A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
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From: PD on 4 Mar 2010 12:44 On Mar 4, 11:09 am, Ste <ste_ro...(a)hotmail.com> wrote: > On 4 Mar, 16:48, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 10:19 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 4 Mar, 12:19, "Inertial" <relativ...(a)rest.com> wrote: > > > > > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > > > > > Not really, because if the total acceleration is small, then so is the > > > > > speed. > > > > > That is a nonsense argument. Acceleration can be small and speeds very > > > > large. > > > > When I went to school, you could not have a large change of speed with > > > only a small amount of total acceleration. > > > The problem is your use of the term "total acceleration". If by total > > acceleration, you mean integral(a dt), then yes, you are correct. > > However, there is already a word for integral(a dt) -- it's called > > "the change in velocity". The term "total acceleration" isn't > > actually defined. Acceleration is defined, velocity is defined, > > deltav is defined. But "total acceleration is not". > > Essentially, I'm defining "total acceleration" as something akin to > total force, so that even though the force may be small, if it > continues for a long time then the total force will be the same as if > a large force was applied for a short period of time. In this way, if > the application of force is what is causing either part or the whole > of the time dilation effect, then it is the final speed that counts, > not how quickly the object reached that speed. Indeed. This should tell you that it is not the details of the acceleration that matter. The overly simplistic statement would be, "Yes, you see that is why SR's effects are based on speed, not on acceleration." In fact, there is a speed time dilation effect on GPS satellites, which are going around in a circular path at constant speed, relative to earth clocks, and accounting for this is crucial to their proper operation. This is the same speed dilation effect, though different size, as seen in muons in a circulating ring. (Since, by the way, the GPS satellites are certainly not inside a magnetic ring but still experience time dilation properly calculated by SR, this is another good way to be sure that the magnetic ring is not what's responsible for the time dilation of the muons.) Regarding something I alluded to earlier, though, what really matters is how straight the path through spacetime is. We're used to thinking that the shortest path through space is the straight one (and that's right), but the straightest path through spacetime yields the LONGEST duration. Any change in motion (such as an acceleration) introduces a kink in this path (something that can be illustrated visually very easily) and so lowers the duration. Why this is, has to do with the structure of spacetime and we could discuss that. But this is perhaps the most intuitive way (once these concepts are explained) to understand why the traveling twin returns younger. > > > Also, you could just be dealing with a system where the velocity > > started out high and you never measured any acceleration. > > Indeed.
From: PD on 4 Mar 2010 12:46 On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > Ok. > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > away from an object. > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > propagation delays would approach zero? > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > moving away the clock would be: > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > and when you move toward the clock > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > so moving away from the clock: > > > > > > > > dt2/dt = 1-v/c > > > > > > > > and toward > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > down as > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > is obviously a contradiction there. > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > SR, not GR. > > > > > > What if they both "break the inertial frame"? > > > > > Then whichever frame they both accelerate into will be the one that > > > > has measured the "correct" time dilation. > > > > So in other words, the clocks will register the same time, but will > > > have slowed in some "absolute sense"? > > > Yes--assuming they both accelerated by the same amount (that is to > > say, assuming they both broke the inertial frame in a symmetric way). > > Otherwise, they will register different times. > > Agreed. > > So let's explore an extension of this scenario. Let's say you have two > clocks, and you accelerate both of them up to a common speed, and > after they have travelled a certain distance, you turn them around and > return them to the starting point. The only difference is that one > clock goes a certain distance, and the other clock goes twice that > distance, but they *both* have the same acceleration profile - the > only difference is that one clock spends more time travelling on > inertia. > > Obviously, one clock will return to the starting point earlier than > the other. But when both have returned, are their times still in > agreement with each other, or have they changed? Agreement. Both of them will agree, but will be showing a time earlier than a third clock that was left behind at the starting point.
From: mpalenik on 4 Mar 2010 12:49 On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > Ok. > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > away from an object. > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > moving away the clock would be: > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > and when you move toward the clock > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > so moving away from the clock: > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > and toward > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > down as > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > is obviously a contradiction there. > > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > SR, not GR. > > > > > > > What if they both "break the inertial frame"? > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > has measured the "correct" time dilation. > > > > > So in other words, the clocks will register the same time, but will > > > > have slowed in some "absolute sense"? > > > > Yes--assuming they both accelerated by the same amount (that is to > > > say, assuming they both broke the inertial frame in a symmetric way). > > > Otherwise, they will register different times. > > > Agreed. > > > So let's explore an extension of this scenario. Let's say you have two > > clocks, and you accelerate both of them up to a common speed, and > > after they have travelled a certain distance, you turn them around and > > return them to the starting point. The only difference is that one > > clock goes a certain distance, and the other clock goes twice that > > distance, but they *both* have the same acceleration profile - the > > only difference is that one clock spends more time travelling on > > inertia. > > > Obviously, one clock will return to the starting point earlier than > > the other. But when both have returned, are their times still in > > agreement with each other, or have they changed? > > Agreement. Both of them will agree, but will be showing a time earlier > than a third clock that was left behind at the starting point. Wait, maybe I'm confused by Ste's setup. Didn't he say that one travels twice as far as the other? But then he also says that you turn them both around and return them to the start after traveling a certain distance. Have they moved different distances in his scenario or not?
