Relativistic Rocket Question [Archive]

pervect

Jun8-06, 02:01 PM

If I'm on a rocket accelerating away from earth, my velocity relative to earth at any given time will be: v = \frac{at}{\sqrt{1+(\frac{at}{c})^2}}

This makes sense, since my velocity relative to earth will be less than the newtonian \Delta v = a \Delta t , due to relativistic effects.

But how about later, if I decelerate relative to earth? Obviously, I can't use the same equation, since my change in velocity relative to earth will now be greater than \Delta v = a \Delta t.

Is there a similar equation I can use to calculate my velocity relative to earth at any given time during deceleration?

Thanks,
Alan

The easiest approach, assuming a constant deacceleration, is to pick the point at which you will stop as a reference.

Let (tstop, xstop) be the coordinates where you come to a stop.

Then (correction) v(t) = -a(t-tstop)/sqrt(1+(a^2(t-tstop)^2)

Thus at t=tstop, v=0
and at t < tstop, v > 0

You are basically time reversing the problem, and accelerating away from your stopping point.

There are a few things about unformly accelerated notion that are extremely useful to know:

1) all such motion is hyperbolic, i.e for the correct choice of origin x^2 - t^2 = constant
(x^2 - c^2t^2 = constant if c is not equal to 1).

for example: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

(d + c^2/a) = (c^2/a) cosh(aT/c)
t = (c/a) sinh(aT/c)

thus (d+c^2/a)^2 - (ct)^2 = (c^2/a)^2

as cosh^2 - sinh^2 =1

[add]
See also, for instance
http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

2) the "lines of simultaneity" for a uniformly accelerated observer all pass through one point (the origin of the coordinate system, if the choice of the origin is made as in #1)

i.e. in the example above, all liens of simultaneity run through the point
x = -c^2/a; t=0

[add] Here is a proof of the later remark: assume c=1, then

x^2 - t^2 = constant

By taking the derivative, we get
2*x*dx - 2*t*dt = 0

therfore v = dx/dt = t/x. The slope of the line of simultaneity is 1/v, i.e. x/t. Thus a line from (0,0) to (x,t) is the line of simultaneity at (x,t) because it passes through (x,t) and it has the correct slope: dx/dt = x/t.

pervect

Jun8-06, 04:56 PM

I've corrected one typo and added some proofs and more references to my previous post, the added content is significant enough that I thought I'd make a note of it.

Al68

Jun9-06, 05:07 PM

What I was really looking for here was an equation for the ship's deceleration relative to earth's apparent position. This would obviously not be the same as the ship's deceleration relative to its destination.

Thanks,
Alan

pervect

Jun9-06, 05:35 PM

The deacceleration relative to Earth will be the same as the deacceleration relative to its destination, as long as inertial coordinate systems are used.

It's a simple space translation.

Suppose the destination is a distance L away from the Earth.

Then the coordiantes relative to the earth will be (t,x) and the coordinates relative to the destination will be (t,x-L).

Thus when x=L, the ship is at its destination. The acceleration of the ship (d^2x/dt^2) will be the same relative to the earth as it will be to the destination, because

d^2(x-L)/dt^2 = d^2x/dt^2

given that L=constant.

This simple "space translation" works the same in SR as it does in non-relativistic mechanics.

Time translation also works in the same way.

Note that we can also formulate the equations of motion using the principle of hyperbolic motion that I mentioned.

Let the destinaton be at (tstop, xstop), and let c=1.

then for x < (xstop/2), the hyperbolic point is at (t=0,x=-1/a) and

(x+1/a)^2 - t^2 = 1/a^2

and for x > xstop/2, the hyperbolic point is at (t=tstop, x=xstop+1/a)

(x-(xstop + 1/a))^2 - (t-tstop)^2 = 1/a^2

We can re-write the second expression to solve for x (using Maple). There are two solutions, one of which is:

{\frac {{\it xstop}\,a+1-\sqrt {1+{a}^{2}{t}^{2}-2\,{a}^{2}t{\it tstop
}+{a}^{2}{{\it tstop}}^{2}}}{a}}

the derivative of this expression is (agian using Maple)

{\frac {a \left( t-{\it tstop} \right) }{\sqrt {1+{a}^{2}{t}^{2}-2\,{a
}^{2}t{\it tstop}+{a}^{2}{{\it tstop}}^{2}}}}

the same as the expresssion I posted earlier.

See the Wikipedia link for the hyperbolic equation (you may have to manually add in the trailing ')' to the URL, for some reason the board omits this symbol and wiki doesn't find the page.

http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

You can confirm that the velocity at t=tstop/2 is the same in both expressions:

Al68

Jun9-06, 05:59 PM

pervect,

If the distance L refers to the distance between earth and the destination, according to the ship's frame, then L will not remain constant. It will get larger as the ship decelerates. Just like L got smaller as the ship accelerated away from earth. My understanding is that the relativistic rocket equation takes this into account, which is why the final velocity relative to earth (after acceleration) is less than it would be if we didn't take length contraction into account, assuming a specified acceleration and time. In other words, velocity is less than a*t.

So, is there an equation that relates to a ship's deceleration relative to earth's apparent position, that takes this length expansion (or de-contraction) into account?

Thanks,
Alan

pervect

Jun9-06, 06:08 PM

The coordinates (x,t) are the inertial coordinates of the ship in the Earth frame.

The destination is a constant distance away in the Earth coordiantes.

Look for instance at

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

(I assume this is where you initally got your equations?)

The proper time as measured by the crew of the rocket (i.e. how much they age) will be denoted by T, and the time as measured in the non-accelerating frame of reference in which they started (e.g. Earth) will be denoted by t. We assume that the stars are essentially at rest in this frame. The distance covered as measured in this frame of reference will be denoted by d and the final speed v. The time dilation or length contraction factor at any instant is the gamma factor γ.

Thus d and t are in the Earth coordinate system. T is in the rocket's coordinate system (proper time). The distance to the destination according to the rocket is not given on the webpage.

JesseM

Jun9-06, 06:13 PM

If the distance L refers to the distance between earth and the destination, according to the ship's frame, then L will not remain constant. It will get larger as the ship decelerates. Just like L got smaller as the ship accelerated away from earth. The distance L is measured in an inertial frame, the frame where the rocket was at rest until it began to accelerate (the rest frame of the earth). You seem to be thinking it's the distance as seen in some non-inertial frame of the ship, or in the ship's instantaneous co-moving inertial rest frame, but that's not correct, the relativistic rocket equation describes the velocity as seen in a single inertial frame. My understanding is that the relativistic rocket equation takes this into account, which is why the final velocity relative to earth (after acceleration) is less than it would be if we didn't take length contraction into account, assuming a specified acceleration and time. In other words, velocity is less than a*t. No, the reason it's less as seen in an inertial frame is because "a" refers to the acceleration experienced by passengers on the ship--constant "a" means that they feel a constant G-force, which means that their acceleration in their instantaneous co-moving inertial rest frame is the same from one moment to another. Because of the way velocities transform in relativity, this would mean that the rate their coordinate velocity increases (their coordinate acceleration, which is different from 'a') as seen in a single inertial frame is constantly decreasing. So, is there an equation that relates to a ship's deceleration relative to earth's apparent position, that takes this length expansion (or de-contraction) into account? What do you mean by "apparent"? Do you mean the distance as seen in the ship's instantaneous co-moving inertial frame?

Al68

Jun10-06, 03:03 PM

OK, then how would the ship's crew calculate their change in velocity? I assume they would have to take into account that as they accelerate, gamma would increase, and maybe we could picture the change in earth's apparent position as a kind of velocity, in the same direction of the ship's velocity. Since the distance from earth, as measured by the ship is increasing due to acceleration, but increasing less due to length contraction.

So, is there an equation that relates to a ship's deceleration relative to earth's apparent position, that takes this length expansion (or de-contraction) into account?
What do you mean by "apparent"? Do you mean the distance as seen in the ship's instantaneous co-moving inertial frame?

Yes, this is what I meant. I'm looking for how to calculate everything from the ship's point of view. And I know the ship's crew cannot observe anything in real time, but they should be able to figure out their velocity, (coordinate) acceleration, and (apparent) distance from earth at any given time. Including during their deceleration. And I'm looking for how they would calculate this, not how they would observe it directly.

Thanks,
Alan

JesseM

Jun10-06, 04:23 PM

OK, then how would the ship's crew calculate their change in velocity? "Velocity" in what coordinate system? If you want to look at a non-inertial coordinate system where they are at rest, then of course their velocity is always zero in this system, although you could calculate the way that the earth's velocity is changing in this coordinate system. But as I've said before, it's not like there's a single non-inertial coordinate system that counts as the "frame" of a non-inertial observer, you could construct such a coordinate system in a variety of ways, I don't think there's any strong physical motivation for preferring one over all the others. What do you mean by "apparent"? Do you mean the distance as seen in the ship's instantaneous co-moving inertial frame? Yes, this is what I meant. I'm looking for how to calculate everything from the ship's point of view. Again, I think you're confusing the issue by using commonsense words like "observe" and "point of view" and "apparent distance" for what is measured in a particular non-inertial coordinate system. Do you think there is any reason to say that this particular type of non-inertial coordinate system--one where the ship's definitions of distance and simultaneity at any given moment match those of their instantaneous co-moving inertial rest frame--represent's the ship's "point of view" in a way that some other non-inertial coordinate system with different definitions of simultaneity and distance does not? And I know the ship's crew cannot observe anything in real time, but they should be able to figure out their velocity, (coordinate) acceleration, and (apparent) distance from earth at any given time. Including during their deceleration. And I'm looking for how they would calculate this, not how they would observe it directly. You only put (coordinate) before acceleration, but you really should put it before velocity and distance from earth as well; what you are really asking is for the value of these things in a particular non-inertial coordinate system (one whose definition of distance and simultaneity at each moment matches that of the ship's instantaneous co-moving inertial rest frame), but there is no reason to say that the coordinate distance of the earth in this non-inertial frame has any more right to be called the "apparent distance" than the coordinate distance in some other frame.

Anyway, like I said earlier, in the coordinate system that I assume you're asking about, their own coordinate velocity and acceleration would be zero, because they're at rest at all moments in this coordinate system. But you can certainly ask about the coordinate velocity, acceleration and position of the earth in this coordinate system. If you can figure out the coordinate position of the earth as a function of time in this system, you could obtain the others in the normal way, by taking derivatives with respect to coordinate time. The calculation of distance as a function of time in this coordinate system seems like it'd be a bit tricky, although the formulas on the relativistic rocket page (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) help...I'll think about it and get back to you.

pervect

Jun10-06, 04:46 PM

To compute or measure in the rockets coordinate system the same velocity that the Earth observer measures, the rocket needs to measure its velocity relative to an object that is stationary with respect to the Earth, but at the same location as the rocket.

"Stationary with respect to the Earth" is really defined in the Earth's frame of referece, not the rockets frame, however. There is a way to do this from the rocket's coordiante system, but it is complicated. It also requires making use of special features of the geomtery. In another post, I think I gave a longish quote about why there is no general notion of "relative velocity" of a distant object, so it should not be a surprise that finding the relative velocity of the rocket to the Earth requires us to make use of some "special features" of the geometry (namely, the fact that space-time is flat in this example).

The advanced way of doing this from the rocket's coordinate system: the rocket "parallel transports" the velocity vector of the Earth along some path to the rocket's current location. The rocket then finds the magnitude of the resulting "parallel transported" velocity to answer the question.

In general, this procedure of parallel transporting would give a result that depended on the exact path over which the vector was transported. But, becuase space-time in this example is flat, this process of parallel transport will give the same results, regardless of the path taken.

To avoid the process of parallel transport, the rocket could make use of (for example) an "Earth outpost" that knows it is stationary with respect to the Earth because the outpost has been exchanging light signals with the Earth for a long time, without noticing any change in the "propagation delay".

This "Earth outpost" appraoch is much simpler, though I don't know if it will satisfy Al68.

If the rocket measures its velocity V with respect to the Earths' outpost, it will measure the same velocity V that the Earth (and the Earth's outpost) measures for the rocket.

A full treatment of parallel transport requires some knowledge of tensors, which I do not believe that AL68 has.

A somewhat popular reference:

http://math.ucr.edu/home/baez/gr/parallel.transport.html

Note that when you parallel transport a vector around a closed path on the Earth's surface, as in the above example, it changes direction.

It is only when the geometry is fundamentally flat that parallel transporting a vector around a closed path does not change it's direction.

Because parallel transporting a vector around a closed path on a flat plane does not change its direction, the result of parallel transporting a vector along a path cannot depend on the path.

Schild's ladder would be an ideal way of explaining parallel transport, but the only reference I have which talks about it is MTW's "Gravitation", and I haven't found much on the WWW, either.

"Gravitation" is a book that is written at the graduate level, but has largish bits that may be comprehensible to someone who is willing to ignore the rest of the book (which won't be comprehensible without graduate level physics).

Al68

Jun10-06, 05:48 PM

Jesse,

I think your last paragraph states my question more clearly than I did. I wanted to figure out the coordinate position of earth as a function of time in the (non inertial) system in which the ship is at rest. And I agree with you that it would be tricky. That's why I was asking for help.

pervect,

You are correct about my knowledge of tensors. I can barely spell tensors. And I worded my question poorly, Jesse worded it better in his last paragraph.

Thanks,
Alan

JesseM

Jun10-06, 07:33 PM

Alan, if I'm understanding you right you want to know how things would look in a particular noninertial coordinate system (call it NI) which has the following properties:

1. If a given tick of the ship's clock is simultaneous with a distant event A in the co-moving inertial rest frame of the ship at the moment of the tick, then A should also be simultaneous with that tick in NI. Also, if D is the distance between the ship at the moment of the tick and A in the co-moving inertial rest frame of the ship at that moment, then the coordinate distance between A and the ship at the moment of the tick should also be D in NI.

2. The time-coordinate assigned to an event on the ship's worldine by NI should be identical to the readings of clocks on the ship at the moment of the event (ie the proper time along the ship's worldine between that event and the event of leaving earth).

I think these conditions should be enough to uniquely specify the NI coordinate system. Note, though, that this will not be a well-behaved coordinate system throughout spacetime--as the lines of simultaneity change angles at different points along the ship's worldline, then there will be points where different lines of simultaneity cross each other, and events at or beyond the crossings will be assigned two or more sets of coordinates by NI. But as long as you consider only the regions of spacetime near enough to the ship's worldline that events within those regions are assigned unique coordinates, it should be OK (although I don't know how to calculate the borders of this region).

