Understanding Time and Space as One entity.

[matheinste]

Hello Antenna Guy.

I think perhaps where you are going wrong is by considering each observer as presiding over his own exclusive reference frame with himself at the origin of any coordinate system which he ascribes to this frame. An observer is free to choose a coordinate system for the reference frame and to consider himself as being at the origin of such a coordinate system, but no other observer in the same frame can also at the origin unless he is colocated with the frst observer.

Matheinste.

 

[Dale]

If you are correct that is fine, we simply use the http://en.wikipedia.org/wiki/Poincar%C3%A9_group" instead of the Lorentz transform. The Poincare group includes translations and spatial rotations in addition to boosts. Although they will then disagree about the coordinates assigned to any given event two observers at rest wrt each other will agree on the time between any two events as well as the spatial distance between them. They will therefore also agree on any speed measurement.

 

[matheinste]

Hello DaleSpam.

Looks good to me but perhaps a bit technical. I think Antenna Guy needs to be clear about what constitutes an inertial reference frame.

Matheinste.

 

[CaptainQuasar]

Two observers that are at rest wrt each other have the same reference frame.

How do you figure that?

Isn't that the exact definition of an inertial reference frame? The set of things that are at inertial rest with respect to each other?

Consider Einstein's clock synchronization procedure for any pair of clocks. Are the two clocks not on a sphere of constant radius about the flash when they set their times? If I add a third clock at the midpoint of the other two and repeat the synchronization procedure, I what frame do the three clocks simultaneously show the same time?

Answer: in the frame of the third clock.

If I understand things properly, distance doesn't have anything to do with SR, just inertial reference frames. Two observers in the same inertial reference frame could be a galaxy's width apart¹ in intergalactic space and they'd be able to see each other (yes, with a delay for the light to travel) but they would not perceive any relativistic effects.

Light cones for talking about causality involve distance and then that subject also discusses SR and GR too… is that what you're thinking of, Antenna Guy? Hey the rest of you guys… when you're looking at the 2D or 3D diagram of a light cone, can't you measure interval vectors within the diagram and based upon whether the interval vector has a steeper or more shallow slope than the surface of the light cone decide some property related to whether it's a spacelike or timelike interval?

¹Not that galaxies have standardized widths.

⚛​

 

[Dale]

can't you measure interval vectors within the diagram and based upon whether the interval vector has a steeper or more shallow slope than the surface of the light cone decide some property related to whether it's a spacelike or timelike interval?

Yes, if it is inside the light cone it is timelike, if it is outside the light cone it is spacelike, and if it is on the light cone it is lightlike or null. Since another way of restating the second postulate is that the light cone is preserved then no change of inertial reference frame will change the spacelike, timelike, or lightlike designation.

 

[belliott4488]

Yes, if it is inside the light cone it is timelike, if it is outside the light cone it is spacelike, and if it is on the light cone it is lightlike or null. Since another way of restating the second postulate is that the light cone is preserved then no change of inertial reference frame will change the spacelike, timelike, or lightlike designation.

To clarify, this is true for a vector between an event at the origin, or apex of the light cone, and a second event that is located as DaleSpam described. You could have two events with a space-like separation that both lay with in a given light cone, provided that the slope of line between them was less than c, i.e. the angle of the cone. Just wanted to be what what "it" referred to in DaleSpam's post ...

 

[Antenna Guy]

let the first observer follow the worldline (t, 1, 2, 3) in "the first" reference frame and let the second observer follow the worldline (t', 3, 6, 9) in "the second" reference frame and let the event of interest be (0, 2, 4, 6) in the coordinates of "the first" reference frame.

t' = t = 0
x' = x = 2
y' = y = 4
z' = z = 6

so the coordinates are also (0, 2, 4, 6) in "the second" reference frame.

Since you seem to have chosen your world-lines arbitrarily within some third coordinate system, is there any reason not to use (t,-1,-2,-3) and (t',1,2,3) as the observer world-lines, and (0,0,0,0) as the event?

anyhoo, each world line [allegedy] describes a direction in 4-space. If I were to substitute 0 for t and t' in your observer world lines, I would have what amounts to three simultaneous positions in 4-space (two observers + one event).

SR implies that translations are not strictly spatial - they are spatio-temporal; and as such, limited in rate (velocity). If "information" (I like that descriptor) is to travel from an event to an observer, it starts at a point in 4-space, and ends on an observer's "world-line". Since the shortest distance between two points is a straight line (please don't bring up spheres, etc. ), let's look at the delta between the event and each observer as you've characterized them.

observer_1-event=(t,1,2,3)-(0,2,4,6)=(t,-1,-2,-3)

observer_2-event=(t',3,6,9)-(0,2,4,6)=(t',1,2,3)

The (average) velocity of information in the respective cases is:

event->observer_1: (1,-1/t,-2/t,-3/t)

event->observer_2: (1,1/t',2/t',3/t')

Thus, information is traveling from our event in opposite spatial directions along a new world-line (that of the event) to get to our two observers. In order to express this transferrence of information in terms of a single world-line, we have t'=-t (n.b. this is a very specific case).

Now, why might these two observers disagree regarding the velocity of some new event that occurs elsewhere in space-time...

Let's arbitrarily choose the location of our new event (denoted event_new) to be (0,1,2,6), and have it traveling along a path parallel to the ray passing through our two observers. Expressed in observer 1's frame, I think (1,1/t,2/t,3/t) will do for the velocity.

As before, let's look at the 4-space displacements first:

observer_1-event_new=(t,1,2,3)-(0,1,2,6)=(t,0,0,-3)

observer_2-event_new=(t',3,6,9)-(0,1,2,6)=(t',2,4,3)

Now, let's look at the displacements after some differential time has passed:

observer_1-event_new=(t+dt,1,2,3)-{(0,1,2,6)+dt*(1,1/t,2/t,3/t)}
=(t,-dt/t,-2dt/t,-3-3dt/t)

observer_2-event_new=(t'+dt',3,6,9)-{(0,1,2,6)+dt'*(1,1/t',2/t',3/t')}
=(t',2-dt'/t',4-2dt'/t',3-3dt'/t')

During this differential time interval, 4-space deltas are then:

observer_1->(event_new'-event_new): (0,-dt/t,-2dt/t,-3dt/t)

observer_2->(event_new'-event_new): (0,-dt'/t',-2dt'/t',-3dt'/t')

This is somewhat problematic from a 4-velocity standpoint (spatial displacement in zero time in either frame).

Near as I can tell, the two spatially displaced observers will agree on the spatial direction of the event, but that's about it (generally).

It appears implicit that some differential amount of time must pass between the event and each observer for the velocity vectors to become meaningful. I think the Lorentz transform accomplishes this by mapping dt from the observer frame to dt in the event frame - leaving a net time displacement in the above intervals.

BTW - since both our observers agree on space, time, and the speed of light, is it not obvious that the two observers will record the 4-space properties of our new event at different positions on the event world-line wherever the observers' spatial displacements from the event differ?

Regards,

Bill

 

[belliott4488]

Bill, you're not going to like this, so I hope you'll take it in the spirit in which it's offered.

