Is Faster than Light travel impossible?

[Doc Al]

OK, so I am going wrong somewhere, this is how I saw it as an error:

What the stationary observer sees is that the moving observers did not measure the actual length of the rod. Suppose there is a backdrop on the other side of the rod from the stationary observers (think of the rod going down a tunnel). The stationary observers see the photon emitted at the front of the rod against the backdrop, watch the rod move forward and the photon move "backward" until it reaches a coordinate where the end of the rod is going to be when it meets the photon. From the stationary observers point of view they definitely did not measure the rod's length, did they ? They measured a distance between two points that they observed and, if they had quick enough eyesight would see the front of the rod has already passed the reference point for the start of the measurement.

From the 'moving' frame's viewpoint, the motion of the backdrop means nothing. They have a perfectly good and stationary frame in which to do their measurements--their own. Of course, being smart fellows they are well aware that the 'stationary' frame measures a different travel distance for the photon. But don't forget that according to the moving frame the clocks used to measure the time interval in the stationary frame are not even synchronized. So who made the 'error'?

Now a stationary observer who knows the value of c and v can use simple math to determine the error in the measurement made my the moving observers.

Again, there's no error. A stationary observer who knows the value of c and v and relativistic physics can translate measurements made in one frame into measurements made in another.

The moving observers, because they don't recognize the stationary frame of reference, simply assume that the photon traveled the length of the rod.

From their viewpoint--which is perfectly legitimate--the photon did travel the length of the rod!

It didn't because by the time the photon reached the back end the back end had moved (they just didn't know it because they were completely absorbed by their own frame of reference).

You still are hung up thinking that one reference frame is the 'correct' one. Either one is perfectly OK to use.

What worries me is how does the photon and the end of the road move relative to each other at faster than the speed of light ?

They don't! In relativity, 'relative speed' means the speed one thing has as measured in the frame of the other. From the frame of the rod, the photon moves at speed c. (What's interesting is that the photon moves at speed c with respect to both frames!)

because when we think of either of those points being the "fixed" point and the other the moving one they are now seeing the other travel faster than the speed of light.

No. Again, viewing things from the frame in which the rod is at rest, the photon moves at speed c, like usual. However, it's also true that viewed from the 'stationary' frame the rod is moving towards the incoming photon, thus the distance between them--according to the stationary frame--is closing at greater than the speed of light. So what?

 

[jmallett]

From the 'moving' frame's viewpoint, the motion of the backdrop means nothing. They have a perfectly good and stationary frame in which to do their measurements--their own. Of course, being smart fellows they are well aware that the 'stationary' frame measures a different travel distance for the photon. But don't forget that according to the moving frame the clocks used to measure the time interval in the stationary frame are not even synchronized. So who made the 'error'?


Again, there's no error. A stationary observer who knows the value of c and v and relativistic physics can translate measurements made in one frame into measurements made in another.


From their viewpoint--which is perfectly legitimate--the photon did travel the length of the rod!

You still are hung up thinking that one reference frame is the 'correct' one. Either one is perfectly OK to use.


They don't! In relativity, 'relative speed' means the speed one thing has as measured in the frame of the other. From the frame of the rod, the photon moves at speed c. (What's interesting is that the photon moves at speed c with respect to both frames!)

No. Again, viewing things from the frame in which the rod is at rest, the photon moves at speed c, like usual. However, it's also true that viewed from the 'stationary' frame the rod is moving towards the incoming photon, thus the distance between them--according to the stationary frame--is closing at greater than the speed of light. So what?

So I have no axe to grind, or pre-conceived notions, I am merely searching for the truth, and until I can understand this, on a personal basis, I would be required simply to take it on faith. I am hoping to achieve an understanding which reaches beyond faith, otherwise it's just a case of pick something at random and believe it. That's not my goal. I'd like to truly understand, and here is what I don't understand:
The math generated by Einstein is based on a moving observer who is denied the knowledge of a key fact - how fast am I travelling, what is my v.

OK, let's accept that for the moment. If I live in a purely relative world then I have no v in my own world and the equations postulated by Einstein should not contain v - my world could well be larger or smaller smaller than the world of someone traveling at a different speed, and since I am not aware of them then I don't know or care.

If I acknowledge I actually do possesses v then I must add it into my statement of measurement and knowing what it is becomes relevant to how large or small I am, and then, like Einstein I must use math that includes it.

Well let's try both in turn.
I live in a relative world. My v is not known to me, or relevant to me because of the frame or reference I live in. I am going to measure this rod in front of my and clearly, in my world, it has a length L. I no longer need to develop math which includes v, because that is not relevant/relative to my world and the equations I am going to use are not Einsteins.

Other World. I think I might have a v. The only way I can determine it's value is by finding something that actually does not have v and I am going to call that a stationary observer, for the sake of argument. Now I have a way to determine my v. I can simply ask him to watch me and to perform some math, and his math is going to use my v and determine my size. But in reality he doesn't have v=0. He simply has negative my v, which is a very special case and hardly a basis for generalization of the universe. I need more observers with a whole variety of v's.

