Риччи Тензор

[Munin]

Why, Susskind's lectures are full of examples, unless you were expecting something like explicit functions solving the discussed equations. Then yes, known explicit solutions are not many. I can quote you one, the embedding of a curved spatial section of a curved spacetime of the 'same' physical system - a ball of constant density (not really same because the ball is distorted) - into Euclidean space (MTU Box 23.2 (4)):

$$z(r)=\left\{\begin{array}{lll} \sqrt{\dfrac{R^3}{2M}}\left[1-\sqrt{1-\dfrac{2Mr^2}{R^3}}\right]&\qquad&\text{for \(r\leq R\),}\\ \mathstrut&&\\ \sqrt{\dfrac{R^3}{2M}}\left[1-\sqrt{1-\dfrac{2M}{R}}\right]+\sqrt{8M(r-2M)}-\sqrt{8M(R-2M)}&\qquad&\text{for \(r\geq R\),} \end{array}\right.$$

You can feed it to a 3d grapher if you want.

For Poisson's equation...
First of all, differential equations don't have some single solution. That's because differentiation lose some information about a function being differentiated. Compare, when you solve an equation $x^2=4,$ you have two roots because squaring loses information about the sign. For differential equations, you have infinitely many solutions. To choose just one of them, you have to add some additional conditions, which are not differential equations by themselves (usually), but help to get rid of that infinity. They are called initial conditions, boundary conditions etc. When the main DE is coupled with proper conditions, it is called a problem. So, when you said that the density is fixed, you didn't yet set the problem, and answers can still be many.

Then, let's move by steps. Poisson's equation is a second-order $n$-dimensional equation, so let's see first, what is a 1-dimentional second-order DE, and what can be its solutions. I'll take that particular DE (it is not that any second-order DE looks like this):

$$\dfrac{d^2\,\phi(x)}{dx^2}=f(x).$$

And let's set $f(x)$ to be a constant $f(x)=f.$ Then, one possible solution is $\phi(x)=\tfrac{f}{2}x^2$ - check it. But it is not unique. If we add 1 to it, then that will be a solution as well, and we can do that as many times as we please. Also we can add $x$ to it, and again arbitrary many times. Actually, the full solution of that equation (not a problem yet) would be

$$\phi(x)=\tfrac{f}{2}x^2+C_1x+C_2,\qquad\text{with arbitrary constants \(C_1,C_2\in\mathbb{R}\).}$$

If we want to get only one function, we have to pose additional conditions, like known values of $\phi$ in some points, or of its derivatives. For example, this is a correct problem for that equation:

$$\dfrac{d^2\,\phi(x)}{dx^2}=f(x),\qquad \phi(x_1)=\phi_1,\quad \phi(x_2)=\phi_2.$$

Now we can substitute $x_1$ and $x_2$ into that solution formula, and find particular values of $C_1$ and $C_2.$ That will make a solution to the problem.

For the actual Poisson's equation, it's $n$-dimensional, and the freedom for its solutions is much larger, and what problems can be set to it - is covered by a whole new chapter of DE theory - called PDE theory, or classical equations of mathematical physics. I won't get into detail, and just say that for the 3-dimensional equation

$$\nabla^2\phi(x,y,z)\equiv\dfrac{\partial^2\,\phi}{\partial x^2}+\dfrac{\partial^2\,\phi}{\partial y^2}+\dfrac{\partial^2\,\phi}{\partial z^2}=\rho$$

with the constant right hand part, the general form of a solution looks like

$$\phi(x,y,z)=\tfrac{\rho}{6}(x^2+y^2+z^2)+h(x,y,z),$$

where $h(x,y,z)$ is a function chosen with very much freedom, though not fully arbitrary, and its particular possible forms are described in textbooks. The set of possible functions $h(x,y,z)$ is defined by that $\nabla^2h(x,y,z)=0,$ and it bears a special name of harmonic functions. A possible example is $h(x,y,z)=\sin x\,\,\exp y$ (check again, if you know partial derivatives). So, these functions can change the form of a solution greatly, making it, for example,

$$\phi(x,y,z)=\tfrac{\rho}{4}(x^2+y^2)$$

(much less symmetric to the eye than the earlier formula, but actually being just the sample of it).

