PJ_Boy made me think of an interesting physics question.
If you derive acceleration, you get the rate at which your acceleration changes. (You could do this nearly infinite times, until you hit a constant, to which the derivative would be zero).
If you integrate acceleration (m/s^2), you get velocity (m/s), i.e. the summation of your acceleration over an interval.
If you integrate velocity (m/s), you get distance traveled (m).
If you were to integrate distance... you'd get (m * s). Perhaps I'm just too far out of physics... but that doesn't seem like it would have any meaning.
Sage advice from a friend of Jim: So put your tinfoil hat back in the closet, open your eyes to the truth, and realize that the government is in fact causing austismal cancer with it's 9/11 fluoride vaccinations of your water supply.
Oh, here's a question I've been wanting to hurt people with for awhile.
Let's say you're in a hybrid automobile at the top of a hill. Your vehicle can convert velocity now to velocity later at a given efficiency rate of A (that is, e.g., if you slow down 5m/s now, then later you can speed up by 5Am/s), with A of course being less than 1. You also have to deal with air resistance, which as usual goes up with the square of velocity. Let's say your drag constant (area * drag coefficient) is B (see Drag Equation).
The question is, when driving down the hill, do you try to conserve as much kinetic energy as possible (i.e. don't use the brakes; just coast), despite increased losses due to air resistance? Or do you brake, losing some energy in the conversion process but also losing less due to air resistance?
Obviously a perfectly-efficient hybrid (A = 1) would go as slowly as possible on the hill, and a non-hybrid (A = 0) would go as fast as possible. It's in the middle where things get muddy.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
PJ_Boy made me think of an interesting physics question.
If you derive acceleration, you get the rate at which your acceleration changes. (You could do this nearly infinite times, until you hit a constant, to which the derivative would be zero).
If you integrate acceleration (m/s^2), you get velocity (m/s), i.e. the summation of your acceleration over an interval.
If you integrate velocity (m/s), you get distance traveled (m).
If you were to integrate distance... you'd get (m * s). Perhaps I'm just too far out of physics... but that doesn't seem like it would have any meaning.
Integrating distance feels strange because you don't normally have situations where distance is a function of time. Velocity and acceleration are often a function of time, but distance?
Think of a particle on the x-axis whose postion at time t is x(t). Then the integral of x(t) is the function whose derivative is x(t); then the definite integral from a to b of x(t) dt is (b-a) times the average position of the particle during that time.
Derakon wrote:
Let's say you're in a hybrid automobile at the top of a hill. Your vehicle can convert velocity now to velocity later at a given efficiency rate of A (that is, e.g., if you slow down 5m/s now, then later you can speed up by 5Am/s), with A of course being less than 1. You also have to deal with air resistance, which as usual goes up with the square of velocity. Let's say your drag constant (area * drag coefficient) is B (see Drag Equation).
The question is, when driving down the hill, do you try to conserve as much kinetic energy as possible (i.e. don't use the brakes; just coast), despite increased losses due to air resistance? Or do you brake, losing some energy in the conversion process but also losing less due to air resistance?
Just to be specific, are we optimizing for total distance or for some other metric (time taken to reach a target, say)?
Oh, dur. Forgot to include the question! The goal is to maximize your usable energy at the bottom of the hill -- thus, your current kinetic energy + A * the energy in your battery.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Integrating distance feels strange because you don't normally have situations where distance is a function of time. Velocity and acceleration are often a function of time, but distance?
Let me rewrite what you just wrote:
Integrating distance feels strange because you don't normally have situations distance is a function ofwhere things move over time. Velocity and acceleration are often a function ofoften change in time, but distance?
See how absurd that sounds now? Yeah, I thought so.
Even for straight motion with constant speed, you have that x(t) = v*t.
With that out of the way, lets see DarkKobold's question:
DarkKobold wrote:
If you were to integrate distance... you'd get (m * s). Perhaps I'm just too far out of physics... but that doesn't seem like it would have any meaning.
As rhebus correctly answered after the above snippet, it is related to position averages: int_a^b{x(t) dt}/(b - a) is the average position over the time interval [a, b].
Derakon wrote:
Oh, here's a question I've been wanting to hurt people with for awhile.[snip]
That is an interesting one. I will be working on it, but it may take some time.
Let's assume that only special relativity is in effect (general relativity probably would make a slight difference because the solar system is a gravity well, and Alpha Centauri is another, but just for the sake of simplification let's consider it negligible).
