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...