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Author Topic: Uncertainty and relativity  (Read 6637 times)

Offline itisus

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Uncertainty and relativity
« on: 20/01/2009 00:17:40 »
Suppose I am stationary relative to a massive object (i.e. accelerating) or, I suppose, simply accelerating in the x direction.  If I attempt to measure position and momentum of a particle in the x direction, my impression is that the minimum uncertainty is greater than hbar, but it's beyond my limited capabilities.  Does anyone know how uncertainty transforms?


 

Offline Vern

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« Reply #1 on: 20/01/2009 01:19:25 »
This link is to wikipedia that will give you a good start.
 

Offline lightarrow

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« Reply #2 on: 20/01/2009 18:20:46 »
Suppose I am stationary relative to a massive object (i.e. accelerating) or, I suppose, simply accelerating in the x direction.  If I attempt to measure position and momentum of a particle in the x direction, my impression is that the minimum uncertainty is greater than hbar,
Why? Can you explain better?
 

Online yor_on

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« Reply #3 on: 22/01/2009 23:52:56 »
"What does it mean physically for a state to have minimum uncertainty?

The correspondent classical physical system (not the statistical ensembles) obey delta x*delta p = 0.
Therefore the minimum uncertainty wave packets are the quantum objects most close to their classical counterparts.
The transition is discontinuous and spontaneous (instant).
It is known as the collapse of the wave packet.

Another remarkable property is that they are stable localized objects.
Notice that it is quite general physical property:
proton is stable, hydrogen atom is stable and isolated classical material point is stable."

And hbar "is equal to the Planck constant divided by (or reduced by) 2π.
It is used when frequency is expressed in terms of radians per second instead of cycles per second.
The expression of a frequency in radians per second is often called angular frequency (ω), where ω = 2πν."

So are you asking where the uncertainty of that particles position and momentum will 'disappear' for you while observing it, accelerating?

 

Offline LeeE

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« Reply #4 on: 23/01/2009 00:49:21 »
There is very little uncertainty in relativity.  Are you not thinking of QM?
 

Offline itisus

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« Reply #5 on: 26/01/2009 22:28:32 »
If the speed of light is constant while both length and time are altered for an accelerating observer, then uncertainty (with or without the 2pi factor) should be affected since it does not have the same units (L/T) as c.  I doubt one could make better measurements by accelerating, but that is the naive dimensional result.  Surely someone has properly derived this.
 

Online yor_on

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« Reply #6 on: 26/01/2009 22:49:33 »
It's an interesting thought you have there.

If one define the Planck length as of a certain 'distance'.
Can one say that it is the same outside your frame, if observed from an accelerating frame for example?

Not that I see how you would be able to measure such a thing:)
But 'distance' seems to be a 'relation' of frames so?
Same as the definition of 'invariant mass'.

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« Last Edit: 26/01/2009 22:51:25 by yor_on »
 

Offline itisus

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« Reply #7 on: 03/02/2009 19:24:31 »
The Planck length certainly is a "distance" since it has dimensions of length.  If you are accelerating (or standing on a massive body) your local length scale may change, depending on how it is defined.  Whether or not units are defined in flat-space terms, there is still the question of how acceleration affects the observational limit hbar.  That in turn affects real-world values like the size of atoms.
 

Offline lightarrow

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« Reply #8 on: 04/02/2009 14:54:04 »
If the speed of light is constant while both length and time are altered for an accelerating observer, then uncertainty (with or without the 2pi factor) should be affected since it does not have the same units (L/T) as c.  I doubt one could make better measurements by accelerating, but that is the naive dimensional result.  Surely someone has properly derived this.
The Heisenberg uncertainty relations are Lorentz invariant so they remain the same changing the frame of reference.
 

Online yor_on

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« Reply #9 on: 05/02/2009 12:31:43 »
Awh Lightarrow, I was just starting up my engines here:)
But you're right, It's Lorentz invariant.

