md65536

"Rest" and acceleration are frame-dependent. A stone thrown upward momentarily comes to rest without proper acceleration. I disagree with the general characterization of the principle you're using. It's not something that applies only in the simplest cases, it always applies. So that includes a free-falling particle with a world line billions of years long, falling past countless moving masses. It can come to rest many times. The principle doesn't say anything about initial conditions, and it doesn't have to because it still applies in all cases. The only restriction is it can't be applied to "distant" (non-adjoining) spacetime patches. To try to paraphrase Taylor/Wheeler ("The Principle of Maximal Aging says that a free stone follows a worldline through spacetime (flat or curved) such that its wristwatch time (aging) is a maximum across every pair of adjoining spacetime patches."): The principle only applies without restriction in "flat enough" spacetime, but it can be applied to an arbitrarily complicated (curved?) free-fall world line by dividing the world line into small enough sections that pass through flat-enough spacetime patches, and applying the principle to each of those sections. If it's complicated enough, there may be other paths that involve greater aging (such as the multiple orbits examples I've given above), but those necessarily involve paths across spacetime patches that are not adjoining a patch through which the world line in question passes. (I think that's what it's saying.)

I'm describing cases where you have two freefall paths that pass through the same pair of events, A and B. A trivial example would be two particles in similar circumpolar orbits leaving together above the north pole and meeting again above the south pole. A more useful example is two particles in eccentric orbits of different sizes, and they meet at one's perihelion and the other's aphelion, and the particle in the smaller orbit makes two orbits for every one of the larger. Since the particle in the larger orbit makes an orbit at lower speeds (in the gravitational mass's reference frame, say) than the one in the smaller orbit, and is also at a higher gravitational potential, it must age more than the one in the smaller orbit. Therefore the principle of maximal aging cannot truthfully say "If a particle traveling between events A and B is in free fall, then its aging is greater than any other path between A and B." I've tried finding the actual definition of the principle, and found several variations, including many like the above which I think are false. I've also seen, "the aging is greater than any other nearby path" which is true, and "The path of maximal aging between A and B is a geodesic", which is true. A possible problem is they're assuming that a geodesic between A and B is unique, when really they can only assume that it is locally unique? The closest to definitive I can find is from Taylor and Wheeler's "EXPLORING BLACK HOLES Introduction to General Relativity Second Edition": That certainly excludes my example. It seems that free fall aging is maximal among "nearby" paths, and the caveat is necessary. A lot of web pages describing the principle of maximal aging are leaving it out and mislead me to the incorrect conclusion that any freefall paths between events A and B will have maximal aging among all possible paths between A and B. Yes they must pass through the same events to compare them, but they definitely don't have to be at rest. If two world lines intersect, that's a single event, regardless of the objects' velocities. If a world line passes through a given event, it does so in every frame of reference.

How is more time accumulated near the massive object? You've just stated that the far away clock appears to go faster. "Accumulation of time" would refer to proper time. Does the principle imply that two objects in different freefall orbits that intersect at two events, must age the same amount between the two intersections? One could not have maximal aging along one freefall path between the two events, yet have the other age more, right? If so, then you could have one clock orbiting a massive object several times at a fixed radius, while another clock orbits once, starting at the same radius but traveling far away from the mass before returning. Both are in freefall, both can start and end together. The "escaping" clock would need a faster initial speed, and would "age less" due to SR time dilation, but would also "age more" while having higher gravitational potential. Do they necessarily age the same between events where they meet? Or are there other caveats or restrictions to the principle? My intuition is that you could make the eccentric orbit so far away and so slow at aphelion that it would have to age more, but the principle seems to say that's wrong.

The principle concerns the path from A to B (two given events on a world line), it doesn't tell you where B must be. Other things tell you that. Eg. drop a stone from rest while standing on Earth, and it will fall downward. Or throw it upward, it will fall upward for some time. B will generally be different in these cases. Either way, if you take two points A and B on the stone's world line, the freefall path between A and B has the greatest proper time of all possible (including non-freefall) paths between A and B. For example, if A is some point on Earth, and B is the same location a few seconds later, a stone thrown upward so it passes through A and B in freefall will age the most. A fly that takes off from A and lands again at B will age less than the stone. A clock sitting on the ground at A (and B) will age less than the stone.

