Conservation of energy-momentum (tensor)

[Apashanka]

For a curve parametrised by ##\lambda## where ##\lambda## is along length of the curve and is 0 at one end point.
At each ##\lambda## say tangent vector V and A be the two possible vectors of the tangent space.
where ##V=V^\mu e_\mu## and ##A=A^\nu e_\nu##, {e} are the basis vectors.
Now ## \nabla_A V=A^\mu \nabla_\mu(V^\nu e_\nu)=A^\mu(\nabla_\mu V^\nu)e_\nu+A^\mu V^\nu(\nabla_\mu e_\nu)##Now if ##\nabla_\mu V^\nu=0## then the covariant derivative is still not zero .
Similarly from the energy-momentum conservation ##\nabla_\mu T^{\mu \nu}=0 ## from this can we say that in general the covariant derivative of this energy --momentum is non-zero by just comparing to rank 1 tensor??
May be there is something wrong above (I apologise) as I am trying to learn these things...
Thank you

 

[PeterDonis]

At each ##\lambda## say tangent vector V and A be the two possible vectors of the tangent space.

This would mean that the manifold is two-dimensional. Is that your intent? If you are asking about energy-momentum conservation, you are asking about spacetime, which is four-dimensional, not two-dimensional.

Now ##\nabla_A V=A^\mu \nabla_\mu(V^\nu e_\nu)=A^\mu(\nabla_\mu V^\nu)e_\nu+A^\mu V^\nu(\nabla_\mu e_\nu)##

Similarly from the energy-momentum conservation ##\nabla_\mu T^{\mu \nu}=0##

You're confusing yourself by writing formulas using two different conventions. If you are going to write a vector as ##V = V^\mu e_\mu##, then you need to also write a tensor as ##T = T^{\mu \nu} e_\mu e_\nu##. And you need to write the covariant derivative of a tensor similarly.

Now if ##\nabla_\mu V^\nu=0## then the covariant derivative is still not zero

If it happens to be the case that ##\nabla_\mu e_\nu \neq 0## in the coordinates you are using, yes. But there are coordinates in which ##\nabla_\mu e_\nu = 0##.

from this can we say that in general the covariant derivative of this energy --momentum is non-zero

Certainly not. The covariant derivative of the energy-momentum tensor is always zero; that is required by the Einstein Field Equation. But you have to write it down properly using the convention you want to use.

 

[Apashanka]

This would mean that the manifold is two-dimensional. Is that your intent? If you are asking about energy-momentum conservation, you are asking about spacetime, which is four-dimensional, not two-dimensional.

You're confusing yourself by writing formulas using two different conventions. If you are going to write a vector as ##V = V^\mu e_\mu##, then you need to also write a tensor as ##T = T^{\mu \nu} e_\mu e_\nu##. And you need to write the covariant derivative of a tensor similarly.
If it happens to be the case that ##\nabla_\mu e_\nu \neq 0## in the coordinates you are using, yes. But there are coordinates in which ##\nabla_\mu e_\nu = 0##.
Certainly not. The covariant derivative of the energy-momentum tensor is always zero; that is required by the Einstein Field Equation. But you have to write it down properly using the convention you want to use.

Is it then ##\nabla_? T=0## where ##T=T^{\mu \nu}e_\mu e_\nu## from the energy momentum conservation ??
What will be then the subscript of ##\nabla##??

 

[PeterDonis]

Is it then ##\nabla_? T=0## where ##T=T^{\mu \nu}e_\mu e_\nu## from the energy momentum conservation ??

Energy-momentum conservation is a covariant divergence; that's why the index ##\mu## is repeated in the expression ##\nabla_\mu T^{\mu \nu}##. It's a sum over all four possible values of ##\mu##. It's not a covariant derivative along a curve.

 

[Apashanka]

Energy-momentum conservation is a covariant divergence; that's why the index ##\mu## is repeated in the expression ##\nabla_\mu T^{\mu \nu}##. It's a sum over all four possible values of ##\mu##. It's not a covariant derivative along a curve.

Ok if I want to compare this with the covariant derivative of a tangent vector V along the tangent vector then ##\nabla_V V=V^\mu \nabla_\mu(V^\nu e_\nu)=V^\mu (\nabla_\mu V^\nu) e_\nu + V^\mu V^\nu (\nabla_\mu e_\nu)## Are the two bracketed terms in the last called the covariant divergence of a vector and similarly for tensors it become ##\nabla_i A^{jk}=0## and for energy momentum tensor ##\nabla_\mu T^{\mu \nu}=0## but why can't it be ##\nabla_\alpha T^{\mu \nu}=0## where all three ##\alpha,\mu,\nu=0,1,2,3##??