An astronaut starts traveling from Earth to Alpha Centauri, a distance of 4.365 light years. The spaceship maintains a comfortable constant acceleration of 9.8 m/s2 until half of this distance has been traveled, after which it makes a 180-degree turn and starts decelerating at the same 9.8 m/s2 so that when it reaches Alpha Centauri, it will stop. The trip back to Earth is started immediately in the same way.
1) How long did the overall trip take from the point of view of the astronaut? (In other words, how much older is he after he's back to Earth?)
2) How long did the overall trip take from the point of view of the Earth? (In other words, how much older is someone on Earth?)
I'm actually curious to see how this calculation is done.
For an extra difficult additional (but optional) assignment: How much of a difference would it make if we took general relativity into account?
That's an interesting question and I'd really like to tackle it, but ironically, I'm studying for a physics test at the moment. Still, I will take the time to point out that to properly define the acceleration, you should refer to the force per unit mass. In particular, the problem is well-stated if you demand that the spaceship's momentum, not its velocity, increases monotonically. Your units imply that the velocity increases monotonically, which is obviously not possible.
Like I said, I'm pretty busy, but I did take the time to calculate that c/g ~ 1 year, so I would guess that the answer to part 1 is roughly 4 years (2 years each way) and the answer to part 2 would be something like 15 years. These are very crude calculations, though, and only reflect ballpark estimates of the answer. I'd love to address this problem in full, but it'll be at least a few days before I have the chance. Someone else will probably beat me to it.
I just finished that problem. I'm no physicist, all I recall is that special relativity is like classic mechanics with some extra square roots in ODEs. A physicist would probably get the solution better, I only know how to solve ODEs.
To model constant acceleration, I considered the quotient of the force to the invariant mass to be constant. I did everything by hand and I'm not sure if it's correct. I don't have time to fully comment now so I'll just post the calculations Warp wants to see (some were omitted, but it's just boring integration/differentiation/algebraic manipulation that I didn't need to write to do).
D is half the distance from Earth to Alpha Centauri in Earth's static frame, T is the time it takes to get to 1/4 of the trip measured in the Earth and T' the time measured in the ship, v is the velocity of the ship and x its position from earth's view. After you pass the D distance, the situation is symmetric (the integrals are the same, just with some changes of variables), so it suffices to do the problem for first 1/4.
Someone please tell me if I did anything wrong. (I did (2) first)
For this problem, k = 2,24, and I found the times:
(1): 4T' = 6,20 years
(2): 4T = 11,95 years
At least for me, this looks coherent :P
Still, I will take the time to point out that to properly define the acceleration, you should refer to the force per unit mass. In particular, the problem is well-stated if you demand that the spaceship's momentum, not its velocity, increases monotonically. Your units imply that the velocity increases monotonically, which is obviously not possible.
The acceleration felt by the astronaut (which can be unambiguously measured by the astronaut using an accelerometer) can be constant forever, even under SR/GR, as far as I understand.
There's a lot of confusion about special relativity and maximum speeds (and I'm in no way claiming I have no such confusions myself, mind you). One such misconception is that you can't eg. travel from here to Alpha Centauri in less than 4 years, no matter how much you try. However, this does not take into account the frame of reference. Certainly you can't launch a probe from Earth to Alpha Centauri and have it come back in less than 8 years. But that's from our frame of reference.
The frame of reference of the probe itself, or in this case the astronaut, is a different matter, though. If I have understood correctly, the astronaut himself, from his own frame of reference, has no lower limit to the time he can reach Alpha Centauri. In principle he could reach Alpha Centauri in 1 minute (disregarding that he would be vaporized by the acceleration and deceleration).
What happens as his speed increases is that the universe contracts and, from the astronaut's point of view, the distance between Earth and Alpha Centauri decreases.
This means that if the spaceship was capable of traveling to Alpha Centauri in 1 minute (from its own frame of reference), a round trip would make the astronaut 2 minutes older, while a bit over 8 years have passed on Earth when he arrives back there.
Someone who actually understands these things please correct me if I'm wrong (because I would really not like to spread BS about this if I am).
(This is actually the subject of the so-called twin paradox.)
The acceleration felt by the astronaut (which can be unambiguously measured by the astronaut using an accelerometer) can be constant forever, even under SR/GR, as far as I understand.
