You probably need to include large scale magnetic fields and forces to get an accurate model.

Virial is zero for circular motion. So, you must be dealing with small deviations fluctuating all over. On the other hand, total kinetic energy and total potential energy are big fat values. So, if you are not interested in an effect connected to fluctuations, the use of virial theorem is justified. 

Dear Igor,

You say: "Virial is zero for circular motion."

However, when we take opposite positioned stars in the same orbit of a disc galaxy, where all the orbits are found to be prograde, the value pi.ri  is of opposite sign for both vectors pi and ri. Hence the scalars that are obtained are adding up, not vanishing.

Hence, the virial is not zero at all, but has a huge value.

Virial is zero for circular motion because velocity is normal to the radius-vector.   That's what the circular motion is. For any number of stars. It is not zero for a deviation from circular motion. But this is small and fluctuating

You keep on getting it wrong: it is not the virial that must be zero for the virial theorem to hold, but its time derivative. It is this time derivative, and not the virial itself, which is time-averaged.

The fact that you cannot get this quite elementary fact into your head in spite of multiple repetitions is truly alarming. Anyway, here a link to a proof of the virial theorem, showing it to be true for arbitrary bound gravitational systems. This may be of interest for those who are interested in finding out the truth about this issue, not merely insisting on propagating lies. The theorem is stated in equation 1.2.16 on p. 5 of the PDF.

An interesting set of considerations which could limit the validity of the virial theorem are stated in Section 5. But they have nothing to do with the shape of the galaxy.

http://ads.harvard.edu/books/1978vtsa.book/Chapt1.pdf

``It is *not* the time derivative as you continue to wrongly say, that is taken, it is the time average. ''

Lagrange's identity (eq. 1.2.13) connects the time derivative of the virial with twice the kinetic energy plus the potential energy. Average both sides and you get the average of the time derivative equal to twice the average kinetic energy and the average potential energy.

You keep referring to ``good textbooks'' without citing anything. The ones you did cite, in a previous thread, merely stated that the virrial theorem was valid for spherical distributions and that more general distributions had a generalised virial theorem. However, they did not anywhere derive the theorem, so it is not possible to say where isotropy would be required. The book I quote presents a clear derivation. If you find it confusing, that reflects on your mathematical abilities. Do say where in Sectoin 1.2, going from 1.2.1 to 1.2.16, you find a problem.

You refer to the author at some point mentioning spherical symmetry: here is the full quote:

``For static configurations exhibiting spherical symmetry these conservation laws take their most familiar form: ''

then follow some equations, unrelated to the virial theorem. There, the author merely states how conservation laws look when the system is spherically symmetric. This is not the virial, and it is a specialisation of universally valid laws (conservation laws) to spherical symmetry. Nowhere is it stated that the virial theorem requies spherical symmetry.

``The derivation in that book is a farce. It says that for a cyclic system, the integral from the begin to the end of the cycle is zero. ''

Another lie: the crucial argument is:

``Indeed, if the motion of the system is bounded [i.e., G(t)

``If you do that at the left hand, you must do that at the right hand too, which makes it trivially 0=0.''

If you average the right hand side over a very long time, you must do the same with the right-hand side. But for the right hand side, the result is zero, but not trivially so: it is the average of the sum of twice the kinetic energy and the potential energy. That this is zero is precisely the statement of the virial theorem.

You keep on appealing to authority ``numerous textbooks'', yet you cite none, and the two you did cite provided no proof of the virial. I have shown you a derivation, and all your comments on ot are so far nonsensical.

``then the right hand would *also* give a trivial zero by setting K(T)-K(0) = 0 and V(T)-V(0) = 0 ''

Integrating V(t) from t=0 to T does not yield V(T)-V(0). You are precisely missing the point. The left-hand side is a time derivative, so its integral can be simplified to a difference, the right hand side is not, so must be integrated as such. The final result is a connection between the time integral of V, the time integral of the kinetic energy and the time integral of the time derivative of the virial, which for a bounded system is zero.

