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I Different forms of energy density in inflation

  1. Jan 13, 2017 #1
    From the second Friedmann equation,

    $$H^2 = \frac{1}{3M_p^2} \rho \quad (k=0, flat)$$

    In warm inflation, radiation is present all the way therefore not requiring proper reheating process, so

    $$\rho = \rho_\phi + \rho_r \, ; \quad \rho_\phi = inflaton, \, \rho_r = radiation$$

    But, $$\rho = \Big(T + V \Big)$$

    What should be the kinetic and potential energy of ##\rho_r## as opposed to ##\rho_\phi = \Big(\frac{1}{2} \dot \phi^2 + V \Big)## (i.e. ##V = \frac{1}{2}m^2\phi^2##)?
  2. jcsd
  3. Jan 16, 2017 #2
    I'm not quite sure what Your trying to do but if I'm not mistaken
    [tex]V_(\phi)[/tex] is your potential


    Looks like you dropped the pressure term of [tex]M_p [/tex]

    [tex]H^2=\frac{8\pi G_n}{3}\rho\rightarrow \frac{\dot{a}}{a}=\sqrt{\frac{8\pi G_nV(\phi)}{3}}[/tex]
    Last edited: Jan 16, 2017
  4. Jan 21, 2017 #3
    How could ##ρ## suddenly become ##V(φ)## in the last equation? Also, there is no pressure term in the second friedmann equation.
  5. Jan 21, 2017 #4
    Are you sure about that? I left you a handy link showing the pressure term in one of your other threads.

    https://redirect.viglink.com/?format=go&jsonp=vglnk_148505730666812&key=6afc78eea2339e9c047ab6748b0d37e7&libId=iy84rrwr010009we000MA49vip5hy&loc=https://www.physicsforums.com/threads/rescaling-the-equation-of-motion-of-inflation.900793/&v=1&out=http://www.google.ca/url?sa=t&source=web&cd=2&ved=0ahUKEwi67MqH0dPRAhVH4mMKHf9vBhgQFgggMAE&url=http%3A%2F%2Fwww3.imperial.ac.uk%2Fpls%2Fportallive%2Fdocs%2F1%2F56439.PDF&usg=AFQjCNFCbq4LLlR6366LhUvr8T_y6_f0eA&sig2=n5C7FRMAGfPcq5gYfq4hMw&ref=https://www.physicsforums.com/forums/cosmology.69/&title=Rescaling the equation of motion of inflation | Physics Forums - The Fusion of Science and Community&txt=http://www.google.ca/url?sa=t&source=web&cd=2&ved=0ahUKEwi67MqH0dPRAhVH4mMKHf9vBhgQFgggMAE&url=http://www3.im...

    see equation 1.2 and 1.3 in the article above

    Start with the equation of state [tex] w=p/\rho [/tex]
    Then look at the scalar field modelling equation


    Though the scalar field modelling equation is better explained in the first link. (I was trying to avoid posting the same two links in two different threads as you have 3 that are related.

    If you don't want to use the first link the Friedmann fluid equation is the second one on this link.

    Note both pressure and energy density in the second equation. If you include the pressure term all three of your threads can be answered.

    As far as the last equation which is under the Newtonian limit there are numerous solutions most involving GR. However chapter 4 here will give you the basics


    Andrew Liddle's "Introductory to Cosmology" though has a decent chapter though done a little too Heuristic. Hope that helps I've usually only enough time for one forum and I usually spend that avaliable time on another as they can use my help more so than this excellent forum.

    Another excellent textbook for introduction level is Ryder Lewis "Introductory to Cosmology " he has a decent chapter on three different inflation models. Lambda based ie ( the equations of state in the links above where w=-1), k inflation and quintessenece
    Last edited: Jan 22, 2017
  6. Jan 22, 2017 #5
    Here is some additional resources you may find handy in your studies

    http://arxiv.org/pdf/hep-ph/0004188v1.pdf :"ASTROPHYSICS AND COSMOLOGY"- A compilation of cosmology by Juan Garcıa-Bellido
    http://arxiv.org/abs/astro-ph/0409426An overview of Cosmology Julien Lesgourgues
    http://arxiv.org/pdf/hep-th/0503203.pdf"Particle Physics and Inflationary Cosmology" by Andrei Linde
    http://www.wiese.itp.unibe.ch/lectures/universe.pdf:" Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis
  7. Jan 22, 2017 #6
    Thank you for your effort to put together some great resources, but what I don't know what you are talking about is how can
    $$H^2 = \frac{8\pi G}{3} \rho$$
    have any pressure term. I want to solve for ##H## so I think the second equation is not so useful.
  8. Jan 22, 2017 #7
    Would it help to know the definition of pressure is force per unit volume?

    So [tex]\frac {8\pi G_n}{3}[/tex] gives you your pressure term for your volume. G is your Newtons gravitational constant. We have set the curvature term k=0 which is why I didn't include k.

    Also as we are working with a homogeneous and isotropic fluid expansion can be described as a perfect fluid following all the gas law rules of an adiabatic fluid ( adiabatic meaning no net inflow/outflow of energy). We are also applying Newtons shell theorem in the above. As we are working with a homogeneous and isotropic fluid.