Nevertheless, I would say that you are incorrect to refer to this as "the (non inertial) system in which the ship is at rest"--as I keep emphasizing, over and over, and you seem to want to ignore, there is no single standard way to define the coordinate sytem of a non-inertial observer, you could certainly come up with other non-inertial coordinate systems in which the ship is at rest that do not have the properties I described for NI above, and there would be no compelling reason to treat NI as having more claim to represent the ship's "point of view" than any of these others. Please tell me, do you understand and agree with this? (and if so, will you quit using phrases like 'apparent distance' and 'the ship's point of view' to describe how things work in the NI coordinate system?)

Anyway, to figure out the earth's distance D as a function of time T within the coordinate system NI, I think the first thing to do would be to figure out D as a function of d and v, the distance and velocity of the ship in the earth's frame. You might think that D would just be d * \sqrt{1 - v^2/c^2} due to Lorentz contraction, but that isn't right, the distance isn't like the length of a rigid object, it's constantly changing so you have to worry about simultaneity issues. (edit: I was wrong about that, see pervect's next post and my response...the distance is like the length of a rigid object, just imagine that there was a rod attached to the earth of length d in the earth's rest frame, and when the ship reaches the end of it, both frames agree that at that moment the length of the rod in their frame is equal to the ship's distance from the earth in their frame.) So let's consider a simple inertial case where the ship is moving away from the earth at constant velocity v instead of accelerating. In the earth's frame, the ship will be at distance d at time d/v, and since in this frame the ship's clocks are running slow by a factor of \sqrt{1 - v^2/c^2}, that means the ship's clock will only read (d/v)*\sqrt{1 - v^2/c^2} at this moment. But in the ship's frame, at the moment its own clock reads this time of (d/v)*\sqrt{1 - v^2/c^2}, it is the earth's clock that reads less than this because it has been slow by a factor of \sqrt{1 - v^2/c^2} since they departed, so according to the definition of simultaneity in the ship's frame the earth's clock reads (d/v)*(1 - v^2/c^2) at this moment. And since in the earth's frame the earth is also moving away at the same speed of v (the speeds with which two inertial observers see the other moving are always reciprocal), the earth's distance at this moment can be found by multiplying velocity and time, giving v*(d/v)*(1 - v^2/c^2), which gives d*(1 - v^2/c^2). So, at the point on the ship's worldline where the distance in the earth's frame is d, the distance in the ship's frame is d*(1 - v^2/c^2).

Now, this conclusion shouldn't be altered if we are talking about the instantaneous co-moving inertial rest frame of an accelerating observer rather than the constant inertial rest frame of an inertial ship. After all, we could imagine that we have both an inertial ship and an accelerating ship moving away from the earth, and that they cross each other's paths at the moment that they are at a distance d in the earth's frame (they must have departed earth at different times for this to work); at the point in spacetime where they cross, they both have the same instantaneous inertial rest frame, so they would both measure the same distance to the earth in this instantaneous frame.

So, we know that at the point on the ship's worldline that corresponds to time t in the earth's frame, when the distance in the earth's frame is d, the NI system should say the distance of the earth at that same point on the ship's worldline is d*(1 - v^2/c^2). And from the relativistic rocket page (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) we have an expression for d as a function of a and t: d = (c^2/a)*(\sqrt{1 + (at/c)^2} - 1). So this means that the distance D in NI as a function of time t in the earth's frame is D = (1 - v^2/c^2)*(c^2/a)*(\sqrt{1 + (at/c)^2} - 1). And the page also gives us a function for the time t as measured in the earth's frame as a function of the time T as measured by clocks on the ship: t = (c/a)*sh(aT/c), where the function sh x is the hyperbolic sine function (also called sinh -- you can find it on most calculators). So, substituting this into the formula for D as a function of t gives D as a function of T:

D = (1 - v^2/c^2)*(c^2/a)*(\sqrt{1 + (a/c)^2 * ((c/a)*sh(aT/c))^2} - 1)

simplifying a bit,

D = ((c^2 - v^2)/a)*(\sqrt{1 + (sh(aT/c))^2} - 1)

and this page (http://www.geocities.com/ResearchTriangle/2363/hyperbol.html) mentions there's a hyperbolic function identity that says (ch(x))^2 = 1 + (sh(x))^2, so this simplifies to:

D = ((c^2 - v^2)/a)*(ch(aT/c) - 1)

There's a good chance I made an error somewhere along the line, but if not, this should be the formula for the coordinate distance of the earth D as a function of coordinate time T in the NI coordinate system. To find the coordinate velocity of earth you'd take the first derivative of D with respect to T, while to find the coordinate acceleration of earth you'd take the second derivative. The page with the hyperbolic function identities above says that d sh(x) / dx = ch(x) and d ch(x) / dx = sh(x). So, taking these derivatives is just a matter of using the chain rule (http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html) of calculus:

dD/dT = ((c^2 - v^2)/a)*(sh(aT/c) * (a/c))

or

dD/dT = ((c^2 - v^2)/c)*sh(aT/c)

Taking another derivative gives:

d^2 D /dT^2 = ((c^2 - v^2)/c)*(ch(aT/c) * (a/c))

or

d^2 D /dT^2 = (a*(c^2 - v^2)/c^2)*ch(aT/c)

Again, buyer beware, but if I haven't made any errors these should be the formulas for the coordinate velocity and coordinate acceleration of the earth in the NI coordinate system.

pervect

Jun10-06, 08:49 PM

I'm think that the distance in the ship's coordinate system should be just sqrt(1-v^2/c^2) times the distance in the Earth's coordinate system.

Basically, the angle betweent the line of simultaneity and the horizontal 'x' axis is just the rapidity, theta, therfore the length in the Earth frame divided by the length in the instantaneous ship frame is just cosh(theta).

Another way of putting this: letting c=1

The slope of the "line of simultaneity" is 1/v , thus delta-t = v*delta-x

If the Earth is at the origin of the coordinate system and the spaceship is at (t,d) in the Earth coordinate system

The point on the Earth's trajectory (t=*, x=0) simultaneous with (t,x) in the ship frame is just

(t-v*d,0)

The lorentz interval betweent these points is d^2 - (v*d)^2

The distance is the square root of the Lorentz interval, i.e d*sqrt(1-v^2)

The following diagram might have helped, if it came out :-(. (I was trying gnuplot this time around). I'll keep it, in case I can get it to work.

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JesseM

Jun10-06, 09:13 PM

I'm think that the distance in the ship's coordinate system should be just sqrt(1-v^2/c^2) times the distance in the Earth's coordinate system. Yes, I think you're correct. I confused myself by thinking about what the earth's clock would read in the ship's frame at the moment that the ship's clock reads (d/v)*\sqrt{1 - v^2/c^2}, but what's important is the time in the ship's own frame at this moment--since in its frame the earth has been moving at velocity v for a time (d/v)*\sqrt{1 - v^2/c^2}, at this time the earth must be a distance v*(d/v)*\sqrt{1 - v^2/c^2} in this frame, or d*\sqrt{1 - v^2/c^2}. So a corrected version of my argument after that point would look like this: So, we know that at the point on the ship's worldline that corresponds to time t in the earth's frame, when the distance in the earth's frame is d, the NI system should say the distance of the earth at that same point on the ship's worldline is d*\sqrt{1 - v^2/c^2}. And from the relativistic rocket page (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) we have an expression for d as a function of a and t: d = (c^2/a)*(\sqrt{1 + (at/c)^2} - 1). So this means that the distance D in NI as a function of time t in the earth's frame is D =\sqrt{1 - v^2/c^2}*(c^2/a)*(\sqrt{1 + (at/c)^2} - 1). And the page also gives us a function for the time t as measured in the earth's frame as a function of the time T as measured by clocks on the ship: t = (c/a)*sh(aT/c), where the function sh x is the hyperbolic sine function (also called sinh -- you can find it on most calculators). So, substituting this into the formula for D as a function of t gives D as a function of T:

D = \sqrt{1 - v^2/c^2}*(c^2/a)*(\sqrt{1 + (a/c)^2 * ((c/a)*sh(aT/c))^2} - 1)

simplifying a bit,

D = \sqrt{1 - v^2/c^2}*(c^2/a)*(\sqrt{1 + (sh(aT/c))^2} - 1)

and this page (http://www.geocities.com/ResearchTriangle/2363/hyperbol.html) mentions there's a hyperbolic function identity that says (ch(x))^2 = 1 + (sh(x))^2, so this simplifies to:

D = \sqrt{1 - v^2/c^2}*(c^2/a)*(ch(aT/c) - 1)

There's a good chance I made an error somewhere along the line, but if not, this should be the formula for the coordinate distance of the earth D as a function of coordinate time T in the NI coordinate system. To find the coordinate velocity of earth you'd take the first derivative of D with respect to T, while to find the coordinate acceleration of earth you'd take the second derivative. The page with the hyperbolic function identities above says that d sh(x) / dx = ch(x) and d ch(x) / dx = sh(x). So, taking these derivatives is just a matter of using the chain rule (http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html) of calculus:

dD/dT = \sqrt{1 - v^2/c^2}*(c^2/a)*(sh(aT/c) * (a/c))

or

dD/dT = (c*\sqrt{1 - v^2/c^2})*sh(aT/c)

Taking another derivative gives:

d^2 D /dT^2 = (c*\sqrt{1 - v^2/c^2})*(ch(aT/c) * (a/c))

or

d^2 D /dT^2 = (a*\sqrt{1 - v^2/c^2})*ch(aT/c)

Again, buyer beware, but if I haven't made any errors these should be the formulas for the coordinate velocity and coordinate acceleration of the earth in the NI coordinate system.

Al68

Jun11-06, 02:14 AM

Jesse,

Thanks for all your work. I have to say, I liked your first answer. It fit in with what I was thinking. But I'd better reread and absorb everything better before I ask any more questions.

pervect, I appreciate your help, too. I just don't have the math skills to use some of it.

Thanks,
Alan

Al68

Jun11-06, 06:12 PM

So, we know that at the point on the ship's worldline that corresponds to time t in the earth's frame, when the distance in the earth's frame is d, the NI system should say the distance of the earth at that same point on the ship's worldline is d*\sqrt{1 - v^2/c^2}.

This makes sense to figure out D as a function of d at any moment. But if we were to have an event, like a signal sent out by the ship's crew, this signal would not be simultaneous in both frames. Is it safe to say, that if we wanted to figure out D, d, T, and t for this event, that your earlier post would be correct? At this event, the earth's clock would read (d/v)*(1 - v^2/c^2), and the rest of your earlier post would apply?

Thanks,
Alan

JesseM

Jun11-06, 06:28 PM

This makes sense to figure out D as a function of d at any moment. But if we were to have an event, like a signal sent out by the ship's crew, this signal would not be simultaneous in both frames. Simultaneous with what? It only makes sense to ask if two events are simultaneous within a single frame, you can't really ask whether a single event is simultaneous in two different frames (you can ask if it occurs at the same time-coordinate in two coordinate systems, but this is different, and it depends on the arbitrary choice of where you put the origin of each coordinate system, not just their relative velocity). Is it safe to say, that if we wanted to figure out D, d, T, and t for this event, that your earlier post would be correct? At this event, the earth's clock would read (d/v)*(1 - v^2/c^2), and the rest of your earlier post would apply? The earth's clock would read (d/v)*(1 - v^2/c^2) at the time of this event in the ship's frame (but you shouldn't say 'At this event, the earth's clock would read...' because an 'event' in relativity refers only to a single point in spacetime, so anything happening elsewhere like on earth would not be 'at this event'). But my calculation of the distance D in the ship's frame was wrong, the distance would be the time of the event according to the ship's clock (ie (d/v)*\sqrt{1 - v^2/c^2}) multiplied by v, since in the ship's frame the ship's own clock is ticking at the normal rate while the earth is moving away at velocity v. Thus the distance D would be v*(d/v)*\sqrt{1 - v^2/c^2}, or d*\sqrt{1 - v^2/c^2}.

To make this more clear, think in terms of the example I mentioned in the edit to that earlier post--suppose there was a rod attached to the earth, of length d in the earth's frame. If we look at the event of the ship passing next to the far end of the rod, then obviously in the earth's frame, the distance between that event and an event happening "simultaneously" on earth (like the event of the earth's clock ticking d/v) would be d, the length of the rod. So if you now look at the same event of the ship passing the end of the rod in the ship's frame, and you want to calculate the distance between this event and an event on earth which is simultaneous with it in the ship's frame (in this case, the event of the earth's clock ticking (d/v)*(1 - v^2/c^2)), then the distance would just be equal to the length of the rod in the ship's frame, or d*\sqrt{1 - v^2/c^2} due to Lorentz contraction. This would be the distance of the earth at the moment the ship passes the end of the rod in the ship's frame.

Al68

Jun11-06, 08:21 PM

JesseM

Jun11-06, 10:47 PM

OK, if the event of the ship sending out the signal would be simultaneous with the earth's clock ticking d/v, in earth's frame, wouldn't this signal also be simultaneous with the event of a rod attached to the ship (length d in earth's frame) reaching earth? In the earth's frame, yes. But these events would not be simultaneous in the ship's frame. And the length of this rod would be d/\sqrt{1 - v^2/c^2} in the ship's frame? That's correct, but again, in the ship's frame the event of the ship reaching the end of the earth's rod and sending out a signal would not be simultaneous with the event of the far end of its own rod passing earth. After all, you agree that in the ship's frame the earth's rod only has length d*\sqrt{1 - v^2/c^2}, right? So as seen in the ship's frame, it must be a lot shorter than the ship's own rod, so at the moment that the ends of the two rods closest to the ship are next to each other, the far ends can't possibly be both next to earth (and we know that the end of the earth's rod is actually attached to the earth, so naturally when the ship passes next to the far end it will say the other end is next to earth, since that end is always next to earth at a given moment in any frame).

If this is still unclear, you might try trying to identify the coordinates of the event of the ship passing next to the end of the earth's rod, and the event of the earth passing next to the end of the ship's rod, as seen in the earth's coordinate system, then use the Lorentz transform to see when these events happen in the ship's coordinate system (and if you have trouble with this I could help out).

