be done here) and you both will gain some warning points.

 

This warning says nothing about who is right and who is wrong (because frankly I don't know). From what I read, this may all be a miscommunication, or it may be a fundamental difference in opinion. I suggest that you start off by trying to figure out what you disagree on, and then move on to answer that. Cooperation in this matter will get you a lot further than this shouting contest you're having now.

 

 

 

Do not respond to this moderator note in this thread. If you have any issues with this note, use the report button at the bottom of this post.

[JVNY]

I am starting to work through the materials. Start with xyzt's question of whether the end to end flashes arrive simultaneously in the accelerating frame, using hyperbolic formulas. Using the formulas, I believe that yes the flashes arrive simultaneously in the accelerating frame. Consider two ships, rear and front, at rest in the inertial lab frame, rear at x=0.5 and front at x=1.0. Simultaneously each flashes a light at the other ship and begins to accelerate so as to maintain their proper distance, rear at proper rate a=2, front at proper rate a=1 (all measures unitless, so c=1, the y axis can be for example years, the x axis light years, and acceleration light years per year^2).

 

Below is a Minkowski diagram from the lab frame, which shows that the flashes strike simultaneously in the ships' frame. The dashed line marks simultaneity in the ships' frame (each ship's velocity is 0.6c there when the flashes strike). The diagram also shows for reference the mid point between the ships, which also maintains proper distance (proper rate a=1.33).

 

The light flashes cross to the left of center, but nonetheless reach the ships simultaneously.

 

Perhaps there is common ground between Iggy and xyzt. It seems that there are two ways to measure speed. One is what Iggy refers to, and xyzt calls coordinate speed. The diagram illustrates that kind of speed -- light goes faster (in both directions) when it is traveling in the area to the right, and it goes slower (in both directions) when it is traveling in the area to the left. This is consistent with the Shapiro delay. The other is the speed that someone measures locally, which is always c.

 

 

Sorry, the post did not take the diagram. Here it is as an attachment.

[xyzt]

 

 

 

 

The diagram illustrates that kind of speed -- light goes faster (in both directions) when it is traveling in the area to the right, and it goes slower (in both directions) when it is traveling in the area to the left. This is consistent with the Shapiro delay. The other is the speed that someone measures locally, which is always c.

 

There is no such thing as "light goes faster" (or "slower").

In the Shapiro delay , light speed is c, constant, do not perpetrate the incorrect claims Iggy has been making. Light follows geodesics, in the presence of gravitational bodies, such geodesics are not lines in the Euclidian sense. See this illustration. Mathematically:

 

[math]c \Delta t_{12}=X_1+X_2+r_s ln \frac{X_1 X_2}{R^2}[/math]

 

Note the "c" in the LHS? This denotes the (invariant) speed of light.

See the extra term [math]r_s ln \frac{X_1 X_2}{R^2}[/math] in the RHS? This denotes the "deviation" of the light path (null geodesic) from an Euclidian (straight) line.

 

 

 

 

Edited November 25, 2013 by xyzt

[JVNY]

OK, let's stay away from the issue of light speed, and just focus on simultaneity of events.

 

Clearly in the lab frame the flash from the front strikes the rear (at lab time t=0.375) before the flash from the rear strikes the front (at lab time t=0.75).

 

Based on the hyperbola formulas and the diagram above, is it the case that in the ships' reference frame the rearward flash arrives at the rear ship simultaneously with the forward flash arriving at the front ship (along the dashed line of ship frame simultaneity)?

[xyzt]

OK, let's stay away from the issue of light speed, and just focus on simultaneity of events.

 

Clearly in the lab frame the flash from the front strikes the rear (at lab time t=0.375) before the flash from the rear strikes the front (at lab time t=0.75).

 

 

Yes, I showed you the calculations.

 

 

 

Based on the hyperbola formulas and the diagram above, is it the case that in the ships' reference frame the rearward flash arrives at the rear ship simultaneously with the forward flash arriving at the front ship (along the dashed line of ship frame simultaneity)?

