Special relativity: Can you explain the paradox?

[ScaryPirateMan]

So then the atomic clocks show different times on two different machines because one thing is accelerating while the other remains in place? But then what makes this true? What is happening within the accleration that is making time slow for the stationary object. Or, is it that it will have appeared to slow for the moving object but appear not to have to the stationary?

 

 

[Sisyphus]

So then the atomic clocks show different times on two different machines because one thing is accelerating while the other remains in place? But then what makes this true? What is happening within the accleration that is making time slow for the stationary object. Or, is it that it will have appeared to slow for the moving object but appear not to have to the stationary?

 

Less time will have passed for the clock that accelerated and was brought back to rest again. It's what actually happens, not just appearances.

 

As for why it happens, the short answer is that all of special relativity can be deduced as logical necessities from the observed fact that light moves at the same speed in all frames of reference. In other words, for that to be possible, a, b, c, and d all have to be true, where a, b, c, and d are some pretty amazing properties of time and space. And, sure enough, a, b, c, and d are experimentally confirmed!

 

For a more specific answer, you're asking a very big question. "Explain relativity to me."

[vuquta]

The distances M-A and M'-A' are different when measured from the M' frame then they are when measured from the M frame. For instance, if the relative velocity difference between the frames is 0.5 c, and and M-A and M'-A' are both 1 km according to M, then according to M', M-A will be 0.866 km and M'-A' will be 1.15 km.

 

If M' and M are co-located at the same instant as A' and A, in the frame of M, they cannot be co-located at the same instant in the M' frame. (In the above example, they will be 0.289 km apart in the M' frame when A' and A are co-located.)

 

So if both frames agree that A'/A occurs at t=t'=0, then only one frame could say that M' and M are co-located at t=t'=0.

 

So what. This is length contraction.

 

You are having a problem not me.

 

Einstein said M and M' are co-located. Yes or no.

 

Then Einstein claimed for all light pulses, you can co-located the light pulses and set the clocks to zero. Yes or no?

If no, then LT construction is impossible and so is his alleged proof of consistency.

 

Now what?

 

So then the atomic clocks show different times on two different machines because one thing is accelerating while the other remains in place? But then what makes this true? What is happening within the accleration that is making time slow for the stationary object. Or, is it that it will have appeared to slow for the moving object but appear not to have to the stationary?

 

Here is the mainstream paper that supports this

 

 

http://arxiv.org/abs/physics/0411233

 

Uniform acceleration under SR is absolute and both frames agree the accelerating frame beats more slowly.

Edited September 22, 2010 by vuquta

[swansont]

So what. This is length contraction.

 

You are having a problem not me.

 

Einstein said M and M' are co-located. Yes or no.

 

Then Einstein claimed for all light pulses, you can co-located the light pulses and set the clocks to zero. Yes or no?

If no, then LT construction is impossible and so is his alleged proof of consistency.

 

Now what?

 

 

And because of length contraction, you won't have clocks in agreement at other points — clock synchronization uses d=ct

[between3and26characterslon]

 

 

Uniform acceleration under SR is absolute and both frames agree the accelerating frame beats more slowly.

 

Are you saying that time dilation only occurs in an accelerating frame?

[vuquta]

Are you saying that time dilation only occurs in an accelerating frame?

 

Yes.

 

And because of length contraction, you won't have clocks in agreement at other points — clock synchronization uses d=ct

 

I mostly agree with you.

 

But, with length contraction, it would be uniform for the A and B lightning strikes points and thus would sync at d'=ct'.

 

So that does not present a problem, but your length contraction does.

 

Let me introduce observers F and R and F' and R' where the primed frame is the embankment.

 

Then, if F and F' are co-located when M and M' are co-located, in the frame of M', then F and F' will not be co-located from the view of M.

 

However Einstein was clear, when a light is emitted, frames can set their clocks to 0 at the common emission point. This is also fundamental to the light cone.

