View Full Version : Speed Of Light Question
zoobyshoe
Jun25-04, 06:11 PM
I'm on a space station in deep space. I depart in a rocket and accelerate to .5 c then level off to that speed. No longer accelerating, I now define myself to be stationary. I define the speed of anything I percieve to be in motion as relative to my own stipulated-to-be-unmoving inertial frame
On board I have a Michelson-Morley (or better) apparatus for measuring the speed of light. However, I am not going to measure the speed of any light that originates on my own ship. I am going to measure the speed of a beam of light that comes from the space station I just left.
At a prearranged time the space station sends this beam of light in exactly the same direction I am traveling.
When it catches up to me I measure it to be going:
A. c?
B. .5 c?
c. other ?
On board I have a Michelson-Morley (or better) apparatus for measuring the speed of light. However, I am not going to measure the speed of any light that originates on my own ship. I am going to measure the speed of a beam of light that comes from the space station I just left.
At a prearranged time the space station sends this beam of light in exactly the same direction I am traveling.
When it catches up to me I measure it to be goingHow, exactly, do you measure the speed of light from the space station?
zoobyshoe
Jun25-04, 07:26 PM
I assumed it could be done since people have measured the speed of light from stars.
You're saying it can't be done?
There are several methods by which one could measure the speed of light. Many of those who find SR hard to understand (or accept) feel that part of the 'problem' is the actual method of measuring the (local) speed of light. :tongue2:
In the case of your hypothetical experiment, I was wondering if you were getting at something special about the space station you had left (or would any distant source do)?
zoobyshoe
Jun25-04, 10:49 PM
Any source would do, actually.
zoobyshoe
Jul3-04, 05:35 PM
In the above situation (space station, rocket ship) my speed away from the station is .5 c. The speed of the light departing the space station in my direction is c.
If I am, say, 300,000 km away from the station ( a distance selected for convenience) when the light is emitted from the station, what math do I use to determine 1.) how long it will take for the light from the station to catch up to me and 2.) How far away from the station I will be when it does catch up?
Assuming you are 300,000 km away from the station as measured in your frame, then:
(1) The light must travel 300,000 km to reach you, so T = D/c = 1 sec.
(2) During that 1 sec the station has traveled a distance D = VT = 0.5cT = 150,000 km. So when the light reaches you you'll be a total of 450,000 km from the station.
In the above situation (space station, rocket ship) my speed away from the station is .5 c. The speed of the light departing the space station in my direction is c.
If I am, say, 300,000 km away from the station ( a distance selected for convenience) when the light is emitted from the station, what math do I use to determine 1.) how long it will take for the light from the station to catch up to me and 2.) How far away from the station I will be when it does catch up?
As measured by you or the space station? If measured by you, the light will reach you in one sec and the station will be 450,000 km away when it reaches you.
If measured by the space station, then the light will reach you in 2 sec and you will be 600,000 km away from the station.
zoobyshoe
Jul3-04, 09:51 PM
Sorry I wasn't clear.
In the second question I meant to change my definition of myself as being motionless to being in motion at .5c relative to the source of the light - the space station.
So, If I am traveling at .5 c away from the station, and the beam of light is emitted from the station when I am a distance of 300,000 km away, how long will the light take to reach me, and how far away will I be from the station?
Sorry I wasn't clear.
In the second question I meant to change my definition of myself as being motionless to being in motion at .5c relative to the source of the light - the space station.
So, If I am traveling at .5 c away from the station, and the beam of light is emitted from the station when I am a distance of 300,000 km away, how long will the light take to reach me, and how far away will I be from the station?
Doc Al and I both gave you the answer already: 1 sec and 450,000 km, assuming that you are making the measurements.
zoobyshoe
Jul3-04, 11:04 PM
Doc Al and I both gave you the answer already: 1 sec and 450,000 km, assuming that you are making the measurements.
OK, What math did you use to arrive at the 2 sec, 600,000 km figures?
OK, What math did you use to arrive at the 2 sec, 600,000 km figures?
From the space stations perpective, the relative speed between the light it emits and you is .5c. Since it the distance between you and the station is 1 light sec at the time of emission, it takes 2 sec for the light to close the distance. If the light travels for 2 secs, it travels 600,000 km.
zoobyshoe
Jul4-04, 01:23 AM
OK the space station viewpoint makes sence to me, but I don't understand how there could be a difference from my perspective.
If the light is emitted from the station when I am 300,000 km away from it, it will take the light one second to reach that point. Once that second has elapsed and the light has reached that point, I will no longer be there. It will be one second later, and at .5c, I will be 150,000 km farther away, the light will not have caught up with me yet. In the half second it takes the light to reach that point 150,000km away, I will have traveled yet further, and won't be there when it arrives. And so on, such that I'm thoroughly confused as to why it isn't also 2 seconds 600,000 km viewed from my perspective as well.
If the light is emitted from the station when I am 300,000 km away from it, it will take the light one second to reach that point.
The light travels at speed c with respect to you. And, with respect to you, "that point" is you! The fact that you are also moving with respect to something else (the spaceship) is irrelevant. No matter what speed you are moving, if the light flashes when it's 300,000 km away from you, it will reach you in one second.
Once that second has elapsed and the light has reached that point, I will no longer be there.
In your frame, you don't move. (You are at rest in your own frame.) In the frame of the station you are moving, but that doesn't matter.
It will be one second later, and at .5c, I will be 150,000 km farther away, the light will not have caught up with me yet.
Yes, you will be 150,000 km further from the station. And yes the light will have caught up with you!
Just remember that in your frame light travels 300,000 km/sec with respect to you.
Hi, I'm new so please excuse me if I'm butting in. This is all very interesting but I'd like to add another observer to the seudo experiment. Assuming what Doc Ai and Janus are saying is correct (as far as I understand they are) then we get the following senario.
Let's say there are 2 space stations 450,000k apart. The rocket leaves the first station (call it "a") as outlined already heading towards space station "b". The light is still turned on when the rocket is 300,000k from space station "a".
Now the observer in the rocket sees the light from space station "a" when he is right beside space station "b" . Space station "b" cannot see the light for another 1/2 second.
Now lets suppose that the observer in the rocket has a mirror set up that aligns the light source from space station "a" with our observer at space station "b" as it passes so that he sees the light reflected from the rocket nearly a 1/2 second before he sees the light from space station "a". :surprise:
Hope I didn't botch that up too bad. It makes perfect sense to me (the experiment not the results that is).
zoobyshoe
Jul4-04, 02:33 PM
The light travels at speed c with respect to you.And, with respect to you, "that point" is you! The fact that you are also moving with respect to something else (the spaceship) is irrelevant.
The light is moving at c in its direction of propagation. That is: at any given time, the tip of the beam of light is propagating away from the point in space from which it was emitted, at speed c.
My speed in the same direction is relevant with respect to the tip of that beam of light. The light cannot reach me in one second, because I am moving away from its approaching tip. And: I cannot be aware of the beam of light to calculate anything about its time untill it first catches up with me. Were I able, by magic, to see the tip of the beam of light, I would not see it closing the distance between us at c. The time it takes for the light to reach me from the time I was at 300,000 km from the station must be greater than 1 sec.
Just remember that in your frame light travels 300,000 km/sec with respect to you.
If, once the light has caught up with me, I measure its speed, I will clock it to be going at c regardless of my speed in any inertial frame. Jcsd confirmed this above, and I believe this is the agreed result. Einstein used this mysterious fact as his second postulate.
That, however, is a separate issue from the fact that the tip of a beam of light is limited to propagation at c from its point of origin. It is for this reason that the space station could determine that there was a difference of .5 c between my speed, and the speed of the beam of light as it is on its way toward me.
The light is moving at c in its direction of propagation.
The light is moving at speed c with respect to the observer.
That is: at any given time, the tip of the beam of light is propagating away from the point in space from which it was emitted, at speed c.
No. From the viewpoint of the spaceship, the separation distance between the tip of the light beam and the space station increases at only 0.5 c. (It is only from the view of the space station that the light travels at speed c with respect to the space station.)
My speed in the same direction is relevant with respect to the tip of that beam of light. The light cannot reach me in one second, because I am moving away from its approaching tip.
Not true. You are moving away from the space station, but the light is still moving towards you at speed c. This is what "invariant speed of light" means.
And: I cannot be aware of the beam of light to calculate anything about its time untill it first catches up with me. Were I able, by magic, to see the tip of the beam of light, I would not see it closing the distance between us at c. The time it takes for the light to reach me from the time I was at 300,000 km from the station must be greater than 1 sec.
The only way to measure the speed at which the light got to you is to know when and where it started. You'll have to imagine an extended space ship frame--or another space ship moving along with yours but located next to the space station at the very moment the light is emitted. Your sister space ship has its clock synchronized to yours, and it is 300,000 km away from you. When the light reaches you, you note the time. Later you compare notes with the other ship, which tells you the time the light flashed: you will find that the light took 1 second to reach you, according to the synchronized clocks in the two ships.
If, once the light has caught up with me, I measure its speed, I will clock it to be going at c regardless of my speed in any inertial frame. Jcsd confirmed this above, and I believe this is the agreed result. Einstein used this mysterious fact as his second postulate.
Exactly right. But in your frame, the light only traveled 300,000 km to get to you. So it took 1 second.
That, however, is a separate issue from the fact that the tip of a beam of light is limited to propagation at c from its point of origin. It is for this reason that the space station could determine that there was a difference of .5 c between my speed, and the speed of the beam of light as it is on its way toward me.
The speed of light will always be c with respect to whatever observer is making the measurement. That does not mean that from your point of view that light cannot separate from a moving object at some speed less than c. Of course it can.
zoobyshoe
Jul4-04, 07:24 PM
Thanks for your patience, Doc Al. I have never explored this in such detail, and these answers are not what I expected.
If I reverse the direction of the rocket in the set up, then, to be headed toward the station, I take it that a beam of light emitted from the station when I am 300,000 km from the station will also reach me in one second, from my viewpoint, however: I will be only 150,000 km from the station when I detect it?
If I reverse the direction of the rocket in the set up, then, to be headed toward the station, I take it that a beam of light emitted from the station when I am 300,000 km from the station will also reach me in one second, from my viewpoint, however: I will be only 150,000 km from the station when I detect it?
Linguistic point:
The phrasing used here (e.g. "headed toward the station") carries the connotation that the rocket is moving and the station is stationary, which is, of course, the familiar setup - vehicles move and destinations don't. However, part of the essence of special relativity is that describing events for a given observer must always presume that observer as being at rest.
Thus while the mindset of a person on the rocket might be "I am approaching the station at .5c", the necessary viewpoint for setting up any special relativistic calculations is "the rocket is at rest, the station is approaching the rocket at .5c".
Once that framework is established, it becomes easier to see that a photon emitted from a station 300k km away takes 1 second to reach the rocket (since the rocket is at rest) no matter the velocity of the station.
My point is that it's easy for intuitive descriptions of a situation to carry assumptions that conflict with the requirements of the formal model.
zoobyshoe
Jul4-04, 08:32 PM
My point is that it's easy for intuitive descriptions of a situation to carry assumptions that conflict with the requirements of the formal model.
Point taken.
So, If I am at rest in my rocket with the station approaching me at .5 c and it emits a beam of light when it is 300,000 km away, I take it I will detect the light one second later, but the station will be 150,000 away at the moment I detect the light?
Yes.
And you now have one second left to figure out how to avoid a collision...
Moral:
Dont try and signal the incoming station with some kind of " Morse-flashlight"!
Switch off all non essential onboard lighting, close one's eyes and hope that Shroedinger's Cat exists as a physical process!
zoobyshoe
Jul4-04, 11:10 PM
Yes.
And you now have one second left to figure out how to avoid a collision...
Not sure it can be avoided:
The one second time for my reception of the light signal seems to me to be directly at odds with Einstein's understanding of the same situation as laid out in chapter IX of SR, The Relativity of Simultaneity
Here Einstein says:
"Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train.? We shall show directly that the answer must be in the negative.
When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A ---->B of the embankment. But the events A and B also correspond to the positions A and B on the train. Let M' be the mid-point of the distance A---->B on the traveling train. Just when the flashes1 of lightening occur, this point M' naturally coincides with the point M, but it moves toward the right in the diagram with the velocity v of the train. If an observer sitting in the position M' in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening toward the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.
1 As judged from the embankment.
If we substitute the situation where the space station is approaching me (from my perspective) for the situation where the observer on the train is approaching the flash of light from point B, you can see that Einstein would not reakon the time between the flash and when I detect it to be one second, rather, less than a second.
Likewise, if we substitute the situation where the station is moving away from me for the situation where the observer on the train is moving away from the flash of light at point A, you can see that Einstein would not reakon the time between the flash of light and when I detected it to be one second, but something greater than one second.
If, I detect the time of the light's travel to be one second coming or going, then Einstein has no basis on which to build his case for what he calls The Relativity of Simultaneity
Chapter 9. The Relativity of Simultaneity. Einstein, Albert. 1920. Relativity: The Special and General Theory
Address:http://www.bartleby.com/173/9.html
The observer in the rocket is the equivalent of the person on the railway embankment who sees the flashes simultaneously.
It is often the case that effects in relativity can appear paradoxical if only some of the effects are accounted for. When the mathematical formalism is used, these problems tend to disappear.
I assume the situation presented by Einstein is thus:
Points A and B are two points on a perfectly straight rail track. Observer X is at the midpoint between A and B. The midpoint of a train which X would measure as having a length equal to the distance from A to B is passing X at .5c. At the same moment (as measured in the rest frame of X and the track) photons are emitted at A and B. Observer Y is positioned at the midpoint of the train.
I will assume it is uncontroversial that these photons reach X simultaneously.
I will also assume that it is uncontroversial that the back of the train is at A when the photon is emitted at A, and that the front of the train is at B when the photon is emitted at B.
Now in order to see why these flashes do not reach Y simultaneously we first note that the train appears length contracted to X. Thus the length of the train at rest is actually longer than distance between A and B. Y, being at rest relative to the train, experiences the train as having that longer length. In addition, an observer on the train would measure the track as length contracted, so A and B are closer together for Y than for X.
So, according to Y, when the front of the train reaches B and the photon is emitted at B, the back end of the train is still (if I'm remembering my formulas right) 1/4 the length of the train away from A.
Thus, though Y would say that the photons traveled equal distances to reach the train's midpoint, Y would not say they were emitted at the same moment.
zoobyshoe
Jul5-04, 02:49 AM
The observer in the rocket is the equivalent of the person on the railway embankment who sees the flashes simultaneously.
This is incorrect. The observer on the train is the counterpart to me in the rocket.
Please go to the link I edited in in my post above for the complete text of that chapter written by Einstein as well as the diagram referred to.
The observer on the train is the counterpart to me in the rocket.
It is, of course, possible to construct an analogy where the observer on the train corresponds to the person in the rocket. In doing this, however, it is also necessary to state what the station(s) and any other relevant objects correspond to in the train scenario.
If you think the scenario I laid out in my previous post is somehow unfaithful to Einstein's thought experiment, could you explain?
I also note that the quote you give from the article describes what X predicts that Y will see (i.e. it is not a statement from the point of view of Y).
Chapters 10-12 of the Einstein article go over more about length contractions.
If I reverse the direction of the rocket in the set up, then, to be headed toward the station, I take it that a beam of light emitted from the station when I am 300,000 km from the station will also reach me in one second, from my viewpoint, however: I will be only 150,000 km from the station when I detect it?
That is absolutely correct. If the light flashes when it's 300,000 km from you, then it will reach you in 1 second regardless of your speed with respect to the light source. (Note that all measurements are made from your rocket frame.)
If we substitute the situation where the space station is approaching me (from my perspective) for the situation where the observer on the train is approaching the flash of light from point B, you can see that Einstein would not reakon the time between the flash and when I detect it to be one second, rather, less than a second.
Right! As measured by observers on the space station, the light reaches you in less than a second.
Likewise, if we substitute the situation where the station is moving away from me for the situation where the observer on the train is moving away from the flash of light at point A, you can see that Einstein would not reakon the time between the flash of light and when I detected it to be one second, but something greater than one second.
Right again! As measured by observers on the space station, in this case the light takes more than a second to reach you.
If, I detect the time of the light's travel to be one second coming or going, then Einstein has no basis on which to build his case for what he calls The Relativity of Simultaneity
Not so fast. Einstein's train example had two events (lightning strikes) happening simultaneously when observed in one frame and he showed that they must happen at different times according to the other frame. Nothing in your example contradicts this.
Everything you've said so far agrees with Einstein. To see how simultaneity fits in, consider how you measured the time between when the light was emitted and when you detected it in your rocket: Remember that you needed a second rocket in your frame (moving at the same speed as you) that just happens to pass the station when the light flashes. That rocket measures the time when the light was emitted, and you find out that the light reaches you 1 second later. But what does the space station think about the clocks in your frame? For one thing, the station frame does not agree that your two clocks are synchronized! According to measurements made in the space station frame: (1) your rocket clocks are out of synch, (2) your rocket clocks are operating slowly compared to station clocks (time dilation), and (3) the distance that you think is 300,000 km is really smaller (length contraction).
To really understand how to compare measurements made in the different frames, you need to consider all of those relativistic effects operating together. But everything is perfectly consistent with the reasoning we used from the rocket frame in determining that the light took 1 second (rocket time) to get from the station to the rocket. We didn't need to take all that stuff into consideration because we assumed that all measurements were made by the rocket frame: the distance the light traveled (measured by rocket rulers), the time light traveled (measured by rocket clocks). Of course these better combine to get the proper speed of light--measured by the rocket!
zoobyshoe
Jul5-04, 08:32 AM
That is absolutely correct. If the light flashes when it's 300,000 km from you, then it will reach you in 1 second regardless of your speed with respect to the light source. (Note that all measurements are made from your rocket frame.)
This seems to be at odds with what Einstein understood would happen, as I pointed out above.
zoobyshoe
Jul5-04, 09:27 AM
Right! As measured by observers on the space station, the light reaches you in less than a second.
Incorrect. The light from point B reaches me in less that one second as measured by me on the train (rocket). Einstein: "Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A." He is talking about the observer on the train (rocket). If you go to the link you'll see that he specifies this in the next sentence: "Obervers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A." The train, moving relative to the flashes of lightning, is the inertial frame which sees the flash it is "hastening toward" before the one it is "riding on ahead of".
Right again! As measured by observers on the space station, in this case the light takes more than a second to reach you.
Incorrect. As percieved by the observer on the train (rocket). You have argued that the observer on the train will percieve both flashes to take the same amount of time to reach him ( 1 sec, coming or going). Einstein is saying something else. Einstein is saying the observer on the train will ascribe a time of less than 1 sec for the flash he is "hastening toward", and a time greater than 1 sec for the flach he is "riding on ahead of".
