0
$\begingroup$

A book I am using for study attempts to provide intuition on Einstein's Equation (${\bf G} = {\bf T}$, where ${\bf G}$ represents the Einstein Curvature Tensor and ${\bf T}$ represents the Stress-Energy-Momentum tensor).

To give an example, a ball filled with test particles that are initially at rest relative to each other is used. Assuming that $V(t)$ is the volume of the ball after a proper time, as measured by the particle at the center of the ball, the equation reads: $$ \frac{\ddot{V}}{V}\biggr\rvert_{t=0} = -\frac{1}{2}(\rho+P_x+P_y+P_z) \\ =-\frac{1}{2}(T_{00}+T_{11}+T_{22}+T_{33}) $$

I (possibly mistakenly) interpret this as the second time derivative of the volume divided by the volume is $-\frac{1}{2}$ times all the pressures and the energy density added up.

It seems that the units for $\frac{\ddot{V}}{V}$ are $(\text{time})^{-2}$. I do not see how this is useful in any physical application.

What does $\frac{\ddot{V}}{V}$ represent from a physical point of view?

As has been mentioned in the comments, the "Volume Acceleration" is normalized by the volume itself. I now understand how the Ricci tensor relates to the change in volume, but is there any intuition for why one needs to divide by volume, or why the following is true? $$\ddot{V} = V(t)R_{\mu\nu} v^{\mu}v^{\nu}$$

$\endgroup$
  • $\begingroup$ Which textbook? Which author? Which page? $\endgroup$ – Qmechanic Aug 21 '17 at 20:25
  • 1
    $\begingroup$ @Qmechanic Vladimir G Ivancevic & Tijana T Ivancevic, APPLIED DIFFERENTIAL GEOMETRY A Modern Introduction, The example is on p.173. It might be a bit unrelated to physics, but I wish to understand. [Moderator redaction: PDF link removed due to copyright violation.] $\endgroup$ – TheBro21 Aug 22 '17 at 9:41
  • $\begingroup$ Concerning the confusion re: units: the book is no doubt using units where $c = G = 1$. This means that length, time, and mass all have the same units, which then implies that $[\rho] = [\text{mass}]/[\text{length}]^3 = [\text{time}]^{-2}$. $\endgroup$ – Michael Seifert Aug 31 '17 at 15:29
1
$\begingroup$

First of all, I would like to thank you for asking this conceptual question and introducing the book.

The idea is to give an intuitive physical (and also geometrical!) meaning of Ricci tensor. And the reason why the second derivative of the volume is divided by the volume itself is just a result of calculation, see page 140 of the book.

If you have problems yet, let me briefly show the result on my way. Let $y^i$ be the vector connecting the center of the ball (the observer) to one of the test particles. The most important equation is the geodesic deviation equation for this vector given by:

$$ \frac{d^2 y^i(t)}{dt^2} =R^i_{\mu\nu\rho} \dot{\gamma}^\mu\dot{\gamma}^\nu y^\rho(t) \,, $$ where $\gamma$ is the geodesic that goes through the point, say $x^i$, where the test particle is and $\mu,\nu\,\rho \in \{0,1,2,3\}$ and $i,j,k, \dots \in\{1,2,3\}$. Now, we can assume $y^0(0)=0$, i.e., there is no time difference between the observer and the test particle at the beginning. Therefore, "$y^\rho = y^j$". Assuming that $y^i$ is analytic function (which is a reasonable assumption) we have

$$ y^i(t) = y^i(0) + \dot{y}^i(0) t+ \frac{1}{2} \ddot{y}^i(0) t^2 + \mathcal{O}(t^3). $$ But, $\dot{y}^i(0)=0$, because particles are initially at rest relative to each other! Hence,

$$ y^i(t) = y^i(0) + \frac{1}{2} R^i_{\mu\nu j} \dot{\gamma}^\mu\dot{\gamma}^\nu y^j(0) t^2 + \mathcal{O}(t^3) \\ =\left( \delta^i_j + \frac{1}{2} t^2 R^i_{\mu\nu j} \dot{\gamma}^\mu\dot{\gamma}^\nu \right) y^j(0) + \mathcal{O}(t^3)\,. $$ We can now determine the volume $V(t)$ by making the determinant of what in bracket is (recall, e.g., how one makes the volume out of a vector which satisfies the relation $\vec{x} = A \vec{y}$, where $A$ is a matrix). Using the fact that

