Speed of the light and dilation of time

[PAllen]

Yes, I did. I hadn't looked at the original problem description for a while.

If the inertial home twin is located somewhere on the circular track, then the traveler can compute the dot product of v and L at any point on his circuit, and then use the CADO equation to compute the home twin's age at that instant in his (the traveler's) life.

...according one of an infinite number of equally valid conventions. The website you linked mentioned two others that that author doesn't like, but expert consensus is those two are popular and equally valid. Further the 'conventionality of simultaneity' says any simultaneity mapping between world lines meeting certain basic conditions is valid. The interesting questions to answer are 'what is true for any valid convention'. A few of my posts in this thread have addressed this, e.g. for the OP scenario with Earth removed, any simultaneity mapping will have simultaneous events where the station observer sees red shift and the traveler sees blueshift; and it will also have paired events where they agree on red shift.

 

[ghwellsjr]

Did you place the son at the center of the Earth instead of in the train station?

Yes, I did. I hadn't looked at the original problem description for a while.

If the inertial home twin is located somewhere on the circular track, then the traveler can compute the dot product of v and L at any point on his circuit, and then use the CADO equation to compute the home twin's age at that instant in his (the traveler's) life.

Can you please do the calculations for several points around the earth, say every 45 degrees? Assume a train speed of .99c, an Earth radius of 4000 miles and a light speed of 186,000 miles per second. (I would really like to see it done every 15 degrees but I don't want to overburden you.)

 

[GrammawSally]

Can you please do the calculations for several points around the earth, say every 45 degrees? Assume a train speed of .99c, an Earth radius of 4000 miles and a light speed of 186,000 miles per second. (I would really like to see it done every 15 degrees but I don't want to overburden you.)

Left as an exercise for the reader.

 

[ghwellsjr]

Left as an exercise for the reader.

Fine, but is this exercise what you meant when you said:

A person who is accelerating is still entitled to his own "point of view" about how other people are aging.

 

[Underwood]

Also, you should be aware that the CADO process is just one of an infinite number of ways to answer the unanswerable question of what is the Current Age of a Distant Object.

I don't think that's true. I think all but one of those answers is wrong. If I was on a rocket trip, and I had a little sister back home, whenever I was thinking about her, I would be sure that she was doing something right then. I wouldn't know then what it was she was doing right then, but I would know that she was doing something definite. And she couldn't be doing something definite without being some particular age right then. I'll never believe that's not true. If I could figure out how old she was right then, I could at least imagine what she might be doing right then. I might not know how to figure out her age, but I would know that she does have that definite age right then.

 

[ghwellsjr]

I don't think that's true. I think all but one of those answers is wrong. If I was on a rocket trip, and I had a little sister back home, whenever I was thinking about her, I would be sure that she was doing something right then. I wouldn't know then what it was she was doing right then, but I would know that she was doing something definite. And she couldn't be doing something definite without being some particular age right then. I'll never believe that's not true. If I could figure out how old she was right then, I could at least imagine what she might be doing right then. I might not know how to figure out her age, but I would know that she does have that definite age right then.

CADO has given you a false sense of security. You have bought into Michael Fontenot's dogmatism. Have you read the paper? Do you understand it? Can you work out the simple problem in post #212 that no one else seems willing or able to work out?

 

[jcsd]

I don't think Fontenot should be given any particular credit for 'CADO' it's just a an overly-simplistic method for creating spatially-extended accelerated frames, which is rarely used because of it's flaws.

The problems with Fontenot's method is that it produces badly-behaved spacetime cooridnates and it doesn't have any particular physical significance. I.e. it has problems mathematically and physically, so why bother with it?

 

[PAllen]

I don't think that's true. I think all but one of those answers is wrong. If I was on a rocket trip, and I had a little sister back home, whenever I was thinking about her, I would be sure that she was doing something right then. I wouldn't know then what it was she was doing right then, but I would know that she was doing something definite. And she couldn't be doing something definite without being some particular age right then. I'll never believe that's not true. If I could figure out how old she was right then, I could at least imagine what she might be doing right then. I might not know how to figure out her age, but I would know that she does have that definite age right then.

This belief is wrong. The idea of uniquely defined 'now' across the universe is anathema to relativity. If you take CADO literally, then in your rocket ship traveling away from Earth (and very far from earth), while you are walking toward the back of the ship, your little sister's son is graduating high school. Turn around and walk toward the front of the ship, and she isn't married yet. Pace back and forth on the rocket and your 'now' on Earth oscillates by decades (or centuries, or millenia, depending on how far away you are).

