# Special Relativity - simple questions?

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The Special Theory of Relativity claims that an observer in a spacecraft  moving with the speed v relative to e.g. Earth, will always see that the clocks on Earth run more slowly than the clock in the spacecraft, according to the formula for time dilation T = To / (1- (v / c) ^ 2)) ^ 0.5.

Let's check the credibility of this claim in the following example:

T  The reference clock on Earth simultaneously generates radio waves with the frequency f = 1000 Hz, and has an external, very large visible from space, digital display of the date measured by this clock. Assume that the spacecraft on 1.01.2020 takes off from Earth and at a speed of v = 180,000 km /s performs a travel to a newly discovered object in space, distant from Earth by 1.5 light years,  in order to photograph it properly.

N    Now, let's answer the following simple questions:

1.    What frequency  f ' generated by the clock on Earth will be measured  by an observer on a spacecraft,  during his return flight towards Earth at a speed  v = 180,000 km /s,  knowing that the light clock in motion runs slower than at rest and  that the Doppler effect occurs here?

2.     What date will indicate the light clock on the spacecraft the day before reaching the Earth?

3.    What date shown on the clock display on Earth, will be seen and photographed by the observer on the spacecraft the day before reaching Earth?

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SR does not claim that such an observer will always see the Earth clock run slow, if by see, you mean what his eyes or instruments directly record.   In this usage of see, he will see it run at a rate of    T =  To ((1-v/c))(1+v/c))1/2

where v is positive if Earth and the Observer are receding from each other and negative if they are approaching each other.

A factor contributing to this observation is the the distance and thus the propagation time for signals is constantly changing, getting longer when receding and getting shorter when approaching.    This factor works out to be c/(c+v)

When you factor this out of the first equation you are left with the time dilation equation.    This means that there are two things to consider: what you see happening to the Earth clock, and what is happening to the Earth clock.

So while while receding from the Earth, the observer will see the the 1000 Hz signal as being 500 hz and the Earth clock as ticking 1/2 as fast as his own,. Taking into account the effect of the increasing distance, he will determine that the Earth clock is ticking 0.8 as fast as his own.  He will meet up with the object when his own clock reads 1.01.2022 (as the distance between Earth will be only 1.2 ly as measured by him and this is how long it takes to traverse this distance at 0.6c.)   He will see the Earth clock reading  1.01.2021, but determine that it is 8.07.2021 on the Earth at that moment.

Now at first, you might be tempted to think " But wait, if he sees 1.01.2021 on the Earth clock, and the Earth is, according to him, 1.2 ly away, wouldn't that mean that it should be 3.15.2022 on the Earth by his reckoning?"  This is not the case.  The light he is seeing at that moment left Earth at a time when the distance between them was less than 1.2 ly, so the time it took the light he is seeing took less than 1.2 years to reach him from the Earth.

Now he accelerates in order to come start the trip back towards Earth.  We will assume a minimal acceleration period.  Now this is the part where people tend to get tripped up.  After he is done and is now approaching the Earth and not receding, we will assume that he still reads 1.01.2021 on the Earth clock by visual means.  However, he will no longer conclude from this that it is 8.07.2021 on the Earth.  Instead he will conclude that it is 6.05.23.     During the return trip he will see a frequency of 2000 hz from the signal and the Earth clock tick twice as fast as his own.  But again, taking into account the decreasing distance effect, he will conclude that the Earth clock is ticking at a rate 0.8 as fast as his own.

Thus he will see the Earth clock tick from 1.01.2021 to 1.01.2025, but conclude that it ticked from 6.05.23 to 1.01.2025 during his return leg. (see will see it tick off 4 years, but conclude that it ticked off 1.6 years.

Again, it all come back to what happens during that acceleration period.   As far as anyone at rest with respect to the Earth is concerned,  nothing special beyond the standard SR effects take place.   But for the observer actually undergoing the acceleration, things aren't this simple.  For him, the rate at which clocks run depend on which direction they are from him relative to the acceleration he is undergoing and the distance from him in that direction.  Clocks in the direction of the acceleration run fast, and those in the opposite direction run slow (beyond what he sees.  This even effects clocks that share his acceleration. A clock in the nose of the Ship will run fast and one in the tail will run slow. ( in this case, since there is no changing distance between himself and the clocks, what he sees, will be in perfect agreement with what is happening to the clocks.