You do not understand Special Relativity. You can continue to argue with people until they go away and leave in frustration, or you can accept the possibility that you've got some misconceptions and maybe try to learn something. I hope you'll try the latter, since you seem genuinely interested and have a basic grasp of some of the concepts. You're simply making too many incorrect statements, though - you need to get clearer on much of this.

Since you seem to have chosen your world-lines arbitrarily within some third coordinate system, is there any reason not to use (t,-1,-2,-3) and (t',1,2,3) as the observer world-lines, and (0,0,0,0) as the event?

anyhoo, each world line [allegedy] describes a direction in 4-space. If I were to substitute 0 for t and t' in your observer world lines, I would have what amounts to three simultaneous positions in 4-space (two observers + one event).

[Edited] I'm not sure what you mean by "simultaneous positions". A "position" in Minkowsky space has four components, which makes it an event. That's what any point in Minkowsky space is. So, to talk about 3 positions, only one of which is an event, makes no sense. Three events can certainly be simultaneous in one reference frame, in which case they will all have the same time coordinate, so by setting t = t' = 0, perhaps you meant simultaneous events?

SR implies that translations are not strictly spatial - they are spatio-temporal; and as such, limited in rate (velocity).

I'm not sure where you get this, but it's not strictly true. There is nothing wrong with spatial translations within a given reference system; they're translations that happen in the 3-d subspace spanned by the spatial axes. As long as you don't translate along the time direction, the translation is spatial. Since the space-like nature of a displacement 4-vector is an invariant, however, this translation will be space-like in all reference frames. It will be simultaneous in only one reference frame, however; in others it will combine a shift in time as well.

In any case, translations are not "limited in rate." You can mathematically translate any point to another by a suitable translation. Whether or not this corresponds to physical motion with a velocity relative to some coordinate frame depends on the translation. A translation consisting of time-like displacements of space-time events could indeed represent motion with velocity, the magnitude of which would depend on the chosen coordinate frame. For one frame, the translation would be a pure time translation, i.e. no spatial motion at all.

If "information" (I like that descriptor) is to travel from an event to an observer, it starts at a point in 4-space, and ends on an observer's "world-line". Since the shortest distance between two points is a straight line (please don't bring up spheres, etc. ), let's look at the delta between the event and each observer as you've characterized them.

This is true in Euclidean space, but not in Minkowsky space. Consider two events with a light-like separation - the "distance" between them, i.e. the space-time interval, is zero, regardless of how much time has elapsed between them in any particular reference frame. Also, as has been pointed out, time-like intervals have negative magnitudes, so talking about the "shortest distance" doesn't have the same meaning it has in Euclidean space.

observer_1-event=(t,1,2,3)-(0,2,4,6)=(t,-1,-2,-3)

observer_2-event=(t',3,6,9)-(0,2,4,6)=(t',1,2,3)

The (average) velocity of information in the respective cases is:

event->observer_1: (1,-1/t,-2/t,-3/t)

event->observer_2: (1,1/t',2/t',3/t')

You have three points in M. space, i.e. three events. You can certainly talk about their vector components in a particular reference frame, as you have, and then you can calculate the interval between any two of them, which you have not. The interval you get in each case is invariant, i.e. it will be the same regardless of which reference frame you choose to express the vectors in - this is the analog of the vector magnitude in Euclidean space.

There is no meaning to the "velocity of information" in this case, since you have not spoken of anything propagating from one event to any other. If you do, however, then by examining the interval between the events, you can say whether it is space-like, time-like, or light-light, which will be invariant, i.e. the same in all reference frames. Only time-like propagation is physical.

I'm going to drop it here, since you can see where this is going - I could continue picking at your statements, but I suspect we'd both lose interest.

Please don't take offense; I don't mean to be insulting. In fact, there is nothing insulting about the suggestion that someone doesn't understand SR, since most people don't. It's just that based on my experience teaching SR, right now I'd say you have not yet mastered it.

 

[Dale]

Since you seem to have chosen your world-lines arbitrarily within some third coordinate system, is there any reason not to use (t,-1,-2,-3) and (t',1,2,3) as the observer world-lines, and (0,0,0,0) as the event?

Not an important reason. It is just that, after proving that it applied to all events in general I didn't want to do an example using a "special" event like the origin.

anyhoo, each world line [allegedy] describes a direction in 4-space. If I were to substitute 0 for t and t' in your observer world lines, I would have what amounts to three simultaneous positions in 4-space (two observers + one event).

Yes.

SR implies that translations are not strictly spatial - they are spatio-temporal; and as such, limited in rate (velocity). If "information" (I like that descriptor) is to travel from an event to an observer, it starts at a point in 4-space, and ends on an observer's "world-line". Since the shortest distance between two points is a straight line (please don't bring up spheres, etc. ), let's look at the delta between the event and each observer as you've characterized them.

observer_1-event=(t,1,2,3)-(0,2,4,6)=(t,-1,-2,-3)

observer_2-event=(t',3,6,9)-(0,2,4,6)=(t',1,2,3)

OK, I follow you so far.

The (average) velocity of information in the respective cases is:

event->observer_1: (1,-1/t,-2/t,-3/t)

event->observer_2: (1,1/t',2/t',3/t')

I don't know how you arrived at this, but it looks wrong to me. If (the usual case) information is sent via an EM pulse then the speed will always be c. If the information is sent via some material object (e.g. via FedEx) then the speed will be the speed of the material object, and if that object is traveling inertially then that speed will be constant. Here, since your velocity is a function of time and not c it seems that you are transfering information via some accelerating material object. Also, the velocity function you are using here is undefined for t=t'=0, and for small values of t it is faster than light.

Thus, information is traveling from our event in opposite spatial directions along a new world-line (that of the event) to get to our two observers.

First, events do not have world-lines. Events are the spacetime equivalent of points. Second, the information will travel (usually at c) along two generally different worldlines to reach each observer. The intersection of the information's worldlines with each observer's worldline are other events (the "signal reception" events). The locations and times of the signal reception events are different from the location and time of the signal transmission event. None of this is problematic since intelligent observers always work backwards to determine when and where the transmission event occured. This is the "correcting for perception" that we discussed earlier.

In order to express this transferrence of information in terms of a single world-line, we have t'=-t (n.b. this is a very specific case).

Yes, this is a special case. Since in general t'=t then the condition t'=-t implies t'=-t=-t'=0. I don't know why you want to restrict your analysis to this specific case, particularly since your velocity is undefined at t=0, but OK.

Now, why might these two observers disagree regarding the velocity of some new event that occurs elsewhere in space-time...

Let's arbitrarily choose the location of our new event (denoted event_new) to be (0,1,2,6), and have it traveling along a path parallel to the ray passing through our two observers. Expressed in observer 1's frame, I think (1,1/t,2/t,3/t) will do for the velocity.

Again, events don't travel, and again observer 1 and observer 2 are stationary wrt each other so they share the same reference frame. But, OK, I follow.