Note: it is not wrong to treat things thus, but is also not necessary at all.

 

[Austin0]

I don't think Fontenot should be given any particular credit for 'CADO' it's just a an overly-simplistic method for creating spatially-extended accelerated frames, which is rarely used because of it's flaws.

The problems with Fontenot's method is that it produces badly-behaved spacetime cooridnates and it doesn't have any particular physical significance. I.e. it has problems mathematically and physically, so why bother with it?

I think you have put it a nutshell here. CADO is essentially nothing but applying the fundamental transformations from a momentarily co-moving inertial frame. With all the resulting weirdness: intersecting planes of simultaneity etc.
Although Mike seems convinced it must represent "reality"

 

[Austin0]

CADO has given you a false sense of security. You have bought into Michael Fontenot's dogmatism. Have you read the paper? Do you understand it? Can you work out the simple problem in post #212 that no one else seems willing or able to work out?

Since the calculation is simply a Lorentz transformation from MCIRF's at various points in the circuit I have no doubt you could easily do it. It would require calculating instantaneous relative velocities and distances to the inertial observer's position orthogonal to the instantaneous velocity vector- (not direct distances) but with the requisite math skills that should be easy. But it is very unlikely that the calculated results could conform to any possible actual observations or have any particular physical meaning.
Still it would be interesting to see and might be instructive.

 

[PAllen]

I think you have put it a nutshell here. CADO is essentially nothing but applying the fundamental transformations from a momentarily co-moving inertial frame. With all the resulting weirdness: intersecting planes of simultaneity etc.
Although Mike seems convinced it must represent "reality"

Just to note that CADO, while not unique or preferred, is not ludicrous: it is just Fermi-Normal coordinates used for a non-inertial observer in SR. In GR, these coordinates are routinely used to describe a local region for an arbitrary observer. The difficulties arise trying to globalize them - but then any coordinate system may run into issues with global extension in GR. (In SR, there are approaches to global non-inertial coordinates that work for the general case excluding rotation). Note, in particular, that Rindler coordinates for an accelerating observer have many properties in common with the CADO scheme applied to a uniformly accelerating observer.

 

[ghwellsjr]

CADO has given you a false sense of security. You have bought into Michael Fontenot's dogmatism. Have you read the paper? Do you understand it? Can you work out the simple problem in post #212 that no one else seems willing or able to work out?

Since the calculation is simply a Lorentz transformation from MCIRF's at various points in the circuit I have no doubt you could easily do it. It would require calculating instantaneous relative velocities and distances to the inertial observer's position orthogonal to the instantaneous velocity vector- (not direct distances) but with the requisite math skills that should be easy.

Since it's so simple and easy, why don't you just work it out?

But it is very unlikely that the calculated results could conform to any possible actual observations or have any particular physical meaning.

But none of the other calculations of remote time conform to any possible actual observations or have any particular physical meaning either, so why pick on CADO? There's nothing wrong with CADO except the opinion that it is better than any of the others, or worse, that it is the only right one as Underwood expressed in post #215.

Still it would be interesting to see and might be instructive.

Go for it.

Please.

 

[jcsd]

Just to note that CADO, while not unique or preferred, is not ludicrous: it is just Fermi-Normal coordinates used for a non-inertial observer in SR. In GR, these coordinates are routinely used to describe a local region for an arbitrary observer. The difficulties arise trying to globalize them - but then any coordinate system may run into issues with global extension in GR. (In SR, there are approaches to global non-inertial coordinates that work for the general case excluding rotation). Note, in particular, that Rindler coordinates for an accelerating observer have many properties in common with the CADO scheme applied to a uniformly accelerating observer.

It's not ludricious, when I first learned about accelerated frames this is how I thoguht they should be constructed. However a little thought led me to realize that you don't get a well-behaved coordinate system as lines of simultaneity can cross and thoguh at first glance they seem like the most physical way to define an accelerated observer's coordinates, in actuality there's nothing special about them physically (Rindler coordinates do have some physical meaning as they are defined by a class of observer).

As Fontenot points out on his website this means that they're not even (necessarily) a coordinate chart (unlike Rindler coordinates).

