Is it possible for relative velocity to exceed the speed of light?

[vilas]

What do you mean by "the same"? The laws of physics work the same relative to both frames, for example if a ship starts out at rest in either frame and does one rocket burst it'll be moving at 0.00099c relative to that frame, and if a ship starts out at rest in either and does two rocket bursts it'll be moving at (2*0.00099c)/(1 + 0.00099^2) relative to that frame. If you think there is some way in which my answers suggest the laws of physics don't work the same relative to each frame, or if you mean something different by "the same" than just the laws of physics working the same, then please explain, because I don't know why you think them being "the same" would conflict in any way with anything I've said.

You are taking velocity addition equation as a base for your argument. I am not. If SR applies to a single object moving in space, then what you say is correct. I am referring to a more fundamental law that all inertial frames are equivalent. This was the view of Newton, Einstein and nobody has any problem with this. Based on this basic principle, for the spaceship, there is no difference between frames #1 and #2. If velocity meter, however calibrated, shows velocity 0.00099c in the frame #2, then it must show velocity 2*0.00099c in frame #3. This is because the spaceship does not recognize frame #1. We recognize it in order to apply velocity addition theorem. For equal amount of fuel burnt, it’s meter must show equal amount of increase in velocity.

If meter is calibrated to record velocity according to SR rule, then, we simply are not treating all frames as equivalent. Application of mechanical law for acceleration between frames #2 and #3 is different than that between frames #3 and #4.

 

[JesseM]

You are taking velocity addition equation as a base for your argument. I am not.

Yes you are, you seem to be taking the Newtonian velocity addition equation w=v+u as a basis for your argument.

If SR applies to a single object moving in space, then what you say is correct. I am referring to a more fundamental law that all inertial frames are equivalent.

You haven't answered my question. What do you mean by "equivalent" and "the same"? Normally this just means the laws of physics work the same in each frame, so if you do an experiment with some apparatus which starts at rest in frame A and get results X described in the coordinates of frame A, then you do an identical experiment with an identical apparatus that starts at rest in frame B, you should also get results X in the coordinates of frame B. As I said above, this is certainly true here--in either frame, if you start with a rocket at rest and measure its velocity after it's fired one burst, you'll find it has a velocity of 0.00099c in your frame, and if you measure its velocity after it's fired two bursts, you'll find it has a velocity of (2*0.00099c)/(1 + 0.00099^2) in your frame. This is true if you are at rest in frame #1, and also true if you are at rest in frame #2, so the two frames are equivalent in the sense I describe above. If you think there is some experiment that according to relativity yields different results when performed in the two frames, please describe it. If you admit that all possible experiments yield identical results in the two frames, but think they are still not "equivalent", then you are using the word "equivalent" in a way that's different from all physicists so you need to give a precise definition (and in any case, if you are using a different definition then physicists probably wouldn't claim different frames are supposed to be "equivalent" in your sense).

 

[DaveC426913]

vilas, from an external frame of reference to the accelerating craft, there would be a difference between frame 1 and frame 2.
In frame 1, starting from v=0, 1 second of burn would raise its velocity from 0 to x.
In frame 2, starting from v=x, 1 second of burn would raise its velocity from x to 1.999...x, not 2x.

 

[JesseM]

vilas, from an external frame of reference to the accelerating craft, there would be a difference between frame 1 and frame 2.
In frame 1, starting from v=0, 1 second of burn would raise its velocity from 0 to x.
In frame 2, starting from v=x, 1 second of burn would raise its velocity from x to 1.999...x, not 2x.

Just so vilas doesn't think this is conflicting with my previous comment, note that this is not a "difference" between frames in the sense of the same experiment in both frames giving different results, since in this case the starting conditions were different in the two frames (in one the ship started at v=0, in the other at v=x)

 

[vilas]

Just so vilas doesn't think this is conflicting with my previous comment, note that this is not a "difference" between frames in the sense of the same experiment in both frames giving different results, since in this case the starting conditions were different in the two frames (in one the ship started at v=0, in the other at v=x)

No! For the ship v=x=0, because any uniform velocity is zero velocity for the ship. Therefore all frames are equivalent for the ship.