$$ \det (\mathbb{1} + A) = 1 + \text{tr}{A} + \mathcal{O}(A^2) $$ we get $$ V(t) = \left( 1 - \frac{1}{2} t^2 R_{\mu\nu} \dot{\gamma}^\mu\dot{\gamma}^\nu\right) V(0) + \mathcal{O}(t^3)\,, $$ where $R_{\mu\nu} =R^\rho_{\mu \rho \nu}= \text{tr} \left(R^\rho_{\mu \sigma\nu} \right)=-\text{tr} \left(R^\rho_{\mu\nu \sigma} \right)$.

We are now almost done. Differentiating the last equation and evaluating at $t=0$ gives:

$$ \ddot{V}|_{t=0} = - R_{\mu\nu} \dot{\gamma}^\mu\dot{\gamma}^\nu V(0) \\ \Rightarrow \quad \left.\frac{\ddot{V}}{V}\right|_{t=0}= - R_{\mu\nu} \dot{\gamma}^\mu\dot{\gamma}^\nu\,. $$ In a local inertial coordinates where $\dot{\gamma}^\mu = (1,0,0,0)$ one obtains the first equation you wrote

$$ \left.\frac{\ddot{V}}{V}\right|_{t=0} = - R_{00} = \frac{1}{2} \left( T_{00}+T_{11}+T_{22}+T_{33}\right)\,. $$

$\endgroup$
0
$\begingroup$

I now understand how the Ricci tensor relates to the change in volume, but is there any intuition for why one needs to divide by volume, or why the following is true? $$ \ddot{V} = V(t)R_{\mu\nu} v^{\mu}v^{\nu} $$

Perhaps a good way to think about this is in terms of Newtonian tidal effects. Roughly speaking, the Riemann tensor encodes the differences between the paths of "nearby" geodesics; in the context of weakly curved spacetimes and timelike geodesics, this translates into a statement about how quickly two nearby objects will diverge due to the effects of the gravitational field. But these effects are precisely the same thing as "tidal forces" in a reference frame that is freely falling under Newtonian gravity.

Viewed in this light, the fact that the second derivative of the volume is proportional to the volume itself becomes a little clearer. For example, imagine two objects freely falling under Newtonian gravity that are initially both at rest. Suppose one of them is at a distance $R$ from a massive point object, while the other is at distance $R + d$. Then the acceleration of the separation objects is just the difference between their accelerations: $$ \ddot{d} = \frac{GM}{(R + d)^2} - \frac{GM}{R^2} = \frac{GM}{R^2} \left[ -2 \frac{d}{R} + 3 \left( \frac{d}{R} \right)^2 - \dots \right] $$ For objects that are sufficiently "close" to each other, we have $d \ll R$ and so the acceleration of the separation is approximately proportional to the acceleration itself. This will be true for a general gravitational field; effectively, we are taking the first-order term in a Taylor series in the separation of the objects.

If we then imagine a parallelepiped of size $d_x$, $d_y$, and $d_z$, the second derivative of each of these quantities will be proportional to the quantities themselves.1 We can call the proportionality factors $\alpha_x$, $\alpha_y$, and $\alpha_z$. Moreover, if we assume that $\dot{d}_x = \dot{d}_y = \dot{d}_z = 0$, then the second derivative of the volume of this region will be $$ \ddot{V} = \frac{d^2}{dt^2} (d_x d_y d_z) = \ddot{d}_x d_y d_z + d_x \ddot{d}_y d_z + d_x d_y \ddot{d}_z = (\alpha_x + \alpha_y + \alpha_z) d_x d_y d_z \propto V. $$ So the fact that $\ddot{V}$ is proportional to $V$ can also be understood from a purely Newtonian perspective: it follows from the fact that to a good approximation, the magnitudes of "tidal forces" is proportional to the separation between two objects.


1 Technically, we could have a situation where $\ddot{d}_x$ is proportional to $d_y$ and $d_z$ as well. But the argument can still be extended to this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.