Quantum particles

[Upisoft]

no, they are real particles. alpha particles are pretty big. 2 protons and 2 neutrons. Basically a helium nucleus. If one were to count that as "virtual" there would be no basis I can think of for anything to not be virtual.

You didn't understand me. The same theory (QM) preducts existance of virtual particles. There have to be a reason to dismiss virtual large objects, like the invisible pink unicorns. The only reason it that they have not been observed. Nothing in the theory forbid their existance.

The thing is, we have the calculations. We can test those formulas across a range of particle sizes. We can verify that those formulas match observations. This is a testable thing.

Yes, it is testable. For particles with very small size. Extrapolating the result outside of the testing range is the same as extrapolating the theory of evolution outside its testing range.

As for the size limitations, we have found no evidence of a universal quantum size or distance. Quarks give us quantum masses, charges and a few other things, but there is no quantum distance or wavelength.

There are number of theories, but I don't remember anything observed. Anyway, there are boundaries everywhere. If you don't suppose a boundary in this case you go against the common case. So, you have to explain why there is no boundary here.

Find the Zeno's arrow paradox and think about it. You may see a reason that the space-time are quantized.

 

[Chalnoth]

You didn't understand me. The same theory (QM) preducts existance of virtual particles. There have to be a reason to dismiss virtual large objects, like the invisible pink unicorns. The only reason it that they have not been observed. Nothing in the theory forbid their existance.

Why? Large objects as virtual particles would only appear with vanishing probability. We're talking tornado through a junkyard producing a 747 type probability here.

Yes, it is testable. For particles with very small size. Extrapolating the result outside of the testing range is the same as extrapolating the theory of evolution outside its testing range.

Except that extrapolating the result to the range of large objects leads to essentially the same exact predictions as classical mechanics. What's the problem?

There are number of theories, but I don't remember anything observed. Anyway, there are boundaries everywhere. If you don't suppose a boundary in this case you go against the common case. So, you have to explain why there is no boundary here.

Boundaries? Like what?

Find the Zeno's arrow paradox and think about it. You may see a reason that the space-time are quantized.

Huh? Zeno's arrow paradox is not a paradox at all, but rather a failure to properly explain the physical scenario.

 

[Upisoft]

Why? Large objects as virtual particles would only appear with vanishing probability. We're talking tornado through a junkyard producing a 747 type probability here.

Not quite. You know the frequency of tornado appearing, say 100000 per year. Or, in other words, very rare.

Except that extrapolating the result to the range of large objects leads to essentially the same exact predictions as classical mechanics. What's the problem?

There is no interference, as you predict.

Boundaries? Like what?

OK. No boundariues. I'm quite happy with that. Therefore there is no upper limit of number of attempts for virtual unicorns to be produced. With unlimited number of attempts and NON ZERO probability(you managed to give an approximate value), there it is: the unvisible pink unicorn. Q.E.D.
By the way that happens in every possible point of space. Since there are no boundaries, even in very small volume, say size of an proton, unlimited number of parallel atempts are being made.

Huh? Zeno's arrow paradox is not a paradox at all, but rather a failure to properly explain the physical scenario.

Whatever pleases you. I just wanted to share thoughts. If you don't want to consider it as paradox, then don't.

 

[[serious]]

You didn't understand me. The same theory (QM) preducts existance of virtual particles. There have to be a reason to dismiss virtual large objects, like the invisible pink unicorns. The only reason it that they have not been observed. Nothing in the theory forbid their existance.

predicted particles in a theory don't translate to unpredicted particles. As for predicted particles, we've figured out ways to test for and confirm the existence of every QM predicted particle but gravitons. I still fail to see how validation of QM predictions would translate to the existence of non-indicated objects.

Yes, it is testable. For particles with very small size. Extrapolating the result outside of the testing range is the same as extrapolating the theory of evolution outside its testing range.

you can extrapolate data outside of what has been directly tested. That is the basis of a great deal of science. What you can't do is extrapolate it to an unrelated field. QM can predict the existance of particles because those particles would naturally follow the equations determined by looking at known particles. We can then test for the existence of these particles as we develop better testing methods and equipment. Likewise, the organization of the elements into the periodic table allowed the prediction of elements not yet discovered. They then went out and looked for those elements and found them.

There are number of theories, but I don't remember anything observed. Anyway, there are boundaries everywhere. If you don't suppose a boundary in this case you go against the common case. So, you have to explain why there is no boundary here.

Find the Zeno's arrow paradox and think about it. You may see a reason that the space-time are quantized.

