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

[kewljerk]

Let there be a space shuttle moving at very high speed relative to Earth so, actually time would be slower in the shuttle than on the earth.But, if we see everything relative to shuttle then, Earth is moving at very high speed that, means Earth is slower.
These two things are contradicting. So, where I am going wrong??

Also, is this true that relative velocity can go beyond speed of light?Let 2 planets are crossing each other in opposite direction with speed=c/2. So, relative to other planet is the other planet is moving with speed of light.How?

 

[Doc Al]

Let there be a space shuttle moving at very high speed relative to Earth so, actually time would be slower in the shuttle than on the earth.But, if we see everything relative to shuttle then, Earth is moving at very high speed that, means Earth is slower.
These two things are contradicting. So, where I am going wrong??

Each observer sees the other's clocks as running slow. Where you are going wrong is thinking that there's a contradiction. (To fully understand how there's no contradiction, you'll need to learn a bit about relativity, especially the relativity of simultaneity.)

Also, is this true that relative velocity can go beyond speed of light?Let 2 planets are crossing each other in opposite direction with speed=c/2. So, relative to other planet is the other planet is moving with speed of light.How?

High speeds do not add the way you think. You need to use the relativistic addition of velocity formula, explained here: http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/velocity.html" )

If two objects traveling at c/2 (with respect to some third frame) approach each other, they each measure the other as approaching at a speed of 0.8c, not c.

 

[Naty1]

Post above is right on.

Everybody's time passes differently meaning that depending on your frame of reference, your local speed, you'll see other's time passing differently from other observers. Time does NOT pass in an absolute fixed interval; only the speed of light is fixed and "absolute" for all observers.

Regarding speed, if two objects approach each other at, say 100 mph, their APPROXIMATE relative speed is 200 MPH. But relativity tells us that at very high speeds, a more accurate formulation is required to take into account changes in observed distance and time...they VARY and so observed speed also varies.

There is a decent explanation here:
http://en.wikipedia.org/wiki/Special_relativity

but you'll have to do some thinking...it is NOT obvious, but it has been proven correct!

 

[HallsofIvy]

Note the "twin paradox" which argues that each twin sees the other as aging slower than himself. So which is younger when they get back together? Part of the resolution is that, if each is moving at a constant velocity with respect to the other, they can't get back together. If they do, then at least one must accelerate which destroys the symmetry.

 

[kewljerk]

thanx a lot friends, now I got it.

 

[vilas]

thanx a lot friends, now I got it.

No! You didn't get it because I didn't get it. Let A and B are two spaceships placed at a great distance. When A burns Q quantity of fuel, it reaches a speed of c/2 with respect to its original reference. So also B. If they move towards each other, there is no third reference frame. Velocity addition theorem is applicable when velcoity measurements are lader type.

 

[Doc Al]

No! You didn't get it because I didn't get it. Let A and B are two spaceships placed at a great distance. When A burns Q quantity of fuel, it reaches a speed of c/2 with respect to its original reference. So also B. If they move towards each other, there is no third reference frame. Velocity addition theorem is applicable when velcoity measurements are lader type.

It seemed clear enough the the two planets were moving at speed c/2 with respect to some common third frame. Otherwise it wouldn't make much sense.

 

[vilas]

It seemed clear enough the the two planets were moving at speed c/2 with respect to some common third frame. Otherwise it wouldn't make much sense.

Yes but that frame is imaginary. Assume that O and O’ are standstill w.r.t. each other. Only frame they have is their own. O’ now accelerates to +v and O to –v. Accelerometers on both indicate these velocities. If these meters are to be believed then they are moving apart with the velocity of 2v.

 

[JesseM]

Yes but that frame is imaginary. Assume that O and O’ are standstill w.r.t. each other. Only frame they have is their own. O’ now accelerates to +v and O to –v. Accelerometers on both indicate these velocities. If these meters are to be believed then they are moving apart with the velocity of 2v.

