# Particle at rest

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In the de-Broglie wave equation, the more momentum a particle has, the less wavelength it has. What happens if a particle is at rest. Can this even happen? Does its spin count towards the equation?

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A particle at rest cannot be detected by measuring equipment that is in the same reference system as the said particle.

A particle is detected if it

- emits photons, rarely other particles, with well-known signatures (i.e., frequencies/wavelengths)

- reflects photons sent toward it

- absorbs photons and ejects photons with lower energies sent toward it

- hits the detector by itself (which undermines "being at rest")

- produces an electric or magnetic field that causes some reaction inside the detector, such as induction.

etc.

In classical physics, "being at rest" is often used to refer to very slow-moving objects, with a Lorentz factor close to 1.0.

Edited by Sensei
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15 minutes ago, Sensei said:

A particle at rest cannot be detected by measuring equipment that is in the same reference system as the said particle.

A particle is detected if it

- emits photons, rarely other particles, with well-known signatures (i.e., frequencies/wavelengths)

- reflects photons sent toward it

- absorbs photons and ejects photons with lower energies sent toward it

- hits the detector by itself (which undermines "being at rest")

- produces an electric or magnetic field that causes some reaction inside the detector, such as induction.

etc.

In classical physics, "being at rest" is often used to refer to very slow-moving objects, with a Lorentz factor close to 1.0.

Thank you for that. But does a particles spin count towards its wavelength?

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39 minutes ago, grayson said:

the more momentum a particle has, the less wavelength it has.

That is not correct as written, but you may have meant something different ( wave number ? )
The wavelength is not involved at all, as it is simply the length of one oscillation.
What is involved  is the number of oscillations in the wave packet; the more compact (less oscillations ), the more defined its position.
Since a compacted wave packet is composed of multiple differing frequency waves, by compacting it, and gaining positional information, we lose momentum ( proportional to frequency ) information.
If, on the other hand, the oscillations comprising the wave packet are large we gain more accurate momentum information, such that, as the number of oscillations approaches infinite ( like a sine wave ) the momentum is almost exact, but the wave ( and particle's position ) is spread to almost infinite extent.

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It can't happen. If a particle is at rest for just the shortest of times, the uncertainty in position becomes zero for just that time, so the uncertainty in momentum goes to infinity in every direction, and the particle immediately flies away. This is kind of a "pictorial" way of talking, of course. One must do the regular Hilbert-space operator procedure.

Particles "at rest" like, eg, in an ion trap, are not really at rest. They're twirling around in some kind of stationary-state confined micro-dance.

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7 minutes ago, joigus said:

the uncertainty in position becomes zero for just that time, so the uncertainty in momentum goes to infinity

We can have it the other way around, i.e., zero momentum and infinitely uncertain position, can't we?

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Sometimes it is best to consider what happens when approaching infinite extremes, not what happens at the infinite extremes.

Edit
Momentum cannot go to infinity.
For massive particles the momentum is constrained by the speed ( must be less than c ).
This then imposes a lower limit to the wave number, extent of the packet, or the 'box' you can put around a particle ( accuracy of position ).

Edited by MigL
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We can have it the other way around, i.e., zero momentum and infinitely uncertain position, can't we?

Absolutely. Both, of course, are mathematical limits. They indicate, AFAIK, that the interpretation cannot be brought to those limits experimentally and something has to give.

Exactly as @MigL says.

That's my understanding, anyway.

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2 hours ago, grayson said:

In the de-Broglie wave equation, the more momentum a particle has, the less wavelength it has. What happens if a particle is at rest. Can this even happen? Does its spin count towards the equation?

The wavelength is h/p, so yes, the wavelength tends to infinity as the momentum tends to zero. It can’t actually happen but can be applied as a thought experiment. There are some quantum implications to having zero momentum, and being in the rest frame is a useful approach to certain problems.

The spin is not part of this; that’s a separate property. p is the linear momentum.

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5 hours ago, joigus said:

It can't happen. If a particle is at rest for just the shortest of times, the uncertainty in position becomes zero for just that time, so the uncertainty in momentum goes to infinity in every direction, and the particle immediately flies away.

Isn't there always a frame where the particle is at rest (unless it is going the speed of light of course).

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11 minutes ago, Boltzmannbrain said:

Isn't there always a frame where the particle is at rest (unless it is going the speed of light of course).

Classically, yes. It loses some meaning in QM, considering the Heisenberg Uncertainty Principle

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5 minutes ago, swansont said:

Classically, yes. It loses some meaning in QM, considering the Heisenberg Uncertainty Principle

But the Heisenberg uncertainty principle has its limits. There is a frame where its momentum would be zero to whatever the uncertainty is.

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Just now, grayson said:

But the Heisenberg uncertainty principle has its limits. There is a frame where its momentum would be zero to whatever the uncertainty is.

You don’t know what the momentum is, so saying it’s zero isn’t strictly possible, though this might be unimportant for certain problems.

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1 minute ago, swansont said:

You don’t know what the momentum is, so saying it’s zero isn’t strictly possible, though this might be unimportant for certain problems.

Well yes, we do not know what the momentum is because of Heisenberg's uncertainty principle. Though from what I heard we do have an average of where it would be. Is this true?

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2 minutes ago, grayson said:

Well yes, we do not know what the momentum is because of Heisenberg's uncertainty principle. Though from what I heard we do have an average of where it would be. Is this true?

Yes.

