Special relativity, gradient of velocity

[Vrbic]

Homework Statement

I have noticed that in some calculations they use $$ \vec{\nabla}\cdot\vec{u}=\frac{1}{V}\frac{dV}{d\tau}$$. I would like to derive it.

Homework Equations

$\vec{u}=(\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})$
$\vec{A}\cdot\vec{B}=A^{\mu}B_{\mu}$
$\vec{\nabla}=(\frac{d}{dt},\frac{d}{dx},\frac{d}{dy},\frac{d}{dz})$

The Attempt at a Solution

$$ \vec{\nabla}\cdot\vec{u}=\frac{d}{dt}\frac{dt}{d\tau}+\frac{d}{dx}\frac{dx}{d\tau}+\frac{d}{dy}\frac{dy}{d\tau}+\frac{d}{dz}\frac{dz}{d\tau}=... $$May I do this?
$$...=\frac{d}{d\tau}(\frac{dt}{dt}+...)??$$
I guess not, because $\frac{d}{d\tau}(\frac{dt}{dt})=\frac{d}{d\tau}1=0$. Please advise.

 

[jambaugh]

Would you please give a reference to one such calculation so that the context is clear? The 4-divergence as you are using it would only be applied to a 4-vector field in which case proper time is not well defined. (I am assuming that you meant to use partial derivatives in the Del operator's definition?) If you mean for $\vec{u}$ to be a four velocity at some point upon a particle's world-line then it has no explicit coordinate dependency and thus the coordinate partial derivatives should all be zero.

As to the step you point to, the fact that the one term is zero does not invalidate the step.

A third issue here is that you are mixing covariant and contravariant vectors in the same parenthetic notation here and you must be careful to use the metric to raise/lower indices or otherwise use column and row vectors (matrices) to distinguish these.

 

[Vrbic]

Would you please give a reference to one such calculation so that the context is clear? The 4-divergence as you are using it would only be applied to a 4-vector field in which case proper time is not well defined. (I am assuming that you meant to use partial derivatives in the Del operator's definition?) If you mean for $\vec{u}$ to be a four velocity at some point upon a particle's world-line then it has no explicit coordinate dependency and thus the coordinate partial derivatives should all be zero.

As to the step you point to, the fact that the one term is zero does not invalidate the step.

A third issue here is that you are mixing covariant and contravariant vectors in the same parenthetic notation here and you must be careful to use the metric to raise/lower indices or otherwise use column and row vectors (matrices) to distinguish these.

I'm sorry I haven't noticed your answer. Thank you for it.
References: Exercise 2.26 b) http://www.pmaweb.caltech.edu/Courses/ph136/yr2011/1102.2.K.pdf and solution http://www.pmaweb.caltech.edu/Courses/ph136/yr2011/ps02-08-1.pdf page 5 on the top.

To a mixing of covariant and contravariant vectors, I a bit know about it, but I think in a special relativity it is not important, yet.

 

[jambaugh]

I begin to see the context now, it is the context of relativistic fluid flow. The $\vec{u}$ quantity is a fluid 4-vector field, i.e. a 4-vector function of the 4-position. It is therefore NOT the expression you used $\vec{u} = \langle \frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau},\frac{dz}{d\tau}\rangle$. Likewise in that context (problem 2.26) It would seem $V$ is a volume element for a local region of fluid in its local rest frame. What you have then is the 4-vector equivalent of a continuity equation.

To derive it consider the world-path of a 3-cube of fluid at a given position and time and with boundaries comoving with the fluid. Let the three dimension of the cubic volume be $\Delta x, \Delta y,$ and $\Delta z$. The volume is then at that instant $V = \Delta x \Delta y \Delta Z$. It is then a matter of considering how opposite faces are moving relative to each other given the fluid's non-uniform flow. The two faces orthogonal to the x-direction will have normal components of their 4-velocities differing in the small scale limit by $\Delta x \cdot \frac{\partial u_x}{\partial x}$. Likewise with the faces in the y and z directions. Thus the total rate of change of volume w.r.t. proper time will be $dV/d\tau = \Delta x \partial_x u_x\cdot \Delta y \Delta z + \Delta y \partial_y u_y \Delta x \Delta z + \Delta z \partial_z u_z\cdot \Delta x\Delta y$ the relative motion of each pair of opposite faces times their areas. Combining terms you get the volume time the three gradient of the spatial components of the 4-velocity.

To complete the argument one needs to argue that, possibly because one is in the fluid's instantaneous rest frame, $\frac{\partial}{\partial t} u_t = 0$ in that frame. (I'm not feeling 100% solid on this part of the argument.) You can thus add this zero term to the equation to get:

$$\frac{dV}{d\tau} = V \partial_\mu u^\mu \quad \partial_\mu = \frac{\partial}{\partial x^\mu}$$
thence the expression you quoted. The 4-divergence of the 4-velocity field then can be calculated in any frame as it is invariant. Its interpretation is likewise given as the proportionate rate of 3-volume expansion of the fluid at a given point and time as measured in its rest frame.

 

[Vrbic]

I begin to see the context now, it is the context of relativistic fluid flow. The $\vec{u}$ quantity is a fluid 4-vector field, i.e. a 4-vector function of the 4-position. It is therefore NOT the expression you used $\vec{u} = \langle \frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau},\frac{dz}{d\tau}\rangle$. Likewise in that context (problem 2.26) It would seem $V$ is a volume element for a local region of fluid in its local rest frame. What you have then is the 4-vector equivalent of a continuity equation.

To derive it consider the world-path of a 3-cube of fluid at a given position and time and with boundaries comoving with the fluid. Let the three dimension of the cubic volume be $\Delta x, \Delta y,$ and $\Delta z$. The volume is then at that instant $V = \Delta x \Delta y \Delta Z$. It is then a matter of considering how opposite faces are moving relative to each other given the fluid's non-uniform flow. The two faces orthogonal to the x-direction will have normal components of their 4-velocities differing in the small scale limit by $\Delta x \cdot \frac{\partial u_x}{\partial x}$. Likewise with the faces in the y and z directions. Thus the total rate of change of volume w.r.t. proper time will be $dV/d\tau = \Delta x \partial_x u_x\cdot \Delta y \Delta z + \Delta y \partial_y u_y \Delta x \Delta z + \Delta z \partial_z u_z\cdot \Delta x\Delta y$ the relative motion of each pair of opposite faces times their areas. Combining terms you get the volume time the three gradient of the spatial components of the 4-velocity.

To complete the argument one needs to argue that, possibly because one is in the fluid's instantaneous rest frame, $\frac{\partial}{\partial t} u_t = 0$ in that frame. (I'm not feeling 100% solid on this part of the argument.) You can thus add this zero term to the equation to get:

$$\frac{dV}{d\tau} = V \partial_\mu u^\mu \quad \partial_\mu = \frac{\partial}{\partial x^\mu}$$
thence the expression you quoted. The 4-divergence of the 4-velocity field then can be calculated in any frame as it is invariant. Its interpretation is likewise given as the proportionate rate of 3-volume expansion of the fluid at a given point and time as measured in its rest frame.

Wrong my point of view. I recount it but it seems reasonable. Thank you.