 

[PeterDonis]

if I want to compare this with the covariant derivative of a tangent vector V along the tangent vector

Then you are doing it wrong. The covariant derivative along a curve is a different thing from a covariant divergence.

Are the two bracketed terms in the last called the covariant divergence of a vector

No. See above.

 

[PeterDonis]

@Apashanka, you say you are trying to learn these things, but making repeated posts here about extremely basic questions--the kind people usually learn by working through a textbook--seems to be a very inefficient way of doing it. What textbooks on differential geometry have you studied?

 

[Apashanka]

@Apashanka, you say you are trying to learn these things, but making repeated posts here about extremely basic questions--the kind people usually learn by working through a textbook--seems to be a very inefficient way of doing it. What textbooks on differential geometry have you studied?

Studying from books I find it confusing and sometimes out of my capacity...that's why I am trying to make these concepts clear by the people of this community

 

[Nugatory]

Studying from books I find it confusing and sometimes out of my capacity...that's why I am trying to make these concepts clear by the people of this community

Much much more effective is to work through a textbook until you find yourself stuck, and then asking for help getting past the hard spot.

 

[Apashanka]

Then you are doing it wrong. The covariant derivative along a curve is a different thing from a covariant divergence.

Ok then the covariant divergence of a vector ##V=V^\nu e_\nu## will be ##\nabla_\mu V^\mu## e.g sum over ##\mu## .
Thanks sir I got it now.

 

[PeterDonis]

the covariant divergence of a vector ##=V^\nu e_\nu## will be ##\nabla_\mu V^\mu## e.g sum over ##\mu## .

Yes.

 

[stevendaryl]

Now ## \nabla_A V=A^\mu \nabla_\mu(V^\nu e_\nu)=A^\mu(\nabla_\mu V^\nu)e_\nu+A^\mu V^\nu(\nabla_\mu e_\nu)##Now if ##\nabla_\mu V^\nu=0## then the covariant derivative is still not zero .

I think you are getting confused by the notation. When people write the expression ##\nabla_\mu V^\nu##, what they really mean is something like:

##(\nabla_\mu V)^\nu##

##\nabla_\mu## operates on a vector field to produce another vector field. Then you take component ##\nu## of that vector field. The result is ##\nabla_\mu V^\nu##.

##\nabla_\mu## already takes into account BOTH (1) the variation of the components ##V^\nu##, and (2) the variation of the basis vectors ##e_\nu##.

In terms of components, you could write:

##\nabla_\mu (V^\nu e_\nu) = (\partial_\mu V^\nu) e_\nu + V^\nu (\nabla_\mu e_\nu)##

By definition, the connection coefficient ##(\nabla_\mu e_\nu) = \Gamma^\lambda_{\mu \nu} e_\lambda##. So we have:

##\nabla_\mu (V^\nu e_\nu) = (\partial_\mu V^\nu) e_\nu + V^\nu (\Gamma^\lambda_{\mu \nu} e_\lambda)##

In the second term, ##\nu## and ##\lambda## are dummy indices, since they are summed over. So switching ##\nu## and ##\lambda## does nothing:
##V^\nu (\Gamma^\lambda_{\mu \nu} e_\lambda) = V^\lambda (\Gamma^\nu_{\mu \lambda} e_\nu)##

Substituting this into our expression for ##\nabla_\mu (V^\nu e_\nu)## gives:

##\nabla_\mu (V^\nu e_\nu) = (\partial_\mu V^\nu) e_\nu + V^\lambda (\Gamma^\nu_{\mu \lambda} e_\nu) = (\partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda) e_\nu##

The expression ##(\partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda)## is what people usually write as ##\nabla_\mu V^\nu##. So you can write, with a little ambiguity of notation:

##\nabla_\mu (V^\nu e_\nu) = (\nabla_\mu V^\nu) e_\nu##

 

[stevendaryl]

This has been my complaint about the notation ##\nabla_\mu V^\nu##. It looks like it's an operator on components of a vector field, but it isn't. It's an operator on vector fields. You take the covariant derivative of the vector field ##V##. That produces a new vector field. Then you take component ##\nu## of that new field. The result is ##\nabla_\mu V^\nu##.