Not so fast. Einstein's train example had two events (lightning strikes) happening simultaneously when observed in one frame and he showed that they must happen at different times according to the other frame. Nothing in your example contradicts this.
Yes, I am demonstrating by direct quotation from Einstein, that you do not agree with him on what the observer on the train will see. You have argued that the observer on the train will see the flashes of lightning as simultaneous: 1 sec coming or going. Einstein said: "Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A." In all cases the light has been emitted at exactly the same distance from the observer, train or rocket.
Everything you've said so far agrees with Einstein. To see how simultaneity fits in, consider how you measured the time between when the light was emitted and when you detected it in your rocket:...
None of that is relevant to which flash of light is seen first.
To really understand how to compare measurements made in the different frames, you need to consider all of those relativistic effects operating together...
Yes, but not relevant here. What is relevant is that Einstein believed that the observer on the train will see the flash he is "hastening toward" before the flash he is "riding on ahead of" despite the fact they were both emitted at the same distance from him.
If you argue that I, in my rocket ship, will see the flash of light 1 sec after it is emitted when the station is 300.000 kms away regardless of whether the station is "hastening toward" me, or whether it is "riding on ahead of" me, then Einstein has no basis on which to build his argument for what he calls The Relativity of Simultanaity. By your argument, the observer on the train will judge the flashes as simultaneous.
zoobyshoe
Jul5-04, 09:54 AM
It is, of course, possible to construct an analogy where the observer on the train corresponds to the person in the rocket. In doing this, however, it is also necessary to state what the station(s) and any other relevant objects correspond to in the train scenario.
I did this in my post above:
If we substitute the situation where the space station is approaching me (from my perspective) for the situation where the observer on the train is approaching the flash of light from point B, you can see that Einstein would not reakon the time between the flash and when I detect it to be one second, rather, less than a second.
Likewise, if we substitute the situation where the station is moving away from me for the situation where the observer on the train is moving away from the flash of light at point A, you can see that Einstein would not reakon the time between the flash of light and when I detected it to be one second, but something greater than one second.
If you think the scenario I laid out in my previous post is somehow unfaithful to Einstein's thought experiment, could you explain?
I didn't respond to the bulk of your post because it started with the false premise that the observer in the rocket corresponded to someone on the embankement in Einstein's version. That is false.
I also note that the quote you give from the article describes what X predicts that Y will see (i.e. it is not a statement from the point of view of Y).
This is false. The quote from Einstein does not contain any predictions by one observer about what another will see.
For some reason, you are mixing up Einstein's train example with your rocket example. They are not identical. (For one, the train embankment observers see two events happen simultaneously.) But if you understand Einstein's reasoning, then you can certainly apply it to your rocket and space station example as appropriate.
Don't make the naive comparison that since the train is "moving" that the train corresponds to the rocket in your example. It's not that simple. Incorrect. The light from point B reaches me in less that one second as measured by me on the train (rocket). Einstein: "Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A." He is talking about the observer on the train (rocket). If you go to the link you'll see that he specifies this in the next sentence: "Obervers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A." The train, moving relative to the flashes of lightning, is the inertial frame which sees the flash it is "hastening toward" before the one it is "riding on ahead of".
There is nothing wrong with Einstein's reasoning, but you misapply it to your rocket example. In Einstein's example, embankment observers see the light from B approach them at speed c. And they also see the train moving towards B. So they agree that the light from B reaches the train before it reaches the midpoint of the embankment. Everyone agrees that the train sees the light from B before the light from A.
What do the train observers think? If the lights were switched on just at the moment they passed the midpoint, then they would agree that the light from A and B would reach them at the same time. (To them, the light approaches at speed c.) After all, it's just the distance that the light source is from them when it flashes that matters, not how fast the light source is moving. But we know that the light from B reaches the train first, so the train observers deduce that in their frame the lights were not turned on simultaneously.
Now let's turn to your rocket example. What do we know? All we know is that according to the rockets the light was switched on when the light was 300,000 km from your rocket. And the rocket measures the speed of light to be c, so it takes 1 second for the light to reach the rocket. Where's the problem? All measurements are in the rocket frame. Now it is certainly true that the space station observers see the rocket approach them while the light moves away from them at speed c. So what? They measure different times and distances as well.
Incorrect. As percieved by the observer on the train (rocket). You have argued that the observer on the train will percieve both flashes to take the same amount of time to reach him ( 1 sec, coming or going). Einstein is saying something else. Einstein is saying the observer on the train will ascribe a time of less than 1 sec for the flash he is "hastening toward", and a time greater than 1 sec for the flach he is "riding on ahead of".
Again, you seem to naively compare Einstein's example to your own. They are not the same. Einstein is merely saying that the observers on the embankment can deduce that the light will take less time to reach the train because the train is moving towards the light source. So what?
Yes, I am demonstrating by direct quotation from Einstein, that you do not agree with him on what the observer on the train will see. You have argued that the observer on the train will see the flashes of lightning as simultaneous: 1 sec coming or going. Einstein said: "Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A." In all cases the light has been emitted at exactly the same distance from the observer, train or rocket.
Again, you take a statement I made about your rocket example and misapply it to Einstein's train example. They are not the same. I assure you that Einstein would agree that any observer will measure light to move at speed c (with respect to themselves) regardless of the motion of the light source.
None of that is relevant to which flash of light is seen first.
Again, you mix examples. In Einstein's example there are two flashes; in yours only one.
Yes, but not relevant here. What is relevant is that Einstein believed that the observer on the train will see the flash he is "hastening toward" before the flash he is "riding on ahead of" despite the fact they were both emitted at the same distance from him.
The reason that the train observers see the lights arrive at different times is that according to the train clocks light B was turned on first!
If you argue that I, in my rocket ship, will see the flash of light 1 sec after it is emitted when the station is 300.000 kms away regardless of whether the station is "hastening toward" me, or whether it is "riding on ahead of" me, then Einstein has no basis on which to build his argument for what he calls The Relativity of Simultanaity. By your argument, the observer on the train will judge the flashes as simultaneous.
Well I certainly make that argument in analyzing your rocket example. But you mistakenly think it directly applies to Einstein's train as well. But the train observers see light B turn on first: when the train passes the midpoint, the light from B is well on its way--and the light at A hasn't even turned on yet!
You may want to strengthen your understanding of Einstein's train gedanken experiment.
zoobyshoe
Jul5-04, 05:35 PM
For some reason, you are mixing up Einstein's train example with your rocket example. They are not identical. (For one, the train embankment observers see two events happen simultaneously.) But if you understand Einstein's reasoning, then you can certainly apply it to your rocket and space station example as appropriate.
There is no mixup: they are identical. The lighning flashes are simultaneous, not to the embankment in general, but at one specific spot: the exact midpoint M between point A and point B. Einstein: "When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A---->B of the embankment."
This mid-point M corresponds to the distance 300,000 kms from the space station in my rocket example because the light is emitted, in the train scenario, exactly at the moment the observer on the train is at the mid-point M between the two points. The observer continues to move and encounters the flash from B before the flash from A by his own reckoning not as seen from the embankment. The flash he percieves from point B corresponds to the flash from the space station as my rocket approaches it, (or it approaches my rocket if you want to define it that way) and the flash from point A corresponds to the flash from the space station as I am moving away from it (or it away from me).
The mid point M in the train scenario corresponds to the distance 300,000 kms in the rocket scenario because it is the same distance in the example where the station is approaching me, and also in the example where the station is moving away from me. In both cases the station emits light when the distance bewteen the station and my ship is 300,000 kms.
To make this as clear as possible lets say that when the distance between point A and point B in the train scenario was measured it proved to be exactly 600,000 kms. The midpoint, therefore, is located 300,000 kms from both A and B. Any value for the distance will do as long as point M is exactly midway between A and B. Using 300,000 kms helps you to see how the rocket example is the same as the train.
In Einstein's example, embankment observers see the light from B approach them at speed c. And they also see the train moving towards B. So they agree that the light from B reaches the train before it reaches the midpoint of the embankment. Everyone agrees that the train sees the light from B before the light from A.
In Einstein's example any observation anyone on the embankment might have about what the observer on the train sees is unimportant, and Einstein doesn't even mention it. What is important about any person on the embankment is that they will see the flashes of light from both directions at exactly the same time. That is important because Einstein wants to contrast it with what the observer on the train will see.
It is important to note that when the flashes occur, the observer on the train is located at the midpoint M between the two flashes. However, he moves away from that spot before the flashes arrive. The result is that he detects flash B before flash A, in spite of the fact he was at the midpoint when they occured!
By this reasoning, I would not detect the light beam from the space station, emitted when the distance between my rocket and the station was 300,000 kms, to be 1 sec, in any case of relative motion toward or away from the station (or it toward or away from me) By Einstein's reasoning, I will detect it in less than a second when the distance between my ship and the station is closing at .5c, and I will detect it in some time greater than 1 sec. when the distance between my ship and the station is widening at .5 c.
Again, If I detect it to be 1 sec coming or going Einstein has no basis upon which to build his case for what he calls The Relativity of Simultaneity.
I don't see ythe problem here, the confusion here is 1 sec for who? and 300,000 km, in whose reference frame? It's clear that the person on the rocket ship who measures the distnace to the spacestaion to be 300,000 km when the light is emitted, will measure the time taken for that light to cover that distance will be 1 second. It doesn't need any calculations or any thought experimnets to prove as it's a postulate of special relativity.
zoobyshoe
Jul5-04, 06:10 PM
I don't see ythe problem here, the confusion here is 1 sec for who? and 300,000 km, in whose reference frame? It's clear that the person on the rocket ship who measures the distnace to the spacestaion to be 300,000 km when the light is emitted, will measure the time taken for that light to cover that distance will be 1 second. It doesn't need any calculations or any thought experimnets to prove as it's a postulate of special relativity.
True, a postulate is a postulate. However, in this case, going by the postulate leads to a situation where there is no Relativity of Simultaneity. Going by the postulate, Einstein's observer on the train should see both light flashes as simultaneous. Einstein says he won't, however.
No, it's not inconsistent with the relativity of simulatneoty I think what you're missing is what the observer on the space station sees:
When the light is emitted, the rocket is approximately 340,000 km away, when it arrives at the rocket it is approximately 170,000 km away as viewed by the space station, and thus took approximately 0.6 secs to arrive at the rocket. You cannot ignore the fact that length and time change with reference frame when comparing the two examples.
There is no mixup: they are identical. The lighning flashes are simultaneous, not to the embankment in general, but at one specific spot: the exact midpoint M between point A and point B. Einstein: "When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A---->B of the embankment."
You are mixed up. In Einstein's example, the lightning flashes are simultaneous in the frame of the embankment. I have no idea what you mean by simultaneous at one specific spot.
This mid-point M corresponds to the distance 300,000 kms from the space station in my rocket example because the light is emitted, in the train scenario, exactly at the moment the observer on the train is at the mid-point M between the two points.
In Einstein's example the midpoint M is a fixed distance from the source B in the embankment frame. So, by analogy, your rocket frame corresponds to the embankment frame. (Sort of.) You've gotten them mixed up.
The observer continues to move and encounters the flash from B before the flash from A by his own reckoning not as seen from the embankment. The flash he percieves from point B corresponds to the flash from the space station as my rocket approaches it, (or it approaches my rocket if you want to define it that way) and the flash from point A corresponds to the flash from the space station as I am moving away from it (or it away from me).
The moving train corresponds to your space station frame. At least it would if your example were comparable to Einstein's, but it's not. :smile:
The mid point M in the train scenario corresponds to the distance 300,000 kms in the rocket scenario because it is the same distance in the example where the station is approaching me, and also in the example where the station is moving away from me. In both cases the station emits light when the distance bewteen the station and my ship is 300,000 kms.
It would make more sense if there was a moving observer at that midpoint. But there isn't is there?
To make this as clear as possible lets say that when the distance between point A and point B in the train scenario was measured it proved to be exactly 600,000 kms. The midpoint, therefore, is located 300,000 kms from both A and B. Any value for the distance will do as long as point M is exactly midway between A and B. Using 300,000 kms helps you to see how the rocket example is the same as the train.
The example with the train is very clear. What's not clear is what it has to do with your rocket? If the "rocket" is the midpoint, where's the moving train that passes by at just the right moment?
In Einstein's example any observation anyone on the embankment might have about what the observer on the train sees is unimportant, and Einstein doesn't even mention it. What is important about any person on the embankment is that they will see the flashes of light from both directions at exactly the same time. That is important because Einstein wants to contrast it with what the observer on the train will see.
No, Einstein uses the observations from the embankment to conclude draw conclusions about what actually happens and what the train sees. He deduces from what the embankment must see that the light reaches the moving train at different times. (Both frames agree with this.) And then the observers on the train use that fact to deduce that those lights cannot have been flashed simultaneously.
It is important to note that when the flashes occur, the observer on the train is located at the midpoint M between the two flashes.
Careful! To the observer on the train the flashes do not occur at the moment he passes the midpoint--they can't, since they happen at different times. Only from the embankment are the flashes observed to be simultaneous.
However, he moves away from that spot before the flashes arrive. The result is that he detects flash B before flash A, in spite of the fact he was at the midpoint when they occured!
It would be a contradiction to the basic assumption of the invariant speed of light if this were true. But it's not. In his frame, the flashes did not occur when he was at the midpoint.
By this reasoning, I would not detect the light beam from the space station, emitted when the distance between my rocket and the station was 300,000 kms, to be 1 sec, in any case of relative motion toward or away from the station (or it toward or away from me) By Einstein's reasoning, I will detect it in less than a second when the distance between my ship and the station is closing at .5c, and I will detect it in some time greater than 1 sec. when the distance between my ship and the station is widening at .5 c.
Your reasoning is incorrect, as I have explained. Don't blame Einstein.
Again, If I detect it to be 1 sec coming or going Einstein has no basis upon which to build his case for what he calls The Relativity of Simultaneity.
Sorry, but you don't seem to understand Einstein's argument. And, on top of that, you are misapplying Einstein's argument in comparing it to your rocket example.
I recommend that you work to understand Einstein's simple argument for the relativity of simultaneity. Maybe this will help: http://physicsforums.com/showpost.php?p=229410&postcount=2
I did this in my post above
If we substitute the situation where the space station is approaching me (from my perspective) for the situation where the observer on the train is approaching the flash of light from point B, you can see that Einstein would not reakon the time between the flash and when I detect it to be one second, rather, less than a second.
Likewise, if we substitute the situation where the station is moving away from me for the situation where the observer on the train is moving away from the flash of light at point A, you can see that Einstein would not reakon the time between the flash of light and when I detected it to be one second, but something greater than one second.
Ok, I see that you want to set stations as being at the points A and B on the train tracks. As Doc Al noted, you have to be careful here. To fill out the analogy, we can stick someone else in a rocket that's half way between the stations, and at rest with respect to the stations. We'll call this observer X. Now, the original specification of the scenario is that the flashes are simultaneous in X's reference frame - not yours.
Suppose you have a long rigid filament stretched out to the front and back of your rocket to correspond to the rest of the train. X says that each station emitted a flash at the same moment that you passed him, and that this is the same moment as the ends of the filament passed the stations.
Now remember, since you're in motion with repect to X, the filament appears length contracted to X. In your frame of reference the filament is not length contracted. Thus if X says that the length of the filament corresponds to the distance between the two stations, then to you, the filament must have a length that is greater than that, and therefore if each flash originates just as the end of the filament passes the station, then to you the flashes could not have originated simultaneously.
(See also below for warnings about the observer on the train regarding themself as being in motion.)
I didn't respond to the bulk of your post because it started with the false premise that the observer in the rocket corresponded to someone on the embankement in Einstein's version. That is false.
The bulk of my post contains no direct comparison to the rocket scenario. Its logic can be analyzed without reference to any outside analogy.
This is false. The quote from Einstein does not contain any predictions by one observer about what another will see.
The form that Einstein's argument takes over the course of the essay is to present the conclusions that can be drawn by a person on the embankment, and to show which common assumptions must be discarded in order to remove apparent contradictions that arise. It is not until the later sections of the essay (starting in Ch 11) that he starts showing what removing these assumptions implies for how time and space measurements differ for the observer on the train compared to the person on the ground.
You quote:
Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A.
Note that the "hastening towards" and "riding ahead" going on here are "considered with reference to the railway embankment". Remember - an observer on the train must regard the train as being at rest. An observer in special relativity can never describe themself as "hastening" or "riding" anywhere!
The quote continues:
Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
These statements are a conclusion (note the "hence") that can be drawn by the observer on the embankment (X) about what the observer on the train (Y) will see.
Thus Einstein is noting that X can conclude from his own space and time measurements that Y will see the photon from B before seeing the photon from A, but there is nothing that yet describes how time and space are arranged from Y's perspective.
One thing to keep in mind here is that observers in any inertial reference frame will agree on the order of events for any given spatial coordinate in other reference frames (thus both X and Y will agree on the order of events that each will see as X occupies a specific coordinate in X's reference frame and Y occupies a specific coordinate in Y's reference frame. What observers disagree on is whether events that are separated spatially in both frames are simultaneous (a difference which can lead to changes in the overall order of events depending on the particular reference frame).
zoobyshoe
Jul5-04, 11:41 PM
No, it's not inconsistent with the relativity of simulatneoty I think what you're missing is what the observer on the space station sees:
In the Einstein train scenario transfered to space, there is no observer at the space station. Nothing observed by any observer in the space stations inertial frame is of importance except to confirm, I suppose, that the observer in the rocket is located at exactly 300.000 kms from the station when the station emits light. We really don't need an observer for that because Einstein stipulates it.
When the light is emitted, the rocket is approximately 340,000 km away...
I have no idea where you are getting 340,000 km. It is stipulated that when the light flashes the observer in the rocket is 300,000 away from the station. It is a stipulation in both Einstein's train scenario (the observer on the train is at the midpoint when the light flashes) and the rocket scenario.
You cannot ignore the fact that length and time change with reference frame when comparing the two examples.
We can't take length and time dilation into account at all yet. Einstein is setting up his case for the existence of time dilation with this very scenario. Relativity of Simultaneity preceeds time and length dilation. He must first demonstrate Relativity of Simultaneity, in order to show why there is a need for time and length dilation. No Relativity of Simultaneity, no different reckoning of time in different reference frames.
Somehow, the postulate that light always propagates at c in all reference frames has gotten in the way of The Relativity of Simultaneity. I am surprised. This is passing strange.
zoobyshoe
Jul6-04, 02:33 AM
Now remember, since you're in motion with repect to X, the filament appears length contracted to X.