Treating this as an experiment, let the ship start from frame #1 and add velocity of 0.001c (my previous assumption of 0.00099c has no meaning as in this case I have just applied correction factor and not SR factor). Suppose we assume it to be true. So velocity meter will show 0.001c. Ship is now in the frame #2. I refuse to buy an argument that the ship has velocity of 0.001c. Of course it has this velocity with respect to frame #1. But it is at rest in the frame #2. Now frame #1 and #2 are equivalent and if they are treated as such, then as an experimental fact, in the frame #3 velocity must be 0.002c. If velocity meter doesn’t show velocity as 0.002c, then experiment conducted in frame #2 is not same as that conducted in frame #1.

We are now talking on subtle concepts. According to you all inertial frames are equivalent and so the ship will get 0.001c, in moving from frame 1 to frame 2. Similarly the ship will gain 0.001c, in moving from frame 2 to frame 3. However, according to you, velocity in the frame 3 will be different than that as measured from frame 1. With this you are presupposing correctness of SR velocity addition equation.

But as I said earlier, for the ship, frame 1 doesn’t exist. Every time it gets a boost, it is in its own rest frame, as if nothing as happened before. No way to calculate its own uniform velocity. Therefore an astronaut in the ship has to rely on the velocity meter. Since every frame is as good as the starting frame, velocity meter must add 0.001c with every boost.

With respect to frame 1, if SR is considered, then the velocity meter reading is false. But there is no reason why it should be false.

 

[Doc Al]

No! For the ship v=x=0, because any uniform velocity is zero velocity for the ship.

Only with respect to itself! Its velocity with respect to something else can certainly be non-zero.

Therefore all frames are equivalent for the ship.

Physically equivalent, but not 'the same'. Velocity matters, especially if you want to get somewhere.

Treating this as an experiment, let the ship start from frame #1 and add velocity of 0.001c (my previous assumption of 0.00099c has no meaning as in this case I have just applied correction factor and not SR factor). Suppose we assume it to be true. So velocity meter will show 0.001c.

Again, your 'velocity meter' only shows what you've programmed it to show. What you actually measure directly is acceleration and time; you've got to calculate your speed with respect to some frame of interest. If you want the correct relative velocity, you'd better use SR formulas.

Ship is now in the frame #2. I refuse to buy an argument that the ship has velocity of 0.001c. Of course it has this velocity with respect to frame #1. But it is at rest in the frame #2.

Velocity is frame-dependent, of course.

Now frame #1 and #2 are equivalent and if they are treated as such, then as an experimental fact, in the frame #3 velocity must be 0.002c.

The velocity with respect to what frame? Since frame #1 and #2 are physically equivalent, the same burst when in frame #2 will give you a velocity of 0.001c with respect to frame #2. You mistakenly think that that means you have a velocity of 0.001c + 0.001c = 0.002c with respect to frame #1. Not so!

If velocity meter doesn’t show velocity as 0.002c, then experiment conducted in frame #2 is not same as that conducted in frame #1.

Assuming you've correctly programmed it--using SR formulas--to show the velocity with respect to frame #1, it will not show 0.002c.

We are now talking on subtle concepts. According to you all inertial frames are equivalent and so the ship will get 0.001c, in moving from frame 1 to frame 2. Similarly the ship will gain 0.001c, in moving from frame 2 to frame 3. However, according to you, velocity in the frame 3 will be different than that as measured from frame 1. With this you are presupposing correctness of SR velocity addition equation.

Of course we are, since we want results that agree with experiment. Note that you had to use SR formulas to get each burst of 0.001c. If you didn't, then those speeds are not reflecting of reality either. It's SR all the way, baby!

But as I said earlier, for the ship, frame 1 doesn’t exist. Every time it gets a boost, it is in its own rest frame, as if nothing as happened before. No way to calculate its own uniform velocity. Therefore an astronaut in the ship has to rely on the velocity meter. Since every frame is as good as the starting frame, velocity meter must add 0.001c with every boost.

Only if you program it wrong!

With respect to frame 1, if SR is considered, then the velocity meter reading is false. But there is no reason why it should be false.

