Re: How can this work in relativity?

russell@xxxxxxxx wrote in
news:1123642251.138169.268380@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

> bz wrote:
>> russell@xxxxxxxx wrote in
>> news:1123623023.028066.63730@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:
>>
>> > bz wrote:
>> >> russell@xxxxxxxx wrote in
>> >> news:1123608433.284461.150150@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:
>> >>
>> >> > bz wrote:
>> >
>> > [snip]
>> >
>> >> >> Since the rod only contracts/expands
>> >> >> *when*looked*at*from*another*frame*of*reference,
>> >> >
>> >> > *This* is the Lorentz-Fitzgerald contraction. Are you paying
>> >> > attention, David? Bz is making a crucial point that you need
>> >> > to learn. You seem to be confused on it.
>> >> >
>> >> > (Of course bz is only talking here of the *Lorentz-Fitzgerald*
>> >> > contraction. I'm sure he would agree that when people pull
>> >> > on a rod, it can stretch physically, and when they release it,
>> >> > it can contract physically. Such physical contraction is *not*
>> >> > Lorentz contraction.)
>> >>
>> >> Of course, but the acceleration would not be uniform. Those on one
>> >> end would have to pull in one direction and those on the other end
>> >> would need to push not quite so hard in the same direction.
>> >
>> > Right, that's what they do.
>> >
>> > (Or pull, depending on the ratio of the rod's density to
>> > its Young's modulus, i.e. on the effect of the tension
>> > in the wire, which is not constant in time.)
>> >
>> > I see you make the mistake of conflating applied force
>> > with acceleration -- a natural enough mistake for someone
>> > brought up on Newton to make.
>>
>> Einstein's SR postulates that the laws of mechanics hold within the
>> FoR.
>
> Yes, that's what he says, but does that mean, for example,
> that as long as you do your calculations in a single frame,
> you can apply F=ma and conclude that any speed can be
> attained by the accelerated mass? No, obviously not.
>
> Einstein was speaking here about the laws of mechanics
> applied *locally*. So, the above F=ma business doesn't
> contradict him because, to accelerate a mass to anything
> like c it will have to travel pretty far, and thus the
> experiment is no longer local.
>
> Similarly, dseppala's experiment is decidedly unlocal.
>
>>
>> > We need to be more cautious
>> > here. In particular, note that *even if* there is tension
>> > in the wire (for whatever reason) we can still give it a
>> > uniform acceleration by applying the appropriate compensating
>> > forces.
>>
>> As long as we apply the accelerating force uniformly, we should not
>> change any current stresses in the wire.
>
> Uniform acceleration (in the original frame) means that the
> rod's measured length is unchanged in the original frame.
> Since the rod's final state in the original frame is *moving*,
> we would have a falsification of SR -- in particular, of
> Lorentz contraction -- if its proper length doesn't increase
> during the acceleration by the exact amount required for its
> measured length to stay the same. Basic SR kinematics.

>From the moving FoR that includes the rod, the rods length never changes.
It is ONLY seen to change when looked at from another FoR that is in motion
wrt the rods FoR. Lorentz considered the contraction 'real', caused by
motion through the ether. Einstein considered the contraction 'apparent',
and only visible from another FoR.

>
> But our rod obeys Hooke's law, i.e. it cannot stretch unless
> it is put under tension.

Not by Einstein-Lorentz's transformation. Only if there is an ether.

> So, we have a choice. Either we apply a subtly varying
> force that both accelerates the rod and puts it under tension,
> or we accelerate it nonuniformly. And since we don't desire
> to do the latter, we do the former.

Of course, we can posit that Lorentz's contraction acts without exerting
any stresses.

> (I phrase it that way to make it clear that the rod is
> not stretched by magic, it's stretched by an applied
> force that is under our control. Sure, if we want
> success in our experiment, our choice is constrained.
> But that's just life.)

Look: Einstein said that there is no way to tell the difference between
accelerating the rod and accelerating our self.

So, instead of accelerating the rod, we accelerate ourself in the opposite
direction. Magic, the rod appears, from our frame of reference, to
contract, just as it did when we accelerated the rod.

And just like magic, the rod should break from the stress caused by our
motion????? I don't think so. But then the postulate of relativity is
invalidated.... or perhaps accelerating the rod uniformly can not induce
stress, provided we really accelerate it uniformly.