Said this way, yes, you are absolutely correct, though I think that speaking of the force felt by the astronaut is even more accurate. In fact, I think the key word there is "felt". The earlier problem came from ambiguity in the word "acceleration" and your choice of units, m/s^2, which, taken in a perfectly valid way, implies a constant dv/dt. In this interpretation, the astronaut would not feel a constant acceleration; he would instead feel an asymptotically increasing force as he approaches the speed of light, with death and annihilation of the spaceship surely occurring sometime before he reaches v=c (in a finite time).
Hope that clears things up. I didn't have time to answer the problem, so I instead settled for a nitpick. I knew what you meant in the original statement, but decided I should point out a minor flaw/ambiguity. At first glance, p4wn3r's math looks at least plausible, but I'd still like to take a crack at the problem myself.
But now, I have a pitiful E&M final to turn in.
Let's assume that only special relativity is in effect (general relativity probably would make a slight difference because the solar system is a gravity well, and Alpha Centauri is another, but just for the sake of simplification let's consider it negligible).
An astronaut starts traveling from Earth to Alpha Centauri, a distance of 4.365 light years. The spaceship maintains a comfortable constant acceleration of 9.8 m/s2 until half of this distance has been traveled, after which it makes a 180-degree turn and starts decelerating at the same 9.8 m/s2 so that when it reaches Alpha Centauri, it will stop. The trip back to Earth is started immediately in the same way.
1) How long did the overall trip take from the point of view of the astronaut? (In other words, how much older is he after he's back to Earth?)
2) How long did the overall trip take from the point of view of the Earth? (In other words, how much older is someone on Earth?)
I'm actually curious to see how this calculation is done.
For an extra difficult additional (but optional) assignment: How much of a difference would it make if we took general relativity into account?
I will give the more modern version of the solution using techniques from differential geometry -- for this case, only knowledge of 4-vectors are enough for it.
I will use light-years (ly) for distance and years (y) for time. The only relevant piece of information needed to convert is the acceleration; the conversion factor is 1 m/s2 = 0.10526033 * 1 ly/y2, making the acceleration a = 1.0315512 ly/y2.
For simplicity, we can assume the movement to be done in one direction. We will work on the reference frame of the Earth/Sun, assuming it is inertial; the coordinates we need to track the rocket in this frame are the time t and the position at this time, x(t). We also need to take care of the rocket's proper time, τ. Now, both t and τ are monotonically increasing -- they mark the passage after all. Moreover, given any time t, we will have a unique corresponding value for τ (the converse also being true) -- hence they are related. We are yet to find this relation, but we can write it in implied form: we can write t and x as functions of the proper time τ: t(τ), x(τ).
This is a mere reparametrization of the rocket's path, by the way: instead of using the current reference frame's time to determine its position and using another equation to determine the relation between frame time and proper time, we mix them both and use proper time to determine frame time and position. This has the advantage that the proper time is an "universal" parameter -- every reference frame (even non-inertial ones) will agree how much proper time elapses between two points in the rocket's trajectory. The proper time has the great advantage, thus, of being a Lorentz-invariant parameter: you can perform Poincaré transformations before or after you apply the operator(s) dn/dτn.
Thus we can define the 4-velocity and the 4-acceleration vectors by
U(τ) = {Ut(τ), Ux(τ)} = {dt(τ)/dτ, dx(τ)/dτ}
A(τ) = {At(τ), Ax(τ)} = {dUt(τ)/dτ, dUx(τ)/dτ} = {d2t(τ)/dτ2, d2x(τ)/dτ2}
Using the chain-rule for the derivative of composite functions, it is easy to see that
(c * Ut(τ))2 - Ux(τ)2 = (c * Ut(τ))2 * {1 - Ux(τ)2 / (c * Ut(τ))2} = (c * Ut(τ))2 * {1 - v(τ)2 / c2}
Moreover, using the time-dilation formula, it is easy to see that
Ut(τ) = dt(τ)/dτ = 1 / sqrt{1 - v(τ)2 / c2}
hence, the Minkowski norm of the 4-velocity is
(c * Ut(τ))2 - Ux(τ)2 = c2
Doing similar steps for A(τ), we reach the Minkowski norm of the 4-acceleration:
(c * At(τ))2 - Ax(τ)2 = -a2
where a is the (constant, in this case) proper acceleration, that is, the acceleration measured by an accelerometer inside the accelerating rocket. These relations also show that the 4-acceleration A and the 4-velocity U are orthogonal in the Minkowski norm:
c2 * Ut(τ) * At(τ) - Ux(τ) * Ax(τ) = 0
(this follows from the constancy of the Minkowski norm of the 4-velocity by doing a proper-time derivative).
