Metric tensor in spherical coordinates

[GRstudent]

So my matrix representation of the Stress-Energy Tensor is right after all?

 

[PeterDonis]

So my matrix representation of the Stress-Energy Tensor is right after all?

Muphrid said "differentials", not "partial derivatives". He meant that you need to define "energy per unit volume" in the limit of very small volumes, not that you should use the rate of change of energy with respect to volume. They're different things.

Also, you have \rho in some components instead of E / V, but aren't those two supposed to be the same thing?

Also, you haven't defined how you are going to actually measure any of these quantities.

Finally, however, you haven't derived any of this from first principles, so it's certainly not a "definition" of what the stress-energy tensor is. At best, what you've written could be a sort of indication of what the tensor components might mean, physically, but even that is strained because the meaning of tensor components is frame-dependent. The actual physical observables are contractions of the tensor with suitable vectors; for example, as pervect pointed out, p^{a} = T_{ab} u^{b}, is a 4-vector describing the energy density and momentum density seen by an observer with 4-velocity u^{b}. *If* it happens that the observer is at rest in the particular coordinates you are using, i.e., if u^{b} = (1, 0, 0, 0), *then* it will turn out that p^{a} = (T_{00}, T_{10}, T_{20}, T_{30}), so T_00 will give the energy density seen by this particular observer, and (T_10, T_20, T_30) will give the momentum density seen by this particular observer. But an observer with a different 4-velocity will see *different* energy and momentum densities, involving other components of the SET.

 

[GRstudent]

Muphrid said "differentials", not "partial derivatives". He meant that you need to define "energy per unit volume" in the limit of very small volumes, not that you should use the rate of change of energy with respect to volume. They're different things.

I have just noticed that. You are correct. I will change it.

Also, you have ρ in some components instead of E/V, but aren't those two supposed to be the same thing?

I used c=1, so from E=mc^2, we have that energy and mass are actually the same.

Finally, however, you haven't derived any of this from first principles, so it's certainly not a "definition" of what the stress-energy tensor is. At best, what you've written could be a sort of indication of what the tensor components might mean, physically, but even that is strained because the meaning of tensor components is frame-dependent. The actual physical observables are contractions of the tensor with suitable vectors; for example, as pervect pointed out, pa=Tabub, is a 4-vector describing the energy density and momentum density seen by an observer with 4-velocity ub. *If* it happens that the observer is at rest in the particular coordinates you are using, i.e., if ub=(1,0,0,0), *then* it will turn out that pa=(T00,T10,T20,T30), so T_00 will give the energy density seen by this particular observer, and (T_10, T_20, T_30) will give the momentum density seen by this particular observer. But an observer with a different 4-velocity will see *different* energy and momentum densities, involving other components of the SET.

I didn't mention about that. Well, that's something trivial here. What important is that we can calculate the energy and momentum of, say, a particle moving through a space.

Here is modified version of my matrix:

T^{\mu \nu}=\left [ \begin{matrix} <br /> \rho &amp; \dfrac{d{E}}{A^x d t} &amp; \dfrac{d{E}}{A^y d t} &amp; \dfrac{d{E}}{A^z dt}\\ <br /> \rho v_{x} &amp; \dfrac{d(mv_{x})}{A^x dt} &amp; \dfrac{d(mv_{x})}{A^y dt} &amp; \dfrac{d(mv_{x})}{A^z dt}\\ <br /> \rho v_{y} &amp; \dfrac{d(mv_{y})}{A^x dt} &amp; \dfrac{d(mv_{y})}{A^y d t} &amp; \dfrac{d(mv_{y})}{A^z d t}\\ <br /> \rho v_{z} &amp; \dfrac{d(mv_{z})}{A^x d t} &amp; \dfrac{d(mv_{z})}{A^y d t} &amp; \dfrac{d(mv_{z})}{A^z d t} <br /> \end{matrix} \right ]

 

[PeterDonis]

Well, that's something trivial here.

No, it isn't; it's critical. See below.

What important is that we can calculate the energy and momentum of, say, a particle moving through a space.

Yes, but what you are writing down for the various SET components *assumes* that the thing we are trying to calculate the energy and momentum of has a particular state of motion relative to us. If the state of motion is different, the SET components will *not* have the meanings you are writing down; T_00 will *not* be the energy density we measure, (T_10, T_20, T_30) will *not* be the momentum density we measure, etc. What you are writing down assumes that T_00 *is* the energy density, etc., etc. So what you are writing down is not generally true; it's only true under a particular set of circumstances.

 

[Muphrid]

Really, I think taking the idea of finding the flux of four-momentum across various hypersurfaces would make this task to find the matrix representation more reasonable and give a better jumping off point for general matter distributions.

