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The is of identity

  1. Feb 25, 2005 #1
    The "is" of identity

    It is one thing to say that "Gravity is caused by the curvature of spacetime" and it is a stronger statement to say "Gravity is the curvature of spacetime".

    Before you run, cringing in semantic horror, how about a more interesting example:

    [tex]G_{\mu\nu} =8\pi G T_{\mu\nu} [/tex]

    This equation can be interpereted in two ways. Either "the curvature at a point of spacetime is equal to the energy-momentum density at the point" or in a stronger sense, "Energy density is the curvature of spacetime". Wheeler said: "Mass tells spacetime how do curve, spacetime tells mass how to move" but I am asking, if Einstien's equals is the "is" of identity i.e. the definition of energy.

    Suppose charge is nothing more than the divergence of the electric field? In other words, take the = in Gauss' law to be the is of identity, rather than a computationally convenient "is equal to".

    Before you say this is crap, I would like you to think about the meaning of "Unified Field Theory". In order to except field as the only entity (no particles at all) how would our physical ideas have to change? Look at the lorentz force law, and think of q as "del dot E". How would we define v and dp/dt ? (a rhetorical question perhaps, but what would kinematical concepts be in a field only universe?"
     
    Last edited: Feb 25, 2005
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  3. Feb 26, 2005 #2

    Andrew Mason

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    Some equations represent a fundamental relationship and some are just useful constructions. One could say that force/acceleration is inertia or mass, because that is really a definition of inertia (or force for that matter). But one could not really say that voltage is current times resistance, because V and I have independent definitions. (But one might be able to say that R is V/I, because that is essentially its definition).

    So perhaps the question is whether the equation expresses a fundamental definition. If it doesn't, it isn't.

    AM
     
  4. Feb 26, 2005 #3

    Tom Mattson

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    My $0.02:

    The "=" in all the equations of physics asserts equality in the numerical sense only. We are, after all, dealing with mathematics here. It should not be confused with the identity operator of logic (eg: Aristotle's "A is A".)
     
  5. Feb 26, 2005 #4

    dextercioby

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    There's a fundamental difference between an equation/egality and a definition/identity.
    To exemplify,let me choose the same example
    Equation/equality:
    [tex] G_{\mu\nu}=8\pi \Theta_{\mu\nu} [/tex]
    Definition/identity:
    [tex] G_{\mu\nu}\equiv R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R [/tex]

    DO YOU SEE THE DIFFERENCE...?

    Daniel.
     
  6. Feb 27, 2005 #5
    Yes Dextercoiby, I see the uninteresting irrelevant point you are making.

    Ovbiously you fall in the camp that says charge is merely equal to the divergence in the electric field. But in a law so general as Gauss's, what is charge other than a divergence in the field? Charge literally does not exist independent of the divergence in the field and the difference between the two is truly no more than a matter of interpretation, because they are equivalent.

    Tom, when you speak of numerical equivalence, what of the divergence in the electric field as related to charge? If you count the charge in a volume, you are realing just counting the divergence of the electric field (in a disguised way), so how can you claim that they are only numerically equal?
     
  7. Feb 27, 2005 #6

    dextercioby

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    I guess you have no idea what i was referring to.Good for you.

    Daniel.
     
  8. Feb 27, 2005 #7
    What you are talking about is the difference between a physical law and the definition of a physical concept.

    I am considering the possibility that equations, such as the Einstien Field Equations, are an identity in this universe. That is, energy and curvature have no independent existence so on the RHS of the equation we call it curvature, and on the LHS we call it energy, but what we are really saying is A = A.
     
  9. Feb 27, 2005 #8

    dextercioby

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    Identity means one thing,equality (which is a part of any equation) means totally different.U cannot IDENTIFY Einstein's tensor with [tex] \frac{8\pi G}{c^{4}}\Theta_{\mu\nu} [/tex],because there are field configurations for which the equality won't hold.On the other hand,you can damn well identify Einstein's tensor to [tex] R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R [/tex],because these 2 are made identical by the very definition of the Einstein tensor...

    Definitions assume identities.Equations/egalities not.

    Daniel.
     
  10. Feb 27, 2005 #9

    Tom Mattson

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    I claim that because it's the only claim that is warranted by the equation. If you use Gauss' law to calculate div E from a charge density, all you know for sure is that div E is numerically equal to ρ/ε0.

    If fact, it is I who should be asking you how you can claim that any more than that is implied.
     
  11. Feb 27, 2005 #10

    Tom Mattson

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    It's not irrelevant. Indeed, once he made his post I saw that I should have said more. Specifically, I should have said that the only true "identities" in physics are definitions. Andrew Mason touched on this as well.

    But as for those relationships which are not a priori, we can't say that those are identities.
     
  12. Feb 27, 2005 #11
    Tom, I claim that there is no independent way to measure the charge and to measure the divergence of the electric field. How can they be numerically equal if this is the case? They are indentical.
     
  13. Feb 27, 2005 #12

    dextercioby

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    Not all electric fields are sollutions to Maxwell's equations.Ergo,not all electric fields have the divergence equal to volumic charge density times the inverse of the vacuum permittivity.

    Therefore,there's a relation of equality & not an identity.The same with Eistein's equations...

    Daniel.
     
  14. Feb 28, 2005 #13

    Tom Mattson

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    You can measure the charge by measuring the force exerted on the charge in a uniform electric field. This can be done by first measuring the weight of the charged body, and then adjusting the strength of the E field until the downward force of gravity is balanced by the upward electric force.

    Independently, you can use a small test charge to map out the E field lines due to a charge density. From there you can compute div E.

    Are you serious??

    The real question is: How can they NOT be numerically equal if this is the case?...!

    That is and was your unsupported claim.
     
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