Al68

Jun12-06, 12:33 AM

JesseM

Jun12-06, 01:47 AM

OK, but if the ship is a distance D from earth (in the ship's frame) when the ship's crew sends the signal, can't we just as easily say that d = D*\sqrt{1 - v^2/c^2} at the time of the signal (in earth's frame), since this distance would be length contracted in earth's frame? No, because if the ship had a rod of length D attached to it, and sent a signal when the far end of its rod passed the earth in its own frame, in the earth's frame the event of the far end of the rod passing it and the event of the ship sending the signal wouldn't be simultaneous. So although it's true that the earth will see the length of the ship's rod as D*\sqrt{1 - v^2/c^2}, that doesn't mean it will judge this to be the distance the ship was when it sent the signal.

If you're worried about reciprocality, it's reciprocal in this sense: if they both want to figure out the distance between the earth and the ship at the moment of an event on the ship's worldline (like the ship sending a signal), then the distance will be shorter in the ship's frame. But if they both want to figure out the distance between the earth and the ship at the moment of an event on the earth's worldline, then the distance will be shorter in the earth's frame. To see this, suppose we have a rod of length D attached to the ship, and the earth sends out a signal at the moment it passes the far end of this rod. Both frames agree that the ship was at the other end of the rod at the moment the far end passed the earth, so in each frame the distance between the earth and the ship at the moment of this event will just be the length of the rod in that frame, which means in the earth's frame the distance must be D*\sqrt{1 - v^2/c^2}.

Another way of thinking about it: if they both want to know the distance between the earth and the ship "at the same moment" as an event on the ship's worldline, their two rest frames will give different answers to which event on the earth's worldline is simultaneous with that event; but if they both want to know the distance between the earth and the ship "at the same moment" as an event on the earth's worldline, their two rest frames will give different answers to which event on the ship's worldline is simultaneous with that event.

Al68

Jun12-06, 02:34 PM

If you're worried about reciprocality, it's reciprocal in this sense: if they both want to figure out the distance between the earth and the ship at the moment of an event on the ship's worldline (like the ship sending a signal), then the distance will be shorter in the ship's frame. But if they both want to figure out the distance between the earth and the ship at the moment of an event on the earth's worldline, then the distance will be shorter in the earth's frame. To see this, suppose we have a rod of length D attached to the ship, and the earth sends out a signal at the moment it passes the far end of this rod. Both frames agree that the ship was at the other end of the rod at the moment the far end passed the earth, so in each frame the distance between the earth and the ship at the moment of this event will just be the length of the rod in that frame, which means in the earth's frame the distance must be D*\sqrt{1 - v^2/c^2}.

OK, and for this example, the earth's signal would be simultaneous (in the ship's frame) with the ship reaching the end of an earth-rod with a length of D in the ship's frame. This earth-rod would be D/\sqrt{1 - v^2/c^2} in earth's frame, but the event of the ship reaching the end of the earth-rod would not be simultaneous with earth's signal in earth's frame.

So if the ship were to send out a signal when it reached the end of the earth rod, and when the end of the ship-rod reached earth, in the ship's frame, this signal would not be simultaneous with the ship reaching the end of the earth-rod in earth's frame. In earth's frame, the ship's signal would occur prior to the ship reaching the end of the earth-rod (length D/\sqrt{1 - v^2/c^2}) in earth's frame. Is this correct?

Thanks,
Alan

JesseM

Jun12-06, 02:48 PM

Al68

Jun12-06, 04:30 PM

OK, and for this example, the earth's signal would be simultaneous (in the ship's frame) with the ship reaching the end of an earth-rod with a length of D in the ship's frame. This earth-rod would be be D/\sqrt{1 - v^2/c^2} in earth's frame, but the event of the ship reaching the end of the earth-rod would not be simultaneous with earth's signal in earth's frame.
Yup, exactly.
So if the ship were to send out a signal when it reached the end of the earth rod, and when the end of the ship-rod reached earth, in the ship's frame, this signal would not be simultaneous with the ship reaching the end of the earth-rod in earth's frame.
In the ship's frame, both rods have length D, so the event of the ship reaching the end of the earth's rod and the event of the earth reaching the end of the ship's rod would be simultaneous. Your question is unclear, are asking whether the signal from the ship in the ship's frame would be simultaneous with the signal from the earth in the earth's frame?
No, I was saying that this ship's signal (when it reached the end of the earth-rod) would be received by earth, but not simultaneously with the ship reaching the end of the earth-rod in earth's frame. In earth's frame, the event of this ship's signal and the event of the ship reaching the end of the earth-rod are not simultaneous. Since the event of the ship reaching the end of the earth rod and the event of the end of the ship-rod reaching earth are simultaneous in the ship's frame, these events cannot be simultaneous in earth's frame.

In earth's frame, the ship's signal would occur prior to the ship reaching the end of the earth-rod (length D/\sqrt{1 - v^2/c^2}) in earth's frame. Is this correct?
My understanding of your scenario was that the ship sends the signal at the time and the place where it reaches the end of the earth's rod--in that case, all frames must agree these two events have the same coordinates, if two events coincided in one frame but not in another that would represent contradictory physical predictions. Imagine the end of the earth's rod was set to explode at the moment the ship passed it--it isn't possible for some frames to say to ship managed to send a signal before it was blown up while others say it was blown up before it could send the signal, that would be an obvious physical contradiction. Similarly, although different frames can disagree on the ages of the twins in the twins paradox as long as they apart, if they reunite at a single time and place all frames must agree on what their clocks read at that point in spacetime.

Why would all frames have to agree that these two events (the ship's signal and the ship reaching the end of the earth rod) would have the same coordinates? I believe you stated earlier that if the event of the earth's clock ticking d/v and the event of a rod attached to the ship (length d in earth's frame) reaching earth were simultaneous in earth's frame, these events would not be simultaneous in the ship's frame. If this is correct, then, in my scenario, the event of the ship's clock reading D/v (and sending a signal) and the event of the ship reaching the end of the earth-rod would be simultaneous in the ship's frame, but not in earth's frame.

I agree this will result in contradictory physical predictions, or paradox so to speak. That's why I'm asking so many questions. I would also note that if we really had rigid rods, earth could observe the ship's position in real time, by observing the ship-rod, and this would clearly violate the laws of physics. This was pointed out to me in an earlier topic. This is why I thought you were right when you said we could not treat distance like the length of a rigid rod.

(edit) Just wanted to add something here. What if, in the Twins Paradox, the destination star explodes when the ship reaches it (in the ship's frame), and destroys the ship. The event of the star's explosion and the event of the ship reaching the star would not be simultaneous in earth's frame. Would the ship be destroyed in one frame but not the other?

Thanks,
Alan

JesseM

Jun12-06, 05:52 PM

No, I was saying that this ship's signal (when it reached the end of the earth-rod) would be received by earth, but not simultaneously with the ship reaching the end of the earth-rod in earth's frame. Oh, I didn't realize you were talking about when the signal would be received, you'd never brought that up before. In earth's frame, the event of this ship's signal Please, if you're going to talk about both the event of the signal being sent and the separate event of the signal being received, you can't use phrases like "the event of this ship's signal" or I won't know which one you're talking about. An event is something that happens at a single point in space and time, so transmission and reception are clearly separate events. and the event of the ship reaching the end of the earth-rod are not simultaneous. In this case, were you talking about the event of the signal being received on earth not being simultaneous with the event of the signal being sent by the ship (which is at the same point in space and time that it reaches the end of the earth-rod)? Since the signal can't travel instantaneously fast but only at the finite speed of c, all frames must agree it took some time to get from the end of the rod with the ship to the other end where the earth is, so all frames will agree the event of reception happened at a later time than the event of transmission. Since the event of the ship reaching the end of the earth rod and the event of the end of the ship-rod reaching earth are simultaneous in the ship's frame, these events cannot be simultaneous in earth's frame. That has nothing to do with the fact that the event of the earth receiving the signal (if that is indeed the event you were referring to above) happens later than the event of the ship sending the signal. Even if the ship's rod were a different length so that the event of the ship reaching the end of the earth's rod did happen at the same moment as the event of the earth reaching the end of the ship's rod in the earth's frame, it would still be true that the event of the earth receiving the ship's signal would happen at a later time than the event of the ship sending it. Why would all frames have to agree that these two events (the ship's signal and the ship reaching the end of the earth rod) would have the same coordinates? Events that happen at the same time and place in one frame must happen at the same time and place in all frames, this is just one of the basic assumptions you need to make when constructing coordinate systems in relativity, otherwise you'd get different physical predictions in different coordinate systems as I mentioned. I think it definitely helps if you think in terms of spacetime diagrams. As an analogy, suppose we're drawing line segments on an ordinary piece of paper, and then using two different sets of x and y axes, with one set's axes rotated with respect to another's, to assign x and y coordinates to the endpoints of different segments. If we have two line segments that meet at a single point, isn't it obvious that no matter which set of x and y axes we use, those two endpoints will be assigned the same x and y coordinates? It's just the same with coordinate systems in relativity, you're just using differently-oriented space and time axes to describe the same pattern of worldlines in spacetime. If each worldine has dots drawn on it to represent successive ticks of a clock moving on that worldline (or any other set of events along it), then if two worldlines cross each other in such a way that two dots coincide, the events represented by those dots must coincide in all coordinate systems, just like dots drawn at the endpoints of the two line segments I desribed would have to coincide in all coordinate systems if they coincided in one.

You might also want to check out a thread I started a long time ago called An illustration of relativity with rulers and clocks (http://www.physicsforums.com/showthread.php?t=59023), where I drew some diagrams of what would happen if you had two rulers moving at relativistic speed with clocks studded at regular intervals along their length--I made a point there of showing how even though they disagree about things like whose clocks are running slower or whose clocks are in sync, they always agree about the readings on a given pair of clocks from each ruler at the moment they pass next to each other. I believe you stated earlier that if the event of the earth's clock ticking d/v and the event of a rod attached to the ship (length d in earth's frame) reaching earth were simultaneous in earth's frame, these events would not be simultaneous in the ship's frame. Never, you must have misunderstood--that would lead to totally contradictory physical predictions in the two frames. I agree this will result in contradictory physical predictions, or paradox so to speak. That's why I'm asking so many questions. I would also note that if we really had rigid rods, earth could observe the ship's position in real time, by observing the ship-rod, and this would clearly violate the laws of physics. When I said "rigid rods", I didn't mean they would remain rigid if the ship accelerated, as that would indeed violate the laws of physics. But as long as the ship and the earth are moving inertially, the rods attached to them will appear "rigid" in the sense that their length is constant in each frame. That was all I meant by "rigid rods". (edit) Just wanted to add something here. What if, in the Twins Paradox, the destination star explodes when the ship reaches it (in the ship's frame), and destroys the ship. The event of the star's explosion and the event of the ship reaching the star would not be simultaneous in earth's frame. Yes, they certainly would! And less violently, if there is a clock at the star, then both frames will predict the same thing about what the star's clock will read and what the traveling twin's clock will read at the moment he reaches the star. To understand this you have to take into account different frames' definitions of simultaneity, as well as their different opinions on the rate that different clocks tick. For example, say the star's clock is synchronized with the earth's clock in their mutual rest frame. Then in this frame, the star's clock read zero at the moment the traveling twin left earth, when his clock also read zero; since his clock is running slow in this frame, they'll predict his clock will be behind the star's clock when they meet. But in the traveling twin's frame, the star's clock was not synchronized with the earth's clock at the moment he left, instead it was significantly ahead of the earth's clock. Thus, even though in his frame the star's clock is ticking slower than his, he'll agree in the prediction that the star's clock will be ahead of his when they meet, because it had that "head start". You might want to try actually plugging some numbers into a scenario like this, using the Lorentz transform to convince yourself that both frames make precisely the same prediction; if you like I could help you out with this.

Al68

Jun12-06, 06:08 PM

Jesse,

You're right, I misread your earlier post completely. That caused a chain reaction that led me to several mistakes before I realized it. I still have questions, but I'd better take my time with them before I post.

Thanks,
Alan

Al68

Jun12-06, 07:40 PM

JesseM

Jun12-06, 07:52 PM

If we say that the distance from the earth to the ship (in earth’s frame) is equal to the length of the ship-rod (in earth’s frame) when the end of the ship-rod reaches earth (in earth’s frame), can we possibly be correct? Since earth can measure the length of the ship-rod before the ship leaves earth, mark the rod along it’s length, and then just watch the ship-rod go by? Then earth could know the ship’s location and velocity in real time. And, of course the ship is not moving inertially in this case, it is accelerating. I was assuming a purely inertial situation, like a ship that moves inertially past the earth rather than one that takes off from earth after originally being at rest with respect to it. If the ship accelerates, the rod can't stay rigid, so in the earth frame the back end of the rod will not begin to accelerate with the ship until well after the ship and the front end begin to accelerate (this is also true in every other inertial frame)--the back end will only start to accelerate once a signal moving at the speed of sound in the rod has reached it from the front end, starting at the moment the front end began to accelerate. Only after all parts of the rod have stopped accelerating and the rod is moving inertially again will its length be constant in the earth's frame, and its new length will be shorter than its length before acceleration, because of Lorentz contraction. But like I said, it's easier just to think in terms of a ship that moves past the earth inertially without ever accelerating.

Al68

Jun12-06, 08:07 PM

I was assuming a purely inertial situation, like a ship that moves inertially past the earth rather than one that takes off from earth after originally being at rest with respect to it. If the ship accelerates, the rod can't stay rigid, so in the earth frame the back end of the rod will not begin to accelerate with the ship until well after the ship and the front end begin to accelerate--the back end will only start to accelerate once a signal moving at the speed of sound in the rod has reached it from the front end, starting at the moment the front end began to accelerate. Only after all parts of the rod have stopped accelerating and the rod is moving inertially again will its length be constant in the earth's frame, and its new length will be shorter than its length before acceleration, because of Lorentz contraction. But like I said, it's easier just to think in terms of a ship that moves past the earth inertially without ever accelerating.

It almost looks like, with an accelerating ship, you were right the first time, when you said that distance isn't like the length of a rigid object. Would this be correct?

