Superconductivity at room temperature??

[johanfprins]

This is precisely what is done. A bunch of muons is indeed a "piece of radioactive material". The half-life of a muon at rest is 2.2 microseconds.

Not the same.

We can take a bunch of muons at rest and measure what fraction are still there 220 22 us later. Experimental answer: 0.1% of them. We take a bunch of muons, send them off around the ring at gamma=29, and measure what fraction are still there 2.2us 22us later. Experimental answer: 80% of them.

Not that you will EVER be able to understand the subtleties in physics, but I must ask the following question: On which clock is the average lifetime of the moving muons being measured in your experiment? A clock moving with the muons, or a clock that remained stationary within the reference frame of the laboratory? It is probably difficult for your dogmatic reasoning to follow and understand, but there is a difference you know. For a change, just TRY and think! Thinking for yourself can be very rewarding you know.

 

[Ziggurat]

But no, you don't add or subtract <V(x)> from the rest-mass term. The missing mass is not missing from the electron, it's missing from the entire system. If you take all of the components of the system and perform a frame-transformation, the fact that "the system's mass includes the binding energy" will pop out naturally; you don't have to take it out by hand.

Sure, but johan doesn't want to deal with the whole system, he wants to assign everything to our electron. So I say let him, in which case we do need to put that change onto the electron. The difference won't matter since we're operating in the Newtonian limit, so we're tossing all the higher-order terms, and it will still work fine. And he will still be wrong.

 

[Ziggurat]

Not the same.

Not that you will EVER be able to understand the subtleties in physics, but I must ask the following question: On which clock is the average lifetime of the moving muons being measured in your experiment? A clock moving with the muons, or a clock that remained stationary within the reference frame of the laboratory? It is probably difficult for your dogmatic reasoning to follow and understand, but there is a difference you know. For a change, just TRY and think! Thinking for yourself can be very rewarding you know.

Wow. It's almost like you don't even know what time dilation is.

There are two clocks in this experiment: the lab clock AND the muon clock. The muon clock is given by the decay rate of the muons. In the muon's frame, their decay rate is constant. But it's not constant when measured from the lab frame. The muon clock runs slow compared to the lab clock when the muons are moving fast. That's time dilation.

 

[ben m]

Not that you will EVER be able to understand the subtleties in physics, but I must ask the following question: On which clock is the average lifetime of the moving muons being measured in your experiment?

The lab clock. There's a computer, sitting in a rack next to the accelerator. That computer sees when the bunch of muons are injected into the storage ring, and it uses its internal clock (at rest in the lab) to note the time. The computer uses its internal clock (at rest in the lab) to wait for some time to elapse. It then queries the detectors (also at rest in the lab) to measure how many muons are still going around.

80% of the muons are still there when the clocks in the lab say they've been going around for 22us.

If you take a bunch of muons at rest in the lab, when the clocks in the lab say 22us have elapsed, only 0.1% of them are still there.

 

[johanfprins]

Wow. It's almost like you don't even know what time dilation is.

There are two clocks in this experiment: the lab clock AND the muon clock. The muon clock is given by the decay rate of the muons. In the muon's frame, their decay rate is constant. But it's not constant when measured from the lab frame. The muon clock runs slow compared to the lab clock when the muons are moving fast. That's time dilation.

At last you have posted something that might prove that you have intelligence (which is superunlikely impossible). Yes, from the perspective of the laboratory, there is time-dilation, but there is no time-dilation when measuring time on the clock travelling with the muons and then later comparing this with the time elapsed on the clock that is stationary within the laboratory. Both clocks keep exactly the same time within their own inertial reference frames. If they do not do this, they violate Einstein's very first postulate on which he based his Special Theory of Relativity.

 

[johanfprins]

The lab clock. There's a computer, sitting in a rack next to the accelerator. That computer sees when the bunch of muons are injected into the storage ring, and it uses its internal clock (at rest in the lab) to note the time. The computer uses its internal clock (at rest in the lab) to wait for some time to elapse. It then queries the detectors (also at rest in the lab) to measure how many muons are still going around.

80% of the muons are still there when the clocks in the lab say they've been going around for 22us.

If you take a bunch of muons at rest in the lab, when the clocks in the lab say 22us have elapsed, only 0.1% of them are still there.

Yes! At last you have posted some sense. Congratulations after 66 pages! Maybe you will learn some physics: Although I despair.