You still seem to think that your accelerometer somehow measures speed directly without the need to program it. But it looks like you've programmed your 'velocity meter' with incorrect Newtonian formulas. Good luck in planning your travels, as your calculated relative velocity with respect to various frames is incorrect.

Of course if you measured your relative velocity directly, then you'd always get the correct SR results. (Look out the window!)

 

[Dale]

Again vilas, the experiment you are proposing cannot distinguish between SR and Newtonian physics. Both theories predict that an accelerometer can read a uniform proper acceleration for an indefinite amount of time.

 

[JesseM]

No! For the ship v=x=0, because any uniform velocity is zero velocity for the ship. Therefore all frames are equivalent for the ship.

You still haven't explained what you mean by "equivalent". Do you mean the same as what I mean by it (which as I'm telling you is what physicists mean by it) or do you mean something different? Simple question.

Treating this as an experiment, let the ship start from frame #1 and add velocity of 0.001c (my previous assumption of 0.00099c has no meaning as in this case I have just applied correction factor and not SR factor). Suppose we assume it to be true. So velocity meter will show 0.001c. Ship is now in the frame #2. I refuse to buy an argument that the ship has velocity of 0.001c. Of course it has this velocity with respect to frame #1. But it is at rest in the frame #2.

Of course it is, no one made the argument that if the ship is moving at 0.001c in one frame it must be moving at 0.001c in other frames. That has nothing to do with my question about running the "same experiment" in different frames, which involves having duplicate versions of the same equipment which each start out at rest in the frame where the experiment is performed (for example if frame #1 has a ship that starts out at rest their and then accelerates to 0.001c relative to frame #1, then frame #2 should have a ship that starts out at rest their and then accelerates to 0.001 relative to frame #2).

Now frame #1 and #2 are equivalent and if they are treated as such, then as an experimental fact, in the frame #3 velocity must be 0.002c.

I'm sorry but you're just making assertions with no rational argument for them (why do you think their being "equivalent" means the #3 velocity should be 0.002c?), you haven't answered my question about what you mean by "equivalent", and if you are trying to use it in the same way I'm using it, you haven't given any specific experiment that's to be performed in both frames. If you refuse to answer my simple question about whether you're using my definition of "equivalent" or a different one, then there's no reason to continue this discussion, you're just making dogmatic assertions based on ill-defined personal terminology which makes sense only to you.

If velocity meter doesn’t show velocity as 0.002c, then experiment conducted in frame #2 is not same as that conducted in frame #1.

The "experiment" just refers to the parts of it under your direct control, the readings on equipment like a velocity meter are the results of the experiment. In a universe where the laws of physics didn't work the same in both frames, you could perform the "same experiment" in different frames but end up with different final results as measured in that frame. The "experiment" refers to the starting arrangement of all the equipment involved (like the fact that you start with a ship at rest in your frame, constructed in some standard manner) and anything you do to that equipment during the experiment (like sending a signal to the ship to cause it to fire one or more rocket bursts). If you do any experiment like this with identical rockets that start out at rest in two different frames, and observers in both frames send the same commands to their respective rockets, then they will both get the same answer for the rocket's final velocity in their frame, and if the "velocity meters" on board the rockets work in the way demanded by relativity, they'll also get the same results in terms of the reading on the velocity meters at the end. Do you disagree? If so, please specify exactly what the experiment is that you want to duplicated (with two different rockets) in each frame, both in terms of the starting conditions and the commands sent to the rockets, and what you think the results will be for each experiment. For example, according to relativity here is a way of performing the "same experiment" which yields the "same results":

1. Experiment: In frame #1, start with a rocket at rest, whose engines are designed to fire "bursts" giving some fixed amount of proper acceleration for a fixed proper time. First we command the rocket to fire one burst and look at its velocity in frame #1, then we command it to fire another burst and look at its velocity in frame #1.

