Is Faster than Light travel impossible?

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SUMMARY

The discussion centers on the concept of faster-than-light (FTL) travel and the implications of Special Relativity (SR). Participants explore scenarios where two spacecraft approach each other at 0.6C, leading to an observed closing velocity of 1.2C from a stationary observer's frame. However, SR dictates that each spacecraft perceives the other approaching at less than the speed of light, reinforcing the principle that information cannot travel faster than light. The conversation also touches on the limitations of rigid bodies and the speed of sound in materials, emphasizing that no physical mechanism allows for FTL communication.

PREREQUISITES
  • Understanding of Special Relativity principles
  • Familiarity with the concept of inertial frames
  • Knowledge of velocity addition in relativistic contexts
  • Basic grasp of the speed of light as a universal constant
NEXT STEPS
  • Study the Lorentz transformation equations in Special Relativity
  • Examine the implications of the Michelson-Morley experiment on the theory of relativity
  • Research the concept of causality in physics and its relation to information transfer
  • Explore resources like "Relativity Simply Explained" by Martin Gardner for a layman's understanding of SR
USEFUL FOR

Physicists, students of theoretical physics, and anyone interested in the implications of Special Relativity and the nature of light speed limitations.

  • #121
Doc Al said:
Nope. The 'stationary' observer does not make use of any length measurements made in some other frame. All his measurements are made in his own frame.


Nope. I've said as many times as possible: Viewed from any frame of reference, the light moves at the same speed. (Again, I have no idea what you mean when you talk of things being in 'the same frame of reference'.)


I've been saying that all along: The 'stationary' observer sees all light travel at the same speed with respect to him (not the train!). A basic premise of relativity is that the speed of light is the same in every frame. Now if you switch to the frame of the train, then the speed of light is also c with respect to the train.

Where you are getting stuck is in not understanding how the stationary observer can see the light and the train close at a rate of 'c + v', even though both the stationary observer and the train both observe the light to move at the same speed c with respect to their own frames. This is tricky stuff.



OK, so we are starting to agree.

"Viewed from any frame of reference, the light moves at the same speed. (Again, I have no idea what you mean when you talk of things being in 'the same frame of reference'.)"


I see that we agree there are various frames of reference.

When I say "same frame of reference" I mean a SINGLE frame of reference, for example the stationary observer has a single frame of reference. He can't see things (like light) behave differently at different times, because he can only exist in a single frame of reference, and everything he observes is now in the same frame of reference, that is to say his. This includes the various cars, trains, tracks, light, etc he has been observing throughout this discussion.

Are we on the same page ?
 
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  • #122
jmallett said:
When I say "same frame of reference" I mean a SINGLE frame of reference, for example the stationary observer has a single frame of reference. He can't see things (like light) behave differently at different times, because he can only exist in a single frame of reference, and everything he observes is now in the same frame of reference, that is to say his. This includes the various cars, trains, tracks, light, etc he has been observing throughout this discussion.

Are we on the same page ?
Nope, not even close. For things to be 'in the same frame of reference' they must be moving together at the same speed. Note that anything--cars, trains, tracks, light--can be observed by anyone in any frame of reference. Just because someone in the track frame (that is, someone at rest with respect to the tracks) sees a train going by does not magically make the train somehow jump into the same frame of reference as the observer. The track frame sees the train moving (thus in a different frame); similarly, the train frame sees the track moving.
 
  • #123
matheinste said:
If you accept the beliefs or postulates of relativity you arrive at certain logically derived consequences. You do not have to accept any postulates if you wish not to. However, you must accept the logical reasoning used. If you do not accept the postulate of light speed, which most agree is counterintuitive, so be it, but at least apply some logic in your arguments against it.

Matheinste.

First accepting the postulates or beliefs of anything is merely taking it on faith. We agree.

I fully accept the postulate of light speed, if we are talking about an observer will see light travel at speed c in his own frame of reference. I think you agree with that too.

Let's try to move to the logic part.

I will, for the moment, accept your stationary observer - we still agree.

We will both accept that the speed of light in the moving body is c.

We will now ask a question. As far as the train is concerned what is his speed ? Well probably he doesn't know, maybe doesn't care, right ? So what do we want him to do about that. The general conclusion is that his velocity relative to himself is 0 and therefore he can ignore his own movement.

Before going too further - any problems ?
 
  • #124
Doc Al said:
Nope, not even close. For things to be 'in the same frame of reference' they must be moving together at the same speed. Note that anything--cars, trains, tracks, light--can be observed by anyone in any frame of reference. Just because someone in the track frame (that is, someone at rest with respect to the tracks) sees a train going by does not magically make the train somehow jump into the same frame of reference as the observer. The track frame sees the train moving (thus in a different frame); similarly, the train frame sees the track moving.


Then the light in the train is not in the same frame of reference as the rod ?
 
  • #125
I do not wish to be drawn into the argument.

Matheinste.
 
