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Energy of a photon in different frames.

  1. Oct 11, 2005 #1
    Just a quick question here, that I've been having a bit of trouble figuring out:
    How do you calculate the energy of a photon in a frame if you know the energy of the photon in another frame as well as the relative velocities of the two frames?

    Thanks a lot.
     
  2. jcsd
  3. Oct 11, 2005 #2

    dextercioby

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    How do you relate the 4-momenta of the photons in the two frames...?

    Daniel.
     
  4. Oct 11, 2005 #3
    There is only one photon, being observed in two different reference frames.
     
  5. Oct 11, 2005 #4
    Write doen the 4-momentum in the first frame and use the Loretz transformation to boost to the new frame. Let P' be the original 4-momenutum in the oiginal frame. Then P = (m'c, p') where the primed quantities represent the quantities in the old frame, i.e. the old photon mass and the onld photon momentum. Now transform to the new frame you'' get P = (mc, p). Suppose you're standing on the Sx axis. Then all we need to is find P0 so we calculate Pu = LuvPv or P0 = L0vPv.

    P0 = L0v[/suo]Pv

    This will give you mc = P0 where m = E/c2 and E = hf. The rest should be easy for you to do since all there is left is calulation and then solve for f.

    Pete

    ps - my HTML is all messed but I'm too tired to fix it. Tommoror perhaps.
     
  6. Oct 11, 2005 #5

    robphy

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    Doppler Effect.
     
  7. Oct 11, 2005 #6

    dextercioby

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    It's actually Doppler-Fizeau effect, though, correctly, it should bear only Armand Hyppolyte Fizeau's name, as he discovered it.

    Daniel.
     
  8. Oct 11, 2005 #7
    Since energy is proportional to frequency you can use the relativistic doppler shift equation. However that doesn't tell you much about what physics is taking place. Since a photon has zero mass you can't ligitimately use [tex]p^{\mu } = (\gamma mc, p)[/tex] for the generalization as was suggested. Instead for generalization you use [tex]p^{\mu } = (\hbar \omega, p^{x}, p^{y}, p^{z})[/tex] where [tex]p^{i} = \hbar k^{i}[/tex]. For any particle with or without mass [tex]v_{g} =d\omega /dk[/tex]. For photons the relation reduces to [tex]v_{g} = c = v_{p} = \omega /k[/tex]
     
  9. Oct 11, 2005 #8
    Trilairian; You've made an erroneous assumption above. Nobody was using the expression

    [tex]p^{\mu} = (\gamma m c, p)[/tex]

    other than yourself. You failed to notice the context in which the quantity "m" was being used you therefore assumed that m was the photon's proper mass, which is indeed zero. Here "m" is not being used to represent proper mass. m = photon's inertial mass = p/v where here v = c (aka m = relatvistic mass). Had I had the case to use proper mass then it would have read m0. It is always wise consider the context a quantity like "m" is being used before you attempt to make a correction in a post such as mine. It was obvious from context what my "m" meant, unless you have a limited knowledge of SR (E.g. see Introducing Einstein's Relativity by D'Inverno, Basic Relativity by Richard A. Mould, Special Relativity by A.P. French, or the Feynman Lectures)

    Especially in relativity since some people use m0 for proper mass and some use m.

    The "m" used in my expression for the 4-momentum was not the photon's proper mass. It was the inertial mass m = p/v = p/c = E/c2 = hf/c2 since E = pc and E = hf. The h would have dropped out in the end of all the calculations and we'd have the relativistic expression for the doppler frequency.

    The expression you used, i.e.

    [tex] m = \gamma m_0[/tex]

    (gamma * proper mass) for P0/c is valid if and only if m0 is not zero which is not the case here. In SR the relation

    m = [tex]\gamma m_0[/tex]

    is derived on the assumption that the particle is a tardyon (a particle which always moves at v < c) which is not the case here. Since in this case it is zero another expression is required and that's m = E/c2 where E = hf and therefore m = hf/c2.

    Pete
     
    Last edited: Oct 11, 2005
  10. Oct 11, 2005 #9

    You are so confused it is funny. Trilairian did not assume that that m was the photon's proper mass. He assumed m was the ridiculous "relativistic mass" (i.e. [itex]\gamma m[/itex]). Are you now suggesting that your relativisitic mass is not always defined as [itex]\gamma m[/itex]? I assume you are going to propose some kind of conditionality.

