I hope it's clear that the intention was to say "After many moons, the rocket stops thrusting and ..."

Further, question c) would have been better stated as "how the atomic clocks will differ in time recording of the start and stop acceleration events, if at all."

Jorrie

You amaze me Jorrie, You come up with some of the dangdest things.

At any rate as I have no training in rocket science I am going to take a shot in the dark. Having observed the trajectory of many rockets launching and then flying one common factor I have noticed in all of them: They all seem to fly with the nose slightly higher than the tail. So as an attempt at stating what is probably obvious to everyone else. Wouldn't the travel distance of the nose be longer over "many moons" than at the tail.

If this is the case and based on a little knowledge of plane travel, in which people who travel by air alot actuall age slightly faster than those who don't. I would assume from this that the the clock at the front would be slightly "older" than the one at the tail.

I have no calculations to back this up just speculation based on observation.

Or I could be completely wrong LOL.

Thanks double_j_b, it sometimes give me a headache, though!

You wrote: "If this is the case and based on a little knowledge of plane travel, in which people who travel by air alot actuall age slightly faster than those who don't. I would assume from this that the the clock at the front would be slightly "older" than the one at the tail."

IF Guest (SL) above is right in his/her calcs, then after one year, the nose clock would be a cool 316 ns older then the tail clock - not quite an accelerated ageing product qualification test!

Your premise about the trajectories of the nose and tail of the rockets as a contributing factor might not be quite right, but your gut-feel answer is!

Regards, Jorrie

OK thanks for at least telling me I am close.

So let me clarify then. I was basing my assumptions on an orbital path. I guess that was incorrect. If we are then speaking of a linear path, then I will reach again.

We have essentially a rigid body rocket with no measurable nose colapse.

Query. Is the rocket stable enough not to elongate on deceleration. And would this cause the same effect.

I would like to add one other piece to my previous post.

The one other thing would be heat. the tail or the rocket would be hotter I would guess and thus extending the overall length of the rocket.

How much affect this would have on the timing I have no way of knowing.

Probably has something to do with quantum physics and alternate realities or some other scientific phenomenon of which I have no understanding.

I just like thinking out of the box. LOL

Hi double_j_b, you asked about the deceleration and the heat of the exhaust.

Firstly, the intention is that the rocket never decelerates - it just stops accelerating (see reply #1).

Secondly, the possible compression and perhaps lengthening of the rocket are transient effects, only happening for a very brief time. Compare that to a year or more of uniformly accelerated motion and those effects are negligible. It is the relativistic effects we are after.

Regards, Jorrie

P.S. I will post a short solution to the challenge here soon, followed by a detailed analysis on my CR4 Blog, so stay tuned!

Always fascinated to learn new things.

Right, here goes. As a rigid structure with a force acting on it as a whole, I can't see any differing influences on the two ends, or how there could be a difference. Between the two. From a viewpoint on the rocket, the rest of the universe would appear to alter, but the conditions on board remain constant. You don't say how long the rocket is, but I would assume a realistic size so distance between the two ends shouldn't play a part. Therefore, I think they should all read the same.

I now think I'm being too simplistic, but as I'm not an expert, stuff it. I'll let my answer stand!

Hi PlbMak, close, darn close, but not quite!

But yep, we are talking about long, but realistic length rockets.

Regards, Jorrie

Fun. Knowing the answers to your challenges are not the ones that jump readily to mind (such as that mind may be), I'm going to go sit in a corner for a while, maybe sleep on it, etc., hoping that something pops into my consciousness.

Hi Ken, yea, I suppose when atomic clocks enter the scene, things become relatively serious! (Pun intentional!). However, good old engineering is part of the problem!

Enjoy the contemplation! Jorrie

Well, I slept on it, and came up with something that was utterly and totally wrong. As I was sitting on the "throne", just after getting out of bed, I thought: Well duh! of course there is a difference in time because the two computers are different distances from the single atomic clock in the nose! Then I re-read the question, and saw that there are actually two clocks... so my post sleep inspiration was useless.

But then I thought, well, what about the Doppler speed indicators? Wouldn't the shift be different at the opposite ends of the rocket? Then I thought, hmmm maybe not.