  • #126
jmallett said:
We will now ask a question. As far as the train is concerned what is his speed ?
With respect to what? With respect to the train itself, the speed of the train is zero of course. The speed of anything is zero in its own frame.
 
  • #127
jmallett said:
Then the light in the train is not in the same frame of reference as the rod ?
The light is not part of any frame of reference. (There is no frame in which the light is at rest.)
 
  • #128
Doc Al said:
The light is not part of any frame of reference. (There is no frame in which the light is at rest.)

Then where is the frame of reference in which the train and the rod are at rest ?
 
  • #129
jmallett said:
Then where is the frame of reference in which the train and the rod are at rest ?
All of a sudden a rod appears. How is it moving with respect to the train?
 
  • #130
jmallett said:
Then where is the frame of reference in which the train and the rod are at rest ?

Stationary with respect to the train, that's where we started and that's the postulate of Einsteins equations, although Einstein didn't actually need a train, he just had the rod floating. The train got introduced somewhere a few pages back, I think, but never mind, let's assume the rod and train are stationary relative to each other. The observer on the train is trying to measure the length of the rod just as Einstein had him do, but measuring the length of the train will do just as well.
 
  • #131
jmallett said:
Stationary with respect to the train, that's where we started and that's the postulate of Einsteins equations, although Einstein didn't actually need a train, he just had the rod floating. The train got introduced somewhere a few pages back, I think, but never mind, let's assume the rod and train are stationary relative to each other. The observer on the train is trying to measure the length of the rod just as Einstein had him do, but measuring the length of the train will do just as well.
OK. Rod and train are in the same frame.
 
  • #132
Doc Al said:
OK. Rod and train are in the same frame.

Good. We are on the same page. Now back a step -

Then where is the frame of reference in which the train and the rod are at rest ?
 
  • #133
jmallett said:
Good. We are on the same page. Now back a step -

Then where is the frame of reference in which the train and the rod are at rest ?

the train/rod are that frame of reference.
 
  • #134
jmallett said:
Good. We are on the same page. Now back a step -

Then where is the frame of reference in which the train and the rod are at rest ?
Moving along with the train, like any good reference frame. :rolleyes: (Seriously, this is getting silly. Why not read that book I gave you the link for? It goes through all this in painstaking detail.)
 
  • #135
jmallett said:
Good. We are on the same page. Now back a step -

Then where is the frame of reference in which the train and the rod are at rest ?

Then the observers on the train are measuring it's length at rest ?
 
  • #136
Doc Al said:
Moving along with the train, like any good reference frame. :rolleyes: (Seriously, this is getting silly. Why not read that book I gave you the link for? It goes through all this in painstaking detail.)

Actually I did read a fair bit of that book, and here's some of the things I read:

- first question the book

- not being able to see your self in a mirror is IMPLAUSIBLE.
This hardly works for me because it is quite plausible, in fact sound works that way. It's possible for the statement to be right or wrong, but to dismiss it as implausible is not a particularly scientific approach, especially if we are going to draw some assumptions from that.
 
  • #137
Doc Al said:
Moving along with the train, like any good reference frame. :rolleyes: (Seriously, this is getting silly. Why not read that book I gave you the link for? It goes through all this in painstaking detail.)

Sorry you see it as silly.

Einstein places the rest frame of the rod as the frame of the stationary observer. It is an explicit assumption that it is not at rest when moving in that frame, so we are stuck with a conclusion that is in it's rest frame when it is both in the train and when it is not moving relative to the stationary observer. Forget all the stuff about light.

He measures the length "at rest" and the length rAB is inherently the length when the "rod is at rest"

The length is therefore, in his experiment, inherently measured by the stationary observer.
 
  • #138
jmallett said:
Sorry you see it as silly.

Einstein places the rest frame of the rod as the frame of the stationary observer. It is an explicit assumption that it is not at rest when moving in that frame, so we are stuck with a conclusion that is in it's rest frame when it is both in the train and when it is not moving relative to the stationary observer. Forget all the stuff about light.

He measures the length "at rest" and the length rAB is inherently the length when the "rod is at rest"

The length is therefore, in his experiment, inherently measured by the stationary observer.

so what.

Make it happen in your mind.

They were at rest and measured the rod and then they were not at rest.
 
  • #139
jmallett said:
Einstein places the rest frame of the rod as the frame of the stationary observer. It is an explicit assumption that it is not at rest when moving in that frame, so we are stuck with a conclusion that is in it's rest frame when it is both in the train and when it is not moving relative to the stationary observer. Forget all the stuff about light.

He measures the length "at rest" and the length rAB is inherently the length when the "rod is at rest"

The length is therefore, in his experiment, inherently measured by the stationary observer.
You're getting all twisted in knots over the term 'stationary' observer.

Rather than use terms such as 'stationary' frame and 'moving' frame, let's call them the track frame and the train frame. No ambiguity there. If the train frame measures the length of the rod, then that's the length of the rod when the rod is at rest (with respect to the observer) since measurements are made in the frame where the rod is at rest. And if the track frame measures the length of the rod, then they measure the length of the moving rod since measurements are made in the frame in which the rod is moving.