Do learn a bit of calculus before posting: it is the integral of a *derivative* that is the difference of initial and final values.

``The argument of time-averaging over a total cycle is worthless in disc galaxies, since none of them made a single winding,''

Untrue, galaxies have made hundreds of windings. But in any case, the issue does not involve cycles, but mere averaging, and the use of the virial's boundedness. As T-> infinity, the left hand side of 1.2.15 goes to zero and the right hand side goes to twice the average kinetic energy plus the average potential energy. Therefore, since the two are equal, the virial theorem holds.

Leyvraz, that is nonsense again! We are talking of physics!

1) Your first mathematical trick for G(T)-G(0) is only zero if the whole system has made a full physical cycle.

This option is of course impossible for disc galaxies, since each orbit rotates at a different rate and will never make a global cycle for all the stars together.

2) Your second mathematical trick divides the whole equation by an infinite time. Do you deny that if you divide the non-zero left hand by an infinite time, you need to do that to the right hand as well?

Well, since the kinetic energy and the potential energy are finite on the right hand, the integration to an infinite time will *not* make their values larger because the system doesn't accelerate nor descelerate, neither widens, nor shrinks.

Hence, dividing as well the kinetic energy as the potential energy after integration, by an infinite time, will turn each of them trivially to zero, on the same way as the left hand!

Hence, neither of the tricks can be used for disc galaxies, and the virial theorem is wrong.

It can only be used for spherical galaxies with random orbits, because there, each prograde orbit can have a symmetric retrograde orbit that compensate each-other for the determination of the virial.

May I suggest to learn using maths in the perspective and at the service of physics, which opens lots of opportunities! Doing the opposite leads to dark insights about physics...

``Since the kinetic energy and the potential energy are finite on the right hand, the integration to an infinite time will *not* make their values larger because the system doesn't accelerate nor descelerate, neither widens, nor shrinks.''

Basic calculus again, Thierry. It is truly useful and you should familiarise yourself with it. Assume, for example, that V is a constant. Then its integral over a large time interval T is the constant VT. If you divide this by T, a process technically known as *averaging*, you get V: perhaps not unsurprisingly, at least to the specialists in the field, the average of a constant is *that same constant*.

So yes, the integration of V to an infinite time *will* make the integral larger. It will not make V larger, of course, but that is neither here nor there: look, if you can, at equation 1.2.14: on the l.h.s. it has an integral of a time derivative, which can be simplified to (G(T)-G(0))/T, on the other it has integrals from 0 to T of V and K, both of which are, by calculus, proportional to T. These integrals are then divided by T, to yield a finite, generally non-zero result, known as the average. Of course, by your own arguments, neither the average of V, nor the average of K vanishes. But V is negative, on average, since the system is bound, and K is positive by definition, so it is possible, and in fact true, that the sum indicated on 1.2.16 vanishes.

The virial theorem wrt disc galaxies is wrong due to at least two major issues, which relate to the maths and to the physics.

1) The maths wrt (G(T)-G(0))/T at equation 1.2.15 and further are wrongly interpreted:

Since you have taken the time derivative at the left hand, you also took the time derivative at the right hand. It is not because you gave names to the terms in the right hand, like kinetic energy or potential energy, that it is no time derivative any more. Hence, when you use the mathematical trick to integrate to infinite time and divide by infinite time, you get the same at the right hand for each of the terms.

Hence, if the left hand becomes trivially zero, each of the terms at the right hand become trivially zero as well!

2) The physics behind are of no importance? Funny! One derivates and integrates physical entities like the momentum and distance of each star. Not just maths.

Look what happens in the equation 1.2.2. of your reference book. That is physically wrong.

It states that for the impulsion, one must take the path r that is radial to the center, hence, it only takes the *radial* impulsion variations. Otherwise, one wouldn't get the moment of inertia wrt the origin in that equation, which is expressed in radii only.

So, they take *only* the radial orbit fluctuations, *not* the orbital speed and its impulsion.