While this may seem to be at odds with common sense, it is how a Relativistic universe works.

A problem with your questions is that they only deal with particular points of the whole scenario without taking in the whole picture.

It like comparing two men walking and only considering where they end up.

Below we have the paths of two men, Red and Blue, over the same interval.  If you just look at where they end up, you would conclude that Blue walked a shorter distance because he ends up closer to the starting point than Red does.  But when you consider the whole interval, it is clear that Blue walked a further distance.

The same thing is true with SR, if you only consider the end results, you are missing what is really  going on.

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Lol have nothing to add to that excellent reply Janus except another up vote

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Excellent explanation, Janus.

The fundamental problem with all threads of this nature is the OP’s tacit assumption that space and time are both separable and absolute. This is the Newtonian paradigm, as established back in the 1600’s. But of course, we know today that space and time are neither separable, nor can they be absolute, since these assumptions lead to contradictions and inconsistencies. Relativity rectifies these problems by recognising that notions of space and time are purely observer-dependent and local. At the frontiers of modern research, we are now also beginning to understand at a fundamental level just why that must be so.

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I have come to really look forward to Janus' clarity of response. +1

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On 8.07.2018 at 9:11 PM, Janus said:

SR does not claim that such an observer will always see the Earth clock run slow, if by see, you mean what his eyes or instruments directly record.   In this usage of see, he will see it run at a rate of    T =  To ((1-v/c))(1+v/c))1/2

where v is positive if Earth and the Observer are receding from each other and negative if they are approaching each other.

A factor contributing to this observation is the the distance and thus the propagation time for signals is constantly changing, getting longer when receding and getting shorter when approaching.    This factor works out to be c/(c+v)

When you factor this out of the first equation you are left with the time dilation equation.    This means that there are two things to consider: what you see happening to the Earth clock, and what is happening to the Earth clock.

......................................................

While this may seem to be at odds with common sense, it is how a Relativistic universe works.

..................................................

Sorry, I probably did’t quite understand your extensive explanation. The questions concerned a very specific theoretical experiment, carried out in the real Universe in which we live, not in some other imaginary universe.

1. What frequency  f ' generated by the clock on Earth will be measured  by an observer on a spacecraft,  during his return flight towards Earth at a speed  v = 180,000 km /s,  knowing that the light clock in motion ticks slower than at rest and  that the Doppler effect occurs here?

The observer will see that the clock on Earth ticks faster. The frequency f ' seen and measured by the observer approaching the Earth with the given speed v, will be 2000 Hz.

What's more, the observer on the spacecraft, knowing that the clock on Earth generates the frequency f = 1000 Hz measured on Earth, is able to determine the speed of his spacecraft  in relation to Earth on the basis of the observed frequency f’.

2.     What date will indicate the light clock on the spacecraft  the day before reaching the Earth?

The light clock on the spacecraft is ticking slower, according to the formula T = To /(1- (v /c) ^ 2)) ^ 0,5, where To  is the tick period of the clock at rest (v = 0). Thus, the clock on the spacecraft  the day before reaching the Earth will show the date 31.12.2003.

3.    What date shown on the clock display on Earth, will be seen and photographed by the observer on the spacecraft  the day before reaching Earth?

Traveling at a speed of v = 180 000 km /s  back and forth to an object  1,5  light years away, will take 5 years according to the clocks on Earth. Thus, the day before the spacecraft  returns, the clock on Earth will show on its screen the date 31.12.2024. And this date will be seen and photographed by the observer on the spacecraft.

This example clearly shows the fallacy of the claim of relativity theory that the observer in the spacecraft  moving with a speed v relative Earth will always see that the clocks on the Earth tick slower than the clock of the spacecraft.

In fact, an observer  on a spacecraft  approaching  Earth at speed  v will always see that the clocks on Earth tick faster than the clock on the spacecraft .