>> > So, it's specious simply to claim that there cannot
>> > be tension in the wire just because the acceleration is
>> > uniform.
>>
>> No tension DUE TO THE uniformly applied accelerating force.
>>
>> > You actually have to work the problem out.
>>
>> Perhaps I missed the place where such a force was non uniformly
>> applied.
>
> As I said before, the original problem did not specify the
> force. It specified the desired acceleration. Given certain
> properties of the rod (Hooke's law) SR predicts what external
> force must be applied to achieve our desire, and it's *not* a
> uniform force. I can see this bothers you. If you work in
> good faith I have no doubt this feeling of uneasiness will
> soon lead you to insight. I was once in exactly your position,
> on this point.

I am working in good faith. The problem is to provide a truely uniform
force along the length, which will provide the uniform acceleration.

On the other hand, if you apply the force at one end or both ends, then the
rod will buckle, bend and/or break and you will be unable to uniformly
accelerate it.

A = dv/dt; v = dx/dt this means that each small particle of the rod must
move the same distance at the same velocity during each instant of time in
order for the acceleration to be uniform. This keeps the distance between
particles constant [from their FoR]. This requires the forces separating
and holding them together to remain constant.

>> >> Then the acceleration would not be uniform[along the length of the
>> >> rod, nor in rate, probably].
>> >
>> > No, it is *specified* to be uniform. It's up to us to
>> > apply whatever forces are needed to achieve that spec.
>>
>> Perhaps I should not have assumed the rod was uniform in composition.
>> If it were not, then in order to apply a uniform acceleration, non
>> uniform forces would be required.
>
> No, I assure you there are no tricks up my sleeve. The
> rod is uniform.

Ok, then uniform acceleration requires uniform force.

>> Reminds me of the two rods in physics lab. Same size, weight, color.
>> One with mass concentrated at the ends of the rod, the other with mass
>> concentrated at the center. You couldn't tell the difference between
>> the rods unless you tried to spin them. The torque needed was much
>> higher for the rod with the high angular momentum.
>>
>> > In a Galilean world, uniform force applied everywhere
>> > simultaneously would do the job. But we do not live in
>> > a Galilean world.
>>
>> Was the rod to be accelerated passing through a gravity field? If it
>> was 'in
>
> No, we are in flat spacetime.
>
>> flat space' and of uniform composition, then it should act Galilean
>> under uniform acceleration cause by uniformly applied force along its
>> entire length simultaniously.
>
> It's the "simultaneously" that is the problem. You get to
> choose *one* inertial frame in which you can apply the force
> simultaneously all along the rod's length.

That is ok. We do it from the Rod's FoR. We have to use Einstein
syncronized clocks to do the timing, however. once clock for each particle
along the length of the rod.

> The application
> of force is not simultaneous in any other frame, due to the
> famous relativity of simultaneity. If we look at it in the
> final inertial frame, one end of the rod accelerates first
> to a standstill while the rest of the rod keeps going, which
> puts the rod under tension. We have to *pull* on the stopped
> end even after it has stopped, to keep it from snapping back
> and following along with the part of the rod that is still
> under motion. Finally, the later end of the rod stops. Note,
> we've had to pull on the earlier end longer and harder than
> we pushed on the later end to stop it. Nonuniform force.

Non uniform force = non uniform acceleration/deceleration = original
problem is broken, not the rod.

>> >> >> and not when looked at from its own frame of reference, there
>> >> >> will be no break.
>> >> >
>> >> > Its breaking or not is determined by the properties of the
>> >> > rod. In dseppala's scenario, the tension in the rod is
>> >> > everywhere nearly zero, so it would have to be a very,
>> >> > *very* weak rod to break.
>> >>
>> >> Perhaps it was sliced into very thin slices with a microtomb.
>> >> Perhaps it only LOOKS like a solid rod.
>> >
>> > You miss the point. And by the way, there is no such
>> > thing as the rod's "own frame of reference" while it is
>> > accelerating.
>> > That is to say, no inertial frame
>>
>> Agreed. But GR would allow such a frame of reference.
>
> No it wouldn't. Not an *inertial* frame. Anyhow, we
> *are* using GR already, in the sense that SR is its
> limit in flat spacetime, and by gum, flat spacetime is
> what we've got.

Then we are allowed to extend the inertial FoR to enclude all objects
undergoing uniform motion, which puts the entire rod [by the 'GIVEN' in the
problem] within the same GR FoR.