This filament and midpoint observer are completely unnecessary since Einstein stipulates that the flashes occur when the train observers position coincides with the midpoint between A and B. Also, as I pointed out to jcsd, at this point we aren't concerned with any time or length dilations. We have no need of them. They are immaterial to his point.
The form that Einstein's argument takes over the course of the essay is to present the conclusions that can be drawn by a person on the embankment,
In this chapter Einstein's point is to show there will be a difference of perception of when the flashes of light occur between the two reference frames: the moving train, and the embankment. He is not describing the embankment observers notion of what the observer on the train will see. He describes what the embankment observer will see, and separately, what the train observer will see. Neither speculates about what the other sees.
You quote:
The quote continues:
Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
These statements are a conclusion (note the "hence") that can be drawn by the observer on the embankment (X) about what the observer on the train (Y) will see.
I am astonished that you can accurately quote him as saying "Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A," and maintain he is talking about something an observer on the embankment is speculating about. Are you just yanking my chain?
Thus Einstein is noting that X can conclude from his own space and time measurements that Y will see the photon from B before seeing the photon from A, but there is nothing that yet describes how time and space are arranged from Y's perspective.
If what you just said here were true Einstein would have had no reason to write this chapter and call it The Relativity of Simultaneity. Where's the story, if all we're talking about is what the embankment observer sees vs what he thinks the train observer sees?
Back to the problem: If the calculation of 1 sec, 150,000 km for one flash, and 1 sec 450,000 km for the other flash is correct, then: 1 sec = 1 sec, and both the train observer and the embankment observer detect the flashes to be simultaneous. This leads to the conclusion there is no Relativity of Simultaneity. How can that be?
In the Einstein train scenario transfered to space, there is no observer at the space station. Nothing observed by any observer in the space stations inertial frame is of importance except to confirm, I suppose, that the observer in the rocket is located at exactly 300.000 kms from the station when the station emits light. We really don't need an observer for that because Einstein stipulates it.
Clearly the train scenario relies on two different inertial observers, otherwise the relativity on simultaneity cannot be demonstrated. Indeed if we're only talking about 1 inertial observer we needn't really bring in relativity to such a simple problem as long as we remember than the speed of light for that inertial observer will be c.
I have no idea where you are getting 340,000 km. It is stipulated that when the light flashes the observer in the rocket is 300,000 away from the station. It is a stipulation in both Einstein's train scenario (the observer on the train is at the midpoint when the light flashes) and the rocket scenario.
A got the the figures of 340,000 and 170,000 by simply applying a Lorentz transformation to the 4-vector postion of the space station a) when the light is emitted b) when it arrives. I think this is the vital point that you're missing that the distance between the rocket and the space station are NOT the same for the two different observers.
We can't take length and time dilation into account at all yet. Einstein is setting up his case for the existence of time dilation with this very scenario. Relativity of Simultaneity preceeds time and length dilation. He must first demonstrate Relativity of Simultaneity, in order to show why there is a need for time and length dilation. No Relativity of Simultaneity, no different reckoning of time in different reference frames.
No time dialtion and length contraction do not require the relativty of simultaneity to be demonstarted, they can be demonstrated with two events that have a definite ordering in 2 different reference frames. Indeed in the situation we describe there is no need to consider the relativity of simulatenoty (and if there was it would arise naturally out of considering time dialtion and length contraction) as both observers observe the light to be emitted before it hits the spaceship.
Somehow, the postulate that light always propagates at c in all reference frames has gotten in the way of The Relativity of Simultaneity. I am surprised. This is passing strange.
It doesn't get in the way it is easy to demonstare that both situations are consistent with relativity.
zoobyshoe
Jul6-04, 04:26 PM
Clearly the train scenario relies on two different inertial observers,
Agreed.
otherwise the relativity on simultaneity cannot be demonstrated.
Indeed if we're only talking about 1 inertial observer we needn't really bring in relativity to such a simple problem as long as we remember than the speed of light for that inertial observer will be c.
Right.
Lets ask though, if we need an observer at the space station? What would the space station correspond to in Einsteins's train scenario? It would correspond to point A or point B. These are the points at which the lightning hits the tracks. As space stations they are the points at which light is emitted. Did Einstein put any observers at points A or B? He did not. This is why I said we don't need an observer at the space station.
We do need an observer in the space station's inertial frame, though, as you pointed out. Where do we put him? Where did Einstein put him? At point M, the mid point between lightning flash A and lightning flash B.
What is that observer going to see? We don't have to calculate this or figure it out in any way. Einstein has carefully controlled everything in order that the observer at point M sees the two flashes of light as simultaneous. That is: he has carefully measured the distance from point A--->M and from B---->M, and determined them to be the same, and then he has stipulated that light will always travel at the same speed, regardless of direction of travel.
Because the two distances the light will travel are the same, and because light has been stipulated by Einstein to always travel at the same speed, there is no doubt whatever that the light from both lightning strikes will arrive at point M simultaneously.
That is why I said we don't need an observer at the space station: for the same reason we don't need an observer at the points where lightning strikes the rails. Einstein didn't have any observers at the point where lightning struck the rails. Why do we need anyone at the space station?
As for the observer who is at point M I said "We don't really need an observer for that because Einstein stipulates it." This was not incorrect. I am saying that Einstein has controlled all the conditions so carefully that there is no doubt about what an actual person at point M will see. He can't see anything but simultaneous flashes of light from A and B. That being the case, we don't actually have to station a person there to know what he'd see.
So we don't need an observer at the space station, and we have taken care of what the potential observer at point M (which would be a point 300,000 km from the space station on the rockets path in that example) will see by stipulation. There we have our two different inertial frames: one observer on the train, and one at point M.
I'll stop here to see if you understand and agree or disagree.
Agreed.
Right.
Lets ask though, if we need an observer at the space station? What would the space station correspond to in Einsteins's train scenario? It would correspond to point A or point B. These are the points at which the lightning hits the tracks. As space stations they are the points at which light is emitted. Did Einstein put any observers at points A or B? He did not. This is why I said we don't need an observer at the space station.
By saying that the light strikes simulatenously on the tracks he tacitly includes an observer in the rest frame of the tracks.
We do need an observer in the space station's inertial frame, though, as you pointed out. Where do we put him? Where did Einstein put him? At point M, the mid point between lightning flash A and lightning flash B.
It's not of great import where the obsrever is spatially locatedd as where only in differences in distances and times rather than definign a co-ordinate system.
What is that observer going to see? We don't have to calculate this or figure it out in any way. Einstein has carefully controlled everything in order that the observer at point M sees the two flashes of light as simultaneous. That is: he has carefully measured the distance from point A--->M and from B---->M, and determined them to be the same, and then he has stipulated that light will always travel at the same speed, regardless of direction of travel.
Because the two distances the light will travel are the same, and because light has been stipulated by Einstein to always travel at the same speed, there is no doubt whatever that the light from both lightning strikes will arrive at point M simultaneously.
That is why I said we don't need an observer at the space station: for the same reason we don't need an observer at the points where lightning strikes the rails. Einstein didn't have any observers at the point where lightning struck the rails. Why do we need anyone at the space station?
As for the observer who is at point M I said "We don't really need an observer for that because Einstein stipulates it." This was not incorrect. I am saying that Einstein has controlled all the conditions so carefully that there is no doubt about what an actual person at point M will see. He can't see anything but simultaneous flashes of light from A and B. That being the case, we don't actually have to station a person there to know what he'd see.
So we don't need an observer at the space station, and we have taken care of what the potential observer at point M (which would be a point 300,000 km from the space station on the rockets path in that example) will see by stipulation. There we have our two different inertial frames: one observer on the train, and one at point M.
I'll stop here to see if you understand and agree or disagree.
Ok, so the obsrever at M is tin the rest frma eof the space staitons right? and in the rest frame of the space staions the two flashes are emitted simulatneously? As I said before though the observer at M will be 340,000 km from the two space stations due to the effects of lenght contraction. Or are you changing it now so that in the space stations' rest frame rtaher than the rocket's the distance is 300,000 km. It cannot be the same for both observers.
zoobyshoe
Jul6-04, 11:36 PM
Ok, so the obsrever at M is tin the rest frma eof the space staitons right?
Right.
and in the rest frame of the space staions the two flashes are emitted simulatneously?
Absolutely. That's the way Einstein worked so hard to set it up.
As I said before though the observer at M will be 340,000 km from the two space stations due to the effects of lenght contraction.
Confusion. The observer at M is the one in the rest frame of the stations. He is at the mid point between the stations. The stations are 600,000 km apart. Therefore the observer at M is always located 300,000 km from both stations in the station's rest frame. (M stands for midpoint)
The observer in the rocket/train is the one in relative motion to the stations. He is different than the observer at M. We shouldn't refer to the guy in the rocket as "the observer at M."
Or are you changing it now so that in the space stations' rest frame rtaher than the rocket's the distance is 300,000 km. It cannot be the same for both observers.
I haven't changed any distance or anything else. What, in fact, I have done is to forget that at .5c the observer on the train will not see the 300,000 km distance as 300,000 kms, and I have erroneously referred to it as 300,000 even when speaking about his perspective. Sorry for that confusion.
Now, my reading of the Einstein leads me to conclude that he believes there is an instant in time when both the observer on the rocket/train and the observer at point M will agree that the observer on the rocket/train is located at point M and that it is at this instant the lights flash. I am not completely sure, but I get the impression that you have an objection to this, and that you don't think there can be an instant where both observers agree that the observer on the rocket is at point M. Do you have such an objection?
[The Einstein quotes in the following are taken from Chapter 9 of the essay Relativity: The Special and General Theory (http://www.bartleby.com/173/).]
When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A--->B of the embankment.
[§ 1]
This reiterates the definition of simultaneity given in Ch 8. As stated here, it uses observations taken at M as an indicator to define simultaneity specifically for events at A and B and only in the embankment frame. Under the definition of "'time' in physics" given in Ch 8, we may also say that any observer who shares the embankment frame (i.e. is at rest with respect to the observer at M) will agree that the flashes are simultaneous in that reference frame by making use of clocks or of some other suitable extension to the mechanism established in the footnote to Ch 8.
But the events A and B also correspond to positions A and B on the train.
[§ 2]
This could be confusing. Let's call the points on the train A' and B' to distinguish them from the points A and B on the embankment.
Further quotes from the Einstein essay are edited to reflect this change.
Let M' be the mid-point of the distance A'--->B' on the travelling train.
[§ 3]
Let's have a diagram:
[motion of train]________---->
[train ] ... ==== A'======= M'======= B'==== ...
[ground] ... ____ A _______ M _______ B ____ ...
Just when the flashes [as judged from the embankment] of lightning occur, this point M' naturally coincides with the point M ...
[§ 4]
The footnote interpolated here in brackets is not strictly necessary as there is only one reference frame (i.e. the embankment frame) for which we have established that the flashes are simultaneous, and thus only one reference frame in which the phrase "Just when the flashes ... occur" has any meaning.
In particular, we have no basis to assert that the three events - 1) flash at A, 2) flash at B, and 3) M' passes M - are simultaneous to an observer in the reference frame of the train.
There are two sets of events that can be asserted as simultaneous in both frames as they have been defined as such -- this definition being possible because the events occur (for the purposes of the argument) at the same location.
The first set is: 1) flash at A, 2) A' passes A.
The second set is: 1) flash at B, 2) B' passes B.
There is, of course, one other set of events which we can assert are simultaneous in the embankment frame: 1) flash from A arrives at M, and 2) flash from B arrives at M. (This assertion follows from the mechanism through which simultaneity of events separated in space was defined.)
...but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M' in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated.
[§ 5]
So, an observer in the reference frame of the train moves with the same velocity as the train, and consequently such an observer cannot be in the frame of the embankment.
Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A.
[§ 6]
Since the observer on the train at M' (the "he" of the preceding quote) is not at rest with respect to the observer at M (i.e. not at rest in the embankment frame), the observer at M will say that the observer at M' moved toward B during the period when light from the flashes was travelling from A and B to M, and thus that the observer at M' was moving toward the flash from B and away from the flash from A.
Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.
[§ 7]
Therefore, the observer at M may conclude, since the flash from B passed M' sometime before reaching M, and the flash from A will pass M' sometime after reaching M, an observer positioned at M' will witness the flash from B prior to witnessing the flash from A.
And if the preceding conclusion about an observer at M' is true, then, using the same procedures from Ch 8 mentioned above, the following conclusion may be made about any observers in the reference frame of the train:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
[§ 8]
Or in other words, events (separated in space) that are determined to be simultaneous by observers in the frame of the embankment could not be determined to be simultaneous by observers in the frame of the train.
If one were to work through all these steps again, only exchanging the assumption that the flashes are simultaneous in the embankment frame for the assumption that the flashes are simultaneous in the frame of the train, the complementary conclusion would be reached: that events (separated in space) that are determined to be simultaneous by observers in the frame of the train could not be determined to be simultaneous by observers in the frame of the embankment.
These two complementary conclusions comprise the definition of Relativity of Simultaneity given by Einstein. In his words:
Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity).
Zoobyshoe:
[§ 9]
I don't know what upsets you about this argument. At this point in history, however, trying to proclaim this argument to be false falls under the rubric of "Extraordinary claims require extraordinary proof".
I doubt that a comparison with the rocket scenario that you worked out is useful - nobody giving answers in the early part of this thread was limiting themselves to the assumptions employed by Einstein in Ch 1-9 of this essay, they were employing whatever knowledge of Special Relativity seemed appropriate. So it is likely that whatever conclusions you have made about the rocket scenario would need to be rejected on the same basis that you reject any mention of length contraction.
I will point out that it was not until post #43 of this thread that you made it explicit enough (at least to me - I can't be sure of what Doc Al and jcsd were thinking, but from their answers I doubt the assumption was clear to them either) that the reason that you insist that the person in the rocket should correspond to the observer on the train is that because the stations are emitting the photons and the lightning strikes the ground in the embankment frame, the stations need to be in the embankment frame.
It is not actually relevant what frame the photon emitters are in, what matters is that the observer in the embankment frame at M sees the flashes at the same moment. The constant velocity of light makes the velocity of the source irrelevant (and the reference frame of the source irrelevant) to when the flashes arrive at M. In addition, Einstein gives this as an empirical result in Ch 7.
It is much more important to have a thorough and explicit listing of the assumptions that are made about each frame. And to make sure that knowledge gained with reference to one frame is not applied to the other in unwarranted ways.
zoobyshoe
Jul7-04, 04:07 AM
Plover,
Excellent, excellent breakdown of the train scenario! You are an extremely patient, persistent, and articulate poster.
Good to have you here.
You have pointed out that the observer on the embankment will suppose the observer on the train sees one flash before the other. However, there is doubt in my mind as to whether you think Einstein is asserting that the man on the train will see one flash of light before the other in his reference frame on the train.
His wording is iffy at many points, but since he comes to the conclusion "Events which are simultaneous with reference to the embankment are not simultaneous with respect to thr train, and vice versa (relativity of simultaneity)." it seems unbelievable to me that he could only be talking about what the observer on the embankment thinks the observer on the train might be experiencing. If he were, why would he go on to conclude "Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of time." It would make no sense whatever for him to draw that conclusion unless he believed an actual observer seated on the train would actually see one flash before the other.
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It never occured to me that the flashes could come from the ends of the train just as well as from fixed points on the rails, but now that you point it out, I see that it is another valid possibility.
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I'm not claiming Einstein's argument is false.
Janus, Doc Al and you answered my questions about the rocket in such a way that it seems to me the observer on the train will see the flashes of light as simultaneous. If there is anything about the train scenario that I failed to carry over to the rocket questions, I am still not aware of what it was. If there is a different method that should be used to calculate how much time the man on the train will judge to have elapsed between the instant he is at the midpoint and when he detects the light and what should be used to determine the time that will elapse for the rocket man between the instant he hits the 300,000 km mark and when he detects the light, I will be very surprised: all the distances and speeds are the same.
I don't object to length contraction or time dilation being brought in if it can be used to explain that the observer on the train will see one flash of light before the other. He seemed to be on his way to use it to say something about how the rocket looks to station.
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I've reread your explanation of the reason the observer on the train might see one flash before the other. It is quite interesting, but is dependent on point M of the track and point M' of the train not actually corresponding when the light flashes for it to work. That would be fine except Einstein is, it seems to me, explicit about the fact that point M' of the train must correspond to point M of the embankment when the flashes occur. (I don't know why he even bothers to mention what the ends of the train are doing when he is so exacting in his description of the midpoints lining up.) It seems to me he is trying very hard to describe a situation exactly like what would happen if a rod sticking out from the outside of the train beside the observer on the train were to trip a switch located at point M on the embankment causing the flashes (the signal getting from the switch to the lights instantaneously, by magic, of course). Does that not seem to be the case to you?
Right.
Absolutely. That's the way Einstein worked so hard to set it up.
Confusion. The observer at M is the one in the rest frame of the stations. He is at the mid point between the stations. The stations are 600,000 km apart. Therefore the observer at M is always located 300,000 km from both stations in the station's rest frame. (M stands for midpoint)
The observer in the rocket/train is the one in relative motion to the stations. He is different than the observer at M. We shouldn't refer to the guy in the rocket as "the observer at M."
I haven't changed any distance or anything else. What, in fact, I have done is to forget that at .5c the observer on the train will not see the 300,000 km distance as 300,000 kms, and I have erroneously referred to it as 300,000 even when speaking about his perspective. Sorry for that confusion.
Right so that is cleared up.
Now, my reading of the Einstein leads me to conclude that he believes there is an instant in time when both the observer on the rocket/train and the observer at point M will agree that the observer on the rocket/train is located at point M and that it is at this instant the lights flash. I am not completely sure, but I get the impression that you have an objection to this, and that you don't think there can be an instant where both observers agree that the observer on the rocket is at point M. Do you have such an objection?
The problem is time is relative, there will be an instant in M's refernce frame where the rocket ship is at M and there will be an instant in the rocket ship's frame when he is at M. The problem is at the instant in M's frame that the rocket ship is at M this will be the instant the two lights flash, BUT in the rocket ship's frame at the instant that he is at M, one of the lights will already of flashed and the other light will not of yet flashed.
Janus, Doc Al and you answered my questions about the rocket in such a way that it seems to me the observer on the train will see the flashes of light as simultaneous.
I'm still not seeing how you conclude this based on what we have explained. You are drawing a confused analogy between Einstein's train and your rocket.
