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Lagrangian density of Newtonian gravity

  1. Feb 4, 2014 #1

    BruceW

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    Hi everyone!

    I've been thinking about a certain problem for a while now. And that is a Lagrangian formulation of Newtonian gravity. I know there is a Lagrangian formulation for general relativity. But I was hoping to find a Lagrangian for Newtonian gravity instead (for some continuous mass distribution). Anyway, the closest thing I have been able to find is this equation (from wikipedia):
    [tex]\mathcal{L} = -\rho \phi - \frac{1}{8\pi G} |\vec{g}|^2[/tex]
    Where ##\rho## is mass density and ##\phi## is the gravitational potential, i.e.
    [tex]\vec{g} = - \nabla \phi [/tex]
    So, anyway, it says on wikipedia that you can vary this Lagrangian density with respect to changes in the gravitational field, to get Newton's law of gravity. This is true. But since we are only varying the gravitational field, and not the mass density, this means that we are keeping the mass density fixed! So this equation is not a true Lagrangian density for Newtonian gravity. In the true Lagrangian density, you would be able to vary both the mass density and the gravitational field, and this would give you both Newton's law of gravity, and some other equations (possibly conservation of momentum and energy?)

    I was looking around for a while, but I haven't been able to find the true Lagrangian density for Newtonian gravity. I guess one way you could derive it, is to simply use the general relativity Lagrangian density, and then make some kind of low-speed, weak-gravity approximations, to get a Lagrangian density for Newtonian gravity. But that seems like quite a long way to go about it.

    Thanks for reading, and hopefully some other members of PF have been thinking about this too. Or maybe know the answer?
     
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  3. Feb 4, 2014 #2

    dextercioby

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    The mass density is fixed in the same way the current 4-vector is 'fixed' in relativistic electrodynamics (in vacuum). From the lagrangian formulation of a field theory, you can only get local (partial) differential equations. That's what you get for your [itex] \mathcal{L} [/itex], no matter which theory you use. What you're saying there is that you want a second set of E-L differentials for the matter distribution, but here you don't vary the lagrangian of the field + lagrangian of the matter wrt the density, but wrt the parameter describing the (Galilean) worldline of an element of matter (think about the similar problem in electrodynamics).

    The lagrangian you wrote is 2/3 of the full Lagrangian, it already has the coupling fields-sources, all it misses is the kinetic term of the source.
     
  4. Feb 4, 2014 #3

    BruceW

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    interesting reply, thanks! Well, our equations so far are:
    [tex]\frac{\partial \mathcal L}{\partial \phi} = \sum_i \frac{\partial}{\partial x_i} \frac{\partial \mathcal L}{\partial \phi_i}[/tex]
    (which is the variation of the gravitational potential). And about notation, I've used ##x_i## as the 3 spatial coordinates and ##\phi_i## is the partial differential of ##\phi## with respect to the i'th spatial coordinate.

    And I would hope that in the full Lagrangian density for Newtonian gravity, we would also have the equations due to varying the mass density:
    [tex]\frac{\partial \mathcal L}{\partial \rho} = \sum_i \frac{\partial}{\partial x_i} \frac{\partial \mathcal L}{\partial \rho_i}[/tex]
    But using the Lagrangian density we currently have, this does not give a sensible answer. So, you were saying that we should add a kinetic term? that sounds reasonable. do you by chance, happen to know it already? Or maybe we will need to do some thinking, to be able to get the right term.
     
  5. Feb 4, 2014 #4

    dextercioby

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    Just write

    [tex] K = \frac{mv^2}{2}= \frac{1}{2} \iiint_{D\subset\mathbb{R}^3}\rho (\vec{x}) \vec{v}^2 (\vec{x}) d^3 \vec{x} [/tex]

    and you end up with the Lagrangian for the ideal Euler fluid
     
  6. Feb 4, 2014 #5

    BruceW

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    I don't understand. are you saying I should add a term ##1/2\rho v^2## to the previous Lagrangian density? That does not help, as far as I can see.
     
