Speed of Light Question

[Mise]

After some googleing:
Some university site
Youtube Video

The video was cool Makes sense now.

 

[lovett]

huh? that doesn't make any sense at all...

Well, the concept of being a widow is human. I.e, widowship was orignally created and then sustained by human society. Thus, surely it relies on being known by society (or a segment thereof) before being 'fact'.

In contrast, physical constants are completely independent of humanity.

 

[xienwolf]

I've been tossing this one through my head all day since reading it this morning. Decided not to post then because I was VERY tired and figured I might be wrong, so wanted to draw up the World Lines for it all.

The problem with this particular case is the wording of the example. The video you linked from You-Tube is how it is typically set up. They are at the equidistant point when the event occurs.

But it was phrased here that they were at the equidistant point when the information for the simultenaity arrived. In that particular case, they will agree that it was simultaneous, because they are both getting the information at the same time, because in that location, at that time, both sets of data exist together.

Now of course you could argue that they are not technically perfectly centered on the 2 events because then they would have to be sitting in the exact same point in space as each other, which is quite catastrophic at best. But in that case you wind up having it so that the first event viewed is the one you are traveling AWAY from, because you were closer to that event and it was traveling faster than you are, so it catches up to you.


So, the proper wording for the statement to have the results intended is that all 3 are at the same point when the deaths occur, and then 2 of them move away at relativistic speeds in the direction of each death (opposite directions)

 

[warpus]

I think that the woman in question is a widow as soon as her husband dies - she just doesn't know it yet.

Say that I'm married.. and my wife is somewhere in Africa doing missionary work. Am I a widower as soon as she is hacked to pieces by a group of crazed monkeys?.. or do I become a widower as soon as I find out? I move that I am a widower the entire time, but just do not know it yet.

Now, since there is no such thing as a universal present, the question of WHEN I become a widower is a bit more complicated, even assuming we all go with my definition above.

It really depends on the velocity through space-time of the observer. So, to one observer I become a widow March 21st, 2008, 5:48pm, to another one, same day at 5:47pm, to yet another one Feb 20th, 2874. I suppose my own reference point is the most important here, so for most intents and purposes we should just go with that.. but yeah, I'd say she's a widow without knowing it.. until she finds out.

 

[Souron]

Words in dispute:
For simplicity, assume that Alice, Bob, and Claire were all equidistant from Mr X and Mrs Y when Alice saw both deaths.

I've been tossing this one through my head all day since reading it this morning. Decided not to post then because I was VERY tired and figured I might be wrong, so wanted to draw up the World Lines for it all.

The problem with this particular case is the wording of the example. The video you linked from You-Tube is how it is typically set up. They are at the equidistant point when the event occurs.

So, the proper wording for the statement to have the results intended is that all 3 are at the same point when the deaths occur, and then 2 of them move away at relativistic speeds in the direction of each death (opposite directions)

That's what I said. They are equidistant when Alice sees both events. "When the events occur" is confusing since the events do not occur at the same time in all reference frames. Heck, same time at all reference frames is not even a meaningful description.

Now of course you could argue that they are not technically perfectly centered on the 2 events because then they would have to be sitting in the exact same point in space as each other, which is quite catastrophic at best.

Alice, Bob, and Claire do not have to be at the same point. They just have to be equidistant from both events. In fact, not only does their position not matter, but any sideways motion would not matter either. Only motion along the up-down axis matters.

 

[Erik Mesoy]

After some consideration, I move that Mrs. X is a widow starting when the [wiki]light cone[/wiki] of Mr. X's death intersects Mrs. X's [wiki]world line[/wiki].

Any objections?

 

[brennan]

Points equidistant from Mr and Mrs X occupy a Plane Xienwolf. There are plenty of equidistant points for different observers.

Really all I was doing was pointing out that Souron's wording was wrong.

Erik, I would say that Mrs X is a widow as soon as her husband dies. She just isn't aware of the fact.

Imagine the legal wrangling that will occur the first time this becomes a real concern.

 

[Erik Mesoy]

Erik, I would say that Mrs X is a widow as soon as her husband dies. She just isn't aware of the fact.

I don't even understand what "as soon as" means in this context. Please be so kind as to spell it out for me, because I'm under the impression that special relativity explicitly denies absolute simultaneity.

My approximate mental state in this thread:

 

[brennan]

Well If Mr and Mrs X are on planets, then they have a roughly similar reference frame to the rest of the galaxy they are in. That makes it fairly easy to establish an arbitrary standard reference frame for all such measurements.

Basically, unless the universe you occupy is empty except for people zipping about at relativistic velocities in all directions, then it's easy to set a standard reference.

 

[Erik Mesoy]

Earth orbits Sol at 30 km/s which gives a Lorentz factor around 1.00005, which adds about 26min to a year (light or otherwise). That's enough to shift widowhood before or after midnight with a good margin of error, so I think it counts as a relativistic velocity when it can mess up bureacracy.

Edit: Hmm, this was my 8888th post.

 

[xienwolf]

After some consideration, I move that Mrs. X is a widow starting when the [wiki]light cone[/wiki] of Mr. X's death intersects Mrs. X's [wiki]world line[/wiki].

Any objections?

This is precisely how the problem must be approached for a correct answer.

Points equidistant from Mr and Mrs X occupy a Plane Xienwolf. There are plenty of equidistant points for different observers.