But in any case, I'd still like to know what you think is wrong with the simple calculation, done by the folks in your rocket, that shows that according to the rocket clocks the light takes 1 second to reach them from the time it leaves the station. You agree that the rocket frame has measured the distance from the station to the rocket to be 300,000 km at the time of emission, right? (That's given in the problem.) And you must also agree that the rocket frame observes the light to travel towards the rocket at speed c, right? (That's a postulate of relativity.) Since all measurements are made in the same frame, the conclusion follows from D = VT: The time (measured by rocket clocks) must be 1 second. There is no need whatsoever to invoke any measurements made in other frames. But you must think something is wrong with this calculation. What is it?
I've reread your explanation of the reason the observer on the train might see one flash before the other. It is quite interesting, but is dependent on point M of the track and point M' of the train not actually corresponding when the light flashes for it to work.
Einstein's conclusion is that it physically impossible for that condition to be satisfied.
That would be fine except Einstein is, it seems to me, explicit about the fact that point M' of the train must correspond to point M of the embankment when the flashes occur.
What Einstein arranges is that M' pass by the midpoint M exactly at the moment that the lights go off according to the embankment clocks. How can this be arranged? Imagine three synchronized clocks in the embankment frame: at A, M, and B. Exactly when those clocks each read 12:00 noon, the lights will flash. So M' is to pass M exactly when the M' clock reads 12:00 noon. That's the scenario.
Einstein argues that the observers on the train must conclude that according to their own clocks, those clocks on the embankment must not be synchronized because those flashes could not have occured simultaneously at the moment M' passed M according to observations made on the train.
Maybe the attached images will help.
Let's assume that you have two space stations and a rocket. the rocket is traveling from one space station to another. Leading and trailing the rocket is a line that extends an equal distance in both directions. on the end of each line is a device that will trigger a flash of light from each station as it passes the station. The length of the lines are such that, from both station's frame of reference, each end of a line touches a station when the rocket is midway between the stations (first image of the attachment).
Now let's look at the same situation from the frame of the rocket. Length contraction has shortened the distance between the stations, so now the ends of the lines extend past the stations positions when the rocket is at the midpoint. (second image) Since the stations will not emit the flashes of light until triggered by the ends of the line(the ends of the line physically trigger the flash), they do not do so when the rocket is at midpoint (according to the rocket).
Instead we see what is shown in the next two images. First the leadiing end of the wire triggers a flash from the station ahead of the rocket, and then sometime later, the trailing end of the line triggers the station behind the rocket.
Thus the events of the station emitting the flashes are simultaneous in the station frame, but not simultaneous in the rocket frame.
Now let's look at the same situation from the frame of the rocket. Length contraction has shortened the distance between the stations, so now the ends of the lines extend past the stations positions when the rocket is at the midpoint. (second image)
Note also that it is the station frame where the length of the line corresponds to the distance between stations. Since the station frame sees the line as length contracted, the rocket frame, in addition to measuring the distance between stations as shorter, also measures the line to be longer.
Note also that it is the station frame where the length of the line corresponds to the distance between stations. Since the station frame sees the line as length contracted, the rocket frame, in addition to measuring the distance between stations as shorter, also measures the line to be longer.
Yep. Exactly so.
zoobyshoe
Jul7-04, 11:41 PM
Maybe the attached images will help.
The situation you describe is extremely interesting, and I find it to be an excellent lesson in length contraction. I haven't had much occasion to think about the length contraction of something large as viewed from something small. I've really only thought about it in terms of Einstein's measuring rods getting shortened. So, by using this example of the ship with filaments you and Plover have gotten me thinking about how, when a ship is traveling at an appreciable fraction of c the very length of the distance it is traveling shrinks from its perspective.
Thank you for putting the graphics together. They, and your verbal description, are quite clear; easy to follow.
However, as I said to Plover when he brought this up, I don't think this particular example can be used to account for why Einstein maintains the observer on his train will see one flash before the other. The reason is that in your example he will not be at point M when the flashes occur, but somewhere else. Einstein is particularly specific in pointing out that the observer on the train will be at point M when the flashes occur: "Let M' be the mid-point of the distance A--->B on the traveling train. Just when the flashes of lightning occur (as judged from the embankment) this point M' naturally coincides with the point M, but it moves toward the right in the diagram with the velocity v of the train."
And if we're in any doubt as to which rest frame Einstein means when he says M' coincides with [/i]M[/i], he makes it specifically clear that he means, at that instant,
in both rest frames:"If an observer sitting at M' in the train did not possess this velocity, then he would remain permantly at [/M] and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated."
In the scenario with the filaments, if the observer at M' "did not possess this velocity then he would remain permanently at"...some point [i]before M.
The fact the light rays would reach him simultaneously if he did not possess this velocity should be completely convincing proof that Einstein wants him at M when the lights flash, and then wants him to clear out before the light arrives.
Doc Al asserts that Einstein later concludes this condition cannot be fullfilled. That would be a critical, crucial piece of imformation for me, but I don't know where to look for it. It isn't in that particular chapter. If Einstein says somewhere that his set up isn't workable, I can stop worrying.
I suggested a means to plover whereby this condition could be fullfilled, to my satisfaction anyway (he hasn't given his reaction) which would be to have a rod sticking out of the side of the train right where the train rider is seated. This rod would make contact with a switch located at point M on the embankment. The switch, when thrown by the rod would send an impulse (of magic, instantaneous energy) to the lights at A and B causing them to flash. In this way, we should be able to be assured that M' is at M when the lights flash.
Zoobyshoes the flashes CANNOT occur simulatnously in both rest frames, indeed in the situation described, no flashes will occur whilst the spaceship is at M', it's fairly trivial to prove this.
The problem is essientially one dimeionsal so if we set that the flashes occur at t = 0 and set M as the x = 0 of our co-ordiante system and L the distance from M to the spacesations (in the rest frame of M) then we get the following (x,t) co-ordinates for the situation as described by M as the flashes occur:
M = (0,0) A = (L,0) B = (-L,0)
but in the rest frame of M':
M' = (0,0) A' = (γL,-γβL/c) B' = (-γL,γβL/c)
Therefore the only way that the lights can be emitted simulatenously in bothe refebrce frames is if L = 0 or the relative velocity, u = 0
zoobyshoe
Jul8-04, 08:01 AM
jcsd,
I'm sorry but I can't follow your last post. I think what you might be doing is setting it up with an x,y,z, and a t (for time) coordinate. I am aware this can be done but I have never done it, and can't follow your logic. (I have worked only with x,y,z coordinates.)
If you are trying to demonstrate that the two observers will not agree about when the flash occured in units of time according a clock one or another has, that I can accept, provisionally.
Likewise, if one catches a glimpse of the others watch he will suppose the others watch is running slow.
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What is your assessment of what will happen in the situation where the rod triggers the flash, as I described to plover and Janus?
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Incidently, you keep turning it into a pair. It's just one shoe.
jcsd,
I'm sorry but I can't follow your last post. I think what you might be doing is setting it up with an x,y,z, and a t (for time) coordinate. I am aware this can be done but I have never done it, and can't follow your logic. (I have worked only with x,y,z coordinates.)
The event M is when the space ship is at the midpoint between the two spacesations I have defined this as happening in the rest frame of M at t =0 and and at point x=0, simlairly I defined the events A and B as the two flashes of light being emitted which also happen at in the rest frame of M at t = 0 (i.e. in this rest frame all 3 events are simultaneous) and at x = L and x = -L respectively.
I then applied a Lorentz transformation to find out what is happening in the rest frame of the spaceship and arrived at the above co-ordinates telling be that none of the evnts are simultaneous in this rest frame (γ and β are 1/√(1 - u^2/c^2) and u/c respectively where u is the relative velocity of the spaceship.[/quote]
If you are trying to demonstrate that the two observers will not agree about when the flash occured in units of time according a clock one or another has, that I can accept, provisionally.
Likewise, if one catches a glimpse of the others watch he will suppose the others watch is running slow.
Yep that's what I'm trying to show, but remember , assuming all clocks are 100% accuarte, no clock is better than any other clock, so both are objectively measuring the 'real distance' in time between the events.
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What is your assessment of what will happen in the situation where the rod triggers the flash, as I described to plover and Janus?
It's pretty much the same situation as the events will still have the same spacetime coordinates
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Incidently, you keep turning it into a pair. It's just one shoe.
I'll try to remember.
zoobyshoe
Jul8-04, 10:59 AM
The event M is when the space ship is at the midpoint between the two spacesations I have defined this as happening in the rest frame of M at t =0 and and at point x=0, simlairly I defined the events A and B as the two flashes of light being emitted which also happen at in the rest frame of M at t = 0 (i.e. in this rest frame all 3 events are simultaneous) and at x = L and x = -L respectively.
OK, I follow this.
I then applied a Lorentz transformation to find out what is happening in the rest frame of the spaceship I think I follow this. Let me check. You are saying you have shifted your perspective to that of the ship and applied the Lorentz transformation to find out what the observer on the ship will say about the timing of the flashes?
and arrived at the above co-ordinates telling be that none of the evnts are simultaneous in this rest frame
(γ and β are 1/√(1 - u^2/c^2) and u/c respectively where u is the relative velocity of the spaceship.
I recognise a Lorentz transformation in all this Greek. Don't know what &gamma and &beta, mean. "&radic" looks like it must mean "square root".
The results you gave earlier:
M = (0,0) A = (L,O) B = (-L,O)
M' = (0,0) A' = (?L, -??L/c) B'= (-?L, ??L/c)
show there is a difference, but I need a bit of an explanation. The parentheses each contain two coordinates. For M and M' are these x and t, respectively? For A and A', and B and B' these are length and time respectively? Also, I don't understand the signifigance of the question marks in the parentheses.
Yep that's what I'm trying to show, but remember , assuming all clocks are 100% accuarte, no clock is better than any other clock, so both are objectively measuring the 'real distance' in time between the events.
OK, you made a point of saying this, so I know it's important to your explanation, but I'm not sure what the phrase "`real distance´ in time" means. I´d appreciate it if you would expand a bit so I don´t miss the signifigance.
Einstein is particularly specific in pointing out that the observer on the train will be at point M when the flashes occur: "Let M' be the mid-point of the distance A--->B on the traveling train. Just when the flashes of lightning occur (as judged from the embankment) this point M' naturally coincides with the point M, but it moves toward the right in the diagram with the velocity v of the train."
Right. Note that Einstein explicitly says "as judged from the embankment".
And if we're in any doubt as to which rest frame Einstein means when he says M' coincides with M, he makes it specifically clear that he means, at that instant,
in both rest frames:"If an observer sitting at M' in the train did not possess this velocity, then he would remain permantly at M and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated."
There is no question that M' passes M at the the exact moment that the lights flashed according to the M-frame observers. Of course, both frames agree that the coincidence of M and M' happens "at the same instant". How could it be otherwise?
The fact the light rays would reach him simultaneously if he did not possess this velocity should be completely convincing proof that Einstein wants him at M when the lights flash, and then wants him to clear out before the light arrives.
Right. M' passes M just as the lights flash according to the M frame.
Doc Al asserts that Einstein later concludes this condition cannot be fullfilled. That would be a critical, crucial piece of imformation for me, but I don't know where to look for it. It isn't in that particular chapter. If Einstein says somewhere that his set up isn't workable, I can stop worrying.
Perhaps I wasn't clear. What I thought you were saying is that M' and M must coincide at the exact moment that the flashes are emitted according to both frames. This is, of course, impossible. (After all, as Einstein shows, the M' observers do not agree that the lights flashed at the same time.) All Einstein requires is that M' pass M just when the embankment frame says the lights flash.
I suggested a means to plover whereby this condition could be fullfilled, to my satisfaction anyway (he hasn't given his reaction) which would be to have a rod sticking out of the side of the train right where the train rider is seated. This rod would make contact with a switch located at point M on the embankment. The switch, when thrown by the rod would send an impulse (of magic, instantaneous energy) to the lights at A and B causing them to flash. In this way, we should be able to be assured that M' is at M when the lights flash.
I assume you are joking. If we start allowing magic, instantaneous messaging between the midpoint and the lights at A and B, then we have left the realm of physics. The way to set it up is as I already described. Three clocks (at A, M, and B) in the embankment frame, synchronized. Prearrange that when the clocks strike a certain time (say 12 noon) the lights will flash. Just have M' pass M exactly when the clock at M reads 12 noon.
OK, I follow this.
I think I follow this. Let me check. You are saying you have shifted your perspective to that of the ship and applied the Lorentz transformation to find out what the observer on the ship will say about the timing of the flashes?
Yep.
I recognise a Lorentz transformation in all this Greek. Don't know what &gamma and &beta, mean. "&radic" looks like it must mean "square root".
The results you gave earlier:
M = (0,0) A = (L,O) B = (-L,O)
M' = (0,0) A' = (?L, -??L/c) B'= (-?L, ??L/c)
That is odd, you should be able to see the html chartacters: here it is again in latex (also it's a zero not an 'O'):
M = (0,0)
A = (L,0)
B = (-L,0)
M' = (0,0)
A' = (\gamma L,\frac{-\gamma\beta L}{c})
B' = (-\gamma L,\frac{\gamma\beta L}{c})
Where:
\beta = \frac{u}{c}
\gamma = \frac{1}{\sqrt{1 - \beta^2}}
Where u is the relative velcoity of the spaceship to the space station
show there is a difference, but I need a bit of an explanation. The parentheses each contain two coordinates. For M and M' are these x and t, respectively? For A and A', and B and B' these are length and time respectively? Also, I don't understand the signifigance of the question marks in the parentheses.
What I've done is basically create a co-ordinate susyetm the co-ordinates for each event are (x,t) where x is the distance from the midpoint of the space station (i.e. x = 0 at M) and t is the difference in time from when the spaceship and M occupy the same spot (i.e. t = 0 as the spaceship arrives at M)
OK, you made a point of saying this, so I know it's important to your explanation, but I'm not sure what the phrase "`real distance´ in time" means. I´d appreciate it if you would expand a bit so I don´t miss the signifigance.
The significance is that both distance and time are not absolute.
I suggested a means to plover whereby this condition could be fullfilled, to my satisfaction anyway (he hasn't given his reaction) which would be to have a rod sticking out of the side of the train right where the train rider is seated. This rod would make contact with a switch located at point M on the embankment. The switch, when thrown by the rod would send an impulse (of magic, instantaneous energy) to the lights at A and B causing them to flash. In this way, we should be able to be assured that M' is at M when the lights flash.
I hope you realise that introducing a method of instantaneous transmission violates the very principles we are trying to clarify. The best we could do is place such rods an equal distance ahead of and behind the train rider, space so that when form the rider's perspctive, these rods trigger switches at the stations themselves which initiate the the flashes at each station.
This just gives us the reverse situation as we had in the last set of images I made.
This time, from the embankment frame, the ends where the poles don't even reach to the stations when the ship is at the midpoint. (image 2)
So the sequence from the embankment would go like this:
First the trailing station is switched on (third image) and then the leading station is triggered (fourth image).
Thus taking both this and the earlier attachment into account, one sees that the stations can either flash simultaneously in the station frame and not so in the train frame, or they can flash simultaneously in the train frame, but not in the station frame. But they cannot flash simultaneously in both frames. (assuming that each staion is only triggered once)
This was a really interesting article to read!
May I ask if the same thing would occur if someone threw a ball (traveling 299,999.99 km per second) at the ship instead of sending a light beam towards it? If the ship was 299,999.99 km away from the station and the ship was moving away from the station at 150,000 km per second, (or the station was moving away from the ship at 150,000 km per second) would the ball take 1 second to reach the ship (relative to the observer on the ship) or would it take well over one second?
May I ask if the same thing would occur if someone threw a ball (traveling 299,999.99 km per second) at the ship instead of sending a light beam towards it? If the ship was 299,999.99 km away from the station and the ship was moving away from the station at 150,000 km per second, (or the station was moving away from the ship at 150,000 km per second) would the ball take 1 second to reach the ship (relative to the observer on the ship) or would it take well over one second?
Same thing. It would take about 1 second (according to observers on the ship) for the ball to reach the ship, assuming that the speed of the ball is 299,999.99 km/s with respect to the ship and that the distance of 299,999.99 km is as measured by the ship. (We'll ignore any difficulties involved in actually throwing a ball with that speed.)
Even if that speed was the speed of the ball with respect to the space station, it would turn out that the speed as seen by the ship would still be pretty close to c. (See relativistic addition of velocities.)
zoobyshoe
Jul9-04, 05:08 AM
Right. Note that Einstein explicitly says "as judged from the embankment".
Noted.
There is no question that M' passes M at the the exact moment that the lights flashed according to the M-frame observers. Of course, both frames agree that the coincidence of M and M' happens "at the same instant". How could it be otherwise?
Super!
Perhaps I wasn't clear. What I thought you were saying is that M' and M must coincide at the exact moment that the flashes are emitted according to both frames. This is, of course, impossible. (After all, as Einstein shows, the M' observers do not agree that the lights flashed at the same time.) All Einstein requires is that M' pass M just when the embankment frame says the lights flash.
What I am saying is that there is an instant when M' and M coincide in time and location, and they would both agree such an instant existed. (This, I believe, is how jcsd has set up his coordinate system. M = (0,0) and M' = (0,0) ).
In the M frame (embankment) the flashes of light are emitted at t=0.
However, at t=0 neither M nor M' knows the light has been emitted. Only Einstein and we know that.
The way to set it up is as I already described. Three clocks (at A, M, and B) in the embankment frame, synchronized. Prearrange that when the clocks strike a certain time (say 12 noon) the lights will flash. Just have M' pass M exactly when the clock at M reads 12 noon.
Your clocks will do wonderfully, thanks.
zoobyshoe
Jul9-04, 05:49 AM
here it is again in latex
Ok. This is much better. Thank you for translating it to LaTex.
What I've done is basically create a co-ordinate susyetm the co-ordinates for each event are (x,t) where x is the distance from the midpoint of the space station (i.e. x = 0 at M) and t is the difference in time from when the spaceship and M occupy the same spot (i.e. t = 0 as the spaceship arrives at M)
Clear. I will plug the .5 c into these and try to make sure I understand what they are saying. Thanks.
zoobyshoe
Jul9-04, 05:56 AM
I hope you realise that introducing a method of instantaneous transmission violates the very principles we are trying to clarify.
At some point in the future I might like to start a thread exploring what criteria must be met by a given suggested gedanken fiction for it to be acceptable. That is moot now since I don't need it any more. Doc Al's three clocks, and 12 noon stipulation work fine.
Good to have you here.
Thanks.
(You gotta love a place where you can get respect for waxing excruciatingly pedantic on a passage from Einstein... :biggrin: )
Note on the following:
For the sake of syntactic (and my own) sanity, I'll refer to my prior post analyzing the Einstein passage as EP (for Einstein Post). I've added section numbers to it, and the notation EP(n) indicates section n of the post.