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58 minutes ago, ravell said:

Sorry, I probably did’t quite understand your extensive explanation. The questions concerned a very specific theoretical experiment, carried out in the real Universe in which we live, not in some other imaginary universe.

You need to answer the comment in Janus' opening line.

What do you mean by see?

Can you describe the mechanism by which you propose the travelling observers monitors the Earth Clock?

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2 hours ago, ravell said:

This example clearly shows the fallacy of the claim of relativity theory that the observer in the spacecraft  moving with a speed v relative Earth will always see that the clocks on the Earth tick slower than the clock of the spacecraft.

In fact, an observer  on a spacecraft  approaching  Earth at speed  v will always see that the clocks on Earth tick faster than the clock on the spacecraft .

As I said in my previous post.  Relativity makes no such claim when it comes to what an observer will visually see.

This is a straw-man argument based on a misrepresentation of Relativity.

To explain the difference between what the observer would visually see vs. what he is conclude is happening, we'll use some space-time diagrams.

First consider two clocks separated by some distance and stationary with respect to each other.

The blue line is our "observed" clock and the green line is our "observing" clock.   The scale is such that light, shown as the yellow lines, is drawn at a 45 degree angle.

Thus our observer will see light that left the blue clock when it read 1 arrive when his clock reads sometime after 3, and he will see the blue clock read 1 at that time. He also will see the light that left the blue clock when it read 2 arrive sometime after his clock reads 4.  However, this does not mean that he will think or conclude that what he sees actually represents what time it is for the blue clock at those moments.  That would be shown by the black horizontal lines, which shows that when the green observer sees the blue clock read 1, he knows that it actually reads the same as his own, or somewhat after 3, and when he sees the blue clock read 2, it actually at that moment reads somewhat after 4.

Now let's add a third clock, one that is moving at 0.6c relative to the both clocks so that it and the blue clock are closing in on each other. This will be the red line in the following diagram.

Now add yet another clock, this time so that it and the observer are receding from each other, as shown by the light blue line.

Again the light leaving when it reads 1 arrives at the green observer when the green clock reads after 3. But the light leaving it when it reads 2 doesn't arrive until the green clock read after 5.  The green observer will see the blue clock ticking at 1/2 the rate of his own. But this time, the light blue clock is further from the green when it reads 2 than it was when it read 1, and when the green observer takes this into account, it will turn out that when compared to his own clock, the light blue clock is ticking slower than his own, and by the same rate as he concluded that the red clock is ticking slow.  The light blue clock exhibits the same time dilation as the red clock.

This is what Relativity says is happening in the real universe, and this is not you you are trying to claim it says ( that an observer will always see a clock as running slow).

If you are going to argue against a theory, you have to argue against the actual theory rather than some imagined version of your own creation.

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On 15.07.2018 at 7:27 PM, Janus said:

"This example clearly shows the fallacy of the claim of relativity theory that the observer in the spacecraft  moving with a speed v relative Earth will always see that the clocks on the Earth tick slower than the clock of the spacecraft "

As I said in my previous post.  Relativity makes no such claim when it comes to what an observer will visually see.

This is a straw-man argument based on a misrepresentation of Relativity

Over a century of discussions and disputes over the "twin paradox" would not exist if Relativity makes no such claim.

One of many examples of these disputes is presented in Herbert Dingle's publication available on the link:

The following is a quotation of the leading view from the Internet discussion regarding this publication:

In special relativity, acceleration is absolute. This is basic, and the fact that Dingle never accepted it is his problem, not special relativity's. The fact that each reference frame sees the other's clocks run slow is no more a paradox than the fact that if you put two rulers at an angle to each other, and look at each along a line perpendicular to the other, each one looks shorter than the other. Dingle's misunderstanding was explained in great detail in the infamous correspondence in the letters page of Nature, though he seemed incapable of understanding that.”

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Ravell,

I asked you  a polite direct question to which you have not responded, contrary to the rules of this forum.

On 15/07/2018 at 3:33 PM, studiot said:

You need to answer the comment in Janus' opening line.

What do you mean by see?

Can you describe the mechanism by which you propose the travelling observers monitors the Earth Clock?