Taking the Minkowski norms above and writing them in terms of derivatives of t(τ) and x(τ), we have
c2 * (dt(τ)/dτ)2 - (dx(τ)/dτ)2 = c2c2 *(d2t(τ)/dτ2)2 - (d2x(τ)/dτ2)2 = -a2
This set of equations is all we need to answer the special relativistic version of the problem. To that, I claim that the following is the solution:
t(τ) = c * sinh{a * (τ - τ0) / c} / a + t0x(τ) = c2 * cosh{a * (τ - τ0) / c} / a + x0
Direct substitution is the best way to check this.
As p4wn3r observed, the special relativistic version has several symmetries: it can be divided in 4 essentially identical subproblems, whose sole difference is the value of the constants t0, x0 and τ0. It is enough to solve one of these sub-problems, so we will take the first -- the constantly accelerating portion for the first half of the trip to Alpha Centauri. Our values for t0, x0 and τ0 are then defined by our boundary conditions that the rocket is at rest at the Earth when t = τ = 0, so that t0 = τ0 = 0 and x0 = -c2 / a; thus, we will work with
t(τ) = (c / a) * sinh(a * τ / c)
x(τ) = (c2 / a) * {cosh(a * τ / c) - 1}
We must calculate the final value of τ we are interested in; this corresponds to the half-way point x(τm) = (4.365/2) ly. To do this, lets invert the equation for x(τ) -- we can do that in this interval because x(τ) increases monotonically with τ in this interval -- to obtain
τ = (c / a) * acosh{(a / c2) * x(τ) + 1}
Plugging in x(τm) = (4.365/2) ly = 2.1825 ly, a = 1.0315512 ly/y2 and c = 1 ly/y, we find
τm = 1.791172306 y
For the Earth reference frame, this corresponds to
t(τm) = 2.999132811 y
Multiplying these by 4 for the four segments of the journey, we see that 7.164689224 y transpire at the rocket, while 11.996531244 transpire on Earth. The difference to p4wn3r's result is probably a result of numerical errors in his part, as his math seems correct.
For the GR version: I started doing some work on it; but it is ugly and messy. FYI, here a few of the confounding factors:
The relative motion of the star systems will greatly complicate matters because of the changing gravitational field;
Alpha Centauri is a binary star system, where each stars have a mass near to the Sun's own mass; not only that causes nonlinearity problems (see next item), but there is also the additional issue of dealing with gravitational radiation;
Due to the non-linear nature of Einstein's field equations, the metric tensor for the 3 stars is not the sum of the metrics of the individual stars;
The exact starting position near the Sun and ending position near the binary Alpha Centauri star is crucial to avoid the Schwarschild radius (unless you model the star interior as well);
The difference in mass between the star systems will remove some of the symmetries that allowed us to calculate 1 leg of the trip and multiply by 4 in the special relativistic version.
There are ways around these issues, using several simplifications; but the math still comes out ugly. I will leave this for the future...
Now for replies to posts made while I was writing the above:
Bobo the King wrote:
Still, I will take the time to point out that to properly define the acceleration, you should refer to the force per unit mass. In particular, the problem is well-stated if you demand that the spaceship's momentum, not its velocity, increasesa monotonically. Your units imply that the velocity increases monotonically, which is obviously not possible.
You haven't yet taken (non-introductory) courses in relativity, have you? If you had, you would already know that "constant acceleration" in relativity is shorthand for "constant proper acceleration", which is perfectly well defined and leads to hyperbolic motion. His choice of units is totally irrelevant for that; he can chose to measure acceleration in m/s2, ly/y2, feet/min2, m-1 (using c as a conversion factor between s and m), s (the reverse conversion), kg-1 (using G/c2 as a conversion factor between Kg and m and c as a conversion factor between m and s) or any other exotic units of his choice and his question would still be well formed. Moreover, his velocity can increase monotonically whether or not this is relativity; in relativity, it also has to grow asymptotically to the speed of light, but that does not prevent monotonic growth.
Warp wrote:
The acceleration felt by the astronaut (which can be unambiguously measured by the astronaut using an accelerometer) can be constant forever, even under SR/GR, as far as I understand.
Absolutely correct.