In the first world I have no need of v and therefore an Einsteinian equation containing v is meaningless.
In the second world v is a very special case and using it in an equation is also meaningless because it will only work if, and when, I can find someone else with a velocity of -my(v)
So in this case why include v if it has no real meaning in this context either ?
Again I have problems justifying the existence of v in an Einsteinian equation.

The big headache, for me, comes when I eliminate v from the equations and now find that I am back at Newtonian math.

My v'ness would be simply a matter of faith, and that's not a step I am quite ready to take, especially since the whole of our modern physics is based on a requirement to believe in my own v'ness without being able to question it.

 

[A.T.]

The math generated by Einstein is based on a moving observer who is denied the knowledge of a key fact - how fast am I travelling, what is my v.

That is actually Galilean invariance, stated long before Einstein:
http://en.wikipedia.org/wiki/Galilean_invariance

In the first world I have no need of v and therefore an Einsteinian equation containing v is meaningless.

You are using 'v' for different things and confusing yourself. Just because you cannot determine an absolute value of your own velocity, doesn't mean that you cannot determine the relative velocity of some object relative to you. The letter 'v' is still quite useful to describe relative velocity.

Try to address more specific questions which have quantitative answers instead of juggling with terms and pseudo philosophical conclusions.

 

[cfrogue]

So I have no axe to grind, or pre-conceived notions, I am merely searching for the truth, and until I can understand this, on a personal basis, I would be required simply to take it on faith. I am hoping to achieve an understanding which reaches beyond faith, otherwise it's just a case of pick something at random and believe it. That's not my goal. I'd like to truly understand, and here is what I don't understand:
The math generated by Einstein is based on a moving observer who is denied the knowledge of a key fact - how fast am I travelling, what is my v.

This is not correct.

The math is generated based on the logic that how fast a frame is actually traveling is not logically decidable.

In order to apply your logic, you must produce this absolute v and then make your decisions.

You do not have a method to produce this v and thus, it is necessary to fix one frame and calculate the motion of other frames in a relative way.

 

[jmallett]

Try to address more specific questions which have quantitative answers instead of juggling with terms and pseudo philosophical conclusions.

Sorry if that's the impression I gave you, it was more of a fairly lame attempt at humor and irony because I recognize that the possession of v'ness is clearly stupid.

However having an absolute v appears to be an inherent assumption of Einstein's paper, based on the way that I am (currently) interpreting it.

He surely can't make that assumption.

 

[Doc Al]

However having an absolute v appears to be an inherent assumption of Einstein's paper, based on the way that I am (currently) interpreting it.

He surely can't make that assumption.

Of course he doesn't make such an assumption. I have no idea where you are getting that. All velocities are relative velocities.

 

[jtbell]

However having an absolute v appears to be an inherent assumption of Einstein's paper, based on the way that I am (currently) interpreting it.

Why do you think that? Can you point us to a specific place where he appears to use that assumption?

 

[jmallett]

Why do you think that? Can you point us to a specific place where he appears to use that assumption?

If there are 2 sets of observers neither can know they are stationary - they are just traveling relative to each other, right ?

In this case any perceived difference by either simply has to be the inverse of the other, right ?

When Einstein states "Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous." he has to make an assumption of which one is stationary, other wise the statement would be:

observers moving in Frame(1) find that the clocks are not synchronous and observers in Frame(2) also find that the clocks are not synchronous.

 

[jmallett]

Of course he doesn't make such an assumption. I have no idea where you are getting that. All velocities are relative velocities.

If they are truly relative then one is simply the inverse of the other.
Einstein states they are not. He states that one observer sees the clocks as synchronous.

 

[Doc Al]

When Einstein states "Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous." he has to make an assumption of which one is stationary,

The designations 'stationary' and 'moving' are arbitrary. (Any observer can choose to view himself as stationary.) In this example, the 'stationary' observers (say they are on a space station) are the ones observing the rod moving past them. The 'moving' observers are the ones moving with the rod.

The point is that whether or not a pair of clocks are synchronized depends on who is doing the observing. For a pair of clocks moving with the rod, the rod observers say they are synchronized but the other frame ('stationary') observers disagree.

Nothing here has anything to do with 'absolute' velocity. All that matters is the relative motion of observer and clock.

 

[Doc Al]

If they are truly relative then one is simply the inverse of the other.
Einstein states they are not. He states that one observer sees the clocks as synchronous.

The clocks are stationary in one particular frame (the one moving along with the rod), not in both. Observers in frames that are moving with respect to the clocks will see those clocks as unsynchronized.

But it certainly works both ways. A pair of synchronized clocks in the 'stationary' frame would be seen as unsynchronized by the observers moving with the rod.

 

[jmallett]

I will go back and study it again, then and think about it some more.

