# Seeming Paradox

1. Jun 3, 2005

### TheAntiRelative

What is the actual solution to the following supposed paradox? I haven't been able to put my finger on it yet.

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The Twins Doppler Paradox:

The following is based upon the axiom that during a specific interval of emission time, a laser beam or some other cohesive reliable source of a given wavelength/frequency will emit a certain number of waves. If that light is doppler shifted, the number of total waves will not change so consequently, the interval of viewing time will be shortened for blue shift and lengthened for red shift in consideration of the constancy of light speed.

In the twins paradox, let us have both twins shining identical lasers at eachother at some exact known wavelength/frequncy. Now, we know for sure that the travelling twin has the slower clock because of acceleration. Lets examine what he sees on his outbound trip given the following test: The earth twin emits a light for 10 minutes at a frequency of 1000 waves/peaks/beats per minute. The travelling twin is moving at a speed close enough to light to cause his time to move at one tenth the time of the earth's.

1) Because he is moving away at some speed close to C he must see a redshifted beam of light that lasts longer than 10 minutes.
2) Because his time is much slower he must see a signal that beats 10,000 times in one of his minutes and is therefore blue-shifted and lasts 1 minute.

These considerations bring up questions such as:
Did the Milky way galaxy get accelerated away from the andromeda galaxy at nearly light speed some time billions of years ago? Are we moving towards or away from Andromeda?

Now use the same circumstances as above except that both twins spend 10 minutes with light on and 10 with it off and repeat this cycle 10 times. Additionally, the travelling twin is now earthbound. In this example we observe the earth twin's perspective.

1) Because the earth twin's time goes so much faster, ten of the travelling twin's minutes lasts 100 of the earth twin's yet the same number of beats will be released from the traveller during that 100 minutes of earth time that would be released from the earth twin's identical laser during only 10 minutes, so the earth twin must see an extremely redshifted signal from the traveller.
2) Because he is traveling towards earth, we know there must be a blue shift for the earth twin observer.
3)If, for earth time, the traveller should only be releasing 100 beats per earth minute, and earth twin sees a blue shifted signal that equates to greater than 1000 beats per minute, earth will see far into the traveller's future. The earth observer twin will see the end of the traveller's 10th burst before 200 minutes. However we know that in 200 minutes of earth time he has only aged 20 minutes. He's only been able to push the button to turn the on laser one time.

NOTE: The beginning and ending location as well as the travel time of the beam can be ignored because the phenomena would result in seeing further and further into the future over time such as to overcome the travel time between emitter and observer given a large enough sample period.

2. Jun 3, 2005

### Staff: Mentor

It's certainly true that all observers will agree that the same number of waves were emitted.

No. No reason to treat the travelling twin as accelerating during the outbound or inbound leg of his journey. In making the turnaround needed for a round trip he certainly accelerates, so we can conclude that the total time elapsed on his clock for the entire trip will be shorter than the time elapsed for his earth twin. But that does not mean that the travelling clock is somehow "absolutely" slower during the constant speed portions of the trip. During the outbound and inbound portions of the trip, both twins will measure the other's clocks as running slowly.

If the earth twin emits light for 10 earth-minutes, the traveling twin will say that the earth light was only on for 100 of his minutes. (Given that gamma = 10 as you propose.) Of course, since the twins are separating, the time it takes for the travelling twin to receive the 10000 waves is longer than 100 minutes. But both agree that 10000 waves were emitted.

Note that this is completely symmetric: If the travelling twin emits light for 10 minutes, the earth twin will say that the light was on for 100 earth minutes, etc.

Right. The light is redshifted and is received for longer than 100 minutes.

Nope.

3. Jun 3, 2005

### Janus

Staff Emeritus
The fact that one twin accelerated has no effect on the time dilation measured by either of two once that acceleration ceases. All acceleration will do is effect how the Accelerated twin measures the non-accelerated twin's clock rate during the period the twin is actually accelerating.
No. As long as the two twins maintain a constant velocity with respect to each other. they each will determine that the other's time is running slower. There is no way that either twin will see the light from the other as blue shifted while they are moving away from each other.
To determine exactly what red or blue shift either will observe use the relativistic doppler shift equation which takes the effects of Relativity into account:

$$f_{observed}= f_{source} \left( \frac{1+ \frac{v}{c}}{1- \frac{v}{c}} \right)$$

where v is considered positive if the objects are approaching each other.

