OK funkstar, it is about time we set up some preliminary concepts for our length contraction experiment.

Since our moving objects are going to need measuring equipment onboard, I think a good size would be 10cm x 4cm. Now if each are 10cm in length, then we can separate their centers (assume the center of mass to be exactly in the center of each object) by 15 cm. This leaves a 5cm clearance between the front and back ends of the two objects.


..........<-5->
----------.....----------
.....<------15----->


Now the objects will always measure their proper length to be 10cm - that is, when the objects are stationary in the Earth frame, they are measured to be 10cm. And when the objects are moving with velocity, .866c, the objects measure their lengths to be 10cm in their frame.

Since the acceleration is measured to be simultaneous (edit in: and identical) for the two objects in the Earth frame, the Earth should always see 15cm between the centers of mass of the objects, even if the objects contract - remember, it is the center of mass that we are measuring to accelerate. If we apply our length contraction formula, we get that the objects will measure 7.5cm between their centers of mass. This means sometime while they accelerated, one bumped into the other and they wrecked. Yet we never see this wreck in the Earth frame, what am I missing?

15 = 5+5+5
7.5 = 2.5 + 2.5 + 2.5

15 = 5+5+5
7.5 = 2.5 + 2.5 + 2.5 So you are saying the objects measure their length to be 5cm in their frame after having accelerated?

From earth, the objects would be observed measuring 5cm and clearance between them 2.5cm

From earth, the objects should look 5cm and clearance between them 2.5cm Yes, everything from the Earth frame is fine (except you have the wrong clearance number). The Earth frame sees the moving objects contracted in length, correct? The objects always measure their own length to be 10cm whether they accelerated or not. The objects will see the distance between their centers of mass length contracted since we stipulated that they both accelerate instantaneously in the Earth frame. That is, in the Earth frame, the length between their centers of mass do not change.

In the objects' frame they still measure 10 - 5 - 10cms. What is happening is the linear space along the direction of motion contracts including any measuring device like scales or whatever.

In the objects' frame they still measure 10 - 5 - 10cms. What is happening is the linear space along the direction of motion contracts including any measuring device like scales or whatever. Then you are saying the acceleration was an instantaneous event in the objects moving frame. And we are back to conflict in the Muon experiment thread.

Actually i modified my 2nd post to be more clear.

Moving muon experiences the length (space/distance between muon and earth) contraction along its path. Earth perspective 'sees' no change in this distance but should apply lorentz formula to get how much length contraction the muon suffers(!).

Actually i modified my 2nd post to be more clear.
If I accept your new interpretation, that means the objects moved closer to each other and did not undergo identical acceleration as seen in the Earth frame. I guess I did forget to mention the Earth sees the two objects undergo identical acceleration in the OP, but that was already established between Funkstar and myself.

I am not sure of what you mean by identical acceleration. both the objects and everything between them, in the moving frame, undergo same acceleration and same scale of length contraction.

Aer,

Just been watching what's going on. Is this a parctical experiment you are talking about or a more elaborate thought experiment?

I am not sure of what you mean by identical acceleration. both the objects and everything between them, in the moving frame, undergo same acceleration and same scale of length contraction.What will the objects measure their length to be in their stationary frame after they accelerated? 5cm?

The acceleration is instantaneous and identical in the Earth frame only. (identical: we measure the acceleration to be the same magnitude at each instant in time until the final velocity is reached)

Just been watching what's going on. Is this a parctical experiment you are talking about or a more elaborate thought experiment?
I'm betting more of an elaborate thought experiment as it would require a vast distance to send off an object at .866c :D

The more I consider it, the more I -believe- length contraction will ultimately be disproven. Although I don't think we'll be able to prove anything with these thought experiments.

What will the objects measure their length to be in their stationary frame after they accelerated? 5cm?

They measure no change if they use any scale in their rest frame.

The acceleration is instantaneous and identical in the Earth frame only. (identical: we measure the acceleration to be the same magnitude at each instant in time until the final velocity is reached)
As i already said, acceleration is common for all the objects in your moving frame. If i scaledown your photograph to half the width will your eyes touch each other?

As i already said, acceleration is common for all the objects in your moving frame. If i scaledown your photograph to half the width will your eyes touch each other? So this length contraction is actual real length contraction? When the objects measure things in the Earth frame, they also see things length contracted, is this length contraction actual and real too? The endless cycle again :( I wish we could have some consistency as to what SRT says.

Aer,


I'm betting more of an elaborate thought experiment as it would require a vast distance to send off an object at .866c

Yes. My calculations indicate that you would have to get a 1m projectile (with transmitter beacons at each end) moving at 0.1c to have even a hope of measuring the length contraction with precision equipment.

Yes. My calculations indicate that you would have to get a 1m projectile (with transmitter beacons at each end) moving at 0.1c to have even a hope of measuring the length contraction with precision equipment. Like I said above, I am only seeking out consistency in the SRT argument you and others present. To me, it doesn't seem consistent.

By the way, did you read the conclusions Pete and I came up with in the Muon experiment thread. :eek: Something surely has to be wrong there.

Cool. Let's see where it goes...

So this length contraction is actual real length contraction? When the objects measure things in the Earth frame, they also see things length contracted, is this length contraction actual and real too? The endless cycle again :( I wish we could have some consistency as to what SRT says.

The length contraction is very much real for the moving frame. Google 'kinematic blue shift' for further proof.

The length contraction is very much real for the moving frame. Google 'kinematic blue shift' for further proof. To get the correct time-dilation according to Special Relativity, yes. But the so-called moving frame can consider the Earth to be the moving frame and thus turn everything on its head.

The Earth and objects can be considered moving at .866c. Then the objects decelerate to rest in the so called "moving frame" which is now our stationary frame. There, the entire experiment is considered in the "moving frame".

OK funkstar, it is about time we set up some preliminary concepts for our length contraction experiment.

Since our moving objects are going to need measuring equipment onboard, I think a good size would be 10cm x 4cm. Now if each are 10cm in length, then we can separate their centers (assume the center of mass to be exactly in the center of each object) by 15 cm. This leaves a 5cm clearance between the front and back ends of the two objects.


..........<-5->
----------.....----------
.....<------15----->


Now the objects will always measure their proper length to be 10cm - that is, when the objects are stationary in the Earth frame, they are measured to be 10cm. And when the objects are moving with velocity, .866c, the objects measure their lengths to be 10cm in their frame.

Agreed.

Since the acceleration is measured to be simultaneous (edit in: and identical) for the two objects in the Earth frame, the Earth should always see 15cm between the centers of mass of the objects,

Yes, I believe so.

even if the objects contract - remember, it is the center of mass that we are measuring to accelerate. If we apply our length contraction formula, we get that the objects will measure 7.5cm between their centers of mass.

No. The objects are moving in the Earth frame, so it is the Earth observer that sees the length contraction of the objects. If the acceleration phase ends with the Earth observer seeing 15 cm between their CMs, and the objects themselves contracted to 7.5 cm, the objects should measure 2*15=30cm between their CMs.

This means sometime while they accelerated, one bumped into the other and they wrecked. Yet we never see this wreck in the Earth frame, what am I missing?
In the length contraction formula, you used the length between the objects when seen from Earth as proper length, when it should have been used as apparent length.