What "meters" do you mean? An accelerometer measures acceleration not velocity, though of course you can use A(T) to calculate v(t) in any frame you like (see the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html equations). I assume when you say they accelerate to +v and -v, you're talking about their new velocities in the frame where they were previously at rest? If so, then if once they finish accelerating they both use rulers and clocks at rest relative to themselves to measure distance/time for the other one, they won't find the other has a velocity of 2v, instead they'll find a value of (2v)/(1 + v^2/c^2) as predicted by the velocity addition formula.

 

[vilas]

What "meters" do you mean? An accelerometer measures acceleration not velocity, though of course you can use A(T) to calculate v(t) in any frame you like (see the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html equations). I assume when you say they accelerate to +v and -v, you're talking about their new velocities in the frame where they were previously at rest? If so, then if once they finish accelerating they both use rulers and clocks at rest relative to themselves to measure distance/time for the other one, they won't find the other has a velocity of 2v, instead they'll find a value of (2v)/(1 + v^2/c^2) as predicted by the velocity addition formula.

How can you apply relativity to prove relativity? No meter can measure uniform velocity so if there has to be a velocity meter then it must show reading based on acceleration and time.
When such a meter shows v for O’ and O then where is the problem in assuming that velocity between O and O’ is 2v.

 

[JesseM]

How can you apply relativity to prove relativity?

I'm not trying to "prove" relativity, I'm just telling you what would be true under relativity. Are you trying to dispute relativity? If so you can't do so with thought-experiments since relativity is an entirely consistent theory, you would have to show some experimental evidence that relativity's predictions about what happens are wrong.

No meter can measure uniform velocity so if there has to be a velocity meter then it must show reading based on acceleration and time.

Yes, but in relativity it would have to do so with calculations that involve "applying relativity", like the relativistic rocket equations which do relate proper acceleration and proper time to velocity in some inertial frame.

When such a meter shows v for O’ and O then where is the problem in assuming that velocity between O and O’ is 2v.

You didn't answer my question about what frame these velocities are supposed to be calculated in. If you think there'd be some procedure for calculating a velocity which didn't require picking a frame, then that wouldn't make sense in relativity. Also I don't see how you each ship could just use its own acceleration and elapsed time to calculate a relative velocity, each ship would have to know something about the other ship's motion to do that.

 

[vilas]

I'm not trying to "prove" relativity, I'm just telling you what would be true under relativity. Are you trying to dispute relativity? If so you can't do so with thought-experiments since relativity is an entirely consistent theory, you would have to show some experimental evidence that relativity's predictions about what happens are wrong.

To this I wish to quote Ian McCausland,
“The heading of the New Scientist article uses the term "scientific malcontents" to refer to those who attack relativity. If being a relativist entails acceptance of all the mutually contradictory arguments (some of which I have recently documented) that have been published in defending special relativity against the criticism of Flerbert Dingle, then I prefer to be a scientific malcontent and I accept that designation with pride. I think every scientist should be a malcontent; after all, what is the value of trying to contribute new knowledge unless one is dissatisfied with the present state of knowledge?”
Thought experiments try to prove inconsistencies in basic arguments. If theory is found to be untenable then scientists should try to find out correct theory behind the experimental results.
Of course O’ and O gain velocities v with respect to the frame in which they were at standstill. However their instruments show velocity v of each. Should they conclude that relative velocity between them is 2v or they should apply SR equation?

 

[JesseM]

Thought experiments try to prove inconsistencies in basic arguments.

Yes, but it's already been shown mathematically that the theory is self-consistent, you might as well try to use thought-experiments to try to find an inconsistency in Euclidean geometry.

Of course O’ and O gain velocities v with respect to the frame in which they were at standstill. However their instruments show velocity v of each. Should they conclude that relative velocity between them is 2v or they should apply SR equation?

The latter, because the "velocity v of each" was itself computed using SR assumptions about inertial frames, why would you suddenly switch to a different notion of velocity not based on SR inertial frames? If it helps though, in relativity there is a separate concept known as "closing speed", which refers to the rate that the coordinate distance between two objects is changing (the rate one is 'closing' on the other, if the distance is shrinking) as seen from the perspective of a third frame in which neither is at rest. In the frame where they were originally at rest and now are moving in opposite directions at v, the closing speed is indeed 2v. But this is different from the speed of either one as measured in the rest frame of the other one.