So then I thought, well, if one clock above another one will run slower by gh/c², then if the clocks were 100 M apart, the nose clock would run maybe 1 part in 10¹⁵ slower (given the constant 1g acceleration). That wouldn't necessarily be detectable by itself in a day or two, given cesium clocks. But after many moons, there'd be a significant difference in the number of cesium heartbeats.

So the clocks should have different times on them.

So, then the acceleration profiles would be the same (always 1G).

The speed profiles (if calculated using the clock differences) would have the back of the rocket going slower.

I'm not sure that's exactly right... But I am being requested to go watch a movie, relatively soon, and relatively soon to one's wife is relatively sooner than one might think...

Hi Ken, you wrote: "So then I thought, well, if one clock above another one will run slower by gh/c², then if the clocks were 100 M apart, the nose clock would run maybe 1 part in 10¹⁵ slower (given the constant 1g acceleration)."

Well argued Ken! Just it's more like 1 part in 10¹⁴ for 100 m height. I think you used 1 for g and not the 9.81 m/s² as required. Not everyone will agree that this applies to a rocket in free space, but let's hear from the others.

I'll have a go at it!

If I remember correctly (which is difficult nowadays), my old physics teacher said that objects traveling near the speed of light would seem to deform, becoming flattened in the direction of travel.

This is the part where I'll be speculating. As the object flattens, the front will be accelerating slower than the rear (which causes the flattening effect). Since acceleration changes, average speed will also change. As for time...well, I don't really know. Relativity being what it is, different speeds also translates to different time rates. To further confuse myself, I remember my teacher saying that such things depend on where you're observing from. I think I tuned out after that...was that the end-of-class bell?

Hi Vulcan, your old physics teacher was right on all counts: objects traveling near the speed of light appear lengthwise compressed, the front appears to move slower and its clock appears to runs faster, but this is all as viewed from the un-accelerated reference of the base-station's, or from any other non-accelerated frame, i.e., any inertial frame.

Here all the sensors are on-board the rocket...

Regards, Jorrie

Assume v=c/10 Length = 100 m at rest (or with a tape measure at any time), relative l is roughly 90 m Over many moons, the nose clock will have traveled a relativistic 10 m less, but for 10/c longer time, resulting in a minuscully lower acceleration over an inversely miniscule time difference, and after acceleration ceases, the v will be identical, albiet the rocket dimension would still be recorded as the 10 m shorter.

Roughly 10 m/s/s, into 3 X 10^9 m/s is roughly 30,000,000 seconds. I guess 350 months is many moons, earth time. And after acceleration ceases, the nose and tail will appear to be moving at the same velocity. The difference in apparent v is only during acceleration. The apparent decrease in length will continue until deceleration occurs. Basically, thats the top of the head Newtonian answer (at 0.1 c, there is yet little difference between Newtonian and Relativistic views) At 30,000,000,000 seconds, the difference starts to show. You did not give us a top v/c, so the 10% was easy to ballpark late at night, without my slipstick.

RichH

Hi and hmmm....

RichH may have point, but I feel Jorrie has some more tricks in mind here, like will the nose and the tail of the rocket start accelerating simultaneously?

This will depend on the speed of sound in the rocket's metal. Has anyone got any idea of what a realistic value for this may be?

Then, if the nose start accelerating later than the tail, the nose must always be moving slower than the tail, so the rocket may really shrink!

SL

Hi Guest, you asked: "This will depend on the speed of sound in the rocket's metal. Has anyone got any idea of what a realistic value for this may be?"

Wikipedia has a value of 5100 m/s for a steel rod. I have no idea what it would be in a real rocket, which is more of a steel cylinder, I think.

One needs to know Young's modulus (E) and the density of the material (ρ); then for a rod, the speed of sound is: c_solids = √(E/ρ).

Regards, Jorrie

Jorrie,

From what you said to Ken and your reply on the speed of sound, I would give my "best buy" as:

The profiles for velocity and acceleration are identical between the nose and the tail of the rocket, just with a time lag. Taking the value of 5100 m/s for sound in steel and a 100 m long rocket, this means the nose profiles start and stop about 20 ms later than the tail profiles.