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17 minutes ago, ravell said:

Over a century of discussions and disputes over the "twin paradox" would not exist if Relativity makes no such claim.

One of many examples of these disputes is presented in Herbert Dingle's publication available on the link:

The following is a quotation of the leading view from the Internet discussion regarding this publication:

In special relativity, acceleration is absolute. This is basic, and the fact that Dingle never accepted it is his problem, not special relativity's. The fact that each reference frame sees the other's clocks run slow is no more a paradox than the fact that if you put two rulers at an angle to each other, and look at each along a line perpendicular to the other, each one looks shorter than the other. Dingle's misunderstanding was explained in great detail in the infamous correspondence in the letters page of Nature, though he seemed incapable of understanding that.”

!

Moderator Note

There's nothing here at odds with the comment by Janus about how the word "see" is being used, and you have done nothing to address that point. You're just trying to use the fallacy of equivocation as a bludgeon.

If you aren't going to address the physics that people bring up, I will close this, and you won't be able to bring it up again. It's your call. (Responding to this modnote would count as not addressing physics)

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21 hours ago, ravell said:

The following is a quotation of the leading view from the Internet discussion regarding this publication:

In special relativity, acceleration is absolute. This is basic, and the fact that Dingle never accepted it is his problem, not special relativity's. The fact that each reference frame sees the other's clocks run slow is no more a paradox than the fact that if you put two rulers at an angle to each other, and look at each along a line perpendicular to the other, each one looks shorter than the other. Dingle's misunderstanding was explained in great detail in the infamous correspondence in the letters page of Nature, though he seemed incapable of understanding that.”

You are reading a lot into that one section of the statement you highlighted in bold.

For one, his use of the word "see" is ambiguous.  Generally, when when physics uses the word "see" when discussing relativity,  they do not mean "visually", but rather after accounting for things like light propagation delay.    An similar thing is done in the ruler example given.  If you were to visually look at the two rulers in the example, one could be closer to you than the other, in which case, perspective would play a role in which ruler "looks" shorter than the other.  In the example, the perspective effect is ignored so that distance between observer and ruler is not considered as a factor.  This usage of "see" is used because it focuses on the primary effects of interest rather than the secondary effects of no interest. When "visually see" is the intent, this is usually specified.

For the other, he doesn't use the word "always", so there is no way to know exactly what exact scenario he is referring to.  Even if he is using "see" as meaning "visually see", there is one situation, where both usages of the word give the same result, this would be at the moment the two observers in each reference frame pass each other and are neither receding nor approaching each other.  In this case, they would both conclude and "see" each other's clocks as running slow.

Your use of this part of his statement to bolster your argument entirely rests on your personal interpretation of what the author means, rather than taking it within the context of the theory of Relativity itself.

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On 15.07.2018 at 4:33 PM, studiot said:

You need to answer the comment in Janus' opening line.

I respect and appreciate Janus's kind comments and I read his posts with pleasure. I fully share his position that it is not true that each reference frame always sees the other's clocks run slow.

If a misunderstanding arose here, it is probably for this reason that I want to draw attention to the fact that such a conlusion results directly from classical physics and the theory of relativity has nothing to do with it.

The formula for the frequency of the light clock on the spacecraft  in motion which is observed on Earth, is exactly the same as for the acoustics and applies to the Doppler effect:

F = F1*c/(c-v), where F1 is the frequency of the observed source of the signal. In our case it is the light clock on the spacecraft  in motion.

The frequency of the light clock in motion is expressed by the formula:

F1 = Fo *(1- (v /c)^2)^0,5  which also results directly from classical physics, where Fo is the frequency of the clock at rest (v = 0).

The formula for the clock frequency on Earth observed on the spacecraft  is also identical to that in the acoustics:

F = Fo*(c + v)/c * Fo/F1, where the added factor  Fo /F1  takes into account  the length of the second of the light clock on board, which is longer on the moving spacecraft than the second on Earth in the ratio  T1 /To  (T=1/F).

In the above formulas the velocity v has a positive sign for the approaching of the spacecraft, and negative when the spacecraft  moves away from Earth.