 

[Doc Al]

Thought experiments try to prove inconsistencies in basic arguments. If theory is found to be untenable then scientists should try to find out correct theory behind the experimental results.

True enough. But the results predicted in a thought experiment depend upon the theory you apply.

Of course O’ and O gain velocities v with respect to the frame in which they were at standstill. However their instruments show velocity v of each.

Let's assume that they somehow end up moving with speed v with respect to that third frame. (Which is where we started, of course.)

Should they conclude that relative velocity between them is 2v or they should apply SR equation?

If they want to predict their relative velocity correctly, they must use the equations of SR. If they use Galilean relativity to deduce a relative velocity of 2v, they would be wrong. Experimental measurement of their relative velocity would settle it in favor of SR.

 

[vilas]

If they want to predict their relative velocity correctly, they must use the equations of SR. If they use Galilean relativity to deduce a relative velocity of 2v, they would be wrong. Experimental measurement of their relative velocity would settle it in favor of SR.

Consider O'. He burns dq quantity of fuel and gains velocity of v. At this velocity, for all mechanical laws, his velocity is as good as 0 and mass remains same, but his instrument shows that he is moving with velocity v. O' again burns dq quantity of fuel and again gains velocity v. Now his meter must show velcoity of 2v. Same with O. Therefore why should instrument show lesser velocity.

 

[JesseM]

Consider O'. He burns dq quantity of fuel and gains velocity of v. At this velocity, for all mechanical laws, his velocity is as good as 0 and mass remains same, but his instrument shows that he is moving with velocity v. O' again burns dq quantity of fuel and again gains velocity v. Now his meter must show velcoity of 2v.

Not true, not if the instruments are calculating those velocities based on relativity. Say that frame #1 is the one where O' was initially at rest before the first burn, and frame #2 is the one where O' was at rest after the first burn but before the second. Then after the first burn he will be moving at v relative to frame #1, and after the second burn he will be moving at v relative to frame #2, but that does not mean he is moving at 2v relative to frame #1, instead he is now moving at (2v)/(1 + v^2/c^2) relative to frame #1.

 

[vilas]

Not true, not if the instruments are calculating those velocities based on relativity. Say that frame #1 is the one where O' was initially at rest before the first burn, and frame #2 is the one where O' was at rest after the first burn but before the second. Then after the first burn he will be moving at v relative to frame #1, and after the second burn he will be moving at v relative to frame #2, but that does not mean he is moving at 2v relative to frame #1, instead he is now moving at (2v)/(1 + v^2/c^2) relative to frame #1.

Instrument will never calculate velocity addition according to SR. For example assume that the current velocity is 3v in frame #3. However the spaceship will behave as if it is in the frame #1 and so will add next v as usual. Therefore meter will show 4v.

 

[JesseM]

Instrument will never calculate velocity addition according to SR.

How does the "instrument" calculate velocity, if not using SR formulas? I said that I was assuming it used SR formulas to calculate velocity in several previous posts, and you didn't contradict me. If I was misunderstanding your thought-experiment, then you need to give an alternate rule for calculating velocity. Suppose the ship's proper acceleration (as measured by an accelerometer) as a function of proper time (as measured by its own clock) is given by some function A(T). If the ship is not using SR equations, then what equations does it use to go from this measured acceleration and time to the velocity, which can't itself be directly measured by any instrument? Please be specific, otherwise your claims about what velocity would be measured/calculated are totally meaningless.

 

[DrGreg]

Instrument will never calculate velocity addition according to SR.

You've never really spelled out exactly how your "instrument" works. If it's an accelerometer that calculates velocity from measured acceleration then the answer is simple:

If the designer programmed the correct (relativistic) formula, the instrument will get the correct answer.