Results: After the first burst, velocity in frame #1 is 0.001c, after the second burst, velocity in frame #1 is (0.002c)/(1 + 0.001^2)

2. Experiment: In frame #2, start with an identical rocket at rest, whose engines are again designed to fire "bursts" giving some fixed amount of proper acceleration for a fixed proper time. First we command the rocket to fire one burst and look at its velocity in frame #2, then we command it to fire another burst and look at its velocity in frame #2.

Results: After the first burst, velocity in frame #2 is 0.001c, after the second burst, velocity in frame #2 is (0.002c)/(1 + 0.001^2)

...so here we do the same experiment in both frames and get the same results. If you think there is some example where we could do the "same experiment" in both frames but get different results according to the predictions of relativity (i.e. you are saying that relativity's own predictions contradict the "equivalence" of different frames, not merely saying you think relativity's predictions are wrong in the real world), then please give the details, using the format above of explaining both the "experiment" to be performed in each frame and the "results" you think relativity would predict in each version of the experiment. If you admit that you can't do this, but still think different frames are not "equivalent", then you are admitting that you are using a different definition of "equivalent" from the one I'm using, so I'd say you're also using a definition that's different from what all physicists mean by "equivalent".

 

[vilas]

.

We must keep out what all physicists think. By this you are influencing the argument.
I think I made it quite clear what I think about what ‘equivalence’ of frame means.
In frame 2, my velocity meter shows 0.001c. I record this velocity and reset the accelerometer. In frame 3, it must and it will show velocity of 0.001c. Based on records and not on any relativistic calculations I add this velocity to old one and I get new velocity as 0.002c.
Our disagreement is here. According to you, VAT must be used and so my record of 0.002c is wrong. According to me there is no reason why I should use VAT and reduce velocity.
In any case velocity addition theorem is not falsifiable and so cannot be relied on.

 

[Doc Al]

We must keep out what all physicists think. By this you are influencing the argument.

By introducing actual physics?

I think I made it quite clear what I think about what ‘equivalence’ of frame means.
In frame 2, my velocity meter shows 0.001c.

With respect to what frame? How did you calculate it? (Answer: With respect to frame 1, using the standard SR formulas.)

I record this velocity and reset the accelerometer. In frame 3, it must and it will show velocity of 0.001c.

Again: With respect to what frame? How did you calculate it? (Answer: With respect to frame 2, using the standard SR formulas.)

Based on records and not on any relativistic calculations I add this velocity to old one and I get new velocity as 0.002c.

Suit yourself, but that's not the velocity with respect to frame 1.

Our disagreement is here. According to you, VAT must be used and so my record of 0.002c is wrong. According to me there is no reason why I should use VAT and reduce velocity.

To get the correct answer you must use the correct equations: SR works, while Galilean addition of velocity does not.

In any case velocity addition theorem is not falsifiable and so cannot be relied on.

Why in the world do you think that? Actually measure your speed with respect to frame 1, instead of relying on incorrect calculations.

 

[vilas]

By introducing actual physics?
I am not alone to criticize SR. I am nobody; but there are thousands of eminent people who have criticized SR. Question is not whether SR is right or wrong. Question is if SR can tolerate criticism.

With respect to what frame? How did you calculate it? (Answer: With respect to frame 1, using the standard SR formulas.)
Forget about frames. These are essential for VAT.

Again: With respect to what frame? How did you calculate it? (Answer: With respect to frame 2, using the standard SR formulas.)
Why should I use standard SR formula?

Suit yourself, but that's not the velocity with respect to frame 1.
That is your point of view.

To get the correct answer you must use the correct equations: SR works, while Galilean addition of velocity does not.
I doubt this assertion

Why in the world do you think that? Actually measure your speed with respect to frame 1, instead of relying on incorrect calculations.

Because I wish to think different way. I am not fundamentalist.

 

[DaveC426913]

No way to calculate its own uniform velocity. Therefore an astronaut in the ship has to rely on the velocity meter.

vilas, please answer me this: how exactly does the astronaut's velocimeter measure velocity of the ship? The ship is stationary in its own reference frame, so what does it use to measure velocity?

 

[jtbell]

Vilas, please read the sticky post "IMPORTANT! Read before posting" at the top of this forum, and don't waste your and our time further by trying to convince us that there's something wrong with relativity.