If I'd wish to set a problem for Poisson's equation, I would have to set a domain, finite or infinite, set a function of density inside that domain, and set values of $\phi$ at the boundaries of the domain (even in the infinite points). This is one possible kind of problems. So, you may ask what if density is constant, and then $\phi$ is zero on the boundary of some ball? Then I would answer unambiguously,

$$\phi(x,y,z)=\tfrac{\rho}{6}(x^2+y^2+z^2-R^2).$$

And you've seen this formula before, in post599164.html#p599164 .

[GRstudy]

Basically, the Poisson's equation says that

"Partial second derivatives of x-y-z components of the gravitational potential are related to the density function by factor of 4 pi G".

Gravitational potential, when differentiated, becomes

$$\frac{\partial^2 \Phi}{\partial x^2}=\frac{GM}{x^3}$$

.

I am just curious if something like Poisson's equation or Ricci tensor has any applications in real-life. I think those are purely theoretical insights.

Speaking about DE, there are many solutions to Einstein Field Equations. I remember proving that Ricci Tensor vanishes in Schwarzschild solution. The most difficult is Kerr metric which describes space time around a rotating spherical object.

Also, are these two equations familiar to you?

$$\frac{d^2x^{\mu}}{d\tau^2}=-\Gamma^{\mu}_{\alpha \beta} \frac{dx^{\alpha}}{d\tau} \frac{dx^\beta}{d\tau}$$

$$\delta V^{\mu}=R^{\mu}_{\sigma \nu \tau}dx^{\nu} dx^{\tau} dV^{\sigma}$$

-- 27.07.2012, 11:57 --

Also, I have a question for you:

It's from Hartle, page 215, pr. 8

"A spaceship without power is moving around a circular orbit about a black hole of mass M. The Schwarzschild radius of the orbit is 7M. What is a period of orbit being measured by a clock on a spaceship and period of orbit as measured by an observer at infinity?"

I know that time measured on spaceship would be less. For example, if orbital period measured on spaceship is 90 minutes, then at infinity it would be like 200 minutes. Time slows down as you get closer to a black hole.

[Munin]

GRstudy в сообщении #599962 писал(а):

Basically, the Poisson's equation says that

"Partial second derivatives of x-y-z components of the gravitational potential are related to the density function by factor of 4 pi G".

Gravitational potential, when differentiated, becomes

$$\frac{\partial^2 \Phi}{\partial x^2}=\frac{GM}{x^3}$$

.

Now we're stuck into the problem when you try to reason and discuss things you are not familiar with. It is a scourge for those who try to self-educate and grasp something without tutors. It is an understandable desire but here hides a pitfall: everything you would thunk up will be erroneous. You just don't know the right road, and wonder nowhere near it. You should learn to adjust your pace to be much slower, with cautious checks for every step, like you're walking in a swamp. Worst of all, you don't notice your errors until someone tells you or you decide to check, and you can wonder too far, memorizing your mistakes as good ideas, and becoming more and more confident that you're right. At some point, no persuasion by other people would be able to change your mind, and you would be sure they are just freaks, not understanding a bit in the things they talk about - at that point you would become freak yourself. I want you to be scared of this way, to always look where you are and know your limits. Your current limits, of course.

Basically, the Poisson's equation says that "A special kind of partial second derivative of the gravitational potential..." and so on. In multivariable calculus, there is not one, but many kinds of derivatives, of first order, of second order, of higher orders, and the Poisson's equation speaks only of one of them - the Laplacian,

$$\Delta\equiv\nabla^2\equiv\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}$$

(written for 3 dimensions, for $n$ dimensions changing appropriately). You should not think of it as of 'just second derivative', you should think of it as of exactly this derivative, and calculate it exactly by this formula.