 

[PAllen]

It's not ludricious, when I first learned about accelerated frames this is how I thoguht they should be constructed. However a little thought led me to realize that you don't get a well-behaved coordinate system as lines of simultaneity can cross and thoguh at first glance they seem like the most physical way to define an accelerated observer's coordinates, in actuality there's nothing special about them physically (Rindler coordinates do have some physical meaning as they are defined by a class of observer).

As Fontenot points out on his website this means that they're not even (necessarily) a coordinate chart (unlike Rindler coordinates).

Fermi-Normal coordinate chart is limited by intersection of simultaneity surfaces. This bounds the region in which it can form a valid coordinate chart. There's nothing wrong with a coordinate chart covering only part of spacetime. Mike Fontenot goes beyond this and tries to use CADO in areas outside the region they form a valid chart - on that score, I would actually call it ludicrous.

Obviously, I agree there is nothing special about them. Even for inertial observers in SR I favor the view that distant simultaneity is purely a matter of convention, not physics.

 

[jcsd]

Fermi-Normal coordinate chart is limited by intersection of simultaneity surfaces. This bounds the region in which it can form a valid coordinate chart. There's nothing wrong with a coordinate chart covering only part of spacetime. Mike Fontenot goes beyond this and tries to use CADO in areas outside the region they form a valid chart - on that score, I would actually call it ludicrous.

Obviously, I agree there is nothing special about them. Even for inertial observers in SR I favor the view that distant simultaneity is purely a matter of convention, not physics.

Yep, I wasn't meaning to imply there's anything wrong with a Fermi-Normal chart. Physically they driectly describe the accelerating observer and the region of spacetime 'near' to his worldline, for which you would expect him to be able to describe sensibly without too many issues.

 

[Austin0]

Since it's so simple and easy, why don't you just work it out?

But none of the other calculations of remote time conform to any possible actual observations or have any particular physical meaning either, so why pick on CADO? There's nothing wrong with CADO except the opinion that it is better than any of the others, or worse, that it is the only right one as Underwood expressed in post #215.

Go for it.

Please.

Note I said easy for YOU with your math skills. I on the other hand have forgotten even basic trig so this is a problem.

But using the previous parameters circle C=314 ls in station frame. v close to c then:

At 90 deg. Calculated time at station is 28.5 sec.

180 deg. 157sec.

270 deg. 285.5 sec.

You can interpolate from this or calculate the distance for the intermediate points and calculate from there.

I was not singling out CADO in this instance , the results of which are the same as normally calculated for specific lines of simultaneity where they intersect another world line.
In many cases they can be correlated to actual observations. Specifically; if the traveling frame is considered extended in space to encompass the object of interest then hypothetical actual observers or virtual observers would agree on the times on respective clocks as calculated.
But I am in complete agreement that this still does not provide any basis for establishing any meaningful temporal relationship between the distant points,,, even if it is not totally devoid of correspondence with possible observation as it is in cases where lines intersect.

 

[PAllen]

Note I said easy for YOU with your math skills. I on the other hand have forgotten even basic trig so this is a problem.

But using the previous parameters circle C=314 ls in station frame. v close to c then:

At 90 deg. Calculated time at station is 28.5 sec.

180 deg. 157sec.

270 deg. 285.5 sec.

You can interpolate from this or calculate the distance for the intermediate points and calculate from there.

I was not singling out CADO in this instance , the results of which are the same as normally calculated for specific lines of simultaneity where they intersect another world line.
In many cases they can be correlated to actual observations. Specifically; if the traveling frame is considered extended in space to encompass the object of interest then hypothetical actual observers or virtual observers would agree on the times on respective clocks as calculated.
But I am in complete agreement that this still does not provide any basis for establishing any meaningful temporal relationship between the distant points,,, even if it is not totally devoid of correspondence with possible observation as it is in cases where lines intersect.

The thing I would emphasize is that a 'line' (or hyper surface in full 4d) of simultaneity is not (ever) a physical observable, and is always a matter of convention. It is simply part of the basis of coordinate chart, which is purely a conventional choice. Observables are things like doppler, the appearance of a distant clock over a specified signal path, clock or velocity comparisons for adjacent objects etc. In flat spacetime, one can talk about relative velocity at a distance because parallel transport is path independent. Simultaneity or current time (or time rate) at a distance are never observables - they are interpretations based on conventions for which there are many (infinite) reasonable choices.

[EDIT: As Gwellsjr never tires of emphasizing, Einstein was very clear that his convention for simultaneity was just that: a stipulation or convention. What matters are the physical predictions computed using it. The rub, again, is you can replace it with other conventions and get all the same physical predictions.]