 

[jmallett]

The clocks are stationary in one particular frame (the one moving along with the rod), not in both. Observers in frames that are moving with respect to the clocks will see those clocks as unsynchronized.

But it certainly works both ways. A pair of synchronized clocks in the 'stationary' frame would be seen as unsynchronized by the observers moving with the rod.

So Einstein is saying there are observers at A and B, both located on the moving rod, right ? and that they see the clocks as being unsynchronized, right ?

I understand that to mean that they see that it takes light longer to go one way when compared with the other, right ? why don't they deduce, then, that they must be moving ?

 

[Doc Al]

So Einstein is saying there are observers at A and B, both located on the moving rod, right ? and that they see the clocks as being unsynchronized, right ?

No. The clocks attached to the rod are synchronized as far as the observers moving with the rod are concerned. (Other observers, who see the clocks as moving with respect to them, will see the clocks as unsynchronized.)

 

[jmallett]

No. The clocks attached to the rod are synchronized as far as the observers moving with the rod are concerned. (Other observers, who see the clocks as moving with respect to them, will see the clocks as unsynchronized.)

Einstein says:

Observers moving with the moving rod would thus find that the two
clocks were not synchronous, while observers in the stationary system would
declare the clocks to be synchronous.

There are three clocks in his experiment. One at A, one at B, and one "stationary" clock.

In his above statement which two clocks is he referring to ?

If he means the two clocks attached to the rod he is saying they are NOT synchronous for the moving observers. If he means one attached to the rod and one "stationary" clock then he is saying those ARE synchronized.
The problem is in in determining which two of the three clocks he is referring to, isn't it ?

 

[Doc Al]

Einstein says:

Observers moving with the moving rod would thus find that the two
clocks were not synchronous, while observers in the stationary system would
declare the clocks to be synchronous.

There are three clocks in his experiment. One at A, one at B, and one "stationary" clock.

In his above statement which two clocks is he referring to ?

The two clocks, at A and B, are in the 'stationary' frame. Thus they are synchronized in the stationary frame. But observers in the 'moving' frame (with the rod) view those clocks as unsynchronized.

 

[jmallett]

Einstein says:



If he means the two clocks attached to the rod he is saying they are NOT synchronous for the moving observers. If he means one attached to the rod and one "stationary" clock then he is saying those ARE synchronized.
The problem is in in determining which two of the three clocks he is referring to, isn't it ?

No, wait, I may have got that wrong. Let me try again. If he means the two clocks attached to the moving rod he is saying they are not synchronous. You are disagreeing with that, therefore the two clocks in qustion must be one on the moving rod and on "stationary" clock, and these are viewed as being not synchronized.

I think that is what you are saying, right ?

 

[jmallett]

No, wait, I may have got that wrong. Let me try again. If he means the two clocks attached to the moving rod he is saying they are not synchronous. You are disagreeing with that, therefore the two clocks in qustion must be one on the moving rod and on "stationary" clock, and these are viewed as being not synchronized.

I think that is what you are saying, right ?

I think first we must define which clocks are being considered. Let's assume there are three:
A, B (both on the moving rod) and S (stationary).

Which ones are synchronized and which are not ?

 

[Doc Al]

No, wait, I may have got that wrong. Let me try again. If he means the two clocks attached to the moving rod he is saying they are not synchronous. You are disagreeing with that, therefore the two clocks in qustion must be one on the moving rod and on "stationary" clock, and these are viewed as being not synchronized.

I think that is what you are saying, right ?

The two clocks are in the 'stationary' frame and thus are synchronized in that frame. Viewed from the frame of the moving rod, they are not synchronized.

(Rather than puzzle your way through Einstein's 1905 paper, I recommend you get one of the many pedagogical books explaining special relativity that have been written over the years. Here's a free one by Dan Styers: http://www.oberlin.edu/physics/dstyer/Einstein/SRBook.pdf" )

[I just noticed that JesseM recommended the same book back in post #16!]

 

[jmallett]

The two clocks are in the 'stationary' frame and thus are synchronized in that frame. Viewed from the frame of the moving rod, they are not synchronized.

(Rather than puzzle your way through Einstein's 1905 paper, I recommend you get one of the many pedagogical books explaining special relativity that have been written over the years. Here's a free one by Dan Styers: http://www.oberlin.edu/physics/dstyer/Einstein/SRBook.pdf" )

thanks, that's my homework for the weekend then, and I won't trouble you until I have studied it carefully, however I still don't quite understand the experiment.

There are 2 clocks in the stationary frame and one on the moving rod ? or is the moving rod considered the stationary frame, in which case the clocks are stationary to each other but moving relative to the stationary clock ?

Is there some wau n which these can be more specifically defined ? Let's say there3 is a clock A which has an observer a, a clock B, which has an observer b, and a stationary clock S, which has an observer s.