I don't know what gravitational potential you took to calculate your derivative, but i can imagine it was a Coulomb potential

$$\phi=-\dfrac{GM}{r}$$

(that's the best guess). Then your calculations after that were totally wrong. First, it should be written in Cartesian coordinates as

$$\phi(x,y,z)=-\dfrac{GM}{\sqrt{x^2+y^2+z^2}},$$

then a partial derivative should be taken, with a great care to make no mistakes, because a mistake in the beginning falsifies all the rest calculations. The result is

$$\dfrac{\partial\,\phi(x,y,z)}{\partial x}=\dfrac{\tfrac{1}{2}GM}{\left(\sqrt{x^2+y^2+z^2}\right)^{3}}\cdot 2x=\dfrac{GMx}{\left(\sqrt{x^2+y^2+z^2}\right)^{3}}.$$

Now the second derivative can be taken:

$$\dfrac{\partial^2\,\phi(x,y,z)}{\partial x^2}=-\dfrac{\tfrac{3}{2}GMx}{\left(\sqrt{x^2+y^2+z^2}\right)^{5}}\cdot 2x\,+\,\dfrac{GM}{\left(\sqrt{x^2+y^2+z^2}\right)^{3}}=\dfrac{GM\left(-2x^2+y^2+z^2\right)}{\left(\sqrt{x^2+y^2+z^2}\right)^{5}}.$$

It looks rather complex, and that's how it is, because the Poisson's equation deals not with just one, but with three such derivatives, with respect to $x,y$ and $z,$ and only after they put together, they cancel each other and make something much simpler (in this case, 0).

-- 27.07.2012 19:42:43 --

GRstudy в сообщении #599962 писал(а):

I am just curious if something like Poisson's equation or Ricci tensor has any applications in real-life. I think those are purely theoretical insights.

Of course they have, and many! Any CAD (computer-aided design system) works with such equations, and solves tons of them, numerically, when a designer draws something and wants to see, what would be stresses in the beam, flows in the volume, currents in the metal, or heat currents around the component, etc. The Poisson's equation is one of many equations to calculate this, and one of the simplest, so some other equations give the Poisson's equation when simplified. Ricci tensor has much narrower applications, of course :-)

GRstudy в сообщении #599962 писал(а):

Also, are these two equations familiar to you?

$$\frac{d^2x^{\mu}}{d\tau^2}=-\Gamma^{\mu}_{\alpha \beta} \frac{dx^{\alpha}}{d\tau} \frac{dx^\beta}{d\tau}$$

$$\delta V^{\mu}=R^{\mu}_{\sigma \nu \tau}dx^{\nu} dx^{\tau} dV^{\sigma}$$

Sure they are. Are you testing me or what? (Though the second one I used to to see in a slightly different form, without $\delta.$)

GRstudy в сообщении #599962 писал(а):

Also, I have a question for you:

It's from Hartle, page 215, pr. 8

"A spaceship without power is moving around a circular orbit about a black hole of mass M. The Schwarzschild radius of the orbit is 7M. What is a period of orbit being measured by a clock on a spaceship and period of orbit as measured by an observer at infinity?"

I know that time measured on spaceship would be less. For example, if orbital period measured on spaceship is 90 minutes, then at infinity it would be like 200 minutes. Time slows down as you get closer to a black hole.

So, what is your question?

Also, don't forget, time slows down as well as when you move with some nonzero velocity.

[GRstudy]

If you remember specific lectures in 18.02 where these ideas are described, I would appreciate you directing me. Only for the fact that I am sort of obsessed in nothing except in these super rare and advanced topics while my classmates are having fun-- I deserve some respect no matter how "incorrect" I may seem.

I believe that you meant was an operator that contained a set of second partial derivatives; you called it

$$\nabla^2$$

. If it a set of just partial derivatives, then it would be divergence operator

$$\nabla$$

. Divergence measures how the field tends to spread out. For example,

1) Proton--Positive Divergence
2) Electron--Negative Divergence
3) Neutron and Magnetic Field--Zero Divergence
4) Gravitational field of the Earth--Negative Divergence

Actually, the divergence of an electric field is charge density divided by some vacuum constant. Also, the Divergence of Magnetic Field is zero as monopoles don't exist in nature (basically, you can't have North Pole separated from South Pole). I believe all of these ideas are composing Maxwell's equations.

Цитата:

then a partial derivative should be taken, with a great care to make no mistakes, because a mistake in the beginning falsifies all the rest calculations. The result is

After that, I think you meant 2nd partial derivative, not an ordinary derivative.

Цитата:

So, what is your question?

Цитата:

"A spaceship without power is moving around a circular orbit about a black hole of mass M. The Schwarzschild radius of the orbit is 7M. What is a period of orbit being measured by a clock on a spaceship and period of orbit as measured by an observer at infinity?"