>> > instantaneously comoves with all parts of the rod at
>> > the same time. You seem unaware of this; it's in effect
>> > just another way of saying all the things I've said in
>> > other words.
>>
>> I am certainly unaware of a lot of things.
>>
>> >> >> In fact, there will be no stress nor strain induced in the rod by
>> >> >> uniform acceleration applied evenly along its length.
>> >> >
>> >> > This is *not* true, unless you are using the word "evenly"
>> >> > in a looser sense than clarity of expression demands.
>> >>
>> >> 'Uniformly accelerated' seems to require that each particle is
>> >> accelerated at the same time and at the same rate, othewise it is
>> >> not uniform, right?
>> >
>> > That's what I thought you meant, but I didn't want to
>> > say categorically that you were wrong, until I asked.
>> > Now I can say it. You were wrong.
>>
>> Clear communications sometimes asking such questions.
>> How am I wrong?
>
> I hope I've answered this sufficiently, above.

Perhaps. So far, I just see that the 'GIVEN' of the problem is violated by
non uniform acceleration, but I am still looking and willing to learn.

>> >> And in both Newtonian and Einstinian universes, if you accelerate
>> >> all portions of something at the same rate, there will be no stress
>> >> on strain from the acceleration.
>> >
>> >>From the *acceleration*?? No. As I said before, you
>> > have the cart before the horse. We *choose* to apply
>> > an uneven force *in order to make* the acceleration
>> > uniform in the original frame.
>>
>> If the rod is uniform, and that is a big IF, and if the original frame
>> has no other accelerting forces then this would see to be a
>> contradiction.
>
> A contradiction of your intuition is not a contradiction
> of logic. This *is* called Bell's *paradox*, remember.
> You are not the first to have found it puzzling.

But, I don't see a paradox, all I see is a violation of the 'GIVEN' in the
problem. If you posit that we can't syncronize clocks along the length of
the rod, then it can not be uniformly accelerated.

>> > The acceleration is the
>> > desired outcome, and the properties of spacetime make
>> > it impossible to achieve that outcome any other way.
>>
>> Any way other than non uniform force? Please elucidate.
>
> Correct, you got my meaning.


>> > Btw I find it telling that in one sentence you use the
>> > phrase "at the same time" but in the next, you omit it
>> > as if unimportant.
>>
>> Just that once it was said, it was to be understood to apply
>> thenceforth.
>>
>> > In SR it is *crucial*. Perhaps you
>> > should begin your further study precisely there.
>>
>> I have been working my way through Einsteins 1905 paper and his 1920
>> book and a few other sources.
>
> I only meant, take a look at the problem from the specific
> standpoint of simultaneity. I did that above, hopefully
> well enough for you to get an "aha" experience. But no
> doubt, Spacetime Physics does it better.

I see that if you break the clocks, you can't uniformly accelerate the rod
and that breaks the conditions of the problem.

>> >> > Have
>> >> > you read the Bell's spaceship FAQ?
>> >> he takes a lot of space to say that "from one's own frame of
>> >> reference, no change will be seen to lengths in your FoR. Another
>> >> ship, accelerating at the same rate, may be separated by space, will
>> >> maintain the same distance and WILL be in the SAME FoR.
>> >
>> > The author of the FAQ is not Bell himself. Bell, long ago,
>> > wrote a scientific paper explaining this now well-known and
>> > entirely uncontroversial prediction of SR. You don't really
>> > know SR until you have understood this prediction.
>>
>> If I read the reference correctly, the ships maintain costant distance.
>> If that is wrong, I need to read more carefully.
>
> Constant distance in the *original* inertial frame. But
> they are further apart in the final frame. Perhaps very
> much further apart, depending on the final speed.

But, the GR says we can treat any FoR undergoing uniform acceleration as an
inertial FoR. So, we carried our original FoR along with us. No change in
the rod's length in that FoR.

And, rememeber, if we move along the rod and the rod stays stationary, the
PoR says the effects are exactly the same. The rod appears to shorten from
our viewpoint, but from it's FoR, it remains constant in length.

Einstein's equivalence principle says we can't even tell if the
acceleration is from gravity or a rocket.

>> > If the
>> > FAQ is not clear enough for you (IMHO it's not as good as it
>> > should be) then look elsewhere; try Taylor and Wheeler for
>> > example.
>>
>> I will look for them.
>
> In other words, "Spacetime Physics", available at Amazon etc.
> They are the authors.
>

Thanks. I love to learn.





--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+sp@xxxxxxxxxxxxxxxxxxxx remove ch100-5 to avoid spam trap
.

Relevant Pages

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  • Re: How can this work in relativity?
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