Theories have been tossed around, but quantum space/time has never been validated.

As for the arrow paradox, it's only a paradox if you start assuming quantum time and extrapolate to quantum space. If you allow for continuous time, there is no need for non continuous space.

 

[Chalnoth]

Whatever pleases you. I just wanted to share thoughts. If you don't want to consider it as paradox, then don't.

It's not that I don't want to consider it a paradox. It just isn't one. It's a failure to understand the physical situation. In particular, there is a claim made that the arrow is stationary at every instant in time: this is patently false (because momentum is a physical quantity). The arrow wouldn't even be stationary in discretized space-time due to the fact that it would still have momentum at every space-time point is crosses.

 

[Upisoft]

predicted particles in a theory don't translate to unpredicted particles. As for predicted particles, we've figured out ways to test for and confirm the existence of every QM predicted particle but gravitons. I still fail to see how validation of QM predictions would translate to the existence of non-indicated objects.

Which is "non-indicated" objects?

you can extrapolate data outside of what has been directly tested.

Yes, you can, but you may not.

What you can't do is extrapolate it to an unrelated field.

The problem is that fields are not properly separated and defined. For my example with abiogenesis, one should have very good knowledge on what is life.

QM can predict the existance of particles because those particles would naturally follow the equations determined by looking at known particles. We can then test for the existence of these particles as we develop better testing methods and equipment. Likewise, the organization of the elements into the periodic table allowed the prediction of elements not yet discovered. They then went out and looked for those elements and found them.
Theories have been tossed around, but quantum space/time has never been validated.

I'm still waiting to see an observation of graviton. And observation of virtual particle.

As for the arrow paradox, it's only a paradox if you start assuming quantum time and extrapolate to quantum space. If you allow for continuous time, there is no need for non continuous space.

Yes, there is. The arrow is never in movement. In every moment the arrow has definite position and in the same moment the arrow is not moving.

 

[Upisoft]

It's not that I don't want to consider it a paradox. It just isn't one. It's a failure to understand the physical situation. In particular, there is a claim made that the arrow is stationary at every instant in time: this is patently false (because momentum is a physical quantity). The arrow wouldn't even be stationary in discretized space-time due to the fact that it would still have momentum at every space-time point is crosses.

The arrow is stationary. It does not travel any distance during the moment. You can't even distinguish two arrows with different speeds in that moment. You'll be unable to say which one is "traveling" faster in this moment.

 

[[serious]]

Which is "non-indicated" objects?

your invisible unicorn

The problem is that fields are not properly separated and defined. For my example with abiogenesis, one should have very good knowledge on what is life.

This would be a case where knowledge of a related field, biochemistry for example, is necessary for understanding aspects of two more specific fields, evolution and origins.

I'm still waiting to see an observation of graviton. And observation of virtual particle.

Every photon you see can arguably be considered virtual. We also have done tests for more classic virtual particles (I'll hunt around for the papers if you want)

Yes, there is. The arrow is never in movement. In every moment the arrow has definite position and in the same moment the arrow is not moving.

this ignores that the arrow has momentum at every point. If you want momentary measurement of this momentum, you can look for red/blue shift, mass/length/time distortions, etc.

 

[Chalnoth]

The arrow is stationary. It does not travel any distance during the moment. You can't even distinguish two arrows with different speeds in that moment. You'll be unable to say which one is "traveling" faster in this moment.

One has more momentum than the other. That's all there is to it.

 

[Upisoft]

your invisible unicorn

It's made of virtual particles. What is the problem?

Every photon you see can arguably be considered virtual. We also have done tests for more classic virtual particles (I'll hunt around for the papers if you want)

Oh, no! Every photon I see is real photon. The virtual photons are being exchanged during the electromagnetic interactions. It doesn't work this way. You cannot use different name and start call real photons "virtual". There are probably theories that say every particle is virtual and probably that our universe is virtual, but we're talking about QM here.

this ignores that the arrow has momentum at every point. If you want momentary measurement of this momentum, you can look for red/blue shift, mass/length/time distortions, etc.

Yes, of course. The momentum, or p = mv. Where v = lim (Δt->0) Δs / Δt. And, the limit is approached by values non-equal to zero. I.e Δt =/= 0. But in this moment in time Δt = 0. Where is your momentum? It is not defined.

