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Modern vs Classical Physics

  1. Apr 26, 2005 #1
    Hoping for some input on the following:

    In the development of what is now known as classical physics, there was a tacit assumption that the universe was governed by laws, which we did not know, but attempted to understand.

    In modern physics, most notably with Relativity and Quantum Mechanics, it seems that this philosophy changed. Instead of assuming that the universe acted in a certain way, regardless of how we observed it, we began treating the universe as acting in a way consistant with how it is measured.

    To more aptly illustrate my point consider the following:
    Werner Heisenberg proposed that we could not measure both the position and momentum of a particle simoultaneously.
    Classically speaking, one might say, a particle has a definite position and momentum at a given moment, but any attempt to measure one will skew the measurement of the other.
    Modern Physics, specifically Quantum Mechanics, seems to insist that the particle does NOT have a definite position and momentum at a given moment.
    Instead, the view became that a particle is actually a "likelihood of positions and momentums."

    In an attempt to simplify my point, I will abuse the language a bit and say:

    Classical View:
    Modern View:
    I don't mean to suggest that either view is invalid. Certainly, physics is a pragmatic science, so we should expect that both of the above statements should hold true. But, if I am correct, it does seem to represent a philisophical difference in regards to the approach.

    Am I mistaking something about the interpretation of modern physics, or overlooking something in the development of classical physics? In any case, input is appreciated.
     
  2. jcsd
  3. Apr 26, 2005 #2
    In fact it seems to me you're speaking about

    1) classical QM
    2) modern QM

    I will first do just classical and quantum mechanics :

    a) Well I think in QM you can measure p and q simultaneously in the sense : if you measure them simultaneously, then you get an uncertainty. Interprete this as : taking the same configuration (wf in qm) if you measure position p', then momentum can have several values q' (uncertainty) in any case (there exist no wf for this is not true)

    b) Classically : if you take the same configuration (same initial conditions, same forces), then at a certain time at a given position you have only one p', there are no other possibility.

    The remarks are :


    1) However if you take commutating observable in QM, you can deduce the value of one knowing the other measurement value. So that in fact the philosophical position of classical mechanics is a special case of the possibilities given by QM....

    2) the time specifically comes into classical mechanics to distinguish them. This should have something to do with the special nature of time in QM

    3) I think QM gives no indication about "classical" interpretation (one measurement disturbs the other, but particle had a precise value for both before measurement). QM just affirms : generally before measurement the particle is in a superposition of states.

    I think there is somewhere in this forum an indication about Feynman Lectures III that gives indication about why in QM a physical quantity has no meaning considered independtly from a measurement (or something equivalent)
     
  4. May 1, 2005 #3
    I think that the real difference between quantum mechanics and general relativity is that quantum mechanics is not only abstract but also deals with the interactions of absolute particles.

    Also quantum mechanics rejects the existence of physical space, existing seperately from absolute matter. I think that's the physical error in relativity. Space cannot exist seperately from matter. The relative is abstract (in our head) and depends on the absolute.

    Still not convinced? Check this site...

    The contradiction is when one introduces a time dimension, the possibility of motion is automically prevented. There is no change/motion in the time axis because time is an invariant by definition.
     
    Last edited: May 1, 2005
  5. May 4, 2005 #4

    Tom Mattson

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    That guy is obviously a crackpot. From his site:

    First of all, the velocity is not [itex]\frac{dt}{dt}[/itex]. It's [itex]\frac{dt}{d\tau }[/itex], which is not 1. Second, the expression [itex]\frac{dt}{dt}[/itex] is not "self referential". That's just stupid.

    Well, he's right about one thing: The units do cancel out when you compute the speed along the time axis. But that's only because we're working in natural units, in which speed is dimensionless! If you prefer to work in SI units, then the speed in the time direction is [itex]\frac{d(ct)}{d\tau}[/itex].
     
    Last edited: May 4, 2005
  6. May 4, 2005 #5
    If you look at the equations, let say Schroedinger for QM : it's an equation for [tex]\Psi [/tex] an abstract mathematical wavefunction which has a probabilistic interpretation....of course it can be defined on space-time.

    GR is an equation for space-time itself (the metric)....no prob. of find anything here.

    So the Graal would be : how to find an equation which governs space-time AND the wave-function depending on this space-time....with of course self-interaction possible....
     
  7. May 4, 2005 #6
    What is the difference between [itex]v = \frac{dt}{dt}[/itex] and [itex]v = \frac{dt}{d\tau}[/itex]?

    Proper time is time measured when the clock is at rest relative to the observer.