You have pointed out that the observer on the embankment will suppose the observer on the train sees one flash before the other. However, there is doubt in my mind as to whether you think Einstein is asserting that the man on the train will see one flash of light before the other in his reference frame on the train.
The short answer: yes, I do think Einstein is asserting that. (I can also agree that the degree to which EP(7) makes that explicit is less than absolute. :smile:)
His wording is iffy at many points,
I agree, it has its ambiguous spots. It would be good if we could know whether it was Einstein or the translator who is responsible for that. Unfortunately, I don't know German. Anyone out there want to comment on the translation of Ch 9 of Einstein's book?
but since he comes to the conclusion "Events which are simultaneous with reference to the embankment are not simultaneous with respect to thr train, and vice versa (relativity of simultaneity)." it seems unbelievable to me that he could only be talking about what the observer on the embankment thinks the observer on the train might be experiencing.
Einstein is making a much stronger statement than this. The underlying logic of the argument for the paragraph analyzed in EP is (as I read it):
Given the rigorous way in which the circumstances have been defined, what conclusions must an observer on the embankment at M deduce about the experience of an observer on the train at M'?
(The conclusion in this case being, of course, the sentence you just quoted: "Events which ... (relativity of simultaneity).")
I'm not claiming Einstein's argument is false.
I did not really expect that you were. But given the number of threads around here called "Special Relativity Must Die" or some such, I thought it best to make completely sure... :smile:
I don't object to length contraction or time dilation being brought in if it can be used to explain that the observer on the train will see one flash of light before the other.
This is not the impression I received from your comments in post #41. Thus the method of EP was to try to use only the premises that Einstein used up through Ch 9 to make the argument. While Einstein doesn't say anything that conflicts with relativity, he does introduce common sense notions and later discards them when they are no longer tenable in the framework he is unfolding, and this might lead to points where certain types of comparisons become muddy.
I've reread your explanation of the reason the observer on the train might see one flash before the other.
Which one are you referring to? The train version in post #25 of the thread, or the rocket/filament one in post #39? (The arguments in these are the same but the context might affect something.)
[It] is dependent on point M of the track and point M' of the train not actually corresponding when the light flashes for it to work.
How so?
That would be fine except Einstein is, it seems to me, explicit about the fact that point M' of the train must correspond to point M of the embankment when the flashes occur.
EP(4) addresses this.
(I don't know why he even bothers to mention what the ends of the train are doing when he is so exacting in his description of the midpoints lining up.)
I suspect that one reason that Einstein chooses a train as the vehicle (no pun inten... oh, who am I kidding?) for his argument is that, as an object that both moves and extends a considerable distance along the line of its motion, it corresponds well to an entire axis of a second reference frame, and thus makes the idea of a moving system of coordinates less abstract than would a moving object that functions in a more point-like way in the imagination.
As the quote preceding EP(1) shows (and Ch 8 goes over in more detail), the point M is not determined a priori but is constructed as part of the mechanism for determining the simultaneity of the flashes at A and B. Similarly, the points A' and B' are used to fix the point M'.
Note that A' and B' are defined in reference to the embankment frame. A' is the point in the train frame that happens to coincide with A when the flash takes place at A. B' is defined similarly.
It seems to me he is trying very hard to describe a situation exactly like what would happen if a rod sticking out from the outside of the train beside the observer on the train were to trip a switch located at point M on the embankment causing the flashes (the signal getting from the switch to the lights instantaneously, by magic, of course). Does that not seem to be the case to you?
As Einstein sets up this example, the point M as I pointed out above, is determined by the points A and B. So if anything it is like the flashes cause someone to be magically teleported to M (ploink!) in order to determine whether or not they were simultaneous.
In the interest of annoying or appeasing everyone evenly (as the case may work itself out), I will suggest that it was injudicious to introduce magical phenomena into a physics discussion that is treating the foundation of relativistic time effects, and which currently seems to turn on the confusion caused by minute differences in definition of terms (and especially when the phenomena introduced are not accompanied by appropriate sound effects); and on the other hand, I will also suggest that commenters could cut some slack for the use of what was essentially (and syntactically marked as) a rhetorical device, and one which, by its own rules (and considered as an "as if"), is not technically incorrect.
(Continued in next post - broken up due to length restriction.)
(Continued from previous post.)
However, as I said to Plover when he brought this up, I don't think this particular example can be used to account for why Einstein maintains the observer on his train will see one flash before the other. The reason is that in your example he will not be at point M when the flashes occur, but somewhere else.
This is an example of why I think that the answers from early in this thread (while correct) may be misleading if introduced to this stage of Einstein's argument.
One thing that needs to be agreed upon is the purpose of Ch 9. I have stated in a couple of places (in slightly different terms) that I believe that the intent is to show what may be deduced about the perception in the train reference frame of events determined to be simultaneous in the embankment frame.
The emphasis of this needs to be changed slightly however: the perceptions that are being examined in the train reference frame are restricted to temporal perceptions. Thus, while a fully worked out framework for Special Relativity offers the result, there is no mention made, or circumstances implied as to spatial relationships as perceived in the train reference frame.
There are several spatial relationships established over the course of the passage, but each is described from the embankment frame. These are (see also EP(4)):
A' coincides with A at the moment of the flashes.
B' coincides with B at the moment of the flashes.
M' coincides with M at the moment of the flashes.
M' is somewhere to the right of M at the moment it passes the flash from B, and somewhere further to the right when the flash from A passes it.
As noted in EP(4) the only frame for which "the moment of the flashes" is defined is the embankment frame. Indeed the conclusion of the argument is that the phrase "the moment of the flashes" turns out to be nonsensical in the train frame.
To make the complaint that an observer at M' "will not be at point M when the flashes occur" is to anticipate the argument by using an assumption about time in the train frame as a premise rather than a conclusion.
And if we're in any doubt as to which rest frame Einstein means when he says M' coincides with [/i]M[/i], he makes it specifically clear that he means, at that instant, in both rest frames:"If an observer sitting at M' in the train did not possess this velocity, then he would remain permantly at [/M] and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated."
There is a moment when M and M' coincide.
In each reference frame it happens at the particular moment when the two points coincide (however that moment is defined by clocks located in the respective frames).
The two points do not continue to coincide because they are in different frames.
In the embankment frame, that moment also happens to be the one when the flashes occur. There is nothing in the sentence you quote that warrants saying anything about the flashes in the train reference frame. The path of the argument is, in fact, to lead you to the conclusion that our knowledge about the flashes in the embankment frame forces the conclusion that they will not be considered simultaneous in the train frame.
At this stage of the argument, there is no point where the position of the observer at M' can be defined - as considered from the train reference frame - when either flash occurs.
The only point under consideration is whether our knowledge of the simultaneity of the flashes in the embankment reference frame allows us to maintain that they will also be simultaneous in the train reference frame. And of course, as it turns out, we may not maintain that.
In the scenario with the filaments, if the observer at M' "did not possess this velocity then he would remain permanently at"...some point before [i]M.
I find this and the points following it rather confusing. Perhaps you could restate them in terms of my comments above. (Some point "before" M? While such must have occurred, it is not mentioned anywhere in the argument.)
zoobyshoe
Jul10-04, 01:39 AM
Plover,
We are definitely at odds about what Einstein has set up in terms of where M' is when the flashes occur, whose perspective is the one he wants us to be aware of in any given sentence, and the function of A' and B' of the train.
Consider the first sentence of the chapter:
"Up to now our considerations have been refered to a particular body of reference, which we have styled a "railway embankment".
By "our considerations" he means his own and his reader's.
He goes on from here to "refer" his and his readers considerations from the view of someone on the railway embankment to that of someone on the train. From this point on, we are in the train frame of reference, except where otherwise noted.
There is no double imagining going on: we are not imagining ourselves to be in the frame of the embankment imagining the frame of the train. We are in one or the other, according to where he asks us to be, but never in one imagining the other.
-------
Also: he never says that A' and B' "coincide" with events A and B. He uses the word "correspond":
"But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A---->B. Just when the flashes (footnote: As judged from the embankment) of lightning occur, this point M' naturally coincides with the point M..."
And he uses the word "coincides" when speaking of the relationship of M to M'. So, "correspond" and "coincide" are not describing the same thing, although I am not at all certain what relationship he means to imply when he uses "correspond". This is where he is "iffy".
-------------------
This sentence is explicit and crucial:
"If an observer sitting in the position M' of the train did not posses this velocity, then he would remain permanently at M', and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated."
That sentence refers to the reference frame of the train. Einstein is speaking to us concerning a point he wants us to understand about the reference frame of the train.
The couple times he deviates from the reference frame of the train are specified: once with the words in parentheses "(considered with reference to the railway embankment)" and another time with a footnote.
zoobyshoe
Jul10-04, 06:04 PM
\beta = \frac{u}{c}
\gamma = \frac{1}{\sqrt{1 - \beta^2}}
Where u is the relative velcoity of the spaceship to the space station
Because I am a math and latex basket case you will, I am sure, be a sport and not make viscious fun of me for approaching this "trivial" problem at a ponderously slow pace.
For \beta I get: .5
u = .5 c = 150,000 km/s
\beta = \frac{u}{c} therefore \beta = 150,000/300,000 = .5
For \gamma I get: 1.1543958
.52 = .25
1- .25 = .75
\sqrt{.75} = .8660254
1/.8660254 = 1.1547005 = \gamma
So, before I proceed, would you be kind enough to glance at this and make sure I've gotten this far correctly using .5c as the ships speed.
Thanks
So, before I proceed, would you be kind enough to glance at this and make sure I've gotten this far correctly using .5c as the ships speed.
Looks good to me.
zoobyshoe
Jul11-04, 01:15 AM
OK. My next step, I believe, should be to look at the distance L. In the inertial frame of the station the distance L = 300,000 km. It will not be the same distance to the ship by virtue of the ships speed. To the ship the length L will be contracted. The ship will find L to measure something less than 300,000 km.
jcsd has provided the formula \gamma L to use to determine what the ship will find L to be from its perspective.
He has provided this formula:
\gamma = \frac{1}{\sqrt{1 - \beta^2}}
to describe "gamma"(\gamma).
The value I got when I solved for gamma earlier using .5 c for the relative velocity of the ship and the station was
1.1547005.
1.1547005 times 300,000 = 346,410.15 km
346,410.15 km > 300,000, not < 300,000.
Instead of contracting, the length has dilated!
jcsd has given us the wrong formula. He has given us the formula for time dilation, not length contraction.
The formula for length contraction is simply:
\sqrt {1 - \frac {v^2}{c^2}}
the result is the percentage of the length that the observer will see that length contracted to.
In our example, the ship will see the 300,000 km to have contracted to .8660254 % of its original length.
That comes out to be 300,000 times .8660254 = 259,807.62 km.
Actually jcsd is correct.
When he writes
M=(0,0)
A=(L,0)
B=(-L,0)
He is stating that the distance to either A or B from M is L and thus each half of the train measured in the embankment frame is also length L (so total length 2L). However, the train is moving in the embankment frame and thus appears contracted.
So the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed \gamma L (an un-contraction, if you will). This is the first value you calculated (~346k).
However, in the train reference frame the track is moving and thus the distance from M to A appears contracted. This length is usually expressed as \frac{L}{\gamma}. This is the second value you calculated (~260k).
OK. My next step, I believe, should be to look at the distance L. In the inertial frame of the station the distance L = 300,000 km. It will not be the same distance to the ship by virtue of the ships speed. To the ship the length L will be contracted. The ship will find L to measure something less than 300,000 km.
Two comments:
(1) Unless I haven't been paying close enough attention, it looks like you have changed the scenario a bit from your original one. I thought that your original scenario had the distance = 300,000 km as measured in the rocket frame, not the station frame. But to get a close analogy to Einstein's train problem, you should have the distance measured in the station frame. That way the station frame corresponds to the embankment frame. (But realize the answer to your original question will be different now.)
(2) Yes, if the distance is L in the station frame, the rocket will measure that distance to be shorter.
jcsd has provided the formula \gamma L to use to determine what the ship will find L to be from its perspective.
I believe you are misinterpreting what jcsd provided. jcsd analyzed the Einstein train example from both the embankment and the train frames. He gave the space-time coordinates of three events: A (the light flash at A), M (the passing of M' by M), and B (the light flash at B). So, when jcsd gave these equations:
M = (0,0)
A = (L,0)
B = (-L,0)
he was giving the location and time coordinates of these events according to the embankment frame. Thus the three events happen at the same time (T=0) and A and B happen at positions -L and +L from the midpoint.
When he gave these equations:
M\'\; = (0,0)
A\'\; = (\gamma L,\frac{-\gamma\beta L}{c})
B\'\; = (-\gamma L,\frac{\gamma\beta L}{c})
he was providing the location and time coordinates of these same events, but according to the train frame. Note that the times are different.
\gamma L is not the distance L as measured by the train! It is the position coordinate of an event.
zoobyshoe
Jul12-04, 04:57 AM
Two comments:
(1) Unless I haven't been paying close enough attention, it looks like you have changed the scenario a bit from your original one. I thought that your original scenario had the distance = 300,000 km as measured in the rocket frame, not the station frame.
If the rocket measures it to be 300,000 km in its own frame, the station will also measure it to be 300,000 in its own frame. Each will measure it to be shorter in the other's frame, however.
(2) Yes, if the distance is L in the station frame, the rocket will measure that distance to be shorter. Kinda goes without saying.
I believe you are misinterpreting what jcsd provided.
What I've done is basically create a co-ordinate susyetm the co-ordinates for each event are (x,t) where x is the distance from the midpoint of the space station (i.e. x = 0 at M) and t is the difference in time from when the spaceship and M occupy the same spot (i.e. t = 0 as the spaceship arrives at M)
He says: "x is the distance from...", so, the L given by jcsd represents the lengths M--->A and M---B respectively.
jcsd analyzed the Einstein train example from both the embankment and the train frames. He gave the space-time coordinates of three events: A (the light flash at A), M (the passing of M' by M), and B (the light flash at B). So, when jcsd gave these equations:
[INDENT]M = (0,0)
A = (L,0)
B = (-L,0)
he was giving the location and time coordinates of these events according to the embankment frame. Thus the three events happen at the same time (T=0) and A and B happen at positions -L and +L from the midpoint.
I would buy what you say about -L and +L as positions, except you can't apply a Lorentz transformation to a point. A point has no dimensions.
When he gave these equations:
M\'\; = (0,0)
A\'\; = (\gamma L,\frac{-\gamma\beta L}{c})
B\'\; = (-\gamma L,\frac{\gamma\beta L}{c})
he was providing the location and time coordinates of these same events, but according to the train frame. Note that the times are different.
Yes, I noticed that the second term in the parentheses for A' and B' were different. I found that mighty peculiar also, since no time has passed yet. We are still at T=0 for both reference frames.
\gamma L is not the distance L as measured by the train! It is the position coordinate of an event.
Whatever it is, he tried to apply the Lorentz tranformation for time dilation to it. There's no good reason to do that to a length or a point in a coordinate system.
zoobyshoe
Jul12-04, 05:37 AM
Actually jcsd is correct.
When he writes
M=(0,0)
A=(L,0)
B=(-L,0)
He is stating that the distance to either A or B from M is L and thus each half of the train measured in the embankment frame is also length L (so total length 2L). However, the train is moving in the embankment frame and thus appears contracted.
So the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed \gamma L (an un-contraction, if you will). This is the first value you calculated (~346k).
However, in the train reference frame the track is moving and thus the distance from M to A appears contracted. This length is usually expressed as \frac{L}{\gamma}. This is the second value you calculated (~260k).
Sophist.
Any length that one observer measures to be contracted will be its original length when "uncontracted, not greater than its original length.
Physicsfreek
Jul12-04, 06:56 AM
Below is an excerpt (Chapter 12) from my upcoming book, "Space, Time and Elementary Particles. (Proof in chapers 13 and 14, not attached)
Chapter 12 The Speed of Light and Space-Time
http://www.physicsforums.com/attachment.php?attachmentid=1164&stc=1
c = d / [(t2-t0) / 2]
The speed of light is measured with the Fizoau procedure (for Armand Hippolyte Louis Fizoau (1819-1896) a French physicist who, as one of his many accomplishments, was the first to measure accurately the velocity of light (on Earth) in 1849. The following is a description of the Fizoau procedure.
Definition: The speed of light is the speed of propagation of electromagnetic waves in a vacuum, which is a physical constant equal to exactly 299 792.458 km/s.
McGraw-Hill Dictionary of Scientific & Technical Terms
The Fizoau procedure takes a beam of light, bounces it off a mirror a measured distance away and then divides that distance by one-half of the measured time for the round trip: d / [(t2 - t0) / 2]. A photon is emitted from RSX at time t0 toward the mirror M that reflects the photon at time t1. RSX sees the photon at time t2. Since the speed of light is a constant, the two world lines of the photon (RSXt0 to Mt1 and Mt1 to RSXt2) are equal.
Fizoau set up an apparatus, which consisted of a rapidly spinning wheel with tooth gaps evenly spaced around the edge of the wheel. At a point (RSX), Fizoau directed a beam of light through the gaps at the edge of the rapidly rotating wheel towards a mirror eight kilometers away. The speed of the wheel was adjusted until the returning light could be seen through the next gap. Fizoau, by measuring exactly how fast the wheel was turning, could calculate an exact time for one gap on the edge of the wheel to be replaced at a specific location by the next gap around the edge of the wheel. After taking many readings, Fizoau calculated the speed of light to be 315,000 kilometers per second. Leon Foucault improved on that measurement a year later by using rotating mirrors and attained a speed of 298,000 kilometers per second. Foucault’s technique was accurate enough to confirm that light travels slower in water than in air.
Different forms of Fizoau’s experiment have been carried out many times over the years with a longer or shorter distance d, using a different frequency of photon, at different locations on the earth, at different times of day, with different positions for the apparatus and with different reference systems in motion relative to each other. The answer is always 299,792.458 kilometers per second exactly (in a vacuum). The speed of light is a constant for all observers that are in motion with a constant velocity.
The speed of light is an exact measurement: 299,792,458 meters per second exactly (in a vacuum) but that exactness is not by accident. Galileo has been credited with being the first scientist to try to measure the speed of light. Galileo used a lamp that he could uncover and when his assistant, at some distance away, saw the light from Galileo’s lamp, the assistant uncovered his own lamp. By knowing the distance between himself and his assistant and by measuring the elapsed time (Galileo had to use a clepsydra, a water clock, to measure the time interval) until he saw his assistant’s lamp, Galileo tried to calculate the speed of the light beams. Galileo concluded that if light is not “instantaneous, it is extraordinarily rapid.” This was in 1667 and Galileo figured that the speed of light was at least ten times faster than the speed of sound. (Actually, the speed of light is almost nine-hundred thousand times faster than the speed of sound.)