By the way, I'm not certain it's a good idea to set up the experiment in the way you propose. I suggest simply accelerating both the objects for a fixed amount of their own proper time. This (I think) means that the Earth observer will not see them accelerating equally, but is should keep both of the objects at rest in each others frame.

It'll also be much easier to design and setup two identical objects than two objects with more complex acceleration schemes that coincide with us seeing them as having identical acceleration and velocity at all times, and turning of their rockets (or whatever) at the same time.

At least, that's what I think. I'm unfortunately not very well versed in acceleration in str. :)

Just a quick question. Why are you guys using two objects?

No. The objects are moving in the Earth frame, so it is the Earth observer that sees the length contraction of the objects. If the acceleration phase ends with the Earth observer seeing 15 cm between their CMs, and the objects themselves contracted to 7.5 cm, the objects should measure 2*15=30cm between their CMs.
If? We defined the acceleration to be instantaneous and identical in the Earth frame. How are we going to measure anything but 15cm between their CM. So everything appears length contracted in the Earth frame. Now lets look at the acceleration phase. Object A will be the one on the left and object B will be the one on the right. When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous. Now that A and B have both accelerated, they slipped into another frame, let's call it their instantaneous reference frame since they technically have an instantaneous velocity. Do A and B always agree that they are both in the same reference frame while accelerating?

Just a quick question. Why are you guys using two objects?
Why not? This is a followup on the length contraction & relativity of simultaneity paper - as the author there used two objects.

By the way, I'm not certain it's a good idea to set up the experiment in the way you propose. I suggest simply accelerating both the objects for a fixed amount of their own proper time. This (I think) means that the Earth observer will not see them accelerating equally, but is should keep both of the objects at rest in each others frame.
There is nothing wrong with the acceleration proposed except that the objects are limited as to how long they can accelerate since they cannot exceed c.

Edit: Actually, I never said the acceleration had to be constant, only that it was measured as identical for all instances of time until the final velocity is reached. So they could accelerate forever under this definition.

But the so-called moving frame can consider the Earth to be the moving frame and thus turn everything on its head.

Yes, according to SRT, the moving frame can consider itself to be at rest & apply lorentz transformation to get the length contraction suffered by the 'moving earth'.

Yes, according to SRT, the moving frame can consider itself to be at rest & apply lorentz transformation to get the length contraction suffered by the 'moving earth'. Unfortunately, I don't think the analysis you provided agrees with funkstar. Who am I to believe on this issue?

I don't know the history of this thread, so i can't say anything. What is your point? You don't agree with SRT length contraction?

Ok Aer.

Why not? This is a followup on the length contraction & relativity of simultaneity paper - as the author there used two objects.

I just thought that a single rod observed in the earth frame with radio beacons at each end might be simpler if you just want to show length contraction. But I don't really know what your ultimate purpose is here, so I'll shut up and watch. :)

But I don't really know what your ultimate purpose is here, so I'll shut up and watch. :)
You assume I have an ultimate purpose here - shame on you! :m:

I know. You are a devious one with an ultimate grand take-over-the-world scheme...

If? We defined the acceleration to be instantaneous and identical in the Earth frame. How are we going to measure anything but 15cm between their CM. So everything appears length contracted in the Earth frame.

Except the distance between them, course.

Now lets look at the acceleration phase. Object A will be the one on the left and object B will be the one on the right. When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous. Now that A and B have both accelerated, they slipped into another frame, let's call it their instantaneous reference frame since they technically have an instantaneous velocity. Do A and B always agree that they are both in the same reference frame while accelerating?
No, not if the Earth observer sees them accelerating equally, i.e. that he always sees 15cm between them.

There is nothing wrong with the acceleration proposed except that the objects are limited as to how long they can accelerate since they cannot exceed c.

The thing is, it's going to be very difficult to constantly monitor acceleration and velocity at a distance, and furthermore make sure that the distance between them stays constant. It's much easier (at least conceptually, to me) to design two identical objects, "launch" them, and see what happens.

Edit: Actually, I never said the acceleration had to be constant, only that it was measured as identical for all instances of time until the final velocity is reached. So they could accelerate forever under this definition.
Of course, they could very well accelerate forever with constant acceleration as well.

I don't know the history of this thread, so i can't say anything. What is your point? You don't agree with SRT length contraction?
I do; Aer reserves judgment until it has been directly experimentally verified. This is a thread designing such an experiment.

I little background info is found in another Aer thread. (http://www.sciforums.com/showthread.php?t=47722)

No, not if the Earth observer sees them accelerating equally, i.e. that he always sees 15cm between them. Let's just look at the initial point of acceleration. Will they each agree that they began the acceleration at the same time as they started in the Earth frame?

If? We defined the acceleration to be instantaneous and identical in the Earth frame. How are we going to measure anything but 15cm between their CM. So everything appears length contracted in the Earth frame. Now lets look at the acceleration phase. Object A will be the one on the left and object B will be the one on the right. When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous. Now that A and B have both accelerated, they slipped into another frame, let's call it their instantaneous reference frame since they technically have an instantaneous velocity. Do A and B always agree that they are both in the same reference frame while accelerating?
Not according to SR.
Why? The short answer is relative simultaneity.

Pick an instant in the Earth frame while the objects are accelerating. Both objects are passing Earth at the same velocity v. The objects at this instant correpond to two events (call them leading-object-atv, and trailing-object-at-v).

Switching to the frame with velocity v, we examine those events and find that at these events, the boxes are stationary at the Earth is passing them at -v... all good so far.

But! In that moving-at-v-with-respect-to-Earth frame, the two events happen at different times. In this frame, the leading-object-atv event happens before the trailing-object-atv event. At any instant in any instantaneous co-moving frame of one of the objects, the leading object is always faster than the trailing object, until the aceleration phase ends.

Of course, they could very well accelerate forever with constant acceleration as well.
Yes, as measured in their instantaneous reference frame. If we defined their acceleration to be constant in this way and the Earth frame says they begin their acceleration simultaneously, then how is the Earth going to measure any differences between the two accelerations of the objects? Shouldn't Earth say their acceleration magnitude is the same for all instances in time?

Not according to SR.
Why? The short answer is relative simultaneity.

We are starting the acceleration in the Earth frame. The Earth frame says the acceleration of the two objects starts simultaneously, how are the objects going to say that this was not instantaneous at just this point?

I wish we could have some consistency as to what SRT says.
It's tough to get consistency in an environment where explanations are wordy.
I know what you mean. That's why I try to spell out the coordinates of events wherever possible, before adding my interpretation.

SR is undeniably consistent in the maths of transforming events between reference frames. However, the interpretation of those transforms isn't always consistent, particularly in this kind of environment. Real difficulties occur when the interpretations are absorbed and used to draw conclusions without understanding the transforms that were originally interpreted.

We are starting the acceleration in the Earth frame. The Earth frame says the acceleration of the two objects starts simultaneously, how are the objects going to say that this was not instantaneous at just this point?
At that instant, the objects are stationary in the Earth's frame. Therefore the objects and Earth agree that the initiation of acceleration was instantaneous.

...unless the acceleration phase is instantaneous (ie discontinuos velocity).

We defined the acceleration to be instantaneous and identical in the Earth frame.
In this case:
In the objects' final velocity frame, the leading object accelerates first, followed by the trailing object some short time later.