Look now at the equation 1.2.5. There, they state that this *radial* impulsion defines the total kinetic energy of the system. Plain wrong!

Hence, in whole that derivation, they neglected the orbit velocities and they only used the radial orbit fluctuations, but they falsely pretended that a part of it represents the total kinetic energy of the system!

So, it is even worse than I first thought. We just found that the whole virial theorem is nonsense wrt galaxies and rotating systems...! The only application is the non-rotating cyclically pulsing star.

Hence, it is not surprizing that the wrongly applied virial theorem lead to wrong mass definitions and wrong evaluations wrt to the luminosity of galaxies! And to the silly "dark matter"!

``Since you have taken the time derivative at the left hand, you also took the time derivative at the right hand.''

What is wrong with you? Lagrange's identity, equation 1.2.13 reads

dG/dt = 2T + Omega

So there is no issue of ``taking the time derivative of the l.h.s. The l.h.s. *is* a time derivative, whereas T is the kinetic energy, with its usual definition (see the last equality of equation (1.2.5) for a definition of T). Similarly for Omega: it is the total potential energy, not its time derivative.

So what you have is that the time derivative of G equals the value ``twice total kinetic + total potential energy''. Then you do the averaging, which means  that the left-hand side vanishes, implying that the average of

2T + Omega

also vanishes. But, of course, as you yourself have pointed out, the average of Omega and of T are non-zero.

The rest of what you say is equally wrong: the virial is indeed defined to be

G = sum_i p_i dotproduct r_i

with a dot product of the *vector* r_i with the *vector* p_i, so the radial component of the momentum is indeed what contributes. As for the dependence on the origin, if you are in a reference frame where the total momentum is zero, then the virial is independent of origin.

dG/dT then contains two terms. The first is

sum_i p_i dotproduct v_i

where we take the dot product of the vectors p_i and v_i. This does give twice the kinetic energy, because in this dot product the full speed, not merely the radial one, is obtained.

Basically you should beware of using mere enthusiasm as a substitute for a basic knowledge of what a vector and what a derivative are. You should first familiarise yourself with these concepts, before you start pontificating away.

After having said yes, 4 times, at the question whether you get the equation you describe, the answer is self evidently no, and shows you have elementary math difficulties: to obtain your equation you stated:

``And, then, I can get the time averaging by performing an integration over the time at both sides of the equation, and divide it by that time?''

Sure,no problem. But *where are the time integrals on the right hand side?*. You took a time integral on both sides, and divided by t on both sides. Then you could simplify the integral on the left-hand side, but not on the right hand side. Indeed

1/t*Integral[(p_i(t') dotproduct dr(t')/dt', dt']

where the integral is taken with t' going from 0 to t, is not equal to

p_i(t) dotproduct [r(t)-r(0)]/t

You took the p_i(t) out of the integral, in a way that is completely unjustified.

The correct way to do things is as in the book: you develop things to the point where they are expressed as in 1.2.15, and then you integrate on both sides. The integral over the kinetic energy will grow linearly in the integration time and so will the integral over the potential energy. But together, the sum of the integrals will not grow linearly, because they add up, by 1.2.15, to the integral of a time derivative of a bounded function.

I hope whoever reads this apart from you has gotten the point, whereas I believe to be sure that you never will.

Clearly you are entirely clueless about how to integrate. The product rule as you state it is correct, but when I say, for example

d/dt(cos(t) sin(t)) = -sin(t) sin(t) + cos(t) cos(t)

it does not mean that any of the factors are constant in the sense of being time independent. So, of course, when you integrate the l.h.s. it is equal to

- Integral[sin(t') sin(t'))]+Integral[cos(t') cos(t'))]

with both of the integrals growing linearly with t, and of course not

-sin(t) Integral[sin(t')] + cos(t) Integral[cos[t']].

So you are entirely wrong, and it is an elementary inability to grasp the most basic differences in math (constant vs. variable, vector vs. scalar) which you combine with an unbounded arrogance to spread the most arrant nonsense around.

