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Natural Science Forum / Physics / Relativity / October 2006mass increase due to speed
If you were in a spaceship with absolutely nothing in your visual range, and another identical spaceship moving in a straight line flew passed you, SR tells us that it would be impossible to know who's moving and who's stationary. But wouldn't measuring the mass of both spaceships tell us who's moving as their spaceship would have a greater mass due to the effect of movement through spacetime? Paul. > If you were in a spaceship with absolutely nothing in your visual range, and > another identical spaceship moving in a straight line flew passed you, SR [quoted text clipped - 4 lines] > > Paul. How do you propose to measure the mass? Whatever method you propose, you will find that each would measure the other as having the greater mass. >> If you were in a spaceship with absolutely nothing in your visual range, >> and another identical spaceship moving in a straight line flew passed [quoted text clipped - 6 lines] > How do you propose to measure the mass? Whatever method you propose, you > will find that each would measure the other as having the greater mass. Say you were in your ship and measured the mass of a basketball. You were then accelerated (say to 99% speed of light). If you now measured the mass of the basketball would you find it 'heavier' or would it appear the same? >>> If you were in a spaceship with absolutely nothing in your visual range, >>> and another identical spaceship moving in a straight line flew passed [quoted text clipped - 10 lines] > then accelerated (say to 99% speed of light). If you now measured the mass > of the basketball would you find it 'heavier' or would it appear the same? You would not notice a difference, as long as you, the ship and the basketball are all moving together. Dear robert: >>> If you were in a spaceship with absolutely nothing in your >>> visual range, and another identical spaceship moving in a [quoted text clipped - 4 lines] >>> greater mass due to the effect of movement through >>> spacetime? >> How do you propose to measure the mass? Whatever >> method you propose, you will find that each would [quoted text clipped - 5 lines] > basketball would you find it 'heavier' or would it appear > the same? All mass measurements in your frame are with respect to other masses in your frame. The basketball in your ship will weigh the same as a basketball, no matter how fast you are moving. Now, you observe a basketball in the other ship, on on the Earth. How do you propose to measure its mass from your frame? Let's make this easier. Let's say you are observing the Moon orbiting the Earth as you fly by at 0.99c. Will the Earth-Moon system curve your path more or less than it would at 0.0001c? Will the Moon suddenly fall into the Earth, will its period suddenly alter beyond what your gamma accounts for? David A. Smith | Dear robert: | [quoted text clipped - 29 lines] | Will the Moon suddenly fall into the Earth, will its period | suddenly alter beyond what your gamma accounts for? Yes. The moon's orbit will be highly elliptical (eccentricity = 0.99) and it's period will be 193.677 days. This is relativity; common sense is not allowed, shithead. Androcles > Dear robert: > [quoted text clipped - 26 lines] > the Moon suddenly fall into the Earth, will its period suddenly > alter beyond what your gamma accounts for? You call this making it "easier"??? These are questions that regularly confuse people who actually know something of relativity, and you're throwing them at a newbie here... Let's take this one: > Will the Earth-Moon system curve your path more or less than it > would at 0.0001c? I'll simplify it. Take two particles tied together with a string (you'll see why there's a string in a moment). Accelerate them to 0.99... C (linear motion, string stays slack). Does their gravitational field get stronger? If so, then light must have a harder time getting away from them, right? 'Cause light's affected by gravity, too, of course. So in that case if we accelerate them to a high enough velocity, light won't be able to get away from them at all. That means our pair of particles will turn into a black hole ... just because we measure it as going really fast. That's obviously wrong! But if their gravitational field does _not_ increase, then what happened to all that energy we pumped in accelerating them to 0.99... C? Let's do it differently -- let's put the same amount of energy in, but this time let's spin the two particles on the end of their string, so they're mutually revolving about each other. Now, does their G field increase? It seems clear that it must. (If that doesn't seem clear, put the two particles in a box so you can't see them any more. Now, all you know is that you've added a pile of energy to the box. Obviously the G field of the box must increase when we add mass/energy to the box.) So, this time, if you spin them hard enough (add enough energy), you will surely find light can't get out of the box (ignore the problem of the string breaking, please, along with the possibility that we're creating a Rindler horizon between the two masses; that messes up the argument...). And in that case we seem to have created a black hole. So why doesn't it work for the straight-line case? Is there something special about rotary motion? If the particles were just bouncing around inside the box, rather than revolving, wouldn't that be the same as the revolutionary case? But that's straight-line motion, too, just like the case of the earth-moon system given above. What really is the correct answer here? Whatever it is, it's confusing. And it probably requires plugging values into the stress/energy tensor and solving the field equations to really see what's what, which is 'way, 'way beyond the level of the OP. > David A. Smith  SignatureNospam becomes physicsinsights to fix the email Dear sal: ... >>> Say you were in your ship and measured the mass >>> of a basketball. You were then accelerated (say to [quoted text clipped - 22 lines] > know something of relativity, and you're throwing > them at a newbie here... The second one is actually a slam dunk, Sal. Think about the situation... The Oh My God particle was travelling at such a speed that it would cross the Milky Way in 1 week proper time (0.999999+c). Did it cause the Earth and Moon to become black holes, or deform their orbits in some catastrophic way? Is it plausible that there was only one of such particles *ever*, or even that this was the fastest particle to have interacted with the Earth? Yes you can duck into the math, or you can just look at what you are expecting of Nature. As to the first question, it is only slightly more complex... Realtivistic mass =/= gravitational mass. And gravitational mass is about the only kind of mass you are going to measure at high speeds. David A. Smith > Dear sal: > [quoted text clipped - 44 lines] > > David A. Smith I think you've answered my next question. If we