As before, let's look at the 4-space displacements first:

observer_1-event_new=(t,1,2,3)-(0,1,2,6)=(t,0,0,-3)

observer_2-event_new=(t',3,6,9)-(0,1,2,6)=(t',2,4,3)

OK

Now, let's look at the displacements after some differential time has passed:

observer_1-event_new=(t+dt,1,2,3)-{(0,1,2,6)+dt*(1,1/t,2/t,3/t)}
=(t,-dt/t,-2dt/t,-3-3dt/t)

observer_2-event_new=(t'+dt',3,6,9)-{(0,1,2,6)+dt'*(1,1/t',2/t',3/t')}
=(t',2-dt'/t',4-2dt'/t',3-3dt'/t')

During this differential time interval, 4-space deltas are then:

observer_1->(event_new'-event_new): (0,-dt/t,-2dt/t,-3dt/t)

observer_2->(event_new'-event_new): (0,-dt'/t',-2dt'/t',-3dt'/t')

OK, both observers agree on your displacement measurements.

This is somewhat problematic from a 4-velocity standpoint (spatial displacement in zero time in either frame).

Of course you get a 0 time difference, you added the dt to the worldline expression (essentially an event on the future of the worldline of the observer). Also, none of your velocities in this post are 4-velocities, they are not Minkowski 4-vectors of any kind. That is not a big problem since you are working in a single frame, but just FYI your "delta displacements" will not Lorentz transform properly. Given that we are working in a single frame and doing no transforms I won't worry about that.

Near as I can tell, the two spatially displaced observers will agree on the spatial direction of the event, but that's about it (generally).

How do you figure that? You haven't shown anything that they disagree on. Even using some strange concepts like your "delta displacement" and your accelerating information.

It appears implicit that some differential amount of time must pass between the event and each observer for the velocity vectors to become meaningful. I think the Lorentz transform accomplishes this by mapping dt from the observer frame to dt in the event frame - leaving a net time displacement in the above intervals.

Events don't have frames. Also, I am not sure what your point about a differential amount of time and "meaningful" is about. The concept of the velocity at an instant is well defined. It is the limit as your differential time goes to zero.

BTW - since both our observers agree on space, time, and the speed of light, is it not obvious that the two observers will record the 4-space properties of our new event at different positions on the event world-line wherever the observers' spatial displacements from the event differ?

Again, events do not have worldlines. The bottom line is that if two observers agree on t, x, y, and z then they must agree on any f(t,x,y,z). This includes any 4-velocities, 4-accelerations, and even your displacements and delta-displacements. This isn't any complicated physics, just very basic math. Your conclusion is completely backwards and has not been substantiated by anything you have done in this post. Actually, everything you have done in this post contradicts this conclusion.

It is honestly difficult for me to know where to go from here in this conversation. You do a bunch of strange work that confirms my claims and then claim that the opposite is obvious. I don't think there is anything I can do to help you out here.

 

[Antenna Guy]

Bill, you're not going to like this, so I hope you'll take it in the spirit in which it's offered.

You do not understand Special Relativity.

Well there's a resounding endorsement!

Seriously - I'm not offended. I've spent the last 20yrs working with antennas, not relativity. I just see a lot of parallels regarding interpretations (even if it is awkward trying to relate how I see them as such). If I learn something new or correct something that is wrong with my understanding while trying to describe these parallels, it's for the better. However, as I'm sure any other "old dog" on this forum will attest, it can be difficult (though not impossible) to teach me new tricks.

BTW - Have you ever thought of a near-field antenna measurement as analogous to an array of observers about a single event? Never mind...

Keep up the good work.

Regards,

Bill

 

[belliott4488]

The (average) velocity of information in the respective cases is:

event->observer_1: (1,-1/t,-2/t,-3/t)

event->observer_2: (1,1/t',2/t',3/t')

I don't know how you arrived at this, but it looks wrong to me. If (the usual case) information is sent via an EM pulse then the speed will always be c. If the information is sent via some material object (e.g. via FedEx) then the speed will be the speed of the material object, and if that object is traveling inertially then that speed will be constant. Here, since your velocity is a function of time and not c it seems that you are transfering information via some accelerating material object. Also, the velocity function you are using here is undefined for t=t'=0, and for small values of t it is faster than light.

No, I think what he's trying to do is to define some kind of rate by dividing the 4-displacement by the time difference between the events at the endpoints, in other words, dividing the components of a vector by the first component. It's not a velocity, but it's not an acceleration either, since he is not taking t (or t') to be a variable, but rather the fixed time difference (in this frame) between the events in question. That's why he calls it an "average velocity." Not that it is a velocity ...

It is honestly difficult for me to know where to go from here in this conversation. You do a bunch of strange work that confirms my claims and then claim that the opposite is obvious. I don't think there is anything I can do to help you out here.

That's a reasonable conclusion, but I'm offering to take up the ball for a post or two, if Bill is open to rethinking his understanding of SR a bit, which I believe to be more than a little flawed.

 

[Dale]

I've spent the last 20yrs working with antennas, not relativity.

Do you have some experience with radar measurements?

 

[Dale]

No, I think what he's trying to do is to define some kind of rate by dividing the 4-displacement by the time difference between the events at the endpoints, in other words, dividing the components of a vector by the first component.

Oh, OK. I couldn't make sense of it so I just thought it was an arbitrary function that he was taking to be the "velocity". (in much the same manner that I arbitrarily picked a couple of worldlines for observers)

The point remains that if they agree on t, x, y, and z they will agree on whatever weird f(t,x,y,z) you care to devise. So the fact that his expression didn't make a lot of sense to me also didn't concern me much.

 

[belliott4488]

Well there's a resounding endorsement!

Seriously - I'm not offended. I've spent the last 20yrs working with antennas, not relativity. I just see a lot of parallels regarding interpretations (even if it is awkward trying to relate how I see them as such). If I learn something new or correct something that is wrong with my understanding while trying to describe these parallels, it's for the better. However, as I'm sure any other "old dog" on this forum will attest, it can be difficult (though not impossible) to teach me new tricks.

Well, thanks for being so reasonable! Most people on the internet would have responded with a diatribe in all CAPS, but I was kind of hoping you weren't one of those ...

BTW - Have you ever thought of a near-field antenna measurement as analogous to an array of observers about a single event? Never mind...

hm ... as a matter of fact ... no. But the wavefront leaving a phased array kind of reminds me of the way a moving rod crosses the x-axes of different observers ... more on that later??

So, while we're all being so reasonable, how about this: Can you tell me how you'd draw the axes for the frame of an observer moving with respect to a stationary observer? In other words, start with a 2-d space-time diagram with the usual x and t axes, and then draw x' and t' axes corresponding to the the rest frame of a moving observer, as given by a Lorentz transformation. I don't mean to be pedantic; I'd just like to confirm that we're all starting from common ground.

 

[Antenna Guy]

I think what he's trying to do is to define some kind of rate by dividing the 4-displacement by the time difference between the events at the endpoints, in other words, dividing the components of a vector by the first component.

That would be accurate.

Regards,

Bill

 

[Antenna Guy]

Do you have some experience with radar measurements?

Not really. I have a basic understanding of how it works, but that's about it.

Regards,

Bill

 

[Antenna Guy]

the wavefront leaving a phased array kind of reminds me of the way a moving rod crosses the x-axes of different observers ... more on that later??