In the non-inertial frame of the leading object, there is a singularity at acceleration where events which were simultaneous in Earth's frame with the acceleration event are not well-defined.

...unless the acceleration phase is instantaneous (ie discontinuos velocity).
That is not what is meant by instantaneous. It just means that the acceleration time interval is insignificant compared to the time interval of travel at constant velocity (that is usually the situation when instantaneous acceleration is used). The time interval is just ignored as the actual change to the end result is negligible.

However, in this case, the acceleration is what we are considering... so it doesn't make sense to say that it was "instantaneous" even if the same acceleration could be considered instantaneous in other situations.

In this case:
In the objects' final velocity frame, We haven't gotten here yet, stop jumping ahead! We have only gotten to the next leg of the acceleration. We agree that the two objects agree their acceleration began instantaneously?

Yes.

Yes. So they are each in a new instantaneous frame?

From this instantaneous reference frame, they each see the other at rest wrt themselves and that their accelerations are identical?

If you wanted to apply special relativity like a mindless robot, you'd have to conclude that the above cannot be true, would you not?

So they are each in a new instantaneous frame?
At the instant acceleration begins, they're both still at rest in the Earth frame (as far as SRT is concerned, anyway... GR says something interesting, but perhaps we'll consider that later).

From this instantaneous reference frame, they each see the other at rest wrt themselves and that their accelerations are identical?
I'm not sure about what they "see" of the other's accelerations... this isn't a straightforward thing to answer in SR. You'd need to consider the velocity at which each sees the Earth going past the other one at time acceleration-begins+&Delta;t, use that to find an average acceleration, and take a limit as &Delta;t approaches zero.
Tedious, but I can do it if you like.

GR might be useful... but since I don't know the formalism of GR I'd be afraid of falling into interpretation traps.

At the instant acceleration begins, they're both still at rest in the Earth frame (as far as SRT is concerned, anyway... GR says something interesting, but perhaps we'll consider that later). No, we do not need to consider GR. We are assuming this experiment is being done away from any massive bodies (i.e., ignore GR).


I'm not sure about what they "see" of the other's accelerations... this isn't a straightforward thing to answer in SR. They have measuring equipment onboard to measure or "see" the others acceleration. That was given in the setup - if SR can't answer to it, then a hypothesis for the experiment cannot be fully developed.



You'd need to consider the velocity at which each sees the Earth going past the other one at time acceleration-begins+&Delta;t, use that to find an average acceleration, and take a limit as &Delta;t approaches zero.
Tedious, but I can do it if you like. No, I don't care about what the Earth sees, only what the objects see.



GR might be useful... but since I don't know the formalism of GR I'd be afraid of falling into interpretation traps. I'm afraid you are wrong, GR has no application here (other than the simplified SR equations we are using).

No, we do not need to consider GR. We are assuming this experiment is being done away from any massive bodies (i.e., ignore GR).
...
I'm afraid you are wrong, GR has no application here (other than the simplified SR equations we are using)
GR isn't just useful in curved space. It also provides useful tools for use in accelerating reference frames, which is what we're looking at here.

My posibly flawed understanding is that in this situation (flat space), GR will give the same results as SR, but the analysis is simpler in GR because GR can handle accelerating frames directly.

An analogy:
For the task of analysing accelerating frame in flat space, GR is like a sliding compound power saw, while SR is a pencil, square, and hand mitre saw. Both will get the job done with the same result, but its a hell of a lot easier with the better tool - as long as you have one and know how to use it.

Unfortunately, I don't know how to use the superior tool. So unless someone with the appropriate knoweldge wants to help, or we wait until I find time to learn (got a year or two??), I'm just speculating wishfully.

They have measuring equipment onboard to measure or "see" the others acceleration. That was given in the setup - if SR can't answer to it, then a hypothesis for the experiment cannot be fully developed.
Can you describe this equipment? Perhaps that will help to determine what SR has to say about it.

No, I don't care about what the Earth sees, only what the objects see.
I didn't say anything about what Earth sees. Here is what I said again:
"You'd need to consider the velocity at which each sees the Earth going past the other one at time-acceleration-begins + &delta;t, use that to find an average acceleration, and take a limit as &delta;t approaches zero."

Can you describe this equipment? Perhaps that will help to determine what SR has to say about it.

If you are not an EE and want help with the realities of electronic measurement equipment, transceivers, etc, feel free to ask, or not. It's an open offer, humbly submitted...

My posibly flawed understanding is that in this situation (flat space), GR will give the same results as SR, but the analysis is simpler in GR because GR can handle accelerating frames directly. If we assume constant acceleration, we can just use the relativistic rocket equations. No need for GR.


Unfortunately, I don't know how to use the superior tool. So unless someone with the appropriate knoweldge wants to help, or we wait until I find time to learn (got a year or two??), I'm just speculating wishfully.
Have you ever seen the general solution to a GR equation? We are talking 12 pages printed for one equation.


Can you describe this equipment? Perhaps that will help to determine what SR has to say about it.
I am just going to assume the equipment fits on board and works :D :m:


I didn't say anything about what Earth sees. Here is what I said again:
"You'd need to consider the velocity at which each sees the Earth going past the other one at time-acceleration-begins + &delta;t, use that to find an average acceleration, and take a limit as &delta;t approaches zero." If they see each other at rest, there is no need to know how fast the Earth is going, what am I not understanding here?

If you are not an EE and want help with the realities of electronic measurement equipment, transceivers, etc, feel free to ask, or not. It's an open offer, humbly submitted... I thank you for the offer, but unless there is some blantantly wrong assumption we are making regarding how good the equipment can measure, then I don't think these details should be neccessary for the discussion at hand.

Yep. No prob.

If we assume constant acceleration, we can just use the relativistic rocket equations. No need for GR.
Like I said. There is no need for GR... but I thought it might make things simpler.
I'm not yet comfortable with the relativistic rocket equation, (I guess I might be by the time we're done here!) but... doesn't it just describe a single accelerating object? Does it give us the tools we need to transform between two accelerating objects?

Have you ever seen the general solution to a GR equation? We are talking 12 pages printed for one equation.
No I haven't. I guess it's not as simple as I wanted to think!

I am just going to assume the equipment fits on board and works :D :m:
But how?
Does it take two velocities at different times and divide the velocity difference by the time?
I'm really not comfortable with the notion of instantaneous acceleration. I'd really like to take an average acceleration over a time period, and find the limit as the time period goes to zero.

If they see each other at rest, there is no need to know how fast the Earth is going, what am I not understanding here?
We're talking about acceleration. Meaning the rate of change in the velocity of the Earth past the objects?

Something else to check:
Is acceleration constant?
If so, is it constant in the Earth frame or the object frame?

Something else to check:
Is acceleration constant?
If so, is it constant in the Earth frame or the object frame? The acceleration is constant in the object's frame. From Earth's perspective, the acceleration will constantly decrease.

I'm really not comfortable with the notion of instantaneous acceleration. I'd really like to take an average acceleration over a time period, and find the limit as the time period goes to zero. We are not considering instantaneous acceleration, only constant acceleration. I don't think we are considering the same experiment.