In our example, the observed frequency of the clock on opposite side is thus 2000 Hz at approach and 500 Hz when the spacecraft  moves away, and is the same for both sides of the observation.

Thus, the whole matter is clearly and unambiguously explained by classical physics.

Quote

What do you mean by see?

Can you describe the mechanism by which you propose the travelling observers monitors the Earth Clock?

Professional observation of the clock frequency on Earth is possible only with the help of a specialized radio receiver with a precise frequency meter.

In addition, at a very short distance from the Earth, you can use a good telescope through which you can see what time it is eg.  on the Big Ben clock in London, with a few seconds delay resulting from the travel time of light from the clock to the spacecraft.

Thanks for Janus and for you for a nice discussion.  Sorry for the late answer.
Best regards,

Edited by ravell

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2 hours ago, ravell said:

Thus, the whole matter is clearly and unambiguously explained by classical physics.

The whole matter is neither clearly nor unambiguously explained by yourself., though Janus has added further meat to his bone for your benefit.

Start with a very simple example that has nothing to do with 'relativity'.

Take a clock with hands and an observer that views the clock via a mirror.

The observer will directly see the clock running backwards (anticlockwise).

So he makes a (perhaps mental) adjustment to the reading to compensate so that he can consider (the posh Physics term is transform) what he actually sees to his own form of reckoning.

In the same way the travelling observer directly sees a clock fact that is acting differently from his own clock, but in your case he is monitoring the changing of the numbers.

So, like any good Scientist, he records his readings.
He will have a table with readings on his clock in one column and tabulated against each entry he will have another reading of what he notes on the Earth clock, which will generally be different.

So your first task is to make clear in which system are we reckoning time?

Because he will need a third column transforming the readings in either the first or second column to the other system as they are not directly comparable.

This would be true in both Galilean and Einstinian relativity as there will alway be a time lag between the Earth clock display and the traveller reading.

Further this delay will be a function of time or separation and not constant.

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2 hours ago, ravell said:

In our example, the observed frequency of the clock on opposite side is thus 2000 Hz at approach and 500 Hz when the spacecraft  moves away, and is the same for both sides of the observation.

Thus, the whole matter is clearly and unambiguously explained by classical physics.

Except that your results are wrong, because you haven't taken relativity into account.

On 15/07/2018 at 3:33 PM, ravell said:

Sorry, I probably did’t quite understand your extensive explanation.

Quite.

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2 hours ago, Strange said:

"In our example, the observed frequency of the clock on opposite side is thus 2000 Hz at approach and 500 Hz when the spacecraft  moves away, and is the same for both sides of the observation."

Except that your results are wrong, because you haven't taken relativity into account.

If the above numbers  are wrong, would you be so kind  to give here the right value of the frequency of the clock on Earth that the observer on the spacecraft would see:

a) when moving away   ???? Hz

b) when approaching    ???? Hz

And the same but for the observer on Earth ?

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4 hours ago, ravell said:

If the above numbers  are wrong, would you be so kind  to give here the right value of the frequency of the clock on Earth that the observer on the spacecraft would see:

a) when moving away   ???? Hz

b) when approaching    ???? Hz

And the same but for the observer on Earth ?

Even though you have failed to answer my questions.

You have chosen a relative velocity of u = 0.6c

The relativistic doppler shift for two bodies moving with an inline relativistic velocity is given by

$\frac{\nu }{{{\nu _0}}} = \frac{{\sqrt {1 - \frac{{{u^2}}}{{{c^2}}}} }}{{1 + \frac{u}{c}}}$

Where the observed frequency is   $\nu$ and the shifted frequency is    ${{\nu _0}}$ and u is the inline relative velocity.

Which I make half and double the original frequencies (no doubt why you chose that value for u).

So what effect does that have on the observed reading on the clocks as tabulated?

Edited by studiot

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9 hours ago, ravell said:

I respect and appreciate Janus's kind comments and I read his posts with pleasure. I fully share his position that it is not true that each reference frame always sees the other's clocks run slow.

If a misunderstanding arose here, it is probably for this reason that I want to draw attention to the fact that such a conlusion results directly from classical physics and the theory of relativity has nothing to do with it.