 

Also, Einstein said the strike were simultaneously emitted from the view of M' when M and M' are co-located.

 

Clearly, this thought experiment is a paradox because we cannot perform all the actions Einstein said we are allowed to do.

[Janus]

 

 

However Einstein was clear, when a light is emitted, frames can set their clocks to 0 at the common emission point. This is also fundamental to the light cone.

 

Also, Einstein said the strike were simultaneously emitted from the view of M' when M and M' are co-located.

 

No he didn't. Any statement made about M and M' being co-located at the time of emission applied to the embankment frame (M) alone.

 

Clearly, this thought experiment is a paradox because we cannot perform all the actions Einstein said we are allowed to do.

 

It is only a paradox in your mind because you are misinterpreting his meaning.

[D H]

Here is the mainstream paper that supports this

 

http://arxiv.org/abs/physics/0411233

 

Uniform acceleration under SR is absolute and both frames agree the accelerating frame beats more slowly.

Are you saying that time dilation only occurs in an accelerating frame?

Yes.

 

vuquta: The paper that you cited does not support your position. Neither does experimentation.

Edited September 24, 2010 by D H

[vuquta]

vuquta: The paper that you cited does not support your position. Neither does experimentation.

 

I have no ides what you are talking about. It is standard mainstream that the accelerating clock beats more slowly.

Here is experimental proof.

 

The effects are emphasized for several different orbit radii of particular interest. For a low earth orbiter such as the Space Shuttle, the velocity is so great that slowing due to time dilation is the dominant effect, while for a GPS satellite clock, the gravitational blueshift is greater. http://relativity.livingreviews.org/Articles/lrr-2003-1/

 

In the GPS experiment, the satellite is in a lower amount of gravity ie less acceleration and hence will beat faster for this GR effect or the more accelerated earth clock beats faster.

 

Now, for the paper, you will see,

t = c/g sinh(gT/c)

 

Where g is the acceleration measured in the accelerated frame and T is the proper time in the context of the accelerated frame. t is the time in the inertial frame.

 

If you plot the graph, you will find t > T or the accelerating frame is time dilated. Oh, and this is an absolute agreement between the frames.

 

Here is a link that gives a simpler analysis.

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

 

In short, the paper I posted backs my position. The accelerating frame is time dilated.

 

No he didn't. Any statement made about M and M' being co-located at the time of emission applied to the embankment frame (M) alone.

 

It is only a paradox in your mind because you are misinterpreting his meaning.

 

First the embankment frame is M'.

 

Now, he said M and M' are co-located when the strikes occured simultaneously in the context of M'. That is exactly what I said.

Now, I said if M' is co-located with M, then M is co-located with M'.

How did I conclude this?

 

Well, in Einstein's "claimed" proof of consistency, he said,

 

At the time t = t' = 0, when the origin of the co-ordinates is common to the two systems

http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

Now, this implies co-location is part of the logic of SR since the origins are common. As such, it is not a theorem of SR that when M is co-located with M', M' is not co-located with M.

 

Therefore, using the Einstein logic above, M and M' are mutually co-located.

 

So, you are wrong.

 

 

 

It is only a paradox in your mind because you are misinterpreting his meaning.

 

I have no idea what you are talking about. You will note I quote Einstein for my positions.

 

So, quote him for yours.

Edited September 25, 2010 by vuquta

[D H]

I have no ides what you are talking about. It is standard mainstream that the accelerating clock beats more slowly.

Here is experimental proof.

 

The effects are emphasized for several different orbit radii of particular interest. For a low earth orbiter such as the Space Shuttle, the velocity is so great that slowing due to time dilation is the dominant effect, while for a GPS satellite clock, the gravitational blueshift is greater. http://relativity.li...les/lrr-2003-1/

 

In the GPS experiment, the satellite is in a lower amount of gravity ie less acceleration and hence will beat faster for this GR effect or the more accelerated earth clock beats faster.