We're talking about acceleration. Meaning the rate of change in the velocity of the Earth past the objects? No, the acceleration is measured in the objects own frame. Think of the objects moving from one inertial reference frame to another inertial reference frame - this is what it means to accelerate (in SR terms).

We are not considering instantaneous acceleration, only constant acceleration. I don't think we are considering the same experiment.
Sorry, by "instantaneous acceleration" I meant "the magnitude of the acceleration at a particular instant". My bad.

No, the acceleration is measured in the objects own frame. Think of the objects moving from one inertial reference frame to another inertial reference frame - this is what it means to accelerate (in SR terms).
But it must be relative to something else? It just seemed to me that the Earth passing by was a convenient reference to measure acceleration against.

I'm really grasping for tools that I can use to analyse this situation, to try to answer your question:

So they are each in a new instantaneous frame?

From this instantaneous reference frame, they each see the other at rest wrt themselves and that their accelerations are identical?

I think this is not a straightforward question, and that we need to proceed carefully, checking all assumptions.

I think that at any time after the objects have begun accelerating (ie at any time that they have some velocity relative Earth) that it is not clear what they see of each other's acceleration.

I think I have an approach that will work... I'll post it up shortly.

But it must be relative to something else? It just seemed to me that the Earth passing by was a convenient reference to measure acceleration against.
So our measuring device is going to measure the acceleration of the Earth? The Earth isn't actually accelerating but the measuring device doesn't know better, so it would work just as well. Fine, the acceleration can be "measured" this way, but our onboard "computers" must be able to conclude that the acceleration is constant at every instant in the objects instantaneous reference frame.


I think this is not a straightforward question, and that we need to proceed carefully, checking all assumptions.
Can we at least agree that both objects are now in a new instantaneous reference frame? That is, they both exist in a single instantaneous reference frame because they see each other at rest wrt themselves.

So our measuring device is going to measure the acceleration of the Earth? The Earth isn't actually accelerating but the measuring device doesn't know better, so it would work just as well. Fine, the acceleration can be "measured" this way, but our onboard "computers" must be able to conclude that the acceleration is constant at every instant in the objects instantaneous reference frame.
Thanks for your patience... after due consideration, I've come around and I'm no longer fretting over how they measure acceleration.

Can we at least agree that both objects are now in a new instantaneous reference frame? That is, they both exist in a single instantaneous reference frame because they see each other at rest wrt themselves.
I don't think so.

At that instant in Earth's rest frame, the two objects have the same velocity - No problem.

But, when you transform to the object frame, you find that you've transformed the objects to different times in that frame!

But, when you transform to the object frame, you find that you've transformed the objects to different times in that frame!
So the relativity of simultaneity requires, for which there is no proof. Like I said at the outset, we are not going to come up with a "thought experiment" that can prove anything.

I'm claiming without proof that length contraction and the relativity of simultaneity are bunk :D :m:

And you're free to do so, of course.

If you wish to tweak your SR skills, I can come up with a SR problem that I like a lot - I'll have to take a little bit of time to dig it out of my memory though.

But only if you are interested of course.

I'm interested!

So am I.

Can we at least agree that both objects are now in a new instantaneous reference frame? That is, they both exist in a single instantaneous reference frame because they see each other at rest wrt themselves.
I don't think so.

At that instant in Earth's rest frame, the two objects have the same velocity - No problem.

But, when you transform to the object frame, you find that you've transformed the objects to different times in that frame!

If 2 objects are at rest (zero relative velocity between the objects) during the whole experiment in one frame how transforamtion to objects' frame is going to change the zero relative velocity between these objects?

If 2 objects are at rest (zero relative velocity between the objects) during the whole experiment in one frame how transforamtion to objects' frame is going to change the zero relative velocity between these objects?
They change frames continuously during the acceleration. While it will appear from either objects view that these frame changes happen simultaneously, giving them no relative velocity, the earth observer will see them happen asynchronously. So the distance will stay the same as view from the objects, but dilate according to Earth observer.

while transformation, acceleration changes the rate of 'scalingdown' the linear space along the direction of movement , not the ratio. the transformed distance between the objects keeps the ratio in tune with the transformed linear space. afterall we need to use transformation to overcome the error in perception.

The major question unanswered in the thread was:
We defined the acceleration to be instantaneous and identical in the Earth frame. How are we going to measure anything but 15cm between their CM. So everything appears length contracted in the Earth frame. Now lets look at the acceleration phase. Object A will be the one on the left and object B will be the one on the right. When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous. Now that A and B have both accelerated, they slipped into another frame, let's call it their instantaneous reference frame since they technically have an instantaneous velocity. Do A and B always agree that they are both in the same reference frame while accelerating?

Clear answer:
They are not in the same inertial frame according to SR, because when they are moving at the same speed in Earth's reference frame they are not at rest in the moving frame at the same time.

They are not in the same inertial frame according to SR, because when they are moving at the same speed in Earth's reference frame they are not at rest in the moving frame at the same time.
I don't think you are reading the same question I asked. Particularly, you are rejecting what I said without stating that what I said is wrong:

When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous.

You are doing the transformation of the reference frame they are in after the acceleration, and not analyzing the acceleration alone.

That is why I rejected your answer and your claim to relativity of simultaneity. I bet you still don't get it: I understand the prediction of special relativity with regard to simultaneity, but that is not what I am asking here. I am asking for the transient slip in simultaneity.

I don't think you are reading the same question I asked....
as this thread is recently "reborn" perhaps Aer wopuld state again the problem. He said:

We defined the acceleration to be instantaneous and identical in the Earth frame. How are we going to measure anything but 15cm between their CM. So everything appears length contracted in the Earth frame. Now lets look at the acceleration phase. Object A will be the one on the left and object B will be the one on the right. When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous. Now that A and B have both accelerated, they slipped into another frame, let's call it their instantaneous reference frame since they technically have an instantaneous velocity. Do A and B always agree that they are both in the same reference frame while accelerating?
To illustrate my lack of understanding what exactly it is Aer is concerned about and what Aer says Pete does not understand, let me ask:

Is this some question about (for example) a meter stick intitially at rest in a canon on earth?
Is the super strong, infinitely rigid, meter sick and super fast burning gun powder for infinite acceleration essential to the question or can the graden variety wooden meter stick with brass end caps be used?
Can Aer's "A" & "B" be the brass end caps or does the wood connnecting them interfer with what ever Aer is asking about?

I.e. if possible Aer, try to illustrate what you are asking about. if my meter stick is not useful, chose what ever is. thanks in advance.

PS I don't know where the 15 cm comes from or if it is essential. i assume the CM is the center of mass but am not sure and don't know of what? etc.

When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous.
This is true according to SR.

You are doing the transformation of the reference frame they are in after the acceleration, and not analyzing the acceleration alone.
This is not true. My statement is only true during acceleration. It is a qualitative description of how the "transient slip in simultaneity" happens.

Crunching the formulas and taking the limit to get a quantitative formulation shouldn't be hard. I'll do it for you if you're unwilling or unable to do it yourself.

When A accelerates, A should see B accelerate as they are starting in the Earth frame which says these two events are simultaneous.

This is true according to SR.

So according to A & B, the accelerations of each, A & B, are simultaneous.



You are doing the transformation of the reference frame they are in after the acceleration, and not analyzing the acceleration alone.