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2 minutes ago, swansont said:

You don’t know what the momentum is, so saying it’s zero isn’t strictly possible, though this might be unimportant for certain problems.

There could never be two physical objects at rest wrt each other could there?

Does a frame of reference have to be applicable to a potential physical  scenario to be physically valid?

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10 minutes ago, geordief said:

There could never be two physical objects at rest wrt each other could there?

No, not to arbitrary precision.

10 minutes ago, geordief said:

Does a frame of reference have to be applicable to a potential physical  scenario to be physically valid?

Not exactly sure what you mean here.

Physicists have no problem approximating things, so something can be treated as being at rest despite all of the caveats we’ve mentioned.

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11 minutes ago, swansont said:

Not exactly sure what you mean here.

Physicists have no problem approximating things, so something can be treated as being at rest despite all of the caveats we’ve mentioned

Ok.I am a bit of a literalist myself.If someone greets me in the street  with a "How are you?" my reflex is to wonder how I am and to communicate my state of being with the person

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Here is a derivation ( some simple maths ) of the HUP as applied to a particle described by a deBroglie wave.
Includes historical perspective and is well illustrated.

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33 minutes ago, geordief said:

Ok.I am a bit of a literalist myself.If someone greets me in the street  with a "How are you?" my reflex is to wonder how I am and to communicate my state of being with the person

I can’t parse “Does a frame of reference have to be applicable to a potential physical  scenario to be physically valid?”

There’s an adage in particle physics that goes “that which is not forbidden is mandatory” so I can’t imagine a frame of reference that’s valid that would not somehow correspond to a physical scenario, but I don’t know if one exists, or what you might have in mind.

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5 minutes ago, swansont said:

I can’t parse “Does a frame of reference have to be applicable to a potential physical  scenario to be physically valid?”

There’s an adage in particle physics that goes “that which is not forbidden is mandatory” so I can’t imagine a frame of reference that’s valid that would not somehow correspond to a physical scenario, but I don’t know if one exists, or what you might have in mind.

It means, does a frame of reference have to be allowed to be real to be real?

And the answer is, I guess not? I do not think you understand reference frames @geordief. They are just frames of reference and don't have to apply to scenarios. You could be floating in the middle of nowhere and be a reference frame.

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14 minutes ago, swansont said:

I can’t parse “Does a frame of reference have to be applicable to a potential physical  scenario to be physically valid?”

There’s an adage in particle physics that goes “that which is not forbidden is mandatory” so I can’t imagine a frame of reference that’s valid that would not somehow correspond to a physical scenario, but I don’t know if one exists, or what you might have in mind.

Well a frame of reference can have its origin tied to a  physical point in space (even though ,I think it applies generally and not to one in particular)

If ,though the frame of reference with a spatiotemporal  origin   coinciding with a quantum object  is populated with other physical quantum objects then it seems to me that it is not simple to map their positions  onto the frame of reference

I did have a look at @MigL video and perhaps I begin to see how some circles can be squared.

So perhaps frames of reference  can apply to physical scenarios where objects' position and momentum are not separately defined?

Might one map objects' combined states  into a Minkowski like frame of reference chart? (Eg position  and momentum combined)

As to the  parsing of my phrase

,"Does a frame of reference have to be applicable to a potential physical  scenario to be  physically valid?” ....

I had thought first of writing

"Does a frame of reference have to be applicable to a potential physical  scenario to be  valid?”

Maybe the second "physically" made it less comprehensible?

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9 minutes ago, geordief said:

Well a frame of reference can have its origin tied to a  physical point in space (even though ,I think it applies generally and not to one in particular)

If ,though the frame of reference with a spatiotemporal  origin   coinciding with a quantum object  is populated with other physical quantum objects then it seems to me that it is not simple to map their positions  onto the frame of reference

I did have a look at @MigL video and perhaps I begin to see how some circles can be squared.

So perhaps frames of reference  can apply to physical scenarios where objects' position and momentum are not separately defined?

Might one map objects' combined states  into a Minkowski like frame of reference chart? (Eg position  and momentum combined)

As to the  parsing of my phrase

,"Does a frame of reference have to be applicable to a potential physical  scenario to be  physically valid?” ....

I had thought first of writing

"Does a frame of reference have to be applicable to a potential physical  scenario to be  valid?”

Maybe the second "physically" made it less comprehensible?

Frames of reference are important in relativity, where they are tied to a velocity, not necessarily a physical point. All points are at rest with respect to each other in that frame.

To rephrase your statement, can you have a valid frame that does not apply to some potential scenario? I don’t know. I can’t think of one at the moment.

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1 hour ago, swansont said:

Frames of reference are important in relativity, where they are tied to a velocity, not necessarily a physical point. All points are at rest with respect to each other in that frame.

To rephrase your statement, can you have a valid frame that does not apply to some potential scenario? I don’t know. I can’t think of one at the moment.

Can a scenario involving only  quantum objects be modeled using spacetime diagrams and their frames of reference?

Is it ever done? Would there be a need?

I understand that special relativity is used in such scenarios.

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28 minutes ago, geordief said:

Can a scenario involving only  quantum objects be modeled using spacetime diagrams and their frames of reference?

Is it ever done? Would there be a need?

I understand that special relativity is used in such scenarios.

I've never seen the SR kind of spacetime diagrams used in QM or in QFT. But a rest frame of a particle or a system is often used and selected in such a way that makes calculations easier. In such a frame, some momentum is zero, which simplifies formulas.

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