You are wrong, for two reasons. The first is selective reading. Read the text that you yourself quoted: For a low earth orbiter such as the Space Shuttle, the velocity is so great that slowing due to time dilation is the dominant effect. It is clearly stating that time dilation due to relative velocity and time dilation due to gravity are two different effects. General relativity adds gravitational time dilation to the mix. Gravitational time dilation is not the sole explanation of time dilation. The second reason you are wrong is that this paper is yet another that does not support your interpretation. In general relativity the GPS satellite is not accelerating. It is gravitational potential, not gravitational acceleration, that causes gravitational time dilation.

[vuquta]

You are wrong, for two reasons. The first is selective reading. Read the text that you yourself quoted: For a low earth orbiter such as the Space Shuttle, the velocity is so great that slowing due to time dilation is the dominant effect. It is clearly stating that time dilation due to relative velocity and time dilation due to gravity are two different effects. General relativity adds gravitational time dilation to the mix. Gravitational time dilation is not the sole explanation of time dilation. The second reason you are wrong is that this paper is yet another that does not support your interpretation. In general relativity the GPS satellite is not accelerating. It is gravitational potential, not gravitational acceleration, that causes gravitational time dilation.

 

I want to make sure I understand you correctly. Are you claiming the accelerating frame beats faster in time?

 

Next, I realize time dilation due to acceleration/gravity and time dilation due to frame speed differentials are different concepts.

That has nothing to do with what we are talking about.

 

We are strictly taking about the effect on time due to acceleration. You are now adding other variables that were not part of the acceleration/decceleration of the OP. The OP did not include a relative motion phase as the normal twins paradox would indicate. So, in the context of this thread, the time dilation is due to acceleration only.

 

But, while we are on paradoxes and you want to talk about time dilation due to inertial velocity differentials in GPS, SR mandates reciprocal time dilation.

Can you explain why GPS proves time dilation is not reciprocal.

 

In other words, the GPS satellite does not view the earth clocks as time dilated as mandated and required by SR.

 

Oh, I forgot to mention gravational mass and inertial mass are not distinguishable. This is why regardless of the mass, all objects move at the same acceleration when dropped toward the earth.

 

Therefore, in GR, time dilation due to a gravitational differentials and time dilation due to acceleration differentials are the same.

 

Further, I produced many papers proving the sin hyperbolic function is used to calculate the time dilation of uniform acceleration relative to an inertial frame.

Do you dispute these mainstream calculations?

Edited September 25, 2010 by vuquta

[swansont]

I want to make sure I understand you correctly. Are you claiming the accelerating frame beats faster in time?

 

I see nothing in D H's post that would lead one to conclude this.

 

Next, I realize time dilation due to acceleration/gravity and time dilation due to frame speed differentials are different concepts.

That has nothing to do with what we are talking about.

 

We are strictly taking about the effect on time due to acceleration. You are now adding other variables that were not part of the acceleration/decceleration of the OP. The OP did not include a relative motion phase as the normal twins paradox would indicate. So, in the context of this thread, the time dilation is due to acceleration only.

 

Yes, it did. It had a 100-day travel period.

 

But, while we are on paradoxes and you want to talk about time dilation due to inertial velocity differentials in GPS, SR mandates reciprocal time dilation.

Can you explain why GPS proves time dilation is not reciprocal.

 

In other words, the GPS satellite does not view the earth clocks as time dilated as mandated and required by SR.

 

SR reciprocity assumes an inertial frame of reference, and SR does not account for gravitational effects. An orbit is not an inertial frame, and gravity is present.

 

 

———

 

 

Incidentally, the measurement of time dilation has been done in an accelerating reference frame, sans gravity. Mossbauer spectroscopy using a rotating cylinder. You can solve it with either the kinematic dilation or use a potential from acceleration, like you do in GR. Either way, you get v^2/2c^2

 

Measurement of the red shift in an accelerated system using the Mossbauer effect in Fe-57

Phys. Rev. Letters. 4, 165 (1960)

H. J. Hay, J. P. Schiffer, T. E. Cranshaw, and P. A. Egelstaff

 

Measurement of Relativistic Time Dilatation using the Mössbauer Effect

Nature 198, 1186 – 1187 (22 June 1963)

D. C. Champeney, G. R. Isaak and A. M. Khan

[vuquta]

I see nothing in D H's post that would lead one to conclude this.