This is not true. My statement is only true during acceleration. It is a qualitative description of how the "transient slip in simultaneity" happens.

Let's take a look at what you said.

They are not in the same inertial frame according to SR, because when they are moving at the same speed in Earth's reference frame they are not at rest in the moving frame at the same time.

How are they moving at any speed in Earth's reference frame? They were -at rest- in Earth's reference frame at the instant of acceleration. You are doing just as I said, assuming the frame after acceleration, in which "they are moving at the same speed in Earth's reference frame".

So according to A & B, the accelerations of each, A & B, are simultaneous.
The start of acceleration is simultaneous. The end is not.

They were -at rest- in Earth's reference frame at the instant of acceleration.
Are you back to truly instaneous acceleration again?
In that case, your answer is here.

But remember this?
That is not what is meant by instantaneous. It just means that the acceleration time interval is insignificant compared to the time interval of travel at constant velocity (that is usually the situation when instantaneous acceleration is used). The time interval is just ignored as the actual change to the end result is negligible.

However, in this case, the acceleration is what we are considering... so it doesn't make sense to say that it was "instantaneous" even if the same acceleration could be considered instantaneous in other situations.

Which is it?
Is the acceleration instantaneous or not?


If it is not, then at any time during acceleration the two clocks are moving at the same speed in Earth's reference frame, but they are not at rest in the moving frame at the same time.

So according to A & B, the accelerations of each, A & B, are simultaneous.

The start of acceleration is simultaneous. The end is not.

We haven't even come close to analyzing the end of the acceleration.




They were -at rest- in Earth's reference frame at the instant of acceleration.

Are you back to truly instaneous acceleration again?
In that case, your answer is here.

Again, you appear to never know when to apply instantaneous situations and when not to. I never mentioned instantaneous acclerelation as it does not apply. The acceleration is constant, not instantaneous. We are however analyzing an instantaneous moment in time. Specifically, since we want to know what A & B see, we are analyzing what they would see in their proper time.

If anyone is to believe what you say, then as A accelerates, A sees B accelerate, but then as A is accelerating, A sees that B has not accelerated yet.

If anyone is to believe what you say, then as A accelerates, A sees B accelerate, but then as A is accelerating, A sees that B has not accelerated yet.
Care to explain how you reached that conclusion?
I have no idea what you think I'm thinking.

Here is your question:
Do A and B always agree that they are both in the same reference frame while accelerating?

Here is my answer:
They are not in the same inertial frame according to SR, because when they are moving at the same speed in Earth's reference frame they are not at rest in the moving frame at the same time.

Note that it is an explanation of what happens while accelerating (like you asked), not at the instant that acceleration begins.

Here is your question:

Do A and B always agree that they are both in the same reference frame while accelerating? ”

Here is my answer:

They are not in the same inertial frame according to SR, because when they are moving at the same speed in Earth's reference frame they are not at rest in the moving frame at the same time. ”

Note that it is an explanation of what happens while accelerating (like you asked), not at the instant that acceleration begins.

I think everyone is well aware that special relativity says there is a simultaneity shift after acceleration. That is not what I intended to ask, nor do I think it is what I asked. I specifically was talking about the instant acceleration began, and the instant immediately following. Not, the instant after acceleration stopped.

If acceleration happens at t=0, then according to A, at t=0-, A & B are in the same reference frame; at t=0 A & B accelerate, at t=0+, A & B are not in the same reference frame.

That is what you (and special relativity) would say, correct?

Aer, you asked what happens during acceleration, and I told you. No one's talking about what happened after acceleration stopped, so I have no idea why you keep croaking about it. Do you understand what during acceleration means? How about while accelerating?

Let me spell it out for you again:
When acceleration begins (at t=0), A and B are in the same reference frame.
At any time t+&Delta;t for &Delta;t>0, A and B are not in the same reference frame.

If acceleration happens at t=0, then according to A, at t=0-, A & B are in the same reference frame; at t=0 A & B accelerate, at t=0+, A & B are not in the same reference frame.
What is "0+", exactly?

Do you mean the limit of 0+&Delta;t as &Delta;t approached zero?
If you do, then A and B are in the same reference frame at t=0+

Or do you mean t=0+&Delta;t for some small &Delta;t > 0?
If you do, then A and B are not in the same reference frame at t=0+

Aer, you asked what happens during acceleration, and I told you. No one's talking about what happened after acceleration stopped, so I have no idea why you keep croaking about it. Do you understand what during acceleration means? How about while accelerating? Yes, as that is what I have been talking about the entire time. You however have brought up:

The start of acceleration is simultaneous. The end is not.


Which is it?
Is the acceleration instantaneous or not?

I was never talking about the end of the acceleration. The acceleration was initally defined as instantaneous, but then I said "let's analyze the acceleration phase".

Now this is key, so read carefully:
In case you don't know, that means that we are no longer defining the acceleration to be instantaneous as we are analyzing the time changing effects of acceleration.


Let me spell it out for you again:
When acceleration begins (at t=0), A and B are in the same reference frame.
At any time t+&Delta;t for &Delta;t>0, A and B are not in the same reference frame. That is exactly what I just said you'd say. That means at t=0+, A & B are not in the same reference frame.

Now this is the key point so again, read carefully:
How do you explain the transient of this? A says B accelerates at the same instant as A itself accelerated. B is not in the same frame as A at t=0+.

And before you bring up the lorentz transformations again, that does not explain the transient. That only gives the simultaneity shift from t=0 to t=a, where a is the time at the end of acceleration.



What is "0+", exactly?

Immediately after t=0. 0- is immediately before t=0.



Do you mean the limit of 0+&Delta;t as &Delta;t approached zero?
If you do, then A and B are in the same reference frame at t=0+

Or do you mean t=0+&Delta;t for some small &Delta;t > 0?
If you do, then A and B are not in the same reference frame at t=0+
t approaching zero is not the same as t=0. Given your first interpetation, am I to assume that if I do an integration, I will not get the correct answer?

Why don't you do the integration and find out?

Why don't you do the integration and find out? Is that all? I have done integrals and they do in fact give the correct answer. However, given this bit of information:

Do you mean the limit of 0+Δt as Δt approached zero?
If you do, then A and B are in the same reference frame at t=0+

I would have to assume that integrals would not give the correct answer.

Aer, I'm not going to be drawn into a discussion of your peculiar brand of mathematics. I find your "0+" idea useless, but if you can get some use out of it, that's fine with me.

So here's the bottom line:

If Earth is at rest in A's reference frame, then B is also at rest in A's reference frame.
If Earth has some velocity with respect to A, then B also has some velocity with respect to A.

You decide whether Earth is at rest in A's reference frame at t=0+.
Then you'll know whether B has some velocity wrt A at that time.

Aer, I'm not going to be drawn into a discussion of your peculiar brand of mathematics. I find your "0+" idea useless, but if you can get some use out of it, that's fine with me.

The idea of 0+ is not my idea. You really think that is some idea that I cooked up? No, I certainly did not make up the notation and it is commonly used.


So here's the bottom line:

If Earth is at rest in A's reference frame, then B is also at rest in A's reference frame.
If Earth has some velocity with respect to A, then B also has some velocity with respect to A.