4. Jun 3, 2005

### pervect

Staff Emeritus
Here's a short way of seeing what's going on.

The Earth observer sends out pulses 1000 per second, continuously.

When the travelling twin takes a round trip, let's assume he goes outwards for 10 minutes at a doppler shift factor of 10, and he goes inbound for 10 minutes at a doppler shift factor of 10

How many total pulses will the travelling twin see on his round trip? We have

10 minutes at 100 pulses/second on the outbound trip, where the pulses are redshifted
10 minutes at 10,000 pulses/second on the inbound trip, where the pulses are blueshifted

The reslut is that he sees 101,000 pulses.

Unsurpisingly, if you work out the appropriate velocity from the doppler shift formula

http://scienceworld.wolfram.com/physics/DopplerEffect.html

eq(1)
10 = sqrt((v+c)/(v-c))

and find the velocity v, then calculate the time dilation factor

gamam = 1 / sqrt(1 - v^2/c^2)

you'll find that the time dilation factor is 5.05 and that 5.05* 20 minutes = 101 minutes elapses on the Earth while the rocket twin is travelling.

So you can predict the exact amount of time dilation from the doppler shift factor k, by equating the total number of pulses emitted to the total number of pulses received over the round trip.

Note that the Earth observe will NOT assign the time coordinate t=10 minutes to the time when the turnaround occurs. You can use the Lorentz transform to determine the space and time coordinates that the earth observer assigns to the turnaround, or you can use the results above and a bit of intuition to realise it occurs at t=50.5 minutes, x = 50.5 minutes * velocity

By contrast, the travelling twin insists that turnaround occured at t' = 10 minutes, x' = 10 minutes * velocity

velocity here will be the solution to eq(1) and the same for both frames of reference

5. Jun 6, 2005

### TheAntiRelative

Great info! Thanks!

I still have a couple questions though to solidify the answer in my mind...

We use acceleration to determine the travelling twin because of his necessity to return to the earth frame. This lets us know whose time is dilated but it seems as though time dilation only actually occurs during acceleration by the above descriptions. Otherwise how is the twins paradox resolved? Both may think that the other is running slower but upon resolution of the question only one can be right. That one is the accelerated one so when he thought that the other's time was slower he must have been incorrect. Otherwise we're back to square one with Twins paradox aren't we?

It just seems like acceleration alone is the culprit and travel time, speed, and doppler shift are only consequeces or effects of the acceleration cause.

Can you explain if and why this isn't true?

6. Jun 6, 2005

### Janus

Staff Emeritus
No. Time dilation happens when the twins have a constant velocity to each other. Tine dilation is what each twin measures as happening to the other's clock (each measure the other's as going slow)
No. Each twin's viewpoint that it is the other twin's clock that is running slow during various points of the trip are equally valid. You cannot say that one is correct and the other isn't. What happens is that the twin that accelerates will see the other Twin's clock shift forward during the period of time that he is turning around at the end of the trip. This is caused by the Relativity of Simultaneity. This shift forward is no less real than the time dilation that either measures.

Thus the twin that does not accelerate sees the following as happening:

His twin travels away, his clock running slow, to a certain distance, turns around and comes back with his clock running slow the whole time, to end up having aged less when he returns. The fact that his brother accelerates during this trip does not add any additional factor.

The twin that accelerates sees the following:

He accelerates up to speed. While he does so, he will see the distance to the turnaround point shrink due to length contraction. As he moves away from the Earth, the Earth clock will be running slow. When he reaches the turn around point ( in less time by his clock than his brother measures for him to reach the turnaround point, since the distance is shorter as measured by him.) he accelerates in the opposite direction to slow down and head back to Earth. As he does so, the Earth clock by his measurement will gain time quickly until it actually reads more time than his own. Once he stops accelerating and is moving back towards Earth, the Earth clock will run slow again. By the time he reaches Earth, the time on Earth's clock will still read more than his own by the same amount that The Earth twin determined.