Thanks,
Alan

JesseM

Jun12-06, 08:21 PM

It almost looks like, with an accelerating ship, you were right the first time, when you said that distance isn't like the length of a rigid object. Would this be correct? Once again, there is no such thing as "the" distance for an accelerating ship, because there is no single standard choice of coordinate systems for accelerating objects. If you use the specific type of coordinate system that I labeled "NI" in that long post from page 1, where the accelerating observer's definition of distance at any moment is defined to match that of their instantaneous inertial rest frame, then the argument I made in that post shows why we can figure out distance at any moment in the NI coordinate system by considering the distance in the rest frame of a passing inertial observer: Now, this conclusion shouldn't be altered if we are talking about the instantaneous co-moving inertial rest frame of an accelerating observer rather than the constant inertial rest frame of an inertial ship. After all, we could imagine that we have both an inertial ship and an accelerating ship moving away from the earth, and that they cross each other's paths at the moment that they are at a distance d in the earth's frame (they must have departed earth at different times for this to work); at the point in spacetime where they cross, they both have the same instantaneous inertial rest frame, so they would both measure the same distance to the earth in this instantaneous frame. Of course, the distance in the NI system would not match that of a physical ruler carried by the accelerating observer, because that ruler would be constantly deformed due to the acceleration and thus wouldn't be a good measure of distance at all. But this is why I keep trying to emphasize that there is no standard way for an accelerating observer to define things like distance, because there's no single obvious physical way to measure them like for the inertial observer, so your choice of which non-inertial coordinate system to use is fairly arbitrary with no reason to say one is a better representation of the accelerating observer's "point of view" than any other.

pervect

Jun12-06, 08:49 PM

But this is why I keep trying to emphasize that there is no standard way for an accelerating observer to define things like distance, because there's no single obvious physical way to measure them like for the inertial observer, so your choice of which non-inertial coordinate system to use is fairly arbitrary with no reason to say one is a better representation of the accelerating observer's "point of view" than any other.

I'm not sure how much this will help the discussion, but I can talk a little bit about the reasons the coordinate system commonly used for an accelerated observer is the coordinate system of an instantaneously co-moving observer.

Let's start with an object at a point. Through that point, there are an infinite number of vectors in 4-d space-time that could be consider the direction of "time".

When we settle on the velocity of an observer at that point, though, that picks out a unique vector that represents that particular's observer's concept of "time".

There will be a 3-dimensional space orthogonal to that time vector. This constitutes that particular observer's notion of "space".

The basis vectors that represent "space" at that point can be picked arbitrarily if one doesn't mind a rotating coordinate system. Rotating coordinage systems are sometimes annoying. To avoid this annoyance, there are procedures for picking out a "non-rotating" space vector. I won't go into the details at this point, I'll just mention that it's something that can be done with enough effort. Of course our choice is not really unique, we have a rotational degree of freedom in picking out our coordinate axes.

We've now set up a purely local coordinate system at that point. We now have to discusss how we extend that coordinate system to cover significant distances, rather than infinitesimal distances.

Relativity has a concept that's roughly similar to a "straight line" - it is called a geodesic.

The previous efforts have given us the information needed to construct the various time-like and space-like geodesics ("straight lines") passing through our initial point. These geodesics form the coordinate axes of our coordinate system. While we don't have to use relativity's notion of a "straight line" (i.e. a geodesic) to define how these coordinates propagate, it is a convenient and naturally choice, to make our axes the best possible equivalent to a "straight line".

The next issue comes up - how do we make "tic marks" on these coordinate axes, to represent equal intervals? We can use the concept of the Lorentz interval along the geodesic path to define where our "tic-marks" should be, such that distance between tic marks, along our geodesic paths, is uniform. This is also a rather convenient choice. (The technically minded might appreciate that this particular choice of tic-mark spacing makes the Christoffel symbol \Gamma^x_{xx} zero.

If we carry out this procedure in flat space-time, the above procedure will necessarily define as our coordinate system the coordinate system of an inertial observer at that point.

The above procedure is a bit more general than this, though, in that it can be carried out in curved space-times (where inertial frames do not really exist) as well as flat space-times (where inertial observers do exist).

So we can see that using the instantaneous inertial coordinates for the coordinate system of an accelerated observer is a rather "natural" choice. But it does have a limitation. Eventually, the different geodesic lines cross. In Euclidean geoemtry, parallel lines never cross, but unfortunately in the more general case, they do, and in particular, for any accelerated observer, a simple diagram shows that the timelike geodesics through different points in the trajectory of the accelerated observer (which we which to specify as the "origin" of our coordinate system) must intersect.

When this happens, when the geodesics ("straight lines") intersect, the coordinate system is no longer a 1:1 mapping. The result is that this general procedure for creating a coordinate system of an accelerated observer has a limited "range" - it is only a "local" coordinate system. The range limit is equal to c^2/a, where a is the accleration of the observer.

JesseM

Jun12-06, 09:58 PM

I realize I was simplifying things a bit for the sake of the argument, in practice there may be many cases in general relativity where one particular coordinate system may be seen as the most "elegant" choice, or the easiest to work with mathematically (aside from the case of an accelerating observer, physicists might also say that the most elegant coordinate system for dealing with the spacetime outside a black hole is Schwarzschild coordinates, or that the most elegant coordinate system for dealing with cosmology is the one where the distribution of matter and energy appears most uniform in space). But the theory of special relativity was originally designed around the idea of inertial reference frames, and they have a very simple physical interpretation in terms of local readings on a network of inertial rulers and clocks; in general relativity there isn't anything like this, GR was actually designed around the idea that you should be able to use any coordinate system you want (this can be true in SR too if you formulate the laws in tensor form, but this wasn't how they were originally formulated or how they'd be described in undergraduate texts). It would actually be incorrect to apply the non-tensor form of SR to anything but an inertial frame, while the laws of GR could be applied in any non-inertial coordinate system you like. And even if the type of coordinate system you describe for an accelerating observer may be the most "natural" in some sense, I don't think there's any simple interpretation of the coordinates assigned to distant events in terms of local readings on measuring-devices like the inertial rulers and clocks of an inertial coordinate system.

Al68

Jun12-06, 11:39 PM

JesseM

Jun13-06, 01:18 AM

Jesse,

OK, I almost forgot the reason I started this thread. I wanted to figure out the coordinate position of the earth as a function of time in system NI. It looks like you've done that. But I also wanted to figure out how to do this for the ship's deceleration and coming to rest relative to earth. Is there a way to figure out the coordinate position of the earth as a function of time in a decelerating system NI? That is in motion relative to earth and starts decelerating at a specified distance D from earth at a specified time T in the ship's frame, and eventually comes to rest relative to earth. In other words, similar to the ship coming to rest at the star system in the Twins Paradox.

Thanks,
Alan If you extend the relativistic rocket equations back through -t, I think it should give you the equation for a ship whose velocity is negative and decreasing until it comes to a stop at t=0 and d=0, then afterwards it turns around and its velocity is positive and increasing, with constant acceleration "a" in a single direction throughout (even if your acceleration is positive, if your current velocity is negative than your speed will be decreasing, so that is 'deceleration'). If this is correct then the equations I derived would be unchanged, you just have to keep in mind that for 'deceleration' you want T to be negative, with the ship reaching the position of the earth at T=0, and D positive but decreasing until that point. Or, you could pick Tstop to be the moment the ship reaches earth, and substitute (T-Tstop) in for T in those equations.

Al68

Jun13-06, 01:52 AM

pervect

Jun13-06, 05:35 PM

If you extend the relativistic rocket equations back through -t, I think it should give you the equation for a ship whose velocity is negative and decreasing until it comes to a stop at t=0 and d=0, then afterwards it turns around and its velocity is positive and increasing, with constant acceleration "a" in a single direction throughout (even if your acceleration is positive, if your current velocity is negative than your speed will be decreasing, so that is 'deceleration'). If this is correct then the equations I derived would be unchanged, you just have to keep in mind that for 'deceleration' you want T to be negative, with the ship reaching the position of the earth at T=0, and D positive but decreasing until that point. Or, you could pick Tstop to be the moment the ship reaches earth, and substitute (T-Tstop) in for T in those equations.

That's the approach I took. Perhaps you could double check the solutions I gave to this at the start of the thread in post #3 and post #6? The only thing I would add is that I've assumed that c=1 in these solutions for simplicity.

i.e. my posted solution

v(t) = -a(t-tstop)/sqrt(1+(a^2(t-tstop)^2)

becomes

v(t) = -a(t-tstop)/sqrt(1+(a^2/c^2)*(t-tstop)^2)

when c != 1

I've also given some solutions for x(t) via a different method (in #6), they should also be able to be obtained by integrating the above expressions. They also assumed that c=1.

As we've discussed, the x(t) I've given is for the Earth coordinate x(t), the ships coordiante x(t) is x_ship(t) = x(t)*sqrt(1-(v/c)^2).

I'm pretty sure Meir Achu'z suggestion is wrong: the velocity around the turnaround point should be an even function of t:

ie. v(t-t_turnaround) = -v(v-t_turnaround)

JesseM

Jun13-06, 06:06 PM

Jesse,

I was looking for equations for a ship that, after leaving earth and travelling away from earth, decelerates to come to rest relative to earth. Like the ship coming to rest at the distant star system in the Twins Paradox. And the coordinate position of earth as a function of time in the ship's frame during this deceleration.

Or, alternatively, the coordinate position of a star system (at rest with earth, that the ship is travelling to) as a function of time in the ship's frame during it's acceleration away from earth.

Thanks,
Alan It should also be possible to use the trick of extending the relativistic rocket equations into negative t and T coordinates to answer this. We can use these equations to figure out the rocket's distance from the earth if it started out from a distant star and accelerated in the direction of the earth; extending this into -t tells us the rocket's distance from the earth as it travels along the path from the earth to the star, decelerating in the direction of the star (which is the same as accelerating in the direction of the earth).

So, the relativistic rocket equation for distance from the star as a function of time, in the star's reference frame, would just be the usual one:

d = (c^2/a)*(\sqrt{1 - (at/c)^2} - 1)

If the earth is located at a distance d_0 from the star, the function for the ship's distance from earth d_e as a function of t in the star's frame would just be:

d_e = d_0 - d = d_0 - (c^2/a)*(\sqrt{1 - (at/c)^2} - 1)

...and this would also be the formula for the ship's distance from earth in the -t when it is moving in the direction of the star and decelerating (Also, if you wanted the ship to accelerate in the direction of the star for the first half of the trip and then decelerate for the second half, then d_e could be taken as the distance from the midpoint, with the distance from earth just being 2*d_e.)

So now we can do the same sort of substitutions that I did originally to translate this into D and T for the NI system. As before, D = d*\sqrt{1 - v^2/c^2}, and likewise the distance to the earth D_e in the NI system would be D_e = d_e * \sqrt{1 - v^2/c^2}. So, plugging in:

D_e = \sqrt{1 - v^2/c^2} * [d_0 - (c^2/a)*(\sqrt{1 - (at/c)^2} - 1)]

And as before, we can plug in t = (c/a)*sh(aT/c) to get:

D_e = \sqrt{1 - v^2/c^2}*[d_0 - (c^2/a)*(ch(aT/c) - 1)]

I realize in retrospect that there's a problem with both this equation and the ones I derived earlier for accelerating away from the earth--they still include v, which is in the original inertial coordinates rather than in the NI system's coordinates. So we should really plug in v(T) from the relativistic rocket page, which is v = c * th(aT/c), giving:

D_e = \sqrt{1 - (th(aT/c))^2}*[d_0 - (c^2/a)*(ch(aT/c) - 1)]

And the page on hyperbolic trig identities (http://www.geocities.com/ResearchTriangle/2363/hyperbol.html) says 1 = (sech(x))^2 + (tanh(x))^2, so 1 - (tanh(x))^2 = (sech(x))^2, so this simplifies to:

D_e = sech(aT/c) * [d_0 - (c^2/a)*(cosh(aT/c) - 1)]

Now just plug in negative values for T (or let Tstop be the time the ship reaches the star and substitute T-Tstop in for T in the above equation), and this should be the distance from the earth in the NI system as the ship decelerates towards the star. As for the velocity and acceleration of the earth in the NI system, I just realized that because I failed to take into account that v was a function of T, my earlier derivations of dD/dT and dD^2/dT^2 were incorrect. To differentiate the expression above you'd use the product rule (http://www.math.hmc.edu/calculus/tutorials/prodrule/) of calculus as well as the chain rule I mentioned earlier, and you'd have to know that the derivative of sech(x) is -tanh(x)*sech(x) while the derivative of tanh(x) is 1 - tan^2(x) (from this page (http://www.math2.org/math/derivatives/more/hyperbolics.htm)). I don't know how interested you are in the velocity and acceleration of the earth in the NI system though.

Al68

Jun13-06, 09:21 PM

Jesse,

Thanks for your response, I will have to take some time to look at it. But I do have another question.

If I'm on a rocket accelerating away from earth, my coordinate velocity relative to earth after acceleration will be v = at/\gamma.
Can this be explained by saying that as my velocity relative to earth increases due to acceleration, \gamma increases, resulting in an increasing amount of length contraction of my distance from earth, which partially counteracts the increase in velocity relative to earth due to proper acceleration. This causes my coordinate acceleration relative to earth to be equal to a/\gamma, and therefore my coordinate velocity relative to earth to be equal to at/\gamma. And my coordinate velocity relative to earth will be v = a_c t, where a_c refers to coordinate acceleration.

Is this correct?

Thanks,
Alan

JesseM

Jun14-06, 12:30 AM

That's the approach I took. Perhaps you could double check the solutions I gave to this at the start of the thread in post #3 and post #6? The only thing I noticed was you had the formula v(t) = -a(t-tstop)/sqrt(1+(a^2(t-tstop)^2), whereas the way I was approaching the problem I don't think that first "a" would be negative, it should be the same as the formula on the relativistic rocket page but with (t-tstop) substituted for t, giving v(t) = a(t-tstop) / sqrt[1 + (a/c)^2*(t-tstop)^2]. Did you change the definition of which direction was positive acceleration? If so, would this also change the definition of positive vs. negative velocity? Other than that minor issue everything in those posts looked good to me, although I couldn't check the equations in #6 that you found using maple. I hadn't known about x^2 - c^2*t^2 being constant for constant acceleration, that's a nice result...sometime I need to go back and study those relativistic rocket equations were derived, I don't remember my college SR course dealing much with accelerations. I'm pretty sure Meir Achu'z suggestion is wrong: the velocity around the turnaround point should be an even function of t:

ie. v(t-t_turnaround) = -v(v-t_turnaround) If it's an even function, and we set t=0 as the turnaround, then wouldn't that mean v(t) = v(-t)? But shouldn't its velocity a given amount of time before the turnaround have the opposite sign as its velocity a given amount of time after the turnaround, since it changed direction when it turned around? It seems to me like the position as a function of time should be an even function of t, but not the velocity (same as for a ball tossed upwards, where t=0 is the maximum height where the ball switches from rising to falling).