You decide whether Earth is at rest in A's reference frame at t=0+.
Then you'll know whether B has some velocity wrt A at that time.

OH Good! It's settled then. A sees B accelerate at t=0. A has some velocity with respect to Earth at t=0+, so B has some velocity with respect to A at t=0+. Conclusion: A did not see B accelerate at t=0.

Let me try to jump in here.

If A and B are two spaceships equipped with identical acceleration programs and start at rest in the Earth frame, then the distance between them as measured in the Earth frame is fixed for all time. However, if one stretched a flimsy string between the two spaceships, it would break as the spaceships and the string speed up in the Earth frame, a consequence of length contraction. From the point of view of the trailing spaceship, the leading spaceship slowly drifts ahead and so the string breaks. From the point of view of the leading spaceship, the trailing space ship slowly lags behind and so the string breaks. This is often called Bell's Spaceship Paradox, and I think this is the issue here, but please correct me if I'm wrong.

Let me try to jump in here.

If A and B are two spaceships equipped with identical acceleration programs and start at rest in the Earth frame, then the distance between them as measured in the Earth frame is fixed for all time. However, if one stretched a flimsy string between the two spaceships, it would break as the spaceships and the string speed up in the Earth frame, a consequence of length contraction. From the point of view of the trailing spaceship, the leading spaceship slowly drifts ahead and so the string breaks. From the point of view of the leading spaceship, the trailing space ship slowly lags behind and so the string breaks. This is often called Bell's Spaceship Paradox, and I think this is the issue here, but please correct me if I'm wrong.
From what I remember of the thread, that is exactly correct. However, the central issue at hand currently is the transient slip of the relativity of simultaneity and not the length contraction issue (although the two go hand in hand).

The idea of 0+ is not my idea. You really think that is some idea that I cooked up? No, I certainly did not make up the notation and it is commonly used.


OH Good! It's settled then. A sees B accelerate at t=0. A has some velocity with respect to Earth at t=0+, so B has some velocity with respect to A at t=0+. Conclusion: A did not see B accelerate at t=0.
Not so fast! You're assuming without foundation that in A's reference frame, B's velocity is the same as Earth's velocity.

If Earth has some velocity in A's reference frame, then B's velocity in A's reference frame is in between zero and Earth's velocity.

However, the central issue at hand currently is the transient slip of the relativity of simultaneity and not the length contraction issue (although the two go hand in hand).
They're effectively the same thing. Two sides of the same coin. Length contraction dictates relative simultaneity and vice versa.

What exactly is the "transient slip of the relativity of simultaneity?"

They're effectively the same thing. Two sides of the same coin. Length contraction dictates relative simultaneity and vice versa.
That is essentially what I said when I mentioned they go hand in hand.


Not so fast! You're assuming without foundation that in A's reference frame, B's velocity is the same as Earth's velocity.

If Earth has some velocity in A's reference frame, then B's velocity in A's reference frame is in between zero and Earth's velocity.
We still have the same problem.

A accelerates at t=0 with magnitude a.
A sees B accelerate at t=0 with magnitude a. A has some velocity, v_A with respect to Earth at t=0+, so B has some velocity, v_B/A with respect to A at t=0+. Conclusion: A did not see B accelerate at t=0 with magnitude a.

What exactly is the "transient slip of the relativity of simultaneity?" A poorly formed phrase. The shift of simultaneity for a noninertial frame.

What exactly is the "transient slip of the relativity of simultaneity?"
While A and B accelerate, their clocks remain synchronized in Earth's frame of reference. They tick 1 simultaneously, 2 simultaneously, etc.

In A's frame and B's frame, the clocks do not remain synchronized during acceleration. The leading clock ticks 1 before the trailing clock.

Aer made up the term "transient slip of the relativity of simultaneity" to refer to the changing difference in the times on the clocks of A and B in A's frame during acceleration.

Ok, thanks for help so far. What then is the issue? I assume there is something here we don't all agree on.

We're talking about the acceleration of B wrt Earth in A's reference frame, right?

A accelerates at t=0 with magnitude a.
A sees B accelerate at t=0 with magnitude a. A has some velocity, v_A with respect to Earth at t=0+, so B has some velocity, v_B/A with respect to A at t=0+. Conclusion: A did not see B accelerate (wrt Earth? - Pete) at t=0 with magnitude a.

You've assumed constant acceleration of B wrt Earth in A's reference frame during the specified time period. Why?

Correct Conclusion:
The average acceleration of B wrt Earth in A's reference frame between t=0 and t=0+ is between 0 and a.

Ok, thanks for help so far. What then is the issue? I assume there is something here we don't all agree on.
Aer seems to think that there's some mathematical inconsistency in this changing simultaneity.

Ok, thanks for help so far. What then is the issue? I assume there is something here we don't all agree on.
The issue is nothing that can be settled here which is why I let the thread go a long time ago. There is a claim to the relativity of simultaneity being factually true when really it is only a prediction of the theory of special relativity.

The issue is nothing that can be settled here which is why I let the thread go a long time ago.
... but insisted on reviving the discussion in another thread.

Correct Conclusion:
The average acceleration of B wrt Earth in A's reference frame between t=0 and t=0+ is between 0 and a.
What is the average of an infinitesimal time period (i.e. immediately after t=0).

... but insisted on reviving the discussion in another thread. Wrong, I just made a comment upon which you told me to start a thread about said comment. I did not go back and bring this thread back - you did.

More information, please. Are you asking what is the average of a changing quantity during an infinitesimal time period?

Instantaneous acceleration, by definition is the acceleration over an infintesimal time period. That is exactly what the value a cited in posts above is. "a" is the acceleration at t=0.

Do you claim that acceleration at t=0+ is exactly the same as acceleration at t=0?

That would be inconsistent with claiming that A's velocity with respect to Earth at t=0+ is not the same as A's velocity with respect to Earth at t=0.

The first implies that 0+ - 0 = 0.
The second implies that 0+ - 0 > 0.

Do you claim that acceleration at t=0+ is exactly the same as acceleration at t=0? No, but the velocity at t=0+ is dependent on the acceleration at t=0, not the acceleration at t=0+.

The velocity at t=0+ is dependent on the average acceleration between t=0 and t=0+

If that's a zero time period, then A is still at rest wrt Earth.

If it is not a zero time period, then the acceleration has changed.

You are trying to set up a situation in which A's velocity can change but the acceleration can't - but it can't be done.

No, I am only taking the limit as t goes to 0. That is not a "zero time period". t=0+ is immediately after t=0, it is shorthand for the limit as &Delta;t goes to 0, also referred to as &delta;t in calculus.

The velocity at t=0+ is dependent on the instantaneous acceleration at t=0. Please note that this usage of "instantaneous acceleration" is not the same as declaring the overall acceleration to be instantaneous.

No, I am only taking the limit as t goes to 0. That is not a "zero time period". t=0+ is immediately after t=0, it is shorthand for the limit as &Delta;t goes to 0, also referred to as &delta;t in calculus.

The velocity at t=0+ is dependent on the instantaneous acceleration at t=0.
You can't have it both ways.

If the time period is greater than zero, then the acceleration changes during that time period, and the velocity afterward depends on the average acceleration.

If you take the limit, then the time period is zero, the acceleration doesn't change, and neither does the velocity.