This being a relativity thread, it could, and generally would, be greater than that. How do you figure it would be greater? You must have conservation of mass+energy. Since we're talking about only mass, the energy of an object or a system of "constituents" wouldn't matter (and is unspecified anyway). I thought maybe you meant there is some new particle that only exists when other particles are combined, like maybe gluons, but those are likely massless. For your statement to be true, I figure either you'd have to have different constituents when an object is considered as a whole vs as parts, or you'd have to change the mass of individual constituents??? What makes up the difference in mass?

No, I never said anything was stationary (I said "a mass" in a stationary system doesn't emit gravitational waves). I went on to explain how Earth contributes to radiated changes in the gravitational field. My point is you can't look at the Earth all by itself and assume it must be making changes. It seems like you're ignoring parts of answers that go against what you say 'you'd have thought', and just repeating your original position. Do you understand that it's asymmetric acceleration that emits gravitational waves? Why not? All it means is putting all of spacetime into an ordered set of hypersurfaces, right? The problem (if I understand it) is that you can do it, in an infinite number of ways, basically requiring an arbitrary choice of decomposition that's not universally meaningful. (Or to try to put it in simple terms, you can arbitrarily define single moments throughout the universe that each include only events that are space-like separated, but your choice won't meaningfully represent a single moment elsewhere.) If it wasn't possible, what would that mean? That a foliation can't include all of spacetime? Or that some space-like hypersurfaces must intersect (which seems to go against causality)?

Sure, why not? A mass doesn't emit gravitational energy just by being there, though, so it doesn't "evolve" in a stationary system. The Earth's not perfectly symmetric, so by simply spinning, it emits a tiny amount of energy as gravitational waves (negligible?). Changes in the gravitational field are more due to Earth's interaction with other masses. For example, according to the gravitational wave wiki, the total energy of the Earth orbiting the Sun loses about 200 watts emitted as gravitational waves. This is an extremely small fraction of the total energy. Regarding your original question, there isn't a decrease in mass in this case, because the energy comes from the kinetic energy and gravitational potential energy of the system, ie. decay in orbit.

See https://en.wikipedia.org/wiki/Gravitational_wave Read the section "Sources", as that explains some cases where gravitational waves are emitted (or "in most cases" aren't). In particular, in the case of black hole mergers, a lot of mass can be lost as gravitational waves. This comes from the extreme acceleration of the masses rotating around each other. That's about 5% of the mass, radiated away. But once merged, the mass stops accelerating and losing mass to gravitational waves, as the animations and charts show.

There are lots of consistent ways to describe what's going on here. See http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html (your video is the second half of a twin paradox experiment so all the analyses apply). Your A and B are symmetrical so describing one describes both. Using Doppler shift analysis, what they observe is: The other's clock appears far behind at the start, and when I move toward it, their clock appears to run faster than mine the entire time until it catches up only when we meet. If you include relativity of simultaneity in your calculation of the other's rate of time, then depending on your acceleration you can describe it as above: the other runs faster during your acceleration, then slower while at higher velocity. Or, you can separate time dilation and relativity of simultaneity and say that the other's clock only runs slower the entire time they're moving (as the video suggests), but the acceleration involves a change in inertial reference frame and corresponding change in relative simultaneity, and the clocks are no longer in sync when they start moving. In this case, our clocks are sync'd while we're at rest, but when I accelerate towards you, events where I am are now simultaneous with events at the other clock where the other clock is far ahead of mine, and I spend the rest of the trip catching up to it. Any analysis works and they're all consistent, but to see it (and believe it) you're probably going to have to do the maths. You haven't calculated anything of what A and B would observe, and yet you're finding imagined paradoxes that aren't in the calculations. So start calculating!, and you'll find exactly where the problem lies (where the results don't add up) and then you'll be able to learn the missing pieces that solves the paradox. If it's your first time with the maths, keep it simple and consider only two periods of inertial motion (when they're at rest, and when they're approaching each other after a quick acceleration). I'm sure you'll get lots of help at every step. Just remember there are a lot of different ways to describe it, and you might not believe they're consistent until you see it in your calculations.

I don't understand why you would write out all those equations when they don't even relate to the case described. Even the classic twin paradox is typically presented with negligible acceleration time, or equivalently considering only 2 legs of inertial motion for the "traveling" twin. Adding in an acceleration phase makes it more complicated, and doesn't help in figuring out the simpler version of the paradox. Why complicate it if a simpler version is specified?