 

I am just trying to clarify this. I think he is claiming I do not believe SR concludes reciprocal time dilation. My post on this very subject for an alternative twins experiment should make that obviously false. So, I have never said anywhere that SR does not conclude reciprocal time dilation.

 

 

Yes, it did. It had a 100-day travel period.

 

OK, reciprocal time dilation for the frames for the 100 days, time dilation only for the accelerated frame when accelerating.

 

 

 

 

SR reciprocity assumes an inertial frame of reference, and SR does not account for gravitational effects. An orbit is not an inertial frame, and gravity is present.

 

 

GPS uses two adjustments, one for GR effects and one for velocity differentials.

 

Now, let's assume a satellite is directly over a unit that is in open space and inertial. This means the unit has a y value only and x = 0.

 

At that instant, the satellite emits a light pulse. The satellite will contend that light travels a distance √[(vt)² + y²] to meet the unit. The unit contends light only has a y component since it is at rest. Given the constancy of light for both frames, the clock of the unit must beat slower from the view of the satellite given the two different frame view on the distance light travels.

 

Now, we put this same logic using the earth.

 

Sure, there exists a gravity component, but there exists a competing SR time dilation component as verified by the logic of Neil Ashby as I posted.

 

However, from the view of the satellite, it turns out to be false that the earth clock beats slower in this context. Only the gravity effects are valid. Hence, the light travel distance differential does not exist. Therefore, light is not behaving by experiment the way SR says it should.

 

You see, this is not strictly are gravity differential issue. The satellite is not moving at the same speed as the earth unit.

 

 

 

 

Incidentally, the measurement of time dilation has been done in an accelerating reference frame, sans gravity. Mossbauer spectroscopy using a rotating cylinder. You can solve it with either the kinematic dilation or use a potential from acceleration, like you do in GR. Either way, you get v^2/2c^2

 

Measurement of the red shift in an accelerated system using the Mossbauer effect in Fe-57

Phys. Rev. Letters. 4, 165 (1960)

H. J. Hay, J. P. Schiffer, T. E. Cranshaw, and P. A. Egelstaff

 

Measurement of Relativistic Time Dilatation using the Mössbauer Effect

Nature 198, 1186 – 1187 (22 June 1963)

D. C. Champeney, G. R. Isaak and A. M. Khan

 

I've never disputed these results.

[D H]

A simple thought experiment for you, vuquta.

 

Ignore that interstellar space is not empty. Pretend that a vehicle plowing through interstellar space at some velocity relative to the solar system will maintain that velocity forever. Ignore that stars have a non-zero velocity with respect to the Sun. Assume humanity has developed some technique that enables it to send probes to other stars. Humanity however has not progressed all that much. The probes still use the burn-coast-burn approach that is used today.

 

We'll send out two probes, one to Alpha Centauri (distance =4.36 light years) and another to Tau Ceti (11.9 light years). Each probe will accelerate toward its target star and 10 g (98.1 m/s²) for 20 days as measured by the probe, coast for some time, and the decelerate at 10 g for 20 days to come to a rest with respect to the target star (and the Sun, per assumptions). In other words, both probes undergo the exact same acceleration profile. The two probes differ only in the direction of travel and the duration of the coast period.

 

My question to you: After correcting for the finite speed of light, what is the difference between the time on Earth and the timestamp on the signals sent back by the two probes? Special relativity has an answer. What does vuquta relativity say the answer is?

Edited September 25, 2010 by D H

[vuquta]

A simple thought experiment for you, vuquta.

