Special Relativity, transversal movement and the first postulate

[Rettich]

I am trying to get grip on the special relativity theory and therefore I have created a thoughtexperiment. Would you be so kind to tell me what mistakes I have made?

 

Imagine the following:

 

There are 3 spacestations (A,B and C) far out in space, not experiencing any apparent gravitational force. These spacestations keep the same distances and angles to each other. A, B and C form an isosceles triangle. The distance between A and B, and also between A and C, is 70 lightyears (ly). The distance between B and C is 14 ly. A, B and C have "synchronised" their clocks. This meaning: A has send a signal to B and C saying "it is now 1-1-2015" (day-month-year) and on arrival of that signal B and C has set their clocks at 1-1-2085. We now know that if B sends a signal to C saying "it is now 1-1-2090" that this signal will reach C at 1-1-2104 local time. The synchronising of the clocks helps us with the obervations done when a rocket R travels from B to C.

 

We will consider the clocks A, B, C and R. These clocks are able to observe each other constantly with a powerful telescope.

 

R leaves B, accelerates very quickly towards a constant speed of 0,96c and on arriving at C decelerates very quickly. According to Special Relativity the time passed at R will be 7/25 th of the time A, B and C have calculated the trip of R will take. The role of the classical Doppler effect will be marginal (i.m.o) in the case of A observing R and R observing A (because the angle between AB and AC is very small).

 

Take the following example: R leaves B at 1-1-2200 (local time B and local time R). At the moment of leaving R observes the clock A indicating 1-1-2130. A observes R leaving B at 1-1-2270 (local time A) and sees the clock R indicating 1-1-2200. R arrrives at C on 1-8-2214 (local time C) (as a result of the time necessary for traveling 14 ly at speed 0,96c). On arriving at C the clock R indicates 1-2-2204 (the time that has passed for R is 7/25 x 25/24 x 14 = 4 years and 1 month) and R observes clock A indicating 1-8-2144. Thus during his trip R observes clock A ticks away 14 years and 7 month. A observes the arrival of R at C on 1-8-2284 (local time A). A observes the clock R when leaving B indicates 1-1-2200 and on arrival at C R indicates 1-2-2204

 

During the trip. A observes clock R going slower than his own clock. R observes clock A i.m.o. all the time going faster than his own clock. This result would violate the first postulate because A and R see each other during the uniform motion in the same manner. Acknowledging the fact that the special relativity theory is working exceptionally well in all experiments and routinely applications this would mean that the special relativity is incomplete.

 

[Janus]

The fact that R observes A differently than A observes R in this situation in no way indicates that SR is incomplete. It is, in fact what SR predicts in this case.

 

You have to take into account that R does accelerate on leaving B. And this acceleration changes how he sees A. It causes an aberration in the light making it seem like it is always coming from in front of him. The amount of aberration depends on his relative speed and the angle at which the light hits him in the frame of A.( which is changing as he moves from B to C)

 

In turn, he sees a transverse Doppler shift which also depends on his speed and the difference between the direction of the light source at emission, and the aberration angle.

In this case, it always lead to the blue shift during the trip as he sees this shift immediately upon acceleration

 

A however, does not accelerate, and only sees a Doppler shift from R some 70 years after R leaves B. He also sees the light from the R coming from first one direction from a 90 degree angle and then from the other direction as R move from B to C. So for half the trip he will see R blue-shifted and for the other half Red-shifted. The total accumulative transverse Doppler shift he sees will match what he would expect time dilation of R during the trip.

 

This does not violate the first postulate, because as I pointed out above, the situations of R and A are not completely symmetrical.

 

Even you remove the acceleration from the problem (R is always moving at 0.96c and just passes B and C), the "view" of A by R as it moves from B to C, will not be the same as the "view" of R by A as R passes from B to C. That is, while there is a period of time for R when A appears to go from being in front to being behind, it is not the period during which R is crossing the distance between B and C.