see the other ship pass us at very near light speed, even though to us it's mass is greatly increased, the gravitational affect it has on us would be the same as if it passed us very slowly? Paul. Dear robert: >> Dear sal: >> [quoted text clipped - 49 lines] > other ship pass us at very near light speed, even though > to us it's mass is greatly increased, It's mass does not increase. Mass is commonly "rest mass", and not "relativistic mass". It is always a good idea to be clear with meanings. What all will agree on is "it's energy is greatly increased". > the gravitational affect it has on us would be the same > as if it passed us very slowly? Energy also has an effect of curving space. So there will be some effect from energy. Now the two paths will be different *mostly* because of the duration the two bodies spend in each other's proximity. You have a "field of force", and two different durations. They result in two different impulses, two different paths (at least the endpoints... which star you are aiming at when done). But inferring a different system mass... might be a tad difficult to measure. David A. Smith Note one thing, before we go on: The ratio of gravitational to inertial mass is fixed, and the same for every material which has ever been tested, to the limits of our ability to test it. To all intents and purposes they are apparently the same thing, and if "inertial mass" changes then so does gravitational mass. So if a moving body becomes more difficult to accelerate (gains "inertia"), it must also gravitate more. Right? Now, let's get on with the comments... (you don't get any brownie points for observing that this post is a mess, I already know that...) > Dear sal: > [quoted text clipped - 22 lines] > > The second one is actually a slam dunk, Sal. Sure, sure, you can't turn the Earth into a black hole by changing your frame of reference ... but what about the **first** one, about the effect of the Earth-Moon system on someone flying by? Read my flounderings below to see why I claim the question of how "massive" a moving object "appears to be" may not be totally obvious. To rephrase the question, if a planet flies by us going "real fast", does it appear to have a stronger gravitational field than it would if it went by going very slowly? After considering the case where we shrink the planet and put it in a box, so it can bounce back and forth, and then add some more tiny planets to the box, until we have a box of gas, I think the answer must be "yes". If we heat the gas (add energy to it) it must gravitate more, and the case of a the free planet is just the simplest possible reduction of that case. This can be tested in various ways and none of them turned out to be as trivial as I expected. (OTOH I didn't try looking it up on Google, which might be the most sensible approach, come to think of it.) The most obvious approach is probably to start with the Schwarzchild metric, transform it to a moving frame, and solve the equation of motion of a particle in that frame. But that's more than I'm going to tackle tonight. Using the Newtonian metric instead of the Schwarzchild metric would most likely simplify that approach a bit but it still seemed like more work than I was willing to put into this. A simpler approach, which certainly only works for weak fields (where we can pretend we've got a Newtonian field overlaid on special relativity), is to observe that an object traveling past a (non-rotating) planet, at the moment when it is traveling perpendicular to its radius vector, ought to "fall" toward the planet with the same (instantaneous) acceleration as an object which is stationary at the same point in space, as measured by an observer who is stationary on the planet. If this weren't true, then hot things would fall at a different rate from cold things, and they apparently don't. An obvious objection to the "hot-things-would-fall-differently" argument is that the particles in "hot" things (of the sort we can actually experiment on!) are moving at non-relativistic velocities, so perhaps the claim is only true in the low-speed limit. That can be tested by solving the geodesic equation for a particle in a Newtonian gravitational field (metric diag[-(1+2phi),1-2phi,1-2phi,1-2phi]) and seeing if, when it is moving "perpendicular" to the field, its acceleration depends on its velocity. That turned out to be messier than I expected; still working on it but it looks like the answer is _NO_ it doesn't necessarily accelerate at a rate independent of velocity, even when it's moving "horizontally", which is too bad but which may explain the bogus result I exhibit a few paragraphs farther along. Let's ignore that and barge ahead, assuming for the moment that the acceleration toward the planet of a passing body is a constant, "a", in the frame of an observer stationary with respect to the planet. Call the moving body "B". Now, the question is: How fast does someone in B's instantaneous rest frame see the _planet_ accelerating? B's proper clock ticks at 1/gamma relative to the planet's clock. B's line of motion is perpendicular to the line along which he/she is accelerating, which means there's no Fitzgerald contraction to worry about: distances along the line of acceleration are the same in both frames. So, we only need to worry about the clock rate difference. After t seconds, in the planet's frame, the distance B moves toward the planet along the line of acceleration would be: (1/2)at^2 In B's instantaneous rest frame, the distance the planet moves is the same, since there's no length contraction along the line perpendicular to the line of flight. But the time is different, by a factor of 1/gamma. So we have (1/2)at^2 = (1/2)a(gamma*tau)^2 or, casting it in terms of the acceleration observed in the moving frame, which we will call a', (1/2)at^2 = (1/2)a'tau^2 or (1/2)at^2 = (1/2)a'(t/gamma)^2 Solving for a', a' = a * gamma^2 Which, unfortunately, looks totally bogus -- a sensible answer should look more like a' = a * gamma but the assumption that B's acceleration doesn't depend on B's velocity as viewed from the planet's frame seems to lead inevitably to this result, as verified by computing it a couple different ways, so it appears that getting a correct answer here really does require solving the geodesic equation for that case. Or alternatively I messed up the transformation of time going between frames, because the planet's moving in B's frame just as B is moving in the planet's frame so the clock issues are subtler than they appear at first. Sigh... Wish I had more time to spend on this but I don't, which is why I'm posting this with all its obvious warts still intact. References to correct explanations of this (trivial?) issue would be appreciated. > Think about the situation... > [quoted text clipped - 14 lines] > > David A. Smith SignatureNospam becomes physicsinsights to fix the email > Note one thing, before we go on: > [quoted text clipped - 9 lines] > Now, let's get on with the comments... (you don't get any brownie