If I understand you correctly, take a line of elements along x and assign a linear phase taper to them, symmetric about the center. Draw a circle about the end-most elements such that r=phase*wavelength/(2*pi) (phase in radians, obviously). Now draw a line tangential to both circles such that positive phase is the "front" of the array, and negative is "behind".

Would that line look like the rod you're thinking of?

FWIW: If you were to compute the composite radiation pattern of this array assuming all elements are in the same frame, and spatial displacements don't matter, you would not get the correct result. Hence my issue with ignoring displacements.

Regards,

Bill

[addendum: Consider that an antenna has the same radiation pattern regardless of whether it is emitting energy (elements as events), or receiving it (elements as observers). Which might help explain why I deal with observers and events the way I do.]

 

[belliott4488]

If I understand you correctly, take a line of elements along x and assign a linear phase taper to them, symmetric about the center. Draw a circle about the end-most elements such that r=phase*wavelength/(2*pi) (phase in radians, obviously). Now draw a line tangential to both circles such that positive phase is the "front" of the array, and negative is "behind".

Would that line look like the rod you're thinking of?

Um ... something like that. But it's more of a cartoon than a real similarity. I'm thinking of how you draw the world sheet of a moving rod and then look it at from the point of view of two different observers to see the length contraction.

FWIW: If you were to compute the composite radiation pattern of this array assuming all elements are in the same frame, and spatial displacements don't matter, you would not get the correct result. Hence my issue with ignoring displacements.

Of course they're in the same frame ... where else would they be? You can express their positions in any frame you like, but there's no special frame that "contains" them ...

Otherwise, I don't see your point. Obviously one can compute the radiation pattern of a phased array radar - how else could they be used? If you don't get the correct result, then you're just not doing the calculation correctly. ?

[addendum: Consider that an antenna has the same radiation pattern regardless of whether it is emitting energy (elements as events), or receiving it (elements as observers). Which might help explain why I deal with observers and events the way I do.]

I understand that you can use the same pattern for radiating or receiving, but no, that does not clarify for me why you talk about observers the way you do.

I think if you want to talk about this you should go back to basics and start with a simple space-time diagram with two inertial observers who have some non-zero relative velocity, and draw each of their reference frames. Then we can talk about how each of them observes events.

 

[DrGreg]

I think if you want to talk about this you should go back to basics and start with a simple space-time diagram with two inertial observers who have some non-zero relative velocity, and draw each of their reference frames. Then we can talk about how each of them observes events.

To help the discussion, here is exactly such a diagram. In fact, here are two views of the same diagram; one has been stretched and squashed relative to the other, but they both show exactly the same space-time content. Time is vertically upward and one dimension of space is horizontal. You can imagine a second dimension of space perpendicular to the diagram if you wish.

The red and blue arrows show the trajectories ("worldlines") of two inertial observers Rachel and Bob. The pink and blue grids are their respective space-time coordinate systems.

The yellow arrows are two photons traveling in opposite directions. Note that they have the same speed (i.e. their worldlines have the same gradient) measured in both the red and blue coordinate systems.

Discuss.

 

[belliott4488]

To help the discussion, here is exactly such a diagram.

Oh ... that's very nice - thanks. I put a link to another site in an earlier post on this thread (on the first page, most likely), which has animations showing how you go from one of these views to the other, but I don't think they can be embedded here.

 

[Antenna Guy]

To help the discussion, here is exactly such a diagram. In fact, here are two views of the same diagram; one has been stretched and squashed relative to the other, but they both show exactly the same space-time content. Time is vertically upward and one dimension of space is horizontal. You can imagine a second dimension of space perpendicular to the diagram if you wish.

Correct me if I'm wrong, but if your photons had a velocity component out of the page, your yellow lines would not be steep enough.

Can I really imagine a second dimension of space perpendicular to the page?

Regards,

Bill

 

[JesseM]

Correct me if I'm wrong, but if your photons had a velocity component out of the page, your yellow lines would not be steep enough.

Can I really imagine a second dimension of space perpendicular to the page?

You can imagine the page as just a cross section of a 3D diagram with two spatial dimensions and one time dimension, in which case the yellow lines would represent the intersection of the light cone of event O with the cross section.

 

[Antenna Guy]

You can imagine the page as just a cross section of a 3D diagram with two spatial dimensions and one time dimension, in which case the yellow lines would represent the intersection of the light cone of event O with the cross section.

Ah... So should I assume that Bob and Rachel are traveling in the same (spatial) direction?

Regards,

Bill

 

[belliott4488]

Ah... So should I assume that Bob and Rachel are traveling in the same (spatial) direction?

Regards,

Bill

Not exactly. In Bob's frame, he is stationary, of course (as is Rachel in hers), so Rachel simply moves in whatever direction that she does. That direction, however, can be used to define an x-t plane, i.e. we choose x to be in the direction of Rachel's motion. (The opposite statements hold in Rachel's frame.) So long as she moves in a straight line (i.e. inertially), then we don't need the second or third spatial dimensions, so we leave them out for simplicity. They won't enter in any transformation between these frames.

Of course, since there is no absolute reference frame, you can't really speak about Bob and Rachel both traveling in any particular direction; their motion will always depend on the frame of reference with respect to which you're describing it. The two frames depicted just happen to be the most natural ones, since they are the ones that Bob and Rachel live in, respectively. Since there is no third observer, there is no other frame in which they're both moving, although we can easily define one if we want to for some reason.

 

[Antenna Guy]

Not exactly. In Bob's frame, he is stationary, of course (as is Rachel in hers), so Rachel simply moves in whatever direction that she does.

Would it be absurd to say that Bob (within Bob's frame) is moving at zero velocity in all directions? If not, then whichever direction Rachel is traveling would/could be parallel to Bob.

I realize this sounds like I'm asking if a line could be construed as parallel to a point - but there's a 4th dimension here.

Regards,

Bill

 

[JesseM]

Would it be absurd to say that Bob (within Bob's frame) is moving at zero velocity in all directions? If not, then whichever direction Rachel is traveling would/could be parallel to Bob.

I realize this sounds like I'm asking if a line could be construed as parallel to a point - but there's a 4th dimension here.

But if you draw the lines showing their position as a function of time, they're clearly not parallel. If Rachel is traveling along the x-axis of Bob's coordinate system, then if you draw the x-axis as horizontal and the t axis as vertical, then Bob's x-coordinate isn't changing with time so his worldline is just a vertical line parallel to the t-axis, while Rachel's x-coordinate is changing with time so her worldline is slanted (for example, if she's moving at 0.6c, and at t=0 seconds she's at x=0 light-seconds, then at t=10 s she's at x=6 l.s., at t=20 s she's at x=12 l.s., and so on. If you can't see this you should try graphing it yourself.

 

[belliott4488]

Would it be absurd to say that Bob (within Bob's frame) is moving at zero velocity in all directions? If not, then whichever direction Rachel is traveling would/could be parallel to Bob.

Well, if not absurd, then certainly unconventional. I don't see how you're defining a direction for a zero-length vector, i.e. his spatial motion in his rest frame, much less how you're concluding that Rachel's motion is parallel.

I realize this sounds like I'm asking if a line could be construed as parallel to a point - but there's a 4th dimension here.