In 1675, the Danish astronomer Ole Roemer measured the speed of light. The astronomer noticed that the eclipses of the moons of Jupiter changed with the relative positions of Jupiter and Earth. When Earth was closer to Jupiter, the eclipses happened sooner than when Jupiter was far from Earth. The time difference between when the moons of Jupiter were expected to eclipse and when they did eclipse was up to sixteen minutes. Roemer reasoned that the change was caused by the greater distance that light had to travel when Earth was farther from Jupiter. Using the then accepted distances between the orbits, he calculated that the speed of light to be 225,000 kilometers per second.
After Fizoau made his measurement in 1849, Foucault and other experimenters began measuring the speed of light with greater precision. As the speed of light was measured with increased precision, the definition of a meter and of a second had to be defined with greater precision.
Historically, the meter was defined by the French Academy of Sciences as 1/10,000,000 of the distance from Earth’s North Pole to the equator on a line that ran through the city of Paris. This definition was replaced 1874 and refined in 1889 by an exact measurement between two marks on a bar of 90% platinum/10 % iridium that was kept in a vault in Paris. Then in 1960, the meter was re-defined as 1,650,763.73 wavelengths of the emissions of the gas Krypton. In 1983, a meter was re-defined again as the distance that light travels in a vacuum in 1/299,792,458th of a second. Then, to define a second, there had to be very accurate clocks. In 1955, the first atomic clock was built and refined in 1958. 1967, the 13th General Conference on Weights and Measures first defined the International System (SI) and a second was defined by 9,192,631,770 periods of the radiation of the cesium-133 atom. These clocks provide an accuracy of 2 nanoseconds per day or one second in 1,400,000 years.
The meter went from being defined as a distance in space (1/10,000,000 of the distance from the North Pole to the equator) to being defined as a measure of time (1/299,792,458th of a second) which can be equated, with the speed of light, into a distance in space.
http://www.physicsforums.com/attachment.php?attachmentid=1164&stc=1
c = d / [(t2-t0) / 2] When t2 – t0 is 2/299,792,458 seconds, d is defined as 1 meter.
The exact length of a meter is a fraction of space-time.
By international standards, a second is defined as the duration of 9,192,631,770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground state of the cesium-133 atom.
Over a distance in time of one meter, the radiation of the cesium-133 atom oscillates 30.663319 times (1/299,792,458 of a second, times 9,192,631,770 Hz/sec. = 30.663319 Hz).
The oscillation of one wavelength equates to a distance of 1.087827757*10-10 seconds on Tk.
Or, one wavelength as measured as a metric distance on Tk is 3.261225571 centimeters.
Measuring Time in Units of Distance
Minkowski described the space-time continuum as an inseperate union and he used the complex configuration (√-1) for time measurement to allow time to be considered as the forth dimension equal to the three spatial dimensions: (ds)2 = x2 + y2 + z2 + tk2. However, space and time are measured in different units. By international convention, the units that are used around the world for scientific discussion are standardized: the meter/kilogram/second standard is used to measure length, mass and time. However, for any unit of distance that is chosen to measure space, a corresponding unit of distance can be used to measured time.
Thus, to change seconds into meters, the time interval in seconds is multiplied by the speed of light measured in meters per second.
c = 299,792,458 meters per second.
▲t = 1 second. c * ▲t = 299,792,458 m/s * 1 second. Then ▲t equals 299,792,458 meters.
▲t = 1/299,792,458th of a second.
c * ▲t = 299,792,458 m/s * 1/299 792 458 second. Then ▲t equals 1 meter.
When t is measured in seconds, a formula to measure a distance, ds, in space-time may be written as ds2 = x2 + y2 + z2 + c2t2 and the time interval conversion is positioned in the formula. In a space-time diagram, when using equal scales of space and time, one unit of space equaling one unit of time, the time units are directly convertible into spatial units. When one unit on the spatial axes is measured as one meter, then ▲t is equal to one meter on the Tk axis (time axis in the forth dimension). Whatever units the speed of light is measured in; meters, miles or cubits, the forth dimension can be measured in exactly the same units.
When ▲t on Tk, is measured as one second, then the passage of space through time, N to N+1sec., has progressed through 299,792.458 kilometers of the forth-dimension.
http://www.physicsforums.com/attachment.php?attachmentid=1165&stc=1
With distance in the forth dimension measured the same as distance is measured in space, a time interval that is measured as one second is equal to 299,792.458 kilometers on Tk. The corollary is that three-dimensional space, N, progresses to N+▲t through the forth-dimension at 299,792.458 km per second measured on Tk. Space, all of space, is physically moving through the forth dimension of time. The speed of light reflects the velocity of space through the forth dimension. Any observer in any reference system is measuring the velocity of N as N progresses to N+▲t when they measure the speed of light.
Space travels through the forth dimension in one direction. This forward direction creates the progression of our existence from the present to the future. As three-dimensional space progresses through the forth dimension from N to N+299 792 458 m, the clocks in RSS (Reference System Space)register one second. Other reference systems that have a velocity reference to RSS will measure space and time differently than RSS (the Lorentz transformations), but every reference system that maintains a continuous existence in space progresses through time with RSS into the future in N to N+▲t.
zoobyshoe
Jul12-04, 07:25 AM
Nay, I think what we are all amazed at is the fact that the moving observer will be gullible enough to think receiving the signal at a different time than the stationary observer is equivalent to the source emitting the signal at a different time.
The observer on the train is meant to be ignorant of the actual timing of the flashes: he doesnt know when the man at the mid point recieves them. All we want to know from him is: "Did you see them at the same time, or one before the other?"
The question, more or less, that Einstein has posed in a previous chapter is "When something seems simultaneous, how do we know if it really is or not?" (Not an exact quote, just a characterization.) He is pointing out in this chapter that you can't tell from the relative timing of the arrival of the flashes, if they were emitted simultaneously in their own rest frame.
If the rocket measures it to be 300,000 km in its own frame, the station will also measure it to be 300,000 in its own frame. Each will measure it to be shorter in the other's frame, however.
So now you have two lengths? One is 300,000 km measured by the rocket? And the other is 300,000 km measured by the station? What are these two lengths? They aren't the same thing, obviously.
Kinda goes without saying.
I would have thought... but here we are.
He says: "x is the distance from...", so, the L given by jcsd represents the lengths M--->A and M---B respectively.
L is the distance between A and M, and B and M, measured in the embankment frame.
I would buy what you say about -L and +L as positions, except you can't apply a Lorentz transformation to a point. A point has no dimensions.
Lorentz transformations are used to convert space-time coordinates of an event (a point in space time) from one frame's measurements to another's.
Yes, I noticed that the second term in the parentheses for A' and B' were different. I found that mighty peculiar also, since no time has passed yet. We are still at T=0 for both reference frames.
You are missing the entire point. The events (1) Light A flashes and (2) Light B flashes do not occur at the same time according to the train frame!
Whatever it is, he tried to apply the Lorentz tranformation for time dilation to it. There's no good reason to do that to a length or a point in a coordinate system.
I suggest that you learn a bit about the Lorentz transformation. :smile:
If A' coincides with A at the moment of the flashes and B coincides with B' at the moment of the flashes and M' coincides with M at the moment of the flashes, how can "the moment of the flashes" be different for either the designated stationary or moving observer? That's like saying given 1=1 and 2=2 and 3 = 3 and 4 = 4, 1 is not 1 and 2 is not 2 and 3 is not 3 and 4 is not 4.
While A' (a point on the train) coincides with A (a point on the embankment) at the moment that A flashes, this does not mean that the train clock at A' agrees with the embankment clock at A. In fact, they obviously don't.
Obviously the the events are absolute in time and place for both inertial observers and the relative simultaneity being argued in that booklet is that of the arrival time of the light reflected from the events.
If that's what you think, then you'd better read it over. The two frames disagree as to when the flashes occured, not just when the light is detected by the observers.
While A' (a point on the train) coincides with A (a point on the embankment) at the moment that A flashes, this does not mean that the train clock at A' agrees with the embankment clock at A. In fact, they obviously don't.
If that's what you think, then you'd better read it over. The two frames disagree as to when the flashes occured, not just when the light is detected by the observers.
I don't quite understand this are you saying the moment when A coincide with A' coincide with A flash are not simultaneous because someone on the embankment is wearing a watch synchronized with the time in Bei Jing while the gal on the train is keeping the time in France?
I don't quite understand this are you saying the moment when A coincide with A' coincide with A flash are not simultaneous because someone on the embankment is wearing a watch synchronized with the time in Bei Jing while the gal on the train is keeping the time in France?
:rofl: Good one, Eyesaw.
When A and A' coincide is a single event in space-time (something that happens at particular place and time). But the time that this event occurs will be measured differently in different frames.
The issue for simultaneity is not: Do A and A' pass each other at "the same instant"? Of course they do! The issue is: does the passing of A and A' happen at the same time as the passing of B and B'? And this question is answered differently in different frames: these two events are observed to happen at different times in the train frame.
This post started out as a simple comment on the last sentence of thread post #73 by Doc Al. The actual result is more of a digression on some of the formal aspects of Special Relativity that, in the context of this discussion, either have only been sketched or have been described piecemeal.
\gamma L is not the distance L as measured by the train! It is the position coordinate of an event.
I agree with what you say here, but as stated (and presuming it is meant as a warning about my previous post [#72]), it may obscure what is correct and what might be misleading about my post.
1) (To clarify for readers unfamiliar with the terminology)
Event is the term used for a given point in the mathematical space used to describe relativity. Specifying an event requires both space coordinate(s) and a time coordinate. The definition covering the current situation, as given by jcsd, is:
"What I've done is basically create a co-ordinate susyetm the co-ordinates for each event are (x,t) where x is the distance from the midpoint of the space station (i.e. x = 0 at M) and t is the difference in time from when the spaceship and M occupy the same spot (i.e. t = 0 as the spaceship arrives at M)."
Each reference frame describes the same set of events. However, different reference frames overlay different systems of coordinates onto these. The differing measurements determined by these different coordinate systems reflect the space and time distortions that occur when measurements are made in different reference frames.
2) The Lorentz transformation is the mathematical operation that maps the coordinates of events in one reference frame to the coordinates in a second frame.
The ordinary Lorentz transformation maps event coordinates in the rest frame to event coordinates in some other inertial frame.
If an event has coordinates (x, t) in the rest frame, then the general transform \Lambda to coordinates (x', t') in a frame with relative velocity u is:
(x', t') = \Lambda (x, t) = \left( \gamma (x - ut), \gamma(t - \frac{ux}{c^2}) \right)
If, however, what is known is the coordinates (x', t') in the moving frame, then to find the coordinates (x, t) in the rest frame what is required is the inverse Lorentz transformation \Lambda^{-1}. This has the general form:
(x, t) = \Lambda^{-1} (x', t') = \left( \gamma (x' + ut'), \gamma(t' + \frac{ux'}{c^2}) \right)
3) Ok, I screwed up. I just realized that jcsd was talking about the spaceship scenario not the train scenario, so the way my previous post (#72) compares what I say to what jcsd said makes no sense, and I should probably edit it. (Without the comparison, the logic still holds - though with the caveats given below.)
4) Just to be clear: I'm discussing the train scenario.
In the embankment frame, coordinates are denoted:
(x_m, t_m)
where xm is a spatial coordinate along an axis defined by the track, and tm is a time coordinate.
For any time tm in the embankment frame, the observer M has coordinates:
M(t_m) = (x_M, t_M) = (0, t_m)
We also have two events - flash at A, and flash at B - that happen at tm = 0. Coordinates for these are:
\textrm{flash}_A = (x_A, t_A) = (-L, 0)
\textrm{flash}_B = (x_B, t_B) = (L, 0)
(I've switched the signs so that the train moves toward positive xm.)
The flashes occur at the same moment that the rear and front of the train pass (respectively) A and B. In other words, in the embankment frame the length of the train equals the distance from A to B.
In the train frame, coordinates are denoted:
(x_{m'}, t_{m'})
The axis along which the spatial coordinate xm' is measured is defined by the track and thus coincides with spatial axis used in the embankment frame (it's coordinate system, however, is, of course, separate).
For any time tm' in the train frame, the observer M' has coordinates:
M'(t_{m'}) = (x_{M'}, t_{M'}) = (0, t_{m'})
The coordinates of flashA and flashB transformed into the train frame (as given by jcsd, but with signs reversed) are:
\textrm{flash}_{A'} = (x_{A'}, t_{A'}) = (- \gamma L, \frac{\gamma\beta L}{c})
\textrm{flash}_{B'} = (x_{B'}, t_{B'}) = (\gamma L, \frac{-\gamma\beta L}{c})
This is an example of the inverse Lorentz transformation as defined in item 2) above. (We consider the train to be at rest so we are going from a moving frame to a rest frame.) Calculating using the formula from item 2) we get:
This has been corrected. See post #99 below for the reasoning behind the use of -u.
\begin{equation*} \begin{split}
(x_{A'}, t_{A'}) &=
\Lambda^{-1}(x_A, t_A) =
\Lambda^{-1}(-L, 0) \\
&= \left( \gamma (-L + 0(-u)), \gamma (0 + \frac{(-u)(-L)}{c^2}) \right) \\
&= ( -\gamma L, \frac{\gamma\beta L}{c} )
\end{split} \end{equation*}
The last point necessary for the set up is that both the observer M and the observer M' set the origin of their respective time axes to the moment that M' passes M.
5) So far quantities such as L and \gamma L are, as Doc Al said, "position coordinate[s] of an event".
To get the spatial distance in one dimension we can use the formula:
\Delta x = |x_1 - x_0|
Thus the distances from the observer M to the points A and B are (in the embankment frame):
\Delta x_A = |x_A - x_M| = |-L - 0| =\ L
\Delta x_B = |x_B - x_M| = |L - 0| =\ L
and thus as I said in the previous post, "each half of the train measured in the embankment frame is also length L (so total length 2L)"
And the distances from the observer M' to the points A' and B' are (in the train frame):
\Delta x_{A'} = |x_{A'} - x_{M'}| = |-\gamma L - 0| = \gamma L
\Delta x_{B'} = |x_{B'} - x_{M'}| = |\gamma L - 0| = \gamma L
and so, again as in the previous post, "the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed \gamma L"
I'll leave off for now, the third assertion of my previous post will have to wait for next time.
zoobyshoe
Jul12-04, 09:46 AM
So now you have two lengths? One is 300,000 km measured by the rocket? And the other is 300,000 km measured by the station? What are these two lengths? They aren't the same thing, obviously.
They are the lengths from the midpoint to the ends. If we're talking about the rocket scenario we can use Janus' rod attached to the rocket, which measures 300,000 km. This will give the rocket a length to measure in its frame if it needs one. The train is already this length from midpoint to the end.
L is the distance between A and M, and B and M, measured in the embankment frame.
Yes.
Lorentz transformations are used to convert space-time coordinates of an event (a point in space time) from one frame's measurements to another's.
Not exactly right. They can't be applied to a single point. In the case of length contraction you need the begining and ending coordinates of the length to be contracted. Einstein gives the begining coordinate as 0 for simplicity in the example he gives. For time dilation you also need two times to compare. T=0 and t=>0. A single point doesn't contain enough information to do anything with.
You are missing the entire point. The events (1) Light A flashes and (2) Light B flashes do not occur at the same time according to the train frame!
I don't know why you think I have missed that point. Jcsd set up coordinates for the M and M' systems at t=0 for both. Nothing has happened yet. I thought it was peculiar that he indicated that something had happened already in the M' system when no time has elapsed yet.
I suggest that you learn a bit about the Lorentz transformation. :smile:
Well, there is no doubt that I have more to learn about them.
This post started out as a simple comment on the last sentence of thread post #73 by Doc Al. The actual result is more of a digression on some of the formal aspects of Special Relativity that, in the context of this discussion, either have only been sketched or have been described piecemeal.
I agree with what you say here, but as stated (and presuming it is meant as a warning about my previous post [#72]), it may obscure what is correct and what might be misleading about my post.
Actually I was just pointing out that I thought zoobyshoe was interpreting the \gamma L in jcsd's equations as the length L as measured by the train. This is not so.
5) So far quantities such as L and \gamma L are, as Doc Al said, "position coordinate[s] of an event".
To get the spatial distance in one dimension we can use the formula:
\Delta x = |x_1 - x_0|
Thus the distances from the observer M to the points A and B are (in the embankment frame):
\Delta x_A = |x_A - x_M| = |-L - 0| = L
\Delta x_B = |x_B - x_M| = |L - 0| = L
and thus as I said in the previous post, "each half of the train measured in the embankment frame is also length L (so total length 2L)"
Maybe I'm misinterpreting your notation, but \Delta x_A = |x_A - x_M| = |-L - 0| = L is the distance--as measured in the embankment frame-- between the events (A) A flashes and (M) M' passes M. Of course, since these events are simultaneous in the embankment frame, then this is a length. It's the length between point A and point M in the embankment frame, which was of course given as L. I don't see this having anything to do with the length of the train. (Was that even specified?)
And the distances from the observer M' to the points A' and B' are (in the train frame):
\Delta x_{A'} = |x_{A'} - x_{M'}| = |-\gamma L - 0| = \gamma L
\Delta x_{B'} = |x_{B'} - x_{M'}| = |\gamma L - 0| = \gamma L
and so, again as in the previous post, "the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed \gamma L"
Now you are measuring the distance between those events (A) & (B) and (M) according to the train frame. \Delta x_{B'} = |x_{B'} - x_{M'}| = \gamma L is not the length L (between A and M) as measured by the train, since these events are not simultaneous in the train frame. I suppose that if (for some unknown reason) the ends of the train happen to coincide with locations A and B when the lights flash, then \gamma L would be the proper length of half the train. (Is that what you are assuming?)
zoobyshoe
Jul12-04, 10:14 AM
So, I just looked through Plover's post about the Lorentz Transforms, and he makes a case I can't follow for jcsd having used the correct formula under the circumstances. Not having the patience to try and inform myself about that particular application of the transforms at this point, I will simply stipulate he is correct, apologise to jcsd for having asserted he used the wrong formula, and try and get back to the question that started this.
Employing jcsds method, and given a speed of .5c for the ship or train, what will be the relative times and distances from t=0 for the detection of the light signals by the observer on the train or ship?