Hahaha, well I look forward to your solution to the problem. It looks like a lot of equations to solve! The solution to the problem of this thread doesn't need the twin paradox to be resolved. You can calculate what's needed from one frame (the inertial one is the easiest). The calculations in any other frame are consistent but there's no need to verify that. "the one that ages differently" makes no sense. They all age differently from each other. That's like taking two frames with relative velocity and saying that only one of them is moving relative to the other. Besides, they all age "similarly", at 1s/s; the differences are only relative to each other. Also you're using 'rapidity' wrong. It's a measure of rate of motion, not an acceleration. If the twins are moving relative to each other, they both have rapidity relative to the other, just like they both have relative velocity.

It gives an exact solution in SR. The problem's been solved. The proper time of the traveling twins can be found with time dilation alone. You're still suggesting complications, but no solutions, and a simple solution is already given. What don't you understand about it?

No, there is still confusion here. From wikipedia: "In relativity, proper time along a timelike world line is defined as the time as measured by a clock following that line." The fact that it is invariant can be paraphrased as: Every observer (every frame of reference) agrees that the clock measured some specific (agreed upon) time at any given event along the path. Eg. everyone agrees that your watch said 12:00 when you were at the base of the mountain and 2:00 when you got to the top. What they won't all agree on is that a clock in town (a non-negligible distance away wrt. speed of light) said 12:00 when you were at the base and 2:00 when you were at the top. (And I add, if you had a drone that followed you at a fixed distance (Born rigid say), and the clock on the drone said 12:00 when you started and 2:00 when you got to the top, not everyone would agree that your watch and the drone were synchronized.) I think "infinitesimal extent" applies to both for the same reasons. No, your world line is defined by the path you take through 4d space, including however you accelerate. You never go off your world line, or have to change worldlines. Say you accelerate by stepping off an inertial train, onto the ground. The world lines of you and the train diverge at that point, but you don't change worldlines. Your worldline is invariant within its spacetime, and its particular coordinate system. The invariance means that everyone agrees that you pass through a particular set of events and that you do so at the time you measure yourself doing so. Not that the coordinates of those events in different reference frames are the same.

Ah thanks, I was worried for a minute there. If I understand that pdf correctly, there was only ever one definition of "proper time", introduced by Minkowski. Other conflicting (but common) uses are described as stemming from confusion. It sounds like you were first taught the incorrect use. From the paper: Just to add 2 cents, you can have a "momentary inertial frame" at any time, as an accelerating object effectively moves between different inertial frames. But you can't have a standard frame of reference with spatial extent follow the accelerating object, because different locations within the frame must accelerate at different times, and you'd need some additional (non-standard) definition of when those other locations follow the object. Does that make sense? It means that proper time is a measure of time at the location of the clock, but not elsewhere. I'm going off on a tangent to the main topic, other than that it's all related.

Where is that defined? Do you mean that it used tau for something other than proper time? Tau (like other letters) is used to mean different things in different contexts. Proper time is defined along the world line of a clock traveling along a spacetime interval, ie. only timelike intervals. It shouldn't be used to describe time away from the clock or throughout the frame, because that's not what it is. The proper time of an inertial clock measures the same time as any other clock elsewhere in the inertial frame, but that doesn't mean they're called the same thing. The proper time measured by an accelerating clock doesn't describe time elsewhere, because you can't synchronize with another clock, due to relativity of simultaneity. Even if you have two coordinated accelerating clocks that measure the same in a given inertial frame, they're not "Einstein synchronized" and won't agree with each other in the clocks' respective reference frames because in one clock's accelerated reference frame, the other clock accelerates at different times and has a non-zero relative velocity. Therefore the proper time measured by one accelerated clock can't remain the same all along its world line, as the proper time measured by any otherwise located clock, right? Edit: I think I'm confused as to what I'm replying to, I got the sense from this thread that "proper time" was incorrectly being used to describe time in a reference frame in which a given clock is at rest, but I can't find what gave me that impression.

Still, there's no "nonsense" whether you have separate length and time measurements or not. It all works out consistently. Add a clock, add a train, look at it from a different frame of reference, make a planet an imaginary particle, use different type of coordinates or measuring tools, all you're changing is how complicated is the experiment that you're describing. RAGORDON2010, since you're saying that using one "fixed" reference frame removes the nonsense, I think that you're missing a lot of special relativity and just avoiding it. Besides, I don't see how "fixed" is meaningful, because to observers with other frames of reference, it's not fixed. Your experiment must make sense from other relatively moving frames of reference. If that leads to nonsense, you've done something wrong. That doesn't matter, it can still be used to make accurate predictions that match real measurements. Besides, it needn't be any of those things to someone familiar enough with it.