As for question 3: if the nose clock runs faster by 1 part in 1014, then it must be ahead of the tail clock by about 316 ns after one year.

BTW, I'm surprised that the speed after 1 year at 1g is only some 76% of light speed. Are you sure about this?

SL

Correction: my "As for question 3: if the nose clock runs faster by 1 part in 1014, then it must be ahead of the tail clock by about 316 ns after one year" must obviously read: "....1 part in 10¹⁴...."!

I'd take Jorrie's words: "not appreciably compressed in the longitudinal direction by the acceleration and the lengthwise transient oscillation and vibrations are negligible, i.e., it's a "near rigid body" problem." as meaning that sound propagation should have no effect (or at least that it is not one of the effects under consideration.)

Hi Ken, you wrote ".... as meaning that sound propagation should have no effect (or at least that it is not one of the effects under consideration.) "

I suppose SL feels that a 20 ms offset vs. a 316 ns clock difference is not an effect to be ignored - it's 2 orders of magnitude larger!

What I think you should sleep about is his acceptance of your premise of a slower nose clock, due to the equivalence between gravity and acceleration - yet he ignores the fact that the 'higher' clock also measures less gravitational acceleration!

Regards, Jorrie

Hi SL, you asked: "BTW, I'm surprised that the speed after 1 year at 1g is only some 76% of light speed. Are you sure about this?"

As the velocity gets relativistic, the relativistic total energy: E = mc²/√(1-v²/c²) shoots up rapidly, so the actual velocity increase per unit time is less. This shows itself in an decreasing acceleration in the original inertial reference frame.

I suppose you have calculated that in Newton dynamics the speed would just exceed the speed of light after one year. In relativistic dynamics, the speed can only approach c asymptotically.

Regards, Jorrie

Hi RichH

You wrote: "Assume v=c/10 Length = 100 m at rest (or with a tape measure at any time), relative l is roughly 90 m Over many moons, the nose clock will have traveled a relativistic 10 m less, but for 10/c longer time, resulting in a minuscully lower acceleration over an inversely miniscule time difference, and after acceleration ceases, the v will be identical, albiet the rocket dimension would still be recorded as the 10 m shorter."

Interesting way to look at it, but I don't think it is quite valid. Remember that all measurements are made onboard the rocket, so the length contraction of special relativity does not play a role here. An observer on the accelerating rocket will find the length to be constant, except for an extremely small compression due to the force applied at the tail.

Regards, Jorrie

P.S: Thanks for your later post on clarifying the "rocket stop" issue! Your assumption is right. I also noticed that, but only after the "edit period" has lapsed, so I posted a clarification in reply no 1 above.

J

I think the contraction does not come into play until well past 90% c. But yes, I was wrong about the contraction. At the low 1 g acceleration, achieving that figure would take about 3 X 10⁷ seconds, or roughly 1 year. Many moons, I guess. Once accelleration has stopped, even if after roughly 2 years to 99% c, the clocks, reporting back to the base for analysis as posed, would reveal no difference between the clocks. except when compared to a third clock that had remained stationary, and even there, the effect would be minimal until if I recall, about 99.99 % c

Thanks for pointing out how dull the top of my head has become in the 45 years since I worked with relativity in school.

RichH

Hi Rich, you wrote: "Once acceleration has stopped, even if after roughly 2 years to 99% c, the clocks, reporting back to the base for analysis as posed, would reveal no difference between the clocks. except when compared to a third clock that had remained stationary, and even there, the effect would be minimal until if I recall, about 99.99 % c."

The effect compared to the base station clock is a bit more severe than that. This table gives the rocket time (t_r) in years, base station time (t_b) in years, the distance travelled (x) in base station's frame in light years and the velocity in the base station frame (v/c).

The effect is significant after 1 year, but not drastically so. However, it's an oder of magnitude different after 3 years of on-board time. The time and distance effects are relatively easy to calculate, but I had to build a long spreadsheet (3 yrs @ .01 yr steps) to numerically integrate for the speed with reasonable accuracy.

Regards, Jorrie

PS: This thing really stresses the remaining old grey cells!