If the designer programmed the wrong (Newtonian) formula, by simply integrating proper acceleration over proper time, the instrument will get the wrong answer. In this case it will calculate a quantity called "rapidity" instead of velocity. The rapidity of light is infinite, so there's no problem in not exceeding it.

 

[vilas]

How does the "instrument" calculate velocity, if not using SR formulas? I said that I was assuming it used SR formulas to calculate velocity in several previous posts, and you didn't contradict me. If I was misunderstanding your thought-experiment, then you need to give an alternate rule for calculating velocity. Suppose the ship's proper acceleration (as measured by an accelerometer) as a function of proper time (as measured by its own clock) is given by some function A(T). If the ship is not using SR equations, then what equations does it use to go from this measured acceleration and time to the velocity, which can't itself be directly measured by any instrument? Please be specific, otherwise your claims about what velocity would be measured/calculated are totally meaningless.

I am sorry if my ignorance has caused misunderstanding. I am talking about un-calibrated instrument that is an instrument that doesn’t take into account SR correction. In any debate that questions validity of the fundamental theory, no presumption about the correctness of the theory can be made.

Still I wonder how the calibration can solve the problem. O’ is a spaceship in which an accelerometer will pass value of force to the digital controller. This controller will take into account SR correction required and the calculated velocity is stored in digital form and also displayed. If Newtonian velocity is 0.001c, let us assume that after correction it will be 0.00099c. Thus when O’ is accelerated from frame #1 to frame #2, record will show 0.00099c.

Since for the spaceship, frame #2 is equivalent to frame #1, there is no need to feed back record of velocity to the accelerometer and therefore after burning equal amount of fuel, next velocity record will be 2*0.00099c. Thus we do not arrive at the gradual decrease of velocity and this does not limit the maximum speed of the spaceship.

Gradual reduction in velocity calibration is permissible if we consider that when spaceship is in frame #2, its velocity is not 0 but it is 0.00099c. As far as my meager knowledge goes, I think that all physical laws-at least all mechanical laws-are same in all inertial systems and so spaceship in frame #2 is the same as a spaceship in frame #1. If this statement is accepted then we must accept that next installment of velocity will be same as before, that is 0.0009c.

 

[Doc Al]

I am sorry if my ignorance has caused misunderstanding. I am talking about un-calibrated instrument that is an instrument that doesn’t take into account SR correction.

The point is that if you want your device to display a speed with respect to some frame, you must program it to do so based on the measured acceleration and elapsed time. If you use pre-relativistic Newtonian formulas, you'll get pre-relativistic Newtonian results for the speed.

Since for the spaceship, frame #2 is equivalent to frame #1, there is no need to feed back record of velocity to the accelerometer and therefore after burning equal amount of fuel, next velocity record will be 2*0.00099c. Thus we do not arrive at the gradual decrease of velocity and this does not limit the maximum speed of the spaceship.

Again, you merely assume that the next velocity record will be twice the first. If you really are using the correct relativistic formulas you will not get that result. (You'd get twice the first speed only if you use pre-relativistic Newtonian formulas.)

 

[JesseM]

I am sorry if my ignorance has caused misunderstanding. I am talking about un-calibrated instrument that is an instrument that doesn’t take into account SR correction.

What does "un-calibrated" mean? There is no way to calculate velocity from acceleration without making some assumption about frames of reference and how coordinate velocity in some frame relates to proper acceleration as measured by an accelerometer. Do you just mean that you want the calculations to be based on Newtonian formulas rather than relativistic ones? But that's not "un-calibrated", assuming Newton is just as much of a calibration as assuming relativity.

 

[Dale]

In any debate that questions validity of the fundamental theory, no presumption about the correctness of the theory can be made.

You also cannot presume that a different theory is correct either, which is what you are doing. You are presuming that Newtonian physics is correct. If you presume that Newtonian physics is correct as you are doing then obviously you will conclude that SR is incorrect, not because SR is incorrect, but because you presumed it was incorrect.