points > for observing that this post is a mess, I already know that...) It's answered my question to the level that I'm 'at' so it's helped me no ends :-) Dear sal: > Note one thing, before we go on: > [quoted text clipped - 3 lines] > purposes they are apparently the same thing, and if > "inertial mass" changes then so does gravitational mass. ... at least until the equivalence principle is broken. Still, there are so many examples of "inertial" being equal to "gravitational"... > So if a moving body becomes more difficult to > accelerate (gains "inertia"), it must also gravitate > more. Right? But it *doesn't* get more difficult to accelerate transverse to the path. That resistance to acceleration stays exactly equal to the rest mass. > Now, let's get on with the comments... (you > don't get any brownie points for observing that this > post is a mess, I already know that...) I'm not keeping score. >> Dear sal: >> [quoted text clipped - 32 lines] > the **first** one, about the effect of the Earth-Moon > system on someone flying by? If you don't turn a massive object into a black hole by changing frames of reference, then how can you alter the mass of a more complex system by merely making a frame of reference change? Sometimes you have to read the whole "test" through before you can answer the first question. > Read my flounderings below to see why I claim the > question of how "massive" a moving object "appears to [quoted text clipped - 7 lines] > some more tiny planets to the box, until we have a box of > gas, I think the answer must be "yes". Any normal process observed in another frame will be affected by the gamma of the frame. Periodic motion or "kinetic motion of gasses" included. > If we heat the gas (add energy to it) it must gravitate more, > and the case of a the free planet is just the simplest [quoted text clipped - 11 lines] > would most likely simplify that approach a bit but it still > seemed like more work than I was willing to put into this. Since "robert" is satisfied... it will be for satisfaction / posterity. > A simpler approach, which certainly only works for > weak fields (where we can pretend we've got a [quoted text clipped - 7 lines] > planet. If this weren't true, then hot things would fall at > a different rate from cold things, Disproven by an eotvos experiment, I believe, for "terrestrial" temperature differences. > and they apparently don't. > [quoted text clipped - 24 lines] > B's instantaneous rest frame see the _planet_ > accelerating? Different rates at different places in its orbit. It might be better / simpler to concentrate on *period*, because the orbiter ends up at the same place wrt the orbitee periodically. Maybe not arriving at an "instantaneous" "continuous" resolution is not necessary. > B's proper clock ticks at 1/gamma relative to the > planet's clock. B's line of motion is perpendicular to the [quoted text clipped - 40 lines] > correct answer here really does require solving the > geodesic equation for that case. I think the orbitting body has: * a *slightly* different gamma with some variation based on position in the orbit (very minor), * the orbit will be seen to be an ellipse, if classically circular, * the orbit will be even more elliptical, if already elliptical, and * the body being orbitted will move from a focus of the ellipse towards the midpoint between the focii, as a function of gamma. > Or alternatively I messed up the transformation of time > going between frames, because the planet's moving in [quoted text clipped - 5 lines] > warts still intact. References to correct explanations of > this (trivial?) issue would be appreciated. relativity flyby "orbital period" site:.edu ... only 78 hits. Someone that can follow your derivation and assumptions might be able to help you. I'm just a duffer. David A. Smith > Dear sal: > [quoted text clipped - 16 lines] > path. That resistance to acceleration stays exactly equal to the > rest mass. Momentum = p^i = m * gamma * v^i (for the spatial components) Suppose a body is moving along the x axis at 3-velocity V, so v^1 = V v^j = 0 (when j <> 1) Then, among other things, d(V*V)/dv^j = 0 for j<>1 and so d(gamma)/dv^j = 0 for j<>1. Now suppose we apply a force along the Y axis (where Y is coordinate #2). Then dp^2/dt = -F = m * gamma * dv^2/dt Hmmm -- Looks like a factor of gamma got in there somehow. Conclusion: Accelerating a moving body transverse to its line of motion _DOES_ get "harder" as the body gains energy. The 3-force needed to achieve a particular transverse acceleration goes up as gamma. The reason is that gamma appears in the formula for momentum, which must be conserved. If you didn't need to "push" any harder to turn an object which was moving really fast, then momentum would not be conserved. There is a reason some people use "mass" to mean "relativistic mass" -- 'cause the mass-energy behaves an awful lot like mass. >> Now, let's get on with the comments... (you don't get any brownie >> points for observing that this post is a mess, I already know that...) [quoted text clipped - 50 lines] > of the frame. Periodic motion or "kinetic motion of gasses" > included. So if a speeding particle doesn't gravitate more due to its speed when it's free, does it gravitate more due to its speed if it's inside a box? If so, why doesn't it when it's "free" -- how does it know it's inside a box? Do two particles in a box gravitate more when they're moving faster than when they're moving slower? If so, why doesn't _one_ particle? If two don't show any effect, how about 200? 2000? 6*10^23? At some point you've got a box of gas, and hot gas gravitates more than cool gas, else something's rotten in stress tensor town. But hot gas is just gas where the particles are moving faster. If one particle moving through space doesn't gravitate more when it moves faster, but a box of gas _does_ gravitate more when its "particles" are moving faster, then just where is the line? How many particles must be in the system before you'll see such an effect? And finally, here's the kicker: I have a u-235 nucleus sitting in space. I measure its gravitational mass by testing its "far field". Suddenly its fissions into two pieces (and a couple neutrons IIRC). The total _rest_ _mass_ of all the pieces is smaller than the _rest_ _mass_ of the original atom. But the pieces are moving hell-bent-for-election, while the original nucleus was stationary (and so the difference is made up). Does the far field of the ensemble _DECREASE_ at the moment of fission due to the loss of rest mass? Or does it remain fixed, because the _energy_ of the pieces still totaled the same? I'd claim it's the latter -- but a lot of that energy now takes the form of linear motion of free particles. >> If we heat the gas (add energy to it) it must gravitate more, and the >> case of a the free planet is just the simplest possible reduction of [quoted text clipped - 24 lines] > Disproven by an eotvos experiment, I believe, for "terrestrial" > temperature differences. Sounds right to