 

[D'rok]

Yes, from the perspective of the laboratory, there is time-dilation, but there is no time-dilation when measuring time on the clock travelling with the muons and then later comparing this with the time elapsed on the clock that is stationary within the laboratory.

Wrong.

http://www.youtube.com/watch?feature=player_detailpage&v=gdRmCqylsME#t=148s

 

[ben m]

Yes, from the perspective of the laboratory, there is time-dilation but there is no time-dilation when measuring time on the clock travelling with the muons and then later comparing this with the time elapsed on the clock that is stationary within the laboratory. Both clocks keep exactly the same time within their own inertial reference frames. If they do not do this, they violate Einstein's very first postulate on which he based his Special Theory of Relativity.

Sorry, your description is not terribly clear. Can you tell me whether *my* description is consistent with (a) your description, and also with (b) mainstream schoolbook physics?

a) define Event A to be "muons enter the ring pass the detector for first time"
b) define Event B to be "muons pass the detector for the 200th time".
c) The traveling muons, using their own decays as a clock, think that 0.35 half-lives elapsed between event A and event B.
d) A collection of stationary muons in the lab, using their own decays as a clock, think that 10 half-lives elapsed between event A and event B.
e) This is called "time dilation", since t' (the A-B time as measured in one frame) differs from t (the A-B time as measured in another frame.

 

[johanfprins]

80% of the muons are still there when the clocks in the lab say they've been going around for 22us.

Correct. But after 22us on the clock travelling with the muons there will only be 0.1% of them still there within the reference frame of the muons.

If you take a bunch of muons at rest in the lab, when the clocks in the lab say 22us have elapsed, only 0.1% of them are still there.

Also correct. Exactly the same as on the clock travelling with the muons. In both reference frames the muons have exactly the same lifetimes.

Good night for now!

 

[ben m]

Correct. But after 22us on the clock travelling with the muons there will only be 0.1% of them still there within the reference frame of the muons.

Yeah. So what's the problem? What's your claim about there being something wrong with the standard approach to relativity? Is there one?

 

[johanfprins]

Yeah. So what's the problem? What's your claim about there being something wrong with the standard approach to relativity? Is there one?

My problem is with the so-called twin paradox. Although each twin is under the impression that the other twin's clock is going slower and has to take it into account when, for example, communicating with the other twin, the clocks with the two twins are actually keeping exactly the same time within their own reference frames respectively. Thus, one twin cannot actually age slower than the other twin. My conclusion is that when they meet up one day they will find that they still have the same age. I might be wrong, and therefore I have proposed a simple experiment:

My proposed experiment, which you claim is the same as the muon experiment, is not the same as the muon experiment since I am using two identical shuttles of radio-active material to act as the two twin-clocks: Then send the one radio-active shuttle on multiple journeys to and fro, and then in the end compare the two shuttles within the reference frame of the shuttle that stayed at home. I predict that they will still have the same age: i.e. that the shuttle that was sent on the journey will not have more undecayed radio-active material than the one that stayed behing: They still have exactly the same amount of non-decayed material within them.

It is a simple experiment that can be done by using a weaving loom.

 

[johanfprins]

Sure you can. There's a well-known and simple procedure for building, and synchronizing, and verifying the synchronization of, clocks at different locations in your inertial frame. This grid of clocks keeps track of the old-fashioned Newtonian time for all observers in this inertial frame. So you can measure the front and rear at the same time.

You are wrong again Ben. Within the reference frame within which the rod is stationary the front end and rear ends are simultaneously stationary at every instant in time. Since two simultaneous events in this reference frame cannot be simultaneous within a passing inertial reference frame, an observer within the latter reference frame cannot see the front and rear ends of the rod at the same instant in time on his clock. So how can he see a length contraction?

 

[johanfprins]

I find the opposite. I also find time dilation, don't you? If you have time dilation but no length-contraction, then the 1st postulate is broken.

If you apply the Lorentz transformation correctly for a pssing rod, you will find time dilation and length-stretching. Say the rod is stationary within the the refrence frame Kp, then at time tp one can choose the rear end at xp=0 and the front end at Lp. Now transform these coordinates into the reference frame K relative to which Kp is moving with a speed v, and derive the transformed length L. You will obtain that:

L=(gamma)*Lp: And that means that L>Lp: not shorter.

 

[johanfprins]

This is precisely what is done. A bunch of muons is indeed a "piece of radioactive material". The half-life of a muon at rest is 2.2 microseconds.