Which clocks are attached to the rod and (I had assumed A & B), which ones are defined as "stationary" ? I had assumed S. Are A & B in fact the "stationary" clocks because they are stationary relative to each other ?

 

[Doc Al]

The key thing to realize is that clocks that are synchronized in one frame are not synchronized in another (moving) frame. Try this. Imagine that instead of a rod there is a huge train traveling along at high speed. Put two clocks on the train, one at the front and the other at the back. Have the people on the train synchronize the two clocks in the usual manner. For example, have a light bulb flash in the middle of the train. When the light reaches each clock, have the clocks set to read 1:00 pm. Since--as far as the train frame is concerned--the light flashes take the same time to reach the two clocks, the clocks are now synchronized.

But now view things from the frame of observers on the ground, who watch the train go by. From their point of view, the front of train moves away from the light flash while the rear of the train moves towards the light flash. Thus they observe that the light reaches the clock at the rear of the train first. So, from the ground observer's viewpoint, those clocks are not synchronized. Make sense?

 

[jmallett]

The key thing to realize is that clocks that are synchronized in one frame are not synchronized in another (moving) frame. Try this. Imagine that instead of a rod there is a huge train traveling along at high speed. Put two clocks on the train, one at the front and the other at the back. Have the people on the train synchronize the two clocks in the usual manner. For example, have a light bulb flash in the middle of the train. When the light reaches each clock, have the clocks set to read 1:00 pm. Since--as far as the train frame is concerned--the light flashes take the same time to reach the two clocks, the clocks are now synchronized.

But now view things from the frame of observers on the ground, who watch the train go by. From their point of view, the front of train moves away from the light flash while the rear of the train moves towards the light flash. Thus they observe that the light reaches the clock at the rear of the train first. So, from the ground observer's viewpoint, those clocks are not synchronized. Make sense?

Not complete sense (yet, at least) because surely what the observer saw was that they measured different distances because during the time they took to make the measurement the point they were measuring to moved, the observer sees that easily.
Using light, the measurement towards the rear of the train simply measured to a point where the end of the train is going to be when the measurement is completed, no ? That's certainly not the distance he intended to measure. It's not the position that the rear of the train was in when he started the measurement, therefore the two positions were calculated at different times. When you take time to make a measurement you are never going to accurately measure the length of that moving object.

Surely the reason the observer sees that the clocks are not synchronized is that it took a different length of time to reach each clock, and that's because the distances measured were different.

OK, what about the observers on the train. Do they actually see the clocks as synchronized ? and why ?

 

[Doc Al]

Not complete sense (yet, at least) because surely what the observer saw was that they measured different distances because during the time they took to make the measurement the point they were measuring to moved, the observer sees that easily.

You're talking about the ground observers. Of course they see the train moving.

Using light, the measurement towards the rear of the train simply measured to a point where the end of the train is going to be when the measurement is completed, no ? That's certainly not the distance he intended to measure.

The ground observers only care about the distance the light flashes travel as seen in their own ground-based frame. All they are interested in (in this thought experiment) is whether the moving clocks (on the train) are synchronized.

It's not the position that the rear of the train was in when he started the measurement, therefore the two positions were calculated at different times. When you take time to make a measurement you are never going to accurately measure the length of that moving object.

The ground observers are not measuring the length of the moving train. That's a different thought experiment. (To measure the length of the train, you'd need to measure the position of both train ends at the same time.)

Surely the reason the observer sees that the clocks are not synchronized is that it took a different length of time to reach each clock,

Of course!

and that's because the distances measured were different.

I would say that according to the ground observers the light hitting the front of the train had to travel a greater distance than the light hitting the rear of the train. So of course it takes a different time to reach each clock.

OK, what about the observers on the train. Do they actually see the clocks as synchronized ? and why ?

Of course. The light travels the same distance according to the train observers.

 

[jmallett]

Of course. The light travels the same distance according to the train observers.

If there are 2 observers, and they see the clocks as synchronized, then, by definition, they are not, because of the time taken to observe the clock. Let's put one at the front of the train and another at the back. Neither can see both clocks showing the same time.
Now let's revert to a single observer and put him midway between the clocks, which, not coincidentally, would be the source of the light flash. In this case he sees them synchronized, but doesn't realize that they are not synchronized because of the exact same problem.
In this case the moving observer sees precisely the same thing as the stationary observer, he just isn't able to detect it because it cancels itself out.

 

[Doc Al]

If there are 2 observers, and they see the clocks as synchronized, then, by definition, they are not, because of the time taken to observe the clock.

Don't get hung up on the word 'see'. Relativistic effects are what is left after taking into account light travel time.

When we say that the two clocks on the train are synchronized we mean that all observers on the train agree that they read the same time. Of course, if an observer at the front of the train were to literally 'see' the clock at the rear (through a telescope, perhaps), then he would have to add the time it took for the light to reach his eyes to figure out what that clock reads 'now'.

 

[darkhorror]

You seem to be taking humans into the equation but that doesn't have anything to do with it.