me. >> and they apparently don't. >> [quoted text clipped - 24 lines] > same place wrt the orbitee periodically. Maybe not arriving at an > "instantaneous" "continuous" resolution is not necessary. The planet example rapidly turns into a mess, I agree. >> B's proper clock ticks at 1/gamma relative to the planet's clock. B's >> line of motion is perpendicular to the line along which he/she is [quoted text clipped - 60 lines] > Someone that can follow your derivation and assumptions might be able to > help you. I'm just a duffer. Moi aussi. Went for about 6 months without touching a physics book and I'm so rusty it's disgusting. SignatureNospam becomes physicsinsights to fix the email Dear sal: >> Dear sal: ... >>> So if a moving body becomes more difficult to >>> accelerate (gains "inertia"), it must also gravitate [quoted text clipped - 31 lines] > of > motion _DOES_ get "harder" as the body gains energy. No. It doesn't take a stronger magnetic field to alter the path of the particle. Only to accelerate it along the path does it take more "force". > The 3-force needed to achieve a particular transverse > acceleration goes up as gamma. The reason is that [quoted text clipped - 6 lines] > "relativistic mass" -- 'cause the mass-energy behaves > an awful lot like mass. It actually doesn't. And there is no good reason to have a "half vector" quantity for mass, when the momentum is entirely capable of filling the need. ... >>> To rephrase the question, if a planet flies by us going >>> "real fast", does it appear to have a stronger [quoted text clipped - 27 lines] > moving through space doesn't gravitate more when it > moves faster, Doesn't it? Even a little? It thought energy also affects spacetime. Just not detectably with "spaceships travelling at 0.99c". > but a box of gas _does_ gravitate more when its > "particles" are moving faster, then just where is the > line? How many particles must be in the system > before you'll see such an effect? The system. In other words, the particle and the rest of the Universe (aka. "the field"). Now remember I have been saying that the effect of the energy is very small, not that it has no effect at all. > And finally, here's the kicker: > > I have a u-235 nucleus sitting in space. I measure > its gravitational mass by testing its "far field". Suddenly > its fissions into two pieces (and a couple neutrons IIRC). Four neutrons I think... > The total _rest_ _mass_ of all the pieces is smaller than > the _rest_ _mass_ of the original atom. But the pieces [quoted text clipped - 6 lines] > but a lot of that energy now takes the form of linear > motion of free particles. As the radiant heat (or particles) passes you, would be my guess... ... >>> Let's ignore that and barge ahead, assuming for the >>> moment that the acceleration toward the planet of a [quoted text clipped - 12 lines] > > The planet example rapidly turns into a mess, I agree. One I am not in the position to either critique, nor assist in developing / cleaving... ... >>> Sigh... Wish I had more time to spend on this but I >>> don't, which is why I'm posting this with all its obvious [quoted text clipped - 9 lines] > Moi aussi. Went for about 6 months without touching a > physics book and I'm so rusty it's disgusting. Yes, I am supposed to be studying for my PE exam on Friday. I don't seem to care about engineering anymore... David A. Smith > It doesn't take a stronger magnetic field to alter the path > of the particle. Only to accelerate it along the path does it > take more "force". No. To achieve a given deflection of a charged particle in a magnetic field one must use a stronger magnetic field for a particle with higher velocity. The radius of curvature of a charged particle moving perpendicular to a magnetic field is proportional to beta*gamma, where beta=v/c and gamma=1/sqrt(1=beta^2). It does require a larger TRANSVERSE force to impart a given acceleration as the object's velocity increases; the required LONGITUDINAL force increases even faster (by an additional factor of gamma). And for angles between those it requires a force in between those. Tom Roberts Dear Tom Roberts: > > It doesn't take a stronger magnetic field to alter the path > > of the particle. Only to accelerate it along the path does it [quoted text clipped - 12 lines] > factor of gamma). And for angles between those it requires a > force in between those. Thanks for the correction, Tom. I hope that fact does no leak out too. David A. Smith >> Dear sal: >> [quoted text clipped - 47 lines] > turn an object which was moving really fast, then momentum would not > be conserved. That's correct, it was even the first prediction of relativity theory, and is rather standard textbook stuff. Feynman used transverse acceleration as an operational definition for determining relativistic mass. Harald > There is a reason some people use "mass" to mean "relativistic mass" > -- 'cause the mass-energy behaves an awful lot like mass. [quoted text clipped - 216 lines] > Moi aussi. Went for about 6 months without touching a physics book > and I'm so rusty it's disgusting. >There is a reason some people use "mass" to mean "relativistic mass" > -- 'cause the mass-energy behaves an awful lot like mass. Except one expects a given object to have a single value for its mass -- "relativistic mass" does not have this property: for forces perpendicular to v "relativistic mass" does indeed correspond to the m in F=ma, but ONLY for that case, and one must use a different value of m for other directions of force. Not very useful. <shrug> The invariant mass of an object has a clear pedigree from earlier meanings of the word "mass", and behaves as one expects (e.g. it corresponds to "how much stuff" is present). "Relativistic mass" is more properly _energy_, and does not have any sensible pedigree at all. Equations written in terms of invariant mass are simple and clear, and the important ones are invariant; when written in terms of "relativistic mass" they are more complicated, obfuscated, and cannot possibly be invariant. > I have a u-235 nucleus sitting in space. I measure its gravitational > mass by testing its "far field". Suddenly its fissions into two [quoted text clipped - 6 lines] > the same? I'd claim it's the latter -- but a lot of that energy now takes > the form of linear motion of free particles. Your claim is wrong, according to GR. This is an impossible measurement, so theory is all we have to guide us. In GR, the field at a distance L away does not change until a delay L/c after the fission, but then it does begin to change. Remember that the "source" of gravity in GR is the energy-momentum tensor, not just mass; after the decay the e-m tensor is quite different from its value before the decay. This should be obvious: consider a long enough time after the decay so all products of the decay are at least a distance 100*L from either the original u-235 location or the point in question -- surely the gravitational field is now much less than originally. >>> A simpler approach, which certainly