We can take a bunch of muons at rest and measure what fraction are still there 220 22 us later. Experimental answer: 0.1% of them. We take a bunch of muons, send them off around the ring at gamma=29, and measure what fraction are still there 2.2us 22us later. Experimental answer: 80% of them.

For the muons to reproduce my experiment, you have to generate two bunches of muons: One bunch staying behind in the laboratory and the other bunch being accelerated within the ring. Then before the muons in the laboratory have all decayed, you must stop the muons moving relative to the laboratory and then, within the laboratory's reference frame measure how many muons have not decayed in both bunches. Only then is your experiment the same as mine.

 

[johanfprins]

Originally Posted by johanfprins:
So you agree that when an electron and a proton bond to form a hydrogen atom, the electron looses mass

Sol: No, I didn't agree with that. I agreed that the mass of an isolated hydrogen atom is less than the mass of an isolated electron plus the mass of an isolated proton.

Are you now arguing that this can happen without the electron and the proton losing mass energy? This will be in violation of the conservation of energy. The fact is that you agreed, and even out of free will, stated that the mass of the hydrogen atom is less than the sum-total of the solitary rest masses of the electron and proton after light has been emitted. Does this not mean that:

(m_P+m_e)-M_H=(h*nu)/c²?

And does this not, in turn, means that both the proton and the electron must have lost mass: Say (delta)m_P and (delta)m_e?

JFP: (delta)me so that its "relativistic" energy is E=mbe=me-(delta)me. So let us substitute it into equation E2=p2*c2 +(me*c2)2 which you agree is generally valid, and solve for p2.. If you can do the mathematics, you will find that p2 is negative: i.e. that p itself is an imaginary number. So can one have an imaginary momentum at any speed and negative kinetic energy at low speed? I do not think soooo!
Sol: Actually, you just asserted that the electron "loses mass" when it's bound to a proton. If so, why are you using the original mass in your calculation? You're not even consistent with yourself.

Where did I use the original mass for the electron's energy E? I explicitly used m_e-(delta)m_e

The expression E²=p²*c² +(m_e*c²)² is "generally valid" only in the sense that it correctly gives the energy E for a free electron in empty space with any momentum p. It manifestly does not give the energy for electrons in other situations.

Then why did Dirac use this expression to postulate his so-called "relativistic wave equation for the electron"? And when he solved this equation he obtained solutions which have negative energies. So you are arguing that Dirac should have discarded his wave equation since E²=p²*c² +(m_e*c²)² is only valid for electrons with energies larger than the rest mass?

For instance, it does not include any reference to the electric potential, which by definition affects the energy of an electron, so it cannot be valid in the presence of a potential.

Will you then kindly write down the correct relativistic equation when there is such a potential? PLEASE.

That's why it doesn't correctly compute the energy of an electron bound to a proton.

Correct, since such an electron cannot have momentum and kinetic energy determined by momentum. This is what I have been stating all along. Now you are stating exactly the same after wasting everybody's time on this thread by first arguing the opposite.

 

[johanfprins]

Yes, the electron loses mass. But how much mass does it lose, johan? Can you calculate that? I can.

Then do it. Calculate how much mass (delta)m_P the proton loses and how much mass (delta)m_e the electron loses.

And if you stick in a mass of 510,961.5 eV, then you DON'T get zero or negative p².

The mass of the electron after it bonded to the proton is m_be=m_e-(delta)m_e. Where does your momentum and "kinetic energy" of the electron come from?

So how do you get the mass, johan? You can't do the calculation, can you?

Neither can you since you claim that the electron can have energy that is less than its rest mass as well as kinetic energy.

You asked us to use quantum mechanics, and now you're avoiding it like the plague.

I am not avoiding it like the plague: Schroedinger's equation is a good approximation for bound electrons as long as one is not so foolish to conclude that the bound electron can have kinetic energy while its bound rest mass energy is less than its free rest mass energy. An electron can only have kinetic energy when its energy E=m*c² is larger than its rest mass energy m_e*c² : i.e. when m>m_e. When m<m_e, as you have for a bound electron, it is impossible for the electron to have kinetic energy.

How can you test a theory if you can't even do the theory?

Well we know that we need a photon of approximately 13.7 eV to break up a hydrogen atom. This means that (delt)m_P*c²+(delta)m_e*c²=13.7 eV. Now go ahead and calculate the masses (delta)m_p and (delta)m_e as you claim that you can do.

How can you disprove a theory if you can't even make a single prediction with it?