JesseM

Jun14-06, 12:33 AM

Jesse,

Thanks for your response, I will have to take some time to look at it. But I do have another question.

If I'm on a rocket accelerating away from earth, my coordinate velocity relative to earth after acceleration will be v = at/\gamma. I think you have the wrong formula there--the relativistic rocket page says that v(t) is at / \sqrt{1 + (at/c)^2}, and anyway it wouldn't make sense to have v be a function of gamma since gamma is itself a function of v.

pervect

Jun14-06, 04:19 AM

I think the formula is actually technically correct, because gamma = sqrt(1+(at/c)^2).

It's not particularly useful, unfortunately: if you don't know v, you don't know gamma either as Jesse points out.

I don't see offhand how the formula can have any physical significance. Letting c=1

1) at can be greater than the speed of light, so it can't be any sort of physical velocity.

2) at is not the rapidity either. (aT is the rapidity).

Perhaps there is some deeper significance to the formula that I'm missing, but I dont' see it.

Al68

Jun14-06, 12:15 PM

Jesse,

As pervect pointed out, \gamma = \sqrt{1 + (at/c)^2} = 1/\sqrt{1 - (v/c)^2}.
It’s obviously more convenient use \gamma = \sqrt{1 + (at/c)^2} when you don’t already know the coordinate velocity. One could also use v = at * \sqrt{1 - (v/c)^2} and solve for v to get v =at/ \sqrt{1 + (at/c)^2}, but this is less convenient.

I have also seen the term “proper velocity” used to describe at. And coordinate velocity = proper velocity divided by \gamma. And I’ve seen proper velocity defined as momentum divided by mass. And of course “proper velocity” can exceed c, but coordinate velocity cannot exceed c. And the velocity restriction in SR applies to coordinate velocity. But that’s not really relevant here.

It just looks to me like the increasing effect of length contraction (during acceleration) is the reason that coordinate velocity (after acceleration) is equal to v = at/\gamma instead of just at. Were it not for this length contraction, velocity would be equal to at, just like it does in classical physics, or at low speeds. Or maybe we could say that the increasing effect of length contraction (during acceleration) is the physical reason that a ship’s coordinate velocity can never exceed c, no matter how long it accelerates.

Is there a better way to physically describe why coordinate velocity does not equal at (at relativistic speeds)?

Thanks,
Alan

JesseM

Jun14-06, 12:35 PM

It just looks to me like the increasing effect of length contraction (during acceleration) is the reason that coordinate velocity (after acceleration) is equal to v = at/\gamma instead of just at. Length contraction of what, in which coordinate system? Remember, we're not dealing with coordinates in the non-inertial NI system here, we're just talking about v and t in the inertial coordinate system of the earth. The earth will see the length of the rocket itself shrink of course, but for the purposes of this formula the rocket is being treated as a point. And in the earth's frame, the earth's own rulers, which are used to measure how the rocket's distance increases, are not shrinking. Were it not for this length contraction, velocity would be equal to at, just like it does in classical physics, or at low speeds. I would say the reason velocity is not equal to at is because "a" does not refer to the coordinate acceleration in the earth's frame, where v and t are being measured; the actual coordinate acceleration is constantly decreasing as the ship approaches the speed of light.

Can you phrase your argument in a way that only refers to what's happening in the inertial frame of the earth, without considering any non-inertial coordinate systems like the NI system?

Al68

Jun14-06, 01:16 PM

OK, in the inertial frame of the earth, a ship’s velocity after acceleration will be v = at/\gamma due to the increasing effect of lorentz contraction of the distance of the ship from earth as velocity increases. This increasing effect of lorentz contraction of the distance to the ship increasingly counteracts the increasing distance of the ship due to acceleration. And if a is the proper acceleration of the ship, and a_c is the coordinate acceleration of the ship, a_c = a/\gamma due to the fact that the effect of lorentz contraction is increasing with time during acceleration. And v = a_c*t, or at/\gamma, instead of v = at as would be the case in classical physics or at low speeds. And the ship’s coordinate velocity will never reach c no matter how long it accelerates due to the increasing effect of lorentz contraction of the ship’s distance from earth. All of this is in earth’s frame.

(edit)And the coordinate acceleration is constantly decreasing as the ship approaches the speed of light because \gamma is constantly increasing.

In other words, were it not for lorentz contraction of the distance between earth and the ship, v would equal at just like in classical physics.

Is this better?

Thanks,
Alan

JesseM

Jun14-06, 03:35 PM

OK, in the inertial frame of the earth, a ship’s velocity after acceleration will be v = at/\gamma due to the increasing effect of lorentz contraction of the distance of the ship from earth as velocity increases. I still don't understand what you mean by this. If distance is measured by physical rulers, what ruler is lorentz-contracting here? The ship's increasing velocity won't cause the earth's own ruler to contract in the earth's frame, and it is this ruler that is being used to measure distance in the earth's frame. The only thing I can think of is that you would be talking about a ruler carried by the ship, but we've already discussed the problems with an acclerating observer trying to measure distances with accelerating rulers, and in any case the length of rulers carried by the ship plays absolutely no role in the earth's calculation of the distance and velocity as a function of time in the coordinates of its own inertial rest frame. And if a is the proper acceleration of the ship, and a_c is the coordinate acceleration of the ship, a_c = a/\gamma due to the fact that the effect of lorentz contraction is increasing with time during acceleration. Are you sure the coordinate acceleration in the earth's frame would be a/\gamma? Did you actually take the derivative of v(t) with respect to t? If not, this would probably be a worthwhile exercise, and if you know enough calculus to use the chain rule and the product rule, it shouldn't be too hard to differentiate the expression for v(t) on the relativistic rocket page (if not I can show you how).

Al68

Jun14-06, 05:52 PM

If we say that, after acceleration, the distance between the earth and the ship is lorentz contracted in the ship's frame, doesn't the distance between the ship and the earth have to be lorentz contracted in earth's frame?
And if two ship's pass each other, and each ship's crew measures the relative coordinate velocity between them, won't this coordinate velocity be based on length contracted measurements of distance, which are reciprocal?
Without lorentz contraction \frac{d^2 x}{dt^2} could be constant, and \frac{dx}{dt} would have no limit. Obviously the variable t does not change, since it is always read on the same clock at rest with the observer.
But with lorentz contraction, \frac{d^2 x_c}{dt^2} decreases at the same rate that 1/\gamma decreases with time, where x_c is the length contracted distance.

As far as a_c = a/\gamma, this is obvious from the relativistic rocket equation, since coordinate velocity equals coordinate acceleration times time. And v = at/\gamma. This is same relativistic rocket equation that is used everywhere for coordinate velocity.

It's my understanding that the relativistic rocket equation is derived by applying the lorentz transformation to acceleration to get a_c = a/\gamma

I can derive the relativistic rocket equation if I assume a_c = a/\gamma, and v refers to coordinate velocity.

v = a_c t
v = at/\gamma
v = at \sqrt{1 - (v/c)^2}, just solve for v, and
v^2 = (at)^2 (1 - (v/c)^2)
v^2 = (at)^2 - (at)^2 (v/c)^2
v^2 + (at)^2 (v/c)^2 = (at)^2
v^2(1 + (at/c)^2) = (at)^2
v^2 = (at)^2/(1 + (at/c)^2)
v = at/\sqrt{1 + (at/c)^2}

And it's obvious that \gamma = 1/\sqrt{1 - (v/c)^2} = \sqrt{1 + (at/c)^2}.
The second form is more convenient to use when you don't already know v.

Now I just have to hope I didn't make a mistake here.

Thanks,
Alan

JesseM

Jun14-06, 08:35 PM

If we say that, after acceleration, the distance between the earth and the ship is lorentz contracted in the ship's frame, doesn't the distance between the ship and the earth have to be lorentz contracted in earth's frame? But what does it mean to say "the distance contracted"? Usually when you use a phrase like that you're comparing one frame to another--for example, in the purely inertial case, we found that if the distance in the earth's frame at the moment of an event on the ship's worldline was d, then the distance in the ship's frame at the moment of that same event was contracted by d*\sqrt{1 - v^2/c^2}. Notice that if we're talking about the distance at the moment of that particular event, this isn't reciprocal, the distance is definitely bigger in the earth's frame than the ship's frame, not vice versa.

Anyway, when you talk about distance contracting "after acceleration" in the ship's frame (again, I would prefer to call it 'the NI coordinate system' to make clear that it isn't the only choice of coordinate system for the ship), it seems like you're not comparing two coordinate systems but talking about the distance at one time vs. another in the ship's coordinate system. But since D(T) actually grows with T, this can't be what you mean by "contracting"--maybe you just mean that a ruler at rest in the earth's frame will be shrinking in length in the NI system? And although we can't really talk about a ruler "at rest" in the NI system since the acceleration would deform it, I suppose we could imagine two small rockets moving in such a way that the coordinate distance between them in the NI system was constant. In this case I think, though I'm not sure, that the distance between them would shrink in the earth frame too...I don't think it'd just be by the gamma-factor of the ship at that moment in the earth's frame, though, because the distance between these two rockets would be constantly changing in the earth's frame.

Anyway, if this is something like what you were getting at, how does the fact that the distance between these two rockets shrinks in the earth's frame tell us anything about why the velocity in the earth's frame is at/gamma? I still can't see any rhyme or reason in your argument, the velocity in the earth's frame doesn't have anything to do with whatever the distance between those rockets may be. Remember, once again, the fact that the NI system isn't fundamentally any more physically valid than some other non-inertial coordinate system you could come up with--you could use some different coordinate system for the ship such that two rockets at constant coordinate distance in this system were actually increasing in distance in the earth's frame, it wouldn't change your conclusions about v(t) in the earth's frame one iota. And if two ship's pass each other, and each ship's crew measures the relative coordinate velocity between them, won't this coordinate velocity be based on length contracted measurements of distance, which are reciprocal? I'm not sure what you mean here either--maybe it would be better to focus on an inertial case like this since it's a lot less complicated than the accelerating case, and would probably help me understand what you're getting at in the accelerating case. The thing is, if two ships pass each other, each ship uses its own rulers and clocks to measure the speed of the other ship, and of course those rulers and clocks aren't contracted or slowed in the ship's own frame! I suppose if each ship wanted to try to figure out how fast the other ship was measuring it to move, then it could look at how fast it was moving past marks and how fast clocks are ticking on the other ship's measuring system...but I'm still not getting what is meant by the phrase "each ship's crew measures the relative coordinate velocity between them, won't this coordinate velocity be based on length contracted measurements of distance". Without lorentz contraction \frac{d^2 x}{dt^2} could be constant, and \frac{dx}{dt} would have no limit. Obviously the variable t does not change, since it is always read on the same clock at rest with the observer. It's not just lorentz contraction though, if you removed lorentz contraction but kept time dilation and the Einstein clock synchronization convention (which would mean there was a preferred frame, but never mind), then "constant a" in the sense that the coordinate acceleration in the ship's instantaneous inertial frame stays the same throughout the journey would still mean non-constant acceleration as seen in a single inertial frame. But with lorentz contraction, \frac{d^2 x_c}{dt^2} decreases at the same rate that 1/\gamma decreases with time, where x_c is the length contracted distance. Your argument about the acceleration is wrong--see below. As far as a_c = a/\gamma, this is obvious from the relativistic rocket equation, since coordinate velocity equals coordinate acceleration times time. Velocity = coordinate acceleration times time only works when coordinate acceleration is constant, not when it's varying over time, in which case velocity is found by integrating a(t) (note that integrating a constant 'a' with respect to 't' gives a*t). You might think that if v = at/gamma, then differentiating with respect to t would give a/gamma...but this is the problem with defining v(t) as a function of both t and gamma, where gamma is itself a function of t. You can't treat gamma as a constant for the purposes of differenting v(t) with respect to t, you have to take into account that it's a function of t as well. What you really need to do to find the coordinate acceleration in the earth's frame is differentiate this expression with respect to t:

v(t) = at / \sqrt{1 + (at/c)^2}

Only when you've differentiated the whole thing will you know the coordinate acceleration as a function of t and the proper acceleration a. Try this and see what you get...again, I can help with the math if you need it.

Al68

Jun15-06, 01:59 PM

JesseM

Jun15-06, 06:45 PM

Jesse,

I should not have said that a_c = a/\gamma, where a_c is coordinate acceleration. I should have said that \overline{a_c} = a/\gamma, where \overline{a_c} is average coordinate acceleration. And that coordinate velocity equals average coordinate acceleration times time. When you say "coordinate velocity equals average coordinate acceleration times time", do you mean the change in coordinate velocity between the two ends of the time-interval you're averaging the acceleration, ie \overline{a_c} = \Delta v / \Delta t? I think it is true that the average value of a function's derivative over some interval is just the difference in values of the function at the two ends of the interval divided by the size of the interval...this is another way of saying that the average value of a function's derivative between two points is the same as the slope of the line connecting the function at those points, and we know that the derivative at any point is like the "instantaneous slope" at that point. So this seems right to me, although my calculus is a little rusty so I'm not sure how you'd prove it. And since v can be written as at/\gamma, that means \Delta v between two times t1 and t2 would be a(t_2 - t_1)/\gamma, or a \Delta t /\gamma for that interval, which means if \overline{a_c} = \Delta v / \Delta t is true, it must also be true that \overline{a_c} = a/\gamma. (edit)And if v = \overline{a_c} t, then: Are you saying that the instantaneous value of v(t) at some time t is equal to the average acceleration times t? That is definitely not right, although as I said above, I think it's true that \Delta v = \overline{a_c} \Delta t. And we do not have time dilation without lorentz contraction in SR. We have both. But if we measure time on a clock at rest with us, it will be proper time. If we measure the length of an object in motion relative to us, it will not be the object's proper length, it will be length contracted according to SR, if we use a ruler at rest with us. True, although strictly speaking you have to use both rulers and clocks at rest with respect to you to measure the length of a moving object--"length" is understood to mean "the position coordinate of one end minus the position coordinate of another end when the positions of each end are recorded at the same time", so how you define simultaneity matters here. So, if we (on earth) were to measure a ship's coordinate velocity by measuring the length of the ship and the time difference between the front of the ship and the rear of the ship to pass a marker (in earth's frame), we know from SR that the proper length of the ship is greater than the coordinate length of the ship we measured. True, although you don't actually need to know the distance between the two ends of the object to measure its speed--an alternate method would just be to measure the position of the front end at one time and compare with the position of the front end at a later time. We know that the distance we measured was length contracted, because we are familiar with SR, not because we ever measured the proper length of the ship. We would have to be at rest with the ship to measure that. But we could have someone on the ship measure it's length for us, with a ruler that was identical to ours (before the ship was in motion relative to us), and tell us the proper length of the ship, and it would match up with our calculation. Or maybe we measured the length of the ship with the same ruler while it was at rest with earth. Either way, the length of the ship will be length contracted when we measure it in motion relative to us. But all of earth's calculations of velocity, and the total distance of the ship from earth at any specified time (by earth's clock), will be based on the earth's measurements of the length of the ship. No, again, it isn't necessary to know anything about the length of the ship, you could just measure the front end of the ship at different times. And in the case of an accelerating ship, at a given moment not every part of the ship is moving at the same speed, and botht the ship and any rulers on board are being distorted by acceleration, an effect separate from length contraction. In practice the relativistic rocket equations are either assuming the ship can be treated as pointlike, or they can be interpreted as only applying to a single point on the ship (the front, the middle, whatever) whose proper acceleration is constant.