Still it leave us with the question of how to define a 'frame'.
Is there any 'least definition' of what might be seen as a 'frame of reference'

Would that to be Planck sized?
For example in that 'not uniformly' accelerating spaceship.
How can one define it?

As 'one' frame, or as a 'infinite' array of frames, macroscopically structured and seen as one entity/frame?
Can one say that this whole object, at all spatial coordinates of its frame, have the same acceleration?

Our universe seems built upon them?
 

Offline JP

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« Reply #10 on: 05/02/2009 16:23:24 »
Lorentz invariance usually deals with something remaining the same when you change between "inertial reference frames," i.e. frames that aren't accelerating or where space isn't curved by gravity.  I'm guessing that's the same with the uncertainty principle.  However, even if space-time is curved, the smaller of a piece you look at, the flatter it is (just like the surface of the earth looks pretty flat, even though the earth itself is curved).  To guess more, I'd think that on the scales we're talking about (Planck lengths), things look pretty flat.  In fact, trying to apply general relativity to those scales is probably going to require a theory of quantum gravity anyway.
 

Offline lightarrow

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« Reply #11 on: 05/02/2009 18:07:32 »
Lorentz invariance usually deals with something remaining the same when you change between "inertial reference frames," i.e. frames that aren't accelerating or where space isn't curved by gravity.  I'm guessing that's the same with the uncertainty principle. 
Yes, the OP wrote: "If the speed of light is constant while both length and time are altered" so I imagined he was actually talking about bodies moving at high speed, not necessarily accelerating (even if he talks about acceleration). My answer relates to SR, not to GR.
 

Online yor_on

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« Reply #12 on: 05/02/2009 18:34:54 »
Nice thinking JPetruccelli.
Just one thing.
You define ""inertial reference frames," ...as... "i.e. frames that aren't accelerating or where space isn't curved by gravity."
That definition seems to exclude our 'spacetime' too?

Or am I getting it all wrong again:)?

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Yes Lightarrow, I see how you thought there.
I'm just trying to see 'frames', correctly?
Sort of:)

If possible?
« Last Edit: 05/02/2009 18:39:10 by yor_on »
 

Offline JP

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« Reply #13 on: 05/02/2009 19:19:05 »
yor_on, I think it's a matter of how accurate you want to be.  It's probably true that most of space-time is curved to some degree (though most of it is probably very very flat since there's lots more vacuum than matter).  If you're far enough from massive objects and you're dealing with relatively small-scales, then it's probably good enough in most cases to assume its flat.  You can draw an analogy with the earth's surface.  If you're drawing a map of something small compared to the earth's curvature, like your neighborhood, then you don't need to worry about the earth's curvature.  It's only when you start drawing maps of larger and larger areas that you need to start worrying about the curvature. 
 

Offline itisus

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« Reply #14 on: 19/02/2009 23:33:00 »
If the speed of light is constant while both length and time are altered for an accelerating observer, then uncertainty (with or without the 2pi factor) should be affected since it does not have the same units (L/T) as c.  I doubt one could make better measurements by accelerating, but that is the naive dimensional result.  Surely someone has properly derived this.
The Heisenberg uncertainty relations are Lorentz invariant so they remain the same changing the frame of reference.

Thanks.  That should have been apparent since SR really is related to relative velocity rather than acceleration, and I was interested in a function of gravitational potential, which is the GR equivalent of velocity, not acceleration.  Now it all makes sense.
 

Offline Mr. Scientist

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« Reply #15 on: 17/10/2009 19:34:07 »
There is very little uncertainty in relativity.  Are you not thinking of QM?

Not a little - there is no uncertainty, which makes relativity a classical theory.
 

Offline LeeE

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« Reply #16 on: 17/10/2009 20:45:05 »
There is very little uncertainty in relativity.  Are you not thinking of QM?

Not a little - there is no uncertainty, which makes relativity a classical theory.

Yes, you are absolutely correct.  I just didn't want to sound too assertive.
 

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« Reply #16 on: 17/10/2009 20:45:05 »

 

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