1. I was wondering since cern is able to measure the velocity of particles before and after a collision their kinetic enegy and mass being known before and after a collision would it not be possible to measure both the velocity and position that an electron had in the nucleus of an atom before a collision using calculus thus disproving heisenberg?  2.

3. The Heisenberg Principle only says that two complimenting values like position and momentum cannot be measured simultaneously. You can always do before-after measurements, however.  4. yes and because you can do the before and after calculations you can find both position and momentum. now repeat the same experiment with no variables and at your calculated instant say 10 mins exactly you will know both velocity and position of the particle therefore both have been measured at the same time and heisenberg is wrong.  5. Wrong. you don't understand the HUP or quantum physics if you can state that.  6. why explain it to me. as i understand it heisenberg said something about the movement of the electron being uncertain and as he explained it if you used something to determine the position of an electron that you would alter its momentum and so knowing one would make the other more uncertain and that all experiments would be subject to a small degree of error. which is alright but it being impossible to know both position and velocity is a bit far i dont like impossibilities. hup works out your degree for error which is okay but i also think it is possible yto design experiments without this degree of error. take the one i imagined as an example only il change it a little and use heisenbergs microscope idea he thought that if you were to look at an electron using a microscope then a photon of energy would hit the particle and because you do not know the position of the electron (or the energy it has) then you cannot work out the momentum and the more you focus in on the particle the more the momentum will change and the greater your degree of error will be in calculating its momentum. All i did is suggest that if you know the position of the particle and the energy that it has then when a photon of known energy and velocuty strikes it it then its momentum will change by a known value if the experiment is set up correctly then there should be zero percentage for error and therefore no uncertainty the only problem i saw with this hypothesis then was about the atom itself if you were to do the experiment the way i suggested then the two atoms you use must be exactly the same in every way to do this i would reckon that the atom would have to be created artificially to ensure that all the variables would remain constant including electon spin,quark composition etc. if i am still wrong please explain i would rather be wrong and learn than never ask questions in the first place and never learn anything.    8. Sure i dont mind im typing this off a phone and the keyboards a pain to use, sorry. Plus the touchscreen is a pain to use when trying to select things in paragraphs and correct spelling.Im asking you to explain this to me because i dont know everthing, i never think of everything but i always question everything. Firstly, heisenberg used the example of the microscope to explain his theory. He suggested that by viewing an electron through a microscope to determine position the momentum of the electon would change when struck by the photon of light. He suggested that the more accurately you measure the position of the particle, the more the momentum of the particle would change and the less accurate your calculation of the momentum would be considering you have no idea of the location of the electon at the time of photon impact or its velocity.What i suggested is that if you devise an experiment. Where you can know everthing about the atom you are studying because it would be artificially created and you know the energy of the electron at the time in question along with its velocity. When you change the electons momentum with a photon of a known amount of light energy. Then you can also work out with certainty the momentum the particle will have after the collision. So long as no other external variable is capable of interfering with your results then you should be able to calculate with certainty both position and momentum (at least in theory anyway).  9. wow this phone took my paragraphs out when i posted sorry i will type it when im near a computer with internet access if you want but it might take me a while to find one.  10. Originally Posted by fiveworlds yes and because you can do the before and after calculations you can find both position and momentum. now repeat the same experiment with no variables and at your calculated instant say 10 mins exactly you will know both velocity and position of the particle therefore both have been measured at the same time and heisenberg is wrong.
You are confusing calculation with measurement. Of course you can calculate anything you wish in terms of probabilities, in that you are correct. However, you cannot physically measure or observe with arbitrary precision. These two are very distinct things, you know. You can only know the exact state of a system once you perform a measurement though, since this is quantum mechanics, and not classical mechanics.  11. Originally Posted by fiveworlds which is alright but it being impossible to know both position and velocity is a bit far i dont like impossibilities.
You are omitting the key point of simultaneity. You can measure position and velocity of a particle with arbitrary precision, no problem, just not simultaneously. Of course you can do calculations, but then again, this being quantum mechanics we are only talking probabilities. You still wouldn't actually know anything until you perform a measurement.  12. You see what i am doing is very simple. I am taking the fact that this is quantum mechanics. Then i am reducing that down to classical mechanics by eliminating all unknown variables and so reducing this experiment down to classical mechanics action-reaction basis. For everyy action there is an equal and opposite reaction. so therefore if i keep my action the same the reaction that happens must also be the same. it is like when looking at a straight-edged triangle you know its angles add to 180 degrees. So logically in my experiment if i take the same atom and hit it with the same photon of energy then my calculations for the probable position of a particle within that atom will be exactly the same. Now if this is a simple question of the exact position of that particle before the collision and its velocity it is simple to continue testing. By continued testing all you would need to do is change the direction of your incident photon of energy keeping all other values the same and you wil get a slightly different value for the position of that particle and the velocity it attains afterward. Do this enough times and you can find the intersection of the probable values which if done enough time will give you its exact position and velocity as the probability of it being anywhere else will become so small it can be ignored. This is like the toss a coin test for the first 10 you might get the probability of tails to be 0.2 but with enough trials your probability will creep closer and closer to the 0.5 variable. Sure you may never actually hit the 0.5 variable but you should be able to get it so close as to be irrelevant.  13. Originally Posted by fiveworlds Then i am reducing that down to classical mechanics by eliminating all unknown variables and so reducing this experiment down to classical mechanics action-reaction basis.
I am afraid that is not possible, because the following premises is wrong :

so therefore if i keep my action the same the reaction that happens must also be the same.
In QM, this isn't necessarily the case because we are now dealing with probability amplitudes. As mentioned before, you do not know what state a system is in until you perform a measurement, thereby collapsing the wave function.

keeping all other values
Again, this bit is impossible since we are dealing with probabilities. In fact, the mere act of sending a photon into an atom will already alter the system, so this can only be done once.

Is your main idea to disprove the HUP ? Surely you do know that this principle is a direct mathematical consequence of Fourier transformations and their properties, do you ? Trying to disprove HUP is equivalent to saying that Fourier transformations are not a suitable representation of the original function. It is a well established mathematical fact that the more concentrated a function f is, the more spread out its Fourier transform becomes, and the HUP is a direct result of this.
Just thought I mention that to you, before you put any more effort in trying to disprove this.  14. Oh alright i will leave it be for now. I will learn a bit about fourier (never heard of him) and i might come back to later. i probably have about 70 years to figure it out so whether i am wrong is irelevant i have plenty of time to figure stuff out thanks for the help.  15. Originally Posted by fiveworlds Oh alright i will leave it be for now. I will learn a bit about fourier (never heard of him) and i might come back to later. i probably have about 70 years to figure it out so whether i am wrong is irelevant i have plenty of time to figure stuff out thanks for the help.
As Markus has said, the HUP derives directly from the fact that momentum and position wavefunctions are (proportional to) Fourier transforms of each other. The uncertainty principle is a fundamental mathematical consequence of this relationship; it has nothing to do with quantum mechanics per se.

Consider a pure sinewave function of time that has always existed and will always exist. We can state its frequency with perfect certainty, but we can't localize where in time it is. Or we could consider a very narrow impulse-like function of time, allowing us to say with great precision when it occurred. However, its spectral content (its Fourier spectrum) is spread out, so we can't assign a frequency to it; we can only say things like "20% of the energy lies somewhere between frequency f1 and f2." This uncertainty is inherent in the mathematics, and is independent of quantum theory.

To argue successfully against the HUP in quantum theory, you would have to show that QM is flawed at such a fundamental level that momentum and position wavefunctions are not Fourier transform pairs. That's a very tall order.  Bookmarks
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