# Torsion Scalar and Symmetries of Torsion Tensor

• nanuba
In summary, the conversation was about the calculation of the torsion scalar in Teleparallel theories and the terms involved in the calculation. The conversation also touched on the topic of totally antisymmetric torsion tensors and how they relate to the calculation. The final part of the conversation was a request for a black coffee.
nanuba
I've started f(T) theory but I have a simple question like something that i couldn't see straightforwardly.
In Teleparallel theories one has the torsion scalar:

And if you take the product you should obtain

But there seems to be the terms like
.
How does this one vanish?

because we know that Torsion tensor is not symmetric in first two indices but antisymmetric in the last two indices.

Often one is interested in torsion tensors that are totally antisymmetric. Is that the case here?

Oscillatory behavior

bcrowell said:
Often one is interested in torsion tensors that are totally antisymmetric. Is that the case here?
Nope. These are defined in terms of Wietzenböck konnections:
and are not totally antisymmetric.

nanuba said:
I've started f(T) theory but I have a simple question like something that i couldn't see straightforwardly.
In Teleparallel theories one has the torsion scalar:

View attachment 85890
View attachment 85891
And if you take the product you should obtain
View attachment 85892
But there seems to be the terms like
View attachment 85893View attachment 85894View attachment 85895.
How does this one vanish?

because we know that Torsion tensor is not symmetric in first two indices but antisymmetric in the last two indices.

Will you buy me strong black coffee if I do it for you? Okay, start by multiplying the two tensor
$$S^{\rho\mu\nu}T_{\rho\mu\nu} = A + B , \ \ \ \ \ (1)$$
where
$$A = \frac{1}{4} T^{\rho\mu\nu}T_{\rho\mu\nu} + \frac{1}{4} T^{\mu\rho\nu}T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu}T_{\rho\mu\nu} , \ \ \ \ (2)$$ and $$B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} + \frac{1}{2}T_{\alpha}{}^{\nu\alpha} \ T^{\rho}{}_{\rho\nu} ,$$ or
$$B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} - \frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} = - T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} ,$$ which is the same as
$$B = T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\rho\mu} = T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} . \ \ \ \ \ (3)$$ That is all for $B$. To work on $A$, we start with the identity
$$T^{\mu\nu}{}_{\rho} \left( T^{\rho}{}_{\nu\mu} + T^{\rho}{}_{\mu\nu} \right) = 0 .$$ In the first term, if we make $\rho \to \nu$, $\mu \to \rho$ and $\nu \to \mu$, we find
$$T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\mu\rho} + T^{\mu\nu}{}_{\rho} \ T^{\rho}{}_{\mu\nu} = 0 .$$ Let us rewrite this as
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} + T_{\mu\nu\rho} \ T^{\rho\mu\nu} = 0 .$$ Using the antisymmetry in the second term, we get
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\rho\mu\nu} \ T_{\mu\rho\nu} .$$ Now, on the RHS we make $\rho \leftrightarrow \mu$ and get
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\mu\rho\nu} \ T_{\rho\mu\nu} .$$ Add $T^{\rho\mu\nu}T_{\nu\mu\rho}$ to both sides,
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T^{\rho\mu\nu} \ T_{\nu\mu\rho} \right) .$$ In the second term on the RHS, lower the indices on the first tensor and raise them on the second tensor
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T_{\rho\mu\nu} \ T^{\nu\mu\rho} \right) .$$ Using the antisymmetry and dividing by 2, we obtain
$$\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = \frac{1}{2} T^{\rho\mu\nu}T_{\nu\mu\rho} = - \frac{1}{2} T^{\rho\mu\nu} T_{\nu\rho\mu}.$$ And finally, we rewrite this identity in the form
$$\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} .$$ Now, substituting this identity in (2) leads to $$A = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} . \ \ \ (4)$$ Now, put (3) and (4) in (1), you get
$$S^{\rho\mu\nu}T_{\rho\mu\nu} = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} + T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} .$$
Okay, don’t forget my coffee.
Sam