The formula for the frequency of the light clock on the spacecraft  in motion which is observed on Earth, is exactly the same as for the acoustics and applies to the Doppler effect:

F = F1*c/(c-v), where F1 is the frequency of the observed source of the signal. In our case it is the light clock on the spacecraft  in motion.

The frequency of the light clock in motion is expressed by the formula:

F1 = Fo *(1- (v /c)^2)^0,5  which also results directly from classical physics, where Fo is the frequency of the clock at rest (v = 0).

The formula for the clock frequency on Earth observed on the spacecraft  is also identical to that in the acoustics:

F = Fo*(c + v)/c * Fo/F1, where the added factor  Fo /F1  takes into account  the length of the second of the light clock on board, which is longer on the moving spacecraft than the second on Earth in the ratio  T1 /To  (T=1/F).

In the above formulas the velocity v has a positive sign for the approaching of the spacecraft, and negative when the spacecraft  moves away from Earth.

In our example, the observed frequency of the clock on opposite side is thus 2000 Hz at approach and 500 Hz when the spacecraft  moves away, and is the same for both sides of the observation.

Thus, the whole matter is clearly and unambiguously explained by classical physics.

While light clocks are a means convenient for illustrating the Time dilation effect, they are not the  end all and be all for Relativity or the physics of the situation.

There are processes and means for measuring time periods that do not rely on light propagation.    Take for example bouncing a particle between the two mirrors at 0.5 c instead of  using light.

In this case, the motion of the clock would have no effect on the observed frequency of the clock, outside that due to Doppler shift.   While the ball would follow a diagonal line according to our stationary observer just like the light  of the light clock does,  its diagonal velocity is not limited to just one speed.  With light it must be c, and if the relative motion of the clock is 0.5c, then   the vertical component of that velocity must be (c2-(0.5c)2)-5 = 0.866...c   and thus the light takes longer to make the round trip.   For the particle, the vertical component stays at 0.5c, and the diagonal velocity is   ((0.5c)2+(0.5c)2)-5 = 0.707...c  by the Newtonian rules of velocity addition.   Not only that, but a light clock aligned such that the light reflected along a path parallel to the motion of the clock would tick at a different rate than one aligned perpendicular to the motion (This was the basis behind the M&M experiment)

Basically what this means is that while our stationary observer would note that the light clock ticks slow, so would someone traveling along with the light clock.    In other words there would be no accumulated age difference between moving and stationary clocks under the application of Newtonian physics. Some physical processes would be effected to certain degrees, but not all of them (for example, radioactive decay does not rely on any mechanism related to the light clock)  You would also be able to determine your absolute velocity by comparing how your local light clock performed compared to other local processes, or how light clocks aligned in different directions compared to each other.  No experiment or measurements made have found such an effect.

Newtonian physics cannot and do not predict the very real Relativistic results we see in experiments.  Newtonian physics can't explain why Muons take longer to decay while traveling at high relative speed wrt the Earth, since muon decay doesn't depend on traveling light signals.

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2 hours ago, studiot said:

Where the observed frequency is   ν and the shifted frequency is    ν0 and u is the inline relative velocity.

I should clarify this line to read that nu nought is the original frequency and nu is the observed frequency.

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On 7/21/2018 at 6:39 AM, ravell said:

I respect and appreciate Janus's kind comments and I read his posts with pleasure. I fully share his position that it is not true that each reference frame always sees the other's clocks run slow.

If a misunderstanding arose here, it is probably for this reason that I want to draw attention to the fact that such a conlusion results directly from classical physics and the theory of relativity has nothing to do with it.

!

Moderator Note

You keep claiming this but I have yet to see you actually derive the expressions from classical physics. Janus has shown you to be incorrect. These two are strongly coupled.

It's something you need to address, rigorously.

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On 21.07.2018 at 10:07 PM, studiot said:

You have chosen a relative velocity of u = 0.6c

The relativistic doppler shift for two bodies moving with an inline relativistic velocity is given by
νν0=1u2c21+uc

v/vo = (1-v^2/c^2)0,5 /(1+u/c)

Where the observed frequency is   ν and the shifted frequency is    ν0 and u is the inline relative velocity.