    [tex]d\tau = \frac{dt}{\gamma}[/tex]

    or

    [tex]d\tau = dt\sqrt{1-[\frac{v(t)}{c}]^{2}}[/tex]

    He writes that tau is also an invariant evolution parameter that is used in physics as a calculational tool with which to describe a rate of motion or change (see time dilation).

    The difference is that spaces in (non-relativstic) quantum mechanics are also abstract. In relativity i'm not sure.
     
    Last edited: May 4, 2005
  8. May 4, 2005 #7

    Tom Mattson

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    The first one equals 1 and the other doesn't.

    Pardon my bluntness, but I could not care less about what he writes. He doesn't know what he's talking about. I strongly recommend you switch to a site such as the following:

    http://farside.ph.utexas.edu/teaching/jk1/lectures/node15.html
     
  9. May 4, 2005 #8
    Sorry and we could care less about you write. What he writes is not only logical but is also supported by evidence from real life. 4-velocity is not even a physical velocity so why do we call it velocity.

    There is no evidence for the physical existence of spacetime and definitely not for motion in spacetime. Get a grip man.
     
  10. May 4, 2005 #9

    Tom Mattson

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    The guy obviously does not understand relativity, which is what he's critiquing.
     
  11. May 4, 2005 #10
    He doesn't have to. All he has is to understand the meaning of time, which you don't.
     
  12. May 4, 2005 #11

    Tom Mattson

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    Of course, one does have to understand a theory in order to make a sensible critique of it.

    And it remains to be seen that he does.

    It's remarkable that you could reach that conclusion, because I have not posted my thoughts on the subject.

    Starship, you are going to find yourself banned from here very quickly if you do not stop posting bald assertions and ad hominem argumentation. Stop it.
     
  13. May 4, 2005 #12
    And he almost surely understands it better than you. Relativity does not allow motion in space-time. Why not? Because time is an abstract invariant.

    It remains to be seen whether time travel is possible, as you say it is.

    How can you say that something is possible if the physical evidence does not support such thing? Everything works on energy, can't you see that?

    The laws of physics (especially thermodynamics) do not even distinguish between our past, present and future. They're all abstract.
     
  14. May 4, 2005 #13

    Tom Mattson

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    If you mean that it is not possible to define 4-velocity in relativity, then you are mistaken. I've already explained two of the fundamental errors in the website you posted, and I posted a website that treats the subject correctly. I can lead you to water, but I can't make you drink.

    You need to stop putting words in my mouth. I never said that. What I will say is that 4-velocity is defined in SR.

    Show me precisely where I made which claim, and I will tell you why I made it. But don't keep blathering on about some claim you think I made.

    This sentence makes absolutely no sense in the context of any known physical theory.

    I wonder why there are irreversible processes then. :rolleyes:

    Kid, you need to bring up the quality of your posts. You aren't making any sense, you are quoting obvious cranks as authorities, and you are making one strawman argument after another. It's really getting boring.
     
  15. May 4, 2005 #14
    It's possible to define a 4-vector in special relativity but it's not a physical velocity:

    [tex]g_{\mu\nu}\eta^{\mu}\eta_{\mu} = \gamma^2 c^2 (1-\frac{v^2}{c^2}) = c^2[/tex]

    If you don't believe in time travel then why are you are arguing with me?

    Particle physics says that there are only particles and their interactions. So why does general relativity say otherwise?

    In the microscopic level almost all processes are time-symmetric. At the macroscopic level it's not so.

    You need to stop dreaming about time travel.

    These cranks obviously make more sense than you.
     
  16. May 4, 2005 #15

    Tom Mattson

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    Why not? It's proportional to the 4-momentum, the conservation of which determines the dynamics of a moving body.

    I'm arguing with you because you put that crackpot website forward as a convincing argument, when the guy obviously does not know what he is talking about.

    Particle physics does not say that. Particle physics describes particles and their interactions yes, but it doesn't also include the statement "and nothing else exists". Furthermore, particle physics doesn't address gravity at all. So why on Earth would you expect that GR says nothing in addition to particle physics?

    And thermodynamics is macroscopic, which is why you are wrong about that.

    There you go again. All I have said is that 4-velocity is a feature of SR. I don't even know what you mean when you say "time travel".

    Perhaps you wouldn't mind explaining why then, just once.
     
  17. May 4, 2005 #16
    Like i said, the velocity 4-vector is not a physical velocity. I guess we just have to agree to disagree.

    You are wrong. What the guy said is that space-time is a mathematical construct and he's definitely not wrong about that. There's no time axis, it's abstract (non-physical).

    Do you have any evidence that something else exists? Is it physical? I don't think so.