 

Ignore that interstellar space is not empty. Pretend that a vehicle plowing through interstellar space at some velocity relative to the solar system will maintain that velocity forever. Ignore that stars have a non-zero velocity with respect to the Sun. Assume humanity has developed some technique that enables it to send probes to other stars. Humanity however has not progressed all that much. The probes still use the burn-coast-burn approach that is used today.

 

We'll send out two probes, one to Alpha Centauri (distance =4.36 light years) and another to Tau Ceti (11.9 light years). Each probe will accelerate toward its target star and 10 g (98.1 m/s²) for 20 days as measured by the probe, coast for some time, and the decelerate at 10 g for 20 days to come to a rest with respect to the target star (and the Sun, per assumptions). In other words, both probes undergo the exact same acceleration profile. The two probes differ only in the direction of travel and the duration of the coast period.

 

My question to you: After correcting for the finite speed of light, what is the difference between the time on Earth and the timestamp on the signals sent back by the two probes? Special relativity has an answer. What does vuquta relativity say the answer is?

 

 

 

I am not going to do a homework problem. I will explain how it is done.

a - acceleration in the accelerating frame.

T - Time in the accelerating frame of the acceleration.

t - Time in the inertial frame that corresponds to the T

d - the distance traveled during acceleration.

v - the resultant relative velocity v achieved by the acceleration.

 

Earth frame proper time

Acceleration phases

 

Hence, t = 2 * (c/a) sinh(aT/c) for those phases

 

Relative motion phase

v = c tanh(aT/c)

 

To calculate the distance traveled by the accelerating frame, it is

d = 2 * (c²/a) (cosh(aT/c) - for both the accel and deccel phases.

 

From the view of the earth, it will be the star distance - d. Hence, t = (SD-d)/v for the relative motion phase.

 

Therefore, the earth clock will elapse (SD-d)/v + 2 * (c/a) sinh(aT/c)

 

Accelerating frame.proper time

T(total) = 2 * T

d = 2 * (c²/a) (cosh(aT/c) - for both the accel and deccel phases.

t = (SD-d)/v

 

Elapsed time = 2 * T + (SD-d)/v

 

 

Earth frame.view of the accelerating frame's elapsed time

2 * T + [ (SD-d)/v ] / γ

 

 

Accelerating frame view of Earth's elapsed time

2 * t + [ (SD-d)/v ] / γ

[D H]

Earth frame.view of the accelerating frame's elapsed time

2 * T + [ (SD-d)/v ] / γ

 

Accelerating frame view of Earth's elapsed time

2 * t + [ (SD-d)/v ] / γ

Wrong. You are ignoring length contraction.

 

 

 

[pioneer]

Special relativity has three aspects, one for mass, one for distance and one for time. Relativistic mass was included to avoid some of the confusion created by relative reference, since you can't have relative mass when looking at two relative references. Rather relativistic mass is connected to energy via E=MC2, therefore needs kinetic energy via velocity. If we could directly measure relativistic mass with a RM meter, one could tell where the energy is and get beyond relative reference.

 

In the original mental experiment, one reference accelerated. We know who has the energy/relativistic mass and therefore the kinetic energy. There is no paradox. The paradox only appears when one avoids an energy balance by using only 2 of 3 SR variables.

[vuquta]

Wrong. You are ignoring length contraction.

LOLOLOL.

 

First, you need to understand acceleration is absolute motion.

 

D H

Posted Today, 05:01 PM

 

vuquta, on 25 September 2010 - 03:44 PM, said:

 

Earth frame.view of the accelerating frame's elapsed time

2 * T + [ (SD-d)/v ] / γ

 

Accelerating frame view of Earth's elapsed time

2 * t + [ (SD-d)/v ] / γ

 

 

Next, given d is absolute and SD is also, SD - d is the correct answer.

 

Then, you can create length contraction as follows.

 

ta = 2 * t + [ (SD-d)/v ] / γ

 

ta - 2t = [ (SD-d)/v ] / γ

 

(ta - 2t)v = (SD-d) / γ

 

There are you happy now?