[Rettich]

Thanks Janus. It's clear that there is a lot more for me to learn about special relativity.

[Rettich]

Janus, you rightly pointed to the red- and blue-shift. For that reason I have slightly modified my thought-experiment. For reasons of clearity I describe my thouhgt-experiment again in full. I would have liked to attache a smal drawing but I didn't succeed.

 

Imagine the following:

 

Three spacestations A, B and C are in deep space, not experiencing any apparent gravitational forces. These spacestations keep the same distances and angles to each other. A, B and C form a right-angled triangle with AB and BC have a right angle. The distance between A and B is 8 lightyears, the distance between A and C is 10 lightyears and therefore the distance between B and C is 6 lightyears. A, B and C have their clocks "synchronised". This meaning: A has send a signal "it is now 1-1-2100" to B and to C. On arrival of that signal B has set his clock at 1-1-2108 and C has, on arrival of that signal, set his clock at 1-1-2110. This "synchronising" helps me with the observations done when a rocket R travels from B to C; during that travel clocks A, B, C and R constantly observe each other.

 

Rocket R leaves B at 1-1-2200 (local time B and R) for a trip to C . R accelerates in 1 second to a stationary speed of 0,8c. On arriving at C R decelerates in 1 second to a hold. So R arrrives on C at 1-7-2207 (locale time C).

 

Clock R has ticked away during his trip 4,5 years (3/5 x 6/0,8).

 

On leaving B R observes clock A gives 1-1-2192. On arrival at C R observes clock A gives 1-7-2197. R has seen clock A tick away 5,5 year. Conclusion: In total R has seen clock A tick away more time than his own clock.

 

A observes R leaving B at 1-1-2208 (local time A) and observes R arriving on C at 1-7-2217 (local time A). Therefore A sees ticking away his own clock 9,5 year.

 

A observes, at seeing R leaving B, clock R gives 1-1-2200 and sees clock R giving 1-7-2204 on arrival at C. Therefore A observes clock R ticking away 4,5 year. In total A has seen clock R tick away less time than his own clock.

 

During the uniform motion of R A observes the clock R moving slower than his own clock and R will also observe the clock A moving slower than his own clock. Considering the whole trip: R has seen clock A ticked away more time than his own clock. Therefore R has to observe clock A moving faster than his own clock during acceleration and/or deceleration. As far as I know experiments have confirmed that acceleration or deceleration has no effect on time dilatation, only speed has.

 

Then how is it possible that R at any moment observes clock A going faster than his own clock? Where did I made a mistake? What did I miss?

 

[Janus]

 

Then how is it possible that R at any moment observes clock A going faster than his own clock? Where did I made a mistake? What did I miss?

 

 

Again, you are not taking Relativistic transverse Doppler effect into account.

 

Again, due to the Relativistic aberration of light, after R reaches 0.8c on leaving B, he will see the light from A coming at him from an angle of 16 degrees from directly in front of him. As he moves towards C, this angle becomes wider.

 

The Transverse Doppler shift depends the difference between the angle at which the he sees the light coming from and the angle of the light at the time of emission.

 

When he is at B and moving at 0.8c, this is a blue shift factor of 1.667 (5/3) for the light coming from A. If he saw this blue shift for the whole trip, he would see A age 7.5 years. However, he doesn't. Since the angle between his path between B and C and the direction to A changes during the trip, so doe the transverse Doppler shift from A. The changing Doppler shift causes the blue shift to drop off during the trip, as a result, total time he sees pass for A during his trip is less than 7.5 years and adds up to 5.5 years, exactly as expected. In fact, in this particular scenario, R does not see A age faster than he does during the total trip. Towards the end of the trip he actually sees A age slower. But what counts is the total time passage that he sees for A during the trip.

 

Again, as long as you apply SR to the problem properly, it is in entire agreement with the final results. SR is completely consistent (this has been generally proven), There is no need to examine each scenario from all the different angles to confirm that they will agree.