All you can do is to compare each theory to the experimental data and devise experiments where their predictions differ. SR agrees with the experimental data to within experimental accuracy, and Newtonian physics does not:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

Your proposed experiment is not an experiment which can distinguish between SR and Newtonian physics because both theories agree that you can have an accelerometer which undergoes uniform proper acceleration for an indefinite amount of proper time.

 

[vilas]

What does "un-calibrated" mean? There is no way to calculate velocity from acceleration without making some assumption about frames of reference and how coordinate velocity in some frame relates to proper acceleration as measured by an accelerometer. Do you just mean that you want the calculations to be based on Newtonian formulas rather than relativistic ones? But that's not "un-calibrated", assuming Newton is just as much of a calibration as assuming relativity.

I am not getting a proper reply. Just let me know in what way spaceship in frame #2 is different than what it was in frame #1.

 

[Doc Al]

I am not getting a proper reply. Just let me know in what way spaceship in frame #2 is different than what it was in frame #1.

There's no difference at all. (Let's ignore mass changes due to fuel expenditure, etc., as they are not essential to the main point.) But your conclusion that the speed of the spaceship (with respect to the original frame) after the second burst will be twice what it was after the first burst is wrong. Velocities do not add that way. Read what I wrote about velocity addition in post #2.

 

[vilas]

There's no difference at all. #2.

Thanks. That is all I wanted to know. And that is all what is important.

 

[JesseM]

Still I wonder how the calibration can solve the problem. O’ is a spaceship in which an accelerometer will pass value of force to the digital controller. This controller will take into account SR correction required and the calculated velocity is stored in digital form and also displayed. If Newtonian velocity is 0.001c, let us assume that after correction it will be 0.00099c. Thus when O’ is accelerated from frame #1 to frame #2, record will show 0.00099c.

Since for the spaceship, frame #2 is equivalent to frame #1, there is no need to feed back record of velocity to the accelerometer and therefore after burning equal amount of fuel, next velocity record will be 2*0.00099c.

You say the controller uses SR to get a "calculated velocity", but after the second acceleration is it calculating the velocity relative to frame #2, or calculating the velocity relative to frame #1? If it's calculating velocity relative to frame #2, then it will calculate 0.00099c (showing that frame #2 is equivalent to frame #1, since the same burst relative to each frame creates a velocity of 0.00099c relative to that frame), but if it's still calculating velocity relative to frame #1, then it will calculate (2*0.00099c)/(1 + 0.00099^2).

 

[vilas]

You say the controller uses SR to get a "calculated velocity", but after the second acceleration is it calculating the velocity relative to frame #2, or calculating the velocity relative to frame #1? If it's calculating velocity relative to frame #2, then it will calculate 0.00099c (showing that frame #2 is equivalent to frame #1, since the same burst relative to each frame creates a velocity of 0.00099c relative to that frame), but if it's still calculating velocity relative to frame #1, then it will calculate (2*0.00099c)/(1 + 0.00099^2).

But that is the point. Why should spaceship calculate velocity with reference to frame #1, if frame #2 is same as frame #1. Just to satisfy SR? In fact any object moving with ANY uniform velocity can be considered as standstill. So it makes no difference to the spaceship whether it accelerates from the uniform velocity of 0.5c or 0c or any other.

If there is any real change to the spaceship in frame #2, so far as mechanical laws are concerned, then alone velocity addition theorem is applicable. Otherwise there is no justification for applying the equation.

 

[JesseM]

But that is the point. Why should spaceship calculate velocity with reference to frame #1, if frame #2 is same as frame #1.

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.

 

[mdmaaz]

Time in the spacecraft would be slower. The time on Earth would not change. But in this case the, the Earth is not going very fast in comparison to the spacecraft , (if we see everything relative to the spacecraft ). If you're driving a car at 100 mph and another car passes by you at 80 mph from the opposite direction, it appears to traveling at 180 mph, but... it is not nearly going that fast. Same thing here, if we see everything relative to the spacecraft , the Earth appears to be moving fast (but it is not moving) But keep in mind that the time inside the spacecraft does not affect the outside world.

 

[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.