 

[[serious]]

It's made of virtual particles. What is the problem?

the physics of such a beast

Oh, no! Every photon I see is real photon. The virtual photons are being exchanged during the electromagnetic interactions. It doesn't work this way. You cannot use different name and start call real photons "virtual". There are probably theories that say every particle is virtual and probably that our universe is virtual, but we're talking about QM here.

every photon is emitted somewhere, absorbed somewhere else, and, from the frame of referance of the photon, traverses the distance instantaneously. "virtual" photons are no different that "real" photons. The only difference is how we are used to treating them. If you see a car backing up instead of moving forward, it's still a car and it's certainly still real. Likewise, if a photon tunneling event occurs between two charged particles, it's still a photon, same as if the tunneling event occurred between two prisms.

Yes, of course. The momentum, or p = mv. Where v = lim (Δt->0) Δs / Δt. And, the limit is approached by values non-equal to zero. I.e Δt =/= 0. But in this moment in time Δt = 0. Where is your momentum? It is not defined.

Ah, fun with math. let's break this up:

p=mv: We will ignore the fine detail provided by QM and stick with the classical physics equation for momentum.

v = lim (Δt->0) Δs / Δt: That will work, I'd simplify to ds/dt,but you are obviously free to muck around with limits if you wish.

Δt = 0: No good. As soon as you assume the limit is what the value is set to, it's no longer a limit and the limit calculation is no longer applicable. Furthermore, since Δs->0 as Δt->0, this would simply yield 0/0 Aside from dividing by zero being disallowed (if you allow it, you can prove 2=1) it would be much easier to get here by simply multiplying by 0/0.

My high school math teacher would drop dead just so he could roll in his grave if I missed a divide by zero error

 

[Upisoft]

the physics of such a beast

every photon is emitted somewhere, absorbed somewhere else, and, from the frame of referance of the photon, traverses the distance instantaneously. "virtual" photons are no different that "real" photons.

Oh, yes, they are different. They cannot be detected.

Ah, fun with math. let's break this up:

p=mv: We will ignore the fine detail provided by QM and stick with the classical physics equation for momentum.

Well, we were talking about an arrow

v = lim (Δt->0) Δs / Δt: That will work, I'd simplify to ds/dt,but you are obviously free to muck around with limits if you wish.

There is no difference. I chose the 'limit' form, because it shows where the idea come from and how it was defined.

Δt = 0: No good. As soon as you assume the limit is what the value is set to, it's no longer a limit and the limit calculation is no longer applicable.

Exactly. However I do not assume that Δt = 0. It is zero. The moment has no length. If Δt =/=0 then there are two different moments t0 and t1, along with all infinite number of moments in between. And then Δt = t1-t0 =/= 0.

Furthermore, since Δs->0 as Δt->0, this would simply yield 0/0 Aside from dividing by zero being disallowed (if you allow it, you can prove 2=1) it would be much easier to get here by simply multiplying by 0/0.

My high school math teacher would drop dead just so he could roll in his grave if I missed a divide by zero error

Correct. 0/0 is undefined. Therefore, you either have to drop the idea that the momentum is defined, or you have to look at the movement in time, not in a single moment.

 

[Chalnoth]

Correct. 0/0 is undefined. Therefore, you either have to drop the idea that the momentum is defined, or you have to look at the movement in time, not in a single moment.

Nope. You just consider momentum to be a conserved quantity, a property of an object at a specific time. This is the way things are done in Hamiltonian dynamics.

 

[Upisoft]

Nope. You just consider momentum to be a conserved quantity, a property of an object at a specific time. This is the way things are done in Hamiltonian dynamics.

And how this property could be measured in a single moment?

 

[Chalnoth]

And how this property could be measured in a single moment?

There are a few ways:
1. By observing the object's redshift/blueshift.
2. By observing the object's length (assuming we know its rest length).
3. By observing the results of collisions with other objects.

Understand that just observing an object's velocity is a very poor measure of momentum when the momentum gets very high. At high momenta, all objects essentially move at the speed of light.

 

[Upisoft]

There are a few ways:
1. By observing the object's redshift/blueshift.

If the object is orbiting your POV what happens?

2. By observing the object's length (assuming we know its rest length).

Needs at least two measurements in different moments. One when the object is in rest.

3. By observing the results of collisions with other objects.

That also requires time.

Understand that just observing an object's velocity is a very poor measure of momentum when the momentum gets very high. At high momenta, all objects essentially move at the speed of light.

Let's first unravel the simpler case with the arrow traveling at normal speeds.

 

[Chalnoth]

If the object is orbiting your POV what happens?

Then that particular form of measurement doesn't work. Just as measuring the position doesn't help if you're only looking at the object at one time.