 

[Apashanka]

This has been my complaint about the notation ##\nabla_\mu V^\nu##. It looks like it's an operator on components of a vector field, but it isn't. It's an operator on vector fields. You take the covariant derivative of the vector field ##V##. That produces a new vector field. Then you take component ##\nu## of that new field. The result is ##\nabla_\mu V^\nu##.

If V be a tangent vector to a curve then the covariant derivative of V along V is ##\nabla_V V## ...(1)
##\nabla_V=V^i\nabla_i##www.physicsforums.com/threads/covariant-derivative-of-tangent-vector-for-geodesic.966893/#post-6138635##
therefore (1) becomes ##V^i\nabla_i V## and the ##\nu^{th}## component of this new vector is ##(V^i\nabla_i V)^\nu=V^i\nabla_i V^\nu## then what will be the next step in expanding this??

 

[stevendaryl]

If V be a tangent vector to a curve then the covariant derivative of V along V is ##\nabla_V V## ...(1)

I think you are misunderstanding what "covariant derivative" means. In general, if you have TWO different vector fields ##U## and ##V## then the meaning of ##\nabla_U V## is this: ##\nabla_U V = U^\alpha (\partial_\alpha V^\beta + \Gamma^\beta_{\alpha \lambda} V^\lambda) e_\beta##. Or in terms of components,

##\nabla_U V^\beta = U^\alpha (\partial_\alpha V^\beta + \Gamma^\beta_{\alpha \lambda} V^\lambda)##

The case where ##V = U## is a special case, but that's not what "covariant derivative" means.

 

[stevendaryl]

If V be a tangent vector to a curve then the covariant derivative of V along V is ##\nabla_V V## ...(1)
##\nabla_V=V^i\nabla_i##www.physicsforums.com/threads/covariant-derivative-of-tangent-vector-for-geodesic.966893/#post-6138635##
therefore (1) becomes ##V^i\nabla_i V## and the ##\nu^{th}## component of this new vector is ##(V^i\nabla_i V)^\nu=V^i\nabla_i V^\nu## then what will be the next step in expanding this??

The meaning of ##\nabla_V V## is the vector field with components ##V^\alpha (\partial_\alpha V^\beta + \Gamma^\beta_{\alpha \lambda} V^\lambda)##

 

[Apashanka]

The meaning of ##\nabla_V V## is the vector field with components ##V^\alpha (\partial_\alpha V^\beta + \Gamma^\beta_{\alpha \lambda} V^\lambda)##

Yaa got it now thanks

 

[Apashanka]

This has been my complaint about the notation ##\nabla_\mu V^\nu##. It looks like it's an operator on components of a vector field, but it isn't. It's an operator on vector fields. You take the covariant derivative of the vector field ##V##. That produces a new vector field. Then you take component ##\nu## of that new field. The result is ##\nabla_\mu V^\nu##.

But I find it still confusing with covariant divergence of a vector field V which is ##\nabla_\mu V^\mu## where ##V^\mu## are the components of the vector and,, ##\mu^{th}## component of covariant derivative of a vector field V which is ##(\nabla_\mu V)^\mu##

 

[Apashanka]

But I find it still confusing with covariant divergence of a vector field V which is ##\nabla_\mu V^\mu## where ##V^\mu## are the components of the vector and,, ##\mu^{th}## component of covariant derivative of a vector field V which is ##(\nabla_\mu V)^\mu##

Yaa got it now in covariant divergence there is a summation over ##\mu## and in covariant derivative it is not.
Is it right??

 

[stevendaryl]

But I find it still confusing with covariant divergence of a vector field V which is ##\nabla_\mu V^\mu## and ##\mu^{th}## component of covariant derivative of a vector field V which is ##(\nabla_\mu V)^\mu##

Let's go through a simple example. Two-dimensional space with Cartesian coordinates ##x## and ##y##. In that case

##\nabla_\mu V^\nu = \partial_\mu V^\nu##

because ##\Gamma^\nu_{\mu \lambda}## is zero for Cartesian coordinates.

The covariant divergence is ##\nabla_\mu V^\mu = \partial_\mu V^\mu = \partial_x V^x + \partial_y V^y##

Now, let's switch to polar coordinates ##r, \theta##.