Thus both tiwns agree as to which one aged more, but they do not agree as to why. Now here is the part which probably the hardest part for some people to grasp. Both are right!. There is no way to say that one twin's view of the sequence of events that led to their difference in ages is more correct than the others. That is what is meant when one say's that according to Relativity, "time is relative". There's is no absolute time reference that we can compare the two clock's to and determine which one was 'really" running slow at any point. All that can be done is compare each clock to the other, and the answer you get depends on what frame you compare them from.

7. Jun 6, 2005

### yogi

Acceleration per se does not enter into the calculation of the age differences in SR. But this statment depends upon the authority upon which you may rely- as I have mentioned previously on these forums, both Max Born and Dennis Scima take the position that the age difference is explained by acceleration (e.g., either turn around or initial acceleration). Most authors however, take the position that the initial acceleration is only incidential to determining which twin moved after their clocks were brought into sync. Einstein clearly rejected the notion that acceleration was responsible for the age difference. However, Einstein did not propose a physical reason for why the two clocks should disagree when they are compared. The only theory that purports to do so is Lorentz Ether Theory - Lorentz proposed that the effects of high speed travel caused physical changes in time and lenght due to the change in the spaceing of interatomic factors such as the size of electron orbits and the like. Some modern ideas have attempted to relate the age differences to the energy of motion (along the lines of the time dilation experienced by clocks in the energy of the gravitational field). In this sense, the clock paradox is still a bewilderment

8. Jun 7, 2005

### pervect

Staff Emeritus
The "physical reason" why one clock reads lower than the other is much like the "physical reason" that if a trinagle copnnects three non-collinear points, the person who walks from A to B walks a shorter distance than the person who walks from A to C to B.

It's a property of geometry. The only slighty odd thing about it is that the geometry of space-time is Lorentzian, not Euclidean, so that there is a minus sign in front of the dt^2 instead of a plus sign. This is what makes the proper time longer for the straight path, rather than shorter.

9. Jun 7, 2005

### TheAntiRelative

So basically, though the period of time traveled determines the amount that the clocks deviate, the difference is only made up during the transition from the moving frame to the stationary one.

During travel, the two apparent reciprocal dilations cancel for calculation of the doppler effect.

That about right?

That would be interesting to see:
Cap Kirk recieves a distress signal from a planet only a light year away that their planet will explode in a year and a half. Travelling at nearly light speed he sees the planet and he thinks he's there in plenty of time. As he starts to decellerate, BOOOM! the planet suddenly seems to age super fast and then explodes in his face.

10. Jun 7, 2005

### TheAntiRelative

the poor people from the ground would see the ship coming and arrive in time but then just sit there nearly unmoving till the planet explodes.... Bummer

11. Jun 7, 2005

### pervect

Staff Emeritus
Nope. Not what would happen, not even close. And we've already posted at length about this same scenario to boot!

Quiz time:

1) If you are approaching an object rapidly (at a good fraction of 'c') is the light from it red-shifted or blue-shifted?

2) If you are approaching an object rapidly (at a good fraction of 'c'), and the object emits pulses at 1000 pulses/second, will you actually receive pulses

a) more rapildy than 1000 pulses/second
b) less rapidly than 1000 pulses/second

12. Jun 7, 2005

### TheAntiRelative

1) Blue shifted

2)Should be A) for Doppler but should be B) for time dilation since they should seem to have slower time than you.

We discussed that the clock suddenly shifts at the acceleration/deceleration point and that both see eachother as slow in the transit.

Theoretically during transit it seems like everything going slower than you would be slightly red-shifted because of the time dilation but then I know there's supposed to be a blue shift and these are apposing facts and not simply numbers that can be summed.

Hopefully you've struck upon what I'm missing and you have the answer that will clear it up for me. I thought I understood that it made up the difference at the ends but apparently that's not right either...

