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HongKongEvil

Time dilation dependence on direction

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I think of this question for a long time too, no one give meaningfully answer to this.....

 

I understand in case of light shoot out and than reflect back the time dilation calculation is correct for all direction

 

But consider different path, In view light speed is constant:-

A object moving towards us, its time should run faster than us

A object moving away us, its time should run slower than us

 

Its a simple calculation but no one can tell what's happening

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I think of this question for a long time too, no one give meaningfully answer to this.....

 

I understand in case of light shoot out and than reflect back the time dilation calculation is correct for all direction

 

But consider different path, In view light speed is constant:-

A object moving towards us, its time should run faster than us

A object moving away us, its time should run slower than us.

Its a simple calculation but no one can tell what's happening

 

What would you consider a "meaningful answer"? You are simply wrong! If a person is moving with speed v relative to you, you will observe time running more slowly for that person whether he is moving toward you or away from you. That has been verified by experiment many times.

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I think of this question for a long time too, no one give meaningfully answer to this.....

 

I understand in case of light shoot out and than reflect back the time dilation calculation is correct for all direction

 

But consider different path, In view light speed is constant:-

A object moving towards us, its time should run faster than us

A object moving away us, its time should run slower than us

 

Its a simple calculation but no one can tell what's happening

Direction does not matter.

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I think of this question for a long time too, no one give meaningfully answer to this.....

 

I understand in case of light shoot out and than reflect back the time dilation calculation is correct for all direction

 

But consider different path, In view light speed is constant:-

A object moving towards us, its time should run faster than us

A object moving away us, its time should run slower than us

 

Its a simple calculation but no one can tell what's happening

It is important here not to confuse what someone visually sees happening to a clock approaching or receding with how fast that clock is actually ticking according to that person.

In the first case you have to account for the Doppler effect which is caused by the decreasing or increasing distance between you and the clock. This would happen even if the clock itself did not tick any faster or slower than your own.

Time dilation of the clock is a measure of how fast the clock is actually ticking compared to your own, and this is independent of its direction of movement.

When you are watching a clock approaching or receding, what you visually see is a combination of these two effects.

The math for what you will see is

 

[math]f_o = f_s \sqrt{\frac{1-\frac{v}{c}}{1+ \frac{v}{c}}}[/math]

Where f0 is the observed tick rate.

fs is the source tick rate

v is the velocity of the clock relative to you(positive if receding and negative if approaching)

c is the speed of light.

 

So let's work out a couple of examples:

Assume you are situated 1 light hr from a stationary(to you) clock that is synchronized to your own clock (when your clock reads 12:00, it reads 12:00)

You looking at this clock when your clock reads 12:00 will visual see the clock as reading 11:00. (since it took the light carrying that image 1 hr to travel from the other clock, to get to you when your clock reads 12:00, it had to leave the other clock when it read 11:00)

 

A third clock travels from your position to the second clock at 0.8c

Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:15. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:15 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

 

Now let's do a reverse trip. you see the second clock read 01:15 and the third clock reads 12:45 when your clock reads 02:15, however this image of those two clocks left 1 hr ago, so the third clock left the second clock when your clock read 01:15 (in other words, if the third clock turns around and heads back the moment it reaches the second clock, by the time you see this, the third clock is already well along its trip back to you.). The return trip takes the same 1.25 hrs, which means it arrive back at you when your clock reads 02:30. This means that you will see its entire return trip occur during the 15 min between 02:15 and 02:30. At a accelerated tick rate of 3 you will see the third clock tick off 45 min during that period, and read 01:30 upon arrival, when your clock reads 02:30. So again, it ticked off 45 min during the 1.25 hr trip, the same as for the outbound trip and accumulated at total of 1.5 hrs for your 2.5 hrs.

 

The direction had no effect on the time dilation or total accumulated time difference even though it did have an effect on what you visually saw happening to the clock.