 

Sure, you could go to all the trouble of confirming that the accumulative transverse Doppler shift seen by R adds up to 5.5 yrs, but it would be an lot of work for nothing.

[Rettich]

Sorry, I just thought you would only get blue shift if you were nearing an object, I didn''t know you could also get blue shift when you are moving away

[DimaMazin]

Sorry, I just thought you would only get blue shift if you were nearing an object, I didn''t know you could also get blue shift when you are moving away

Only when your mass is very great and your speed of moving away is very low.

[Rettich]

In my second though-experiment the rocket R (traveling from B to C) moves at any time further away from A with high speed. That is why I thought that during the stationary movement there would always be a red shift. Therefore I can't understand the reason for R observing clock A have moved faster than his own clock at the end of the trip.

[DimaMazin]

In my second though-experiment the rocket R (traveling from B to C) moves at any time further away from A with high speed. That is why I thought that during the stationary movement there would always be a red shift. Therefore I can't understand the reason for R observing clock A have moved faster than his own clock at the end of the trip.

Maybe I was wrong earlier.

http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

f_{s /} f_{0 =}gamma(1+ v _* cos theta /c)

blue shift is when f_s/f₀ < 1

then:

gamma(1+v_*cos theta/c) < 1

v_*cos theta < (1-gamma)c/gamma

[Rettich]

DimaMazin, I looked at the link, but rocket R leaves B at a right angle to A. When I look at the formula I think R observes A therefore always redshifted. I am afraid Janus somehow misread the text of my second thought-experiment.

[DimaMazin]

DimaMazin, I looked at the link, but rocket R leaves B at a right angle to A. When I look at the formula I think R observes A therefore always redshifted. I am afraid Janus somehow misread the text of my second thought-experiment.

I am afraid we misunderstood Janus. See relativistic explanation of light aberration:

http://en.wikipedia.org/wiki/Aberration_of_light#Relativistic_Explanation

 

tan(phi)=sin(theta)/gamma(v/c+cos(theta))

 

You can see deflection of perpendicular lines at different speeds and change of color shift from "red shift" to "blue shift".

http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

Visualisation

Diagram 2. Demonstration of aberration of light and relativistic Doppler effect.

Edited January 21, 2014 by DimaMazin

[Rettich]

I don't see it. Could of course be my lack of understanding or the fact that english is not my mother language.

[Janus]

DimaMazin, I looked at the link, but rocket R leaves B at a right angle to A. When I look at the formula I think R observes A therefore always redshifted. I am afraid Janus somehow misread the text of my second thought-experiment.

There are a couple of different formulas to that can be used for transverse Doppler shift. Which one you use depends on which frame you are measuring the angle in.

 

The first is

 

[math]f_o = \frac{f_s}{\gamma (1+ \frac{v}{c} \cos \theta_o)}[/math]

 

where theta is the angle to the source at the time the light was emitted ( according to the observer).

 

Note, that in this case the light being received by the observer, when R is next to B, would have have left A, years prior to then. So in this case, it is easier to thing in terms of R having been moving for some time before reaching B. IOW, the light he sees when passing B is light emitted from A before R reaches B.

 

Now I know that you have R starting at rest at B and then accelerating, but this doesn't change what he would see in terms of Doppler shift after accelerating. He would see the same thing as someone at the same spot moving the same speed who had always been moving.

 

But you still have to figure out the B-R-A angle at the time that the light left A.

 

So in this case, it is easier to measure angles form A.

 

The Doppler shift formula then becomes

 

[math]F_o = \gamma \left ( 1-\frac{v}{c} \cos \theta_s \right )[/math]

 

And theta is measured by the source (A) at the time of emission.