PWiz, nanuba, vanhees71 and 1 other person
samalkhaiat said:
Will you buy me strong black coffee if I do it for you? Okay, start by multiplying the two tensor
$$S^{\rho\mu\nu}T_{\rho\mu\nu} = A + B , \ \ \ \ \ (1)$$
where
$$A = \frac{1}{4} T^{\rho\mu\nu}T_{\rho\mu\nu} + \frac{1}{4} T^{\mu\rho\nu}T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu}T_{\rho\mu\nu} , \ \ \ \ (2)$$ and $$B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} + \frac{1}{2}T_{\alpha}{}^{\nu\alpha} \ T^{\rho}{}_{\rho\nu} ,$$ or
$$B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} - \frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} = - T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} ,$$ which is the same as
$$B = T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\rho\mu} = T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} . \ \ \ \ \ (3)$$ That is all for $B$. To work on $A$, we start with the identity
$$T^{\mu\nu}{}_{\rho} \left( T^{\rho}{}_{\nu\mu} + T^{\rho}{}_{\mu\nu} \right) = 0 .$$ In the first term, if we make $\rho \to \nu$, $\mu \to \rho$ and $\nu \to \mu$, we find
$$T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\mu\rho} + T^{\mu\nu}{}_{\rho} \ T^{\rho}{}_{\mu\nu} = 0 .$$ Let us rewrite this as
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} + T_{\mu\nu\rho} \ T^{\rho\mu\nu} = 0 .$$ Using the antisymmetry in the second term, we get
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\rho\mu\nu} \ T_{\mu\rho\nu} .$$ Now, on the RHS we make $\rho \leftrightarrow \mu$ and get
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\mu\rho\nu} \ T_{\rho\mu\nu} .$$ Add $T^{\rho\mu\nu}T_{\nu\mu\rho}$ to both sides,
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T^{\rho\mu\nu} \ T_{\nu\mu\rho} \right) .$$ In the second term on the RHS, lower the indices on the first tensor and raise them on the second tensor
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T_{\rho\mu\nu} \ T^{\nu\mu\rho} \right) .$$ Using the antisymmetry and dividing by 2, we obtain
$$\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = \frac{1}{2} T^{\rho\mu\nu}T_{\nu\mu\rho} = - \frac{1}{2} T^{\rho\mu\nu} T_{\nu\rho\mu}.$$ And finally, we rewrite this identity in the form
$$\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} .$$ Now, substituting this identity in (2) leads to $$A = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} . \ \ \ (4)$$ Now, put (3) and (4) in (1), you get
$$S^{\rho\mu\nu}T_{\rho\mu\nu} = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} + T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} .$$
Okay, don’t forget my coffee.
Sam

I owe you a black coffee :)

## 1. What is a Torsion Scalar?

A Torsion Scalar is a mathematical quantity that describes the amount of twisting or rotation in a given space. It is commonly used in the study of spacetime curvature, where it represents the amount of torsion (twisting) in the fabric of spacetime.

## 2. How is the Torsion Scalar related to the Torsion Tensor?

The Torsion Scalar is derived from the Torsion Tensor, which is a mathematical object that describes the amount of twisting in a given space. The Torsion Tensor contains information about the direction and magnitude of the twisting, which is used to calculate the Torsion Scalar.

## 3. What is the significance of symmetries in the Torsion Tensor?

The Torsion Tensor is a mathematical object that has certain symmetries, meaning that it remains unchanged under certain transformations. These symmetries are important in understanding the underlying structure of spacetime and can provide insights into the nature of gravity and other fundamental forces.

## 4. How are Torsion Scalar and Symmetries of Torsion Tensor used in physics?

The Torsion Scalar and Symmetries of Torsion Tensor are used in various areas of physics, including general relativity, quantum field theory, and particle physics. They are important in understanding the behavior of spacetime and the fundamental forces that govern the universe.

## 5. Are there any real-world applications of the Torsion Scalar and Symmetries of Torsion Tensor?

While the Torsion Scalar and Symmetries of Torsion Tensor are primarily used in theoretical physics, there are some potential real-world applications. For example, torsion could be used in the development of new materials and technologies, such as more efficient engines or advanced energy storage systems.

Replies
6
Views
5K
Replies
1
Views
932
Replies
19
Views
755
Replies
22
Views
2K
Replies
1
Views
580
Replies
10
Views
2K
Replies
8
Views
3K
Replies
1
Views
1K
Replies
13
Views
4K
Replies
3
Views
2K