I presume your statement that "the rocket stops" refers to it ceasing it's burn/acceleration, not decelerating to zero relative velocity. You know how picky some people here can be.

RichH

I'm not sure that I understand how a Doppler speedo would work on a spacecraft. Would it: 1. sense some background radiation, select a coherent frequency from that, and then compare frequencies "in front" vs "behind", or 2. or send out a signal and detect a reflection off of some celestial body, or 3. do something else?

Hi Ken, you asked: "Would it: 1. sense some background radiation, select a coherent frequency from that, and then compare frequencies "in front" vs "behind",... "

The simplest would be for base station to continuously beam a coherent signal with a stable frequency to the rocket. Doppler shift can then be extracted simply and reasonably accurately.

Spectral shift in the light of a well-chosen star is also possible, but complex.

Regards, Jorrie

Since this particular challenge was posted on a Friday and appeared in the "CR4 Daily Digest" email on Saturday, it didn't do too well on the reply front - I guess by Monday it has moved over the "radar horizon" and few even noticed (or maybe it was just too crazy?). I am posting a brief 'solution' here, followed by a more detailed one on my CR4 Blog.

Guest SL (reply #18) had it right that the time of the nose clock would be ahead by about 316 ns after one year of acceleration - tiny, but quite measurable with atomic clocks. This is the same time that a clock on a 100 m high tower will gain in one year on a clock at ground level.

About the acceleration SL was wrong. One can again turn to a height of 100 m in a gravitational field and call up the equivalence principle. Here the effect is also 1 part in 10¹⁴, but unlike the case with time, this is not a cumulative effect and hence it is so tiny that no accelerometer could ever hope to measure it. So, apart from the time offset in the profiles, due to the speed of sound in the rocket's structure, there will be no other noticeable difference in the acceleration profiles. For severe acceleration of a very long structure, the difference in the accelerations between the front and the rear will however become quite measurable.

Now, what about the recorded speed profiles? Almost paradoxically, the recorded speed profiles will be identical - not even a 1 part in 10¹⁴ difference (although one can argue that no speedometer can be that accurate either). The reasoning behind it is quite complex and I will defer that to a detailed analysis that will soon be posted on my CR4 Blog, so watch out for it!

Regards, Jorrie

i.e., it's a "near rigid body" problem.

Length of the rocket is the only real variable between the two ends, the givens and assumptions are all constants for both....so my answer is.

Given that their is no compression or exstension to deal with that means both tail and nose units will start and stop at the same time and transmitt their data at the same time. The only thing that will change is the time at which the receiving station will receive each data pack. So assuming the tail is closer than the nose then the tail data will be received first, which means the "Nose" data will be older but only from the assumed receivers perspective. A receiver that is perpenticular will have a different perception as would a receiver out in front.

Now that I have read what other people have said...

Problem is you are using gravity in your thinking. The rocket is in "gravity free" space as per the assumptions. Rocket thrust while is measured in terms of gravity or accerlation can expressed in g's it is not the same as Gravity due to mass.

IF the rocket was sitting on earth with tail touching the ground and we assumed that the tail felt 1.0000000000000 g's, we could then calculate the force of gravity on the nose (lets say is 1000 feet up) and since I don't feel like doing the math lets assume the force of gravity would be 0.999999999999999 g's, then some of your math would work.

Hi, this is Guest SL.

I think the other guest is wrong. If someone would be riding with the rocket in a closed container, say as long as the rocket, there would be no way for this person to tell if the container is in gravity field or is being accelerated.

I do not think you can get away from the equivalence principle (between gravity and acceleration), so the front clock will gain time just as on a tower on earth.

SL

Einstein would argue, I think, that for the things in the rocket, the effect of 1G caused by earth or 1G caused by the rocket's acceleration is the same.