Are you thinking about parallel world-lines? See JesseM's response for that.

 

[Antenna Guy]

I don't see how you're defining a direction for a zero-length vector, i.e. his spatial motion in his rest frame, much less how you're concluding that Rachel's motion is parallel.

What does positive space in the two diagrams mean regarding direction?

Near as I can tell, Rachel could be traveling in the direction of either positive or negative "space" with respect to Bob's increasing time. The same holds for the second diagram (with the names swapped).

Regards,

Bill

 

[belliott4488]

What does positive space in the two diagrams mean regarding direction?

Near as I can tell, Rachel could be traveling in the direction of either positive or negative "space" with respect to Bob's increasing time. The same holds for the second diagram (with the names swapped).

Regards,

Bill

That's arbitrary, just as it is any time we pick a direction for a spatial axis. In this case, Bob picks his +x axis to point in the direction of Rachel's motion, so that her velocity is positive. For consistency, Rachel picks her +x axis to point in the same direction relative to Bob, so that she observes his velocity to be in the negative direction. She could have picked it the other way, but then the Lorentz transformations between their two reference frames would pick up negative signs in front of the x's, which would not be a problem mathematically but would look funny.

It's like doing a coordinate transformation in an ordinary set of 2-d Cartesian axes. If you define a second set of axes x',y' that are parallel to x,y but displaced along the x-axis, you could also flip the x' axis so that it pointed the opposite direction, and that would work but would be weird.

 

[belliott4488]

By the way, the "+x" and "-x" are just labels, right? The world lines must be drawn where they are regardless of how we label the axes. They are physical, and they just go where they go.

 

[Antenna Guy]

By the way, the "+x" and "-x" are just labels, right?

Not how I see it - they are two valid answers to the same question:

"Where is Rachel with respect to Bob at time t?"

At time t, the 4-space displacement in Bob's frame reduces to a 3-space displacement of radius +/- r (or "space") in Bob's frame.

The difference between the two being a 180deg rotation about Bob's 4-space axis (aka "time" in Bob's frame).

Is that not true?

Regards,

Bill

 

[belliott4488]

Okay, I think this is worth a careful look, since we've hit the first spot where our understandings are diverging.

Not how I see it - they are two valid answers to the same question:

"Where is Rachel with respect to Bob at time t?"

First, what kind of answer are you looking for? Spatial coordinates? That's how I would expect to answer the question, "Where?" But any such answer is frame-dependent, so we expect to get different answers in different frames, i.e. for different observers, right?

At time t, the 4-space displacement in Bob's frame reduces to a 3-space displacement of radius +/- r (or "space") in Bob's frame.

Nope, you're losing me. By "4-space displacement" I assume you mean the vector difference between Rachel's space-time coordinates and Bob's - right? Since Bob is at rest, his coordinates are (t,0) everywhere on his world line (or (t,0,0,0) if you insist on a 4-vector, but since we have motion in one spatial dimension, that's really not necessary). Rachel's coordinates are (t,x) = (t,vt), where v is her velocity as measured by Bob. The difference between the two is just (0,x) = (0,vt).

I don't follow your "3-space displacement of radius +/- r (or "space") in Bob's frame" at all, I'm afraid.

The difference between the two being a 180deg rotation about Bob's 4-space axis (aka "time" in Bob's frame).

No, I don't follow this either, but it can't be right since you can't define a rotation in 4-d with respect to an axis the way you can in 3-d. In 3-d if you pick an axis, the remaining space is 2-d, i.e. a plane, so one angle defines a rotation in that plane. In 4-d if you pick one axis, the remaining space is 3-d, so you must specify a 3-d rotation with three angles to define a rotation "about that axis". If you really mean a rotation by one angle, then you can specify the plane of that rotation.

 

[DrGreg]

I've attempted a 3D spacetime diagram (2D space + 1D time) which might help further.

The centre and right pictures show Bob's and Rachel's spacetime coord systems separately, and the left diagram shows both superimposed.

Both Bob and Rachel have to choose spatial x,y,z axes, and it's conventional to assume

(a) they align their space-axes to be parallel to each other
(b) they line up their space x-axes so that their relative motion is along their mutual space x-axis

Neither of these assumptions are necessary, but they make the maths a whole lot easier, and the standard "Lorentz transform" equations that you see make this assumption, as does my diagram. Under these assumptions, both Bob and Rachel agree on the y- and z-coords of any event.

Yellow is the shared "vertical" plane y_R = y_B = 0 (and the plane of my previous diagram).
Blue is Bob's "horizontal" plane t_B = 0.
Pink is Rachel's "tilted" plane t_R = 0.

 

[belliott4488]

Nice work DrGreg.

There are also 3-d (2 space + time) diagrams at the site I linked earlier in this thread:
http://casa.colorado.edu/~ajsh/sr/sr.shtml

He shows the light cone emanating from a source and superimposes the coordinates of two different observers. It's animated, so you can see how each observer sees the propagation, including the length contraction in the direction of motion. On one of the pages he also has an animation that goes continuously from one frame of reference to the other (Bob's to Rachel's and back, in our scheme, although he uses different names).

 

[Antenna Guy]

Okay, I think this is worth a careful look, since we've hit the first spot where our understandings are diverging.

I'm glad to hear that - because I think we're making progress.

"Where?" But any such answer is frame-dependent, so we expect to get different answers in different frames, i.e. for different observers, right?

When I wrote "with respect to Bob", I meant to imply "within Bob's frame". Sorry for not being more clear.

By "4-space displacement" I assume you mean the vector difference between Rachel's space-time coordinates and Bob's - right?

Yes. Let's limit ourselves to Bob's frame for now, and draw a horizontal line at some time t. The distance between Rachel and Bob along that line of constant time is their (radial) separation at time t.

Since Bob is at rest, his coordinates are (t,0) everywhere on his world line (or (t,0,0,0) if you insist on a 4-vector, but since we have motion in one spatial dimension, that's really not necessary).

This is where you lose me. If I can't make a distinction between (-t,-x) and (t,x) as directions, I don't know why these diagrams have four quadrants. I'd like to think that I could multiply a 4-velocity by -1, and assume that "t" now represents some (positive) time displacement toward the past (i.e a 180deg rotation about the "space" axis of the diagram).

I don't follow your "3-space displacement of radius +/- r (or "space") in Bob's frame" at all, I'm afraid.

Does x=\pm \sqrt{x^2} help?

The diagrams collapse three spatial dimensions onto one, but I believe this is still the general form of the distance expression. "x", as an absolute (positive) "distance", has a direction in 3-space that could be either (1,0,0), or (-1,0,0). The absolute value of the distance times the direction yields the displacement (n.b. "direction" implies a unit vector as far as I'm concerned).

The difference between the two being a 180deg rotation about Bob's 4-space axis (aka "time" in Bob's frame).

No, I don't follow this either, but it can't be right since you can't define a rotation in 4-d with respect to an axis the way you can in 3-d.

You're on shakey ground with "it can't be right" as your proof.

In 3-d if you pick an axis, the remaining space is 2-d, i.e. a plane, so one angle defines a rotation in that plane.