Really all I was doing was pointing out that Souron's wording was wrong.

Erik, I would say that Mrs X is a widow as soon as her husband dies. She just isn't aware of the fact.

Imagine the legal wrangling that will occur the first time this becomes a real concern.

Yes, points that are equidistant from them both do occupy a plane. However, the light will expand from each event in a sphere, and the spheres will intersect each other in an expanding Circle (which is in that plane). So I was incorrect to state that they would have to be in the exactly same individual point because that is only true to see the events as simultaneous at the soonest possibile moment. To see the events as simultaneous at any other point, they may occupy any position in the expanding circle of intersecting light cones.

Words in dispute:
For simplicity, assume that Alice, Bob, and Claire were all equidistant from Mr X and Mrs Y when Alice saw both deaths.

That's what I said. They are equidistant when Alice sees both events. "When the events occur" is confusing since the events do not occur at the same time in all reference frames. Heck, same time at all reference frames is not even a meaningful description.

"When the events occur" is defined by tracing the light back in the reference frame of your stationary observer. Since you have defined them as your stationary frame, you have defined that as your overall reference frame, and thus you are allowed to state that both events occured in that frame at the same time just to set the initial conditions. It will not matter which person's reference frame you actually define as your stationary observer, you will get the same results for every drawing of the world lines.

Alice, Bob, and Claire do not have to be at the same point. They just have to be equidistant from both events. In fact, not only does their position not matter, but any sideways motion would not matter either. Only motion along the up-down axis matters.

Sideways motion does matter. It is TANGENTIAL motion that doesn't matter. And it is not possible to move tangential to both spheres at once. That is part of what led me to stating incorrectly that they would have to be at the exact same point as each other because I had them traveling along the axis connecting the two events to avoid tangential elements.

 

[Souron]

Sideways motion does matter. It is TANGENTIAL motion that doesn't matter. And it is not possible to move tangential to both spheres at once. That is part of what led me to stating incorrectly that they would have to be at the exact same point as each other because I had them traveling along the axis connecting the two events to avoid tangential elements.

That makes so much sense! Thanks for correcting me.

 

[Tank_Guy#3]

Here's another speed of light question I've pondered:

It's stated that nothing with mass can't accelerate to the speed of light. However, would it be possible (if not concievable) to have an instantaneous jump from motionless to light speed?

Let's for a moment completely disregard the stresses applied to the object that is doing the "jumping".

 

[xienwolf]

Well, any change in velocity is called an acceleration. An non-continuous shift would simply be an infinte acceleration (definably with a Dirac Delta for different levels of jumping Velocity), not a lack of it.

 

[Souron]

"When the events occur" is defined by tracing the light back in the reference frame of your stationary observer. Since you have defined them as your stationary frame, you have defined that as your overall reference frame, and thus you are allowed to state that both events occured in that frame at the same time just to set the initial conditions. It will not matter which person's reference frame you actually define as your stationary observer, you will get the same results for every drawing of the world lines.

But see, I did not define a stationary observer. Fifty defined Mr and Mrs X as being stationary, but that has no impact on the scenario.

 

[xienwolf]

You said Alice was in that reference frame, so while not neccessarily completely stationary, she would be at the least non-relativistically mobile in relation to the event. Which is good enough to define that as the stationary reference frame, no?

 

[Souron]

I did not say Alice was in that reference frame. I only said that Alice saw both events happen at the same time, and that she was equidistant from the two events. Neither of those imply that she is in a non non-relativistically mobile plane relative to Mr X and Mrs Y. It is entirely possible that Claire or Bob is in that reference frame.

 

[Souron]

After some consideration, I move that Mrs. X is a widow starting when the [wiki]light cone[/wiki] of Mr. X's death intersects Mrs. X's [wiki]world line[/wiki].

Any objections?

While this will yield consistent results (I think), that fact is if Mr X's death's light code intersects Mrs's Y's light cone, Mr X must already have died at a time proportional to the distance between the intersection, and the point of death (and the speed of light).

 

[xienwolf]

Ok, looked back and read closer, you did indeed never set a reference frame for a stationary observer as the one which includes the 2 people's own reference frames during their lives.

Assume that there exists a reference frame in which they both die at the same time. Alice happens to live in this reference frame. Bob is in a different reference frame.

But that does state a reference frame, and the very nature of establishing a reference frame is to set your "stationary observer" because that is just another way to state that it is your reference frame. I do not remember precisely why we are debating the reference frame anyway at this point, and am disinclined to check Needless to say, you are correct in what you stated

While this will yield consistent results (I think), that fact is if Mr X's death's light code intersects Mrs's Y's light cone, Mr X must already have died at a time proportional to the distance between the intersection, and the point of death (and the speed of light).

Correct, and the same can of course be stated about Mrs. Y (that she has to have already died at a time proportional to the distance between the intersection and the point of death).

And the light cone/World-line approach will indeed yield cinsistent results, it was designed for just these kind of thought experiments.


If you want a really fun one, you look at a train entering into a tunnel that is the same length as the train itself in the Earth Frame. Set the train moving at relativistic speeds and compare the sequence of events (front of train enters tunnel, back of train enters tunnel, front of train leaves tunnel, back of train leaves tunnel), as seen from an observer on the train and an observer standing outside the tunnel (equidistant from the 2 ends).

This is the standard thought experiment to demonstrate mass contraction, which is far more enjoyable than Time dilation anyday.