    [tex] m = \frac{p}{v} = \gamma m_0[/tex]
     
    Last edited: Oct 11, 2005
  11. Oct 11, 2005 #10
    I'm known for having a great sense of humor but I doubt that's how you meant that to be read. So while you're here please kill the poor attitude. This is a moderated board.
    I'm sorry to inform you, pyro, that it is you who's confused. I said that the only person who was using the expression
    P = [tex](\gamma m c, [bp)[/tex]
    was Trilairian. When he saw my "m" in my 4-momenta, i.e. in my expression (mc, p) he must have thought it was proper mass since he stated "Since a photon has zero mass you can't ligitimately use [tex]p^{\mu } = (\gamma mc, p)[/tex] for the generalization as was suggested. He tossed in the gamma himself thus changing the context from "m" being inertial mass to "m" being proper mass.
    This follows since for my expression (mc, p) to be true then m must be the particle's inertial mass (what you enjoy refering to as "relativistic mass" in your condescending manner). And that is exactly what I meant. I'm well known on this and every forum for using m to mean inertial mass. But since you're new here you wouldn't know that yet. It may be that Trilairian doesn't know the difference but that remains to be seen if and when he chooses to respond.
    Many people use "m" to mean proper mass. I do not. Trilairian does. Regardless, what a quantity is can be deduced from its usage or its context.
    You're being too vauge. What "m" are you refering to? The one when he read my 4-momentum and later wrote "....for the generalization as was suggested."? It was he who claimed that my "m" was zero and thus assumed that my "m" was proper mass when it was really inertial mass (aka relativistic mass)
    If you think that relativistic mass is "ridiculous" then you don't know SR as well as you'd like to think you do.
    Mine? Since when did relativistic mass become mine? I wasn't even alive when the quantitiy was defined.
    If "m" is the proper mass of the particle then [itex]\gamma m[/itex]The so-called "relativistic mass" of a particle is the quantity "m" such that the quantity "mv" is a conserved quantity. If the particle is a luxon (I.e. a particle which always moves at v < c. I made an error above on terminology regarding "luxon" which has now been corrected). In such a case the particle's (relativistic) mass is a constant. Otherwise its a function of speed. One only need look it up to see this is the case. I know of no SR text which uses relativistic mass that says anything different than I just told you.
    Well at least you're sort of on the right track. You can think of inertial mass as the ratio of momentum to speed. See footnote in French's text on bottom on page 16 "By inertial mass we mean the ratio of the linear momentum to speed"or for a complete coverage see D'Inverno's text section 4.5 Photons on pages 49->50. See Eq. (4.27) if nothing else.
    And it wouldn't hurt you to kill the poor attitude.
     
    Last edited: Oct 11, 2005
  12. Oct 11, 2005 #11
     
    Last edited: Oct 11, 2005
  13. Oct 11, 2005 #12

    JesseM

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    How do you define "inertial mass"? A mirrored box filled with photons would be slightly more resistant to acceleration than an empty but otherwise identical box, no?
     
  14. Oct 11, 2005 #13

    JesseM

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    Your source was a usenet post, hardly authoritative. I'd like to see a definition of what you mean by "inertial mass", because "inertia" is usually defined in terms of resistance to acceleration, and as I said before a box filled with photons will be more resistant to acceleration that an empty but otherwise identical box.

    It's true that you can't use Newtonian formulas to predict the path of photons, but I don't see where pmb_phy ever said you could. Also, note that if you take the weight of a box filled with photons and subtract the weight of the box when empty, the extra weight will be equal to the total energy of the photons divided by c^2. You can see this based on the equivalence principle, and the fact that the inertia of a bound system is proportional to the system's total energy (in its rest frame) divided by c^2.
     
  15. Oct 11, 2005 #14
    That's easily shown. "Inertial mass" means that quantity which resists changes in momentum. As I explained to pyro (who wasn't able to follow) was that inertial mass (what he calls 'relativistic mass') is defined as the m such that mv is conserved.

    Contrary to pyro I can back everything I say up either by straight forward calculation or a reference to the physics literature.

    Nope. Since when did Newton come into the picture???? A particle follows the path of a parabola in Newtonian gravity in a uniform g-field regardless of its mass. As the mass goes to zero in a uniform g-field the spatial portion of the geodesic approaches that of a semi-circle.

    Jesse - Please note that when pyro demonstrated that he was one of those people who can't disagree with someone without flaming then I plonked him right away (i.e. he's on my ignore list) so I won't be able to understand what you're replying to if its from pyro unless you quote him.

    To be 100% precise for a definition of mass (aka inertial mass) please see - http://www.geocities.com/physics_world/sr/inertial_mass.htm

    I believe I put all the derivations in this paper

    http://www.geocities.com/physics_world/mass_paper.pdf

    However from the part of pyro's post that you quoted its quite evident that pyro doesn't have a clue as to what he's talking about. Its a fact of nature that a photon has mass but zero proper mass. When it is said that a photon has "mass" it means that it has inertial properties, active gravitational properties and passive gravitational properties. All of which are well known facts in GR.