From: mpalenik on 4 Mar 2010 12:55 On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > Ok. > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > away from an object. > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > moving away the clock would be: > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > and when you move toward the clock > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > so moving away from the clock: > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > and toward > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > down as > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > is obviously a contradiction there. > > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > SR, not GR. > > > > > > > What if they both "break the inertial frame"? > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > has measured the "correct" time dilation. > > > > > So in other words, the clocks will register the same time, but will > > > > have slowed in some "absolute sense"? > > > > Yes--assuming they both accelerated by the same amount (that is to > > > say, assuming they both broke the inertial frame in a symmetric way). > > > Otherwise, they will register different times. > > > Agreed. > > > So let's explore an extension of this scenario. Let's say you have two > > clocks, and you accelerate both of them up to a common speed, and > > after they have travelled a certain distance, you turn them around and > > return them to the starting point. The only difference is that one > > clock goes a certain distance, and the other clock goes twice that > > distance, but they *both* have the same acceleration profile - the > > only difference is that one clock spends more time travelling on > > inertia. > > > Obviously, one clock will return to the starting point earlier than > > the other. But when both have returned, are their times still in > > agreement with each other, or have they changed? > > Agreement. Both of them will agree, but will be showing a time earlier > than a third clock that was left behind at the starting point. If they have the same acceleration profile but one travels twice as far as the other because it is allowed to move at a constant speed longer (as I understand his question--although he seems to contradict this at the beginning), they will NOT be in agreement. If he accelerates them up to a common speed and turns them both around at the same time and stops them at the same time, so they both travel the SAME distance, which is what he says at the beginning (but then contradicts this later on), then the WILL be in agreement.
From: Ste on 4 Mar 2010 13:00
On 4 Mar, 17:23, mpalenik <markpale...(a)gmail.com> wrote: > On Mar 4, 12:17 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR. > > > > > > > > > > > > You must just go through the entire thread and not pay any attention > > > > > > > > > > > to what anybody says. Ever. > > > > > > > > > > > > 1) What you've stated above is not an effect of SR. It is an effect > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of > > > > > > > > > > > the moons of jupiter hundreds of years ago. > > > > > > > > > > > Ok. > > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run > > > > > > > > > > > faster. But SR says nothing about whether you are moving toward or > > > > > > > > > > > away from an object. > > > > > > > > > > > <suspicious eyebrow raised> Ok. > > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down > > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown? > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity, > > > > > > > > > > would SR still predict an "actual" slowdown, even though the > > > > > > > > > > propagation delays would approach zero? > > > > > > > > > > With what you have described, I checked just to be sure, even though I > > > > > > > > > was already pretty sure what the answer would be, the time you read > > > > > > > > > moving away the clock would be: > > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x > > > > > > > > > > and when you move toward the clock > > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x > > > > > > > > > > so moving away from the clock: > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > and toward > > > > > > > > > dt2/dt = 1-v/c > > > > > > > > > > Special relativity predicts that the moving clock will always slow > > > > > > > > > down as > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2) > > > > > > > > > > What you *measure* is a combination of the actual slow down predicted > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation > > > > > > > > > delays (which depend on the direction of motion). > > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a > > > > > > > > certain distance, synchronise them when they are both stationary, and > > > > > > > > then accelerate them both towards each other (and just before they > > > > > > > > collide, we bring them stationary again). Are you seriously saying > > > > > > > > that both clocks report that the other clock has slowed down, even > > > > > > > > though they have both undergone symmetrical processes? Because there > > > > > > > > is obviously a contradiction there. > > > > > > > > Yes, that is correct. Both will report a slow down. And in fact, > > > > > > > which ever one breaks the inertial frame to match speed with the other > > > > > > > is the one that will be "wrong". This is still within the realm of > > > > > > > SR, not GR. > > > > > > > What if they both "break the inertial frame"? > > > > > > Then whichever frame they both accelerate into will be the one that > > > > > has measured the "correct" time dilation. > > > > > So in other words, the clocks will register the same time, but will > > > > have slowed in some "absolute sense"? > > > > Yes--assuming they both accelerated by the same amount (that is to > > > say, assuming they both broke the inertial frame in a symmetric way). > > > Otherwise, they will register different times. > > > Agreed. > > > So let's explore an extension of this scenario. Let's say you have two > > clocks, and you accelerate both of them up to a common speed, and > > after they have travelled a certain distance, you turn them around and > > return them to the starting point. The only difference is that one > > clock goes a certain distance, and the other clock goes twice that > > distance, but they *both* have the same acceleration profile - the > > only difference is that one clock spends more time travelling on > > inertia. > > > Obviously, one clock will return to the starting point earlier than > > the other. But when both have returned, are their times still in > > agreement with each other, or have they changed? > > They are not in agreement because one was traveling faster longer. It does not travel faster, only longer. I assume that is what you meant. So what you're saying is that mere movement causes a time dilation effect? > Acceleration only appears indirectly in SR. Velocity is what appears > in all the SR equations, it just happens that velocity is a function > of acceleration, which is why acceleration is important. > > Also, I would avoid using the term "traveling on intertia". It just > physically isn't a very meaningful thing to say. You could simply say > "moving" are "traveling at a constant velocity," or something like > that. Indeed, it was just convenient to emphasise that no force was being applied, whereas "constant velocity" seems to beg the question of "relative to what".. |