##\nabla_\mu V^\mu = \nabla_r V^r + \nabla_\theta V^\theta = \partial_r V^r + \Gamma^r_{r \lambda} V^\lambda + \partial_\theta V^\theta + \Gamma^\theta_{\theta \lambda} V^\lambda##

where ##\lambda## is summed over. So expanding the sum over ##\lambda##:

##\nabla_\mu V^\mu = \partial_r V^r + \Gamma^r_{r r} V^r + \Gamma^r_{r \theta} V^\theta + \partial_\theta V^\theta + \Gamma^\theta_{\theta r} V^r + \Gamma^\theta_{\theta \theta} V^\theta##

For polar coordinates, the only nonzero connection coefficients are ##\Gamma^r_{\theta \theta} = -r## and ##\Gamma^\theta_{r \theta} = \Gamma^\theta_{\theta r} = \frac{1}{r}##. So

##\nabla_\mu V^\mu = \partial_r V^r + \partial_\theta V^\theta + \frac{1}{r} V^r ##

 

[Apashanka]

Let's go through a simple example. Two-dimensional space with Cartesian coordinates ##x## and ##y##. In that case

##\nabla_\mu V^\nu = \partial_\mu V^\nu##

because ##\Gamma^\nu_{\mu \lambda}## is zero for Cartesian coordinates.

The covariant divergence is ##\nabla_\mu V^\mu = \partial_\mu V^\mu = \partial_x V^x + \partial_y V^y##

Now, let's switch to polar coordinates ##r, \theta##.

##\nabla_\mu V^\mu = \nabla_r V^r + \nabla_\theta V^\theta = \partial_r V^r + \Gamma^r_{r \lambda} V^\lambda + \partial_\theta V^\theta + \Gamma^\theta_{\theta \lambda} V^\lambda##

where ##\lambda## is summed over. So expanding the sum over ##\lambda##:

##\nabla_\mu V^\mu = \partial_r V^r + \Gamma^r_{r r} V^r + \Gamma^r_{r \theta} V^\theta + \partial_\theta V^\theta + \Gamma^\theta_{\theta r} V^r + \Gamma^\theta_{\theta \theta} V^\theta##

For polar coordinates, the only nonzero connection coefficients are ##\Gamma^r_{\theta \theta} = -r## and ##\Gamma^\theta_{r \theta} = \Gamma^\theta_{\theta r} = \frac{1}{r}##. So

##\nabla_\mu V^\mu = \partial_r V^r + \partial_\theta V^\theta + \frac{1}{r} V^r ##

Yaa similarly I have thought for a 2-D space the covariant divergence of a vector field V is ##\nabla_i V^i=\nabla_1 V^1+\nabla_2 V^2## where ##V=V^\mu e_\mu##
And ith component of covariant derivative of V is ##(\nabla_i V)^i## which is either of ##\nabla_1 V^1## or ##\nabla_2 V^2## ,thats how the two term differs

 

[stevendaryl]

Yaa similarly I have thought for a 2-D space the covariant divergence of a vector field V is ##\nabla_i V^i=\nabla_1 V^1+\nabla_2 V^2## where ##V=V^\mu e_\mu##
And ith component of covariant derivative of V ##(\nabla_i V)^i## which is either of ##\nabla_1 V^1## or ##\nabla_2 V^2## ,thats how the two term differs

Usually, when someone writes ##\nabla_i V^i## they mean the sum over ##i##. So the result is not a vector any more. So it doesn't make sense to take a component of it.

Going back to coordinates ##x## and ##y##:

##\nabla_x V## is a vector, with components ##\nabla_x V^x## and ##\nabla_x V^y##
##\nabla_y V## is a different vector, with components ##\nabla_y V^x## and ##\nabla_y V^y##.

When people write ##\nabla_i V^i## they typically mean the sum: ##\nabla_x V^x + \nabla_y V^y##. That is not a vector. It is the sum of components of a vector. [edit] It is the sum of components of two different vectors.

 

[Apashanka]

Usually, when someone writes ##\nabla_i V^i## they mean the sum over ##i##. So the result is not a vector any more. So it doesn't make sense to take a component of it.

Going back to coordinates ##x## and ##y##:

##\nabla_x V## is a vector, with components ##\nabla_x V^x## and ##\nabla_x V^y##
##\nabla_y V## is a different vector, with components ##\nabla_y V^x## and ##\nabla_y V^y##.

When people write ##\nabla_i V^i## they typically mean the sum: ##\nabla_x V^x + \nabla_y V^y##. That is not a vector. It is the sum of components of a vector.

Yes ,,now got it...