They are the lengths from the midpoint to the ends. If we're talking about the rocket scenario we can use Janus' rod attached to the rocket, which measures 300,000 km. This will give the rocket a length to measure in its frame if it needs one. The train is already this length from midpoint to the end.
In the train gedanken, L is the length from one light to the point M in the embankment. It is not the length of the train. (I guess it could be, but that would be just an arbitrary coincidence.)
Not exactly right. They can't be applied to a single point. In the case of length contraction you need the begining and ending coordinates of the length to be contracted. Einstein gives the begining coordinate as 0 for simplicity in the example he gives. For time dilation you also need two times to compare. T=0 and t=>0. A single point doesn't contain enough information to do anything with.
Thanks for the lecture, but once a common origin is defined the LT can certainly be applied to a point (an event in space-time).
I don't know why you think I have missed that point. Jcsd set up coordinates for the M and M' systems at t=0 for both. Nothing has happened yet. I thought it was peculiar that he indicated that something had happened already in the M' system when no time has elapsed yet.
The two systems agree that the M clock and the M' clock both read t = 0 at the instant that M' passes M. That's all. They certainly do not agree that the lights flashed at the same time. At time t=0, the train frame says light B already flashed and light A didn't flash yet. But the embankment insists that both lights flashed at t = 0.
zoobyshoe
Jul12-04, 10:39 AM
In the train gedanken, L is the length from one light to the point M in the embankment. It is not the length of the train. (I guess it could be, but that would be just an arbitrary coincidence.)
The length of the train is 2L. Always has been.
Thanks for the lecture, but once a common origin is defined the LT can certainly be applied to a point (an event in space-time).
Point, as defined by plover, yes, you may well be right. However, he/she may simply have made that whole post up. I certainly don't trust anything he/she says farther than I can throw it. The "uncontraction" post where the length "uncontracts" to one larger than it had before has made me suspicious.
The two systems agree that the M clock and the M' clock both read t = 0 at the instant that M' passes M. That's all. They certainly do not agree that the lights flashed at the same time. At time t=0, the train frame says light B already flashed and light A didn't flash yet. But the embankment insists that both lights flashed at t = 0.
Yes. This is absolutely what Einstein has said. However, I still don't know by what math we discover the difference in the reception time of the flashes by M'.
Maybe I'm misinterpreting your notation, but \Delta x_A = |x_A - x_M| = |-L - 0| = L is the distance--as measured in the embankment frame-- between the events (A) A flashes and (M) M' passes M. Of course, since these events are simultaneous in the embankment frame, then this is a length. It's the length between point A and point M in the embankment frame, which was of course given as L.
Yes, this was my intention. Since L was introduced as a coordinate, I was just making the derivation of how it also ends up as a length.
I don't see this having anything to do with the length of the train. (Was that even specified?)
Over the course of the thread, the assumption that the length of the train as measured in the embankment frame is the same as the distance from A to B has drifted in and out of use. The assumption is not present in Einstein's original set up, and in fact is implicitly denied by both Einstein's diagram and the one I gave in an earlier post. I should have been explicit that I was using it here.
Now you are measuring the distance between those events (A) & (B) and (M) according to the train frame.
If it appears that way, I expect its because I forgot to make the assumption about the length of the train explicit.
\Delta x_{B'} = |x_{B'} - x_{M'}| = \gamma L is not the length L (between A and M) as measured by the train, since these events are not simultaneous in the train frame.
Yes.
I suppose that if (for some unknown reason) the ends of the train happen to coincide with locations A and B when the lights flash, then \gamma L would be the proper length of half the train. (Is that what you are assuming?)
Well, there's someone standing at each end of the train holding a pointy metal pole, and as it's the middle of an electrical storm, these someones must be fairly gullible, and, well, being one of those ubiquitous "assistants" that show up in gedanken-experiments always has been a thankless job... :wink:
However, I still don't know by what math we discover the difference in the reception time of the flashes by M'.
Ask nicely and we'll tell you. :smile:
Let's be clear. I'll define two events:
(1) photons from light B arrive at M'
(2) photons from light A arrive at M'
According the to the train, these events occur at:
(x_1', t_1') = (0,\frac{\gamma L}{c} - \frac{\gamma v L}{c^2})
(x_2', t_2') = (0,\frac{\gamma L}{c} + \frac{\gamma v L}{c^2})
Note added: The order of the lights in my set up is A, M, B. M' moves towards B. So event #1 occurs before event #2 in both frames.
zoobyshoe
Jul12-04, 12:22 PM
Incidently, when we multiply gamma times length and get 346,410.15 and -346,410.15, what do these numbers tell us about A and B in relation to M, other than that they are still equidistant from M?
If the rocket measures it to be 300,000 km in its own frame, the station will also measure it to be 300,000 in its own frame. Each will measure it to be shorter in the other's frame, however.
Kinda goes without saying.
No, That's not right.
Let's try this again with the following attachment.
Assume two very long trains. each car is 20,000 km long, as measured from the train to which it belongs.
Train 1 has three observers, M, A and B, A and B are 300,000 km from M, according to M, which puts them 15 cars away from M.
Train 2 is moving at .5c relative to train 1. It has observers M', A' and B'.
A" and B' are each placed an equal distance from M' such that from the frame of M, A and B, when M and M' pass each other, A' passes A and B' passes B. since train 2 is length contracted according to train 1, this places A' and B' about 17 cars away from M' on train 2. This means that the distance between A' and M' (or B' and M') is 346420 km according to train 2.
Now look at what happens according to train 2. A' and B' are a little over 17 cars from M' (this has not changed) putting them 346420 km from M'
A and B are still 15 cars from M, but since train 1 is length contracted according to train 2, each of these cars is only 17320 km long, meaning that A and B are, according to train 2, only 259800 km from M.
When M passes M' A' does not pass A and B' does not pass B.
so if the rocket measures the distance as 300,000km in its frame, the station does not measure it to be 300,000km in its frame.
Incidently, when we multiply gamma times length and get 346,410.15 and -346,410.15, what do these numbers tell us about A and B in relation to M, other than that they are still equidistant from M?
I assume you are talking about the \gamma L that appears in these equations:
B\'\; = (\gamma L,\frac{-\gamma\beta L}{c})
A\'\; = (-\gamma L,\frac{\gamma\beta L}{c})
(I just noticed that jcsd defined the lights opposite to the way I would. In my set up, the lights are arranged in order A, M, and B. M' moves towards B. I hope this doesn't add any confusion.)
If so, then you must understand what they mean. These are the space-time coordianates according to the train frame of the events (A) light flashes at A and (B) light flashes at B. They tell us that:
- when light A flashed it was at position x' = -\gamma L in the train frame
- when light B flashed it was at postion x' = \gamma L in the train frame
Yes, those flashes occur at postions that are equidistant from the midpoint M' of the train (but they occur at different times).
zoobyshoe
Jul12-04, 02:25 PM
so if the rocket measures the distance as 300,000km in its frame, the station does not measure it to be 300,000km in its frame.
Somewhere back there I gave the distance A---->B as 600,000 km. That's in the embankment frame.
I have also given the train a length of 600,000 km in its own frame whereever I mentioned a length for the train.
This means that when the train is at rest in the embankment frame they both measure 600,000 km.
When the train is moving at .5 c, it still measures itself to be 600,000 km long. The distance A--->B in the embankment frame, is still measured to be 600,000 km long in its frame.
If the train measures the embankment while moving at .5 c, it is, of course a different story. The moving train will see the distance A--->B contracted from 600,000 to .8660254% of that which is 5196152.4 km.
If an observer in the embankment frame measures the train, it will find the train to have contracted to 519,615.4 km, as well, from the embankment viewpoint.
If we need to demonstrate the same for the ship example for any reason we can do so by attaching your rods, provided the rods give it a length of 600,000 km in its own frame.
zoobyshoe
Jul12-04, 02:41 PM
I assume you are talking about the \gamma L that appears in these equations:
B\'\; = (\gamma L,\frac{-\gamma\beta L}{c})
A\'\; = (-\gamma L,\frac{\gamma\beta L}{c})
That's right.
(I just noticed that jcsd defined the lights opposite to the way I would. In my set up, the lights are arranged in order A, M, and B. M' moves towards B. I hope this doesn't add any confusion.)
Nope.
If so, then you must understand what they mean. These are the space-time coordianates according to the train frame of the events (A) light flashes at A and (B) light flashes at B. They tell us that:
- when light A flashed it was at position x' = -\gamma L in the train frame
- when light B flashed it was at postion x' = \gamma L in the train frame
OK. This is exactly what I thought. However, what confuses me is that 346,410.15 km is a greater distance than 300,000 km. I would expect it to be contracted, not lengthened. Of what use is it to apply the time dilation equation to a length? What does this help us understand about the situation?
Yes, those flashes occur at postions that are equidistant from the midpoint M' of the train (but they occur at different times).
OK. This is exactly what I thought. However, what confuses me is that 346,410.15 km is a greater distance than 300,000 km. I would expect it to be contracted, not lengthened.
You are still thinking that \gamma L is the length L as measured by the train. It's not. According to the train, things go like this: Light B flashes. Then, some time later, light A flashes. During that time the train is moving (of course, in the train frame, the train sees the platform move).
Of what use is it to apply the time dilation equation to a length? What does this help us understand about the situation?
I'm not sure what you mean by "apply the time dilation equation" to a length, as I did no such thing. I applied the Lorentz transformation to the space-time coordinates of an event.
If you wish to think of sticks shrinking and clocks moving slowly, then you apply the LT to sticks (lengths) and clocks. For example, a length of L in the embankment is measured to be L/\gamma from the train (yes, the train measures the lights A and B to be only 2L/\gamma apart); and the train frame will measure a time of \gamma \Delta t when an embankment clock measures only a time of \Delta t.
To analyze a problem correctly you need to apply length contraction, time dilation, and the relativity of simultaneity all together. That's tricky. The Lorentz transformation takes account of all of that for you.
zoobyshoe
Jul12-04, 04:36 PM
In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation as:
x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}
y'=y
z'=z
t'=\frac{t-\frac{v}{c^2}•x}{\sqrt{1-\frac{v^2}{c^2}}}
The two interesting ones, of course, are those for x' and t'.
In the next chapter, Einstein gives a simpler version of x':
^x(beginning of rod)^=0\sqrt{1-\frac{v^2}{c^2}}
^x(end of rod) ^1•\sqrt{1-\frac{v^2}{c^2}}
"The distance between the two points being
\sqrt{1-\frac{v^2}{c^2}}
If we want to express A or B as A' or B', it seems to me this is the one to use, multiplying it by the length, L, which is the distance between the two points (M--->B, or M--->A) in the embankment frame that we want to locate in the train frame.
Instead, jcsd used the next one refered to by Einstein; the one he uses to find the time dilation between two successive ticks of a clock. The first tick is t = 0. The second tick is:
t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
If the 1 in the numerator of the equation given by jcsd for \gamma is not the one used by Einstein in the time dilation version, I am quite confused about how jcsd arrived at the 1 from x-vt, which is the numerator in the more general equation for x'.
zoobyshoe
Jul12-04, 05:55 PM
Those are from this chapter:
Chapter 11. The Lorentz Transformation. Einstein, Albert. 1920. Relativity: The Special and General Theory
Address:http://www.bartleby.com/173/11.html
And the following one:
Chapter 12. The Behaviour of Measuring-Rods and Clocks in Motion. Einstein, Albert. 1920. Relativity: The Special and General Theory
Address:http://www.bartleby.com/173/12.html
In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation as:
x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}
y'=y
z'=z
t'=\frac{t-\frac{v}{c^2} x}{\sqrt{1-\frac{v^2}{c^2}}}
The two interesting ones, of course, are those for x' and t'.
Yes, those are the same Lorentz transformations that jcsd and I both used. :smile:
In the next chapter, Einstein gives a simpler version of x':
^x(beginning of rod)^=0\sqrt{1-\frac{v^2}{c^2}}
^x(end of rod) ^1 \sqrt{1-\frac{v^2}{c^2}}
"The distance between the two points being
\sqrt{1-\frac{v^2}{c^2}}
Careful. This is not a "simpler version of x' ", it is a specific application to find how a length gets transformed when observed from a moving frame. Note that each end must be located at the same time.
If we want to express A or B as A' or B', it seems to me this is the one to use, multiplying it by the length, L, which is the distance between the two points (M--->B, or M--->A) in the embankment frame that we want to locate in the train frame.
If the problem is: Find out the distance from A to B as seen by the train frame, then that would be correct. But that's not what jcsd was calculating. Instead he (and I) calculated when and where those two flashes occured according to the moving frame. That's a different problem. For one thing, those flashes occur at different times--so that length contraction formula doesn't apply.
jcsd used A to represent the coordinates of the event "Light A flashes" in the embankment frame, and A' to represent the coordinates of that same event in the train frame. (I'm not crazy about this notation, but that's what he used.)
How did we get the answers we did? It's easy. We know the coordinates of the events in the embankment frame, that's a given: x = -L, t = 0 (for A flashing) and x = +L, t = 0 (for B flashing). Now to find out when and where the train observers determine these flashes took place: apply the LT. Try this for yourself!
Instead, jcsd used the next one refered to by Einstein; the one he uses to find the time dilation between two successive ticks of a clock. The first tick is t = 0. The second tick is:
t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
This is not what jcsd did. This particular equation describes the time dilation of a moving clock.
In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation
This is a four dimensional version of the transformation I gave above. As it shows, spatial distortions happen only along the direction of motion (in this case along the x-axis). This is why the two dimensional system suggested by jcsd is useful.
Note: When Doc Al says: "jcsd defined the lights opposite to the way I would", he makes precisely the same change as I did where I said "I've switched the signs so that the train moves toward positive xm".
The equation I gave was:
(x', t') = \Lambda (x, t) = \left( \gamma (x - ut), \gamma(t - \frac{ux}{c^2}) \right)
This is entirely equivalent to the separate equations:
x' = \gamma (x - ut)
t' = \gamma \left(t - \frac{ux}{c^2} \right)
(I expect this goes without saying -- I'm just ensuring the intention is clear.)
The notations introduced by jcsd (and which I should perhaps have reiterated):
\beta = \frac{u}{c}
\gamma = \frac {1} { \sqrt{1 - \beta^2} } = \frac {1} { \sqrt{1 - \frac{u^2}{c^2}} }
These notations are quite standard. Using u for the velocity in SR calculations also seems to be a current standard practice.
Again just to be clear, I'll apply these definitions to the separated equations above:
x'\ =\ \gamma (x - ut)
\ =\ \frac{1}{\sqrt{1 - \beta^2}} (x - ut)
\ =\ \frac{x - ut}{\sqrt{1 - \frac{u^2}{c^2}}}
t'\ =\ \gamma \left( t - \frac{ux}{c^2} \right)
\ =\ \frac{1}{\sqrt{1 - \beta^2}} \left(t - \frac{ux}{c^2} \right)
\ =\ \frac{\left(t - \frac{ux}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}}
These are obviously identical to the equations given by Einstein.
Note: There is a typo in Ch 12 of the Einstein. Near the end of paragraph 1 a sentence reads:
"For the velocity v = 0 we should have
\sqrt{1 - v^2/c^2} = 0
and for still greater velocities the square-root becomes imaginary."
This sentence only makes sense if it is changed to read: "For the velocity v = c we should have...". (Someone should probably back me up on this.)
There is one issue that everyone has neglected so far (I think). Einstein's coordinate system diagram (in Ch 11) shows that the k system views the k' system as moving to the right, i.e. toward positive x. From the viewpoint of k', however, k is moving to the left, i.e. toward negative x. Thus if k measures the velocity of k' to be u then k' will measure the velocity of k to be -u. Similarly, if the x axes of the train and embankment frames are oriented to be positive in the same direction, then if an embankment observer measures the velocity of the train as u, an observer on the train measures the velocity of the embankment to be -u.
If I'm reading everyone correctly, jcsd and Doc Al took this into account (though I don't think they mentioned it). I, on the other hand, um, managed to make the issue disappear by making a calculation error when I applied the inverse Lorentz transform in my previous post, which I should fix (and which is probably why zoobyshoe found my argument for jcsd's values ambiguous...).
I assume it is agreed that the coordinates in the embankment frame for the event "flash at B" (which we are also taking to be the coordinates of "front of train passes B") are:
x = L
t = 0
If we insert these values into Einstein's equations (and reapply the standard notations), we get:
x'\ =\ \frac{x - ut}{\sqrt{1 - \frac{u^2}{c^2}}}
\ =\ \frac{L - u \cdot 0}{\sqrt{1 - \frac{u^2}{c^2}}}
\ =\ \frac{1}{\sqrt{1 - \beta^2}}\ \cdot\ L
\ =\ \gamma L
t'\ =\ \frac{\left(t - \frac{ux}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}}
\ =\ \frac{\left(0 - \frac{uL}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}}
\ =\ \frac{-1}{\sqrt{1 - \beta^2}}\ \cdot\ \frac{u}{c}\ \cdot\ \frac{L}{c}
\ =\ \frac{-\gamma\beta L}{c}
So if we set u to .5c our value for x' is:
\beta\ =\ .5c/c\ =\ .5
\gamma\ =\ \frac{1}{\sqrt{1 - (.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15
x'\ =\ \gamma L\ =\ 1.15L
Now since:
the origin of the x' axis of the train frame is the center of the train, and
the front of the train is at the point (x, t) that we transformed from the embankment frame
we know that the value of x' also equals half the length of the train. So at rest the length of the train is: 2 * 1.15 * L. Now why should this be so?
The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.
Sophist.
Any length that one observer measures to be contracted will be its original length when "uncontracted, not greater than its original length.
_______________________________________________
he/she may simply have made that whole post up. I certainly don't trust anything he/she says farther than I can throw it. The "uncontraction" post where the length "uncontracts" to one larger than it had before has made me suspicious.
:uhh: :confused: :uhh:
What are you talking about?
I used the word "uncontraction" to refer to the purely mathematical operation of finding the original length given a contracted length. When did I ever claim that something would be greater than its original length?
And what have I done that would lead you to countenance that I might intend such a reading? That makes a suspicion of willful deceit or "making whole posts up" more plausible than a mistake on one or both of our parts?
What's really going on here? (If you've got some beef, my pm box is open...)
zoobyshoe
Jul13-04, 05:52 AM
Careful. This is not a "simpler version of x' ", it is a specific application to find how a length gets transformed when observed from a moving frame.
Good point. "Simpler version" is not particularly accurate. "Specific application" is much more to the point, much more accurate. I appreciate the distinction.
Note that each end must be located at the same time.
By "at the same time" do you mean at a specific time coordinate such as T=0? Or something else?
