Does the gravitational field have energy like the electric field ?

[Bergstein]

We know that the energy density of the electric field is:
\frac{1}{2}*\epsilon*E²
then, can we infer that the energy density of the gravitational field is:
1/(8\piG)*g²? Here, g is the gravitational field intensity

 

[conway]

It's a nice thought, but there are problems with it. In electrostatics, as two charges are allowed to come together, the field energy collapses, and we can say that the potential energy of the field goes into the kinetic energy of the charges. If you try to analyze gravity this way, the sign comes out negative. It's a problem.

 

[Bergstein]

I can't get it. How does the sign turn to negative?

 

[Division]

Do you mean the gravitational field of a planet?

 

[Dale]

We know that the energy density of the electric field is:
\frac{1}{2}*\epsilon*E²
then, can we infer that the energy density of the gravitational field is:
1/(8\piG)*g²? Here, g is the gravitational field intensity

I don't think so. Consider the energy in the E field of a positive and negatively charged point charge. The total energy of this field (proportional to the integral of E²) is less than the energy of either field alone. As the charges are brought closer together the total energy decreases further as work is done on the charges. When the points are brought together the field dissapears and there is no more potential energy.

In the case of two masses the integral of g² would be greater than the integral of either alone, as they are brought closer together the total integral increases further despite that work is being done on the masses. When the points are brought together the integral is a maximum despite that the potential energy is gone.

Perhaps as conway suggested the negative of that integral could work.

 

[Bergstein]

I don't think so. Consider the energy in the E field of a positive and negatively charged point charge. The total energy of this field (proportional to the integral of E²) is less than the energy of either field alone. As the charges are brought closer together the total energy decreases further as work is done on the charges. When the points are brought together the field dissapears and there is no more potential energy.

In the case of two masses the integral of g² would be greater than the integral of either alone, as they are brought closer together the total integral increases further despite that work is being done on the masses. When the points are brought together the integral is a maximum despite that the potential energy is gone.

Perhaps as conway suggested the negative of that integral could work.

I agree with what you say, but does the integral of g² have physical interpretation?

 

[Dale]

I agree with what you say, but does the integral of g² have physical interpretation?

Not that I know of. But the more I think about it the more I suspect that my objection above is fixed simply by a minus sign.

 

[granpa]

but what is the physical significance of a minus sign? subtracting a positive quantity and adding a negative quantity may be mathematically equivalent but they are not conceptually equivalent. negative sheep do not exist in reality no matter how convenient they may be mathematically.

remember too that in this model of potential energy we don't have the luxury of designating an arbitrary potential with an arbitrary number as we do in the usual usage of 'potential enegy'. when the field is zero the potential MUST be zero too. the only question is the sign itself and its significance.

 

[Bob S]

The sign is very important. As pointed out before, if you bring two equal masses M together, they attract one another and the total field energy (integral of field energy density over all space) doubles.
However, if you bring two equal positive charges Q together, they repel one another and the total field energy (integral of field energy density over all space) doubles. This is reasonable, because the work done to bring the two charges together increased the total stored energy.
How can gravitational force between two masses be attractive, when the total gravitational field energy increases?

 

[Dale]

when the field is zero the potential MUST be zero too.

No, when the field is zero the derivative of the potential must be zero.

 

[granpa]

but what is the physical significance of a minus sign? subtracting a positive quantity and adding a negative quantity may be mathematically equivalent but they are not conceptually equivalent. negative sheep do not exist in reality no matter how convenient they may be mathematically.

remember too that in this model of potential energy we don't have the luxury of designating an arbitrary potential with an arbitrary number as we do in the usual usage of 'potential enegy'. when the field is zero the potential MUST be zero too. the only question is the sign itself and its significance.

when the field is zero everywhere[/color] the potential MUST be zero.

 

[Dale]

When the field is zero everywhere the potential must be constant. 0 is certainly the constant I would pick, but it is a choice of convention not a physical necessity.

 

[granpa]

then your definition of potential energy is not the energy stored in the field. that is the idea that we are discussing in this thread.

in this model of potential energy

 

[Dale]

In a conservative field like gravity only the change in potential has any physical significance.

Remember, if gravity is like EM then the energy density is the square of the field (dot product), and the field is the gradient of the potential. The gradient of any constant is the zero vector, and the square of the zero vector (dot product) is zero. Therefore any constant potential has zero energy.

 

[granpa]

yes the math is trivial. you don't need to remind me of it. we were discussing the physical significance of the math not the math itself. I define potential energy as (positive) energy stored in a field. which leads to the conclusion that energy is not conserved in gravitational interactions. if you prefer a different definition then so be it.

the Hamiltonian (kinetic + potential) would not be conserved but the Lagrangian (kinetic - potential) would be.

 

[Dale]

I define potential energy as (positive) energy stored in a field. which leads to the conclusion that energy is not conserved in gravitational interactions. if you prefer a different definition then so be it.

I assume we are talking about Newtonian gravity, not GR or some fringe quantum gravity, correct?

Every high-school level physics course teaches that Newtonian gravity is conservative. If you are in disagreement please explain carefully and in detail how you reached that conclusion.

I think you are probably confusing energy, the potential, and potential energy.

 

[Bergstein]

let's consider the potential energy of one positive charge and one negative charge, it's:
-k*q*q/r,which is negative.in this model, we assume that the potential which is infinitely far away from the charge is zero.However, if we use the integral of the field intensity to calculate the potential energy,we will find that the energy is always positive, and the assumption in this model is same with the one above.So, I suspect that the energy of field and the potential energy is not same as we think.