Although I can follow your logic, the difference in time from front to back (or top to bottom if you think of clocks on earth) is not due to the difference in gravitational attraction as it changes with distance. Instead, it is due to a difference in gravitational potential. The analogy in Newtonian physics would be the difference between the potential energy of a brick on a 100 meter tower and one sitting at ground level. (The potential energy normally calculated would be a large value, whereas the adjustment for body distance would be very tiny.) *

Using my favorite calculator, google, (and my usual approximation for the acceleration due to gravity) the difference in time calculates like this

((10 ((m / second) / second)) * 100 m) / (c^2) = 1.11265006 × 10^-14

^* Even in doing the Newtonian calculation for potential energy, you can't escape relativity, in a way. Suppose you have a brick at the top of a 100m tower. If it falls on your head, you will be convinced it had a lot of potential energy, when it was up there, swapped for kinetic on its way down. Suppose, then, after regaining consciousness, you see the brick on the ground. Clearly, it now has 0 potential energy. But then you dig a hole 100 meters deep next to the brick. Still no potential energy, right? Then you start to think about pushing the brick over the edge. Suddenly, if you are going to do the math, then the brick now has the same potential energy as it had at the top of the tower. Potential energy: it's all in your head.

Hi Ken, I like your parable of the "Newton brick"!

This whole thing can give one quite an headache, because there are relativists that preach that the time in front of the rocket will differ, but the proper (accelerometer) accelerations will be the same - which I think is a minority position, but just enough to confuse us laymen even further...

If the accelerations were the same in a 'rigid' rocket, then Bell's spaceship paradox would not break the string!

Also look at my Blog post on the solution to the rocket challenge and some replies.

Regards, Jorrie

Hi Jorrie:

Re headaches, you are so right. I continue to think about the doppler speed indicator, going around in circles re its own time base shifting from the gravitational effect vs the rocket's major time dilation relative to the surrounding space, etc.

But in any case, I always enjoy your posts, and any headaches are easliy cured with aspirin.

Thanks,

Ken

I'm going to go to Hollywood and ask a question from a movie (can't remember which one it was though).

If one of the clocks is ahead or behind in time. Does that mean that we won't be able to see it? For that matter, can the computer even get data from it, since it's in a different time zone?

This is apart from the subject of the post but I'm asking if we can even get the data. Another thought is that since time in the rocket itself has changed from our own time clock on earth, might the rocket seem to have been lost or disappeared?

The triplet paradox to keep aspirin in business. Our sensation of the clock in motion is the ticking, beating, pulsing that we simpletons use to track that 4th dimension in it's one way travel. (So far). Assume that you are aboard the rocket, under acceleration. Your sensation of time would not vary in relation to your clock. Your time would also slow. (What of an interferometer?)

So you have your doppler accelerometer receiving a signal from your base, to assess your rate of acceleration. Because your time has slowed, the base transmitter frequency would be generated at a higher time rate than you are. That means that there would be an increase in frequency, relatively speaking, in your time frame.

Quickie, would the increase rate from time dialation at some point equal or exceed your Hubble doppler effect? Would your red shift eventually become a blue shift?

Back to the future, So, from your frame. the incoming frequency that should be higher, is reduced by your v/c, you would be basing your speed determination on a higher frequency (at base frame), resulting in a falsely (relatively) slower that actual rate. The blue shift from the forward would be greater than the calculated value due to a higher frequency generation (your frame) than your on-board reference. Then add your true blue shift. Your interpretation of the data says you are traveling faster than you actually are. Even faster than your calculated rate. The effect is that the faster you go, the greater speed you go (from the front), and the faster you go, the slower you go (to the rear).

I realize that there is a bit of Newton there, and a bit of general and special. The contractile effect, (getting closer to the front without leaving the back so far behind) (equivalent to lengthening of your rocket.)

I begin to see why Albert couldn't control that hair.

RichH

Hi NoSciFi, interesting post! You wrote: "So you have your doppler accelerometer receiving a signal from your base, to assess your rate of acceleration."

The idea was a normal accelerometer and a Doppler speedometer, but yes, one can get acceleration if you have speed change and time change. The accelerometer is better for acceleration, because it is coordinate system independent.

"Quickie, would the increase rate from time dilation at some point equal or exceed your Hubble Doppler effect? Would your red shift eventually become a blue shift?"

Yes, if you keep on accelerating, everything, including the cosmic microwave background will become blue shifted - in fact all radiation from the front would get white hot at super-high frequencies! As good a reason as any not to try and travel near the speed of light...