[Munin]

GRstudy в сообщении #600194 писал(а):

If you remember specific lectures in 18.02 where these ideas are described

I haven't even see these lectures (unless Susskind's)! :-) So please give me a more exact reference.

GRstudy в сообщении #600194 писал(а):

Only for the fact that I am sort of obsessed in nothing except in these super rare and advanced topics while my classmates are having fun-- I deserve some respect no matter how "incorrect" I may seem.

In no way I wanted to be disrespectful. I only want to direct you away from known pitfalls, and warn againts mistakes.

GRstudy в сообщении #600194 писал(а):

I believe that you meant was an operator that contained a set of second partial derivatives; you called it $\nabla^2$.

Yes. That's the shorthand of taking the $\nabla$ twice. We take the scalar field $\phi(x,y,z),$ then apply $\nabla$ once, and get what is called the gradient of the field, $\nabla\phi\equiv\operatorname{grad}\phi.$ The gradient shows where the scalar field tends to grow, measures the speed of growth, and points in the direction of the fastest growth, 'up the slope'. The gradient is a vector, and that's why the next $\nabla$ applied to it can have a sense of the divergence. It is written as $\nabla\cdot(\nabla\phi)\equiv\operatorname{div}\operatorname{grad}\phi,$ and since no ambiguity can arise, it is interpreted as $\nabla\cdot(\nabla\phi)=(\nabla\cdot\nabla)\phi=\nabla^2\phi,$ where the power 2 means squared as a vector. So actually this is an operator of the divergence of the gradient, written in a handy short way.

GRstudy в сообщении #600194 писал(а):

After that, I think you meant 2nd partial derivative, not an ordinary derivative.

Yes.

GRstudy в сообщении #599962 писал(а):

Also, I have a question for you:

It's from Hartle, page 215, pr. 8

Do you want me:
to tell you how it is solved,
to solve it myself,
or to take the answer from somewhere and just calculate it?
I'd rather not do the second :-) It's a piece of work, and it is intended to be done by student, to get more familiar with the way of calculations. And the first is probably too early for you now.

[GRstudy]

I would like to have a complete solution to that problem.

[Munin]

Can you read the section 9.3 "Particle Orbits", and say if you can find there formulas to calculate the speed, or the angular velocity, of the spaceship, knowing that the orbit is circular, and its radius? Actually I found a more direct formula, and I wonder if you would find it too.

[GRstudy]

I have actually no idea. All I know is that the speed of a satellite is given by a square root of GM/R.

[Munin]

Ok, do you understand anything that is said in the section 9.3?

Maybe there would be a good idea to read a Newtonian theoretical analysis of Kepler's laws before. Landau and Lifshitz 'Mechanics' §§ 14, 15 'Motion in a central field' and 'Kepler's problem' would do.

[GRstudy]

How to solve this equation?

$$\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}=4 \pi G 5520$$

It's 2D space...

[Munin]

First you have to learn how to solve 1D equations. This one equation (a Poisson's equation) is described in the courses of Mathematical Physics or PDE, that is a more advanced course than ODE.

Why didn't you answer my last questions?

-- 31.07.2012 17:12:40 --

I am not familiar with textbooks in English, so the best I can suggest is that you read
Morse, Feschbach 'Methods of Theoretical Physics'
chapters 1, 2, 5, 6, 7, 10. The chapter 10 is named 'Solutions of Laplace's and Poisson's equations'.

[GRstudy]

Nope, I don't know 9.3.

But I figured out Poisson's equation's solution.

$$k=4 \pi G$$

$$\phi (x,y,z) = \frac{k \rho (x^2 + y^2 + z^2) }{6}$$

, when multiplied by operator of second partial derivatives, it should reduce to

$$k\rho$$

You are correct, when twice differentiated some terms may vanish.

Also,

$$\phi (x,y) = -\frac{GM}{\sqrt{x^2+y^2}}$$

when put into 3d plotter yields something that looking very similar to space time curvature of a very dense object. Looks like a geometry around a supermassive black hole. I don't get why does it bend so badly...check for yourself...

-- 01.08.2012, 13:42 --

Also, please give me a real example where Einstein Tensor has non-zero components.