 

I answered this question as well: we don't know until we calculate the travel times. You know the (unequal) travel times in the inertial (what you call "lab") frame. Using the hyperbolic motion formulas, you can calculate the travel times in the accelerated frame of the rocket.

[JVNY]

OK, here are the calculations of the proper times for the ships. The rearward flash hits the rear ship at rear ship's proper time 0.346, and front ship's proper time 0.693. The forward flash hits the front ship at the rear ship's proper time 0.346, and the front ship's proper time 0.693. So the flashes arrive simultaneously in the ships' frame. Riders on each ship agree that the flashes arrive simultaneously, even though they disagree on what time the flashes arrived. The rear clock runs slower than the front clock in the ships' own frame (the rear at half the rate of the front), so although they agree on simultaneity, if they set their clocks at zero upon initiating acceleration then a rider on the rear will say that the flashes arrived at time 0.346, whereas the rider on the front will say that the flashes arrived at time 0.693.

[xyzt]

OK, here are the calculations of the proper times for the ships. The rearward flash hits the rear ship at rear ship's proper time 0.346, and front ship's proper time 0.693. The forward flash hits the front ship at the rear ship's proper time 0.346, and the front ship's proper time 0.693. So the flashes arrive simultaneously in the ships' frame. Riders on each ship agree that the flashes arrive simultaneously, even though they disagree on what time the flashes arrived. The rear clock runs slower than the front clock in the ships' own frame (the rear at half the rate of the front), so although they agree on simultaneity, if they set their clocks at zero upon initiating acceleration then a rider on the rear will say that the flashes arrived at time 0.346, whereas the rider on the front will say that the flashes arrived at time 0.693.

I make a habit of not following any numerical calculation (because it is not tractable). Can you put what you just said in the form of a symbolic calculation, please?

[JVNY]

Here are the formulas, from http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

 

Inertial frame time: t = sqrt[(d/c)² + 2d/a] eq. 1

Inertial frame distance traveled: d = (c²/a) (sqrt[1 + (at/c)²] - 1) eq. 2

Ship's elapsed proper time: T = (c/a) ch^-1 [ad/c² + 1] eq. 3

 

Inertial frame velocity: v = at / sqrt[1 + (at/c)²] eq. 4

 

 

Each flash must close 0.5 distance in the inertial frame. The rearward flash closes with the rear ship (which accelerates at a=2), and the flash travels one unit of d for every unit of t (because c=1), so the flash strikes the rear ship when d + t = 0.5.

 

d + t = 0.5

 

Using equations 1 and 2 above:

 

(c²/a) (sqrt[1 + (at/c)²] - 1) + sqrt[(d/c)² + 2d/a] = 0.5

 

Solving that equation, t = 0.375 and d = 0.125

 

Solving eq. 3 above using a = 2, t = 0.375 and d = 0.125, yields T = 0.346.

 

The forward flash must cover 0.5 distance plus the distance that the front ship travels, so that the flash strikes the front ship at t = 0.5 + d, or t - d = 0.5.

 

Using equations 1 and 2 above, where a = 1 for the front ship,

 

sqrt[(d/c)² + 2d/a] - (c²/a) (sqrt[1 + (at/c)²] - 1) = 0.5

 

Solving that equation, t = 0.75 and d = 0.25

 

Solving eq. 3 above using a = 1, t = 0.75 and d = 0.25, yields T = 0.693

 

These are simultaneous in the ships' frame because each is at rest with respect to the other. Their velocities are equal, 0.6c, determined using eq. 4 above for each ship.

 

Another way to confirm that the two ships agree on simultaneity is to determine the line of simultaneity, the dashed line in the diagram, which has the slope 0.6. One determines the locus of the line as follows.

 

If a point undergoes "constant proper acceleration a₀," then the "locus of simultaneity constantly passes through the point (-1/a₀, 0), and it maintains a constant absolute spacelike distance of -1/a₀ from that point." See http://mathpages.com/rr/s2-09/2-09.htm.

 

For the rear ship, that means that the locus of simultaneity is (-1/2, 0), or (-0.5, 0). Since the rear ship begins at 0.5, that means that its locus of simultaneity is at (0.5 - 0.5, 0), that is (0, 0).