"Your interpretation of the data says you are traveling faster than you actually are. Even faster than your calculated rate. The effect is that the faster you go, the greater speed you go (from the front), and the faster you go, the slower you go (to the rear)."

I do not quite follow you here - it is true that the blue shift due to time dilation opposes the Newtonian Doppler shift when you look to the rear, but the time dilation is a second order effect and is always 'beaten' by the straight Doppler shift. The faster you go, the more red shifted the signal from the rear, approaching infinite red shift (i.e., zero received frequency) when you approach the speed of light relative to the transmitter. It just shifts towards the red at a slower rate than in the straight Newton case.

Regards, Jorrie

"'Your interpretation of the data says you are traveling faster than you actually are. Even faster than your calculated rate. The effect is that the faster you go, the greater speed you go (from the front), and the faster you go, the slower you go (to the rear).'

I do not quite follow you here - it is true that the blue shift due to time dilation opposes the Newtonian Doppler shift when you look to the rear, but the time dilation is a second order effect and is always 'beaten' by the straight Doppler shift. The faster you go, the more red shifted the signal from the rear, approaching infinite red shift (i.e., zero received frequency) when you approach the speed of light relative to the transmitter. It just shifts towards the red at a slower rate than in the straight Newton case."

On board the rocket, my time is slowing, let's say to 10 seconds base frame equaling 1 sec of my frame time. My reference instruments are also sharing my time frame. My base doppler feed is being sent at 10^10 hz base frame. At the point where my frame is 10:1, that means that the base refernece is being transmitted at 10^11 hz my frame. At a 100:1 time ratio, the base signal would be sent at 10^12 hz my frame.

The relativistic effect would be far greater than the Newtonian, but would the Doppler frequency shift in the base frame signal (>hz) be, at some point <c, become so compounded by the increasing base frequency of time dilation that the Newtonian doppler effect begins to possibly even reverse (my frame, time dilation versus doppler freq loss)? Add to the Newtonian blue shift the effect of time dilation such that the 10^10 hz line is being generated at 10^12 my frame, plus the 2X doppler representing v=c, my instruments would interpret v=1.9999999 doppler, plus the 100X dilation effect. Would not my instruments read 200 c when the blue shift is 20000 % ? And the destination would read my blue shift as what? Minus 99.999999% freq loss due to time dilation plus 2X from near c? Would that not become a red shit to the destination?

A ref signal to my base stationary destination, in subtraction to the 1.99999 hubble shift, would find the -100X frequency retreating into a dramatic red shift at some point, because the faster I approach, the slower my time goes, as does my freq generator's time frame. Do you see the point of inversion, in both directions, at some value below c? Add the same effects to the base and destination reference beams, and I am truely becoming confused. At some point (.999999999999c), 1 m/s more would double my time dilation, thus my forward reference shift, according to my frame/base signal, and also the base frame/rocket signal would pass the point wherein the effect of time dilation becomes greater than the reference shift to the rear, and the forward could even appear to exceed c. Blue shift is a direct relationship, whereas red shift is an inverse, becoming infinite at c. and who knows what after. Of course, my time rate becomes so close to 0 approaching c, that I would have a different view than the base would.

???

Hi again NoSciFi; you wrote: "The relativistic effect would be far greater than the Newtonian, but would the Doppler frequency shift in the base frame signal (>hz) be, at some point <c, become so compounded by the increasing base frequency of time dilation that the Newtonian doppler effect begins to possibly even reverse (my frame, time dilation versus doppler freq loss)?"

No, I do not believe the relativistic effect ever becomes large enough to reverse good old Newton! The opening velocity relativistic one-way Doppler factor can be written:

f/fo = (1-v/c)/√(1-v²/c²), where f is the rocket's received frequency and fo the base frequency. Above the line is Newton and below the line time dilation. This function always decreases for increasing v, obviously within the limits v < c, meaning increasing redshift. My website's download from this page: inertial movement has more on the relativistic Doppler effect.

Regards, Jorrie

I wrote: "f/fo = (1-v/c)/√(1-v²/c²), where f is the rocket's received frequency and fo the base frequency."