 

[Austin0]

The thing I would emphasize is that a 'line' (or hyper surface in full 4d) of simultaneity is not (ever) a physical observable, and is always a matter of convention. It is simply part of the bases of coordinate chart, which is purely a conventional choice. Observables are things like doppler, the appearance of a distant clock over a specified signal path, clock or velocity comparisons for adjacent objects etc. In flat spacetime, one can talk about relative velocity at a distance because parallel transport is path independent. Simultaneity or current time (or time rate) at a distance are never observables - they are interpretations based on conventions for which there are many (infinite) reasonable choices.

[EDIT: As Gwellsjr never tires of emphasizing, Einstein was very clear that his convention for simultaneity was just that: a stipulation or convention. What matters are the physical predictions computed using it. The rub, again, is you can replace it with other conventions and get all the same physical predictions.]

I am not sure what you are objecting to here. Certainly clock synchronization is purely a convention and carries no implication of actual simultaneity of separated clocks in the same frame.
But a surface of simultaneity is simply the frame of the point in question.
So the point of intersection of that line with another world line in many cases represents an actual observer with a clock showing the same time as the clock at the origen of that S line at that moment SO it represents an actual observation. In cases where the frame itself does not physically extend to the other world line eg. the twins scenario , then it is a matter of extending a virtual frame for virtual observations.
As I said this does not in any way imply actual simultaneity between the two clocks with equal proper time readings and therefore does not support any implication of an actual temporal relationship between the two separate points.

 

[PAllen]

I am not sure what you are objecting to here. Certainly clock synchronization is purely a convention and carries no implication of actual simultaneity of separated clocks in the same frame.
But a surface of simultaneity is simply the frame of the point in question.
So the point of intersection of that line with another world line in many cases represents an actual observer with a clock showing the same time as the clock at the origen of that S line at that moment SO it represents an actual observation. In cases where the frame itself does not physically extend to the other world line eg. the twins scenario , then it is a matter of extending a virtual frame for virtual observations.
As I said this does not in any way imply actual simultaneity between the two clocks with equal proper time readings and therefore does not support any implication of an actual temporal relationship between the two separate points.

A surface of simultaneity is just the instantiation of a simultaneity convention - it is every bit an artifact of convention. What it intersects has no bearing on the issue. Examine the bolded statement above. This is claiming simultaneity at a distance is an observable, and that is just wrong.

 

[Austin0]

A surface of simultaneity is just the instantiation of a simultaneity convention - it is every bit an artifact of convention. What it intersects has no bearing on the issue. Examine the bolded statement above. This is claiming simultaneity at a distance is an observable, and that is just wrong.

Not at all. At no point did I imply that a surface of simultaneity was anything but a convention. Nor do I think so. I also did not claim that simultaneity at a distance was observable as I don't think simultaneity at a distance is even remotely determinable by any means whatsoever.
I simply noted that conventional simultaneity as determined by system clocks was an observable under certain conditions of motion and not under other conditions of acceleration and changes of direction.
Now if I, in any way implied there was any real temporal meaning to those observations (i.e. simultaneity)
you would certainly be justified in your reaction but none was intended or explicit.

 

[PAllen]

Not at all. At no point did I imply that a surface of simultaneity was anything but a convention. Nor do I think so. I also did not claim that simultaneity at a distance was observable as I don't think simultaneity at a distance is even remotely determinable by any means whatsoever.
I simply noted that conventional simultaneity as determined by system clocks was an observable under certain conditions of motion and not under other conditions of acceleration and changes of direction.
Now if I, in any way implied there was any real temporal meaning to those observations (i.e. simultaneity)
you would certainly be justified in your reaction but none was intended or explicit.

and I'm expressing some disagreement: even for inertial motion in flat spacetime, distant simultaneity is never an observable, only a convention. In other words, I disagree with the bolded statement above. Of course, I think there is a very useful convention for this case, but it is still not an observable, in any sense.

 

[Austin0]

and I'm expressing some disagreement: even for inertial motion in flat spacetime, distant simultaneity is never an observable, only a convention. In other words, I disagree with the bolded statement above. Of course, I think there is a very useful convention for this case, but it is still not an observable, in any sense.