Replace "remaining" with "affected" and I follow.

In 4-d if you pick one axis, the remaining space is 3-d

Once again, replace "remaining" with "affected" and I follow.

Regards,

Bill

P.S. Kudos to DrGreg for all the snazzy diagrams.

 

[belliott4488]

Yes. Let's limit ourselves to Bob's frame for now, and draw a horizontal line at some time t. The distance between Rachel and Bob along that line of constant time is their (radial) separation at time t.

I'm not sure what you mean by "radial." You've got two space-time points, and you construct the vector between them, i.e. the space-time interval. It has both time and space components, of course, and only in Bob's frame is it purely spatial, i.e. the time component is zero. But I don't see how that make it a "radius" of anything in particular.

This is where you lose me. If I can't make a distinction between (-t,-x) and (t,x) as directions, I don't know why these diagrams have four quadrants.

Who said you can't distinguish between (t,x) and (-t,-x)? I'm not sure why you call them "directions", since those look like the coordinates of two points, reflected through the origin in the x-t plane. Perhaps you're referring to the -x and -t axes? Again, perfectly okay to distinguish between those and the +x and +t axes ...

I'd like to think that I could multiply a 4-velocity by -1, and assume that "t" now represents some (positive) time displacement toward the past (i.e a 180deg rotation about the "space" axis of the diagram).

Well, you could do that, although I'm not sure what you're trying to accomplish. Also we have not yet introduced any 4-velocities in this discussion, and we should probably talk about what that means. You need to be clear on what is meant be the proper time and differentiation with respect to it.

Also - again, you shouldn't speak a rotation by some angle about an axis here. If we're sticking to the 2-d picture, then all rotations are in the x-t plane and are not about either axis. If we're thinking of the full 4-d space, then rotations cannot be defined about an axis, at least not by one angle.

I don't follow your "3-space displacement of radius +/- r (or "space") in Bob's frame" at all, I'm afraid.

Does x=\pm \sqrt{x^2} help?

Not much, I'm afraid. Once we've picked the direction of the +x axis, then Rachel's displacement along that axis is either positive or negative (it's positive, using the convention that we've been using). I don't see the ambiguity in sign that you see. At t=0, they're both at the same point. At any later time, i.e. any t>0, Rachel has moved to a position on the positive x-axis, i.e. x=vt (this is all expressed in Bob's frame).

The diagrams collapse three spatial dimensions onto one, but I believe this is still the general form of the distance expression.

Whoa! Lots of trouble here ... let's take this one item at a time:

"x", as an absolute (positive) "distance", ...

Okay, in "normal" Euclidean space, you can talk about the magnitude of a displacement vector, and yes, that's a positive distance. So we can talk about the absolute value of x as the distance between Bob and Rachel, in Bob's frame only. But be careful when we're talking about space-time intervals, because then the vector magnitude need not be positive-definite; in fact, time-like and space-like intervals have opposite signs.

... has a direction in 3-space that could be either (1,0,0), or (-1,0,0). The absolute value of the distance times the direction yields the displacement (n.b. "direction" implies a unit vector as far as I'm concerned).

Okay, I think I've figured out what you mean, although your language is a little imprecise (a capital offense in math, of course ). A distance does not have a direction - it's a scalar value. That's what it means that it's positive-definite, i.e. the absolute value. You can talk about the displacement vector as a signed quantity, but not the distance. If you do change the sign of x, then you're just changing between saying "Rachel's position relative to Bob" to "Bob's position relative to Rachel" but expressing each still in Bob's reference system.

You're on shakey ground with "it can't be right" as your proof.

Ha. I went on from that line to explain why your statement could not be right!

Replace "remaining" with "affected" and I follow.
Once again, replace "remaining" with "affected" and I follow.

Well, I think the correct mathematical term is "complementary space", but I think you understood that I meant the remaining dimensions of the space after eliminating one. In other words, what freedom do you have left once you decide to hold one axis fixed? In 4-d, you still have 3 dimensions to play with, so saying "rotate by angle theta about the x-axis" is undefined.

 

[Noone]

Nature's Parables

I ( Swaminathan) started studying theory of relativity. I need some help in understanding the following:

Special relativity says that what looks like time to one observer(in a moving frame) looks like space to another observer(in a stationary frame).

I assume that Two events are happening. An observer views these two events happening at same space but at different times. Another observer views these two events happening at same time but at different space.

Is my assumption is correct. If so, can someone give real world example for the above situation. I searched for examples in textbooks and it is not available.

- Swaminathan.

a great example of were this naturely happens would be here on earth. We all see the same light from the sun, But if you were to wacth a sun rise on the water front you would see a stright line of light reflecting to your location, and if there was a person standing nexted to you they wouldn't see the line that gose to you but they would see a line that gose to them. hence same start light and time yet diffrent vantage points and perception of the same start light and time. no matter were you place point B (which would be you) point A (that would be the start of light aka the sun) that is a fixed location or not, would allways make a stright line to the observer but another observer wouldn't be able to see the line of reflection or the perception of the first observer. So to recap if the observers would be at diffrent points of time :D well each person would be standing there at the same time on a clock but there perception of the time and light would be diffrent. I won't say your right or wrong, that would be a miss-judgement on my part for lack of information of what your assumption is :D i try to stay away from words that define actions. for i speak in actions no long words that the masses mainly don't know :D

 

[Noone]

i suggest wacthing a sun rise with some people and take the time to think about your concepts and the concepts of others on the matters. its best to see with your own eyes, that's why math ways also made, to prove things to peoples eys and minds. observation don't we all love it :D

 

[Antenna Guy]

I'm not sure what you mean by "radial." You've got two space-time points, and you construct the vector between them, i.e. the space-time interval. It has both time and space components, of course, and only in Bob's frame is it purely spatial, i.e. the time component is zero. But I don't see how that make it a "radius" of anything in particular.

Let's say, for sake of argument, I displace Rachel by one unit space within Bob's frame. Does Rachel's velocity in Bob's frame change, or is it still \frac{\delta s(t)}{\delta t} at any given time? I assume you will agree that spatial displacement of Rachel within Bob's frame has no effect on Rachel's velocity in Bob's frame.

If we map the above into Rachel's frame, Bob is now offset by some space-time interval that is equivalent to one unit space within Bob's frame. It is not clear to me how Rachel can determine whether the time component of that interval is positive or negative without making some additional determination: Is Bob approaching, or receding (radially)? In order to make that determination, Rachel must make two separate (in Rachel's time) measurements of the interval between her and Bob.

Agree?

Who said you can't distinguish between (t,x) and (-t,-x)?

I'm pretty sure I can, but I think I'd need more information than the diagram provides.

again, you shouldn't speak a rotation by some angle about an axis here. If we're sticking to the 2-d picture, then all rotations are in the x-t plane and are not about either axis.

O.K. We'll limit ourselves to rotations about an axis that doesn't exist.

If we're thinking of the full 4-d space, then rotations cannot be defined about an axis, at least not by one angle.

Not to be facetious, but is there some reasoning behind that assertion? Although it's not obvious to me how a rotation about a 4-vector might manifest itself, I certainly do not know that it cannot be characterized by a single angle.