 

[Dale]

I think you are making the same mistake as granpa. There is no meaning to the sign of a potential, only a potential difference has physical significance.

 

[granpa]

for 2 charges is the Hamiltonian (kinetic + potential) not conserved? I thought it was. does the potential energy not convert to positive kinetic energy? I thought kinetic energy was always positive.

for simple things like predicting the path of a cannonball (where the mass of the cannonball is negligible compared to the mass of the earth) we can pretend that the Hamiltonian is conserved (in gravitational interactions) because the error is trivial but in reality it is the Lagrangian (kinetic - potential) that is actually conserved. (where potential energy is calculated by integrating the square of the field intensity over all space)

edit:actually, in a 2 body problem, the Hamiltonian wouldn't produce any error. in many body problems though I think you would have to use the lagrangian.

for electric charges it is the Hamiltonian that is conserved (where potential energy is calculated by integrating the square of the field intensity)

 

[Dale]

in reality it is the Lagrangian (kinetic - potential) that is actually conserved.

The Lagrangian is not conserved. If the Lagrangian, L, of some system were conserved that would mean that dL/dt = 0. In fact dL/dx - d/dt(dL/dx') = 0.

Consider for example a mass m dropped from the origin in a uniform gravity field g pointing in the -x direction. We know from classical mechanics that the solution is x = -1/2 gt². The kinetic energy is T = 1/2 mx'² and the potential energy is U = mgx. The Lagrangian is L = T - U = 1/2 mx'² - mgx. Substituting in the solution for x gives us L = g²mt² which is not constant. The Lagrangian is not conserved.

The utility of the Lagrangian is not that it is conserved (since it is not), but that it gives a way of deriving Newton's laws of motion in terms of generalized coordinates that may be more appropriate for a given system than the classical Eulerian formulation. It is most emphatically not an additional conserved quantity.

 

[granpa]

didnt I just say:

for simple things like predicting the path of a cannonball (where the mass of the cannonball is negligible compared to the mass of the earth) we pretend that the Hamiltonian is conserved in gravitational interactions

for the lagrangian you must calculate the potential by integrating the square of the field over all spaceedit:actually, in a 2 body problem, the Hamiltonian wouldn't produce any error. in many body problems though I think you would have to use the lagrangian.
for a 2 body problem the field that each body moves in doest not change with time. that's not true in many body problems.

for electric charges it is the Hamiltonian that is conserved (where potential energy is calculated by integrating the square of the field intensity)

 

[granpa]

consider the case of a star that collapses to a black hole releasing an infinite amount of energy. the Lagrangian is still conserved because the field now contains an infinite amount of potential energy (where potential energy is calculated by integrating the square of the field intensity over all space)

 

[D H]

The Lagrangian is not a conserved quantity.Where do you get these nonsense ideas?

 

[granpa]

do the math. if I'm wrong then prove me wrong. otherwise I have nothing to say to you.

 

[Dale]

do the math.

dL/dx - d/dt(dL/dx') = 0
Math done.

 

[granpa]

The Lagrangian is not conserved. If the Lagrangian, L, of some system were conserved that would mean that dL/dt = 0. In fact dL/dx - d/dt(dL/dx') = 0.

Consider for example a mass m dropped from the origin in a uniform gravity field g pointing in the -x direction. We know from classical mechanics that the solution is x = -1/2 gt². The kinetic energy is T = 1/2 mx'² and the potential energy is U = mgx.[/color] The Lagrangian is L = T - U = 1/2 mx'² - mgx. Substituting in the solution for x gives us L = g²mt² which is not constant. The Lagrangian is not conserved.

The utility of the Lagrangian is not that it is conserved (since it is not), but that it gives a way of deriving Newton's laws of motion in terms of generalized coordinates that may be more appropriate for a given system than the classical Eulerian formulation. It is most emphatically not an additional conserved quantity.

you arent using the definition of potential energy that I have clearly and repeatedly stated that you must use if the Lagrangian is to be conserved. did you not read my posts at all or was it too complicated for you to understand? for the Lagrangian to be conserved you must calculate the potential energy by integrating the square of the field intesity over all space.

this only works for gravity. for electric fields it is the Hamiltonian that is conserved

 

[Dale]

Mine is the standard definition of potential energy available in any freshman textbook. If you wish to introduce new concepts please do not re-use existing terms which are clearly defined and in general use. Because of your non-standard usage, you are confusing the concepts of the energy of a field, a conservative field's potential, and potential energy.

None of that is relevant to whether or not the Lagrangian is conserved. It is not. In general dL/dt does not equal 0. The Lagrangian is not a conserved quantity, that is not its purpose. For you to state otherwise is an eggregious misunderstanding of mechanics.

 

[granpa]

um. if L=K-P
where K=kinetic energy
and P=potential energy
then dL=dK-dP
dK=-dP for a conservative field
therefore dL=dk-(-dK)=2dK
so the change in the lagrangian is nothing more than just twice the change in the kinetic energy? no wonder the formula works.

 

[granpa]

if L=K-P
then dL/dx - d/dt(dL/dx') = 0 reduces to
dP/dx - d/dt(dK/dx') = 0

near as I can figure it the lagrangian itself doesn't even enter into that formula. the only use I can find for the lagrangian itself is that it is the 'rate of change of action' whatever that is.

but according to http://en.wikipedia.org/wiki/Action_(physics ) even the need for the 'action' to be minimized just reduces to the formula above.

 

[D H]

You are thinking way to Cartesian, granpa. The power of the Lagrangian approach only becomes evident when you switch to generalized coordinates.