Lets say you have 2 clocks on a train that is moving close to the speed of light with respect to the train tracks. The two clocks are started by a beam of light being sent in both directions from the middle of the train. The clocks will be synchronized when compared to the train as the beam of light is moving the same distance in both directions. Relative to the train tracks the beam of light that moving forward takes long than the beam of light moving backwards to get to each end of the train. This is because the train is moving with respect to the train tracks. So with respect to the train tracks they are not synchronized.

 

[jmallett]

You seem to be taking humans into the equation but that doesn't have anything to do with it.

Lets say you have 2 clocks on a train that is moving close to the speed of light with respect to the train tracks. The two clocks are started by a beam of light being sent in both directions from the middle of the train. The clocks will be synchronized when compared to the train as the beam of light is moving the same distance in both directions. Relative to the train tracks the beam of light that moving forward takes long than the beam of light moving backwards to get to each end of the train. This is because the train is moving with respect to the train tracks. So with respect to the train tracks they are not synchronized.

I don't understand how can the beam of light be said to be moving the same distance ? Someone on the train believes that, because he is denied the knowledge of his speed, but in fact the light moving backwards, for example, is not moving to the position of the end of the train as it was when he started the measurement. He is simply measuring to a point in space which is where the end of the train is going to be when the end of the train gets there at some later time. The opposite is true for the front. Therefore the clocks are not synchronized. I think we agree and I think we have the same reasoning, right ?

However when they are observed (on the train) the time taken for the observation is the inverse, so it looks, to the observer on the train, that they are.

The observer who is not moving can see this and can determine the problem because he has an extra piece of information. He knows the speed of the train.

The clocks are, indeed, not synchronized. The "tracks" can see this, the train cannot because it is making 2 errors of measurement which cancel each other out.

 

[jmallett]

"Of course. The light travels the same distance according to the train observers."

because they made two errors in measurement which cancel each other out. It doesn't actually travel the same distance.

 

[Doc Al]

"Of course. The light travels the same distance according to the train observers."

because they made two errors in measurement which cancel each other out. It doesn't actually travel the same distance.

What errors? You're saying that the two halves of the train are different distances?

 

[jmallett]

What errors? You're saying that the two halves of the train are different distances?

No, not at all. We can easily agree on that. I am saying they didn't measure to the end of the train as it was at the time that the measurement was started.

 

[Doc Al]

No, not at all. We can easily agree on that. I am saying they didn't measure to the end of the train as it was at the time that the measurement was started.

What are you talking about? From the train's point of view, the train isn't moving. Who cares if the ground rushes by? All measurements are done inside the train.

 

[Doc Al]

I don't understand how can the beam of light be said to be moving the same distance ? Someone on the train believes that, because he is denied the knowledge of his speed, but in fact the light moving backwards, for example, is not moving to the position of the end of the train as it was when he started the measurement.

You still seem to be thinking that the train is 'really' moving and the ground is 'really' at rest. Perhaps if we replaced the ground observers by another long train it might be easier to understand. Now let there be two giant space trains floating in outer space. Train S (what we used to call the ground) sees train S' moving by. Of course, train S' also sees train S moving by. Who is really moving and who is really at rest? That's a meaningless question--only relative motion makes any sense. Either train is perfectly justified in treating themselves as being at rest. (For mechanical things, like moving trains, this was well known long before Einstein. This is called 'Galilean relativity', after Galileo, who used moving ships in his examples.)

 

[jmallett]

What are you talking about? From the train's point of view, the train isn't moving. Who cares if the ground rushes by? All measurements are done inside the train.

Yes, that's true. I agree. Now here comes the problem. Einstein cares (or cared anyway) so he placed an observer outside of the train. As soon as he does that he places them all in the same frame of reference and then proposes that light is traveling at two different speeds in this single frame of reference.

Is it possible, then, to derive Einstein's equations without the observer ? No it's not because the speed of the light in the train traveling at (c + v) is not possible when it is placed in the frame of reference of the observer.

Try deriving Einsteins equations without having light traveling at two different speeds in the same inertial frame.

 

[Doc Al]

Yes, that's true. I agree. Now here comes the problem. Einstein cares (or cared anyway) so he placed an observer outside of the train. As soon as he does that he places them all in the same frame of reference and then proposes that light is traveling at two different speeds in this single frame of reference.

Please show exactly where Einstein proposes that light travels with two different speeds in a single frame.

Is it possible, then, to derive Einstein's equations without the observer ? No it's not because the speed of the light in the train traveling at (c + v) is not possible when it is placed in the frame of reference of the observer.

You misunderstand the meaning of (c + v). In any given frame, the speed of light is c, as always. From the view of the outside observer, the speed of the train is v. "c + v" is the rate at which the light catches up with the oncoming train, as seen from the frame of the outside observer--it's not the speed of light in that frame.

Try deriving Einsteins equations without having light traveling at two different speeds in the same inertial frame.