 

[GRstudent]

Yes, but what you are writing down for the various SET components *assumes* that the thing we are trying to calculate the energy and momentum of has a particular state of motion relative to us. If the state of motion is different, the SET components will *not* have the meanings you are writing down; T_00 will *not* be the energy density we measure, (T_10, T_20, T_30) will *not* be the momentum density we measure, etc. What you are writing down assumes that T_00 *is* the energy density, etc., etc. So what you are writing down is not generally true; it's only true under a particular set of circumstances.

In my matrix, I assume that two observes are looking at each other--they are moving at the same speed. And I wanted to make sure that I have given the right formulas for each component--that's all. One of the guys, pointed out that I should have written the differential differences as opposed to delta which is a finite difference. I am open to other comments about formulas which correspond to each components of the stress-energy tensor matrix that I wrote.

So if we are talking about a particle which is moving with velocity v with respect to us, then we should multiply each component by v?

 

[PeterDonis]

In my matrix, I assume that two observes are looking at each other--they are moving at the same speed.

Ok, so it *is* restricted to a particular set of circumstances.

And I wanted to make sure that I have given the right formulas for each component--that's all.

Right formulas in terms of what? You still haven't described how you are going to actually measure any of the quantities in your formulas. Without that, no one can tell whether the formulas are "right".

So if we are talking about a particle which is moving with velocity v with respect to us, then we should multiply each component by v?

No. It's nowhere near that simple. But in any case I don't think you're ready for that exercise yet; first I think you need to explain how you are going to measure all the quantities appearing in your formulas for the special case where the observer and the substance being observed are at rest with respect to each other, which is what you say your formulas apply to.

 

[GRstudent]

how you are going to measure all the quantities appearing in your formulas for the special case where the observer and the substance being observed are at rest with respect to each other, which is what you say your formulas apply to.

I would ask you the same question. How?

 

[PeterDonis]

I would ask you the same question. How?

You're the one that wrote down the formulas, so it's up to you to explain what they mean and how to measure the quantities in them. I didn't write them down, so I don't know what you meant by them.

 

[micromass]

GRstudent, what book are you using to learn relativity?? You are using a book, right??

 

[GRstudent]

You're the one that wrote down the formulas, so it's up to you to explain what they mean and how to measure the quantities in them. I didn't write them down, so I don't know what you meant by them.

Ok, let's take T^{32}. It is the flow of the z-component of momentum in y direction. So I have \dfrac{d(mv_{z})}{A^y d t}. Also, the flow of some quantity is given by \dfrac{\Delta Q}{A \Delta t}, where Q is the quantity, A is an area through which the Q flows and t is a time interval.

GRstudent, what book are you using to learn relativity?? You are using a book, right??

I use a combination of different sources of information.

 

[Dale]

GRstudent, I would recommend a different approach for understanding the SET. I would start with some standard SETs, like the perfect fluid, and transform them to other coordinate systems to see what they look like.

 

[GRstudent]

^
OK, write down your own matrix.

 

[Dale]

Here it is written out
http://en.wikipedia.org/wiki/Stress–energy_tensor#Stress-energy_of_a_fluid_in_equilibrium

 

[GRstudent]

^
so what difference did you find between those components and my components?

 

[GRstudent]

Also how would you intuitively distinguish Riemann from Ricci tensor?

 

[Dale]

^
so what difference did you find between those components and my components?

A lot of mine are zero

GRstudent, I am not trying to correct your components; I am just suggesting an alternative approach for learning what a SET is. If you have already learned enough with your approach and feel comfortable with the SET and don't want to try my suggestion then that is fine by me.

 

[Dale]

Also how would you intuitively distinguish Riemann from Ricci tensor?

Riemann is a rank 4 tensor and Ricci is a rank 2 tensor. I never felt that there was any confusion, so I am not sure what you are asking.

 

[GRstudent]

^
I wasn't asking about "mathematical" difference between those tensors. I was asking about "intuitive" difference between them.

GRstudent, I am not trying to correct your components; I am just suggesting an alternative approach for learning what a SET is. If you have already learned enough with your approach and feel comfortable with the SET and don't want to try my suggestion then that is fine by me.

First, I was looking for comments about my SET (matrix). And second, if you think that other SETs would help me better understand my matrix then you, of course, can write down (here!) your own SET instead of linking me the link which I myself linked here.

 

[Dale]

I wasn't asking about "mathematical" difference between those tensors. I was asking about "intuitive" difference between them.

I just don't see how there is any room for intuitive confusion between them, they are tensors of different ranks. They are as different as scalars and vectors, or vectors and matrices. Can you describe better what it is about them that confuses you?