    The physical mechanism of gravity is still unknown but we'll find it out sooner or later. But in order to that we'll have to free ourselves from abstract concepts such as space and time.

    If the laws of thermodynamics are irreservable (like conservation of energy), what makes you think there is an arrow of time at all? In fact, irreversibility proves there is no arrow of time. It's abstract.

    How was Savain wrong when he said that motion in space-time is impossible?

    In fact, i think he was right in everything he said. There is no evidence for the physical existence of a time axis.
     
    Last edited: May 4, 2005
  18. May 5, 2005 #17

    Tom Mattson

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    You can say it all you want, but without one iota of valid argumentation to back it up all you've got is a bald assertion.

    He is wrong about the things I pointed out, which are the invalidity of the unitless velocity, motion along the time axis = dt/dt, and self-reference.

    Yes, the space-time of SR is a mathematical construct. So is any other theoretical device used in physics. The question is, "How well does that mathematical construct map onto observable reality?" The answer is that SR maps very well onto it.

    Guess what? The x-, y-, and z-axes are abstract, too. The time axis is no more or less real than the others.

    Something else besides what is included in the standard model of particle physics? Of course I do. It's called "gravity".

    Oh for Pete's sake! Please look up "irreversible"! Irreversible procceses suggest that there is an arrow of time. And by the way, I'm not talking about abstract laws. I'm talking about actual physical process.

    Assuming Savain is the guy who wrote that website, I've already explained why his argument is faulty. I have neither the time nor the inclination to repeat it.

    If you think that he's right in everything he said, then you are either unwilling or unable to understand the obvious mistakes I pointed out, which is too bad.
     
  19. May 6, 2005 #18
    Where is the valid experimental argumentation for space-time, time travel etc?

    First of all velocity is a vector and a vector has both magnitude and direction but distance & direction are both abstract (non-physical).

    Then you understand that gravity has nothing to do with the curvature of a physical space-time?

    What do you mean by devices? The devices we use are absolute.

    The theory is mathematically accurate and makes correct predictions but the flaw in it is that it describes metric spaces as physical, instead of abstract.

    Yes, because metric tensors are abstract.

    There is no problem i think. Gravity is an effect generated by a series of particles interaction with the CMBR. Particles on an elementary level have intrinsic vibration states, that is internal motion (change in location) capability. This is confirmed with the http://frontiernet.net/~mgh1/ [Broken] of the electrons.

    How so? In fact the irreversibility of physical processes prohibits a physical arrow of time.

    And we cannot reverse physical processes right? Matter on earth cannot fall up, can it?

    Louis is completely correct. Particles, their properties and their interactions is all that exists. All the rest is abstract (non-physical).
     
    Last edited by a moderator: May 2, 2017
  20. May 6, 2005 #19
    So you mean space and time are non-physical in theirselves, (without actually having a ruler and a watch ?)

    So a particle can be somewhere in space at some time only if it has a watch and a ruler ?

    Or you go back to the dilemma among scientists during WWII ?
     
  21. May 6, 2005 #20
    What i meant is that space-time is abstract. In relativity, we make extensive use of metric tensors. A tensor is a generalization of a vector to higher dimensions. Vectors have both magnitude and direction. In QFT spinors are used instead of tensors.

    In general relativity, metric spaces (like Hilbert spaces in QM) are only globally definied. They are abstract and therefore are subject to different interpretations. See also global analysis.
     
    Last edited: May 6, 2005
  22. May 6, 2005 #21
    Because you mean mass is not subject to different interpretations (Newton : inertia towards movement or quantity of matter ?)...Those diverging interpretation were solved by the weak equivalence principle which leads to GR...so what is your point ?
     
  23. May 6, 2005 #22

    Tom Mattson

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    In case you've lost track of this part of the discussion, it pertained to the existence of the 4-velocity, which is all I've claimed so far. I cannot believe that you are still asking me about "time travel", when you are (still!) the only one of us to have ever brought it up. :rolleyes:

    As for the 4-velocity, I say that the evidence for it is one and the same with the evidence for relativistic dynamics (which explictly contains the first derivative of the 4-velocity) or for relativistic energy-momentum conservation (which explicitly contains the 4-velocity itself).

    Your comment doees not address the cited errors in the slightest.

    Non-sequitir.

    By "theoretical device" I mean "mathematical construct". That's the bread and butter of theoretical physics. And I have no idea of what you mean by "absolute" in this context.

    No, what SR/GR does is makes falsifiable predictions about the real universe, and it does so exceedingly well. That's science, so you should get used to it.

    ...as are all of the other concepts of theoretical physics.