  7. Feb 5, 2014 #6

    samalkhaiat

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    Do you know that the gravitational “field” in Newton’s theory is not dynamical quantity. The theory is based on “action at a distance” not action principle. So, it does not make a lot of sense to ask for Lagrangian formalism. Mathematically, you can derive the Poisson’s form of Newton law from “the Lagrangian” function
    [tex]
    \mathcal{ L } ( \phi , \partial \phi ) = - \frac{ 1 }{ 8 \pi G } \partial_{ i } \phi ( x ) \partial^{ i } \phi ( x ) + \rho ( x ) \phi ( x ) ,
    [/tex]
    if you use the “Lagrange equation”
    [tex]
    \frac{ \partial \mathcal{ L } }{ \partial \phi } - \partial_{ i } \left( \frac{ \partial \mathcal{ L } }{ \partial \partial_{ i } \phi } \right) = 0 .
    [/tex]
    This leads to the Poisson’s equation for the instantaneous potential [itex]\phi ( x )[/itex]
    [tex]
    \nabla^{ 2 } \phi ( x ) = - 4 \pi \rho ( x ) = - 4 \pi \int \ d^{ 3 } y \ \rho ( y ) \ \delta^{ 3 } ( x - y ) .
    [/tex]
    This equation can be solved if you use the relation
    [tex]
    \nabla^{ - 2 } \delta^{ 3 } ( x - y ) = \frac{ - 1 }{ 4 \pi } | \frac{ 1 }{ x - y } | .
    [/tex]
    The solution is
    [tex]
    \phi ( x ) = G \int \ d^{ 3 } y \ \frac{ \rho ( y ) }{ | x - y | } + \mbox{ const. } \ \ (1).
    [/tex]
    From this you recover Newton law for point mass, if you use
    [tex]
    \rho ( y ) = M \delta^{ 3 } ( y ) , \ \ \mbox{ and } \ \ F = - m \nabla \phi ( x ) .
    [/tex]

    Question for you (since I have just seen Dexter's post): How can you put "my Lagrangian" in the form [itex]T - V[/itex]?


    Sam
     
    Last edited: Feb 5, 2014
  8. Feb 5, 2014 #7

    BruceW

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    hi, thanks for the reply. But this is what we already had so far. the Lagrangian density you wrote is exactly the same as the one I wrote in the first post, except it is negative of the one I wrote. (I'm guessing the partial derivative you are writing is just summing over the spatial coordinates, and not time).
    [tex]- \frac{ 1 }{ 8 \pi G } \partial_{ i } \phi ( x ) \partial^{ i } \phi ( x ) + \rho ( x ) \phi ( x ) = \rho \phi + \frac{1}{8\pi G} |\vec{g}|^2[/tex]
    And I understand that varying the Lagrangian with respect to the gravitational field gives Newton's law of gravity. But I was hoping for a 'total' Lagrangian of some sort, where we can also vary the mass density. I'm thinking this would give us another equation, maybe showing how energy and momentum is converted from/to the gravitational field.

    edit: whoops, actually your equation is not negative of mine. I think it should instead be
    [tex]
    \mathcal{ L } ( \phi , \partial \phi ) = - \frac{ 1 }{ 8 \pi G } \partial_{ i } \phi ( x ) \partial^{ i } \phi ( x ) - \rho ( x ) \phi ( x ) ,
    [/tex]
    Since the Poisson equation for gravity is:
    [tex]\nabla^{ 2 } \phi ( x ) = 4 \pi \rho ( x )[/tex]

    edit again: aha, and there should also be a ##G## in the Poisson equation for gravity, too. But yeah, anyway, the point is that the equation is the same as the one in the OP.
     
    Last edited: Feb 5, 2014
  9. Feb 5, 2014 #8

    samalkhaiat

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    Did you read the first paragraph in my post? There is NO time dependence in Newton’s gravitational potential and that is the problem. If the gravitational field is not dynamical, you cannot you talk about “converting” energy and momentum “from/to the gravitational field”. This is one of the differences between GR and Newton’s gravity. So, the question you are asking is ill-defined. Plus, does the mass density, momentum or energy satisfy equations other than their conservation equations? No, even if you consider Newtonian self-gravitating fluid you will have no chance of having Lagrangian that gives you equations other than the following:
    1) Mass conservation
    [tex]
    \frac{ \partial \rho }{ \partial t } + \vec{ \nabla } \cdot ( \rho \vec{ v } ) = 0 .
    [/tex]
    2) Conservation of momentum which is equivalent to the Euler equations of motion:
    [tex]
    \frac{ D }{ D t } \vec{ v } = - \frac{ \vec{ \nabla } p }{ \rho } - \vec{ \nabla } \phi ,
    [/tex]
    where
    [tex] \frac{ D }{ D t } = \frac{ \partial }{ \partial t } + \vec{ v } \cdot \vec{ \nabla } ,[/tex]
    is the so-called convection operator and [itex]p[/itex] is the pressure. To see how much you know, I ask you to prove that the Euler equation is equivalent to momentum conservation.

    3) The Poisson equation for the potential (putting [itex]G = 1[/itex]):
    [tex]\nabla^{ 2 } \phi = 4 \pi \rho .[/tex]

