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Is that Energy Density equation make sens?

  1. Aug 27, 2003 #1
    Posing that O4 and O3 are energy density vectors, and Rp is momentum density value.
    Could I write:
    O4 = ( O3, icRp )
    and therefore
    (O4)2 = (O3)2 - (cRp)2
    where (O4)2 is invariant under Lorentz transformation.

    The point here, from what I learned from preceding posts, is to write something like:
    div4(O4) = div3(O3) + dRp/dt (partial)

    Rp has (kg-m/s)/(m3)
    O4 and O3 have (kg-m/s)/(m2-s)

    I would physically interpret this like:
    Considering a small volume, T.
    The rate at which the momentum T*Rp, enclosed in the volume T increase with time is T*dRp/dt (partial)
    The rate at which the enclosed momentum decrease with time is T*div3(O3) since the energy density, O3, is the momentum flowing out per unit time and per unit area.
  2. jcsd
  3. Aug 27, 2003 #2


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    Hello imagine,

    I think that the 3-vector part is related to momentum
    and the solo component to adjoin to it is the energy-related part.

    I am having a little difficulty understanding your notation.
    I will get back to you in a moment after another look.

  4. Aug 27, 2003 #3
    Bonjour Marcus,

    I understand that the well known energy-density equation uses the 3-vector part as related to momentum-density and the solo component part as related to energy-density.

    In my equation, I vectorized the energy and soloized the momentum. I have a final purpose to this but I have to confront that equation to your well appreciated comments.

    P.S.: I agree with you, I have difficulties with my notation but I try to fix it.
  5. Aug 27, 2003 #4


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    Imagine, it would be a pity if my replying to you PREVENTED some mentor such as Janus or Tom from replying, because they think one reply sufficient! Your posts are both logical and adventurous, and merit serious attention. In a certain sense your confusion can be more interesting than the clarity of some other people. There is more here than I can address.

    I hope you do get others' attention. And soon.

    Making up a 4-vector seems to be like packing a suitcase, you must get the right things in it in the right places

    You use the phrase: "...the momentum flowing out per unit time and per unit area."

    This is an excellent image
    I would call it the MOMENTUM FLUX along a certain direction

    I would think of the momentum flux as having 3 components,
    the one along the x direction is the momentum flowing (per unit time)through a
    small square dydz--------including the per unit time it is a "dydzdt" thing. I am speaking very informally and hope you understand.

    the one along the z direction is the momentum flowing (per unit time) thru a small square dxdy----and so it is a "dxdydt" thing

    the energy density can have only ONE component, it is not a "per unit time" thing. the energy density is a "dxdydz" thing. It is energy just SITTING in a small cube-----it is not FLOWING spatially. It is just residing, while time passes. Perhaps it gets larger or smaller but it is not directional

    Now let us look at the units.
    We have three momentum fluxes
    and we have one energy density

    the energy density is simple "joule per cubic meter"
    and joule is equivalent to newtonmeter
    so energy density is "newtonmeter per cubic meter" which
    is algebraically the same as "newton per square meter"

    I will stop for the moment. Isnt this a bit confusing? Do you suppose that the unit of momentum flux will also turn out
    to be "newton per square meter". That would be very strange but also absolutely necessary!!!! We cannot pack the suitcase
    unless all four quantities have the same unit!
    Last edited: Aug 27, 2003
  6. Aug 27, 2003 #5


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    I see you are offline for the moment, so i will conclude without waiting to see what you have to say about momentum flux having the same unit as pressure

    In fact it does. so one can assemble a 4-vector in the way I was suggesting.

    there are other ways to do it too and your investigation may turn up one of these other 4 vector types.

    small lemma: momentum flux also has unit "newton per square meter"

    one version of the momentum unit is "newtonsecond"
    (algebraically equiv to kilogram meter per second)

    so momentum thru a square meter per second has unit
    "newtonsecond per second per square meter"
    and this is "newton per square meter"

    So Blaise Pascal was a very lucky man since his name is applied not merely to pressure but to energy density and momentum flux!
  7. Aug 27, 2003 #6
    Am I wrong but "newton per square meter" could be surface tension, which contains volume like a water drop? Is it depending upon the point-of-view? Could it Please to Blaise?:wink:
  8. Aug 27, 2003 #7


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    I am becoming increasingly convinced that style is important in Physics (as it is in mathematics).

    Indeed the unit of surface tension is JOULE PER SQUARE METER and for an excellent reason! It is not "newton per square meter" the unit of pressure to which the philosopher/essayist Blaise Pascal generously contributed his name, and which is also the unit of energy density and momentum flux.

    I seems that I must answer the question of why is the unit of surface tension joule per sq. meter instead of something else like newton per sq. meter?

    Tyger, for instance, could do this very well----and of course any PF mentor. I think sometimes I am going to disappear simply to force you to get to know more people. Such as dhris----he was a good one.

    When you stretch out a surface (that has a constant surface tension) the WORK that you do is proportional to the amount by which you increase the area.

    Doesnt it seem natural that the surface tension should be
    this proportion of work to ΔA?

    Well, it seems so to me----surface tension is W/ΔA
    the work you do per square meter of addtional surface area you make
    and so the unit is joules per square meter

    A water molecule likes to have friends on all sides
    If you deform a water drop so that more molecules are forced to be on the surface (because there is more surface) then
    those partly lonely molecules are breaking contact with some of their companions and it costs energy (work) to break those connections-------the molecule at the surface is in contact with the others only on 5 sides instead of the usual 6 sides, so energy has been paid out to deprive him of fellowship on one side.

    If left undisturbed (and with nothing to adhere to) the water drop will minimize its surface area and thus its energy

    feeling surface tension is a way of feeling the millions of intermolecular forces---but you know this so why am I saying it?
    Last edited: Aug 27, 2003
  9. Aug 27, 2003 #8

    Tom Mattson

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    Marcus was telling you that you got it backwards:

    The 3-momentum are the spacelike components and the energy is the single timelike component. The 4-momentum p is:


    If you recall the Lorentz group, you will see that this has to be the case. The spacelike components are subject to rotations in both Euclidean 3-space and in Minkowski 4-space. We know that momentum is affected by rotations, and that energy is not.
  10. Aug 28, 2003 #9
  11. Aug 28, 2003 #10
    From my small knowledge: (note: θ is the four-dimensional operator)

    p = ( p, i E/c ) where p2 - ( E/c )2 is invariant
    θ*p = ∇*p + ∂ E/c2 ∂t have units ( kg/s )

    I try to find υ and μ as per:

    θ*υ = ∇*υ + ∂ μ/∂t have units ( kg/m2-s2 )
    therefore, υ units are ( kg/m-s2 ) and μ units are ( kg/m2-s )
    to form υ = ( υ, i c μ ) where υ2 - ( c μ )2 is invariant

    My goal is to interpret:
    υ units as ( Momentum/m2-s ) and μ units as ( Momentum/m3 )

    What are υ and μ ? That is the question?
    Last edited by a moderator: Aug 29, 2003
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