OK it is now clear . This is PURELY a semantic quibble over the meaning of the word observable.
AS the bolded statement included the conventionality of the observed clock readings it was only the word observable you questioned. correct?
So then conventional simultaneity is based on observations but is not observable.
By this then length measurements of moving objects are also not observables being based on conventionally synchronized clocks, right??
I am certainly aware of the conceptual distinction so will now be careful of the terminology.

 

[PAllen]

OK it is now clear . This is PURELY a semantic quibble over the meaning of the word observable.

So then conventional simultaneity is based on observations but is not observable.
By this then length measurements of moving objects are also not observables being based on conventionally synchronized clocks, right??
I am certainly aware of the conceptual distinction so will now be careful of the terminology.

Exactly. Determination of the distance between two moving objects depends on simultaneity convention, and thus is not really an observation as I understand it - it is an interpretation of a series of observations mediated by a convention.

 

[ghwellsjr]

Since it's so simple and easy, why don't you just work it out?

But none of the other calculations of remote time conform to any possible actual observations or have any particular physical meaning either, so why pick on CADO? There's nothing wrong with CADO except the opinion that it is better than any of the others, or worse, that it is the only right one as Underwood expressed in post #215.

Go for it.

Please.

Note I said easy for YOU with your math skills. I on the other hand have forgotten even basic trig so this is a problem.

But using the previous parameters circle C=314 ls in station frame. v close to c then:

At 90 deg. Calculated time at station is 28.5 sec.

180 deg. 157sec.

270 deg. 285.5 sec.

You can interpolate from this or calculate the distance for the intermediate points and calculate from there.

According to the CADO reference for his equation:

CADO_T = CADO_H - v * L

But in any case, once those three quantities have been determined (for any given age of the traveler), the quantity CADO_T can be determined from the same CADO equation, with (as always) only a single multiplication and a single addition or subtraction.

The way the three quantities v, CADO_H, and L can be determined, for each instant of the traveler's life, will only be very briefly described here. Since all three quantities correspond to the conclusions of a perpetually-inertial observer (the "home-twin"), their determination is fairly widely known.

He makes it sound so easy but I haven't been able to figure it out from a brief reading of the paper. I don't know if your calculations are compatible with CADO, but either way, I don't know what you did and I don't know what he's doing.

 

[Austin0]

According to the CADO reference for his equation:

CADO_T = CADO_H - v * L

He makes it sound so easy but I haven't been able to figure it out from a brief reading of the paper. I don't know if your calculations are compatible with CADO, but either way, I don't know what you did and I don't know what he's doing.

CADO_H is simply the coordinate time at the location (in the station frame in this case)

SO at 90deg. the local time is 78.5
L is the distance in the station coordinates
so at 90deg.the velocity vector would orthogonally intersect the position of the station observer (the tangent at the station point) so the distance in the station frame would be equal to the radius 50 ls ...multiplied by v =approx 50 sec this is just the lorentz math for determining relative clock desynchronization.between frames.
subtracting this from 78.5 gives CADO_T=28.5 ..the calculated simultaneous time at the station.
At 180 deg. the traveler shares simultaneity for an instant (equivalent to the moment of zero velocity at turnaround in a normal linear situation) so the CADO_T is just the local time.
Or alternately it is the calculated station time to reach that point D/v in station coordinates
This should be enough for you to recognize the underlying transformation with the relabeling.

 

[GrammawSally]

SO at 90deg. the local time is 78.5
L is the distance in the station coordinates
so at 90deg.the velocity vector would orthogonally intersect the position of the station observer (the tangent at the station point) so the distance in the station frame would be equal to the radius 50 ls ...multiplied by v =approx 50 sec [...]

L is the vector position of the traveler, relative to the inertial person, according to the inertial person. v is the vector velocity of the traveler, relative to the inertial person, according to the inertial person.

So if the inertial person's position on the circle is taken as zero degrees, and if the traveler's position is momentarily at 90 degrees (CCW), then the L and v vectors will be neither perpendicular nor parallel, and L will have a magnitude greater than the radius of the circle. The dot product of L and v will be nonzero, and will have a magnitude less than the product of the magnitudes of L and v.

Maybe I misunderstood you, but it didn't sound to me like that's what you were doing.

 

[ghwellsjr]

L is the vector position of the traveler, relative to the inertial person, according to the inertial person. v is the vector velocity of the traveler, relative to the inertial person, according to the inertial person.