If you try to extend the method of measuring speed by measuring the time between the front end and back ending passing a particular fixed point to the case of the relativistic rocket, you run into trouble because there is some finite time interval between the two measurements while the speed of the rocket is changing (and the speed of different parts of the rocket is not even the same at a single moment in your frame), and also because, again, the rocket is being physically distorted by the acceleration. You could try to eliminate the issue of distortion by the idea I suggested in my last post, where you have two tiny rockets programmed to accelerate in such a way that their separation is constant in the NI coordinate system, and they are treated like the "front" and "back" of an imaginary rocket which would appear to be at rest in the NI system. But the first issue won't go away here, you're still measuring the time between the front rocket and the back rocket passing you, but they are both constantly changing speed, and their speeds won't even be the same at a single moment in your inertial frame.

In contrast, when making two measurements of the front of the rocket, you could make the time-interval between the measurements as small as you wish, thus getting as close as you want to a measurement of the front's instantaneous velocity in your frame. So this is really a much better way to think of measuring the speed of an accelerating object. This leads me to believe that \Delta v = at/\gamma instead of \Delta v = at (like in classical physics) because of length contraction, not because of time dilation. OK, just to run through it again to check that I've got this right: you're saying that one way of measuring the speed of an inertial object would be based only on two items of knowledge, 1) the time between the front and back of an object passing a single clock at rest in our frame, and 2) knowledge of the proper length of the object, which we can use to figure out its length in our frame. You want to extend this procedure to an accelerating object in some way. An accelerating object does not really have a "proper length", but I suggested that we could at least add the condition that its length in the object's own instantaneous co-moving inertial frame is constant (which is the same thing as saying its length in the NI coordinate system is constant), perhaps just by creating an imaginary object whose front and back ends are mini-rockets programmed to accelerate in such a way that this works out.

As I said, there is the problem that the front and back ends/mini-rockets will not have the same instantaneous velocity in our frame, but I suppose that in the limit as the distance between the two ends in the co-moving frame goes to zero, the difference in instantaneous velocities in our frame would go to zero too. It could be that in this limit your argument makes sense, I'm not quite sure.

Here's another way I might make sense of the argument. Instead of having the rocket accelerating continuously, have its velocity increase in a series of discrete steps--ie have the rocket move at constant velocity, then instantaneously accelerate, then move at constant velocity again, etc. (with the time the rocket stays at constant velocity being constant in its co-moving inertial frame for that segment of the trip, and the amount that the velocity increases on each instantaneous acceleration also being constant in the co-moving inertial frame before that acceleration). In the limit as the time between accelerations goes to zero, I think this would become identical to the regular accelerating rocket. But with discrete accelerations, there would be no problems in using your method of measuring speed during each constant-velocity segment. Would it then make sense to say that the reason that the increase in velocity seen in our frame gets smaller with each jump, even though the increas in velocity is constant in the co-moving frame before each jump, is just because the length of the ship in our frame gets smaller after each jump? I'm not sure if I can see how that explains it, but if I could then I think the same explanation would make sense in the continuously-accelerating case, since that should just be the limit of this discrete-jump case. I'll have to think about it...can you think of a way of explaining yourself how the smaller change in velocity with each jump could be understood exclusively in terms of the ship's length in our frame getting smaller with each jump, and the fact that the change in velocity is constant in the ship's co-moving frame before each jump? It seems to me that explaining how the changes in velocity get smaller in our frame would also require knowing how each co-moving frame's clock speed and definition of simultaneity is different from ours, but I'm open to being convinced otherwise.

Al68

Jun16-06, 06:11 AM

(edit)And if v = \overline{a_c} t, then:
Are you saying that the instantaneous value of v(t) at some time t is equal to the average acceleration times t? That is definitely not right, although as I said above, I think it's true that \Delta v = \overline{a_c} \Delta t.

I meant that the final velocity (v) after a period of acceleration (t) would equal the average coordinate acceleration times time. I was assuming that v = 0 at t = 0. Just like the relativistic rocket equation, v = at/\gamma is only true if we assume that v = 0 at t = 0. I know it is a little sloppy to use v and \Delta v interchangeably, and t and \Delta t interchangeably, even though they are not interchangeable. But everybody else does it.

And, you are right that my method of measuring velocity would not be accurate (except in the limit as we used a smaller and smaller ship), but it would be pretty close, since presumably the length of the rocket would be very small compared to the total distance involved. And the change in velocity during the time interval would be very small. And your idea of measuring velocity would be more accurate.

But my main question now is, if we always use clocks at rest with us to measure time, is there any other explanation for why v = at/\gamma is correct? Is there any other specific reason that (change in) coordinate velocity does not equal proper acceleration times (change in) time, like it does in Newtonian physics?

Thanks,
Alan

JesseM

Jun16-06, 01:27 PM

I meant that the final velocity (v) after a period of acceleration (t) would equal the average coordinate acceleration times time. I was assuming that v = 0 at t = 0. Just like the relativistic rocket equation, v = at/\gamma is only true if we assume that v = 0 at t = 0. I know it is a little sloppy to use v and \Delta v interchangeably, and t and \Delta t interchangeably, even though they are not interchangeable. But everybody else does it. OK, I see what you mean. v(t) for any time t is always equal to \Delta v for the interval between t=0 and that t.

However, I realize in retrospect there was a problem with the argument that \overline{a_c} = a/\gamma. I had said earlier that \Delta v between two times t1 and t2 would be equal to a(t_2 - t_1)/\gamma, but I forgot that gamma would be changing from one time to another, it would actually be a* t_2 /\gamma(t_2) - a* t_1 /\gamma(t_1). So if you plug that into \overline{a_c} = \Delta v / \Delta t, what you actually get is \overline{a_c} = a* t_2 /[\gamma(t_2) * (t_2 - t_1)] - a* t_1 /[\gamma(t_1) * (t_2 - t_1)], not \overline{a_c} = a/\gamma. Analogous to what you say above about v, if we let t_1 = 0 and t_2 = t in the above, then we get \overline{a_c}(t) = a * t /[\gamma(t) * t], or \overline{a_c}(t) = a / \gamma(t). This might be what you meant anyway, but the idea that \overline{a_c} is a changing function of t is important, as I'll explain below.

Although it's true that v(t) = \overline{a_c} t can be derived from \Delta v = \overline{a_c} \Delta t, that only works if you assume \overline{a_c} to be a constantly changing function of t, like my \overline{a_c}(t) = a / \gamma(t)--for any given t, you can look at the change in velocity from 0 to t and that's v(t), but the average of a_c also has to be taken from the time interval of 0 to the t you're looking at, so it won't be the same from one value of t to another. So the rest of your argument, where you said: \frac{dv}{dt} = \overline{a_c}, since at any t, \overline{a_c} will be a constant. ...is incorrect, because \overline{a_c} is not constant. But my main question now is, if we always use clocks at rest with us to measure time, is there any other explanation for why v = at/\gamma is correct? Is there any other specific reason that (change in) coordinate velocity does not equal proper acceleration times (change in) time, like it does in Newtonian physics? Well, the way I think of it is based on the velocity addition formula. Like I said, constant acceleration can be thought of as a limiting case of making a series of instantaneous jumps in velocity, such that each time you make a jump, the magnitude of the increase is constant in your last co-moving rest frame, and the amount of time spent cruising inertially between jumps is also the same in each frame where you're at rest during the cruising period. But if you think about the way velocities add in relativity, it should be clear that an inertial observer will see the increase in velocity with each jump being smaller on successive jumps, Because if something is moving with respect to me with velocity v1 (in this case, the velocity before the jump), and then it increases in velocity by v2 in the frame where it was at rest when moving at v1, I will not see the velocity increase to v1+v2 but only to (v1+v2)/(1 + v1*v2/c^2), according to the velocity-addition formula. And the larger v1 is, the less this formula says the new velocity will have increased from v1, given constant v2. In classical mechanics, if the object was jumping according to this rule, then after 3 such jumps starting from rest I'd see the velocity as v2 + v2 + v2, but in relativity it'd only be v2 + (v2 + v2)/(1 + v2*v2/c^2) + (v2 + [(v2 + v2)/(1 + v2*v2/c^2)])/(1 + v2*[(v2 + v2)/(1 + v2*v2/c^2)]/c^2). And besides the fact that the increase in velocity is getting smaller with each jump as seen in my frame, there's also the fact that the time between jumps is getting longer and longer since the time is only supposed to be constant in the last co-moving rest frame before the jump, and each successive co-moving rest frame's clocks are ticking slower and slower in my frame. So this further decreases the rate that the velocity seems to increase in my frame, when compared with classical predictions. I think if you consider continuous acceleration as a limiting case of this sort of series of discrete jumps--and I think you could actually show rigorously that the accelerating case works out as the limit of the time between jumps and the size of each jump approaching zero, although I can't guarantee this--then you can see more easily why the acceleration in an inertial frame is continuously decreasing even if the proper acceleration "a" is constant.

Al68

Jun16-06, 08:31 PM

Jesse,

Again, you are right, the correct form of the equation shoud be \overline{a_c}(t) = a / \gamma(t). And the correct form of the relativistic rocket equation should be v(t) = at/\gamma(t). I was being a little sloppy again. (edit) And you are correct that \overline{a_c} would not be constant with respect to t. That was a mistake on my part.

I don't think we could say that the velocity addition formula actually causes v = at/\gamma to be correct. One might even say that Lorentz contraction causes the velocity addition formula to be correct.

But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today. He even said that the ship's twin should be able to consider himself at rest the entire time, and that the Twins Paradox could not be resolved in SR. This was supposedly one of his reasons for pursuing GR. He tried to resolve it in GR by considering the ship's twin to be at rest, but that resolution is considered faulty today. If one could argue that Einstein understood SR as well as anyone, why would he still view the Twins Paradox as a problem for SR?

Thanks,
Alan

George Jones

Jun17-06, 06:13 AM

But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today. He even said that the ship's twin should be able to consider himself at rest the entire time, and that the Twins Paradox could not be resolved in SR. This was supposedly one of his reasons for pursuing GR. He tried to resolve it in GR by considering the ship's twin to be at rest, but that resolution is considered faulty today. If one could argue that Einstein understood SR as well as anyone, why would he still view the Twins Paradox as a problem for SR?

This is an example of a logical fallacy - proof by authority. Einstein was a flesh-and-bllood human being who, like all flesh-and-blood human beings, made mistakes.

In any case, if Einstein really did hold these views, I would like to see some references.

jtbell

Jun17-06, 12:10 PM

But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today.

Today? Einstein has been dead for over fifty years. Surely people have thought of new ways to look at the twin paradox since he died, and refined earlier solutions.

If one could argue that Einstein understood SR as well as anyone,

I am not familiar with the detailed history of the development of our understanding of SR, during Einstein's lifetime and afterwards. Nevertheless, I find it very plausible that there are many people today who understand SR better than Einstein ever did. Many people have been refininig our knowledge of the consequences and implications of SR, and testing its predictions experimentally.

Al68

Jun17-06, 04:58 PM

This is an example of a logical fallacy - proof by authority. Einstein was a flesh-and-bllood human being who, like all flesh-and-blood human beings, made mistakes.

In any case, if Einstein really did hold these views, I would like to see some references.

What logical fallacy are you talking about? I never said anything about proof or authority. And if you assumed that I intended to challenge SR, you are not correct.

Mostly I was refering to Einstein's paper "Dialogue about objections to the theory of relativity". In this paper, he presents a resolution of the Twins Paradox. He says that special relativity is not suitable for resolving the issue. He then says the cause of the asymmetry between the twins is a pseudo-gravitational field created during the acceleration of the ship. He then uses the gravitational time dilation of GR to try to resolve the paradox.

The sources I have seen say that most physicists consider this resolution faulty. I would have to agree with that. I do in fact believe Einstein was wrong here.

This paper is available in libraries, but as far as I know, it is not available on the internet. I have looked. I only find other papers that reference it. And Wikipedia's article on the Twins Paradox references this and talks about it a little. http://en.wikipedia.org/wiki/Twin_paradox

(edit) there is also a link at the bottom of the Wikipedia article that discusses Einstein's resolution of the paradox.

And of course Einstein has been dead a long time, but the resolutions presented in textbooks today were around long before that.

And while there has been more experimental evidence to support SR since Einstein, he did not doubt SR, anyway, so this evidence would not be relevent to why Einstein believed what he did about the Twins Paradox.

And I believe most of the resolutions accepted today use the same version of SR found in Einstein's original 1905 paper. Although they make assumptions that are not explicitly stated in this 1905 paper, they normally reference Einstein's SR as their source.

(edit) It also seems strange that Einstein's thought's on the Twins Paradox are rarely mentioned in discussions of the Twins Paradox.