Bobo the King wrote:
Said this way, yes, you are absolutely correct, though I think that speaking of the force felt by the astronaut is even more accurate.
So far so good. Irrelevant, but correct nevertheless. Sadly, the rest of the paragraph, does not have any single correct sentence:
Bobo the King wrote:
In fact, I think the key word there is "felt". The earlier problem came from ambiguity in the word "acceleration" and your choice of units, m/s^2, which, taken in a perfectly valid way, implies a constant dv/dt.
See above.
Bobo the King wrote:
In this interpretation, the astronaut would not feel a constant acceleration; he would instead feel an asymptotically increasing force as he approaches the speed of light, with death and annihilation of the spaceship surely occurring sometime before he reaches v=c (in a finite time).
This "interpretation" is not a result of any real understanding of relativity. The force he feels is the force measured in his reference frame, period (same for acceleration). If the force is constant in this reference frame, so is the acceleration (and the converse also holds). If you were to demand constant acceleration as seen by any other reference frame, the ship would break apart quickly due to a matter of simultaneity (that is, enormous stresses due to parts of the ship being accelerated before others in the ship's reference frame); but no one but you is talking about that.
Bobo the King wrote:
Hope that clears things up. I didn't have time to answer the problem, so I instead settled for a nitpick.
You can clear things up with a confused explanation about an invalid nitpick. Try harder.
We must calculate the final value of τ we are interested in; this corresponds to the half-way point x(τm) = (4.365/2) ly. To do this, lets invert the equation for x(τ) -- we can do that in this interval because x(τ) increases monotonically with τ in this interval -- to obtain
τ = (c / a) * acosh{(a / c2) * x(τ) + 1}
Plugging in x(τm) = (4.365/2) ly = 2.1825 ly, a = 1.0315512 ly/y2 and c = 1 ly/y, we find
τm = 1.791172306 y
For the Earth reference frame, this corresponds to
t(τm) = 2.999132811 y
Multiplying these by 4 for the four segments of the journey, we see that 7.164689224 y transpire at the rocket, while 11.996531244 transpire on Earth. The difference to p4wn3r's result is probably a result of numerical errors in his part, as his math seems correct.
Thanks for looking over my "solution". I didn't do the results very precisely, but still the time for the rocket is differing by a large value. Now I have a question, why can you you plug half the distance from the star directly into that formula? Shouldn't the distance for the ship's frame shrink because of the Lorenz factor?
EDIT: Never mind, I've read it properly now, I evaluated my results in a calculator and got the same thing, it's not easy to evaluate an ln by hand...
We must calculate the final value of τ we are interested in; this corresponds to the half-way point x(τm) = (4.365/2) ly. To do this, lets invert the equation for x(τ) -- we can do that in this interval because x(τ) increases monotonically with τ in this interval -- to obtain
τ = (c / a) * acosh{(a / c2) * x(τ) + 1}
Plugging in x(τm) = (4.365/2) ly = 2.1825 ly, a = 1.0315512 ly/y2 and c = 1 ly/y, we find
τm = 1.791172306 y
For the Earth reference frame, this corresponds to
t(τm) = 2.999132811 y
Multiplying these by 4 for the four segments of the journey, we see that 7.164689224 y transpire at the rocket, while 11.996531244 transpire on Earth. The difference to p4wn3r's result is probably a result of numerical errors in his part, as his math seems correct.
Thanks for looking over my "solution". I didn't do the results very precisely, but still the time for the rocket is differing by a large value. Now I have a question, why can you you plug half the distance from the star directly into that formula? Shouldn't the distance for the ship's frame shrink because of the Lorenz factor?
The distance for the ship's frame indeed gets reduced; but the formulas I obtained at the end, for t(τ) and for x(τ), give (respectively) the time and position of the rocket on Earth's reference frame as a function of the rocket's proper time τ. Thus, when the rocket is halfway to Alpha Centauri (according to Earth), x(τ) will be half the distance to Alpha Centauri (as seen by Earth).
OK, now let me try to understand what you did. I read some things on the internet, but most of marzojr's solution I ended up deducing myself, I have little knowledge of special relativity besides high school introduction. See if my considerations are correct.
First of all, you define the motion of the particle in a 4-D vector space containing time and its x,y,z axis positions. Then, instead of using the Euclidean norm, you introduce some form of Riemannian-like metric (which is technically not a metric, because it can be negative sometimes, but I think this fact shouldn't be a problem here) that is invariant of the coordinate system provided you take a base that's orthonormal with regard to this "inner product".