 

For the rear ship, the locus of simultaneity is (-1/1, 0), or (-1, 0). The front ship starts at 1.0, so its locus of simultaneity is (1.0 - 1.0, 0), that is (0, 0).

 

Therefore the dashed line of simultaneity starts at (0, 0), and with a slope of 0.6 crosses the rear and front world lines as the flashes strike each.

 

 

[xyzt]

Here are the formulas, from http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

 

Inertial frame time: t = sqrt[(d/c)² + 2d/a] eq. 1

Inertial frame distance traveled: d = (c²/a) (sqrt[1 + (at/c)²] - 1) eq. 2

 

Ship's elapsed proper time: T = (c/a) ch^-1 [ad/c² + 1] eq. 3

 

Inertial frame velocity: v = at / sqrt[1 + (at/c)²] eq. 4

 

 

Each flash must close 0.5 distance in the inertial frame. The rearward flash closes with the rear ship (which accelerates at a=2), and the flash travels one unit of d for every unit of t (because c=1), so the flash strikes the rear ship when d + t = 0.5.

 

d + t = 0.5

 

Using equations 1 and 2 above:

 

(c²/a) (sqrt[1 + (at/c)²] - 1) + sqrt[(d/c)² + 2d/a] = 0.5

 

Solving that equation, t = 0.375 and d = 0.125

 

This is not quite right, the units don't match and the equation you ended up with is not what you want. The correct equation is:

 

[math]\frac{c^2}{a}(\sqrt{1+(at/c)^2}-1) + ct=L/2[/math]

 

The above has a very nice solution:

 

[math]t=\frac{L}{2c} \frac{1+aL/(4c^2)}{1+aL/(2c^2)}[/math]

 

So, the answer should be very , very close to [math]\frac{L}{2c}[/math], so your t=0.375 doesn't seem correct.

Edited November 25, 2013 by xyzt

[JVNY]

OK, so let's start with the rear ship, and an example from Dr. Baez's website http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html.

 

With the rear ship, we use c=1, a=2 and t=0.375. Then you just plug c, a and t into the formula for d from Dr. Baez's website. The result is distance traveled for the rear ship of 0.125 to the right. In t=0.375, the rearward flash travels a distance of 0.375 to the left. Combined, that is 0.125 + 0.375 = 0.5, and the flash has closed the 0.5 distance between the ships in the ground frame.

 

Then do the same with c=1, a=1.03 (the equivalent of 1g, per that website), and t=1.19. The result is distance traveled of 0.56, which is what the website gives in the first example of typical results for a=1g travel (excerpted here):

 

These equations are valid in any consistent system of units such as seconds for time, metres for distance, metres per second for speeds and metres per second squared for accelerations. In these units c = 3 × 10⁸ m/s (approx). To do some example calculations it is easier to use units of years for time and light years for distance. Then c = 1 lyr/yr and g = 1.03 lyr/yr². Here are some typical answers for a = 1g.

T t d v γ
1 year 1.19 yrs 0.56 lyrs 0.77c 1.58

 

 

Here is a chart showing the two results from the formula, d=0.125 in the case of the rear ship, and d=0.56 in the website example:

 

[xyzt]

OK, so let's start with the rear ship, and an example from Dr. Baez's website http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html.

What "rear ship"? There is only one ship and you are sending a signal towards the front and one signal towards the rear of the same ship.

[JVNY]

I posited two ships, front and rear, separated by 0.5. It works just the same for a single ship as you posit. Then one would posit a single ship of proper length 0.5. The front accelerates at a=1 and sends a flash to the rear, simultaneous with the rear accelerating at a=2 (and all points in between accelerating between 1 and 2 such that the whole ship maintains its proper length of 0.5). In that case, in inertial frame time t=0.375 the flash from the front of the ship travels 0.375 to the left. The rear of the ship travels 0.125 to the right, and the flash strikes the rear of the ship.

 

Are we in agreement on that?

[xyzt]

I posited two ships, front and rear, separated by 0.5. It works just the same for a single ship as you posit. Then one would posit a single ship of proper length 0.5.