13. Jun 7, 2005

### Janus

Staff Emeritus
No. For the Earth twin, the difference accumlates during the whole of of the trip due to his twin's clock running slow.
For the "traveling" twin, the age difference is due to the fact that the Earth clock ran slow during the inbound and outbound trips, and briefly runs fast during the turnaround phase.

Relativistic Doppler effect takes time time dilation into account. They do not cancel out.
No. The planet would age quickly from Kirk's view when he starts to accelerate towards the planet. The distance to the planet would also shrink. Assume that he travels at .99c , the distance will contract to 1/7 of a light year for him. He will take some 1.7 months to cross that distance by his clock and during that time the planet will age some .24 months. But, As his accelerates up to .99c the planet will age some 11.76 months. For Kirk, the planet will explode while he is still accelerating up to speed. He won't see the planet until later when the light reaches him, but it will be long before he starts to slow to a stop.

14. Jun 7, 2005

### Janus

Staff Emeritus
No, they would never see the Enterprise coming. From their perspective, the Enterprise traveling at .99c would only have covered a little less than one half the distance separating them when their planet explodes. The light carrying the info that the Enterprise is coming travels at c and thus would have only crossed 1/2 the distance. This light would not reach the planet until 6 months after the planet explodes, closely followed by the Enterprise.

Note that while according to the planet, the Enterprise had reached the halfway point when the planet explodes, according to the Enterprise, they were no weren't at the halfway point when the planet exploded. This is the Relativity of Simultaneity coming into play. The event of the Enterprise reaching the halfway point and the planet exploding are simultaneous according to the Planet, but are not so according to the Enterprise.

Last edited: Jun 7, 2005
15. Jun 7, 2005

### pervect

Staff Emeritus
The first answer is correct. The second answer isn't what I was looking for, so it reveals some possible grounds for misunderstanding.

The answer I was looking for in the second answer is A) - faster than 1000 pulses per second.

Regardless of the time dilation effect, the light emitted by any source moving towards you will be higher in frequency (that's what blue-shifted means) than the light emitted by a source that is not moving. Both the carrier frequency of the light, and any modulation of it (such as the rate of pulsations) will be shifted upwards, by the same numerical factor.

The formula for "relativistic doppler shift"

sqrt((v+c)/(v-c))

already incorporates relativistic time dilation factor in the formula. There is no separate multiplication of the above formula for relativistic doppler shift by a time dilation factor.

So let's revist what Captain Quark sees in his viewscreen. (I always think of Kirk as flying FTL, and since our captiain is travelling slower than light, I'm using a different name).

What Quark actually sees on his viewscreen (after his viewscreen shifts the incoming light into the visible range) is the image of a planet and star that are aging faster than normal. The only place that any "slowing down" occurs is when he applies mathematics to his images and compensates for the changing distance between himself and the planet by subtracting out the travel time. This slowing down occurs not in what he sees, but in a metaphysical construct that he creates.

When Quark deaccelerates to match velocities, nothing dramatic happens at all. There will never be any dramatic physical effect of deceleration on what he actually sees in the viewscreen, all that happens is that the doppler shift factor smoothly drops towards one.

There are some places where dramatic things happen to the mathematical construct of "what time is it on the planet now". The point of view I am trying to get across is that "what time is it now" does not have any really well-defined meaning for two events that are not at the same point in space.

I suspect I haven't gotten this point across, fully. But for a start, we can get across what everyone actually "sees". Hopefully the metaphysical issues will clarify themselves in time.

16. Jun 7, 2005

### yogi

How does the moving clock know that its moving - wrt space .. so that it can slow down??? Where is the Physics?

17. Jun 8, 2005

### pervect

Staff Emeritus
The physics is very simple, really

There is no unique mapping of the time coordinate t1 measured by one observer to the time coordinate t2 measured by another observer - except when the two observers are located at the same point in space-time.

Instead of one unique mapping, there are many possible mapping - an infinite number

m1 : t1->t2, m2 : t1->t2, m3 : t1 ->t2, etc etc etc.