Einstein doesn't do that. On the contrary, when viewing things from a particular frame the speed of light is always the same with respect to that frame. That's one of the premises used in deriving the relativistic effects.

 

[jmallett]

If the observer can see the train then the train must be in his frame of reference. In that case the light in the train is also in his frame of reference. Light in his frame of reference cannot travel at (c +v)

 

[Doc Al]

If the observer can see the train then the train must be in his frame of reference.

Not true. Anyone in any frame can see the train. If the train is seen as moving, then it clearly is not at rest in the observer's frame.

In that case the light in the train is also in his frame of reference. Light in his frame of reference cannot travel at (c +v)

As measured in any frame, the light travels at speed c. Not at speed "c + v".

Try this. Imagine a road with two cars separated by 100 miles. Let one car travel east at 50 mph as seen by observers at rest on the ground. Let the other car travel west at 50 mph as seen by observers at rest on the ground. In one hour, they will collide. Thus they close the distance between them at a rate of 100 mph--but they still only travel at 50 mph with respect to the ground.

The same logic applies to beams of light. If someone shines a beam of light to the east while someone else shines a beam of light to the west, the leading edge of those beams will close the distance between them at a rate of twice the speed of light according to an observer on the ground--yet the speed of each light beam is still just c with respect to the ground.

 

[darkhorror]

Look at it this way instead of a train and train tracks turn those into 2 space ships in an otherwise empty universe. Have the 2 spaceships approaching each other at close to the speed of light.

 

[jmallett]

Not true. Anyone in any frame can see the train. If the train is seen as moving, then it clearly is not at rest in the observer's frame.

As measured in any frame, the light travels at speed c. Not at speed "c + v".

Try this. Imagine a road with two cars separated by 100 miles. Let one car travel east at 50 mph as seen by observers at rest on the ground. Let the other car travel west at 50 mph as seen by observers at rest on the ground. In one hour, they will collide. Thus they close the distance between them at a rate of 100 mph--but they still only travel at 50 mph with respect to the ground.

The same logic applies to beams of light. If someone shines a beam of light to the east while someone else shines a beam of light to the west, the leading edge of those beams will close the distance between them at a rate of twice the speed of light according to an observer on the ground--yet the speed of each light beam is still just c with respect to the ground.

I tried it and concluded:
1. While you started out with good intentions by considering the purely relativistic mechanics this theory, within nano-seconds, reverted back to the observer on the ground. This theory apparently needs someone "on the ground", which is where exactly - sitting in the ether ? is it a go9d-like Einstein. Mitchelson Morely showed us that the place where that observer is situated simply doesn't exist.
The train and track, moving relatively towards each other, that I was looking for suddenly morphed into two cars (and we dismiss their relativistic mechanics quickly) and the tracks, which have simply been renamed as a road. This is not relativity. It considers only the relative motion of objects in an absolute space. Try using ony relativity to develop your theories.

2. Let the cars not collide, but pass each other very closely traveling along the same axis. The observer in Car A looks out his window and can see the light in the other car. At this point the light in car B, Car B and the observer in Car A are now all in the same frame of reference.
How fast is the light traveling - and remember that, by definition, it must be the same for both Observers A & B because they are in same same frame of reference.

Here's the bigt problem. No-one seems to be interested in developing theories which are acyually based on relativistic mechanics. It's like the whole community just gave up looking into the subject and worshipped at the altar of Einstein, the observer who can exist in a position we have proven does not exist, sees all, knows all and has no impact on the cosmos.

We can all repeat and explain Einstein's approach. That's not the point, and I don't believe he wanted inquiry to stop there and be studied o9n faith like some kind of holy book.

If we are truly interested then we need to develop the mathematics of relativity using only relative mechanics. Let's go back the road with the 2 cars and completely remove the road as a concept. Now how do we derive the laws of mechanics. I cannot, and so far I have not yet met anyone who can, without reverting back to Einstien's god-like observer.

I believe you are explaining Einsteins theory in the way he explained it, but the theory, and the math, just doesn't properly deal with relativity when it falls backn to the crutch of the stationary road, tracks, eatrth, universe.

Let's seek the mathematics of relativity by considering only relative motion.

 

[jmallett]

Please show exactly where Einstein proposes that light travels with two different speeds in a single frame.


You misunderstand the meaning of (c + v). In any given frame, the speed of light is c, as always. From the view of the outside observer, the speed of the train is v. "c + v" is the rate at which the light catches up with the oncoming train, as seen from the frame of the outside observer--it's not the speed of light in that frame.


Einstein doesn't do that. On the contrary, when viewing things from a particular frame the speed of light is always the same with respect to that frame. That's one of the premises used in deriving the relativistic effects.

in that case there is, by definition, no observer, because as soon as you place an observer in the picture he forces the light in his frame of reference to be the same in all the frames of reference he is observing.