No, it doesn't. I showed you how to do the calculations symbolically. Try doing the same thing (please stop using tables and numbers) and you might find your error.

 

 

 

The front accelerates at a=1 and sends a flash to the rear, simultaneous with the rear accelerating at a=2 (and all points in between accelerating between 1 and 2 such that the whole ship maintains its proper length of 0.5). In that case, in inertial frame time t=0.375 the flash from the front of the ship travels 0.375 to the left. The rear of the ship travels 0.125 to the right, and the flash strikes the rear of the ship.

 

Are we in agreement on that?

 

This is not equivalent with sending a pulse from the center of one rocket towards its front/rear end. You are dealing with TWO accelerations, this is much more complicated than what you set to solve, not to mention that your equations do not reflect the above. Try solving the simple case (one rocket) first.

Edited November 26, 2013 by xyzt

[JVNY]

OK, thanks.

[Iggy]

Here are the formulas, from http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

 

Inertial frame time: t = sqrt[(d/c)2 + 2d/a] eq. 1

Inertial frame distance traveled: d = (c2/a) (sqrt[1 + (at/c)2] - 1) eq. 2

 

Ship's elapsed proper time: T = (c/a) ch-1 [ad/c2 + 1] eq. 3

 

Inertial frame velocity: v = at / sqrt[1 + (at/c)2] eq. 4

 

 

Each flash must close 0.5 distance in the inertial frame. The rearward flash closes with the rear ship (which accelerates at a=2), and the flash travels one unit of d for every unit of t (because c=1), so the flash strikes the rear ship when d + t = 0.5.

 

d + t = 0.5

 

Using equations 1 and 2 above:

 

(c2/a) (sqrt[1 + (at/c)2] - 1) + sqrt[(d/c)2 + 2d/a] = 0.5

 

Solving that equation, t = 0.375 and d = 0.125

 

Solving eq. 3 above using a = 2, t = 0.375 and d = 0.125, yields T = 0.346.

 

The forward flash must cover 0.5 distance plus the distance that the front ship travels, so that the flash strikes the front ship at t = 0.5 + d, or t - d = 0.5.

 

Using equations 1 and 2 above, where a = 1 for the front ship,

 

sqrt[(d/c)2 + 2d/a] - (c2/a) (sqrt[1 + (at/c)2] - 1) = 0.5

 

Solving that equation, t = 0.75 and d = 0.25

 

Solving eq. 3 above using a = 1, t = 0.75 and d = 0.25, yields T = 0.693

 

These are simultaneous in the ships' frame because each is at rest with respect to the other. Their velocities are equal, 0.6c, determined using eq. 4 above for each ship.

 

Another way to confirm that the two ships agree on simultaneity is to determine the line of simultaneity, the dashed line in the diagram, which has the slope 0.6. One determines the locus of the line as follows.

 

If a point undergoes "constant proper acceleration a0," then the "locus of simultaneity constantly passes through the point (-1/a0, 0), and it maintains a constant absolute spacelike distance of -1/a0 from that point." See http://mathpages.com/rr/s2-09/2-09.htm.

 

For the rear ship, that means that the locus of simultaneity is (-1/2, 0), or (-0.5, 0). Since the rear ship begins at 0.5, that means that its locus of simultaneity is at (0.5 - 0.5, 0), that is (0, 0).

 

For the rear ship, the locus of simultaneity is (-1/1, 0), or (-1, 0). The front ship starts at 1.0, so its locus of simultaneity is (1.0 - 1.0, 0), that is (0, 0).

 

Therefore the dashed line of simultaneity starts at (0, 0), and with a slope of 0.6 crosses the rear and front world lines as the flashes strike each.

That looks spot on to me. Very good use of the relativistic rocket equations.

 

Another way to verify the dotted line is simultaneous for both ships, and that they maintain constant proper distance is with rindler coordinates.

