0
$\begingroup$

Wikipedia says-

Penrose's theorem is more restricted and only holds when matter obeys a stronger energy condition, called the dominant energy condition, in which the "energy is larger than the pressure".

(inverted commas mine)

How can we compare 2 very different physical quantities of pressure and energy? I have thought some possibilities: 1. The wikipedia page has a mistake. 2. It's simply energy is in large amount for the black-hole , bur pressure is low. 3.Some counterintuitive higher- level physics relation between those 2 quantities (It's very counterintuitive to me because my physics knowledge is very much limited to A level physics.)

But which of the 3 possibilties is true? Any wise words?

$\endgroup$
  • $\begingroup$ More restrictive than what? Is it the penrose theorum page - include a link? $\endgroup$ – JMLCarter Mar 19 '17 at 13:47
  • $\begingroup$ Someone edited the question. $\endgroup$ – Mockingbird Mar 20 '17 at 6:19
2
$\begingroup$

There is a realationship between the two quantities of energy and pressure, but, fortunately for you, it is neither counter-intuitive or higher level. When it says "energy" it really means "energy density". Now notice that the dimensions actually work out: from a relation such as $W=P\Delta V$, you deduce that $$[\textrm{Pressure}]=\frac{[\textrm{Energy}]}{[\textrm{Volume}]},$$

while clearly energy density has the same dimensions.

In fact, the dominant energy condition for perfect fluids is evaluated here in the fourth bullet point, and you can see that the condition becomes

$\rho \ge |p|,$

where $\rho$ is the energy density, and $p$ is the pressure.

$\endgroup$
  • $\begingroup$ Whoop! Someone should edit that wikipedia page. $\endgroup$ – Mockingbird Mar 20 '17 at 6:21
2
$\begingroup$

The actual condition is about energy density, which as you may check, has the same dimension as pressure.

$$\rho > |p|$$

$\endgroup$

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.