[Munin]

The most important thing you should get is that

$$\phi(x,y,z)=\dfrac{\rho(x^2+y^2+z^2)}{6}$$

is not the only solution. There are many others.

GRstudy в сообщении #601814 писал(а):

Looks like a geometry around a supermassive black hole.

Actually, no. This potential has a singularity only in the point $r=0.$ The black hole has a singularity on an $r_g>0.$

GRstudy в сообщении #601814 писал(а):

I don't get why does it bend so badly...

Why not? Seems that simple electrostatics can impress you not less that general relativity, so I can't get why you narrow your interests that much.

GRstudy в сообщении #601814 писал(а):

Also, please give me a real example where Einstein Tensor has non-zero components.

Ok, I'll get an example from MTU Box 23.2 one more time: the space-time metric in $r<R$ is

\[\begin{split} ds^2=-\Bigl(\tfrac{3}{2}\sqrt{1-\tfrac{8\pi}{3}\rho R^2}-\tfrac{1}{2}\sqrt{1-\tfrac{8\pi}{3}\rho r^2}\,\Bigr)^2dt^2+{}\\ {}+\dfrac{dr^2}{1-\tfrac{8\pi}{3}\rho r^2}+r^2(d\theta^2+\sin^2\theta\,\,d\phi^2), \end{split}\]

that is, the metric tensor has non-zero components

$$g_{tt}=-\left(\dfrac{3}{2}\sqrt{1-\tfrac{8\pi}{3}\rho R^2}-\dfrac{1}{2}\sqrt{1-\tfrac{8\pi}{3}\rho r^2}\,\right)^2$$

$$g_{rr}=\dfrac{1}{1-\tfrac{8\pi}{3}\rho r^2}$$

$$g_{\theta\theta}=r^2$$

$$g_{\phi\phi}=r^2\sin^2\theta$$

Calculation of components of the Einstein tensor I leave to you. Don't expect them to be simple: they are simple in principle but that is masked by not-so-natural coordinates.

[GRstudy]

That's Schwarzschild interior solution! It's a big mess to solve it. I would like to get an answer (components of Einstein Tensor). I have managed previously to prove that Schwarzschild metric has vanishing Einstein Tensor--that was enough! It took me a ridiculous amount of time to show that. So please, write down the components of Einstein Tensor of interior solution.

I agree, differential equations have infinite number of solutions. I guess, when defining the potential, you missed "k", the constant, which is 4 pi G.

I find General Relativity very fascinating!

Waiting for Einstein Tensor!

Thanks!

[Munin]

GRstudy в сообщении #601969 писал(а):

That's Schwarzschild interior solution!

No, not at all. The Schwarzschild's solution is found for empty space (even the interior part, which is black hole's interior), while here the star's interior is considered.

GRstudy в сообщении #601969 писал(а):

It's a big mess to solve it.

There's nothing to solve here. The solution is already found. Now only some conversions of the answer could be done.

And actually, no, the Schwarzschild's solution is not a big mess to find, it is rather basic.

GRstudy в сообщении #601969 писал(а):

I would like to get an answer (components of Einstein Tensor).

NO. Either you find it yourself (and I can help you in the case you get stuck), or you learn how to find it in textbooks (where I have taken it). Giving you the result would be worthless.

GRstudy в сообщении #601969 писал(а):

I have managed previously to prove that Schwarzschild metric has vanishing Einstein Tensor--that was enough! It took me a ridiculous amount of time to show that.

Congratulations getting to the answer at all! And this was not the wasted effort on no account. You have made some valid calculations, and you have learned that you are still too unskilled to make them easily. Now you know that you need much more practice, and that should motivate you to dig into textbooks and problem books on basic subjects, such as multivariable calculus and linear and tensor algebra.

GRstudy в сообщении #601969 писал(а):

I agree, differential equations have infinite number of solutions. I guess, when defining the potential, you missed "k", the constant, which is 4 pi G.

I omitted it as being absolutely irrelevant. I could either solve $\nabla^2\phi=4\pi G\,\rho$ and write down the solution along with $4\pi G,$ or I could solve $\nabla^2\phi=\rho$ and write down the solution without any unnecessary multipliers. Note that $\tfrac{1}{6}$ is not unnecessary.