    :rofl:

    So now you know what gravity is? And you know that your view is backed up by the Young's experiment with electrons?

    Send it to Physical Review Letters. Then we'll talk.

    Didn't I urge you to look up "irriversible"? :confused:

    Irreversibility is the indication of temporal asymmetry in nature, which is colloquially referred to as "the arrow of time".

    Bingo. Hence the temporal asymmetry!

    Well then I regret to inform you that you are every bit the crackpot he is.

    For the last time, he is not "completely correct", and the following are evidence of that obvious fact:


    1. Contrary to what Louis says, SR does not predict that the velocity of a body along the t-axis is dt/dt=1.

    2. Contrary to what Louis says, it is not the case that there is a violation of any principle of physics to refer to a unitless velocity. A trivial choice of units can achieve this.

    3. Contrary to what Louis says, the statement dt/dt=1 is not self-referential. Louis (and Starship) desperately needs to look that term up in a textbook on logic.

    The simple truth is that not one thing that Louis uses to support his "arguments" is a fact.
     
    Last edited by a moderator: May 2, 2017
  24. May 6, 2005 #23

    Tom Mattson

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    Gee, what spinor is used to describe photons? :rolleyes:

    (Hint: Spinors are used for fermions, but good old tensors are still used to describe bosons).
     
  25. May 6, 2005 #24

    dextercioby

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    This one

    [tex] \left[\xi(x)\right]_{\alpha} \ ^{\dot{\beta}}\rightarrow^{\Lambda}\left[\xi'(x')\right]_{\alpha'} \ ^{\dot{\beta}'}=\left[\mathcal{D}^{\left(\frac{1}{2},0\right)}(\Lambda)\right]_{\alpha'} \ ^{\alpha}\left[\mathcal{D}^{\left(0,\frac{1}{2}\right)}(\Lambda)\right]^{\dot{\beta}'} \ _{\dot{\beta}}\left[\xi(x)\right]_{\alpha} \ ^{\dot{\beta}} [/tex]

    [tex]=\left\{\mathcal{D}^{\left(\frac{1}{2},0\right)}(\Lambda)\xi (x)\left[\mathcal{D}^{\left(0,\frac{1}{2}\right)}(\Lambda)\right]^{T}\right\}_{\alpha'} \ ^{\dot{\beta}'} [/tex]

    The [itex] 2\times 2[/itex] components of the spinor field [itex] \xi [/itex] can be mapped into the 4 components of a vector field by forming

    [tex] \left[\xi(x)\right]^{\mu}=:c_{\dot{\beta}\dot{\beta}'}\left(\tilde{\sigma}^{\mu}\right)^{\dot{\beta}' \alpha}\left[\xi(x)\right]_{\alpha} \ ^{\dot{\beta}}=\mbox{Tr}\left[c\tilde{\sigma}^{\mu}\xi(x)\right] [/tex] (*)

    It can be shown that the [itex] \left[\xi(x)\right]^{\mu} [/itex] transforms like a vector field under Lorentz transformations.The definition (*) means that a vector field is equivalent to a [itex] \left(\frac{1}{2},\frac{1}{2}\right) [/itex] irreducible representation of the restricted homogenous Lorentz group.Actually,this definiton induces a isomorphism of vector spaces:the tangent vector space to the flat [itex] \mathbb{M}_{4} [/itex] (which contains contravariant vector fields,such as the photon field) and the vector space of the [itex] \left(\frac{1}{2},\frac{1}{2}\right) [/itex] irred.linear rep.of the restricted hom.Lorentz group.


    Spinors can be used to describe any tensor field,because tensor fields are obtained by taking Kronecker products of irreducible reps of the restricted hom.Lorentz group... :wink: Sides,how about the Rarita-Schwinger (gravitino) field...?5/2,7/2,etc. fields...?

    Daniel.

    P.S.I'm really surprised you haven't seen Maxwell's equations in spinor form. :surprised
     
  26. May 6, 2005 #25

    dextercioby

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    And one more thing:i know that in a GR course you're being taught that the vector field [itex] A_{\mu}(x) [/itex] which classically describes the electromagnetic field is a covector/1-form/covariant/dual vector field and is defined on the cotangent bundle to a curved [itex] \mathbb{M}_{4} [/itex]...But in my previous post i meant the flat [itex] \mathbb{M}_{4} [/itex] and in that case the cotangent bundle is a infinite reunion of identical cotangent spaces.The fact that the metric allows us to pass from the tangent to the cotangent bundle makes us also speak of [itex] A^{\mu} [/itex]...

    Daniel.
     
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