So if the inertial person's position on the circle is taken as zero degrees, and if the traveler's position is momentarily at 90 degrees (CCW), then the L and v vectors will be neither perpendicular nor parallel, and L will have a magnitude greater than the radius of the circle. The dot product of L and v will be nonzero, and will have a magnitude less than the product of the magnitudes of L and v.

Maybe I misunderstood you, but it didn't sound to me like that's what you were doing.

Your explanation I can understand, but is it the same as Michael Fontenot's CADO scheme? He nowhere mentions the dot product of any vectors or even that L and v are vectors.

 

[ghwellsjr]

Can you please do the calculations for several points around the earth, say every 45 degrees? Assume a train speed of .99c, an Earth radius of 4000 miles and a light speed of 186,000 miles per second. (I would really like to see it done every 15 degrees but I don't want to overburden you.)

Left as an exercise for the reader.

If I did it right, here is a graph depicting what I think you described as the CADO scheme. It will take the father 135 milliseconds to go around the Earth once. But according to CADO, he will not determine that his son is aging hardly at all as he departs him and as he approaches him at the end of the loop. He determines that his son ages most quickly when he is on the opposite side of the Earth and it is just enough so that when he gets back to his son, they agree that he has aged 135 milliseconds.

Does this look correct?

 

[PAllen]

If I did it right, here is a graph depicting what I think you described as the CADO scheme. It will take the father 135 milliseconds to go around the Earth once. But according to CADO, he will not determine that his son is aging hardly at all as he departs him and as he approaches him at the end of the loop. He determines that his son ages most quickly when he is on the opposite side of the Earth and it is just enough so that when he gets back to his son, they agree that he has aged 135 milliseconds.

Does this look correct?

Looking at the linked web site, yes that looks correct to me. If you change the horizontal from CADO_H to t (time as experienced for the traveler), the shape would be the same, but greatly squeezeed - still near horizontal at beginning and end, but very steep in the middle.

 

[Austin0]

L is the vector position of the traveler, relative to the inertial person, according to the inertial person. v is the vector velocity of the traveler, relative to the inertial person, according to the inertial person.

So if the inertial person's position on the circle is taken as zero degrees, and if the traveler's position is momentarily at 90 degrees (CCW), then the L and v vectors will be neither perpendicular nor parallel, and L will have a magnitude greater than the radius of the circle. The dot product of L and v will be nonzero, and will have a magnitude less than the product of the magnitudes of L and v.

Maybe I misunderstood you, but it didn't sound to me like that's what you were doing.

Well you may be right but try this:

At 90 deg assume a complete coordinate chart for the traveler. One axis tangent at that point and another orthogonal through the center of the circle.

The relative simultaneity is what would be observed by a traveling observer at the station at that instant. This is determined by the distance between the two travelers along the vector of motion in their frame. Simply dx , not the direct distance sqrt(dx^2+dy^2)

Another way to look at it is:
The lines of the travelers simultaneity are orthogonal to the instantaneous direction of travel so at 90 deg.the relevant line would be congruent to the tangent at the station point. So this dx (in the traveler frame) times gamma gives L
Or alternately it is clear that geometrically L =the radius in station coordinates.
Remember this is all calculating simultaneity from the traveling frame.
it is just convenience that values are those of the inertial frame. The magnitude of clock desynchronization (relative simultaneity) for a distant clock is v times the proper distance in that frame.
Suppose the traveler was on a linear course tangent to 90 deg. At that point sees the local time there. Based on this how would you calculate the simultaneous time at 0 deg according to the traveler
. ??
Would you think it relevant to consider the straight line distance to 0 deg?
Or the instantaneous vector velocity as calculated by an observer at 0 deg.?

So if I am wrong in this approach I hope the correct approach is explained.

 

[Austin0]

If I did it right, here is a graph depicting what I think you described as the CADO scheme. It will take the father 135 milliseconds to go around the Earth once. But according to CADO, he will not determine that his son is aging hardly at all as he departs him and as he approaches him at the end of the loop. He determines that his son ages most quickly when he is on the opposite side of the Earth and it is just enough so that when he gets back to his son, they agree that he has aged 135 milliseconds.

Does this look correct?

Looks fine . What was your procedure for calculating L and v ??

 

[ghwellsjr]

Looks fine . What was your procedure for calculating L and v ??

I just followed GrammawSally's super clear directions:

L is the vector position of the traveler, relative to the inertial person, according to the inertial person. v is the vector velocity of the traveler, relative to the inertial person, according to the inertial person.