Thanks,
Alan

jtbell

Jun17-06, 05:25 PM

Mostly I was refering to Einstein's paper "Dialogue about objections to the theory of relativity."

The Wikipedia article on the Twins Paradox gives the date of this article as 1918. This is almost ninety years ago, and rather early in the history of relativity theory. At this time, I wouldn't expect even Einstein to have grasped SR as it relates to this paradox, as deeply as many people do now.

I have the impression that much of the current thinking about the Twin Paradox dates from the 1950s or 1960s, and it wasn't until a bit later that the currently popular explanations started to appear in introductory textbooks. When I was an undergraduate in the early 1970s, my sophomore-level introductory modern physics book (whose first edition dated back to the mid 1950s, I think) still claimed that resolving the paradox requires GR. However, none of the similar textbooks that I've seen since I started teaching in the mid 1980s does this.

Al68

Jun17-06, 06:11 PM

The Wikipedia article on the Twins Paradox gives the date of this article as 1918. This is almost ninety years ago, and rather early in the history of relativity theory. At this time, I wouldn't expect even Einstein to have grasped SR as it relates to this paradox, as deeply as many people do now.

I have the impression that much of the current thinking about the Twin Paradox dates from the 1950s or 1960s, and it wasn't until a bit later that the currently popular explanations started to appear in introductory textbooks. When I was an undergraduate in the early 1970s, my sophomore-level introductory modern physics book (whose first edition dated back to the mid 1950s, I think) still claimed that resolving the paradox requires GR. However, none of the similar textbooks that I've seen since I started teaching in the mid 1980s does this.

Well, you are probably right that most of the current thinking about the Twins Paradox has been after 1918. But Einstein certainly had a good grasp on SR, and obviously spent time thinking about the Twins Paradox. As had other physicists of the time.

But none of the resolutions I have seen even address Einstein's thoughts. Especially the question of why can't the ship's twin consider himself to be at rest the entire time. Note that if we did consider acceleration to be irrelevent, and the ship's twin to be at rest, then we would have to say that the earth's twin changed reference frames relative to the ship. And the ship would never change reference frames. And I have heard that we cannot consider the ship to be at rest because the ship changes reference frames. And the ship changes reference frames because the ship accelerated (did not stay at rest). But this sounds like circular logic to me. Einstein thought that there was no reason (in SR) that we could not consider the ship to be at rest the entire time, resulting in the earth twin aging less.

I am not saying Einstein was right or wrong about this part. I just haven't seen it proven either way. And all the explanations I have seen don't attempt to prove anything. They just show the math and diagrams that match up with the assumptions made.

Thanks,
Alan

robphy

Jun17-06, 10:28 PM

Well, you are probably right that most of the current thinking about the Twins Paradox has been after 1918. But Einstein certainly had a good grasp on SR, and obviously spent time thinking about the Twins Paradox. As had other physicists of the time.

I think it's fair to say that, early on, Einstein did not make use of the geometric interpretations suggested by Minkowski. Formulated geometrically, the "paradox" is easily resolvable... in particular, by direct calculation (i.e. spacetime arc-length) without issues of "transformations" and "reference frames".

One should also note that modern relativists (who emphasize the geometric structure and not much on "issues of reference frames") interpret "SR" and "GR" differently from the early physicists and relativists and from many textbooks and pop-sci books that haven't caught up yet.

But none of the resolutions I have seen even address Einstein's thoughts. Especially the question of why can't the ship's twin consider himself to be at rest the entire time. Note that if we did consider acceleration to be irrelevent, and the ship's twin to be at rest, then we would have to say that the earth's twin changed reference frames relative to the ship. And the ship would never change reference frames. And I have heard that we cannot consider the ship to be at rest because the ship changes reference frames. And the ship changes reference frames because the ship accelerated (did not stay at rest). But this sounds like circular logic to me. Einstein thought that there was no reason (in SR) that we could not consider the ship to be at rest the entire time, resulting in the earth twin aging less.

I am not saying Einstein was right or wrong about this part. I just haven't seen it proven either way. And all the explanations I have seen don't attempt to prove anything. They just show the math and diagrams that match up with the assumptions made.

Thanks,
Alan

The key word missing in your post is "inertial" (which is not the same as "at rest"). The travelling twin can "regard himself at rest"... but he cannot regard himself as "inertial".

Al68

Jun18-06, 10:09 AM

I think it's fair to say that, early on, Einstein did not make use of the geometric interpretations suggested by Minkowski. Formulated geometrically, the "paradox" is easily resolvable... in particular, by direct calculation (i.e. spacetime arc-length) without issues of "transformations" and "reference frames".

One should also note that modern relativists (who emphasize the geometric structure and not much on "issues of reference frames") interpret "SR" and "GR" differently from the early physicists and relativists and from many textbooks and pop-sci books that haven't caught up yet.

The key word missing in your post is "inertial" (which is not the same as "at rest"). The travelling twin can "regard himself at rest"... but he cannot regard himself as "inertial".

I'm not sure what you mean by direct calculation. I've seen it explained many ways. But in every case I have seen, the explanation, the drawings, and the calculations only describe the assumptions and conclusions made. For example, one could easily just assume that the ship was at rest, and draw a spacetime diagram just like the normal ones, except labeled the other way around. The same could be done with the calculations. I'm not saying that the accepted explanations are wrong. Just that the ones I've seen explain what happens, but not why it happens that way instead of a different way. And of course you're right, the ship's twin is not inertial during the turnaround. But, if one object accelerates away from another, the mathmatics and kinematics look the same on paper regardless of which object is said to accelerate relative to the other.

I'm not saying that it's not relevent that only one twin accelerates. Maybe I just haven't seen the right explanation. That's why I ask so many questions. Is there an explanation available on the internet that shows why it's important that only one twin accelerates? Instead of just assuming that it's important.

Thanks,
Alan

robphy

Jun18-06, 11:57 PM

I'm not sure what you mean by direct calculation.

By direct calculation, I mean the computation of the path-dependent integral \int_A^B ds , where A and B are events and ds is a timelike arc-length element in spacetime [analogous to the Euclidean arc-length used when measuring the length of a curve]. This always works. In certain simplified situations, like a piecewise-inertial path, one often computes proper-time intervals by Lorentz-boosting displacements between end-events (endpoints)...then summing. In the latter case, one often makes use of "time dilation" formulae or "length contraction" formulae, which are okay.... although incomplete in my opinion if the key idea of spacetime-arc-length is not addressed.

I've seen it explained many ways. But in every case I have seen, the explanation, the drawings, and the calculations only describe the assumptions and conclusions made. For example, one could easily just assume that the ship was at rest, and draw a spacetime diagram just like the normal ones, except labeled the other way around. The same could be done with the calculations.

I'm not saying that the accepted explanations are wrong. Just that the ones I've seen explain what happens, but not why it happens that way instead of a different way. And of course you're right, the ship's twin is not inertial during the turnaround. But, if one object accelerates away from another, the mathmatics and kinematics look the same on paper regardless of which object is said to accelerate relative to the other.

If all you drew were the worldlines of the two twins, you could be misled into thinking they were equivalent. The key point is that "the inertial observer between the two events logs the most proper time". [Mathematically, inertial-worldlines are geodesics.] To see this in your diagrams, you should draw in (say) a family of inertial-worldlines traveling with the same velocity. In the inertial-twin's diagram those worldlines are all straight (and are past and future extendible as such). In the rocket-twin's diagram, those inertial-worldlines will have a kink, break, or other irregularity in them... which no Lorentz transformation can transform away. [Draw the worldline of a ball sitting on the frictionless floor of the rocket. At the turn around event, according to the law of inertia, does the ball's worldline follow the rocket's worldline?] Another way to express this is to say that the coordinate system of the rocket-twin (which needs to be fully specified) does not have the same properties of the coordinate system of the inertial-twin.

I'm not saying that it's not relevent that only one twin accelerates. Maybe I just haven't seen the right explanation. That's why I ask so many questions. Is there an explanation available on the internet that shows why it's important that only one twin accelerates? Instead of just assuming that it's important.

Thanks,
Alan

The bottom line is that the "proper-time logged between two events" depends on the spacetime path under consideration... In Galilean/Newtonian relativity, for two given events, those time-intervals are equal regardless of the choice of path. In relativity, the clock effect could be seen even when each twin undergoes some acceleration... certainly there may situations when they are equal... but in general, they will differ. It's just easier to analyze if one case is inertial... as well as potentially confusing because now one has an opportunity to misuse and misinterpret the SR results up to that point... e.g. the "twin paradox".

Al68

Jun19-06, 12:34 AM

The key point is that "the inertial observer between the two events logs the most proper time".

Well, it looks like if we assume that the inertial observer will always log the most proper time, then of course we will find that the observer who logs the most proper time will turn out to be the inertial observer. And if we assume that the observer who logs the most proper time is the inertial observer, we will find that the inertial observer will turn out to be the one who logs the most proper time.

And if we draw inertial worldlines as straight, and put kinks in non-inertial worldlines, we will end up with inertial worldlines that are straight and non-inertial worldlines that are kinked.

My question is, if the relative velocity between two objects changes, how do we know that the inertial worldline should be drawn as straight, and the non-inertial worldline should be drawn as kinked?

I'm not saying you are wrong, I'm just saying that the mathematics and drawings don't really seem to answer the question. They just seem to explain the answer afterward.

(edit) There are also examples where the twins could both be inertial and experience differential aging. The twins could be in intersecting orbits around earth, one nearly circular and the other very elliptical. Each twin would be following a geodisic the entire time, yet everytime they met, they would see that they experienced different lapses of proper time since the last time they met, even though they are both in freefall.

Thanks,
Alan

jtbell

Jun19-06, 02:00 AM

Well, it looks like if we assume that the inertial observer will always log the most proper time,

No, we do not assume this. It is a consequence of our definition of an inertial observer as one who remains at rest (or moving at constant velocity) in an inertial reference frame. You can test it by calculation on some simple examples.

For example, consider a simple twin paradox scenario in which the traveling twin goes out in a straight line for 5 years at a speed of 0.8c, traveling a distance of 4 light-years, turns around "instantaneously" and returns at the same speed. In the earth's reference frame (assumed to be inertial), let us calculate the total spacetime path length for the stay-at-home twin (A) and the traveling twin (B).

Event #1 is the traveling twin's departure, at x = 0 and t = 0.

Event #2 is the traveling twin's turnaround, at x = 4 ly and t = 5 yr.

Event #3 is the traveling twin's return, at x = 0 and t = 10 yr.

The spacetime path of twin A has just one straight-line segment, with length

[tex]\Delta s_A = \sqrt{(t_3 - t_1)^2 - (x_3 - x_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10[/itex]

The spacetime path of twin B has two straight-line segments, with total length

[tex]\Delta s_B = \sqrt{(t_2 - t_1)^2 - (x_2 - x_1)^2} + \sqrt{(t_3 - t_2)^2 - (x_3 - x_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6[/itex]

This calculation is easiest in the inertial reference frame in which twin A is at rest throughout, but we can use any other inertial reference frame, and get the same values for \Delta s_A and \Delta s_B.

If we let twin B follow any other path we like, except of course one which simply duplicates twin A's path, we will likewise calculate that \Delta s_B < \Delta s_A. I'm sure this can be proven mathematically, but I'm not up to mathematical proofs at 3:00 am. :tongue:

My question is, if the relative velocity between two objects changes, how do we know that the inertial worldline should be drawn as straight, and the non-inertial worldline should be drawn as kinked?

It seems to me that you're asking basically, "why are inertial reference frames special?" or "why is inertial motion special?"

These two assumptions go all the way back to Newton and his laws of motion. They're not a new invention in special relativity. In classical mechanics, we define an inertial reference frame as one in which Newton's laws of motion hold, e.g. the First Law which states that an object that has zero net external force acting on it moves in a straight line at constant speed. In special relativity we do pretty much the same thing, except that we have to tweak the Second Law (F = ma) a bit in order to get it into relativistic form.

In a non-inertial reference frame, Newton's Laws don't work. We have to add terms corresponding to "fictitious" or "inertial" forces that arise from our non-inertialness. (For example, the force that apparently pushes you into your car seat when you press your car's accelerator pedal.)

robphy

Jun19-06, 09:34 AM

Well, it looks like if we assume that the inertial observer will always log the most proper time, then of course we will find that the observer who logs the most proper time will turn out to be the inertial observer. And if we assume that the observer who logs the most proper time is the inertial observer, we will find that the inertial observer will turn out to be the one who logs the most proper time.

It's not an assumption.... it's a key result obtained from Einstein's postulates for Special Relativity, which can be mathematically formalized as a Minkowskian geometry (the geometry of an R4 with a Minkowskian metric... such a geometry is preserved under Lorentz Transformations, whose eigenvectors are the lightlike vectors).

And if we draw inertial worldlines as straight, and put kinks in non-inertial worldlines, we will end up with inertial worldlines that are straight and non-inertial worldlines that are kinked.

We do this in Euclidean geometry and in Galilean/Newtonian relativity.
In Euclidean geometry, a straight line is a geodesic.
In relativity, inertial observers (which are "free particles", not being subjected to forces... and thus have no acceleration and thus no worldline curvature) travel on a geodesic (http://en.wikipedia.org/wiki/Geodesic) in spacetime.

While geodesics are often characterized by minimizing (or in the Minkowskian case, maximimzing) the arc-length, there is another way to characterize a geodesic. Along a geodesic, the tangent vector is parallelly transported along itself. Crudely, "it doesn't turn" according to its geometry.

My question is, if the relative velocity between two objects changes, how do we know that the inertial worldline should be drawn as straight, and the non-inertial worldline should be drawn as kinked?

As a said earlier, if all you draw is the worldlines, you can draw them how you like. However, if you wish to attach a coordinate system to span spacetime (so that observer can assign a unique set of coordinates for each event in spacetime), then you'll find that the inertial observer can do it, however, a non-inertial one can't generally do it... in some cases, the non-inertial one will not be able to assign coordinates to an event (e.g. the uniformly accelerated observer has a horizon), or else will have multiple assignments to an event (e.g. associated with the instantaneous turn-around event in the twin paradox, events beyond the turn-around event may have multiple coordinate assignments [draw the lines of simultaneity to see]).

I'm not saying you are wrong, I'm just saying that the mathematics and drawings don't really seem to answer the question. They just seem to explain the answer afterward.