Here is a graph of the above Doppler function:

The purple line is the linear, first order Newtonian factor; The blue graph is the second order time dilation factor (below the line). It is clear that when they are divided, giving the red line, the Doppler factor is always decreasing for increasing v.

Regards, Jorrie

Hi Jorrie,

That blue "time dilation" line of your graph looks like a semi-circle. Is it, and if it does, any special significance?

SL

hi, Jorrie.

First, your f/fo should be fo/f. At 0 v/c, fo/f = 1, at v/c <1, f_o observed frequency would always be less than f.

As I recall relativity, at c, time stops. At .9c, I read a t_r/t_b of 4.36. Newton would read 0.1 f_o. Add my relativistic dilation, and my f_o in my time frame would be the 0.1 Newtonian multiplied by the 4.36 dilation factor, 0.436 f, or an apparent v = 0.564 c, according to my slowed instruments. Move on to 0.99 c, and t_r/t_b becomes 14.2. Newtonian doppler would read 0.01 f, with dilation, 0.142 f, .858 c. Let's go to .999 c. Newton would read 0.001f, dilation would bring this to 44.7 X Newton, or 0.0447 f, .9553 c

Let's go one more. 0.9999 c. Newtonian = 0.0001 f. Dilate by a t_b/t_r 141.4, and suddenly f_o/f_b = 0.0141 , .9859 c

0.99999 c, f_newton =0.00001 f_base, dilate by 447, f_o = 0.0047 f, 9953 c

0.999999 c, f_newton =0.000001 f_base, dilated by 1414, f_o = 0.001414 f, .9986 c

Lets go to the front f_d = destination frequency. At v/c =1 (limit) f_o/f_d = 2, Newtonian. At v/c= .4, f_o Newton = 1.4 f_d At v/c .4, dilation of 1.52 results in 1.4 X 1.52, f_o of 2.13 f_b, faster than c.

c would appear (f_o/f_d =2) at about v=.32 c

At f_o/f_d 3, v = 2 c, time dilation at about .53 c.

Maybe wavelength would work better forward.

Aspirin, anyone?

Bugger the aspirin, my headache is so bad I'm going strait for the morphine!

Hi again NoSciFi

You wrote: "First, your f/fo should be fo/f. At 0 v/c, fo/f = 1, at v/c <1, f_o observed frequency would always be less than f."

Huh? OK, I see, but you just read my definitions of frequency the wrong way round - I, perhaps unwisely, used fo (f-zero) for base frequency and f for received frequency, so the formula is right. And as I read your calcs, we get the same answers as well! My graphs also confirm that, except that I plot 1/sqrt(...), i.e., the inverse of the time dilation factor.

The only thing I differ on is your conversion of the relativistic Doppler factor back to a Newtonian velocity again. What meaning does that have? In the forward direction, the relative speed always stay below c.

Regards, Jorrie

I realize, and have some appreciation for relativity. Time dilation is the source of many "illogical" paradoxes. I just don't recall anyone pointing out the effect of reaching c to a destination while traveling at 1/2 c leaving the base. Since the fastest man has gone to date is about 0.00004 c, we've probably got a few decades before anyone has to worry about the relativisitic effects of 0.01 c.

So, assuming your 1 g acceleration, reaching about 0.5 c (leaving speed) after about a year, what is our approximate travel time to reach our neighbors 5 light years away?Our instrumentation would tell us that we are approaching at .999 c, which should give us 4 years of inertial travel, plus 1 year of deceleration, which would appear to us to be 3 years base time, or 6 months destination time.

The greatest relativistic effect we've achieved for a man is a 0.001 % dilation, and even that for a relatively brief week or 2. That would mean that the lunar mission could have lost as much as 5 seconds on the round trip. How much did their chronometers actually lose?

I'm not sure that even relativity is applicable forward. The concept of traveling > c is about running away from light, either traveling with the photons, or dropping the frequency of the beam to 0. That would not be the same effect as approaching something at > c. Probably another exclusionary effect. I really don't remember the exclusion principle. Maybe I tripped on it.

