Orthogonal Projection of Perfect Fluid Energy Momentum

In summary: That should give you the missing term. In summary, the derivation of the relativistic Euler equation involves contracting the conservation law with the projection tensor and considering each term separately. However, in the last term, the contribution from the second term of the projection tensor is missing, resulting in a discrepancy with the result stated in Carroll's textbook.
  • #1
GL_Black_Hole
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Homework Statement


Derive the relativistic Euler equation by contracting the conservation law $$\partial _\mu {T^{\mu \nu}} =0$$ with the projection tensor $${P^{\sigma}}_\nu = {\delta^{\sigma}}_\nu + U^{\sigma} U_{\nu}$$ for a perfect fluid.

Homework Equations


$$\partial _\mu {T^{\mu \nu}} = \partial_\mu {(\rho +p)}U^{\mu} U^{\nu} + (\rho +p)(U^{\nu} \partial_\mu U^{\mu} + U^{\mu} \partial_\mu U_{\nu}) + \partial^{\nu} p$$

The Attempt at a Solution


Going term by term:
$$P^{\sigma}_\nu \partial_\mu (\rho +p) U^{\mu} U^{\nu} = \partial_\mu (\rho +p) U^{\sigma} U_{\mu} + U^{\sigma} U_{\nu} \partial_\mu (\rho +p)U^{\mu} U^{\nu}$$ but $$U^{\nu} U_{\nu} =-1$$ so this term is zero.

Next,
$$P^{\sigma}_\nu (\rho +p) U^{\nu} \partial_\mu U^{\mu} = 0$$
Then:
$$P^{\sigma}_\nu (\rho +p) U^{\mu} \partial_\mu U^{\nu} = (\rho +p)U^{\mu} \partial_\mu U^{\sigma}$$,
and finally:
$$P^{\sigma}_\nu \partial ^{\nu} p = \partial ^{\sigma} p$$.
But if I compare my answer to the result Carroll states on pg. 36 on his GR text I am off by a term:
$$U^{\sigma} U^{\mu} \partial_\mu p$$.

Where did I lose this term?
 
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  • #2
GL_Black_Hole said:
$$P^{\sigma}_\nu \partial ^{\nu} p = \partial ^{\sigma} p$$
On the right hand side, where's the contribution from the second term of the projection tensor?
 

1. What is the concept of orthogonal projection in the context of perfect fluid energy momentum?

Orthogonal projection in the context of perfect fluid energy momentum refers to the process of breaking down the energy and momentum of a fluid into two components - one parallel to a specified direction and one perpendicular to it. This allows for a more detailed understanding of the fluid's behavior and energy distribution.

2. How is orthogonal projection used in fluid mechanics?

Orthogonal projection is commonly used in fluid mechanics to analyze the behavior and properties of fluids. It allows for the separation of energy and momentum into different components, making it easier to study the fluid's flow patterns and energy distribution.

3. What is the importance of studying orthogonal projection in perfect fluid energy momentum?

Studying orthogonal projection in perfect fluid energy momentum is crucial in understanding the dynamics of fluids and their impact on various processes, such as fluid flow in pipes or channels, fluid-structure interactions, and energy transfer. It also plays a significant role in the development of engineering applications and technologies, such as turbines and pumps.

4. How is orthogonal projection calculated in perfect fluid energy momentum?

The calculation of orthogonal projection in perfect fluid energy momentum involves using vector calculus and fluid mechanics equations to determine the parallel and perpendicular components of energy and momentum. This can be done through mathematical equations or computer simulations.

5. What are some real-world applications of orthogonal projection in perfect fluid energy momentum?

Orthogonal projection in perfect fluid energy momentum has various real-world applications, including aerodynamics, hydrodynamics, and heat transfer. It is used in the design and analysis of aircraft, ships, and other fluid-based systems. It is also essential in weather prediction and climate modeling, as well as in the development of renewable energy technologies such as wind turbines and water turbines.

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