A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
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From: Ste on 4 Mar 2010 12:09 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. > Also, you could just be dealing with a system where the velocity > started out high and you never measured any acceleration. Indeed.
From: Ste on 4 Mar 2010 12:17 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?
From: mpalenik on 4 Mar 2010 12:20 On Mar 4, 12:02 pm, Ste <ste_ro...(a)hotmail.com> wrote: > On 4 Mar, 16:29, 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. > > > I don't know what happened to this thread, or why google registers > > several different copies of this thread, each with only a few messages > > in it, but I'm going to describe the thought experiment I posted > > before in greater detail. > > > In order for the ^ to receive the pulse of light from both emitters at > > the same time and half way between the two emitters, the light must be > > emitted BEFORE the ^ is half way between the two emitters. > > Indeed. > > > From the stationary frame, the two emitters emit light at the same > > time and the ^ moves vertically until it is half way between the two > > of them, when it receives both pulses. > > Ok. > > > But from the moving frame, the ^ sees the emitters emit light when it > > is not half way between them, and then it sees the emitters move until > > he is exactly half way between them. However, the speed of light is > > the same in all frames, and unaffected by the motion of the emitters. > > > So the the emitter emit while one is closer to him than the other. > > The fact that they are moving does not affect the way light > > propagates. If the light signal from the closer one reaches him at > > the same time as the light signal from the farther one, that means the > > farther one must have emitted first. The fact that he is half way > > between the emitters at the end doesn't matter, because in his frame, > > it is the emitters that are moving, which does not affect the speed of > > light. > > This seems implausible. You cannot possibly (that is, physically) have > the situation you describe, in the way you describe it. The receiver, > when it receives the pulse of light, cannot possibly be in more than > one position relative to the two sources. It isn't. > It either receives when the > sources are equidistant, or it receives when one source is closer. The sources are equidistant. > It > cannot receive when it is equistant *and* when one source is closer, > for that would be an obvious contradiction. What you don't understand is the fact that the sources were not equidistant when they emitted the pulse. From the frame of reference of the moving ^, THAT is the location that is important. Even though it is equidistant from the emitters when it RECIEVES the pulse, only the location of the emitters when they SENT the pulse matters in the ^ frame, because in the ^ frame, it is the emitters that are moving, which does not affect the speed the light is propagated.
From: mpalenik on 4 Mar 2010 12:23 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. 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.
From: mpalenik on 4 Mar 2010 12:28
On Mar 4, 12:09 pm, 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, The only meaningful definition I know of total force is net force, which is simply the sum of all the forces acting on the object. Time doesn't enter into that. > 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. What you are describing is called impulse, which is the change in momentum (integral(F*deltat)). > 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. I understand what you're saying, but we wouldn't call this "total acceleration." We would simply call this the change in velocity. There's nothing exactly wrong with the way you've defined total acceleration, it's just that there is no standard definition for the term "total acceleration" so people might infer several different things by that term--such as the sum of the accelerations due to several different forces acting on an object. |