By taking note of this fact, if you're analyzing movement from a certain frame, you can take the norm of the time-position vector (I'll call it X), defined as the square root of the Minkowski product of the vector with itself. The postulates of relativity imply that this product is always positive and the norm is the time that passed for the mover, called proper time. Thus, its well defined for every coordinate system.
Because differentiation and base changes are linear operators and more specifically, the operator changing from and to orthonormal bases is a self-adjoint operator, one can differentiate X as many times as he wants and all d^nX/dT^n vectors will have the same norm for any coordinate system.
This is specially useful for velocity and acceleration with respect to T. Because the norm (or maybe the square of the norm?) will always be the same, one can compare it with a trivial case to find its value. The most simple is a frame that's moving together with the particle. To find the norm of proper velocity, you simply look at the vector (c,0,0,0) which has norm c^2.
For the norm of proper acceleration, it's trickier because you have to differentiate more, but for the frame of the ship, one can find -a^2.
Thus, on Earth's frame, if you take the 4-D velocity and acceleration vectors and use the results that their norms must equal c^2 and -a^2, you arrive at a system that can be transformed into one of non-linear first order ODE's, which can be solved by undetermined coefficients.
Also, I believe this is a typo, because it should have a square root, but didn't influence the posterior steps:
marzojr wrote:
marzojr, you are much more learned in physics than I am. Nevertheless, I have to ask: what is wrong with interpreting a as dv/dt? Okay, if a is commonly understood to be the "constant proper acceleration", I suppose I can't argue with that, though physicists are generally more precise than most with their nomenclature. If we cannot discard the possibility of a=dv/dt (which seems quite sensible to me), then we must at least hint to our interpretation of acceleration by choice of units and words. I still maintain that velocity cannot increase monotonically indefinitely (and if it does not increase monotonically indefinitely, it technically does not increase monotonically at all).
And as a physicist, I'm not afraid to be wrong or to concede valid points to someone who knows better than I do-- you're probably correct and I am out of my league. But please, no need to be so snarky about it.
The way I'm reading this it seems to me that it's possible for a body to exceed the speed of light from an outside reference frame (when neglecting space contraction?). The space surrounding it would contract so that it wouldn't exceed the speed of light from its own point of reference? Last time I checked, such things were just pure speculation. My intuition says that once it becomes necessary to contract space a huge deal a constant force shouldn't be enough for the body's accelaration to remain constant. Previously I had always thought that the accelerating body would have to start expanding at some point. Is there any difference at all except the point of reference? Wouldn't this introduce many side-effects, essentially making it impossible? Is it a misconception that contracting the surrounding space should require extraordinary forces, is this just a pure thought experiment or am I terribly misinterpreting all of this? Can somebody tell me some keywords that I should search for or direct me to a book (I suck at math and physics and hate sensationalism) I could buy to get a better understanding of all this?
Sorry for the wall of text.
The way I'm reading this it seems to me that it's possible for a body to exceed the speed of light from an outside reference frame (when neglecting space contraction?).
It cannot in any circumstance exceed the speed of light (c). If you think about it, if it was possible it would mean that it would actually outrun photons traveling in vacuum (which is what "traveling faster than light" literally would mean), which is impossible. Photons (and any massless particles for that matter) in vacuum will always move faster than any massive body (and if you measure the speed of those photons, it will always be exactly c, regardless of where you are measuring from).
If you send a probe from Earth to Alpha Centauri and have it come back, it's not possible for it to come back in less than 8.6 years. (GR might allow some bending of space which would literally reduce the distance. Even then the speed c would have not been breached. It's just that the physical geometry of space has been modified and the distance reduced. However, actually bending space like this might be practically impossible.)
The traveling astronaut can himself experience the round trip in less than 8.6 years (in other words, when he arrives back to Earth he would have aged less than that amount). I suppose that if you put a clock on the spaceship, have it go there and come back, and use that clock as your measurement of traveling speed, you would get a value which is larger than c. However, that would be a miscalculation because you would be taking time from one frame of reference (the spaceship) and distance from another (the Earth). This would be equivalent to watching two objects approach each other at 0.8c and concluding that they are seeing each other approaching at 1.6c. Obviously that's not the case.
The acceleration felt by the astronaut (which can be unambiguously measured by the astronaut using an accelerometer) can be constant forever, even under SR/GR, as far as I understand.