Okay, I think I've figured out what you mean, although your language is a little imprecise (a capital offense in math, of course ).

I'm just a dumb engineer - don't bother suing me for heresy.

A distance does not have a direction - it's a scalar value. That's what it means that it's positive-definite, i.e. the absolute value.

The term "radius" is commonly construed as being positive-definite, but it is not. Positive radius in one direction is equivalent to negative radius in the opposite direction. I'll have to take another look at what I wrote to try and figure out why you don't think I understand this...

what freedom do you have left once you decide to hold one axis fixed? In 4-d, you still have 3 dimensions to play with, so saying "rotate by angle theta about the x-axis" is undefined.

Hey - I like freedom. In fact, I think we need more! (knyuk-knyuk)

Rather than lump my assumptions into some [hopefully not] amorphous blob, let me try to put my "math-hat" on and be as explict as I can with what follows.

Let us start with the diagram of Bob's frame that DrGreg so graciously provided.

Let me define a quantity "R" that represents a positive-definite (ack!) quantity composed by taking the RSS of the "space" and "time" components of some event in Bob's frame (relative to O, at some time t).

Let me define a quantity "r" that represents the absolute value of "space" relating some event (Rachel) to Bob's frame at time t.

Let me define a pseudo-spherical \theta - \phi coordinate system about O where:

\theta=sin^{-1} (\frac{\delta r}{\delta R})

\phi=tan^{-1} (\frac{\delta R}{\delta r})

From the diagram, I infer that Rachel's velocity (v) is 0.5 (c) relative to Bob since she exhibits half the displacement of a photon over any given time interval (someone should make it clear that "time" = ct rather than t in the diagram, but I digress...). Hence, I can assume that R_R=0.5ct (versus R_B=t). Using the definitions of \theta and \phi supplied above, I infer that r_R=R_R sin(\theta), or r_R=\frac{R_R}{tan(\phi)}. Substituting for r_R yields:

sin(\theta)=\frac{1}{tan(\phi)}.

Although it is obvious that \theta is the angle between Bob and Rachel's (velocity) vectors on the diagram, it may be less clear what the \phi I've defined relates to. To illustrate, draw a horizontal line at some non-zero t across the diagram. Set a compass to a radius equal to the "space" between Rachel and Bob at time t, and then draw an arc about Bob's position that passes through Rachel's world-line at two different points. The more negative time point is where Bob observes Rachel at time t, and the more positive point is Rachel's Lorentz corrected position. More succintly, \phi is the angle between the observed position and projected position at time t in Bob's (time-travelling) frame.

-----

"math-hat" off...

In relation to Maxwell's equations, there are two operators that I feel are relavant to this discussion: curl and divergance. If I were to draw a circle of radius R=ct about O in Bob's frame, divergence could be represented as a radial vector toward any point on the circle. The curl at any point can then be construed as a unit vector tangential to a circle of radius R, and perpendicular to \vec R. In order to express divergence as a unit vector, we have div=\frac{\vec R}{ct}. Hence, if I posit that Maxwell's equations apply to closed spherical surfaces, then "time" and "div" are parallel - and curl is perpendicular to "time".

BTW - You may note that drawing a circle corresponding to Bob's distance from O bisects the two points on Rachel's world line identified earlier...

Regards,

Bill

P.S. I hope I don't look too much like the tropicana girl with my "math-hat" on.

 

[Antenna Guy]

Would it be absurd to say that Bob (within Bob's frame) is moving at zero velocity in all directions? If not, then whichever direction Rachel is traveling would/could be parallel to Bob.

I realize this sounds like I'm asking if a line could be construed as parallel to a point - but there's a 4th dimension here.

But if you draw the lines showing their position as a function of time, they're clearly not parallel.

True.

If Rachel is traveling along the x-axis of Bob's coordinate system, then if you draw the x-axis as horizontal and the t axis as vertical, then Bob's x-coordinate isn't changing with time so his worldline is just a vertical line parallel to the t-axis, while Rachel's x-coordinate is changing with time so her worldline is slanted

If Rachel were traveling along the x-axis of Bob's coordinate system, wouldn't she be traveling perpendicular to Bob's time?

I realize this is a nit-pick, but by what differential quantity does x change if time is perpendicular to it?

Regards,

Bill

 

[belliott4488]

Hi, Bill - I've been distracted for the past couple of days, so I haven't had a chance to respond to your last response to me. I was going to ... but then I saw this:

If Rachel were traveling along the x-axis of Bob's coordinate system, wouldn't she be traveling perpendicular to Bob's time?

I realize this is a nit-pick, but by what differential quantity does x change if time is perpendicular to it?

No, no, no! And again, I say, NO! This is NOT a nit-pick - it's the very crux of what we're discussing, so I'm deferring a response to your last response to me until we've straightened this out. (Note that the quotation above was a response to JesseM.)

We arbitrarily define the x-axis in Bob's frame to be in the direction of Rachel's motion, since we are free to do that. Therefore she travels along Bob's x-axis by definition - but this does NOT mean she is traveling perpendicular to Bob's time axis in space-time. The points on the x-axis all have the same time coordinate (t=0), so "motion" - i.e. a world-line - perpendicular to the t-axis is not possible - all space-time points on the x-axis are simultaneous. As Rachel moves along the direction of the x-axis (in 3-d space), she is at a succession of (x,t) points, where the ratio of the changes in each coordinate , dx/dt, is simply her velocity. That ratio also gives the slope of her path on the space-time diagram (inverse slope, actually, since slope would be dt/dx).

Bob is stationary in his frame, so his x-coordinates don't change, so his world line consists of the points along his time axis (this is true for all observers in their own rest frames). Rachel moves along a slanted path somewhere between vertical (v=0) and 45 degrees (v=c=1, since we've been implicitly assuming that we're using units where c=1).

It's critical that you see this. If you already do, then please excuse my pounding the table, but otherwise let's try to reach agreement on this before moving on.

 

[Antenna Guy]

Hi, Bill - I've been distracted for the past couple of days, so I haven't had a chance to respond to your last response to me. I was going to ... but then I saw this:

If Rachel were traveling along the x-axis of Bob's coordinate system, wouldn't she be traveling perpendicular to Bob's time?

I realize this is a nit-pick, but by what differential quantity does x change if time is perpendicular to it?

No, no, no! And again, I say, NO! This is NOT a nit-pick - it's the very crux of what we're discussing, so I'm deferring a response to your last response to me until we've straightened this out.

Well.. I didn't think "nit-pick" was quite the right term, but I see you understand what I was getting at.

We arbitrarily define the x-axis in Bob's frame to be in the direction of Rachel's motion, since we are free to do that.

Well, since there are three elements in Bob's frame (Rachel + two photons), am I not to assume that everything is traveling along Bob's "x" (where "x" is arbitrary)? If one of the two photon's were traveling along Bob's "y", would that change how it maps into Bob's "space" if all three start at O?

Therefore she travels along Bob's x-axis by definition - but this does NOT mean she is traveling perpendicular to Bob's time axis in space-time. The points on the x-axis all have the same time coordinate (t=0), so "motion" - i.e. a world-line - perpendicular to the t-axis is not possible - all space-time points on the x-axis are simultaneous.