 

Now it is obvious that it is the light emitted at 90 degrees to R's motion, tht reaches R when it is next to B, so this makes theta =90 in this case, and the formula reduces to

 

[math]f_o = \gamma f_s[/math]

 

[math] \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

which equals 5/3 at 0.8c

 

So the observed frequency by R when it is next to B and moving at 0.8c is 5/3 that of the frequency that leaves A. (R sees a blue shift).

 

The B-C-A is 53.13 degrees, so this the angle that light leaving A must travel to reach C and thus is the value for theta when R reaches C.

 

The above formula then give an answer of 0.866, which means that when R is at C, he sees a red-shift from A.

 

Thus, As R goes from B to C, he sees the Doppler shift in the light from A go from a blue shift to a red shift, The accumulative time he sees pass for A will be what it should be.

Edited January 21, 2014 by Janus

[Rettich]

Thanks Janus. With all my poor knowledge of physics, it looks like a clear explanation.

[Rettich]

So the theory predicts a blue shift, but, is there any conformation (experimental or otherwise) on this alleged blue shift?

[Rettich]

Also if the conclusion is that R has seen clock A tick away more time than his own clock during constant movement, than A has to see clock R tick away more time than his own clock during constant movement (due two the second postulate). This does not look to be the case. I think that this problem is at any case not solvable without taking acceleration/deceleration in acount.

Sorry, I mean the first postulate

[xyzt]

 

The Doppler shift formula then becomes

 

[math]F_o = \gamma \left ( 1-\frac{v}{c} \cos \theta_s \right )[/math]

 

And theta is measured by the source (A) at the time of emission.

 

Small correction:

 

[math]\frac{f_o}{f_s}= \gamma (1-\beta cos \theta_s )[/math]

 

Based on the above, redshift occurs for all:

 

[math]\frac{f_o}{f_s}<1[/math]

 

meaning for all

 

[math]\theta>\theta_{critical}=arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math]

Edited January 23, 2014 by xyzt

[md65536]

Also if the conclusion is that R has seen clock A tick away more time than his own clock during constant movement, than A has to see clock R tick away more time than his own clock during constant movement (due two the second postulate). This does not look to be the case. I think that this problem is at any case not solvable without taking acceleration/deceleration in acount.

Janus explained this. Observers R and A each see the other blue-shifted (clocks appearing to tick faster) but for different amounts of time. R sees it for a longer duration, because the effect that it sees immediately is delayed by the travel time of light before A sees it. Thus both can agree that observer A is seen to tick away more time overall. You can work it out considering only durations of constant motion, but those durations won't appear to be the same for the different observers.

[Janus]

Also if the conclusion is that R has seen clock A tick away more time than his own clock during constant movement, than A has to see clock R tick away more time than his own clock during constant movement (due two the second postulate). This does not look to be the case. I think that this problem is at any case not solvable without taking acceleration/deceleration in acount.

Sorry, I mean the first postulate

 

No, The difference here is that you are comparing the light seen by R as its passes from B to C, to the light seen by A as it watches R move from B to C.

 

Again, this is easier to visualize if you consider R as having been moving at 0.8c for sometime before passing B. The light he sees coming from A as he passes B, is light that was emitted from before he was next to B. So when he gets to B, he sees light that was emitted from a point that was in front of him when it was emitted.

 

Now consider the light seen by A as he watches R pass from B to C. The light he sees showing R passing B, was emitted at a right angle to R's path and at a time when R was neither "in front" or "behind" of A. And he will not even see that light until some time after R reaches C.

 

IOW, R is measuring the light from coming from A starting with light that was emitted A from before he was even with A and A is measuring the light coming from B starting with light that was emitted from R when R was even with A. They are comparing different relative segments of each others relative paths.

 

The acceleration has no real effect on this. I can consider two rockets, R1 and R2. R1 is like above where it has always been moving at 0.8c. R2 waits at B until R2 reaches B and then instantly accelerates to match speeds with R1 so that they are flying side as R1 passes B. The instant R2 and R1 match speeds they will see the same Doppler shift from A, and A will see the same Doppler shift from R1 and R2 as it watches them travel from B to C.