Over and out!

Thierry, here is my simple advice.  Just reread carefully what François wrote, and realize that before you thousands of learned people have studied and understood the virial theorem and applied it to disk galaxies without your objections.  

Daniel, we are talking about physics and about maths, not about popularity. If you find any mistake in my findings, just show them, and I will give you my humble reply...

I found many mistakes in your arguments, but won't spend time repeating essentially what François wrote.   Just study seriously what François  wrote before replying.

I have to add the following, Daniel, without adressing any personal attack whatsoever.

You are a specialist in galaxy dynamics, and so, my quest on the virial theorem is very sensible to you.

However, I think your research can possibly fit very well with some of my findings,

So, I invite you to assist to the debate, and to not put your head in the sand...

Dear all,

The virial theorem, mathematically derivated, seems to not be contradictory for rotating systems, since the equation v^2 = G M /R is valid.

Hence, mathematically, 2K + V = 0

The problem resides in the definition of the averages for the velocity (Kinetic energy) and the average mutual distance between all the masses (Potential energy).

For a spherical 3D system with randomly oriented orbits, that average distance R will be expected to be quite different from that of a thick 2D disc that has all prograde orbits.

How to define this average distance R? Isn't it than much simpler to perform the Newtonian integration over the system for each given orbit radius?

The annexed link shows a peer-reviewed paper that explains a gravity difference of 600% at some places of the 2D system compared with the 3D system, explaining the overestimation of the mass by a factor 6.

How do you understand that paper?

http://www.worldscientific.com/doi/abs/10.1142/S2424942417500049

Thierry:  There is much confusion in the discussion, so let me state a few points which I hold as true.

0.1) The reference Chap. 1 comes from George W.  Collins II book, which is regarded as an excellent reference in astrophysics, besides Chandrasekhar monographs and articles.  Collins keeps calling Clausius "Claussius" though.

0.2) The virial term introduced by Clausius is not G = sum(p_i . r_i)  but the second term in the virial theorem,  the time-average of sum(f_i . r_i), where f_i is the force on particle i.

1) At the basis of the virial theorem is the assumption of global equilibrium.  What disk galaxy dynamics requires for accounting for rotation curves is more, a local radial equilibrium.    Of course if a disk satisfies local equilibrium at each radius, it must satisfy the virial theorem, which results from averaging local equilibriums.

2) More general than the virial theorem is Lagrange's identity which works for any system of particles, steady or not, in any geometry,

   (1/2) dG / dt =  sum(m_i v_i . v_)  + sum(m_i r_i . a_i) .  

No time average needs to be taken, just that instead of assuming strict global equilibrium one needs only to suppose that the terms dG / dt  stays small with respect to each of the right hand side terms to get an approximate steady dG/dt term (the first kinetic energy term is always positive, so the second term should be negative).

3) If |dG / dt| =  0, the system is not necessarily in equilibrium, while the converse is true, so the virial theorem applies to uniformly expanding of contracting systems too. G=0 is a stricter condition, distinct from the virial theorem, which tells that the moment of inertia I  = sum(m_i r_i . r_i)  is constant.   But still some systems may satisfy G ~ 0 and be in local disequilibrium (at the very local particle level this is the case, since usually each particle moves).  

4) What is needed in disk galaxies is to verify local equilibrium at the level of a ring  of small width.  The virial theorem is thus insuffcient and can only be used for order of magnitude global check.   To find the local gravitational acceleration, one needs to solve Poisson's equation,

    div . a = -4 pi G rho,

for a the local acceleration due to gravity, rho is the local mass density (see e.g.  Binney & Tremaine "Galactic Dynamics" textbook about efficient methods to solve Poisson's equation) .   Assuming purely circular motion and local equilibrium, the local attraction should be balanced by centrifugal force, that is, a . r = v . v.   But in real galaxies, and especially in elliptical galaxies, circular motion is only a part contributing to equilibrium,  another part is due to the radial velocitiy dispersion, a kind of pressure term.   For solving the more realitstic case one needs to solve Jeans' equations (see again Binney & Tremaine for the details) which are the hydrodynamical equations for collisionless fluids.   This is much more elaborated though than the pure circular rotation case.