Which I make half and double the original frequencies (no doubt why you chose that value for u).

So what effect does that have on the observed reading on the clocks as tabulated?

Why no one answered clearly to the simple question previously asked to Strange:
If my calculations are incorrect then please show what specific frequency values F will see, according to relativity, an observer on Earth and on a spacecraft for a given speed v = 180 000 km /s, when approaching, and  when the spacecraft is moving away, if  Fo =1000Hz?

If the speed v = 180 000 km /s seems to someone to be deliberately chosen by me, then please provide these values  F for any other speed  v you choose.

The formula given by Studiot as the correct one according to the theory of relativity, is exactly the same (!)  which I presented in classical  physics:                              F = F1 * c /(c + v)   where   F1 = Fo * (1- (v /c))^2)^0.5,  and it is only in a slightly modified mathematical form. Please check it out.

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1 minute ago, ravell said:

Why no one answered clearly to the simple question previously asked to Strange:

They have. You have ignored them. Because, as you admit, you didn't;t understand it.

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42 minutes ago, ravell said:

Why no one answered clearly to the simple question previously asked to Strange:
If my calculations are incorrect then please show what specific frequency values F will see, according to relativity, an observer on Earth and on a spacecraft for a given speed v = 180 000 km /s, when approaching, and  when the spacecraft is moving away, if  Fo =1000Hz?

!

Moderator Note

Janus has addressed this in the very first response to you. You can't just pretend that the answer was not given.

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I am at a complete loss to understand where you are coming from in this thread.

You correctly state Einsteinian relativistic formulae about time (saying they are relativistic) and then ask questions about frequency.

You then directly contradict yourself by saying the formulae are 'classical'., but do not say why.

I don't know what you mean by 'classical' since modern Physics regards Einsteinian relativity as classical, but no matter I assume that since you wish to make a distinction you mean theory preceding this.

So far as I am aware c and square roots do not enter into any preceding theory.

I assume you know the correct classical formulae, but we can discuss them if you wish and compare them with the Einsteinian ones.

You are also wrong to say that no one has offered you the correct changes of frequency for the relative speed you specified.

These appeared in my post that you quoted.

Finally I asked you simple question more than once.

What does the frequency either perceived or as generated)  have to do with the reading on the clock?

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4 hours ago, swansont said:

Janus has addressed this in the very first response to you. You can't just pretend that the answer was not given.

You're right. I know that Janus has already given this in his first post. But as I showed on the basis of classical physics, that in our example, we get axactly the same values as in Janus’ post, ie.  the values of  F = 2000 Hz for approaching and  F = 500 Hz for receding, then Strange commented  it as follows:
"Except that your results are wrong, because you have not taken relativity into account."
And no one commented on his statement.

1 hour ago, studiot said:

You correctly state Einsteinian relativistic formulae about time (saying they are relativistic) and then ask questions about frequency.

You then directly contradict yourself by saying the formulae are 'classical'., but do not say why.

I don't know what you mean by 'classical' since modern Physics regards Einsteinian relativity as classical, but no matter I assume that since you wish to make a distinction you mean theory preceding this.

The formulas given by me concern the Doppler effect  and the delay of the light clock in motion, which results exclusively from classical physics. I was hoping that my post from Saturday would make it clear. If, however, this is still a bit incomprehensible, then I am not able to present this case more simply.

Classical physics and theory of relativity are separate matters even on this Science Forum.

2 hours ago, studiot said:

Finally I asked you simple question more than once.

What does the frequency either perceived or as generated)  have to do with the reading on the clock?

If, for example, you measure the frequency F = 1000 Hz generated by the clock A (at rest) with a meter clocked by another clock B, which for whatever reason (wrongly tuned, sensitive to motion, temperature or shock, etc.) is at a given moment 10% slower than clock A (about which you do not know), you read on your clock meter B that the clock A is in a hurry, because in one second of your clock B, you receive 1.1 x 1000 Hz = 1100Hz from the clock A.  During the movement of  clocks, the Doppler effect must be additionally taken into account.

I can not explain it more clearly.

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