 

So again, the difference in Doppler shift seen by and A and R come from the difference between whether the Light being measured reaches the observer when source and observer are at their closest to each other when it arrives at the observer(R's view) and when the source and observer were at their closest when the light leaves the source (A's view).

[DimaMazin]

Small correction:

 

[math]\frac{f_o}{f_s}= \gamma (1-\beta cos \theta_s )[/math]

 

Based on the above, redshift occurs for all:

 

[math]\frac{f_o}{f_s}<1[/math]

 

meaning for all

 

[math]\theta>\theta_{critical}=arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math]

beta=v/c

theta_s - theta_o = arctan(v/c)

What is theta_critical ?

[xyzt]

beta=v/c

theta_s - theta_o = arctan(v/c)

What is theta_critical ?

The angle where blueshift changes into redshift. The point was that it is smaller than [math]\frac{\pi}{2}[/math].

Edited January 27, 2014 by xyzt

[md65536]

[math]\frac{f_o}{f_s}= \gamma (1-\beta cos \theta_s )[/math]

 

Based on the above, redshift occurs for all:

 

[math]\frac{f_o}{f_s}<1[/math]

 

meaning for all

 

[math]\theta>\theta_{critical}=arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math]

That should be all

[math]\theta_s < \theta_{s\mathrm{\ critical}}=\arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math].

 

Your theta_s is the angle in the source frame. At 90 degrees light will be blueshifted. The angle in the observer's frame (theta_o, using http://en.wikipedia.org/wiki/Relativistic_aberration_formula) will be greater, due to aberration... it seems the critical theta_o will always be greater than 90 degrees (light coming from the side is red-shifted).

Edited January 27, 2014 by md65536

[xyzt]

That should be all

[math]\theta_s < \theta_{s\mathrm{\ critical}}=\arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math].

 

Your theta_s is the angle in the source frame. The angle in the observer's frame (theta_o, using http://en.wikipedia.org/wiki/Relativistic_aberration_formula) will be greater, due to aberration... it seems the critical theta_o will always be greater than 90 degrees (light coming from the side is red-shifted).

Incorrect, at [math]\theta>\theta_{critical}[/math], the blueshift changes into redshift, as I said.

 

 

 

At 90 degrees light will be blueshifted.

 

Incorrect as well, at [math]\theta=\frac{\pi}{2}[/math] light is redshifted, not blueshifted. In the notation used by the formula [math]f_0=f \gamma (1-\beta cos \theta)[/math] , [math]f_0[/math] represents the proper frequency (the frequency in the frame of the emitter). See Moller, "Theory of Relativity", top of page 62, formula (90). The confusion may be caused by the fact that this notation is opposite to the one used by wiki.

Edited January 27, 2014 by xyzt

[md65536]

Incorrect as well, at [math]\theta=\frac{\pi}{2}[/math] light is redshifted, not blueshifted. In the notation used by the formula [math]f_0=f \gamma (1-\beta cos \theta)[/math] , [math]f_0[/math] represents the proper frequency (the frequency in the frame of the emitter). See Moller, "Theory of Relativity", top of page 62, formula (90). The confusion may be caused by the fact that this notation is opposite to the one used by wiki.

Check again, have you copied the wrong formula?

 

f_o increases as theta increases. Higher angles means increasingly blue-shifted. Using the formulas in this thread, an angle of pi means the light source is directly approaching, and is blue-shifted.

 

The angle theta equals theta_s in your formula. At angle pi/2, the light is blueshifted, not redshifted.

[xyzt]

Check again, have you copied the wrong formula?

No, you should try reading the book, it is very good.

 

 

 

The angle theta equals theta_s in your formula. At angle pi/2, the light is blueshifted, not redshifted.

 

From whose perspective? think about it.

Edited January 27, 2014 by xyzt