Absolutely correct.
This is the point I have trouble understanding. Constant acceleration forever means that its speed will eventually exceed the speed of light, doesn't it? Would a space traveller then be able to exceed the speed of light from his own reference frame?
Edit: Is it that time freezes for him (of course without him noticing) the instant he reaches c (or rather it almost freezes as he approaches c)? (So he'd essentially measure that last moment of constant acceleration forever, or until he arrives)
Oh, I see, I'm an idiot then... It was essentially a misconception of mine about how acceleration is measured then that has lead me to misinterpret this discussion.
The acceleration felt by the astronaut (which can be unambiguously measured by the astronaut using an accelerometer) can be constant forever, even under SR/GR, as far as I understand.
Absolutely correct.
This is the point I have trouble understanding. Constant acceleration forever means that its speed will eventually exceed the speed of light, doesn't it? Would a space traveller then be able to exceed the speed of light from his own reference frame?
Constant acceleration as the acceleration felt/measured by the astronaut (eg. using an accelerometer). In other words, no matter how long the spaceship travels, its acceleration is always such that the astronaut experiences the same "gravity" as on Earth. As the King pointed out, it might be less confusing to talk about a constant force felt by the astronaut (due to the speed of the spaceship always increasing) than acceleration.
However, no matter how much time the spaceship accelerates this way, if the astronaut measures his own speed (with respect to anything), it will always be less than c, and if he measures the speed of light in vacuum (anywhere), that will always be exactly c. (Again, general relativity makes things a bit more complex because the geometry of space can change in such ways as distances truly increasing faster than c, but we are restricting ourselves to special relativity for simplicity.)
As you might remember, mass increases as velocity increases, which AFAIK is related to the explanation of why the astronaut can feel a constant acceleration forever.
Bobo the King wrote:
I still maintain that velocity cannot increase monotonically indefinitely (and if it does not increase monotonically indefinitely, it technically does not increase monotonically at all).
I think that the point was that m/s2 is a unit of acceleration (it's what eg. an accelerometer could report) and hence it's completely valid to use it. Your objection seems to be that using that unit would mean that the speed of the spaceship (as measured by some frame of reference) would literally increase by 9.8 m/s for each second that passes, while in SR the speed can only asymptotically approach c, and hence using a different, less confusing unit of acceleration would be better. Perhaps for the sake of clarity that could be the case. However, that doesn't make m/s2 any less valid of a unit for the measurement of acceleration performed by the astronaut.
Nevertheless, I have to ask: what is wrong with interpreting a as dv/dt?
The word "interpreting": you don't interpret something to be what it is defined to be. A definition in science has to be unambiguous and as free of interpretation as possible or it is useless. This is why the meaning of "v" and "t" have to be specified also when defining acceleration (whether in classical or relativistic physics) -- for example, 't' being 'time' is unambiguous in classical mechanics (universal time), but ambiguous in relativity (because time is relative); the most straightforward measure to remove ambiguity is to use the proper time of the thing whose velocity is being measured (because this isn't contingent on the observer).
Here is a perfect way of explaining the difference between definition and interpretation: quantum physics. You have several definitions (what a wave function is, mathematically speaking, and how you extract observable values from it) and several interpretations (Copenhagen, Bohm, consistent histories, many worlds). In practical terms, the interpretations are irrelevant -- they all give the same experimental results and are indistinguishable (so far), while the definitions are all-important.
Bobo the King wrote:
Okay, if a is commonly understood to be the "constant proper acceleration", I suppose I can't argue with that, though physicists are generally more precise than most with their nomenclature. If we cannot discard the possibility of a=dv/dt (which seems quite sensible to me), then we must at least hint to our interpretation of acceleration by choice of units and words.
Bobo the King wrote:
I still maintain that velocity cannot increase monotonically indefinitely (and if it does not increase monotonically indefinitely, it technically does not increase monotonically at all).
For t >= 0:
v(t) = c(1 - e-t)
dv(t)/dt = ce-t > 0 for all t
This asymptotically approaches c and is monotonically increasing throughout the domain of t (from 0 to infinity).
If you mean physically, the answer is 'duh': it would need an infinite power supply to do it, which not only does not exist but would collapse the entire universe if it did (infinite Schwarzschild radius and all).
Warp wrote:
As you might remember, mass increases as velocity increases, which AFAIK is related to the explanation of why the astronaut can feel a constant acceleration forever.