OK. Now what?
 
  • #140
Doc Al said:
You're getting all twisted in knots over the term 'stationary' observer.

Rather than use terms such as 'stationary' frame and 'moving' frame, let's call them the track frame and the train frame. No ambiguity there. If the train frame measures the length of the rod, then that's the length of the rod when the rod is at rest (with respect to the observer) since measurements are made in the frame where the rod is at rest. And if the track frame measures the length of the rod, then they measure the length of the moving rod since measurements are made in the frame in which the rod is moving.

OK. Now what?

Now what is the equation the track uses to measure the length of the rod ?
 
  • #141
jmallett said:
Now what is the equation the track uses to measure the length of the rod ?
How would you like them to measure it? I suspect you'd like them to use a beam of light, else there wouldn't be much to talk about. So if they send a beam of light from one end (A) to the other (B), and they measure the travel time to be Δt, then the length of the rod in the track frame will be: L = cΔt.
 
  • #142
Doc Al said:
How would you like them to measure it? I suspect you'd like them to use a beam of light, else there wouldn't be much to talk about. So if they send a beam of light from one end (A) to the other (B), and they measure the travel time to be Δt, then the length of the rod in the track frame will be: L = cΔt.

Well, that's what Einstein said, so in terms of understanding Einstein's Special Theory of relativity this is going well.

But, the "stationary observer", the track observer we are talking about, sees something quite different from that, and relatively more complex, because he knows the speed v, and because he knows the speed of light c, therefore he needs to make 2 calculations, one to the front end, and one to the back end, and he has to take account of the fact that the rod is moving and that the speed of light is not instantaneous.
 
  • #143
jmallett said:
But, the "stationary observer", the track observer we are talking about, sees something quite different from that, and relatively more complex, because he knows the speed v, and because he knows the speed of light c, therefore he needs to make 2 calculations, one to the front end, and one to the back end, and he has to take account of the fact that the rod is moving and that the speed of light is not instantaneous.
And so?
 
  • #144
Doc Al said:
And so?

:eek: you don't see it ?
 
  • #145
jmallett said:
:eek: you don't see it ?
See what? :wink:

Depending on what the track frame measures, they can calculate the length of the rod with just one equation. If the train and the light beam are both moving to the right, the length of the moving rod according to the track frame will be Ltrack = (c - v)Δttrack.

See if you can derive that equation.
 
  • #146
Doc Al said:
See what? :wink:

Depending on what the track frame measures, they can calculate the length of the rod with just one equation. If the train and the light beam are both moving to the right, the length of the moving rod according to the track frame will be Ltrack = (c - v)Δttrack.

See if you can derive that equation.

That particular equation only works if the track can use the train's light, so it can't be valid in this useage. That's a simplification of what occurs. The track must use the track's light.
 
  • #147
jmallett said:
That particular equation only works if the track can use the train's light, so it can't be valid in this useage. That's a simplification of what occurs. The track must use the track's light.
Nope. There is only one beam of light that travels from one end of the rod (A) to the other (B). It doesn't 'belong' to the train or the track. Both frames can analyze the motion of that same beam of light.
 
  • #148
jmallett said:
That particular equation only works if the track can use the train's light, so it can't be valid in this useage. That's a simplification of what occurs. The track must use the track's light.

You can only say this if you have another theory of light.

DA presented a valid equation.

What is your theory?
 
  • #149
cfrogue said:
You can only say this if you have another theory of light.

DA presented a valid equation.

What is your theory?

That particular theory was the basis of the ether. Mitchelson Morely and a number of others disproved it. Einstein went forward with the assumption that light is relative to the frame of reference in which it exists, i.e. the train's light is not the track's light.

I don't yet have a full theory, just a whole lot of questions which I can't answer, but I am putting together some math on the subject.

Certainly in the situation Doc Al proposes we need to add in the distance deltaL which the rod passes through during the duration of the measurement. That's simple math and I will write it up for you. I hope soon to be able to share those equations, and I hope that someone will point out any errors I might have made. In addition to that I am working on an a demonstration with two bodies moving without reference to anything, i.e. truly relative, that use their own electromagnetic radiation (maybe light) to make a common measurement.
 
  • #150
jmallett said:
Einstein went forward with the assumption that light is relative to the frame of reference in which it exists, i.e. the train's light is not the track's light.
Nope--Einstein makes no such assumption. As I've already stated, light does not 'belong' to any reference frame. What Einstein does assume is that all light moves at the same speed with respect to any reference frame.

Certainly in the situation Doc Al proposes we need to add in the distance deltaL which the rod passes through during the duration of the measurement.
The equation I provided already takes into consideration the fact that the rod moves a distance vΔttrack during the time it takes for the light to travel from one end to the other.

(Note: Discussion of personal theories is not permitted.)
 

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