 

Special relativity has three aspects, one for mass, one for distance and one for time. Relativistic mass was included to avoid some of the confusion created by relative reference, since you can't have relative mass when looking at two relative references. Rather relativistic mass is connected to energy via E=MC2, therefore needs kinetic energy via velocity. If we could directly measure relativistic mass with a RM meter, one could tell where the energy is and get beyond relative reference.

 

In the original mental experiment, one reference accelerated. We know who has the energy/relativistic mass and therefore the kinetic energy. There is no paradox. The paradox only appears when one avoids an energy balance by using only 2 of 3 SR variables.

 

What?

 

If the relative motion period adjustment exceeds the acceleration adjustment, then the paradox exists regardless of mass.

 

Do you have math to back up this assertion?

[D H]

Measurement of the red shift in an accelerated system using the Mossbauer effect in Fe-57

Phys. Rev. Letters. 4, 165 (1960)

H. J. Hay, J. P. Schiffer, T. E. Cranshaw, and P. A. Egelstaff

 

Measurement of Relativistic Time Dilatation using the Mössbauer Effect

Nature 198, 1186 – 1187 (22 June 1963)

D. C. Champeney, G. R. Isaak and A. M. Khan

I've never disputed these results.

Yes, you have. You did so when you said that acceleration is the cause of time dilation.

 

 

First, you need to understand acceleration is absolute motion.

If by that you mean that proper acceleration is Lorentz invariant, I agree. If you mean something else you need to elaborate.

 

 

Next, given d is absolute and SD is also, SD - d is the correct answer.

d is not absolute. It isn't even absolute in vuquta relativity. You have time dilation, and time dilation and length contraction go hand in hand. Your result is mathematical nonsense and also is in contradiction to experimental results.

 

 

 

 

Edited September 26, 2010 by D H

[D H]

BTW, the correct answers to my question in post #89:

In both cases, the probes will accelerate for 20 days as measured by the spacecraft clock, or 21.08 days on Earth. By this time the probes will have covered a distance of 0.015898 light years as measured from the Earth and achieved a velocity of 0.512008 c. The probes will then drift for some time at this velocity until it is time to accelerate (decelerate) to come to a stop (relative to the Earth) at the target star.

So, how long do the probes need to drift? The acceleration at the end of the flight is simply the reverse of the acceleration undergone at the start. The probe will need to start firing once again when it detects that the distance to the target star is equal to the distance to the Sun at the end of the initial boost phase. That distance is not 0.015898 light years. It is instead 0.0136561 light years thanks to length contraction. The distance to the target star will also undergo length contraction. At the point when the boost phase ends, Alpha Centauri will appear to be only 3.7315 light years away as opposed to 4.3441 light years away (the original 4.36 light years less the 0.015898 light years traveled during the boost phase). The coast phase needs to end when Alpha Centauri appears to be 0.0136561 light years distant. The coast phase thus lasts 7.2613 years from the perspective of the probe but 8.45339 years from the perspective of the Earth.

Putting it all together, the trip to Alpha Centauri takes 8.56884 years as measured by someone on the Earth but only 7.37082 years as measured by the probe, a difference of 1.19802 years.

The calculation of what happens to the probe sent to Tau Ceti is similar. In this case the trip takes 23.2952 years from the perspective of an Earth-bound observer but only 20.0205 years per the spacecraft clock, a difference of 3.27472 years.

Edited September 26, 2010 by D H

[between3and26characterslon]

OK, reciprocal time dilation for the frames for the 100 days, time dilation only for the accelerated frame when accelerating.

 

Earth frame.view of the accelerating frame's elapsed time

2 * T + [ (SD-d)/v ] / γ

 

 

Accelerating frame view of Earth's elapsed time

2 * t + [ (SD-d)/v ] / γ

 

My understanding of relativity is somewhat different to yours.