Do you have any confusion between the stress energy tensor and the four-momentum? The relationship between Riemann and Ricci is similar. They may have similar units, but they have different ranks.

 

[GRstudent]

^
For example, we say that intuitively Gammas represent how the metric tensor g_{mn} varies from place to place. So in this sense, I have some intuition for Gammas. The Riemann tensor is made of second derivatives of the g_{mn} and product of the Gammas; so, in this sense I have a slightly shady view of what Riemann tensor really is; and Ricci tensor is made up of a sum of the Riemann tensor. So what's the intuition here?

 

[Dale]

I have a slightly shady view of what Riemann tensor really is

Why don't we start here then. Do you know what parallel transport is and how parallel transport around a closed loop doesn't bring a vector back to itself in curved spaces?

If not, then I would recommend watching this lecture:
http://www.youtube.com/watch?v=Pm5ROyoaMZA&feature=relmfu

If so, then you can start with the next one:
http://www.youtube.com/watch?v=xOzVmC-4gpw&feature=relmfu

 

[GRstudent]

If not, then I would recommend watching this lecture:
http://www.youtube.com/watch?v=Pm5RO...feature=relmfu

That's Susskind! Haha! Yeah, I was following these lectures. I have managed to understand first five but stuck on this 6th lecture. I'll give it a try today.

 

[Dale]

OK, that is a good starting point. If you get stuck somewhere in the lecture, please let me know and we can go from there.

 

[GRstudent]

^
OK, I'll definitely let you know. Thanks!

 

[PeterDonis]

Ok, let's take T^{32}. It is the flow of the z-component of momentum in y direction.

This part is ok.

So I have \dfrac{d(mv_{z})}{A^y d t}. Also, the flow of some quantity is given by \dfrac{\Delta Q}{A \Delta t}, where Q is the quantity, A is an area through which the Q flows and t is a time interval.

You're assuming that momentum = mv, which is only true in the non-relativistic limit. You're also assuming that whatever you are measuring the stress-energy of can be assigned a rest mass m, but in other components you assign an energy density rho instead. For something like a fluid, for example, rho is well-defined, but how would you define m? And what about the stress-energy tensor for an electromagnetic field, which doesn't even have a rest mass to begin with?

I would agree with DaleSpam's recommendation to work this from the other direction: look at some actual SETs for different kinds of things and develop an intuition for what the components mean from that.

 

[GRstudent]

OK, that is a good starting point. If you get stuck somewhere in the lecture, please let me know and we can go from there.

Ok, I listened to the first link that you gave (lec 6). The main point of the conversation was about parallel transport and the curvature in general sense. So the formula was that d \theta = R d a. Basically, the smaller the sphere is, the larger the value of curvature is. And this makes sense because if Earth were smaller in its Volume then we would feel that it is move curved. In the same way, if the Earth were large in its Volume then we would more and more perceive that it is flat. Therefore, this formula makes a lot of sense.

 

[Dale]

Do you feel like you have a good understanding of what parallel transport is and how parallel transport along closed loops gets you back to the same vector in flat space but not in curved space?

 

[GRstudent]

Do you feel like you have a good understanding of what parallel transport is and how parallel transport along closed loops gets you back to the same vector in flat space but not in curved space?

Sort of. I understand why in parallel transport around the closed loop the vector in curved space don't come back to its initial form. Susskind, in Lec 6, gave a tip of the cone as an example. Around the cone, the final and initial vector are the same; however, at the tip, the final vector deviates by \theta from its original form. And the \theta is a deficit angle. In this case, the d\theta=R dA, so R=\dfrac{d\theta}{dA}. Here I tread R as sort of a curvature (not curvature tensor!) value at some point on the surface of a particular object. Parallel transport is just keeping the vector parallel to itself. The parallel transport is a test to find out whether the space has curvature or not. The mathematics is d V^{m}=-\Gamma^m_{np}{} V^{n} dx^p. It is derived from \dfrac{dV^m}{ds}+\Gamma^m_{np}{} V^n \dfrac{dx^p}{ds}=0. So we have \dfrac{d V^{m}}{ds}=-\Gamma^m_{np}{} V^{n} \dfrac{dx^p}{ds}. And we multiply both sides by ds to get to my equation.

 

[Dale]

I understand why in parallel transport around the closed loop the vector in curved space don't come back to its initial form.

OK, good. That is the basis of defining the Riemann curvature tensor. Susskind will give a much better and more detailed explanation in the 7th lecture, but basically, the Riemann curvature tensor describes how much and what direction a vector deviates when it is parallel transported around an infinitesimal loop in a particular orientation. You need a total of four indices to describe that completely.