This is basic calculus, Aer. Back to grade school for you!

OK, then.

The limit of v_A as t approaches zero equals zero.
The limit of &Delta;v_A approaches 0 as &Delta;t approaches 0. The velocity at t=0+ is not equal to 0.


Do you know what "taking the limit" means?
Yes, do you? And I see that you had to delete your original message as it was wrong as I noted above.


You can't have it both ways.

If the time period is greater than zero, then the acceleration changes during that time period, and the velocity afterward depends on the average acceleration.

If you take the limit, then the time period is zero, the acceleration oesn't change, and neither does the velocity.

This is basic calculus, Aer. Back to grade school for you!

You still don't get it I see. The limit of &Delta;v_A approaching 0 is not the same as v_A(0+)=0.

Your "average acceleration" is true, however, we are not taking discrete intervals. The intervals are infinitesimal intervals, so the instantaneous acceleration at t=0 is the same (fundamentally) as the average acceleration of a discrete interval model.

Looks like we're down to a fundamental disagreement. We're not going to get beyond "yes it it" "no it's not" without appeal to some authority.

Wikipedia (http://en.wikipedia.org/wiki/Limit_%28mathematics%29):
Consider http://en.wikipedia.org/math/09368bc6428ff35acd59669b91f2714d.png as x approaches 2. In this case, f(x) is defined at 2 and equals its limit of 0.4

See also the formal definition (http://en.wikipedia.org/wiki/Limit_%28mathematics%29#Formal_definition) of a limit (also at Mathworld (http://mathworld.wolfram.com/Limit.html)), with which it's easy to prove that lim_t&rarr;0 a.t = 0

The limit of ΔvA approaching 0 is not the same as vA(0+)=0.
Those two authorities say it is. You're welcome to disagree, but forgive me if I don't.

Looks like we're down to a fundamental disagreement. Yes it is fundamental.


Consider http://en.wikipedia.org/math/09368bc6428ff35acd59669b91f2714d.png as x approaches 2. In this case, f(x) is defined at 2 and equals its limit of 0.4

Please note that what you are applying is the following:
<table border=0 cellspacing=0 cellpadding=0><tr><td align=center>_lim</td><td rowspan=2>v(t) = a(t)&middot;t</td></tr><tr><td>^t&rarr;0</td></tr></table>
to get v(0)=0 which we all agree to.

What we really want is v(0+)

Here is the acceleration from the definition of dv/dt
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

v(t+dt) = v(t) + &int;a dt
v(0+) = v(0) + &int;a dt
v(0+) = &int;a dt

In the interval from t to t + dt, the velocity changes by dv = a(t*) dt. The mean valua theorem tells us that for some t* in the interval, this equation is exact no matter what size dt is. Of course we have in mind that dt is small (compared, say, to other timescales in the system). If we neglect terms of order dt&sup2; and higher, then the it doesn't matter where we evaluate the acceleration in the interval (t,t+dt). In other words, a(t+dt) dt = a(t) dt = a(t+.5dt) dt, etc and the error in each equal sign is of order dt&sup2;. Thus as long as the trajectory is differentiable, it doesn't matter where you evaluate the acceleration when computing the change in velocity in the limit i.e. when you do an integral. This is a fundamental result of the theory of integration, but it is nontrivial as can be seen in stochastic differential equations were the point of evaluation in the interval does matter (leading to two inequivalent prescriptions which are the Stratonovich and Ito calculi).

Physics Monkey, are the following equivalent?

<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t-&Delta;t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

Anyway, my point is, that v(0+)=v(t+&Delta;t) as &Delta;t&rarr;0. and from the definition of a(0), we see that v(0+) cannot be zero, or else a(0) must be zero.

v(t+dt) = v(t) + &int;a dt
v(0+) = v(0) + &int;a dt
v(0+) = &int;a dt

You need some bounds on that integral. If you do it right, I believe you'll find that v(0+) = 0

Which equals zero. Definately not!

What does it equal, then?

Anyway, my point is, that v(0+)=v(t+&Delta;t) as &Delta;t&rarr;0. and from the definition of a(0), we see that v(0+) cannot be zero, or else a(0) must be zero.
Correction:
v(0+)=v(0+&Delta;t) as &Delta;t&rarr;0

Your conclusion is a crock.
Look at the definition of a limit, and try again.

What's the limit of a&Delta;t as &Delta;t&rarr;0?

This is basic stuff.

v(0+)=v(0+Δt) as Δt→0

Good job, yes, t=0 was a given here.


What's the limit of aΔt as Δt→0?

v(0+) &ne; a&Delta;t as &Delta;t&rarr;0.
v(0+) = &int;a dt



What a crock.
Look at the definition of a limit, and try again. I did and I explained to you that your limit was finding v(0), not v(0+).

What does it equal, then?
Have we defined any values? I don't believe we have other than to say that a(0)&ne;0.

Do you agree that:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

I did and I explained to you that your limit was finding v(0), not v(0+).
Yes, you explain a lot of things. Poorly and incorrectly.

Congratulations on discovering that v(0) = v(0+).


Do you agree that:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>
Yep

Yes, you explain a lot of things. Poorly and incorrectly.

Congratulations on discovering that v(0) = v(0+).

Wrong, my explanation is correct and yours is wrong.


Do you agree that:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>
Yep
Which can be written in the following notation:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(0) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(0+)-v(0) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

So given your conclusion that v(0+)=v(0), a(0)=0. Are you saying now, that B has not accelerated according to A?

Aer,

The two expressions are not the same. The first of your expressions is the definition of the derivative, and the second is actually twice the derivative. To see this you can add and subtract v(t) and group terms: v(t+dt)-v(t) gives one copy and v(t) - v(t-dt) gives another.

Aer,

The two expressions are not the same. The first of your expressions is the definition of the derivative, and the second is actually twice the derivative. To see this you can add and subtract v(t) and group terms: v(t+dt)-v(t) gives one copy and v(t) - v(t-dt) gives another.
OK, I didn't think they were the same but wasn't sure since they are limit cases.

What about the following (they may seem equivalent but I don't think they are, or rather, t in a(t) is slightly differently defined for each):

<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t)-v(t-&Delta;t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

If one defines v(0+) as the limit dt -> 0 of v(dt) then v(0+) = v(0), this is essentially the definition of continuity. Of course, v(dt) is not zero for any finite dt (provided a is monotone), but in the limit it is zero if v(0) is zero.

According to the mean value theorem, and neglecting higher order terms in dt, all the following are true v(dt) = a(0)dt = a(dt)dt = a(.5dt)dt, etc.

Wrong, my explanation is correct and yours is wrong.
"Yes it is!"
"No it isn't!"
"Yes it is!"
"No it isn't!"

Which can be written in the following notation:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(0) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(0+)-v(0) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

I really don't think so. You're trying to take the limit on the top half of the equation before taking the limit on the bottom half.

This form is closer to the truth:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=3>a(0) = </td><td colspan=2 style='border-bottom:1px solid black'>v(0+)-v(0)</td></tr><tr><td align=center>_lim</td><td rowspan=2 align=center> &Delta;t</td></tr><tr><td align=center>^{&Delta;t&rarr;0}</td></tr></table>

It's still not completely correct (it is no longer possible to evaluate the function). I have no idea why you'd want to use that 0+ notation - it just doesn't seem useful.