 

Einstein said, "That if we start with certain axioms then propositions are true if they can be logicaly derived in the recognised manner from those axioms. This then reduces to a question as to the truth of the axioms. This question is not only unanswerable but is entirely without meaning, we can not ask if it's true that only one line goes through two points. We can only say that Euclidian geometry deals with things called straight lines to each of which is ascribed the property of being uniquely determined by two points situated on it"

 

There are very specific rules to follow in SR one of which is it only considers frames which are Cartesian coordinate systems based on Euclidean geometry, inertial, non rotational and in a state of uniform rectilinear translation.

 

If frames have the properties described above then they will see each other as time dilated and length contracted. Read this though.

 

http://www.einstein-...ights/dialectic

 

However if frames start of as stationary then they do not meet the specific rules of relativity and if you change the axioms then you will have different propositions which are true. So if two frames are stationary relative to each other and one of them experiences a force (as per the law of inertia) it accelerates. It experiences time dilation due to acceleration (like you say). It now has a velocity which the other frame does not have (you can not accelerate A by applying a force to B )

 

In this situation one frame does have speed and the other does not. The one with speed will experience time at a slower rate than the stationay one and they will agree on this. If B is accelerated, and therefore has velocity then it will have to assign a (-)negative value to A's velocity in the Lorentz transformation when working out A's time, B will see A's time go faster.

 

Watch this lecture, it's full of maths, you'll love it. The point of interest is between 1h 30min and 1h 40min

 

http:// www. youtube. com/watch?v=BAurgxtOdxY (take out the spaces, I put them in cos this is only 1 of 8 vids and they're better to watch on youtube)

[swansont]

My understanding of relativity is somewhat different to yours.

 

 

!

Moderator Note

An understatement, to be sure. FYI, vuquta has been banned (details in the banned user log post), so there is no further point in responding to him

[Heretik]

As soon as you say the word, "accelerate", the scenario is not subject soley to Special Relativity. Special Relatively only pertains to the special case of objects travelling in a straight line at a constant speed. As soon as the speed is altered General Relativity is invoked. Seeing as Brian is the one being accelerated relative to Andy, he is the one to experience time dilation. Both he and Andy on his return agree that Brians clock is slow but the math to figure out by how much is beyond me, i'm afraid.

[swansont]

As soon as you say the word, "accelerate", the scenario is not subject soley to Special Relativity. Special Relatively only pertains to the special case of objects travelling in a straight line at a constant speed. As soon as the speed is altered General Relativity is invoked. Seeing as Brian is the one being accelerated relative to Andy, he is the one to experience time dilation. Both he and Andy on his return agree that Brians clock is slow but the math to figure out by how much is beyond me, i'm afraid.

 

You can use SR to account for accelerations, as long as you are doing so an an observer in an inertial reference frame. You'll get the same answer you get by using GR. GR is necessary when you include gravity, because being at rest in a gravitational field is an accelerated reference frame, and freefall in a uniform gravitational field is an inertial frame.

[Craig Dilworth]

Special relativity says that a clock on a spaceship moving relative to a stationary observer appears to be running more slowly than if it was stationary. So if this is true, what would you get in the following scenario?

 

Stationary observer Andy and moving observer Brian are both initially stationary and in the same location, and each has an atomic clocks measuring time in days, and displaying what they can see on a large monitor visible to both. In addition both clocks emit an electronic "pip" whenever one day passes to the next, and each records the number of pips made since reset. Day = 0, and the pip counter is set to 0.

 

Brian rapidly accelerates towards a distant star, and it takes him one second to reach around half the speed of light. After 100 days of travelling Brian stops in 1 second and accelerates to the same speed back towards Andy (in one second), and arrives back at Andy in a further 100 days.

Before they got out their clocks to compare what they showed, they each gave their opinion as to what they were expecting them to show.

 

 

Brian reports that Brian's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Brian reports that Andy's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

Andy reports that Andy's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Andy reports that Brian's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

When they both look at their clocks, and read their pip counters, what do they see?

 

Thanks