Conservation of energy-momentum (tensor)

In summary, the conversation discusses the concept of a curve parametrized by ##\lambda## and the possible tangent vectors V and A along the curve. It also mentions the covariant derivative and energy-momentum conservation, but there is confusion regarding the notation and conventions being used. The covariant derivative along a curve is a different concept from the covariant divergence, and the latter is defined as a sum over all possible values of an index, not just one. There is also confusion about the indices used for the covariant derivative of a tensor.
  • #1
Apashanka
429
15
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
 
Physics news on Phys.org
  • #2
Apashanka said:
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.

Apashanka said:
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)##

Apashanka said:
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.

Apashanka said:
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##.

Apashanka said:
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.
 
  • #3
PeterDonis said:
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##??
 
  • #4
Apashanka said:
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.
 
  • #5
PeterDonis said:
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##??
 
Last edited:
  • #6
Apashanka said:
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.

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

No. See above.
 
  • #7
@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?
 
  • #8
PeterDonis said:
@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
 
  • #9
Apashanka said:
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.
 
  • #10
PeterDonis said:
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.
 
  • #11
Apashanka said:
the covariant divergence of a vector ##=V^\nu e_\nu## will be ##\nabla_\mu V^\mu## e.g sum over ##\mu## .

Yes.
 
  • #12
Apashanka said:
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##
 
  • #13
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##.
 
  • #14
stevendaryl said:
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??
 
  • #15
Apashanka said:
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.
 
  • #16
Apashanka said:
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)##
 
  • #17
stevendaryl said:
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
 
  • #18
stevendaryl said:
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##
 
Last edited:
  • #19
Apashanka said:
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??
 
  • #20
Apashanka said:
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 ##
 
  • #21
stevendaryl said:
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
 
  • #22
Apashanka said:
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.
 
  • #23
stevendaryl said:
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...
 

Related to Conservation of energy-momentum (tensor)

1. What is the conservation of energy-momentum (tensor)?

The conservation of energy-momentum (tensor) is a fundamental law in physics that states that the total energy and momentum of a closed system remains constant over time. This means that energy and momentum cannot be created or destroyed, but can only be transferred or transformed from one form to another.

2. What is the significance of the conservation of energy-momentum (tensor)?

The conservation of energy-momentum (tensor) is significant because it is a fundamental principle that governs many physical processes and phenomena. It is essential for understanding and predicting the behavior of particles and systems, and has been verified by numerous experiments and observations.

3. How is the conservation of energy-momentum (tensor) related to the laws of motion?

The conservation of energy-momentum (tensor) is closely related to the laws of motion, specifically Newton's laws of motion. These laws describe the relationship between forces acting on an object and its resulting motion. The conservation of energy-momentum (tensor) is a consequence of Newton's third law, which states that for every action, there is an equal and opposite reaction.

4. Can the conservation of energy-momentum (tensor) be violated?

No, the conservation of energy-momentum (tensor) is a fundamental law of nature and cannot be violated. However, in certain situations, it may appear that energy or momentum is not conserved. This is often due to incomplete or inaccurate measurements, or the presence of external forces that are not accounted for.

5. How does the conservation of energy-momentum (tensor) apply to real-world scenarios?

The conservation of energy-momentum (tensor) applies to all real-world scenarios, from the motion of planets in our solar system to the collision of subatomic particles in particle accelerators. It is a universal law that governs the behavior of all matter and energy in the universe and is essential for understanding and predicting the behavior of physical systems.

Similar threads

  • Special and General Relativity
Replies
2
Views
1K
  • Special and General Relativity
4
Replies
124
Views
7K
  • Special and General Relativity
Replies
19
Views
1K
  • Special and General Relativity
Replies
7
Views
2K
  • Special and General Relativity
Replies
17
Views
1K
  • Special and General Relativity
Replies
4
Views
220
  • Special and General Relativity
Replies
6
Views
5K
  • Special and General Relativity
2
Replies
62
Views
4K
  • Special and General Relativity
Replies
5
Views
1K
  • Special and General Relativity
Replies
6
Views
2K
Back
Top