If any theory leads to the conclusion that an entity with a mass that is less than its rest mass has kinetic energy, then I know that either the theory is wrong, or that the interpretation of this theory is wrong. The solution derived from Schroedinger's equation for Sol's famous standing waves in a box gives the energies E_n=(hbar*k_n)²/m_e)-V, which must be interpreted as an approximation for (delta)m_P+(delta)m_e: So where is your kinetic energy?

My offer still stands, despite your repeated refusal: I'll do half the calculation for you if you promise to do the other half. But you won't, because you can't. Admit you can't and I'll even do the whole thing for you.

What calculation do you want me to do? You want me to calculate <p²>. But for what purpose if a bound electron cannot have any momentum since its energy E=m*c² is less than its rest mas energy m_e*c². As Sol had to concede above, the momentum term is only physically possible when m*c² is larger than m_e*c².

 

[johanfprins]

I disagree. Remember that "electric potential" has an arbitrary zero point, and it's only for convenience that we so often say "let U=0 at infinity". Rest mass is an absolute, Lorentz-invariant, gauge-invariant quantity.

Bravo!!

This is one of the things I was trying to illustrate by bringing up the Van de Graaff. If you are in a Faraday cage stuck to the terminal of a Van de Graaff, there is no local experiment you can do to tell whether the terminal is charged up or grounded. Electrons around you appear to weigh 511 keV (and there's no spacetime transformation to problematize that statement)

I disagree. If the cage is positively charged the electrons will all be at a higher energy than they will be outside the cage and will threfore move out of the cage. If it is negatively charged, the electrons within the cage can decay into lower energy states than what they will be at outside the cage. They will then form standing waves, like the one-dimensional ones that Sol so proudly posted without understanding what he is trying to say.

These states will all have lower energies than outside the cage and will thus have lower masses than the rest mass uitside the cage.

Even though you even though they're "in fact" at the bottom of a twenty-megavolt potential well with -40 electron masses of binding energy.

If there were free electrons within the cage before you created the twenty megavolt potential well, they will all decay to form lower energy standing waves, each wave filling the cage and each wave will have a lower energy than the rest mass energy outside the cage.

 

[sol invictus]

Are you now arguing that this can happen without the electron and the proton losing mass energy?

Mass is a characteristic of isolated particles or objects. Particles like electrons cannot lose mass any more than they can lose charge - the mass of the electron is one of its defining characteristics.

Isolated hydrogen atoms have a mass too, and the mass of a H atom is less than the mass of an electron plus a proton. How to account for the difference? It's not because the electron and/or proton lost mass - that's impossible by definition. It's because of the electromagnetic interaction between the two, which, because it's attractive, necessarily contributes a (negative) energy.

But that's not the whole story - if it was, you wouldn't get the correct answer of -13.7 eV (you'd get something more negative). There's another term in the binding energy - the kinetic energy of the electron, which is (obviously) positive, but smaller in magnitude than the negative electric potential term.

Do the calculation, Johan. Wait, I forgot - you can't. You don't know how. If you could do it, you'd see precisely what I'm telling you - the difference between the energy of hydrogen at rest and the energy of an isolated electron and proton at rest comes from two terms: the electric potential times the charge, which is negative, and the kinetic energy of the electron and proton, which is non-zero and positive, but smaller than the electric term.

Go ahead, do the calculation and prove me wrong.

 

[johanfprins]

Isolated hydrogen atoms have a mass too, and the mass of a H atom is less than the mass of an electron plus a proton.

Correct! Thus the hydrogen atom's total energy is M_H*c² which is less that the sum of the total energies of the electron m_e*c² and the proton m_p*c². Thus during the formation of the hydrogen atom, mass energy equal to (m_e*c²+m_P*c²)_ MINUS_ M_P*c² has changed into light energy. So where does momentum and kinetic energy come in after this has happened? NOWHERE.

How to account for the difference? It's not because the electron and/or proton lost mass - that's impossible by definition.

Nonsense: Mass is energy and we know that when entities bond quantum mechanically mass energy is radiated away as light energy. The latter energy is the ONLY kinetic energy involved in the whole process.

It's because of the electromagnetic interaction between the two, which, because it's attractive, necessarily contributes a (negative) energy.

Correct! Mass is non-kinetic electromagnetic energy and for this reason the mass energy, which has no kinetic component internally, decreases.

Go ahead, do the calculation and prove me wrong.

What calculation? You keep on asking me to do a calculation. Tell me which one!

 

[sol invictus]

What calculation? You keep on asking me to do a calculation. Tell me which one!

Calculate the energy of hydrogen.