The only restriction on mappings m from t1->t2 is that if a light beam can travel from event #1 to event #2, event 2 occurs later than event 1 according to all observers. When two events are separated by a distance so large that light from the first cannot reach the second, some observers (mappings) will place event #1 as comiing "first", other observers will place event #2 as coming first.

When two clocks are separated and then re-united so that they are at the same location in space, there is a unique reading on clock 1, and a unique reading on clock 2, because there is no spatial separation (the two clocks are at the same point in space when they are compared).

To say this in less technical language:

Q: When it's noon on Jan 1, 2000 on the planet, what time is it on the space-ship a light year away?

A: The question is ambiguous

Q: Yes, but what time is it really
A: The question doesn't have an answer. Really.

Q: But I really want to know!
A: The question is not a question about physics, it is a question about coordinates. There are as many answers as there are coordinate systems. If you specify a specific coordinate system, the question may have an answer specific to that coordinate system, but a different coordinate system will give a different answer.

18. Jun 8, 2005

### learningphysics

But the planet will only explode when it has aged 18 months. So it seems like Kirk gets there 6 months early and saves the people right?

Just want to make sure I didn't miss anything.

19. Jun 8, 2005

### Janus

Staff Emeritus
It took 12 of those 18 months just for the distress signal to reach the Enterprise.

20. Jun 8, 2005

### TheAntiRelative

Now the twins paradox is back though isn't it? If Quark sees the planet aging really fast the entire trip then the planet sees Quark aging really fast. If they see the ship aging really fast then then he would appear to arrive before he does or some other strange effect.

according to the explanation of when the time shifts, clocks moving forward quickly during the deceleration (or acceleration)as percieved from the decelerating perspective would mean an extreme blueshift as well. More waves per local second produced by the planet.

Last edited: Jun 8, 2005
21. Jun 8, 2005

### TheAntiRelative

Though I may have not had good example durations and speeds, isn't there a set up where the planet explodes during deceleration? I thought clocks shifted at all points of acceleration.

In the twins excersize, concerning the forward shift of the Earth clock during acceleration period. What is the shift proportionate to? Is the local time interval of acceleration a factor? Or is it that .1G for 10 minutes is equal to 1G for 1 minute? If that is the case, one inertial system would always require the exact same amount of acceleration to line up with another inertial system regardless of the period of time used to apply it.

With this newer set of posts, I'm back to the original dilemma in the first post that on the outbound trip, the twin should SEE and redshifted image. Yet he should also SEE a rapidly aging planet behind him. To rapidly age would mean that any light source on the planet would appear to the travelling twin to produce more waves per one of his local seconds than it would if he were in the same inertial system. This would be a blue shift.

These two effects cannot be added. They are opposite final conclusions.

Last edited by a moderator: Jun 8, 2005
22. Jun 8, 2005

### learningphysics

Yes.
No. Why is this?

Are you referring to the non-accelerating parts of the trip?

In the outbound trip the moving twin sees the planet aging more slowly that the rate at which the planet ages in its rest frame (redshift). Also the planet is aging slowly in the moving twin's inertial frame than the planet ages in its rest frame.

In the inbound trip the moving twin sees the planet aging faster than the rate at which planet ages in its rest frame (blue shift). However in the moving twin's inertial frame the planet is actually aging more slowly than the planet ages in its rest frame. These are not in contradiction. It's just the effect of the doppler shift, and the light reaching the moving twin from the planet.

Suppose we're in the rest frame of the planet. The ship is moving towards us at a speed v. We fire
a photon towards the ship every m seconds as measured in our rest frame starting at t=0.

