By energy conservation, show the 4-velocity of dust satisfies this....

In summary, using conservation of energy, we can show that the 4-velocity ##u^{\mu}## of dust satisfies ##u^{\mu}\nabla_{\mu}u^{\nu} = f u^{\nu}##, where ##f## is a function of ##u^{\mu}##, ##\rho## and their derivatives. In a comoving observer in a homogeneous Friedmann-Robertson-Walker Universe, this equation becomes ##\dot{\rho} + 3 \left(\frac{\dot{a}}{a}\right)\rho = 0##. The first part involves manipulating the energy-momentum tensor and using the product rule, while the second part involves
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Homework Statement


Using conservation of energy, show that the 4-velocity ##u^{\mu}## of dust satisfies:
##u^{\mu}\nabla_{\mu}u^{\nu} = f u^{\nu}##

Explicitly identify ##f##, which is some function of ##u^{\mu}##, ##\rho## and their derivatives. And show that this equation becomes

##\dot{\rho} + 3 \left(\frac{\dot{a}}{a}\right)\rho = 0##

For comoving observers in a homogeneous Friedmann-Robertson-Walker Universe.

Homework Equations

The Attempt at a Solution


I don't think I've done the first bit right, I'm not really sure how to treat the ##\rho##. For dust, ##T^{\mu\nu} = \rho u^{\mu}u^{\nu}## and energy conservation means:
##\nabla_{\mu}T^{\mu\nu} = 0##
So
##\nabla_{\mu} (\rho u^{\mu}u^{\nu})##
Using the product rule,
##\rho u^{\mu} (\nabla_{\mu}u^{\nu}) + \rho u^{\nu} (\nabla_{\mu}u^{\mu}) + u^{\mu}u^{\nu}\nabla_{\mu}\rho = 0##
I think ##\nabla_{\mu} \rho =0##, but if it is then I could just divide through by ##\rho## and I could rearrange to get the LHS
I want:
##u^{\mu}\nabla_{\mu}u^{\nu}##
But the RHS wouldn't be a function of ##\rho##. So if ##\nabla_{\mu} \rho## isn't zero, then I get:
##u^{\mu}\nabla_{\mu}u^{\nu} = -u^{\nu}\nabla_{\mu}u^{\mu} - \frac{u^{\nu}u^{\mu}}{\rho} \nabla_{\mu}\rho##
##= \left(-\nabla_{\mu}u^{\mu}-\frac{u^{\mu}}{\rho}\nabla_{\mu}\rho \right)u^{\nu}##
Is that right? Is ##\nabla_{\mu}\rho## not zero?
 
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For the second part, I'm not really sure how to do it. I think it's a case of using the FLRW metric and the equations of motion for an FRW universe?
 

What is energy conservation?

Energy conservation is a principle in physics that states energy cannot be created or destroyed, but can only be transferred or converted from one form to another.

What is the 4-velocity of dust?

The 4-velocity of dust is a mathematical representation of the motion of a particle in four-dimensional spacetime, taking into account both its spatial velocity and its time component.

How does the 4-velocity of dust relate to energy conservation?

The 4-velocity of dust satisfies energy conservation because it is a conserved quantity, meaning that its magnitude remains constant throughout a particle's motion. This is a result of the dust particle not experiencing any external forces.

Can you show mathematically how the 4-velocity of dust satisfies energy conservation?

Yes, the 4-velocity of dust satisfies energy conservation through the use of the Euler-Lagrange equations, which show that the derivative of the dust's energy with respect to time is equal to zero.

What are some real-life applications of energy conservation and the 4-velocity of dust?

Energy conservation and the 4-velocity of dust have applications in various fields such as astrophysics, fluid mechanics, and thermodynamics. They are used to understand and analyze the behavior of particles and systems in motion, and to make predictions about their future behavior.

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