I was studying a book called Theoretical Astrophysics by Prof T Padmanavan, and in the very first chapter, i find the expression of Net Contributing Pressure ($P$) as:

$$P = \frac{1}{3} \int_{0}^{\infty}n(\epsilon)p(\epsilon)v(\epsilon)d\epsilon$$

followed by relativistic momentum $P = \gamma mv$ and kinetic energy, $\epsilon = (\gamma-1)mc^2$ with $\gamma$ as the relativistic factor. By far this I understood it, but then, substituting these values in the above given expression, as:

$$P = \frac{1}{3} \int_{0}^{\infty}n\epsilon\left(1+\frac{2mc^2}{\epsilon}\right)\left(1+\frac{mc^2}{\epsilon}\right)^{-1}d\epsilon$$

is not understood by me. Can anyone please help me out with proper steps to achieve this expression. Thanks in advance !

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  • 1
    $\begingroup$ You should express the identities in $1.2$ in terms of $\epsilon$; solve $1.2b$ for $\gamma(\epsilon)$, rewrite $1.2c$ for $v(\gamma)$ and substitute the previous result to get $v(\epsilon)$, lastly substitute the former two result into $1.2a$ to get $p(\epsilon)$. Substitute them into the $1.1$ $\endgroup$
    – nluigi
    Jun 26, 2020 at 15:04
  • $\begingroup$ oh yes! You're right. But that's a little lengthy. But doesn't need concepts of Hyperbolic Coordinates, and curved space-time, and more general and self-sustained w.r.t. the given informations. Thanks ! $\endgroup$ Jun 26, 2020 at 18:30
  • 1
    $\begingroup$ seems a more direct route to me than using hyperbolic functions. It is not my field of expertise so the hyperbolic functions come out of the blue for me whereas the substitution method seem clear to me that the author intended that route to get from 1.1 to 1.3 $\endgroup$
    – nluigi
    Jun 26, 2020 at 19:26
  • $\begingroup$ yes! This method is better and easier actually. $\endgroup$ Jun 26, 2020 at 19:28
  • $\begingroup$ Hyperbolic trigonometry often suggests how to proceed, in analogy to ordinary trigonometry, since formulas in terms of are not intuitive and don’t suggest their geometric interpretation. (Look at the identity in physics.stackexchange.com/a/510201/148184 ). Maybe one just cares about the algebra (not the geometrical meaning)... that’s fine. $\endgroup$
    – robphy
    Jun 26, 2020 at 20:44

3 Answers 3


Here are some useful relations for relativistic kinetic energy

$E=m\cosh\theta=m(\cosh\theta-1)+m= T+m $




from my post https://physics.stackexchange.com/a/551250/148184

  • $\begingroup$ Yes, I understand this, but this doesn't answer my question. Can you please be more specific regarding the exact steps of the problem I am facing. $\endgroup$ Jun 26, 2020 at 12:09
  • 1
    $\begingroup$ Can you try some algebraic substitution? You may have use different variable names. $\endgroup$
    – robphy
    Jun 26, 2020 at 12:34
  • $\begingroup$ yeah, you were right, and doing a little algebraic manipulation I got it. Thanks ! $\endgroup$ Jun 26, 2020 at 14:24

First from the second identity, let solve it in terms of $\epsilon$ and generate some other useful equations:




So now notice, if we substitute $p$ into the integral (ignoring arguments, assumed to be $\epsilon$) we get:

$$P=\frac{1}{3}\int_{0}^{\infty}n\gamma mv^{2}d\epsilon$$

which depends on the square of $v$. We solve the third identity for $v^2$ in term of $\gamma$:


Now we can solve for the term inside the integral, which gives: $$\begin{aligned} \gamma mv^{2} &=\gamma mc^{2}\left(1-\frac{1}{\gamma^{2}}\right) \\ &=\gamma^{-1}mc^{2}\left(\gamma^{2}-1\right) \\ &=\epsilon\left(1+\frac{mc^{2}}{\epsilon}\right)^{-1}\left(1+2\frac{mc^{2}} {\epsilon}\right) \end{aligned} $$ where in the last step the previously determined identities were used: $$\begin{aligned} \gamma^{-1}\left(\gamma^{2}-1\right) &=\gamma^{-1}\left(\gamma-1\right)\left(\gamma+1\right)\\ &=\left(\gamma-1\right)\left(\frac{\gamma-1}{\gamma}\right)\left(\frac{\gamma+1}{\gamma-1}\right)\\ &=\frac{\epsilon}{mc^{2}}\left(1+\frac{mc^{2}}{\epsilon}\right)^{-1}\left(1+2\frac{mc^{2}}{\epsilon}\right) \end{aligned}$$ Substituting this result into the integral gives the form of the integral found in the book.

  • $\begingroup$ yes I did that already, but thanks for adding that as an answer to the post ! $\endgroup$ Jun 27, 2020 at 19:13
  • $\begingroup$ Forming $\epsilon=\gamma-1$ is a natural step since you are dealing with relativistic kinetic energy? So, what is your motivation for writing an expression for $\gamma+1$ or $(\gamma-1)/\gamma$ [and not others]? Is it just making list of possibly useful equations? Using rapidity in the expression for momentum, $\gamma+1$ arises naturally from the difference of squares. $\endgroup$
    – robphy
    Jun 27, 2020 at 19:52
  • $\begingroup$ I am not claiming the derived identities mean anything physically, they are just algebraic manipulations. $\endgroup$
    – nluigi
    Jun 27, 2020 at 20:10

Thanks to @robphy for providing the answer, and with some algebraic manipulation, I got this:

$$P = \frac{1}{3}\int_{0}^{\infty}np(\epsilon)v(\epsilon)d\epsilon$$ $$= \frac{1}{3}\int_{0}^{\infty}np\frac{p}{\epsilon+m}d\epsilon$$

As, $v(\epsilon) = \frac{p}{E} = \frac{\sqrt{\epsilon(\epsilon+2m)}}{\epsilon+m}$ $$= \frac{1}{3}\int_{0}^{\infty}n\frac{p^2}{\epsilon+m}d\epsilon$$ $$= \frac{1}{3}\int_{0}^{\infty}n\frac{\epsilon(\epsilon+2m)}{\epsilon+m}d\epsilon$$

As, $p=\sqrt{\epsilon(\epsilon+2m)}$

Upto this, it was normalized as $c=1$, and reconsidering that, we get:

$$P = \frac{1}{3}\int_{0}^{\infty}n\frac{\epsilon(\epsilon+2mc^2)}{\epsilon+mc^2}d\epsilon$$ $$= \frac{1}{3}\int_{0}^{\infty}n\epsilon(1+\frac{2mc^2}{\epsilon})(\frac{\epsilon}{{\epsilon+mc^2}})d\epsilon$$ $$= \frac{1}{3}\int_{0}^{\infty}n\epsilon(1+\frac{2mc^2}{\epsilon})(\frac{1}{1+\frac{mc^2}{\epsilon}})d\epsilon$$ $$\therefore P = \frac{1}{3}\int_{0}^{\infty}n\epsilon(1+\frac{2mc^2}{\epsilon})(1+\frac{mc^2}{\epsilon})^{-1}d\epsilon$$


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