 

Position and acceleration will be related by x = 1/a, so having your aft ship at x = .5 and a = 2 works with the forward ship at x = 1 and a = 1. We will have the forward ship be the timekeeper, in which case the metric is:

 

[math]ds^2 = -(ax)^2 dt^2 + dx^2[/math]

 

In the lab frame the forward observer observes the ray of light at t=0.75, x=1.25. Using the transforms:

 

[math]t_A = \frac{1}{g} \mbox{arctanh}\left(\frac{t}{x}\right),\; x_A= \sqrt{x^2-t^2}[/math]

the Rindler coordinates become the following:

 

[math]t_A[/math] = (1/1)*arctanh(0.75/1.25) = 0.6931

 

[math]x_A[/math] sqrt(1.25^2-0.75^2) = 1.0

If the observers maintain a constant proper distance and the aft observer receives the ray of light simultaneously then transforming the rear observation from the lab frame to rindler coordinates will give the same t (0.6931) and x = .5

 

In the lab frame the aft observer observes the ray of light at t=0.375, x=0.625

 

[math]t_A[/math] = (1/1)*arctanh(0.375/0.625) = 0.6931

 

[math]x_A[/math] sqrt(0.625^2-0.375^2) = 0.5

 

So I have you exactly right. In accelerated coordinates where the ships consider themselves at rest with respect to one another and with respect to themselves the light rays arrive simultaneously at the numbers you give.

[Iggy]

 

...It works just the same for a single ship as you posit...

No, it doesn't. I showed you how to do the calculations symbolically. Try doing the same thing (please stop using tables and numbers) and you might find your error.

 

...The front accelerates at a=1 and sends a flash to the rear, simultaneous with the rear accelerating at a=2..

This is not equivalent with sending a pulse from the center of one rocket towards its front/rear end. You are dealing with TWO accelerations, this is much more complicated than what you set to solve, not to mention that your equations do not reflect the above. Try solving the simple case (one rocket) first.

 

I'm afraid I don't see the error you're referring to, and I don't understand how an equation could demonstrate an error in those statements.

 

To accelerate anything rigidly (be it one ship or two, or any number of particles and/or objects), each point being accelerated must have a magnitude of proper acceleration that is inversely proportional to the position along the rigid system's axis of motion. The front and the rear can maintain constant proper distance and different proper accelerations simultaneously -- no matter if a single ship separates the front and rear, or if a void between two ships separates them.

 

"Born Rigidity" is the principle by which this is very well established in relativity.

 

May I ask you to demonstrate the error for the sake of my curiosity?

 

[xyzt]

This is not quite right, the units don't match and the equation you ended up with is not what you want. The correct equation is:

 

[math]\frac{c^2}{a}(\sqrt{1+(at/c)^2}-1) + ct=L/2[/math]

 

The above has a very nice solution:

 

[math]t=\frac{L}{2c} \frac{1+aL/(4c^2)}{1+aL/(2c^2)}[/math]

 

So, the answer should be very , very close to [math]\frac{L}{2c}[/math], so your t=0.375 doesn't seem correct.

In the opposite direction, the light front "chases" the front of the rocket that recedes from it, so the equation is:

[math]\frac{c^2}{a}(\sqrt{1+(at/c)^2}-1) = ct+L/2[/math]

 

Solving for t, you get that the fronts of light hit the front and the aft of the rocket at different times.

[Iggy]

In the opposite direction, the light front "chases" the front of the rocket that recedes from it, so the equation is:

[math]\frac{c^2}{a}(\sqrt{1+(at/c)^2}-1) = ct+L/2[/math]

 

Solving for t, you get that the fronts of light hit the front and the aft of the rocket at different times.

 

using the equation in the context you have given would require signifying who is measuring t, L, and a, and what t, L, and a are specifically measuring, and how that disagrees with any number given in the derivation to which you are objecting, otherwise it is too ambiguous to solve.

 

I can tell you I traced every step of the derivation JVNY did with the relativistic rocket equations and each number (also labeled on the diagram he made) was correct. I also individually solved them via an independent method and got the same numbers, as shown 2 posts ago.

[md65536]

This is not equivalent with sending a pulse from the center of one rocket towards its front/rear end. You are dealing with TWO accelerations, this is much more complicated than what you set to solve, not to mention that your equations do not reflect the above. Try solving the simple case (one rocket) first.