Instead he (and I) calculated when and where those two flashes occured according to the moving frame.
This is what I have understood you to be saying about this since I raised my objection. However, I haven't been able to understand from you how the answer that is produced: -346, 410.16 and + 346,410.16, contains any useful information about the when and where according to the moving frame. If I observe that the A or B event took place in the embankment frame at t=0, x=+/-300,000 km. What does a value of -/+346,410.16 tell me about the moving frame?
According to you, this number says something useful about the when and where of event A' or B' in the moving frame. If this number actually has any use, I can't see it. In fact, to me it looks like it is saying something both incorrect and useless about these events in the A' and B' frame. These numbers seem to me to say nothing at all about the when in the A' and B' frame, and to say something outright incorrect about the position A' and B' will have in the moving frame.
From the general direction and tone of your answers, it seems clear that you understand the difficulty I am having with these numbers. I am trying to phrase my difficulty more specifically in the hope you'll see how to solve it.
jcsd used A to represent the coordinates of the event "Light A flashes" in the embankment frame, and A' to represent the coordinates of that same event in the train frame.
Yes, this is what I have understood him to be doing.
How did we get the answers we did? It's easy. We know the coordinates of the events in the embankment frame, that's a given: x = -L, t = 0 (for A flashing) and x = +L, t = 0 (for B flashing). Now to find out when and where the train observers determine these flashes took place: apply the LT. Try this for yourself!
Specify which of the four you wish me to try for myself. I can use the one for x' and I can use the one for t' but there is no one of the four that gives an answer for both x' and t'. You are giving me one equation and claiming the answer is telling me both the x' and t' coordinates with one number.
This is not what jcsd did. This particular equation describes the time dilation of a moving clock.
jcsd gave:
\gamma=\frac {1}{\sqrt{1-\beta^2}}
where:
\beta = \frac{u}{c}
This is exactly the same equation as:
t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
solving for u/c separately and then squaring the result
\beta^2
gives the same answer as: the square root of u2/c2.
I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.
Zoobyshoe:
I'll explain the exact method I used: I treated the Lorentz tensor and 4-vector postions as matrices and then simply used matrix mutiplication to perform the Lorentz transformation.
One worth pointing out s that we've all apporached the problem in slightly different ways, but still obtained equivalent results which is a strong indication in itself that we are correct.
It is worth learning 4-vector algebra as it makes problems like these extremely simple to solve and also helps to eliminate errors.
I think the answer is A.c . This is because the speed of light is measured as constant in all reference frames. you cant add a sublight speed to a light, or translight speed! :smile:
I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.
What we've done is perform the Lorentz transformation completely, to do this you must consider all four dimesnions (though saying that we've cut out two because they don't change as the direction of motion is orthogonal to them, for confirmation of this in the method examine the Lorentz transformation in full), the metod I used as outlined in my last post you don't even need to consider each directrion independently.
t doesn't equal gamma, gamma is just a unitless quantity that is useful in special relativity and it comes into the equation for length contraction too.
zoobyshoe
Jul13-04, 06:47 AM
Zoobyshoe:
I'll explain the exact method I used: I treated the Lorentz tensor and 4-vector postions as matrices and then simply used matrix mutiplication to perform the Lorentz transformation.
Fascinating!
One worth pointing out s that we've all apporached the problem in slightly different ways, but still obtained equivalent results which is a strong indication in itself that we are correct.
Somehow, I missed any statement of the results. What actual times do you get for the flashes from the train frame? Please show how you moved from - or + 346,410.15 to a value in seconds from t=0 for each flash. As I mourned to Doc Al, this number, 346,410.15 carries no meaning for me, + or -.
Somehow, I missed any statement of the results. What actual times do you get for the flashes from the train frame? Please show how you moved from - or + 346,410.15 to a value in seconds from t=0 for each flash. As I mourned to Doc Al, this number, 346,410.15 carries no meaning for me, + or -.
As I said (see previous post for derivation):
So at rest the length of the train is: 2 * 1.15 * L. Now why should this be so?
The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.
So if we're setting L = 300,000km then the rest length of half the train is: x' = (1.15)(300,000km) = 345,000km.
That's one significance of the number.
So far no one's calculated a value for the time of the flashes in the train frame. It would be (for the flash at B):
\beta\ =\ .5c/c\ =\ .5
\gamma\ =\ \frac{1}{\sqrt{1 - (0.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15
t'\ =\ \frac{-\gamma\beta L}{c}
\ =\ \frac{-1.15 \cdot 0.5 \cdot 300,000\ \textrm{km}}{300,000\ \textrm{km}/\textrm{s}}
\ =\ -0.575\ \textrm{s}
So the event "flash at B" has coordinates
(x, t) = (L km, 0 s) = (300000 km, 0 s)
in the embankment frame, and coordinates
(x', t') = (345000 km, -0.575 s)
in the train frame.
Below is an excerpt (Chapter 12) from my upcoming book...
Please don't do that again. Your post is tangential to this topic. Please start a new topic if you are looking for feedback on your writing.
By "at the same time" do you mean at a specific time coordinate such as T=0? Or something else?
That's exactly what I mean by "at the same time". If the train frame wishes to measure the length of something that's moving with the embankment frame, then measurements of the positions of the ends of that length must be made at the same time according to the train frame. Of course, since we already figured out length contraction, we know the answer will be L/\gamma if L is the length as measured in the embankment frame.
This is what I have understood you to be saying about this since I raised my objection. However, I haven't been able to understand from you how the answer that is produced: -346, 410.16 and + 346,410.16, contains any useful information about the when and where according to the moving frame. If I observe that the A or B event took place in the embankment frame at t=0, x=+/-300,000 km. What does a value of -/+346,410.16 tell me about the moving frame?
It tells you where the event took place according to the train frame. The events are the flashing of the lights. We know where they took place according to the embankment: at x = +/- L = +/- 300,000 km. We use the LT to find out where they took place according to the train. To find out where, use the LT equation for x': note that x' depends on both x and t. It tells us that x' = +/- 346,410.16 km. That's where these flashes take place according to the train frame. That simply means that train observers measure the location of these flashes to occur at a distance of 346,410.16 km from the origin (M'). I believe you think this is somehow wrong because you are unconsciously thinking that the two frames are somehow "lined up" at T = 0. But, no, the only point they have in common is at 0,0: the point (in space-time) where M' passes M.
According to you, this number says something useful about the when and where of event A' or B' in the moving frame. If this number actually has any use, I can't see it. In fact, to me it looks like it is saying something both incorrect and useless about these events in the A' and B' frame. These numbers seem to me to say nothing at all about the when in the A' and B' frame, and to say something outright incorrect about the position A' and B' will have in the moving frame.
To find the when, use the LT for t': note that t' depends on both x and t.
From the general direction and tone of your answers, it seems clear that you understand the difficulty I am having with these numbers. I am trying to phrase my difficulty more specifically in the hope you'll see how to solve it.
I can see that you are serious about learning this stuff, so I am happy to help. But please realize that jcsd and plover already know this stuff and they are trying to help. You are the one who needs to learn it. :smile: You will get there!
Specify which of the four you wish me to try for myself. I can use the one for x' and I can use the one for t' but there is no one of the four that gives an answer for both x' and t'. You are giving me one equation and claiming the answer is telling me both the x' and t' coordinates with one number.
Here's how it goes. An event is specified by a set of coordinates in the embankment frame: x, t. That tells where and when the event occured in the embankment frame. Now use the LT to find where and when that same event occured according to the train frame. You will need two equations: one that gives you x' and one that gives you t'. Here they are again:
x' = \gamma (x - vt)
t' = \gamma (t - vx/c^2)
jcsd gave:
\gamma=\frac {1}{\sqrt{1-\beta^2}}
where:
\beta = \frac{u}{c}
These are just the definitions of \gamma and \beta.
This is exactly the same equation as:
t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
No it's not. This is just a version of the the time dilation formula. It should be written as:
\Delta t=\frac{\Delta t'}{\sqrt{1-\frac{v^2}{c^2}}} = \gamma \Delta t'
This is a specialized formula meant for a specific use: it is not a general Lorentz transformation. (For one thing, it only applies when \Delta x' = 0, which is the case for a moving clock.) This formula means: When a moving clock measures a time of \Delta t', how much time will pass according to the observer's clocks \Delta t.
I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.
Apply the real Lorentz transformation as given above. Don't try to take short cuts by applying other formulas that are derived from the LT, but only apply in specific situations. (Like the "time dilation" formula or the "length contraction" formula.)
zoobyshoe
Jul13-04, 08:34 AM
What we've done is perform the Lorentz transformation completely, to do this you must consider all four dimesnions (though saying that we've cut out two because they don't change as the direction of motion is orthogonal to them, for confirmation of this in the method examine the Lorentz transformation in full),
I follow this logic without problem. y'=y and z'=z so we don't need to treat these two coordinates. We are concerned only with the two that will change from one system to the other. These two are x' and t'.
Yet, at the moment we are concerned about, t=0 and t'=0' there is nothing in the train frame coordinate for time that is different from the embankment frame coordinate for time. x' is different. At this point x is the only coordinate that is different between the two frames due to relative motion at speed .5 c.
When you and Doc Al say you are considering time and position both at once, I grasp the concept, but since the time coordinate is 0 for both frames, the net result of considering both time and location for this moment t' = 0 should be that only position, the x coordinate, is left to treat in any way.
The position of A' from M' or B' from M' should turn out to be closer together by any means of calculation than it is in the M frame. Length contracts. Instead, to my disbelief and confusion, they seem to be farther apart.
Positions A and B, are indeed, located by means of length. To say they have a coordinate is meaningless unless we locate that coordinate with repect to some point =0. This you have done, by using L, length, as the distance from M--->A and M--->B. We have designated that length to be 300,000 km.
The Lorentz transformation you gave:
\gamma = \frac{1}{\sqrt{1-\beta^2}}
gives us the figure 1.1547005 when we put .5 c in as the relative velocity of the train and the embankment.
1.1547005 times the L, 300,000, = 346,410.16
Since the coordinate system requires a length in order to define a position, it seems obvious to me that this number: 346,410.16 refers to a length. You may argue it also refers to a time because you have used the tensor Lorentz transformation, but since the time component at this point is 0, we are left only with length to transform.
Therefore, I am confused by the fact the length 300,000 km has become 346,410.16, i.e. a larger figure.
t doesn't equal gamma, gamma is just a unitless quantity that is useful in special relativity and it comes into the equation for length contraction too.
Bear the concept of context in mind. In the context of my post, the statement: t = \gamma was completely accurate.
Just a quick comment, as I know all will be clear when you read my previous post! :rolleyes:
Yet, at the moment we are concerned about, t=0 and t'=0' there is nothing in the train frame coordinate for time that is different from the embankment frame coordinate for time. x' is different. At this point x is the only coordinate that is different between the two frames due to relative motion at speed .5 c.
When you and Doc Al say you are considering time and position both at once, I grasp the concept, but since the time coordinate is 0 for both frames, the net result of considering both time and location for this moment t' = 0 should be that only position, the x coordinate, is left to treat in any way.
No no no! You assume that the times are the same in both frames. Not so! They only have one point where they agree: that point where/when M' passes M. That point is 0,0 in both frames. But imagine a clock at A and a clock at B. While the embankment frame will say that the flashes occured at t = 0 according to their clocks, the train frame sees those clocks as being out of synch! That's a key result of the "Einstein train gedanken", quantitatively explained by the LT. In the train frame, those flashes do not occur at the same time.
The position of A' from M' or B' from M' should turn out to be closer together by any means of calculation than it is in the M frame. Length contracts. Instead, to my disbelief and confusion, they seem to be farther apart.
Positions A and B, are indeed, located by means of length. To say they have a coordinate is meaningless unless we locate that coordinate with repect to some point =0. This you have done, by using L, length, as the distance from M--->A and M--->B. We have designated that length to be 300,000 km.
Again, you are talking about the length between A and M, but we are talking about the distance between events that happen at different times! Not the same thing at all!
The spped of light is independant of motion and the same for all observers.
i.e no matter how fast you travel it will always catch up with you at the same rate.
for example the speed of light is 670 million mph(not S.I units i know) if you try to outrun the light at 600million mph it will still gain on you at a rate of 670million mph
I hope this helps
zoobyshoe
Jul13-04, 10:06 AM
Gentlemen,
I do not believe, given the amount of math a person seems to need to know to understand this situation, that I'm going to be able to grasp it.
It would be nice if I understood Relativity as you do, but since I have no actual need of it in my life, this is turning out to be more trouble than it is worth simply to satisfy a curiosity.
I think that if four intelligent, educated people working patiently over several days, can't get me over my confusion, then it is probably a sure diagnosis of a defective Relativity lobe in my brain. I'm disapointed to discover I suffer from this malady, but there is nothing to be done.
So, I am retiring from this thread. I would like to thank you all for your efforts.
Zooby
Doc Al makes a vital point t= 0 and t'=0 are not equivalent only (x,t) and (x',t') are equivalent that is because we are dealing with two different co-ordinate systems.
imagine the transformation of a co-ordiante system on a piece of graphing paper:
x' = x + 5
y' = y + 1
The Lorentz tranfomration is simlair to this, by changing refernce frames we are just choosing a new co-ordinarte system in spacetime. I choose the above example for simplicity but in actual fact the Lorentz transformation is more like rotating the axes about a point and you won't go far wrong as visualizng the Lorentz transformation as a rotation about a point in 4-dimensional spacetime.
I do not believe, given the amount of math a person seems to need to know to understand this situation, that I'm going to be able to grasp it.
To derive and understand the LT and the famous SR effects (time dilation, lenght contraction, and simultaneity) all you need is a bit of algebra. (Forget about 4-vectors and tensors... for now. :smile:)
I think that if four intelligent, educated people working patiently over several days, can't get me over my confusion, then it is probably a sure diagnosis of a defective Relativity lobe in my brain. I'm disapointed to discover I suffer from this malady, but there is nothing to be done.
I think part* of the problem is that you are jumping into the middle of a problem before learning the basics. What I would recommend--the next time the relativity bug bites you--is to pick up a book geared to the beginner. One that takes you step by step, anticipates student misconceptions and thinking traps, and is focused on teaching the basics, not covering everything there is to know about SR. Here are a few that I like:
Space and Time in Special Relativity by N. David Mermin. This is probably the easiest place to start.
Space Time Physics by E. Taylor and J. Wheeler. A classic.
A Traveler's Guide to Space Time by Thomas Moore. I only recently discovered this one (even though it's 10 years old), but I like it.
Check these out in your library and see if one suits your learning style. (Every one will require work.) Based on the questions you'ved asked, I recommend Mermin.
Good luck to you. We'll be here when you're ready.
* another part of the problem is that each of us "helpers" is approaching the problem slightly differently--that can only add to the confusion. Some of us are so used to doing these kinds of problems, that it is hard to remember how a complete SR novice feels. There are "advanced" (not really, but I'm sure it looks that way) methods for calculating SR effects--using 4-vector algebra and tensors--that make these problems trivial. But you don't need that to do these problems or to learn the basics.
Yep, I should of said that 4-vectors and tensors aren't necessary in anyway to understanding SR, they're just useful tools (infact SR is a good way of introducing tensors).
Zoobyshoe SR is not beyond you, infact it's a very logical and easy to follow theory provided you follow it correctly.
* another part of the problem is that each of us "helpers" is approaching the problem slightly differently--that can only add to the confusion. Some of us are so used to doing these kinds of problems, that it is hard to remember how a complete SR novice feels. There are "advanced" (not really, but I'm sure it looks that way) methods for calculating SR effects--using 4-vector algebra and tensors--that make these problems trivial. But you don't need that to do these problems or to learn the basics.Another challenge, not necessarily in this thread, is the somewhat disjointed nature of the 'conversation', as well as an unavoidable reliance on a somewhat limiting medium (PF is great for in many ways, but real-time discussion of good diagrams, for example, isn't one of them). There's also - necessarily - a limitation in being able to match the 'teaching' to each learner's own best learning styles.
I was trying to work out a way to set up a list of agreed upon points and assumptions for the thread so people wouldn't have to construct a different context for each poster, but I never finished writing up the suggestions. Plus, of course, there's no real support for that kind of thing in the forum software.
I agree with jcsd, there's no reason this should be beyond you. The scenario that's been worked out here may have become too tangled to be useful at the moment, but writers have come up with many different examples to illustrate the foundations of SR. Working through the logic of one of these other examples might be all that's necessary to bring SR into focus for you.
If you are a visual learner, the Einstein essay, good as it is, is probably not the best place to start as it is so sparsely illustrated. More recent treatments tend to have a lot of diagrams to show the different steps of the logic, e.g. the Wheeler book mentioned above does this. I'm not familiar with the Mermin book, but Mermin is a terrific writer - so even without Doc Al's recommendation I would expect a treatment of these issues by him to be worth a look.
Richard Harris
Jul14-04, 05:35 AM
Hmmm, I see i'm not needed anymore, I got to page three, and i saw what i think then was the aspect that zoobyshoe seemed to miss.
I came up with the following, before it got to all that mathmatics ( i couldnt follow as clearly as i'd have liked)
A & B are the spaceships. to maintain compatability, just drop A and only look at references to B
Lightning flashes at points A & B when Rocket is at point M as seen from Observer at point X ( at rest with A & B, located Parallel to M, at some distance to A-M-B)
the time for light from A & B to reach point M is 1 second.
Rocket is moving away from point A, towards point B at .5c (stated because it's the most important fact.
in 1 second, light from B has passed the rocket and reaches point M
in 1 second, Light from A has reached point M, the rocket is another .5 Second away in terms of distance to travel at light speed, so the Observer on the rocket has seen the flash from B, but has yet to see the flash from A.
from this I conclude the observer sees the flash from B at .66 Seconds from M and Flash from A at 1.66 seconds from M
while the Observer at X see's both flashes after one second (slightly longer than one second, as X is part of a Rightangle triangle AMX)
Richard Harris
Jul14-04, 05:39 AM
Forgot to add, an observer at point X would see the Flashes at the same instant s/he saw the rocket halfway between M & B. I got the impression Zoo was thinking they all saw the flashes when the rocket was at point M.
the lightning hit when both saw the rocket at point M, not both saw the light when the rocket was at point M.
Chronos
Jul27-04, 05:25 AM
Ouch. I tried to follow the needlepoint in this thread. My head hurts. Simplification. When a photon is emitted, everthing in its path 'freezes'. No matter what speed the receiver is traveling, the photon will reach it in exactly d/c seconds [according to the receiver's clock]. Call it time dilation, length contraction, or whatever. It will get there right on 'time'.
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