 

[jmallett]

So let's try this from a purely relativistic point of view. All ye who enter here first abandon stationary observers, ether, gods, train tracks, roads or other devices created for the sole purpose of being independent of relative motion.

Two space ships are floating through the cosmos. Each has an undetermined speed both by themselves and by the other. for simplicity, and by sheer luck (for you and me but only because it simplifies our math), they are traveling along the same axis.

Both send out a beam of light of the same frequency in the forward direction. Will those beams synchronize and will they synchronize independently of the speed of the two ships and without knowing anything else about the ships ?

This is a simple, first step in relativistic mechanics and easily provable if we could just find light emission from bodies moving arbitrarily in a universe. Let's try some assumptions and then test them by the experiment of looking out the window.

I'll let you go first in making the first assumption.

 

[darkhorror]

I tried it and concluded:
1. While you started out with good intentions by considering the purely relativistic mechanics this theory, within nano-seconds, reverted back to the observer on the ground. This theory apparently needs someone "on the ground", which is where exactly - sitting in the ether ? is it a go9d-like Einstein. Mitchelson Morely showed us that the place where that observer is situated simply doesn't exist.
The train and track, moving relatively towards each other, that I was looking for suddenly morphed into two cars (and we dismiss their relativistic mechanics quickly) and the tracks, which have simply been renamed as a road. This is not relativity. It considers only the relative motion of objects in an absolute space. Try using ony relativity to develop your theories.

I can't really see what you are trying to say here.

2. Let the cars not collide, but pass each other very closely traveling along the same axis. The observer in Car A looks out his window and can see the light in the other car. At this point the light in car B, Car B and the observer in Car A are now all in the same frame of reference.
How fast is the light traveling - and remember that, by definition, it must be the same for both Observers A & B because they are in same same frame of reference.

ok you say the observer in car A looks out the window and can see the light in the other car. This does NOT put them in the same frame as Car B. Car A and Car B are the 2 frames, if you are in Car A you are in that frame. If you are in Car B you are in that frame. Observing doesn't change anything.

Here's the bigt problem. No-one seems to be interested in developing theories which are acyually based on relativistic mechanics. It's like the whole community just gave up looking into the subject and worshipped at the altar of Einstein, the observer who can exist in a position we have proven does not exist, sees all, knows all and has no impact on the cosmos.

We can all repeat and explain Einstein's approach. That's not the point, and I don't believe he wanted inquiry to stop there and be studied o9n faith like some kind of holy book.

If we are truly interested then we need to develop the mathematics of relativity using only relative mechanics. Let's go back the road with the 2 cars and completely remove the road as a concept. Now how do we derive the laws of mechanics. I cannot, and so far I have not yet met anyone who can, without reverting back to Einstien's god-like observer.

I believe you are explaining Einsteins theory in the way he explained it, but the theory, and the math, just doesn't properly deal with relativity when it falls backn to the crutch of the stationary road, tracks, eatrth, universe.

Let's seek the mathematics of relativity by considering only relative motion.

There is no god-like observer that is the very basic premise of relativity. Look at what I was trying to get with when I said there are only 2 space ships moving twards each other at close to the speed of light in an otherwise empty universe.

 

[matheinste]

in that case there is, by definition, no observer, because as soon as you place an observer in the picture he forces the light in his frame of reference to be the same in all the frames of reference he is observing.

What force does the observer use to do this, and what does the light do when no observer is observing it.

Matheinste.

 

[jmallett]

Look at it this way instead of a train and train tracks turn those into 2 space ships in an otherwise empty universe. Have the 2 spaceships approaching each other at close to the speed of light.

DarkHorror, I like this as place for starting to think, I don't feel adequate in stopping my explorations at that.
Please develop this idea further and consider the frequency (color) of the light emanating from those ships.

 

[Doc Al]

I tried it and concluded:
1. While you started out with good intentions by considering the purely relativistic mechanics this theory, within nano-seconds, reverted back to the observer on the ground. This theory apparently needs someone "on the ground", which is where exactly - sitting in the ether ? is it a go9d-like Einstein. Mitchelson Morely showed us that the place where that observer is situated simply doesn't exist.
The train and track, moving relatively towards each other, that I was looking for suddenly morphed into two cars (and we dismiss their relativistic mechanics quickly) and the tracks, which have simply been renamed as a road. This is not relativity. It considers only the relative motion of objects in an absolute space. Try using ony relativity to develop your theories.

There's nothing special about using the ground or anything other reference frame for describing the relative motion of things. Nothing 'god-like' or absolute about it.

2. Let the cars not collide, but pass each other very closely traveling along the same axis. The observer in Car A looks out his window and can see the light in the other car. At this point the light in car B, Car B and the observer in Car A are now all in the same frame of reference.

I don't know what you mean when you say that Car A and Car B are 'in the same reference frame'. They are certainly not moving together. From Car A's reference frame, Car B is moving. And vice versa.