How is this simpler? Are you trying to calculate relativistic effects at the front vs back of the ship, while assuming that the ship accelerates as one uniform thing (simultaneously all over the ship for everyone)? If so this is not physically realistic.

 

Otherwise, even if you model the ship as simply as you suggest --- perfectly rigid but still obeying relativity, accelerating from a single point --- you can still model its various parts using separate rockets. If you treat the various parts of the rocket as the same, the rocket has negligible length, then you can't continue calculating relativistic effects along its length.

Edited November 26, 2013 by md65536

[xyzt]

How is this simpler? Are you trying to calculate relativistic effects at the front vs back of the ship, while assuming that the ship accelerates as one uniform thing (simultaneously all over the ship for everyone)? If so this is not physically realistic.

 

 

Nope, it is a very simple scenario, two pulses of light is emitted from the center of an uniformly accelerated rocket , one towards the front and the other one towards the rear. Find the elapsed times when the two pulses hit the two rocket ends, as expressed in what the OP calls "the lab" (i.e. the inertial frame of the Earth). This was repeated quite clearly throughout the thread by the OP author and by me. The math I posted describes his scenario, did you read the whole set of exchanges?

 

 

 

Otherwise, even if you model the ship as simply as you suggest --- perfectly rigid but still obeying relativity, accelerating from a single point --- you can still model its various parts using separate rockets. If you treat the various parts of the rocket as the same, the rocket has negligible length, then you can't continue calculating relativistic effects along its length.

 

The discussion has nothing to do with rigidity.

[md65536]

Nope, it is a very simple scenario, two pulses of light is emitted from the center of an uniformly accelerated rocket , one towards the front and the other one towards the rear.

The front and rear of your single rocket can be modelled by two separate rockets configured properly. It is equivalent to a single rocket. If your one-rocket example is simpler than is possible with 2 rockets, then you're doing something wrong. It doesn't matter if the force accelerating the front of a rocket is transmitted through the structure of the ship, or comes from a separate nozzle at the front.

 

Edit: I looked back and realize the 2-rocket description is describing something entirely different. I see now, the problem is with using 2 different accelerations, not with treating a rocket as separate parts.

Edited November 26, 2013 by md65536

[JVNY]

Some of this confusion is due to my having posted two separate scenarios. Apologies. In the first scenario, the front ship sends a flash toward the rear ship, and the rear ship sends a flash toward the front ship. In the second, there is also a center ship. The center ships sends flashes forward and rearward; they reflect off of the front and rear ships, then return to the center ship.

 

The bulk of the discussion thread has covered the first scenario. It seems from the calculations, confirmed by Iggy, that the flash from the front reaches the rear simultaneously (in the ships' frame) with the flash from the rear to the front.

 

So perhaps now let's turn to the second scenario.

 

 

Nope, it is a very simple scenario, two pulses of light is emitted from the center of an uniformly accelerated rocket , one towards the front and the other one towards the rear. Find the elapsed times when the two pulses hit the two rocket ends, as expressed in what the OP calls "the lab" (i.e. the inertial frame of the Earth). This was repeated quite clearly throughout the thread by the OP author and by me. The math I posted describes his scenario, did you read the whole set of exchanges?

 

 

 

 

The discussion has nothing to do with rigidity.

 

So here is the scenario and a Minkowski diagram. The three ships start out at rest in the inertial frame, rear at x=0.5, center at x=0.75, and front at x=1.0. Simultaneously, center sends flashes rearward and forward, and the ships begin constant proper acceleration of rear a=2, center a=1.33, and front a=1. This maintains their proper distance in their own accelerating reference frame. When a flash hits a ship, it reflects and returns to the center ship. The results, which are based again on the relativistic rocket formulas from Dr. Baez's website, are as follows. In the inertial frame, the order of light flashes striking ships is: rearward flash hits rear ship, then forward flash hits front ship, then forward flash returns to center ship, then rearward flash returns to center ship. The diagram shows the inertial frame times and locations for each strike.