In summary don't think Thierry that astronomers are unable to use the virial theorem correctly.  What is involved in comparing the rotation curves to the detecetd forms of matter is much more elaborated than what you were considering.   Spherical geometry is only assumed in the simplest works.

5) Finally the article by E. Li is wrong because the author assumes that the acceleration is only produced by the inner mass at a given position, so the author does not solve Poisson's equation correctly (in a disk the local acceleration is composed of both the attraction made by the inner particles, pulling inwards, and the attraction made by outer particles, pulling outwards). 

http://bifrost.cwru.edu/personal/collins/virial/

Dear Daniel,

Thank you very much for this extended explanation, with which I agree for the largest part, and thanks for the link.

There are however two issues:

1) The global check of the system's mass (global check) by the virial theorem (see first link annexed).

If we look at the extreme limit for elliptic galaxies, which are disc galaxies, the way of evaluating the virial theorem is said to depend from the average distance between the masses, two by two, which defines the potential energy in the virial theorem.

For the evaluation of the kinetic energy, the position vector is used.

As I said, thick prograde 2D systems with orbits that move at the same speed have numerous stars that are sometimes very close, at a fraction of the radius, while 3D systems with random orbits always have all the stars being most of the time (mutually) far away.

The method of the virial theorem (as annexed) takes the overall radius for any system.

So, that method must be improved. However, for elliptical galaxies, a larger part of the orbits, much more than 50% will be globally prograde. Hence, also there, the "corrections" must be significant, and not just a shape correction between spherical and ellipsoidal.

2) I agree that “in a disk the local acceleration is composed of both the attraction made by the inner particles, pulling inwards, and the attraction made by outer particles, pulling outwards”, and I appreciate that you remarked that. However, the author's argument is that in hollow *spheres* the outer shell doesn't influence the inner parts. Since the error for disc galaxies is proportional to the matter behind (farther than) the circle-chord through a given star (tangent to the orbit), the error is larger for more inner orbits, and become smaller for more outer orbits, even more because the outer part of the disc is also much more 3D than 2D.

Now, Li finds the huge difference for the outer edges, and I don't think there can be any cumulative error in this reasoning.

I did myself a Newtonian calculus (Poisson's equation) for the disc of a disc galaxy, with a density slope ~1/R, but this must be understood as follows: it is calculated for the spirals, and I consider all the stars in the thick 2D disc (of a given thickness), while in reality, the disc becomes thicker, which is lowering the observed density.

The global results are complying well with observation. (see second link herewith).

http://www.astro.cornell.edu/academics/courses/astro201/vt.htm

https://gsjournal.net/Science-Journals/Research%2520Papers-Astrophysics/Download/2245

Dear all,

For your information, my latest ongoing insights pushed me to severely modify the details of my question formulation of this thread.

The issue of the virial theorem has been greatly solved thanks to the clear explanations of F. Leyvraz and others.

Now, it reduces to the following:

1) The definition of the mass evaluation of disc galaxies is wrong, because the overall diameter is used in the equation of the mass evaluation, while in reality, there is a big difference in the distances between the stars, two by two, when finding the potential energy equation in the virial theorem.

The distances of stars in disc galaxies are in many cases very close for a long time, while the distances in spherical galaxies are in almost all cases much further, and that is certainly so in average.

2) The infinite time averaging for the non-rotational component of the masses (like in pulsing systems) is non-physical, because it leads either to a collapse in a singularity, or in an infinite widening of the system. So, the left side hand of the virial theorem can physically not become zero.

The first issue causes "dark matter" in the disc galaxies, while there is no dark matter when integrating the forces over the disc.

The second issue causes "dark matter" in Zwicky's galaxy cluster issue.