Needs at least two measurements in different moments. One when the object is in rest.

So what?

That also requires time.

As far as we know, particle interactions happen instantaneously. Measuring the output of a particle interaction gives us information about the energy and momentum of the particles that went into the interaction.

But regardless, you're moving the goalposts here. Your original assertion was that an object at a particular instant of time doesn't have a property called velocity, that one needs to look at a finite span of time to observe such a thing. I (and dantose) gave a number of ways which you could use to find the velocity of a particle at an instant in time, and you totally changed the definition of the problem.

Let's first unravel the simpler case with the arrow traveling at normal speeds.

That's a pointless case for discretized space-time, though. It'd rather be like using a thought experiment involving bees to try to find out about lions. If space-time is discretized, the discretization is so minute that an arrow will never notice it: the arrow is much, much larger than the discretization. Heck, an atomic nucleus is much larger than the discretization (by a factor of about 10^20, actually).

If you're going to worry about how to describe discretized space-time, worry about single particles, not big composite objects like arrows moving at nonrelativistic speeds.

 

[Upisoft]

So what?

There is no difference. Two measurements in different moments are required to have the knowlege of momentum.
And, by the way, what means 'at rest' in your definition? Does it mean 'with relative speed = 0'? I thought you tried to explain how one could measure momentum independent of speed.

As far as we know, particle interactions happen instantaneously. Measuring the output of a particle interaction gives us information about the energy and momentum of the particles that went into the interaction.

You transfer the problem here. Instead measuring the momentum of original particle now you have to measure momentum of the particles collides with the original one. And that also does not save you from having to measure the new momentum of the original particle.

But regardless, you're moving the goalposts here.

You're doing it. By trying to show how one could observe differential value by single measurement. Thus, you even try to hide one of the goalposts.

Your original assertion was that an object at a particular instant of time doesn't have a property called velocity, that one needs to look at a finite span of time to observe such a thing. I (and dantose) gave a number of ways which you could use to find the velocity of a particle at an instant in time, and you totally changed the definition of the problem.

Hiding one of the measurements does not make it irrelevant.

If space-time is discretized, the discretization is so minute that an arrow will never notice it: the arrow is much, much larger than the discretization. Heck, an atomic nucleus is much larger than the discretization.

Probably this is the case. For example, let's look at an random process, like favorite of Creationists radioactive decay. If one supposes linear time, then each decaying atom must have some information about some time interval to "decide" if it should decay. Quantized time explanation is simpler. Just in each time step each atom has intrinsic probability to decay.

If you're going to worry about how to describe discretized space-time, worry about single particles, not big composite objects like arrows moving at nonrelativistic speeds.

I'm not worried about single particles.They would be just fine.

 

[[serious]]

Exactly. However I do not assume that Δt = 0. It is zero. The moment has no length. If Δt =/=0 then there are two different moments t0 and t1, along with all infinite number of moments in between. And then Δt = t1-t0 =/= 0.

Δt cannot be arbitrarily set to 0. Δt represents the instantaneous change in time. This is the whole function of taking the limit. Remember, the limit equation is the whole "infinitely small slice" idea. It does not require two moments to have a value, it's whole point is to NOT require two moments for calculation. In fact, separating an arrow from the flow of time would likely only possible if the arrow was traveling at the speed of light. Moving at the speed of light would, of course, mean that it had quite a bit of momentum and would also require the more detailed momentum calculation. If we are dipping into this level of physics, we will have to also deal with the uncertainty principle.

Correct. 0/0 is undefined. Therefore, you either have to drop the idea that the momentum is defined, or you have to look at the movement in time, not in a single moment.

now you're fighting all of calculus! any equation can be brought to 0/0 if dividing by 0 is allowed. That's why dividing by 0 isn't allowed.

Let me give an example:
A=X set some number X equal to A
A+X=2X add x to both sides
X-A=2(X-A) subtract 2A from both sides
1=2 divide both sides by (X-A)

 

[Chalnoth]

Upisoft, you're totally missing the point. You attempted to argue that an object does not have a property called velocity at one instant in time. This is false. It is false because there are effects of velocity (more specifically momentum) that can be observed at single instants in time.

It doesn't matter that one has to know other things about the object at different times to make sense of the measurement of the instantaneous velocity of the object in question. Your argument was that one cannot measure the position more than once at the same point in time and infer any information about the velocity. There are other ways to measure the momentum/velocity that are completely independent of position. Even if these methods require more than one measurement of the object at different times, they aren't comparing positions at different times, and thus are genuinely observing instantaneous velocity.