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*To*: KellySt@aol.com, hous0042@maroon.tc.umn.edu, stevev@efn.org, rddesign@wolfenet.com, RUSSESS@cellpro.cellpro.com, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl*Subject*: Engineering Newsletter*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Sun, 26 Nov 1995 17:39:02 +0100

Reply To : Steve Reply From : Timothy >> I think I can't agree with that Steve, when the photon enters the first >> stage of the reflection, i.e. absorption, it adds some momentum to the ship. >> So the velocity of the ship increases. Now we enter the second stage of >> reflection, re-transmission. Relative to the ship the outgoing wavelength of >> the photon is the same as the incoming photon (because of invariance). But >> the observer at rest sees that that transmitted photon has dopplershifted >> (nice verb) and thus lost some energy. > >And I don't disagree with that analysis; it's correct in a frame >where the ship is moving. In that frame, the photon has more >energy and momentum, and while it changes momentum by the same >amount as in the frame where the ship is "stationary", this >necessarily results in a different-looking scenario. I was wrong to compare it with the police radar, so forget that part. But I still don't see what was wrong with part preceding it. o you say the photon looses energy if it accelerates the ship or do you say that the energy of the photon stays the same? What does that different scenario look like? >> Assuming these mirrors have mass, they indeed will get some velocity that >> correspondends to the same momentum as the photon. By each reflection the >> photon will loose some momentum and lower its wavelength. > >This is exactly the analysis I came up with. In the limit the >mirrors carry the momentum of the light pulse, and the light >pulse fades to nothing. Somehow we seem to agree on this, but I'm not sure why. >The way I was taught to measure mass is as sqrt(E^2 - p^2). So >the mass of a particle with energy E = gamma * m_rest and >momentum p = gamma * m_rest * v (where v is a vector) is > >m^2 = E^2 - p^2 > = gamma^2 * m_rest^2 - gamma^2 * m_rest^2 * abs(v)^2 > = gamma^2 * m_rest^2 * (1 - abs(v)^2) > = m_rest^2 > >(gamma is 1/sqrt(1 - abs(v)^2)). So no matter how the particle >moves it has the same mass. Mass is not energy. What you call >"relativistic mass" I call energy; what I call "mass" you seem to >want to call "rest mass", yet the quantity "mass" that I am using >doesn't vary with motion. This is what I was feeling; We think the same but use different terminology. What you call mass doesn't vary. I call that quantity rest-mass and that doesn't vary either. Taylor and Wheeler tell us that rest-mass revokes a potential misunderstanding so they call it invariant mass. I don't see why rest-mass would change when a body starts moving, I guess I've grown above that misunderstanding. You say the energy changes while I say the relativistic-mass or the kinetic energy changes. This kinetic energy is not the classical 0.5mv^2 but m_rest*c^2(gamma-1) Or more clear the total energy minus the rest-energy. > > >If, on the other hand, a moving object did exert greater > > >gravitation than a stationary object of the same mass, I'd > > >probably be looking for a relation to a quantity that did change, > > >like the object's total energy. > > > > So that would mean that some of the change in energy is change of > > gravitational energy, from which I would conclude that it comes from extra > > mass (or bending of space time). > >First you have to convince me that a moving object really does >exert more gravitational force. Let me quote a few sentences of "Introducing Einstein's relativity" by Ray d'Inverno in the paragraph "The principle of equivalence": Next, we wish to make explicit the assumption that matter both responds to, and is a source of, a gravitational field. However, we have seen in special relativity that matter and energy are equivalent, so the statement about the gravitational field applies to energy as well. This means that even photons excert and react to gravity as well as all other kinds of energy. I would conclude from this that moving bodies excert greater gravitation either because of gain of mass or gain of energy. I feel both can be used, its a bit like the wave-particle duality. >> Is it this translation of energy to mass that gives the trouble? > >It is that my studies of relativistic kinematics do not allow for >the treatment of energy and mass as identical quantities. They are indeed not identical but equivalent, meaning they can be be interchanged. =============================================================================== ReplyTo : Kevin ReplyFrom : Tim >BTW, in english, a nitwit is a stupid person. (someone with the brains >of a nit <a small insect>) in so naming myself, I admit my previous errors Yes, I knew, I looked it up in a dictionary, it was just a repeat of your own words. >actually, it's Kevin, I know it can be confusing when three or more >people are engaged in debate, but try to keep us straight. It's not nice >to insult Steve's intelligence like that. I guess, it was a bit late when I wrote that just after I finished the letter to Steve which occupied my mind. >not very, considering i lost it (or it was stolen), I've been using EXCEL >for my calculations, also, I rounded the actual angle of the sail >elements from 21.47 to 21 I figured you used tables or something like that. Anyway here are some extra corrections of my calculations: >> Red x-comp = b cos(-48) = -18.55 (you got -18.9) ^ should be positive And I made mistakes in the general formulas, so here are the right ones: Px=2b sin(t) Py=2b cos(t) Rx=b sin(2t-180)=- b sin(2t) Ry=b cos(2t-180)=- b cos(2t) x=Px+Rx=2b sin(t) - b sin(2t) y=Py+Ry=2b cos(t) - b cos(2t) t=90 gives y=b which is logical because that means that if there is no mirror the energy is captured right away. Every other angle makes the problem worse!

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