Of course, the result of mathematical formulation is just that:
the result of a mathematical formulation. The one used in special relativity is used because that seems to correspond well with the experimental result (e.g. the muon lifetime experiments) as well as lead to further predictions that can be tested in the real world. Had that muon experiment and others like it not yielded their non-Newtonian results, we probably would not be discussing special relativity now.

(edit) There are also examples where the twins could both be inertial and experience differential aging. The twins could be in intersecting orbits around earth, one nearly circular and the other very elliptical. Each twin would be following a geodisic the entire time, yet everytime they met, they would see that they experienced different lapses of proper time since the last time they met, even though they are both in freefall.

Thanks,
Alan
Due to the presence of spacetime curvature, this is not special relativity any more. In Minkowski spacetime, distinct timelike geodesics can't intersect more than once. In your case, you seem to be describing a situation with conjugate points (http://planetmath.org/encyclopedia/ConjugatePoints.html). (Look at #14 in this thread (http://www.physicsforums.com/showthread.php?t=104508).)

jtbell

Jun19-06, 10:51 AM

From a purely mathematical, descriptive point of view, non-inertial reference frames are just as "valid" as inertial reference frames, except perhaps that in non-inertial frames you can have two or more sets of coordinates for the same event, as robphy points out. But it might be possible to invent rules for choosing consistently which set of coordinates to use in such cases.

But from a physical point of view, we have the experimental observation that inertial frames are simpler to work with than non-inertial ones. Not only do inertial reference frames not have the problem of multiple coordinate sets (in relativistic situations), we also don't have to invent "fictitious" forces that don't have any apparent physical source, in order to explain certain changes in velocity of objects. In an inertial reference frame, any change in an object's velocity, so far as we know, can be explained as the result of a physical interaction with some other object. And there are a relatively small number of possible physical interactions, each with relatively well-defined rules.

Going back to the twin-paradox scenario, in the stay-at-home twin's inertial reference frame, the traveling twin's worldline develops a "kink" because his rocket engines fire, thereby creating a reaction force on the rocket which changes its velocity. We can easily calculate that reaction force and the resulting change in velocity, in terms of the nozzle speed of the rocket engine's exhaust, and the amount of fuel burned. The traveling twin knows this is happening, even if the engines were fired by remote control without his advance knowledge, because he can feel it happening.

In the traveling twin's non-inertial reference frame, when he fires his rocket engines, the stay-at-home twin suddenly changes his motion. What is the physical cause of that change? What force acts on the earth to change its velocity in that non-inertial reference frame? What laws allow us to calculate that force? And why doesn't the stay-at-home twin feel any effect from this?

JesseM

Jun19-06, 11:16 AM

I don't think we could say that the velocity addition formula actually causes v = at/\gamma to be correct. I also mentioned time dilation though, which means that in the stepped-velocity-increase scenario I was discussing, if each constant-velocity interval lasts for the same amount of time in the co-moving inertial rest frame of the ship during that interval, each step will appear longer and longer in the earth's inertial frame. With the velocity addition formula and the time dilation formula I think it would be possible to figure out just how my stepped-velocity-increase scenario would look in the earth's frame, and I think in the limit of smaller and smaller time-intervals (and velocity increases) this scenario reduces to the continuous acceleration scenario. Do you disagree? One might even say that Lorentz contraction causes the velocity addition formula to be correct. No you couldn't. The velocity-addition formula depends on Lorentz contraction and on time dilation and the relativity of simultaneity. If you imagined an alternate universe with no time dilation and no relativity of simultaneity, but where there was still Lorentz contraction relative to a preferred frame, the relativistic velocity-addition formula would not be correct in such a universe. And I don't think the constant proper acceleration scenario would lead to the same functions as on the relativistic rocket page in this universe either, which is why I don't think it makes sense to explain things like v(t) = at/ \gamma (t) and a_c (t) = a / \gamma (t) solely in terms of Lorentz contraction. But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today. He even said that the ship's twin should be able to consider himself at rest the entire time, and that the Twins Paradox could not be resolved in SR. This was supposedly one of his reasons for pursuing GR. He tried to resolve it in GR by considering the ship's twin to be at rest, but that resolution is considered faulty today. If one could argue that Einstein understood SR as well as anyone, why would he still view the Twins Paradox as a problem for SR? I think you may have misunderstood what Einstein was arguing. I am sure he would have agreed that 1) the standard non-tensor laws of special relativity only work in inertial frames, and 2) as long as you stick to an inertial frame, you will always get the same prediction about the ages of the twins when they reunite. He probably did not see this as satisfactory because the fundamental distinction between inertial and non-inertial frames in 1) went against his "Machian" view of physics (are you familiar with Mach's principle (http://en.wikipedia.org/wiki/Mach's_principle) and the influence it had on Einstein's thinking?), so although he wouldn't disagree that 1) and 2) would "resolve" the twin paradox in SR, I think he saw SR itself as problematic for this reason, and wanted a new theory that would apply the same laws to non-inertial observers as inertial ones. He was ultimately successful in this, since GR has the same laws in all coordinate systems, inertial and non-inertial alike. And with GR you can understand the twin paradox in terms of a coordinate system where the travelling twin is at rest throughout the journey, it just means that the travelling twin will see a gravitational field during the period where inertial observers see him accelerating--see this section (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html) of the Twin Paradox page (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html).

jtbell

Jun19-06, 11:39 AM

And with GR you can understand the twin paradox in terms of a coordinate system where the travelling twin is at rest throughout the journey, it just means that the travelling twin will see a gravitational field during the period where inertial observers see him accelerating--see this section (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html) of the Twin Paradox page (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html).

I was about to jump on you with the question "but where does the gravitational field come from?", but then I saw that the page that you linked to does address this point, very well. In the spirit of that page, both "GR" and "gravitational field" would be in scare quotes in your statement.

George Jones

Jun19-06, 03:08 PM

What logical fallacy are you talking about? I never said anything about proof or authority. And if you assumed that I intended to challenge SR, you are not correct.

Are you saying that the logical fallacy of proof by authority can occur only when someone uses the words "proof" and "authority"?

I've gone back and reread the passage of yours that I quoted, and, even if you did not intend it that way, I don't think it's too big a leap to intrepret what you wrote as

[(Einstein understood SR as well as anyone) /\ (Einstein viewed the twin paradox as a problem for SR] => (the twin paradox is a problem for SR)

He then says the cause of the asymmetry between the twins is a pseudo-gravitational field created during the acceleration of the ship. He then uses the gravitational time dilation of GR to try to resolve the paradox.

It might be the case that, in spite of what he writes, Einstein's analysis never leaves the cozy confines of SR.

there is also a link at the bottom of the Wikipedia article that discusses Einstein's resolution of the paradox.

Unfortunately, I can't get this link to work.

It also seems strange that Einstein's thought's on the Twins Paradox are rarely mentioned in discussions of the Twins Paradox.

Which thoughts? Einstein also analysed the twin paradox in terms of the relativity of simultaneity, and this concept is often used in discussions of the twin paradox.

It seems that you want an answer to the question, "Why are some frames of reference inertial?"

I think this is similar in kind to the question "Why is the magnitude of force in Newton's law of gravity given by F = GmM/r^2?"

My answer to both questions is "I don't know."

Why questions tend to be very deep and difficult, but this doen't mean that we shouldn't ask them.

pervect

Jun19-06, 03:28 PM

By direct calculation, I mean the computation of the path-dependent integral \int_A^B ds , where A and B are events and ds is a timelike arc-length element in spacetime [analogous to the Euclidean arc-length used when measuring the length of a curve].

This is a very important point. I don't know if Al68 appreciates how the Lorentz interval is calculated (as above), the significance of this calculation, or the fact that this quantity (unlike time, or distance) is the true invariant that is shared between different observers. He (Al68) seems to be very focused on coordinates and not particularly interested in the Lorentz interval.

The Lorentz interval defines "proper time" and "proper length", i.e. how much time a clock reads, and how long a ruler measures. So when one wants to calculate the amount of time that a clock takes to traverse its path, you just calculate the Lorentz interval of that path. This automatically gives you the time that a clock travelling that path would read, by definition.

I think that calculating the Lorentz interval has mainly been presented in terms of SR. It's possible to generalize this to GR by adding in the concpet of a metric (another rather important idea that hasn't been talked about much). The metric is the key to calculating the Lorentz interval in arbitrary/accelerated/noninertial coordinate systems.

But I don't want to go into this too much (the metric) unless I'm reassured that Al68 understands the Lorentz interval and why we want to calculate it in the first place.

Al68

Jun20-06, 12:12 AM

I don't think it makes sense to explain things like v(t) = at/ \gamma (t) and a_c (t) = a / \gamma (t) solely in terms of Lorentz contraction
Jesse, I guess I have to agree, since length contraction and time dilation are so intertwined, that we really can't separate them.
I think you may have misunderstood what Einstein was arguing. I am sure he would have agreed that 1) the standard non-tensor laws of special relativity only work in inertial frames, and 2) as long as you stick to an inertial frame, you will always get the same prediction about the ages of the twins when they reunite. He probably did not see this as satisfactory because the fundamental distinction between inertial and non-inertial frames in 1) went against his "Machian" view of physics (are you familiar with Mach's principle (http://en.wikipedia.org/wiki/Mach's_principle) and the influence it had on Einstein's thinking?), so although he wouldn't disagree that 1) and 2) would "resolve" the twin paradox in SR, I think he saw SR itself as problematic for this reason, and wanted a new theory that would apply the same laws to non-inertial observers as inertial ones. He was ultimately successful in this, since GR has the same laws in all coordinate systems, inertial and non-inertial alike. And with GR you can understand the twin paradox in terms of a coordinate system where the travelling twin is at rest throughout the journey, it just means that the travelling twin will see a gravitational field during the period where inertial observers see him accelerating--see this section (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html) of the Twin Paradox page (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html).
Yes, I agree that Einstein saw SR as problematic because inertial frames were considered different from non-inertial frames. But according to several sources on the net, including this one, http://en.wikipedia.org/wiki/Twin_paradox Einstein's GR resolution of the Twins Paradox is considered to be faulty today. So I would disagree that he was ultimately successful. At least in this respect. It's my understanding that he never considered any of the SR resolutions of the Twins Paradox to be satisfactory.

Are you saying that the logical fallacy of proof by authority can occur only when someone uses the words "proof" and "authority"?

I've gone back and reread the passage of yours that I quoted, and, even if you did not intend it that way, I don't think it's too big a leap to intrepret what you wrote as

[(Einstein understood SR as well as anyone) /\ (Einstein viewed the twin paradox as a problem for SR] => (the twin paradox is a problem for SR)

That is not what I intended, although I can understand why it could be interpreted that way, given all the internet discussions on this issue. I am trying to understand the Twins Paradox better. I consider SR to be correct, so even if I were to ultimately choose to disagree with the accepted Twins Paradox resolution, it will not cause me to view the Twins Paradox as a problem for SR. It would cause me to view SR as a problem for the accepted Twins Paradox resolutions.

And I certainly don't believe in proof by authority, if I did, I wouldn't have so many questions.

Why questions tend to be very deep and difficult, but this doen't mean that we shouldn't ask them.

Thanks, I agree.

No, we do not assume this. It is a consequence of our definition of an inertial observer as one who remains at rest (or moving at constant velocity) in an inertial reference frame. You can test it by calculation on some simple examples.

For example, consider a simple twin paradox scenario in which the traveling twin goes out in a straight line for 5 years at a speed of 0.8c, traveling a distance of 4 light-years, turns around "instantaneously" and returns at the same speed. In the earth's reference frame (assumed to be inertial), let us calculate the total spacetime path length for the stay-at-home twin (A) and the traveling twin (B).

Event #1 is the traveling twin's departure, at x = 0 and t = 0.

Event #2 is the traveling twin's turnaround, at x = 4 ly and t = 5 yr.

Event #3 is the traveling twin's return, at x = 0 and t = 10 yr.

The spacetime path of twin A has just one straight-line segment, with length

[tex]\Delta s_A = \sqrt{(t_3 - t_1)^2 - (x_3 - x_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10[/itex]

The spacetime path of twin B has two straight-line segments, with total length

[tex]\Delta s_B = \sqrt{(t_2 - t_1)^2 - (x_2 - x_1)^2} + \sqrt{(t_3 - t_2)^2 - (x_3 - x_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6[/itex]

This calculation is easiest in the inertial reference frame in which twin A is at rest throughout, but we can use any other inertial reference frame, and get the same values for \Delta s_A and \Delta s_B.

I think this last statement is the key. You will only get these values if you use the numbers from an inertial frame for the calculation. If we were to assume that the ship could be viewed the same way, we would get different values.

Like this: We consider the ship's frame (B) to be preferred instead of the inertial frame (A). And we say that the distance is given as 4 ly in the ship's frame. And:

Event #1 is the traveling twin's departure, at x' = 0 and t' = 0.

Event #2 is the traveling twin's turnaround, at x' = 4 ly and t' = 5 yr.

Event #3 is the traveling twin's return, at x' = 0 and t' = 10 yr.

Now the spacetime path of twin B has just one straight-line segment, with length

\Delta s_B = \sqrt{(t'_3 - t'_1)^2 - (x'_3 - x'_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10

And the spacetime path of twin A has two straight-line segments, with total length

\Delta s_A = \sqrt{(t'_2 - t'_1)^2 - (x'_2 - x'_1)^2} + \sqrt{(t'_3 - t'_2)^2 - (x'_3 - x'_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6.

And of course this would add up, since the distance travelled would now be 4/\gamma or 2.4 light years each way in earth's frame.

The only difference here is, I considered the non-inertial frame as the "preferred frame" and said that the earth turned around relative to the ship's frame at a distance specified in the ship's frame.

I'm not saying this is correct, just that the math can be done either way. Of course, I know that I did not use an inertial frame for my numbers. And by definition, the inertial frame is the one with the longest spacetime interval. And this will always result in the most time passing for the inertial frame (between any two events). I have to repeat that I do not consider my above calculations to be correct. I just wanted to illustrate that mathematics doesn't provide the answer, it only describes the answer. And it is very easy to use mathematics to describe a wrong answer, as I think I have shown above.

It seems to me that you're asking basically, "why are inertial reference frames special?" or "why is inertial motion special?"

Yes, that is my question.