    This is another exercise for you. Show that the above system of equation is derivable from the Lagrangian density
    [tex]
    \mathcal{ L } = \frac{ 1 }{ 2 } \ \rho \ v^{ 2 } - \rho \phi - p - \frac{ 1 }{ 8 \pi } \ \vec{ \nabla } \phi \cdot \vec{ \nabla } \phi .
    [/tex]
    For a pressureless fluid, show that the total energy of the fluid plus gravitational field is given by the integral
    [tex]
    E_{ \mbox{ tot }} = \int d^{ 3 } x \left( \frac{ \rho }{ 2 } v^{ 2 } - \frac{ 1 }{ 8 \pi } ( \vec{ \nabla } \phi )^{ 2 } \right) .
    [/tex]
    And lastly, explain why the gravitational energy integral of Newton theory:
    [tex]
    E_{ \mbox{ g } } = - \frac{ 1 }{ 8 \pi } \int d^{ 3 } x \ ( \vec{ \nabla } \phi )^{ 2 } ,
    [/tex]
    is equal to the Newtonian limit of Einstein-Hilbert Lagrangian
    [tex]
    L_{ \mbox{ g } } = - \frac{ 1 }{ 8 \pi } \int d^{ 3 } x \ ( \vec{ \nabla } \phi )^{ 2 } ,
    [/tex]
    but not to the [itex]\mathcal{ T }^{ 0 0 }[/itex] component of the gravitational energy-momentum pseudotensor? If you can solve these exercises (plus the one from the previous post), then we can talk physics. If not, then you need to spend some time learning about the canonical formalism.
    By the way, you can work with either ([itex]\phi[/itex]) or ([itex]- \phi[/itex]) because it does not change the sign of the energy integral.

    Sam
     
    Last edited: Feb 5, 2014
  10. Feb 6, 2014 #9

    BruceW

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    I guess, but then the gravitational field points away from masses. So we'd need to say the force of gravity is opposite to the direction of the gravitational field, which is less 'nice'. Also, the Poisson equation for the potential would need to have a negative sign.

    yep, I know Euler's equations. They're one of my favourite set of equations. Even nicer is the general form, with the Cauchy stress tensor. Maybe only Maxwell's equations I like more.

    I have no idea how to do that. literally no clue. I'm pretty sure I could do it for a point particle, but the continuous mass distribution is the interesting problem. Do you know the answer to this? This seems pretty close to the kind of equation I had in mind when I made this thread.

    oh, this is an interesting question. I guess it could be a first step to using the general relativity Lagrangian, to find a Newtonian gravity approximation. uh, anyway, for this question. OK, the Lagrangian density for the Hilbert-Einstein action is ##1/(2\kappa )R## (not including the contribution of the lagrangian density due to matter). So, anyway, ##1/(2\kappa )R = -1/(2 \kappa)G## and ##G=\kappa T## So the Lagrangian density is ##-(1/2)T##. (the symbols ##R## ##G## and ##T## are the trace over the related tensor, with the metric). Continuing onward, (I think) if we make the time-independent,low-speed, weak gravity, approximations, then we get:
    [tex]T_{\mu \nu} = diag(\rho c^4,0,0,0)[/tex]
    [tex]g_{\mu \nu} = diag(1/(-c^2 -2 \phi),1,1,1)[/tex]
    So, taking the trace of ##T_{\mu \nu}## using ##g_{\mu \nu}##, we would get
    [tex]T = \frac{\rho c^4}{-c^2-2\phi} [/tex]
    So this means the Lagrangian density is:
    [tex]\frac{\rho c^4}{2c^2 + 4\phi} [/tex]
    So, from here, you were saying this should be equal to ##- |\vec{g}|^2## right? I guess that works if we say that the denominator ##2c^2+4\phi## is roughly equal to ##2c^2##, and not caring about the constants, we just end up with ##\rho## And we know ##\rho## is proportional to ## - |\vec{g}|^2## due to the Poisson equation. So... yeah, I guess that works. Pretty nice. One question, though: does this mean that the energy density in Newtonian theory would be ##|\vec{g}|^2## ? That would make more sense to me, since the gravitational field is usually a kind of potential energy, and the Lagrangian is made up of kinetic energy terms, minus the potential energy terms. I suppose to answer this properly we need to start talking about canonical momenta, so that we can define the energy density, by using the Lagrangian density.
     
  11. Feb 6, 2014 #10

    samalkhaiat

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    You made all incorrect choices and conclusions. No, you only need to take
    [tex]\vec{ g } = \vec{ \nabla } \phi[/tex]

    Then you should waite until you learn about functional differential calculus.

    You are no way near the right path. I said the gravitational energy-momentum pseudotensor which one can extract from the Einstein tensor. Any way I believe you need to learn a lot in this subject before tackling this sort of problems.

    Sam
     
  12. Feb 7, 2014 #11

    BruceW

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    oh yeah, that's true. I see what you mean now.

    The maths is not the problem. I'm pretty sure I know enough about functional differentials. The problem is that it is quite a specific topic that seems to be rarely discussed. I haven't seen it in any textbooks, even though many of them mention a lot about Lagrangian formalism.

    well, you said: "explain why the gravitational energy integral of Newton theory: ... is equal to the Newtonian limit of Einstein-Hilbert Lagrangian" Which is why I started with the Einstein-Hilbert Lagrangian. It looked to me like the gravitation energy-momentum pseudotensor was a follow-up question of some kind.

    p.s. your writing is not very encouraging. you should work on that maybe.
     
  13. Mar 2, 2015 #12
    Please, could you help us with some references?
     
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