only works for weak fields (where we >>> can pretend we've got a Newtonian field overlaid on special relativity), [quoted text clipped - 9 lines] > > Sounds right to me. It's not right in GR. The acceleration of the traveling object depends on its speed. This is only apparent for speeds approaching c. For instance, light (traveling at c) has double the "acceleration" [#] you expect for an object at rest at the point "perpendicular to its radius vector". [#] Measured by deviation of the path from a straight line. And, of course, if you heat up a cold object then the rest mass of the object increases. This, too, is a very small effect, and is too small to measure terrestrially; indeed all known solids will vaporize long before this effect would be measurable (that makes it impossible to keep the constituents of the object bound together, which is necessary to consider it an "object" Tom Roberts >>There is a reason some people use "mass" to mean "relativistic mass" >> -- 'cause the mass-energy behaves an awful lot like mass. [quoted text clipped - 4 lines] > for that case, and one must use a different value of m for other > directions of force. Not very useful. <shrug> Yes, I realize that; I'm aware of the old transverse versus longitudinal mass notions. I was just trying, however fuzzily, to emphasize the point that something besides the difficulty of accelerating a body along its line of motion appears to change as we observe it speed up. > The invariant mass of an object has a clear pedigree from earlier > meanings of the word "mass", and behaves as one expects (e.g. it [quoted text clipped - 19 lines] > > Your claim is wrong, according to GR. Perhaps it was just too badly stated. Again the point I was trying to make is that, if the gravitation of a body depended only on its rest mass, then the far field of a particle would change _abruptly_ at observer's time L/c, as the aggregate rest mass of the system dropped _abruptly_ at the moment when some rest mass was converted into kinetic energy. Of course, I realize the field will certainly change and weaken as the newly fissioned particles move apart from each other, but that happens as a continuous change, subsequent to time L/c, not a step function. It's similar to the question I posed here a long time back about sealing an entire laboratory in a box (so neither mass nor energy can enter or leave ... except gravity waves and maybe a stray neutrino or two) and putting the box on a spring balance. Will the measured weight of the box change as a result of mechanical, chemical, or nuclear operations taking place inside the box? My understanding at that time was that the answer was "no" (if we ignore gravitational radiation which can fly out through the box walls) despite the fact that the details of the stress/energy tensor for the box must change as the mass and energy distribution within the box changes. > This is an impossible measurement, > so theory is all we have to guide us. In GR, the field at a distance L [quoted text clipped - 7 lines] > original u-235 location or the point in question -- surely the > gravitational field is now much less than originally. Again, yes, this is clear and it was not what I was thinking of; my original statement was unclear. >>>> A simpler approach, which certainly only works for weak fields (where >>>> we can pretend we've got a Newtonian field overlaid on special [quoted text clipped - 11 lines] > > It's not right in GR. Er ... which is not right? I was agreeing that hot things fall at the same rate as cold things. Are you saying that is false? > The acceleration of the traveling object depends > on its speed. Aha! Thank you. That's what it looked like when I was fiddling with the geodesic equation for the object flying by but I didn't get far enough to be sure. > This is only apparent for speeds approaching c. For > instance, light (traveling at c) has double the "acceleration" [#] you > expect for an object at rest at the point "perpendicular to its radius > vector". Good point! It didn't occur to me to relate that to the behavior of objects moving at close to C. > [#] Measured by deviation of the path from a straight line. > [quoted text clipped - 6 lines] > > Tom Roberts SignatureNospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org >>There is a reason some people use "mass" to mean "relativistic mass" >> -- 'cause the mass-energy behaves an awful lot like mass. [quoted text clipped - 4 lines] > for that case, and one must use a different value of m for other > directions of force. Not very useful. <shrug> Tom, in p=mv, m corresponds to relativistic mass. It is not invariant, but it has a single value value for a certain speed. You confuse the obsolete "longitudinal" and "transverse" mass of Lorentz and Einstein with the "relativistic mass" of Feynman and as you can find in textbooks. BTW, this has also been explained several times on this newsgroup. > The invariant mass of an object has a clear pedigree from earlier meanings > of the word "mass", and behaves as one expects (e.g. it corresponds to > "how much stuff" is present). That's also erroneous: the number of protons and neutrons represents "how much stuff is present", which doesn't accurately correspond to any use of the word "mass". It's a failed attempt to return to Newtonian concepts. For example, the "invariant mass" is (in theory) a function of temperature and thus it's an impure version of "relativistic mass". If anything is "obfuscated" as you state below, it's certainly "invariant mass". Harald > "Relativistic mass" is more properly _energy_, and does not have any > sensible pedigree at all. Equations written in terms of invariant mass are [quoted text clipped - 59 lines] > > Tom Roberts >> Except one expects a given object to have a single value for its mass -- >> "relativistic mass" does not have this property [...] > > Tom, in p=mv, m corresponds to relativistic mass. Sure. But I was discussing the m in F=ma. But those are THREE-VECTORS, which are essentially useless in SR (except for exploring relationships with Newtonian physics). The relativistic analog is P=mU, where P is the FOUR-momentum, U is the FOUR-velocity, and m is the INVARIANT mass. >> The invariant mass of an object has a clear pedigree from earlier meanings >> of the word "mass", and behaves as one expects (e.g. it corresponds to [quoted text clipped - 3 lines] > much stuff is present", which doesn't accurately correspond to any use of > the word "mass". Neutrons+protons is nowhere close to the "amount of stuff" -- you cannot even count them, multiply by the masses of neutrons and protons, and obtain the mass of a macroscopic object -- there are several other contributions that are important. In Newtonian mechanics, "mass" has a number of properties; in SR no single quantity can have all of them. The invariant mass retains the most important properties of mass in Newtonian mechanics: it is a single value for each and every object, it is an invariant, it is intrinsic, etc. "Relativistic mass" has