Actually, those two are equivalent. In general anything of the form v(t+ c dt) - v(t- (1-c) dt) gives the derivative dv/dt at t (c in [0,1]).

Aer,
You still haven't figured out the limit of aΔt as Δt→0.

The idea of 0+ is not my idea. You really think that is some idea that I cooked up? No, I certainly did not make up the notation and it is commonly used.
Can you give me a reference to somewhere it is used?
I'd be interested in seeing it in a useful context.

Pete,

You will find the notation 0+ and 0- in the very famous book "A Course of Modern Analysis" by Whittaker and Watson. It was conventionally used to denote one sided limits and otherwise employed in the analysis of discontinuous functions. More recent treatments tend to use the notation &epsilon; and - &epsilon; for 0+ and 0-. In either case, the notation conventionally means that one should take the limit at the end of the day i.e. f(0+) = f(&epsilon; ) is short for lim x -> 0 (from above) f(x).

v(0+) &ne; a&Delta;t as &Delta;t&rarr;0.
v(0+) = &int;a dt

This is informative. In this statement, you're implying that v(0+) depends not only on a(0), but on a changing value of a.

But that disagrees with what you said earlier:
The velocity at t=0+ is dependent on the instantaneous acceleration at t=0.

It looks like you don't know what you think.

"Yes it is!"
"No it isn't!"
"Yes it is!"
"No it isn't!"
I've not said one of the above phrases. Are you quoting yourself perhaps?

I really don't think so. You're trying to take the limit on the top half of the equation before taking the limit on the bottom half.

This form is closer to the truth:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=3>a(0) = </td><td colspan=2 style='border-bottom:1px solid black'>v(0+)-v(0)</td></tr><tr><td align=center>_lim</td><td rowspan=2 align=center> &Delta;t</td></tr><tr><td align=center>^{&Delta;t&rarr;0}</td></tr></table>

It's still not completely correct (it is no longer possible to evaluate the function). I have no idea why you'd want to use that 0+ notation - it just doesn't seem useful.
That is a weak argument. You take the limit of the top and bottom separately in this situation anyway.

You are too boggled down with the 0+ notation. It is just shorthand when you are talking about limits (And is useful when the concept is not limits itself which unfortunately this thread has fallen too).

v(0+) ≠ aΔt as Δt→0.
v(0+) = ∫a dt

This is informative. In this statement, you're implying that v(0+) depends not only on a(0), but on a changing value of a.

But that disagrees with what you said earlier:

The velocity at t=0+ is dependent on the instantaneous acceleration at t=0.


It looks like you don't know what you think.
Are you seriously suggesting that a, because I wrote it on the right side of the integral means it cannot be a constant? If you don't think it is a constant, then what is a, as a function of t? In this case, for v(0+), I think a is constant from the definition of a.

Pete,

You will find the notation 0+ and 0- in the very famous book "A Course of Modern Analysis" by Whittaker and Watson. It was conventionally used to denote one sided limits and otherwise employed in the analysis of discontinuous functions. More recent treatments tend to use the notation &epsilon; and - &epsilon; for 0+ and 0-. In either case, the notation conventionally means that one should take the limit at the end of the day i.e. f(0+) = f(&epsilon;) is short for lim x -> 0 (from above) f(x).
Thanks PhysicsMonkey, that makes sense.

So it's immediately obvious that if v is continuous, v(0+) = v(0), right?

Aer,
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(t) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(t+&Delta;t)-v(t) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

... can be written in the following notation:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(0) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(0+)-v(0) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

Not correct, since you're mixing notations and taking a limit twice.
Try this:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(0) = </td><td style='border-bottom:1px solid black'> v(0+)-v(0) </td></tr><tr><td align=center> 0+</td></tr></table>

Actually, those two are equivalent. In general anything of the form v(t+ c dt) - v(t- (1-c) dt) gives the derivative dv/dt at t (c in [0,1]).

Yes, but what about when v(t=0)=0, v(0-)=0, but v(0+)=c? One definition will give a(0)=0 and the other a(0)=d where c and d are constants.

So it's immediately obvious that if v is continuous, v(0+) = v(0), right?

Right. So long as v(0+) is defined as I have said, any contnuous function satisfies v(0-)=v(0+)=v(0) as a matter of definition.

Are you seriously suggesting that a, because I wrote it on the right side of the integral means it cannot be a constant? If you don't think it is a constant, then what is a, as a function of t? In this case, for v(0+), I think a is constant from the definition of a.
If a is a constant, then your first statement is false:
v(0+) ≠ aΔt as Δt→0
and the second is unnecessary.

Have you figured out the limit of aΔt as Δt→0 yet?

<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(0) = </td><td align=center>_lim</td><td style='border-bottom:1px solid black'>v(0+)-v(0) </td></tr><tr><td>^{&Delta;t&rarr;0}</td><td align=center> &Delta;t</td></tr></table>

Not correct, since you're mixing notations and taking a limit twice.
Try this:
<table border=0 cellspacing=0 cellpadding=0><tr><td rowspan=2>a(0) = </td><td style='border-bottom:1px solid black'> v(0+)-v(0) </td></tr><tr><td align=center> 0+</td></tr></table>
I never intended to expand v(0+) back into a limit, but only to say that it is a value and plug that value in. In that respect, I am not taking the limit twice.

Aer, if v(0+) does not equal v(0) then the velocity is discontinuous and the derivative is not defined. Alternatively, if you're a physicist who likes to play it fast and loose, one can define the derivative to be a delta function. Either way, the derivative is highly singular and then it does matter where you evaluate your limits as you have pointed out.

I never intended to expand v(0+) back into a limit, but only to say that it is a value and plug that value in. In that respect, I am not taking the limit twice.
Then you're using the notation incorrectly. The notation 0+ directly implies taking the limit.

I think this discussion is done. It's been nice chatting with you, Aer. You too, Physics Monkey.

Aer, if v(0+) does not equal v(0) then the velocity is discontinuous and the derivative is not defined. Alternatively, if you're a physicist who likes to play it fast and loose, one can define the derivative to be a delta function. Either way, the derivative is highly singular and then it does matter where you evaluate your limits as you have pointed out.
Well, we do have a discontinuous function. dv/dt is 0 for t<0 and dv/dt is a constant for t&ge;0. Is that not a discontinuous function?

If we take the limit from separate sides of t=0, we get different results. How can we get this if as you just claimed, v(0+)=v(0)=v(0-).

Is that not a discontinuous function?
It's a continuous function with a discontinuous derivative.
If we take the limit from separate sides of t=0, we get different results.
You get different results for dv/dt (which is discontinuous), but not for v (which is continuous).

It's a continuous function with a discontinuous derivative.

dv/dt is the function I was talking about, the function that is defined from the limit I gave which is merely the definition of a derivative. dv/dt is a discontinuous function as I said the first time. It is not "a continous function with a discontinuous derivative" although, I am sure you mean, v is a continous function with a discontinous derivative..


You get different results for dv/dt (which is discontinuous), but not for v (which is continuous).

Precisely what I said, dv/dt is discontinous.

Yes, dv/dt is discontinuous in your idealization, but v is still continuous and so v(0) = v(0+) = v(0-).