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[re HongKongEvil #1]

If direction made a difference then if 2 clocks get closer & then meet (cross) & then get further apart, then the relative ticking would change abruptly as they crossed. But no theory predicts this -- Einsteinians don't, aetherists don't, everybody don't.

 

As u say, its a simple calculation -- but contrary to what u say everyone can tell what happens (u say no one can tell) -- but unfortunately the "what happens" don't all agree.

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Someone redirect my post so you don't see my previous message

 

Imagine a space craft is moving towards me with speed 0.6C, I shoot a laser with frequency "f" to a spacecraft,the reflected laser frequency should be 2.56f, right?

 

But for the people inside spacecraft, the frequency of original and reflected laser must be the same

 

 

The what happened?

Someone redirect my post so you don't see my previous message

 

Imagine a space craft is moving towards me with speed 0.6C, I shoot a laser with frequency "f" to a spacecraft,the reflected laser frequency should be 2.56f, right?

 

But for the people inside spacecraft, the frequency of original and reflected laser must be the same

 

 

The what happened?

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Someone redirect my post so you don't see my previous message

 

Imagine a space craft is moving towards me with speed 0.6C, I shoot a laser with frequency "f" to a spacecraft,the reflected laser frequency should be 2.56f, right?

 

But for the people inside spacecraft, the frequency of original and reflected laser must be the same

 

 

The what happened?

Someone redirect my post so you don't see my previous message

 

Imagine a space craft is moving towards me with speed 0.6C, I shoot a laser with frequency "f" to a spacecraft,the reflected laser frequency should be 2.56f, right?

 

But for the people inside spacecraft, the frequency of original and reflected laser must be the same

 

 

The what happened?

Let's assume that you send 1 sec sec laser pulse towards the ship at a frequency of n hertz. This means you sent n waves of light during that 1 sec.

Also assume that the at the moment the front edge of your laser reaches the ship, it is 1 light sec from you as measured by you. Thus the front edge had been traveling for 1 sec upon reaching the ship before being reflected and heading back to you. arriving another 1 sec later or a total of 2 seconds after it left you.

At the same moment the leading edge of the laser is reaching the ship, you are shutting off the laser at your end. the trailing edge of the laser is rushing towards the ship at c and the ship is coming towards you at 0.6c, so they will meet at a point closer to you than 1 light sec, 0.625 light sec from you to be exact. It will take another 0.625 sec for the trailing edge to reflect back to you, for a total round trip time of 1.25 sec. This means you see the trailing edge of the laser return .25 sec after the leading edge does, so the the returning laser pulse lasts 1/4 sec, during that 1/4 sec you will measure the same n number of waves returning, meaning you will measure the frequency of the returning laser as having been multiplied by a factor of 4 at 4n hertz.

 

Now what would you expect the ship to measure? The time difference between the front of the laser and the trailing edge reaching the ship is 0.625 sec by your clock. Due to time dilation, according to you the Ship's clock will be running at a rate 0.8 that of your own, so by the ship clock, the arriving pulse lasts 1/2 second at double the frequency. If you compare this answer to the one arrived at by the formula I gave earlier, you get the same answer. At a relative velocity of 0.6c, the observer will measure a doubling of the source frequency from a source approaching at 0.6c.

 

So the ship gets a 1/2 sec pulse at a frequency of 2n hertz and from its perspective reflects back a 1/2 sec pulse at 2n hertz. Now from his perspective, it is you that is rushing towards him at 0.6c and it is your distance from him that changes between the front end and trailing end of the pulse reaching you. The pulse travels at c relative to him. So by the same argument as above, as seen from your perspective, he would measure 0.3125 sec between the front and trailing edges reaching you and your clock would be time dilated and running 0.8 as fast as his own. This means he would expect your clock to measure the length of the pulse as being 0.25 sec at 4 hertz, which is exactly what you expect to measure the length of the returning pulse as being.

 

Everything matches up and everyone agrees as to what everyone else measures.

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