NONE of those 3 properties. > It's a failed attempt to return to Newtonian concepts. Not at all! The invariant mass is the most useful of all the candidates, and makes the fundamental equations of physics be the simplest. E.g. P=mU above -- that is an invariant equation between 4-vectors, which simply cannot be expressed in an invariant manner using "relativistic mass". > For > example, the "invariant mass" is (in theory) a function of temperature and > thus it's an impure version of "relativistic mass". You are confused. For a composite object, both invariant mass and "relativistic mass" are functions of the object's temperature. This is inescapable (not "impure", whatever you mean by that). <shrug> Tom Roberts >>> Except one expects a given object to have a single value for its mass -- >>> "relativistic mass" does not have this property [...] >> >> Tom, in p=mv, m corresponds to relativistic mass. > > Sure. But I was discussing the m in F=ma. Indeed F=ma doesn't apply, that's correct. > But those are THREE-VECTORS, which are essentially useless in SR (except > for exploring relationships with Newtonian physics). "Essentially useless" is a rather big exaggeration for an approach that was used to develop SRT and which is perfectly suited for calculating relativistic collissions in particle accelerators... > The relativistic analog is P=mU, where P is the FOUR-momentum, U is the > FOUR-velocity, and m is the INVARIANT mass. [quoted text clipped - 11 lines] > obtain the mass of a macroscopic object -- there are several other > contributions that are important. Generally "energy" is not what is meant with "amount of stuff"; the rest of your argument is identical to mine. > In Newtonian mechanics, "mass" has a number of properties; in SR no single > quantity can have all of them. Again we agree... > The invariant mass retains the most important properties of mass in > Newtonian mechanics: it is a single value for each and every object, it is > an invariant, it is intrinsic, etc. "Relativistic mass" has NONE of those > 3 properties. Neither have relativistic "length" and "time" any of those properties. "Proper" is the proper term for such things (pun intended). >> It's a failed attempt to return to Newtonian concepts. > > Not at all! The invariant mass is the most useful of all the candidates, > and makes the fundamental equations of physics be the simplest. E.g. P=mU > above -- that is an invariant equation between 4-vectors, which simply > cannot be expressed in an invariant manner using "relativistic mass". For that equation correct; for calculations in general it is debatable if it's the simplest. It's a matter of taste. >> For example, the "invariant mass" is (in theory) a function of >> temperature and thus it's an impure version of "relativistic mass". > > You are confused. For a composite object, both invariant mass and > "relativistic mass" are functions of the object's temperature. That's what I wrote above and is my argument. No confusion on my part here. > This is inescapable (not "impure", whatever you mean by that). <shrug> Invariant mass is simply relativistic mass as measured in the proper frame - it pretends to be what it isn't. Regards, Harald >>>> Except one expects a given object to have a single value for its mass -- >>>> "relativistic mass" does not have this property [...] [quoted text clipped - 69 lines] >Regards, >Harald All the splainin' can't conceal the fact that the equation for relativistic energy is suspicious. This relativistic mass topic comes up repeatedly and the inquirers are put off: " no there's no relativistic mass, But, what about...".etc. It all becomes clear if we factor the equation for total energy as given by relativity, E =sqrt(m2c4 + p2c2) where p = mv, bcomes E = mc*sqrt(c2 + v2) = mcC The bracketed quantitiy defines Big C, a velocity greater than c. Can we justify that? The factoring lets m remain m, but puts the burden on c. In truth there cannot be a big C (can there?). The OP's suspicion is well grounded, that it simply becomes less and less efficient to impart energy as velocity approaches c, and as I have mentioned before, the easy citation of 100 GeV particles for example, equivocates 100 units of effort with 100 units of result and they are not the same. John >>>>> Except one expects a given object to have a single value for its >>>>> mass -- [quoted text clipped - 101 lines] > > John John, your statement that imparting energy becomes "less efficient" at higher speeds is erroneous, as was demonstrated in the Bartocci experiment: all the imparted energy ("effort") is given off as heat on impact ("result"). The experiment neatly showed that imparted energy is carried along with the particles and in part expressed as increase of inertia ("relativistic mass"). No reduced efficiency for imparting energy. Cheers, Harald >>>>>> Except one expects a given object to have a single value for its snipped to reduce bulk >> All the splainin' can't conceal the fact that the equation for >> relativistic energy is suspicious. This relativistic mass topic comes [quoted text clipped - 26 lines] >Cheers, >Harald Harald, thank you for taking the time to reply. I used Google Scholar to study C. Bartocci, but only found abstracts (not even) that referred to his book on supermanifolds. Please give a citation with numbers if you care to, but the total energy equation is just simple geometry. You used the "R" word: relativistic mass. Relativistic mass is not in the factored equation above and as you should know, the very expression has incendiary qualities plus, it is never improper to admonish that there is no relativistic mass. (Any time I see the sum of squares, I feel duty bound to put the components on a right triangle and see if they make sense, or lead to other findings. This is one of the few places where relativity can be modeled, and the insight afforded is disheartening in this case). Big C = sqrt(c2 + v2) would clearly be the hypotenuse of a right triangle with c the adjacent side, so big C is greater than c. The magic of RM evidently lies in exceeding the velocity of light and v/c is the tangent of an angle A. Dual Space theory has a different construction for the combination of v and c in which v/c is the sine of the angle A, and the effect of v is to simply rotate c from c to c' by angle A = asin(v/c). You get two vectors c and c' separated by A, with v normal to c. From this diagram you see that the effective v-force can only be that which is normal to the c' vector as it rotates, and the force is subject to degradation by cosine A. That's what I meant by losing efficiency. The other component would lie along c and is a vain effort to increase c, which can't be done and so is lost. I might make a sketch on my website, because who can (or cares to) follow the above labyrinthine prescription? Incidentally with the latter diagram I can get most of special and general relativity. John Polasek On Fri, 20 Oct 2006 18:42:04 -0700, "N:dlzc D:aol T:com \(dlzc\)" <N: dlzc1 D:cox T:net@nospam.com> wrote: >>>> If you were in a spaceship with absolutely nothing in your >>>> visual range, and another identical spaceship moving in a [quoted text clipped - 29 lines] > >David A. Smith Another thought then (which is probably wrong!): If, as you approach the speed of light, it takes exponentially more energy to accelerate, wouldn't /this/ be a way of telling if you were stationary or moving (to an ether). Couldn't you add a set amount of energy to the ship (via engines) and see how much it accelerates by? If you were 'still' the ship should accelerate by a maximum amount. If you were going close to light speed, would the resultant acceleration applying the same energy be a lot lower? So if you measured a maximum acceleration, you'd know that you were stationary and ship b was 'moving' (and if ship b did the same experiment at the same time their acceleration would be less than a). Dear padremo: > On Fri, 20 Oct 2006 18:42:04 -0700, "N:dlzc D:aol T:com \(dlzc\)" <N: > dlzc1 D:cox T:net@nospam.com> wrote: ... > >Let's make this easier. Let's say you are observing the Moon > >orbiting the Earth as you fly by at 0.99c. Will the Earth-Moon [quoted text clipped - 4 lines] > Another thought then (which is probably wrong!): If, as you > approach the speed of light, How would you know? > it takes exponentially more energy to accelerate, wouldn't /this/ > be a way of telling if you were stationary or moving (to an ether). No. No process internal to the ship can tell you how fast *someone else* measures you moving. You could make measurements at 1 G, just a few seconds after achieving it, and again a million years later. Same results for processes internal to your ship. > Couldn't you add a set amount of energy to the ship (via > engines) and see how much it accelerates by? If you were > 'still' the ship should accelerate by a maximum amount. If > you were going close to light speed, would the resultant > acceleration applying the same energy be a lot lower? No. Time dilation and length contraction will affect your "impulse". > So if you measured a maximum acceleration, you'd know that > you were stationary and ship b was 'moving' (and if ship b did > the same experiment at the same time their acceleration would > be less than a). Nature will not let us look up her skirt in this way. We are required to use our imaginations. Would you really want it any other way? David A. Smith > Dear padremo: > [quoted text clipped - 11 lines] > > How would you know? 'Cause the blue-shifted CMBR, aka the CXBR (Cosmic Xray Background Radiation), would ionize everything in sight which would most likely get your attention. The CMBR actually provides a pretty reasonable universal rest frame ... though, of course, there's no way to prove experimentally that the CMBR itself is "really" at rest, so to speak. If you somehow overlooked the CXBR you'd still probably notice the erosion of the nose of your ship due to interaction with various stray hydrogen atoms and other cosmic detritus which, under other conditions, we think of as "ignorable". Some time back somebody posted something in this NG alleging that exceeding about 1/10 C was likely to be fatal due to those effects. [And so ends the dream of ever getting to the stars, short of using a sleeper ship.] Sorry, I'm just picking nits here. >> it takes exponentially It's not an exponential, because it hits a vertical asymptote at c. Exponentials go up fast but never hit any brick walls. Perhaps saying it takes "hyperbolically" more energy would provide a better notion of what's going on. >> more energy to accelerate, wouldn't /this/ be a >> way of telling if you were stationary or moving (to an ether). [quoted text clipped - 20 lines] > > David A. Smith SignatureNospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org Dear sal: >> Dear padremo: >> [quoted text clipped - 19 lines] > CMBR actually provides a pretty reasonable universal > rest frame ... There is no way to know your velocity, by measurements within your ship, or processes in your ship. For a short period of time, you could be approaching a white dwarf and get half the sky heavy in X-radiation and the other half not. > though, of course, there's no way to prove experimentally > that the CMBR itself is "really" at rest, so to speak. You could use one particular GPS satellite, as I recall. It spends about half an orbit moving towards the "hot pole" (of the anisotropy), and half an orbit moving away from it. Then you could see if the CMBR (and the visible Universe's contents) were closer to rest wrt the source of maximal aging. > If you somehow overlooked the CXBR you'd still probably > notice the erosion of the nose of your ship due to [quoted text clipped - 5 lines] > NG alleging that exceeding about 1/10 C was likely to > be fatal due to those effects. I posted this too. There is some data that agrees with such a low speed being a problem. > [And so ends the dream of ever getting to the stars, > short of using a sleeper ship.] Run behind a slowly rotating asteroid. Let it run interference for you. It radiates most stuff at 90 degrees to you (if you get the right asteroid as baseline). You can harvest materials from the side nearest you for fuel or whatever. It won't be thrusters and shields that gets us to the stars. If civilization holds out long enough. it will be some variant of "folding space". Either that or our DNA gets sprayed across the stars and we hope for fertile ground in a few billion years. > Sorry, I'm just picking nits here. Its OK. I still don't see how the OP would measure the mass of a basketball in another frame... David A. Smith Dear padremo: > On Fri, 20 Oct 2006 18:42:04 -0700, "N:dlzc D:aol T:com \(dlzc\)" <N: > dlzc1 D:cox T:net@nospam.com> wrote: ... > >Let's make this easier. Let's say you are observing the Moon > >orbiting the Earth as you fly by at 0.99c. Will the Earth-Moon [quoted text clipped - 4 lines] > Another thought then (which is probably wrong!): If, as you > approach the speed of light, How would you know? > it takes exponentially more energy to accelerate, wouldn't /this/ > be a way of telling if you were stationary or moving (to an ether). No. No process internal to the ship can tell you how fast *someone else* measures you moving. You could make measurements at 1 G, just a few seconds after achieving it, and again a million years later. Same results for processes internal to your ship. > Couldn't you add a set amount of energy to the ship (via > engines) and see how much it accelerates by? If you were > 'still' the ship should accelerate by a maximum amount. If > you were going close to light speed, would the resultant > acceleration applying the same energy be a lot lower? No. Time dilation and length contraction will affect your "impulse". > So if you measured a maximum acceleration, you'd know that > you were stationary and ship b was 'moving' (and if ship b did > the same experiment at the same time their acceleration would > be less than a). Nature will not let us look up her skirt in this way. We are