This is where we diverge. Rachel's time is not equivalent to Bob's time, and the difference between the two is the mechanism whereby Rachel becomes displaced (in the direction of "space") in Bob's frame. If you have some other rationalization of this, I'm all ears.

As Rachel moves along the direction of the x-axis (in 3-d space), she is at a succession of (x,t) points, where the ratio of the changes in each coordinate , dx/dt, is simply her velocity.

If I were to generalize velocity to dr/dt (and refer to r as "space"), which direction must time go for me to get the correct dt to go with my dr?

That ratio also gives the slope of her path on the space-time diagram (inverse slope, actually, since slope would be dt/dx).Bob is stationary in his frame, so his x-coordinates don't change, so his world line consists of the points along his time axis (this is true for all observers in their own rest frames).

Since we still seem to be operating on different wavelengths, let me draw the diagram a different way. We have an x-axis that Bob, Rachel and the two photons might travel along, and all four start at O at t=0 - proceding at constant velocity. Each velocity vector has a (positive-definite) magnitude in either the +x or -x direction. At any time t the velocity vectors have the same magnitude they had at t=0, so I can simply lay them on the x-axis in the appropriate direction to show how they relate at any given time. Let's say that I'd like to see how those velocity vectors relate to x at t=1. For Rachel and one of the photons, t=1 is in the direction of +x, and the other photon observes time to procede in the direction of -x. Thus, from a (positive-definite) velocity times time equals distance standpoint, the diagram contains two different senses of time relative to O (clarification: recall that I define v=dx/dt).

Like you, my available time has been somewhat limited lately. Compounding this is the fact that I'll be traveling over the next week - so please be patient waiting for further replies from me. I'm curious read what you had going before I got you side-tracked.

Regards,

Bill

 

[belliott4488]

Well, since there are three elements in Bob's frame (Rachel + two photons), am I not to assume that everything is traveling along Bob's "x" (where "x" is arbitrary)? If one of the two photon's were traveling along Bob's "y", would that change how it maps into Bob's "space" if all three start at O?

I'm not sure why you're introducing photons (the two lines on the diagram are usually there as references to define Bob's future space cone), but it's no problem to include them.

To answer your question, if one photon travels in the x direction and the other in the y direction, then we need a 3-d space-time diagram, i.e. spanned by t, x, and y. Rachel and one photon have world lines in the x-t plane, and the other photon has its world line in the y-t plane. The two photon lines are at 45 deg. to the t axis, and Rachel's line is steeper.

Therefore she travels along Bob's x-axis by definition - but this does NOT mean she is traveling perpendicular to Bob's time axis in space-time. The points on the x-axis all have the same time coordinate (t=0), so "motion" - i.e. a world-line - perpendicular to the t-axis is not possible - all space-time points on the x-axis are simultaneous.

This is where we diverge. Rachel's time is not equivalent to Bob's time, and the difference between the two is the mechanism whereby Rachel becomes displaced (in the direction of "space") in Bob's frame. If you have some other rationalization of this, I'm all ears.

We've said nothing about Rachel's time or any other observations she makes, so let's leave them out for now; we're just concerned with Bob's observations and how we can deduce them from the space-time diagram.

As for the "mechanism whereby Rachel becomes displaced," I'd say that mechanism is simply her motion. She is moving, i.e. her x-coordinate in Bob's frame is changing as a function of time, so therefore she is "displaced" over time. I don't know why you bring in the difference between Rachel's and Bob's frames of reference; they're certainly different, but that is a result of her motion, not the cause of it.

If I were to generalize velocity to dr/dt (and refer to r as "space"), which direction must time go for me to get the correct dt to go with my dr?

I don't think I understand this r variable. What do you mean r is "space"? x is the variable the defines Rachel's position in Bob's frame, and it does so unambiguously - why introduce anything else? The only meaning I can see for r is that it is the magnitude of x (or the absolute value of x if we're using just one space dimension), but why introduce it? You throw away the sign of x, and thus introduce an unnecessary ambiguity in your knowledge of Rachel's position. In any case, however, since we do know Rachel's position, we can still discuss the r variable. Since we picked the +x direction to be in the direction of her motion, however, then her x positions after t=0 will all be positive (assuming she starts at O at t=0), so r is simply equal to x.

As for the direction of t, +t is in the direction of the future, by almost universal convention. Why change that? Once you've picked your directions for +x and +t, you have labeled all space-time points unambiguously, and you can trace Rachel's history and future (assuming she continues her inertial motion) in an equally unambiguous way.

Since we still seem to be operating on different wavelengths, let me draw the diagram a different way. We have an x-axis that Bob, Rachel and the two photons might travel along, and all four start at O at t=0 - proceding at constant velocity. Each velocity vector has a (positive-definite) magnitude in either the +x or -x direction. At any time t the velocity vectors have the same magnitude they had at t=0, so I can simply lay them on the x-axis in the appropriate direction to show how they relate at any given time. Let's say that I'd like to see how those velocity vectors relate to x at t=1.

When you talk about velocity vectors, are you talking about three-vectors? It sounds like it. Keep in mind, though, that plotting a velocity vector against axes that have units of length is kind of arbitrary, since you have to specify how you're drawing the velocity units relative to the length units. Generally you'd draw velocity vectors in velocity space, where the axis units are units of, well ... velocity. We can do so, and align the velocity axes with the space axes, which is natural, but understand that we're overlaying two different spaces: one where vectors between points correspond to displacements, and one where vectors between points correspond to velocities.

For Rachel and one of the photons, t=1 is in the direction of +x, and the other photon observes time to procede in the direction of -x.

What do you mean "t=1 is in the direction of +x"? Are you going to a space-time diagram now? If so, then there is no point "t=1"; there are an infinite number of points with t-components equal to 1: they form a line parallel to the x-axis and correspond to all the simultaneous events in Bob's frame that occur at t=1.

Now, if what you meant is the direction of the three-velocity vectors at time t=1, well - why would they be any different than they were at any other time? There's no acceleration here, right? so no one's velocity is changing. You defined two of them to point in the +x direction and one to point in the -x direction, so they still do.

Thus, from a (positive-definite) velocity times time equals distance standpoint, the diagram contains two different senses of time relative to O (clarification: recall that I define v=dx/dt).

Okay - (positive-definite) velocity magnitude times time does give you distance, i.e. the magnitude of the position vector, but again, why are you limiting yourself to this when we have all the information we need to determine position unambiguously?

Rachel moves in the +x direction at velocity +v, so her position at any time t is simply x = vt. Similarly, photon 1 (the one that moves in the same direction as Rachel) moves at velocity +c, so its position is given by x=ct. Photon 2 has velocity -c, so its position is x = -ct.

This is easy to draw in the space-time diagram as two 45 degree lines from O for the two photons - one to the left and one to the right, and a steeper slanted line going up to the right (positive slope) for Rachel. (Again, the photon lines are at 45 degrees because we are using units where c=1.)

So far, there's really no relativity in this at all - it's all just basic Physics. The relativity will come in when we talk about how Rachel sees all this - but in Bob's frame there's nothing unusual at all.