I still don't know if this is the same as Michael Fontenot's scheme. Can anybody confirm that?

 

[Austin0]

I just followed GrammawSally's super clear directions:

I still don't know if this is the same as Michael Fontenot's scheme. Can anybody confirm that?

Super clear to you but much less so to me. Hopefully you will help shed some light.

If the inertial home twin is located somewhere on the circular track, then the traveler can compute the dot product of v and L at any point on his circuit, and then use the CADO equation to compute the home twin's age at that instant in his (the traveler's) life.

Here she is talking about calculations from the traveler frame
She doesn't detail how v and L are determined.
I am just starting to study dot products so could you help put the use here in a geometric conceptual context?
In a linear situation vL is a simple product of values.

L is the vector position of the traveler, relative to the inertial person, according to the inertial person. v is the vector velocity of the traveler, relative to the inertial person, according to the inertial person.

Here she is now talking about calculations from the inertial frame.
When she says vector velocity is this the direction of the tangent to the circle at that point and the magnitude of speed relative to the inertial observer ?
If not what?
Is L the direct line distance along the vector connecting them according to the inertial observer??

So if the inertial person's position on the circle is taken as zero degrees, and if the traveler's position is momentarily at 90 degrees (CCW), then the L and v vectors will be neither perpendicular nor parallel, and L will have a magnitude greater than the radius of the circle. The dot product of L and v will be nonzero, and will have a magnitude less than the product of the magnitudes of L and v.

since she outlined two apparently different approaches which one did you use?
Any clarification would be appreciated.

 

[PAllen]

Austin0:

In Euclidean geometry, dot product of two vectors is simply product of their lengths with cosine of angle between them. If two vectors are parallel, it is thus the product of their lengths; if perpendicular, it is zero. More generally (in non-Euclidean geometry, e.g. for 4-vectors in SR spacetime), it is contraction of both vectors with the metric tensor. The CADO website specifies that L, v, and CADO_H are all as measured by the distant inertial observer. The CADO website requires only the Euclidean definition.

 

[ghwellsjr]

Super clear to you but much less so to me. Hopefully you will help shed some light.

Here she is talking about calculations from the traveler frame
She doesn't detail how v and L are determined.
I am just starting to study dot products so could you help put the use here in a geometric conceptual context?

It seems from your comments below, along with PAllen's explanation of the dot product, that you should have a good handle on this.

In a linear situation vL is a simple product of values.

In Michael Fontenot's CADO paper, he first discusses the "standard twin paradox":

The traveler then instantaneously reverses course, and spends the next 20 years of his life returning to his home-twin. The magnitude of his velocity is still 0.866 ly/y, but since he is now moving toward his twin, by convention his velocity is now negative, -0.866 ly/y.

I couldn't find where he explained why the velocity takes on a negative value at the turn-around point but GrammawSally's explanation of the dot product makes it automatic because the velocity vector points in the opposite direction so it isn't merely "a simple product of values".

Here she is now talking about calculations from the inertial frame.

Yes, she made that very clear. They're all from the inertial frame. If Michael ever says the same thing, it is not clear at all, partly because he is doing the same calculation for the "standard twin paradox" in the conventional way from the frame of the home twin where he introduces gamma but as far as I can tell, it is never used in his CADO equation. I could never tell whether he was discussing the home twin's frame just for the conventional explanation or also for the CADO explanation.

When she says vector velocity is this the direction of the tangent to the circle at that point and the magnitude of speed relative to the inertial observer ?

Yes.

Is L the direct line distance along the vector connecting them according to the inertial observer??

Yes. You then have to take the difference between the two vector angles to use in the dot product which simply means you multiply the magnitudes of the two vectors together and then multiply by the cosine of the delta angle.

So if the inertial person's position on the circle is taken as zero degrees, and if the traveler's position is momentarily at 90 degrees (CCW), then the L and v vectors will be neither perpendicular nor parallel, and L will have a magnitude greater than the radius of the circle. The dot product of L and v will be nonzero, and will have a magnitude less than the product of the magnitudes of L and v.

since she outlined two apparently different approaches which one did you use?

I don't know why you think this is a different approach. She is simply providing some background to the setup of the scenario and then pointing out why you cannot just multiply L and v together (as Michael Fontenot implies, if not directly states, that you can) but rather why you need to treat L and v as vectors and use the dot product.

Any clarification would be appreciated.

Hopefully this has helped but if you need more clarification, just ask.