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*To*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Subject*: Engineering Newsletter*From*: Steve VanDevender <stevev@efn.org>*Date*: Sat, 25 Nov 1995 18:48:43 -0800*Cc*: 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*In-Reply-To*: <199511252351.AA19695@student.utwente.nl>*References*: <199511252351.AA19695@student.utwente.nl>

Timothy van der Linden writes: > > > I have a lot of problem bying the idea that the light bouncing > > > off a sail doesn't lose energy proportional to the kinetic or > > > heat energy gain of the ship. Th power has to be coming from > > > somewhere. > > > >If the sail is reflective to the incident photon, then the photon > >doesn't lose energy as a result of reflection. The ship changes > >velocity because the photon changes direction. That's where the > >"power" comes from. > > 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. > > This is the same principle the police uses to measure speeding cars with radar. 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. > >What happens to the mirrors? What happens to the light pulse? > >An answer describing the limit state of the mirrors and light > >pulse is acceptable; you don't have to perform a step-by-step > >analysis of each reflection. > > 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. > >> I've always looked at is as follows: When you move faster and faster, part > >> of the energy is transformed into mass, the other part is used to get the > >> extra momentum. > > > >I used to look at it this way, but Taylor and Wheeler talked me > >out of it (see chapter 8 of _Spacetime Physics_ for a lengthy, > >careful discussion on "Use and Abuse of the Concept of Mass"). > >The problem here is that relative velocity or acceleration do not > >cause any fundamental changes in the structure of the moving > >object. Where is this extra mass? If it's really stashed on the > >ship somewhere then the people on the ship could measure it. But > >they don't feel the ship getting heavier or see any increase in > >the mass of the ship in their frame. > > I wrote it wrong the first time, indeed then the mass must be somewhere on > the ship. But look at it this way: When an object starts moving it deforms > space-time in such a way that the object looks heavier to the outside world > and the other way around, i.e. the object "notices" the outside world to be > heavier. > This means that locally no mass increase is measured, for the same reason > that locally no length contraction or time dilation is measured. Taylor and Wheeler say it better than I can. Their approach is based on years of teaching special relativity to new students, which is why I give it credence. So, to quote Taylor and Wheeler: Q: If the factor c^2 is not the central feature of the relationship between mass and energy, what _is_ central? A: The distinction between mass and energy is this: Mass is the magnitude of the momenergy 4-vector and energy is the time component of the same vector. Any feature of this discussion that emphasizes this contrast is an aid to understanding. Any slurring of terminology that obscures this distinction is a potential source of error or confusion. Q: Is the mass of a moving object greater than the mass of the same object at rest? A: No. It is the same whether the object is at rest or in motion; the same in all frames. To explain some of their terminology: Taylor and Wheeler teach relativistic kinematics using the notion of "momenergy". Since energy is conserved and the vector quantity of momentum is conserved in a system, they express the entire state of an object or system using a 4-vector that combines both quantities, which they call a "momenergy vector". An object's state of motion can be expressed as a vector of the form [ E px py pz ], where E is the object's energy and [ px py pz ] are the components of the object's momentum. Since vectors are normally added component by component, adding momenergy vectors preserves the conservation of the individual components. The other notion that they introduce at the very beginning of the book is that relativistic spacetime is non-Euclidean, even in special relativity. A vector of the form [ t x y z ] is defined to have magnitude sqrt(t^2 - x^2 - y^2 - z^2), which results in non-zero vectors that have magnitude zero (or even imaginary magnitude). The magnitude of a momenergy vector is exactly the mass of the system described by that vector. The only time that mass is equivalent to energy is when momentum is zero, or in a frame in which the object in question is at rest. > >Taylor and Wheeler's wisdom on the subject is that the definition > >of mass is sqrt(E^2 - p^2); then every observer sees the same > >mass for the same object, no matter what their relative motion. > > How do you measure E and p? E seems to be relativistic mass and p > relativistic momentum. (E=gamma*m_rest p=gamma*m_rest*v) > I would measure the perceived relativistic mass and the relative velocity. > As far as I can see, doing that all observers will also agree about the rest > mass. Another quote from Taylor and Wheeler: Q: In order to make this point clear, should we call invariant mass of a particle rest mass? A: That is what we called it in the first edition of this book. But a thoughtful student pointed out that the phrase "rest mass" is also subject to misunderstanding: What happens to the "rest mass" of a particle when the particle moves? In reality mass is mass is mass. Mass has the same value in all frames, is invariant, no matter how the particle moves. [Galileo: "In questions of science the authority of a thousand is not worth the humble reasoning of a single individual."] 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. > >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. > 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.

**References**:**Engineering Newsletter***From:*T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)

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