Assume that at t=0, the ship is at a distance L.
The equation of motion of the ship: x = L - vt
The equation of motion of the first photon. x = ct

Now what time does this photon hit the ship? Solve the equations, we get t1 = L/(v+c)

Now we fire a second photon at t = m
Equation of motion of the second photon: x=c(t-m)

What time does it hit the ship? Solve the two equations and we get. t2 = (L + mc)/(v+c)

So we have 4 spacetime events of interest here:

A) firing of photon 1: (x,t) = (0,0)
B) firing of photon 2: (x,t) = (0,m)
C) arrival of photon 1 at ship: (x,t) = (cL/(v+c) , L/(v+c))
D) arrival of photon 2 at ship: (x,t) = (c(L-mv)/(v+c), (L+mc)/(v+c))

Use the lorentz transformation to see how these 4 events occur in the ship's rest frame:
$$x' = \gamma(x + vt)$$
$$t' = \gamma(t + vx/c^2)$$

$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

A) $$(x',t') = (0,0)$$
B) $$(x',t') = (0, \gamma m)$$
C) $$(x',t') = (\gamma L, \gamma(L/c))$$
D) $$(x',t') = (\gamma L, \gamma(L/c) + (mc/(v+c))/\gamma))$$

Notice the time dilation effect of events A and B. The time difference has increased by the gamma factor.

Notice that the time difference between the two photon arrivals in the moving ship's rest frame is:
$$\frac{mc}{(v+c)\gamma}$$

This reduces to:
$$\frac{m \sqrt{1-v/c}}{\sqrt{1+v/c}}$$

The time between photon arrivals has decreased even though the time between photon emmissions has increased.

In the earth's rest frame the rate of photon emmission was 1/m. In the ship's rest frame the rate of photon arrivals is 1/(above expression) =
$$\frac{1}{m}*\frac{\sqrt{1+v/c}}{\sqrt{1-v/c}}$$

That's the doppler shift.

Last edited: Jun 8, 2005
23. Jun 8, 2005

### pervect

Staff Emeritus
What makes you think he should SEE a rapidly aging planet behindhim on the outbound trip?

We just got through arguing that he sees a rapidly aging planet on the INBOUND trip where there is BLUESHIFT. The same argument shows that he sees a slowly aging planet on the outbound trip where there is REDSHIFT.

I am going to describe in more detail how we know that redshift corresponds to seeing imagines in slow motion, while blueshift corresponds to "seeing" imagines in fast motion, in the hopes that this will clear up your confusion.

The logic is quite simple. Suppose we have a 1 Mhz radio signal that we are using to communicate with spaceship. (This isn't a realistic example, but it's easy to compute the numbers.)

We send the signal in 1 second bursts. This means that we send 1 million cycles of the signal before we turn it off.

Now, the signal gets redshifted to 100khz. How long is the pulse that the moving spaceship recives?

We know that we sent 1 million cycles during the pulse.

We know that the receiver recives the same number of cycles that we sent.

We know that the radio signal is red-shifted to 100 khz.

Therfore we know that the spaceship receives 1 million cycles at 100 khz.

We compute the time this takes - it's

1 million cycles * 1 second / 100,000 cycles

Thus the signal is 10 seconds long, rather than 1 second long.

This means that the traveller "sees" things happen in slow motion on his trip out (because there is redshift), not speeded up.

Now, let's compute the results for the blueshifted signal. The length of the pulse is now

1 million cycle * 1 second / 10,0000 cycles

So the blueshifted spaceship sees the pulse as only a tenth of a second long. This is watching what happens in fast motion. This is what the traveller "sees" on his inbound trip (because it is on the inbound trip that there is blueshift).

24. Jun 8, 2005

### Janus

Staff Emeritus
The rate of shift is proportional to the local acceleration and the distance to the undergoing the preceived shift. Also of importance is in what direction the clock lies with respect to that acceleration. If you are accelerating towards the clock (this includes decelerating while moving away, such as when the twin brakes to a stop at the turn around point.), the clock will fast forward. If you are accelerating away from the clock ( same as decelerating when moving towards the clock, as when the twin is braking to a stop at the end of the return leg), the clock will slow down.

The duration you accelerate and the factors above determine the total difference you'll see in the clock. (of course as you are accelerating towards or away from the clock your distance will change, so that factor will always be changing.)

25. Jun 9, 2005

### yogi

Pervect

If I sync two adjacent clocks on the planet and move one a fixed distance away as measured in the system in which the clocks were synchronized - there is an unambiguous relationship between the time on the two clocks - they will not be the same - but the temporal difference is precisely determined. They do not have to be proximate to be compared.

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