How fast is the light traveling - and remember that, by definition, it must be the same for both Observers A & B because they are in same same frame of reference.

Measured from Car A's frame (meaning: from a frame in which Car A is at rest) the speed of light is c. And from Car B's frame (a different frame from Car A's frame) the speed of light is also c.

Here's the bigt problem. No-one seems to be interested in developing theories which are acyually based on relativistic mechanics. It's like the whole community just gave up looking into the subject and worshipped at the altar of Einstein, the observer who can exist in a position we have proven does not exist, sees all, knows all and has no impact on the cosmos.

We can all repeat and explain Einstein's approach. That's not the point, and I don't believe he wanted inquiry to stop there and be studied o9n faith like some kind of holy book.

If we are truly interested then we need to develop the mathematics of relativity using only relative mechanics. Let's go back the road with the 2 cars and completely remove the road as a concept. Now how do we derive the laws of mechanics. I cannot, and so far I have not yet met anyone who can, without reverting back to Einstien's god-like observer.

I believe you are explaining Einsteins theory in the way he explained it, but the theory, and the math, just doesn't properly deal with relativity when it falls backn to the crutch of the stationary road, tracks, eatrth, universe.

Let's seek the mathematics of relativity by considering only relative motion.

I have no idea what you are talking about here. The only motion considered in developing relativity--and certainly in the examples we've discussed here--is relative motion. The cars move relative to each other; the train moves relative to the tracks. What's the problem?

 

[Doc Al]

in that case there is, by definition, no observer, because as soon as you place an observer in the picture he forces the light in his frame of reference to be the same in all the frames of reference he is observing.

Now what are you talking about? Of course there is an observer--the one who measures the train as moving at speed v.

 

[jmallett]

What force does the observer use to do this, and what does the light do when no observer is observing it.

Matheinste.

Matheinste, Great questions, and this is something for the defenders of Einstein. By placing an observer in the picture they immediately specify that with, or without, force it must be so. The dictate is that light travels at a single and constant speed in any given frame of reference. It may, or may not, be true, but if it is true then you are right to ask - by what law, or force can the observer do this ?

The next part is the exciting part. No-one seems to be considering this, and Einstein certainly didn't, so you are beginning to think beyond the rote learning of the last 100 years.

 

[Doc Al]

So let's try this from a purely relativistic point of view. All ye who enter here first abandon stationary observers, ether, gods, train tracks, roads or other devices created for the sole purpose of being independent of relative motion.

There is nothing about train tracks or roads that makes them 'independent of relative motion'.

Two space ships are floating through the cosmos. Each has an undetermined speed both by themselves and by the other. for simplicity, and by sheer luck (for you and me but only because it simplifies our math), they are traveling along the same axis.

OK. I assume that they have some speed relative to each other.

Both send out a beam of light of the same frequency in the forward direction. Will those beams synchronize and will they synchronize independently of the speed of the two ships and without knowing anything else about the ships ?

What do you mean 'synchronize'? Note that relativity assumes that any frame of reference (each of the two ships, in this case) will measure the speed of any beam of light as moving with the same speed c with respect to that frame.

This is a simple, first step in relativistic mechanics and easily provable if we could just find light emission from bodies moving arbitrarily in a universe. Let's try some assumptions and then test them by the experiment of looking out the window.

I have no idea what you are looking for.

 

[jmallett]

Now what are you talking about? Of course there is an observer--the one who measures the train as moving at speed v.

Then by definition he sees the light in the train, and by seeing it and the train at the same time, then the speed of the light in the train is the same in the train as it is for him - (c + v) disappears and the observer in the train measures the train's length incorrectly.

 

[Doc Al]

Matheinste, Great questions, and this is something for the defenders of Einstein. By placing an observer in the picture they immediately specify that with, or without, force it must be so. The dictate is that light travels at a single and constant speed in any given frame of reference. It may, or may not, be true, but if it is true then you are right to ask - by what law, or force can the observer do this ?

The next part is the exciting part. No-one seems to be considering this, and Einstein certainly didn't, so you are beginning to think beyond the rote learning of the last 100 years.

I think it's becoming clear that you are not interesting in learning about relativity and that you have some sort of axe to grind. Please take a look at the sticky labeled "IMPORTANT! Read before posting" at the top of this forum before continuing.

 

[darkhorror]

Ok let's say they are space ships A and B moving twards each other close to the speed of light. On spaceship A a beam of light in the middle of the ship causes two clocks to start on both ends of the ship when the light hits them. In that frame both clocks are synchronized because the light travels the same distance to the back as it does to the front and light is traveling at the speed of light obviously.

With respect to spaceship B the beam of light in A has to travel different distances. This is because spaceship A is moving with respect to spaceship B. But you can do this on both spaceships and to A the B clocks are out of sync but to B they are in sync. Just as to A the A clocks are in sync, but to A the B clocks are out of sync.

I left out the word observer as it seems to be adding to confusion, having observers does nothing to change what is actually happening.