required to use our imaginations. Would you really want it any other way? David A. Smith > If you were in a spaceship with absolutely nothing in your visual range, and > another identical spaceship moving in a straight line flew passed you, SR > tells us that it would be impossible to know who's moving and who's > stationary. But wouldn't measuring the mass of both spaceships tell us > who's moving as their spaceship would have a greater mass due to the effect > of movement through spacetime? Each ship will measure its own mass to be M. Each ship will measure the mass of the other ship to be M again, because 'mass' is an invariant as defined. You are probably referring to 'relativistic mass'. If that is the 'mass' you are refering to in your question, then each ship will 'measure' the mass of the other to be m > M. >> If you were in a spaceship with absolutely nothing in your visual range, and >> another identical spaceship moving in a straight line flew passed you, SR [quoted text clipped - 11 lines] >'mass' you are refering to in your question, then each ship will >'measure' the mass of the other to be m > M. I was confused because I've read in popular science books (come back!!!) that as an object approaches light speed, it takes more and more energy to get it there, so I presumed that this meant that if you were the object there'd come a point when you realised that your acceleration was decreasing, even thought the energy put into the acceleration was the same. After reading the replies to the post, I think I'm right in saying that you can accelerate a spaceship for as long as the universe will be in existence and yet you'd never reach light speed (though I need to learn the maths to understand the concept that you can accelerate forever and yet never reach a certain speed). On Oct 24, 11:18 am, padr...@hotmail.co.uk wrote: > On 20 Oct 2006 19:04:42 -0700, "rot...@gmail.com" <rot...@gmail.com> > wrote: [quoted text clipped - 25 lines] > to learn the maths to understand the concept that you can accelerate > forever and yet never reach a certain speed Indeed. Most folks have learned that if a ball is thrown forward inside a car at 15 mph (relative to the car seat), and the car is moving 40 mph forward (relative to the road), then someone standing on the side of the road would measure the speed of the ball to be 55 mph (relative to the road). This is, unfortunately, a lie. What is true is that this rule is *approximately* true but not exactly true. And the faster the ball and the car go, the more apparent the error involved in the approximation. That simple addition of the velocities seems intuitively plain doesn't actually count for much. Experiment says that what is intuitive is wrong in this case. PD | If you were in a spaceship with absolutely nothing in your visual range, and | another identical spaceship moving in a straight line flew passed you, SR | tells us that it would be impossible to know who's moving and who's | stationary. No, not al all. Galilean Relativity says that, SR says you DO know who is moving. | But wouldn't measuring the mass of both spaceships You can't. There is "absolutely nothing in your visual range" to measure it against (your rules). You don't even need a spaceship for this, an ordinary ship in the middle of the ocean will do. You can only tell which ship is moving by the bow wave and the wake it makes, and that is relative to the water. How would you measure the mass of an ocean liner, or even a rowing boat? Androcles tell us | who's moving as their spaceship would have a greater mass due to the effect | of movement through spacetime? | | Paul. > | If you were in a spaceship with absolutely nothing in your visual range, > and [quoted text clipped - 5 lines] > No, not al all. > Galilean Relativity says that, SR says you DO know who is moving. how? | > | If you were in a spaceship with absolutely nothing in your visual range, | > and [quoted text clipped - 5 lines] | > No, not al all. | > Galilean Relativity says that, SR says you DO know who is moving. Sorry, what was that? Did you snip something to say? > > | If you were in a spaceship with absolutely nothing in your visual range, > > and [quoted text clipped - 7 lines] > > how? Because you said so in your question. You say that "another identical spaceship moving in a straight line flew passed you". If it flew passed you then it is moving relative to you. You said it! One does not move relative to himself. If you see something approaching you then that thing is moving, not you. You can measure its position, speed, relativistic mass, energy to know 'how much' it is moving. And again, carefull when you use the word 'mass'. It does not have the meaning you think it has. Mass is invariant, its the propoer mass, rest mass, which does not change. > If you were in a spaceship with absolutely nothing in your visual range, and > another identical spaceship moving in a straight line flew passed you, SR [quoted text clipped - 4 lines] > > Paul. ----------------------- mass is consatnt and nothing can change it ( fortunately enough ** !!!-- nothing can create mass and nothing can eliminate it even if transffered to energy the mass eas transffered to the energy!! it is not the mass that is being increased it is the ENERGY NEEDED TO ADD ACCELERATION that is inceasing it is not E= gama m c^2 it is E/Gama = mc^2 and m relaines constant!!! and there is only one kind of mass REST MASS!! ATB Y.Porat ----------------------- > If you were in a spaceship with absolutely nothing in your visual range, and > another identical spaceship moving in a straight line flew passed you, SR [quoted text clipped - 4 lines] > > Paul. Ah, but you see, mass is an observer-dependent property. You are assuming that the mass increase is intrinsic to the object, that something is *physically* happening to the object to increase its mass, and that all observers would agree that the mass has increased by so-and-so much. That's not the case. Same is true for length, by the way. Length is not an intrinsic property of the spaceship. This is the part that most casual readers have the hardest time getting a grip on. PD robert schrieb: > If you were in a spaceship with absolutely nothing in your visual range, and So if you are alone, velocity has no meaning, there is no reference to measure it. Even you have a, from your acceleration history, calculated speed >c . You have only calculated speed, but still no direction. > another identical spaceship moving in a straight line flew passed you, SR > tells us that it would be impossible to know who's moving and who's > stationary. Yes, see above. Only with Radar you can measure your rel. velocity before and after the rendezvous.. Newtonians predict , depending on speed, loss of radio contact after the passpoint. Rudi But wouldn't measuring the mass of both spaceships tell us > who's moving as their spaceship would have a greater mass due to the effect > of movement through spacetime? > > Paul. |
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