 

In the ships' frame, the order is different, as you can see from the lines of simultaneity. The forward flash hits the front ship first, then the rearward flash hits the rear ship, then the forward flash returns to the center, and finally the rearward flash returns to the center.

 

In their proper times (not shown on the diagram), the rearward flash hits the rear at rear ship time Tr=0.203; the forward flash hits the front ship at front ship time Tf=0.288; the forward flash returns to the center at center ship time Tc=0.432, and finally the rearward flash returns to the center at center ship time Tc=0.608.

 

This seems consistent with radial light travel and the Shapiro delay, if one sends signals in opposite directions, one toward the gravitational mass, and one away from it.

 

[md65536]

In the ships' frame, the order is different, as you can see from the lines of simultaneity. The forward flash hits the front ship first, then the rearward flash hits the rear ship, then the forward flash returns to the center, and finally the rearward flash returns to the center.

 

In their proper times (not shown on the diagram), the rearward flash hits the rear at rear ship time Tr=0.203; the forward flash hits the front ship at front ship time Tf=0.288; the forward flash returns to the center at center ship time Tc=0.432, and finally the rearward flash returns to the center at center ship time Tc=0.608.

I haven't followed the whole thread and it's over my head, but are you saying that you've set up the accelerations so that the 3 ships all have an equivalent accelerating frame, which they can treat as a single frame of reference? There is no time dilation between any two ships, they're able to keep their clocks synchronized according to an observer on each ship?

[JVNY]

Yes and no. I believe that the ships remain at rest with respect to each other, maintain their proper distance, and agree on the simultaneity of events, all in their own single accelerating reference frame. However, their clocks do not run at the same rate. The rear ship's clock runs at 50% of the rate of the front clock, and the center at 75% of the rate of the front clock. This is the equivalence principle in action, just as clocks closer to a gravitational mass run slower than those farther away.

 

The ships can always compensate for this and show the same clock times. The rear ship, for example, can program its clock to record 2 units of time for every 1 unit that elapses at the rear. Then its clock will always show the same time as the front clock (simultaneously in the ships' frame). I think that this is what Iggy means in post 40 when he says "We will have the forward ship be the timekeeper." This does not change the passage of time. Time continues to elapse more slowly the farther back one is from the front. But it does allow the ships to label events that they agree are simultaneous with the same clock time.

Here are some links from earlier posts to explanations.

 

For setting up the hyperbolas, see http://en.wikipedia.org/wiki/Rindler_coordinates

 

For details on the locus of simultaneity and drawing the line of simultaneity, see http://mathpages.com/rr/s2-09/2-09.htm

[Iggy]

The ships can always compensate for this and show the same clock times. The rear ship, for example, can program its clock to record 2 units of time for every 1 unit that elapses at the rear. Then its clock will always show the same time as the front clock (simultaneously in the ships' frame). I think that this is what Iggy means in post 40 when he says "We will have the forward ship be the timekeeper."

In a manner of speaking. I really was only thinking that the lead clock would measure [math]\tau = 0.6931[/math] between both the two red events and the two blue events:

 

 

making the top red and top blue events simultaneous.

 

 

The variable t in the metric [math]ds^2 = -(ax)^2 dt^2 + dx^2[/math] is the proper time of a clock at [math]x_A = 1[/math] (the lead clock), so the dotted line would be called t = 0.6931, but only the lead clock measures that proper time on the dotted line.

 

Proper time in that Rindler metric is given by [math]d \tau = a x dt[/math], so the time between red events as measured by the aft clock (at [math]x_A = 0.5[/math]) would be [math]d \tau = (1)(0.5)(0.6931) = 0.3466[/math] (as you say, 50%).

 

It is convention for the coordinate acceleration, a, and the coordinate time, t, in the metric to correspond to the proper acceleration and proper time for the observer at x = 1. If a different observer were chosen then the metric would look slightly different -- it would have a translation of the origin along the x axis. But, physically, it doesn't matter which observer is chosen.

 

 

Rereading what I just wrote, I think I just only managed to obfuscate what you explained very well.

Edited November 27, 2013 by Iggy