That explanation has long fallen in disfavor and is no longer used. The more correct explanation takes into account the contraction of space and dilation of time (or rather, the changing hyperplanes of simultaneity) -- the proper acceleration remains constant from the rocket's point of view, but because of the contraction of space and dilation of time, it is always getting smaller from the point of view of an external observer.
Warp wrote:
I think that the point was that m/s2 is a unit of acceleration (it's what eg. an accelerometer could report) and hence it's completely valid to use it.
And it is, indeed, used by professional physicists working with relativity. Although for expediency when doing math by hand, it is usually measured in m-1 using c to convert between time and distance units (measuring time in m, with speed being adimensional and c then being set to 1), with dimensional analysis to recover the missing lost factors of c.
This got me thinking: The path taken by the astronaut probably doesn't matter much on the basic phenomenon that he ages less than a twin on Earth. He could make a perfectly circular path, with a diameter of 4.3 light years, and the same thing would still happen: When he comes back to Earth he would be younger than his twin.
The diameter itself shouldn't matter either. Whether it's 4.3 light years or 1 meter, it should still happen: After returning to the starting point, the traveler should have aged less than someone who remained stationary at that starting point (even if in this case the difference is minuscule).
This got me the idea that if you spin a wheel very fast, shouldn't the wheel age more slowly than its surroundings? If you put a clock on the rim of the wheel (a clock which working is not hindered by a strong acceleration), shouldn't it run more slowly than a clock that is outside, stationary?
In fact, the aging of the wheel is not constant throughout. The rim ages more slowly than the center (which experiences no acceleration and thus ages at the same speed as the surroundings).
(Of course the faster the wheel spins, the closer we start getting to the so-called Ehrenfest paradox, but that's another question entirely.)
I like classical mechanics more. I'll post a problem I could solve with some considerable amount of work. I think people here will find it interesting.
Alice is a student that doesn't like paying attention to her classes. Instead, she likes to do physics experiments while her teachers are speaking. In one occasion, she placed one of her pencils (which had the form of a regular hexagonal prism) in the top of her table, an inclined plane, and started giving minor impulses to the pencil so that it rolled down the table. Eventually, she realized that when her initial impulse was enough to make it roll once, the pencil always rolled down non-stop to the end of the table. She concluded that this happened because her table was inclined by a significantly high angle and if the table's inclination were lighter, the pencil could stop before reaching the end. Since she wasn't paying attention to mechanics class, she asks you to analyze this phenomenon for her.
Consider a long, solid, rigid, regular hexagonal prism with uniform mass distribution. The prism is initially at rest with its axis horizontal on an inclined plane which makes a small angle \theta with the horizontal. Assume that the surfaces of the prism are slightly concave so that it only touches the plane at its edges. The effect of this concavity on its inertial properties can be ignored.
The prism is now displaced from rest and starts an uneven rolling down the plane. Assume that friction prevents any sliding and that the prism does not lose contact with the plane. Calculate the smallest angle \theta, for which the uneven rolling, once started, will continue indefinitely.
There must be some energy loss for the pencil to stop. In an ideal setting (no losses from friction, air resistance, etc. and no outside inputs after the initial impulse), I'm not seeing where this energy would go. Thus in such a setting, the pencil should continue rolling indefinitely even on a level surface. As soon as you slope the surface up, though, the pencil will start making net KE -> PE trades with each revolution, and will eventually lack the KE to push itself over the energy barrier to the next revolution.
My question then is, what is the uphill slope at which the pencil will start rolling back down instead of coming to a stop? :)
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
There must be some energy loss for the pencil to stop. In an ideal setting (no losses from friction, air resistance, etc. and no outside inputs after the initial impulse), I'm not seeing where this energy would go.
I think you also have to assume non-elasticity (iow. the prism doesn't bounce when it hits the surface). An elastic collision could make a difference. (When the next edge of the prism hits the surface, it would bounce back, making the prism to jump, after which... it becomes complicated.)
My question then is, what is the uphill slope at which the pencil will start rolling back down instead of coming to a stop? :)
Wouldn't that depend on the initial orientation of the prism and the initial force?
I'm assuming the prism is initially aligned so each point is at a constant elevation, one face is flat against the surface, and the impulse is straight up the slope with exactly enough force to cause a single revolution.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
For this problem, k = 2,24, and I found the times:
(1): 4T' = 6,20 years
(2): 4T = 11,95 years
At least for me, this looks coherent :P