Cheers,

RichH

Hi RichH

I'm trying to understand how you got 5 seconds lost for a Lunar mission. The average speed for a swing around the moon and back (like the unfortunate Apollo 13), would have had an average speed in the order of 6 km/s, or ~0.00002c.

This translates to a time dilation factor sqrt(1-v²/c²) = 0.9999999998, or 1 part in 5 x 10¹⁰. This works out to about 100 μs per week. Where did I go wrong?

SL

The greatest relativistic effect we've achieved for a man is a 0.001 % dilation, and even that for a relatively brief week or 2. That would mean that the lunar mission could have lost as much as 5 seconds on the round trip. How much did their chronometers actually lose?

What about the two Voyager spacecraft that were sent off into the outer reaches of the solar system over a quarter of a century ago. They are traveling at an average speed of around 20 Kms^-1 and toped out at around 45 Kms^-1. If we work on the slower speed the dilation would be around 0.007% so over the 25 years the clocks would be out by something like 14 or 15 hours by now.

Great accomplishments, all of them, but they lacked the "man" part I was strictly referring to. I was giving the Apollo's an over-estimate of 40,000 km/hr. As I recall, they had to do substantial skipping into the atmosphere to lose speed, but I'd guess that they would have had to gain 25,000 km/hr gravitational velocity from the lunar orbit. I'd guess that that would have been about their V_max on the trip.

RichH

Hi Masu

You wrote: "If we work on the slower speed the dilation would be around 0.007% so over the 25 years the clocks would be out by something like 14 or 15 hours by now."

I'm afraid I can't buy that! According to my calcs, the time dilation alone (0.00000022222%) could only produce some 1.75 seconds difference (at 20 km/s for 25 years). Remember time dilation is a factor √(1-v²/c²), or [1-√(1-v²/c²)]/100%,

But that's not the real point - the gravitational blue-shift of those craft, being much farther from the Sun than us, would more than compensate for the velocity time dilation. In fact, the clocks on the Voyager twins should be ahead of our clocks! I haven't made the sums, but it will probably be in the single digit seconds as well.

Hope this makes sense, but I may be wrong - I hope someone will check my calcs.

Regards, Jorrie

Hi Jorrie,

I'm afraid I can't buy that! According to my calcs, the time dilation alone (0.00000022222%) could only produce some 1.75 seconds difference (at 20 km/s for 25 years). Remember time dilation is a factor √(1-v²/c²), or [1-√(1-v²/c²)]/100%,

You are absolutely correct, please forgive my brain stall, it was 03:00 in the morning when I did the calculation and I was obviously running on empty.

Those tiny craft are now over 15,000,000,000 Km out and still going which in itself is an incredible achievement. The round trip for a radio signal is now close to 28 hours and they are using a transmitter that is pumping out less than 5 watts for the return signal.

To put that in perspective Voyager can get is message across 15 Tm with a 5 W transmitter and the average Rock band need 10 Kw to get their message across 50 m. Makes you wonder doesn't it

Hi Masu, you wrote: "Those tiny craft are now over 15,000,000,000 Km out and still going which in itself is an incredible achievement."

Those are the Voyagers, I guess. Interestingly, although the Pioneer craft (10 and 11) are now silent, they are still creating a stir in the relativity world, by means of the so-called "Pioneer anomaly". The two craft both travelled a bit slower than expected, with an anomalous acceleration in the direction of the Sun of about 0.1 nano-g (~10^-9 ms^-2).

This may be just a mundane effect, e.g., the nuclear powerplant scattering particles forward, or it may show a misunderstanding of gravity at large distances. There are various rival gravitation theories that claim to explain it, but the jury is still out...

The effect could apparently not be extracted from Voyager data due to the way they were stabilized, swamping any possible effect. I believe the same is true for Cassini, but the New Horizons mission to Pluto may help to verify the Pioneer anomaly.

Regards, Jorrie

Or the Pioneers, 5 years earlier, and 30 years of being the most distant man made object, until Voyager passed them 5 years or so ago. The computer you are working from is 10 times what any of them have in capacity.

RichH

I think that the ability to sift the signal from the noise factor is still remarkable with the old Voyagers, and lack of signal in all of the noise of the other is of some significance here.

RichH