What happens if you simplify all of this and get rid of Earth, moon, trains, and just make them named inertial frames? As described, it makes no difference whether it's set up for an observer on Earth or a test particle in empty space. The Earth frame can be made symmetric to a train frame in experiments like these. Making a moving frame a train gives a concrete example that's easier to think about, but it can also give false impressions of material differences between reference frames. Do your insights remain unchanged if you make all the frames generic, and get rid of the weekdays? You mention "initial velocity [...] playing the role of relative velocity", but all velocity is relative velocity. Velocities are relative to frames of reference.

Ah I made a mistake, I get several answers that fit. I get one solution if I assume that Mr. Johnson knows the answer in the end.

I have two possible answers that seem to fit, one for whether Mrs. Albert replies "no" and one for "yes." Is there supposed to be a single solution? I'm assuming 1) "smallest" means smaller than the number of Charles family children, though technically the statements would be true if they were the same, 2) Mr. Johnson only asks questions he can't already know the answer to.

You left out that they measure the distance in their respective frames, which are different.

Sorry to have to confuse you, but the light that makes up an image, and the subject represented by the light, are two different things. Suppose you print a photograph of yourself. The photo will age: discolor, disintegrate, etc. but that doesn't mean that it will show you aging in the image. In the case of beaming light across some great distance, neither would you age in the image (same as with a photo), nor would the light itself change with respect to time (as measured by an observer).

Just to be clear, this is entirely due to delay time of light. Your reflection travels at a speed of c and doesn't age. Time dilation applies to a moving mirror's age, but that doesn't affect the timing (or speed) of the reflection. As for the equations, sqrt((1-v/c)/(1+v/c)) = sqrt((c-v)/(c+v)) is the inverse Doppler factor, which is how many times the source frequency is multiplied to get what you see. (sqrt((c+v)/(c-v)) - 1) I think is the redshift, which is what fraction of the source wavelength is added to get what you see. Maybe someone can correct this or state it better. The source moving away from you has a positive velocity.

The "perhaps" is right, the mirror doesn't have to be moving. If it's stationary relative to you it will only make you appear half your age at one specific age. If you set up a series of mirrors each 1/4 light years further away from you, then once every year, one of the mirrors would show you at half your current age. (1/4 LY away takes half a year for light to make a 2-way trip, so at age 1, light that left you half a year ago returns. At age 2 it takes 1 year return trip to a mirror 1/2 LY away, etc.) If a mirror is moving away from you at any speed, your reflection is actually doubly Doppler shifted. If you have any object moving away from you, and it is emitting light (eg. if it is self-lit or lit by a source in its own inertial frame) then it is red shifted according to the relativistic Doppler shift (even at non-relativistic speeds). Symmetrically, an observer in the object's frame sees you red-shifted by the same amount. As for the reflection, it is equivalent to an observer receiving a red-shifted image of you, then projecting that image back to you, which you see as red-shifted again. So say it's a mirror with a wooden border and a light on it. The wood appears at an intensity and frequency that is sqrt((1-v/c)/(1+v/c)) times its emitted intensity and frequency. Your reflection appears at intensity and frequency of (1-v/c)/(1+v/c) times what a relatively stationary mirror shows. The formulae will differ depending on whose frame you're talking about (and who has +v or -v in your setup), whether you're talking about source or observer, whether you're talking wavelength or frequency, etc (all these cases means is an inverse in the formula).

It you had a mirror that left you at birth and traveled at a speed of c/3 away from you, you would always look half your age in the mirror. Your reflected image would appear to be n/2 light years away when you see it at age n. Pop quiz! This mirror's moving at a relativistic speed. Why am I not using gamma or relativistic composition of velocities? What's the Doppler shift of your image, and why isn't it the relativistic Doppler shift?

Choosing a reference frame doesn't determine what happens. His trip happens in all reference frames. You choose a reference frame to measure/describe what happens, and the measurements from both reference frames implied here are valid. Only relative to various observers. For the traveling twin, the biological processes occurred at the usual rate (1 s/s). The traveling twin *is* aware of the difference in aging. She can calculate the aging of Earth, and she can also see it happening. No. This is just justifying misunderstanding SR.