    Remember what I said about quotring the references? Then recall this. From The Feynman Lectures Vol -I page 7-11, section entitled Gravitation and Relativity
    That's a quote. Here's a calculation

    http://www.geocities.com/physics_world/gr/grav_light.htm

    This was first done by Tolman et al but is so easy that anyone who knows GR can do it as a matter of course.

    Pete
     
    Last edited: Oct 11, 2005
  16. Oct 13, 2005 #15
    The relation E0 = m0c2 means that an increase in the proper energy of the bnody (aka "rest energy") wil give yield to an increase in proper mass. This also says that since the proper inertial mass of a body is proportional to the proper passive gravitational mass of the body then that too will increase with an increase in an increase in rest energy.

    Pete
     
  17. Oct 13, 2005 #16
    You didn’t “explain” anything to Pyro. In fact he was the one who explained why you were wrong. Since your own slanted papers are in fact just plain wrong referencing them proves nothing. You don’t even know what mass is, inertial gravitational etc-all the same thing, i.e. equivalent, and invariant which I can prove easily if anyone is intereseted. Gravitational mass passive or active are Newtonian gravitation concepts. In relativity even massless entities such as electromagnetic plane waves gravitate and deviate from constant motion in the presence of gravitation. In relativity gravitation is expressed by Einstein’s field equations, not the Poisson equation that yields Newtonian gravitation. In relativity the source term is the stress-energy tensor, not the mass density as in that Poisson equation. In relativity all particles massive or massless follow geodesics under the influence of gravitation alone, not the paths of motion as forced about by Newtonian gravitation. The photon is massless. The mass m, the proportionality between the conserved momentum and the four-vector velocity [tex]p = mU[/tex], is invariant. It is also the m in the relation to rest energy given by [tex]E_{0} = mc^{2}[/tex]. In short you need to catch up with the real world of physics or at least stop misrepresenting it.
     
  18. Oct 13, 2005 #17

    JesseM

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    Can you address my question to Pyro above, since he has not replied? I'd like to know what mathematical definition you are using for "inertial mass" in relativity, since the only types of mass I have seen defined are rest mass and relativistic mass. Do you agree that the inertia of a bound system in its rest frame is proportional to its total energy in that frame (including the kinetic energy of its constituents) divided by c^2? Do you agree that according to the equivalence principle, inertial mass and gravitational mass must be equal? (ie if the bound system is sitting on a scale being accelerated at 9.8 m/s^2 in deep space, the reading on the scale will be the same as if it is placed on a scale on earth)
     
  19. Oct 13, 2005 #18
    Both are bad terminology. Mass is rest energy. Inertial mass means the measure of resistence something has to change in motion. In Newtonian physics this means that it is the m in f = ma, and in relativistic physics this means it is the m in four-vector force equals mass times four-vector acceleration
    F = mA. This mass m is invariant. It does not change with speed, no not in relativity.
    No.
    Yes, although gravitational mass is a purely Newtonian quantity and is not the source term in the more general case of general relativity. Instead, the stress energy tensor is. The m which is invariant and is the inertial mass is also the [tex]m_{g}[/tex] called gravitational mass which only is defined in the case of the Newtonian applicability of [tex]|g| = Gm_{g}/r^{2}[/tex].
     
    Last edited: Oct 13, 2005
  20. Oct 13, 2005 #19

    JesseM

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    Then your understanding of inertia in relativity is incomplete. This issue was discussed at length on this thread--for example, someone posted a quote from this page, where Einstein explains that all forms of energy contribute to an object's resistence to motion (inertia) in the same way, so that a hot iron is harder to accelerate than a cold one (the only difference being the average kinetic energy of the atoms that make it up):
    Also posted was a link to this paper, which shows some experimental evidence that the kinetic energy of the parts of a bound system contribute to its inertial and gravitational mass. And potential energy certainly has to be taken into account in calculating inertial or gravitational mass too--as this page which was linked to earlier in that thread points out, the mass of a deuteron is less than the sum of the masses of the proton and neutron which make it up, because the potential energy of a proton and neutron bound into a deuteron is lower than the potential energy of the same proton and neutron kept a large distance apart.
     
    Last edited: Oct 13, 2005
  21. Oct 13, 2005 #20
    No it isn't.
    prior to the formulation of the modern version of the law, the four-vector equation I gave - so it is your understanding that isn't complete. I understand the dynamics much better than you and the inertia is not proportional to the energy as you assert. In fact the kind of inertia you are referring to, the proportionality between ordinary force and coordinate acceleration even has a directional dependence. You're so far off in assuming that it is the energy that it isn't even funny.
     
    Last edited: Oct 13, 2005
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