    As you are working with two time derivitaves you must describe each volume seperately the pressure at R and the pressure at [tex]\dot {R}[/tex] in order to derive H^2 or alternately the two time derivitaves of [tex]\frac {\dot {a}}{a}[/tex] as according to the ideal gas laws an increase in volume decreases pressure/density and temperature. So for H you cannot use 1 value for pressure and energy/mass density but the time derivitaves of both at each time derivitave of your commoving volume

    So using R for volume my two time derivitaves between two commoving volumes becomes
    [tex]R =\frac {8\pi G_n}{3}\rho [/tex]
    [tex]\dot {R }=\dot {\frac{{8\pi G_n}}{3}}\dot {\rho} [/tex] where [tex]\frac {8\pi G_n}{3}[/tex] is in essence the force per unit volume of time derivitave R

    [tex]H^2=\frac {\dot {R}}{R}[/tex] you can literally plug the two time derivitaves of pressure and energy/density into the last formula to get H

    The same time derivitave rules apply to your scalar field equations in my first reply

    [tex]\dot {\phi}[/tex] denotes the potential is a time derivitave and needs to be treated as a time derivitave. The curvature term can be safety assumed as constant between the time derivitaves
    Does that help?

    Edit ;important side note Do not think of expansion as due to pressure/(force per unit volume) as were dealing with a homogeneous and isotropic fluid there is no net flow of force as there is no pressure gradient at a particular time
    Last edited: Jan 22, 2017
  9. Jan 22, 2017 #8


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    Science Advisor

    There is no pressure term in this equation.
  10. Jan 22, 2017 #9


    Staff: Mentor

    Pressure is force per unit area, not volume.
  11. Jan 22, 2017 #10
    Doh right lol my bad working too late in am
  12. Jan 22, 2017 #11
    That is what I'm wondering about, as Mordred posted.
  13. Jan 23, 2017 #12
    I should have said derive not give. Did you look at equations 1.1 to 1.3 of the first link yet?
    1.1 your Friedmann metric
    [tex]H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{k}{a^2}[/tex]
    [tex]\frac{\ddot{a}}{a}=-\frac{-4\pi G}{3}(\rho+3p)[/tex]
  14. Jan 23, 2017 #13
    Yes, I'm using (1.2) and there is no pressure term there.
  15. Jan 23, 2017 #14
    I take it you missed the sentence above relating those two equations then?

    "These are related to the energy density and pressure terms through the Friedmann equations 1.2 and 1.3."

    key word AND not one or the other but both
  16. Jan 23, 2017 #15
    Yes but what is your point? I'm only interested getting ##H##, so I can obtain it through $$H^2 = \frac{1}{6M_p^2}(\dot \phi^2 + m^2\phi^2)$$
  17. Jan 23, 2017 #16
    what does the subscript p stand for on M?
  18. Jan 23, 2017 #17
    ##M_p## stands for the reduced Planck mass.
  19. Jan 23, 2017 #18
    right and your reduced planck mass formula is?
    [tex]m_p=\sqrt{\frac{\hbar c}{8\pi G}}[/tex]
  20. Jan 23, 2017 #19
    Yes, and ##\hbar = c = 1##.
  21. Jan 23, 2017 #20

    and in your OP you want your temperature right ? ie[tex]\rho=T+V[/tex] right?

    How do you solve the temperature without knowing the pressure?

    Particularly since your starting with the scalar field equations
  22. Jan 23, 2017 #21
    The kinetic ##T## and potential energy ##V## of the radiation density; but aside from that I have one more question, the equation of motion of inflation is
    $$\frac{d^2\phi}{dt^2} + 3H\frac{d\phi}{dt} + V,_\phi = 0$$

    where ##ϕ## is the inflaton, ##H## is the Hubble parameter, and ##V,_ϕ## is the derivative of the potential with respect to ##ϕ##. To solve this differential equations, we need two initial value conditions, ##\phi(0)\,## and ##\dot \phi(0)\,##. I'm using Mathematica to solve for ##\phi##, and in order to solve the above DE we need to eliminate the unknown function ##H## and from the Friedmann equations, $$H^2 = \frac{1}{3M_p^2}\Big(\frac{1}{2}\dot \phi^2 + V\Big)$$

    For a potential given by ##V = \frac{1}{2}m^2\phi^2##, we have
    $$\ddot \phi^2 + 3\sqrt{\frac{1}{6M_p^2}\Big(\dot \phi^2 + \phi^2\Big)} \dot \phi + m^2\phi = 0$$

    I can set ##\phi(0) \approx M_p## and ##\dot \phi(0) \approx 0.1## (any small value just to get started).
    But in order to get an exact numerical value for ##\phi(t)## after solving this DE numerically, I need to set some ##t##. But I'm not sure about what ##t##'s to set during the inflationary stage, or in other words, I want to know the proper ##t## to use during the evolution of ##\phi##?
  23. Jan 23, 2017 #22
    lol glad I stated previously the above in your OP. I misread the T for temp not kinetic energy.

    I'm not sure what the appropriate values for t on your time derivatives are most appropriate in this case. Bapowell will probably best answer that
  24. Jan 23, 2017 #23
    When you solve a DE with ##t## as the independent variable, you need to state at what ##t## you want to solve for the dependent variable right? It's just the same case with what I'm doing, here the dependent variable is ##\phi##.
  25. Jan 23, 2017 #24
    I'm still looking for your t question. There is a section that applies in the first link I supplied see chapter 2.with regards to the rates involved in inflation in terms from the potentials. I can't quite remember this so have to dig deeper than the article I just mentioned.
  26. Jan 23, 2017 #25
    I'm not sure if ##t## should be the Hubble time which during inflation, the Hubble parameter is fairly constant or ##t## should be the time period for inflation.
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