# Time dilation dependence on direction

## Recommended Posts

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

##### Share on other sites

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.

##### Share on other sites

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.

##### Share on other sites

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

$f_o = f_s \sqrt{\frac{1-\frac{v}{c}}{1+ \frac{v}{c}}}$

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.)

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.

##### Share on other sites

[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.

##### Share on other sites

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?

##### Share on other sites

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.

##### Share on other sites

To Janus:

Thx for answering, I know what's wrong wiht me, I have a clearer understanding

##### Share on other sites
On 4/26/2017 at 6:43 PM, Janus said:

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

Why 01:15 and not 01:25?

##### Share on other sites
On 4/26/2017 at 6:43 PM, Janus said:

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 01:25. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:25. 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 1:25 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.)

You are considering as if the travel was 1 hour long, but the travel was 1.25 long.

And you have explained to the community how to travel 1 LightHour in 45 minutes.

##### Share on other sites
1 hour ago, michel123456 said:

And you have explained to the community how to travel 1 LightHour in 45 minutes.

A light hour in one frame during a duration of 45 minutes in another frame.

We're all doing it right now.

##### Share on other sites
4 hours ago, michel123456 said:
Quote

Why 01:15 and not 01:25?

Because one and a quarter hours is an hour and 15 minutes, not an hour and 25 minutes.

Edited by Halc

##### Share on other sites
1 hour ago, michel123456 said:

You are considering as if the travel was 1 hour long, but the travel was 1.25 long.

And you have explained to the community how to travel 1 LightHour in 45 minutes.

It's one light-hour only in the rest frame.

##### Share on other sites

I am confused.

first of all, hours have 60 min, not 100. So 60/0.8=75 or 1h15min (and not 1h 25 min.). Correct?

On 4/26/2017 at 6:43 PM, Janus said:

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.)

Generally I am confused with the mix of 1.25 & 1.15 in the text above.

##### Share on other sites

Yes, it takes 1:15 in the frame of 1st/2nd clock for the 3rd clock to get one LH away.

In the frame of that 3rd clock, it is stationary, so it isn't traveling anywhere in the 45 minutes it logs, so it isn't moving faster than light.

Edited by Halc

##### Share on other sites
11 minutes ago, Halc said:

Because one and a quarter hours is an hour and 15 minutes, not an hour and 25 minutes.

Oh, I got it: you mean 1.25 hour means 1h and a quarter? That's a bad mix of notation.

2 hours ago, michel123456 said:
On 4/26/2017 at 6:43 PM, Janus said:

if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:25 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.)

You are considering as if the travel was 1 hour long, but the travel was 1.25 long

Still not understanding this: if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45.

##### Share on other sites
3 hours ago, michel123456 said:

Oh, I got it: you mean 1.25 hour means 1h and a quarter? That's a bad mix of notation.

Still not understanding this: if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45.

How is using 1.25 hrs any different than saying 1 and a quarter hrs?  or 1 1/4 hrs.  The fact that the number is followed by "hrs" indicates that the numerical value is applies to the one unit, hours, while with 1:15  the ":" denotes a marker between units, or 1 hr, 15 min.

I'm sorry if this confuses you, but when working with such problems it is easier to the work the math out in decimal hrs and then convert back to Hr and min afterwards if you need to.

If the clock left you moving at 0.8c to a point 1 light hr away ( as measured by you), while both it and your clock both read 12:00, then it will take, by your determination, 1 hr 15 min to reach that point. You, however, will not see him arrive when your clock reads 1:15, as It will take an another hr for the light from his arrival to reach you, so you will see the image of his arrival 2 hr and 15 min after he left, when your clock reads 2:15. In other words, you will "see" his entire trip as being stretched out over 2 hrs 15 min.

As you watched him recede from you, you will see him via relativistic Doppler shift. This does not only effects the measured frequency of the light you get from him, but also the rate at which you would "see" his clock tick.  The Relativistic Doppler shift rate for 0.8c is 1/3.   Since you see this happening for the entire 2 hr and 15 min, you will see that clock ticking off only 1/3 of that, or 45 min.  Thus when you see the clock arrives at that point 1 light hr away, you will see it as reading 12:45.

##### Share on other sites
4 hours ago, michel123456 said:

I am confused.

first of all, hours have 60 min, not 100. So 60/0.8=75 or 1h15min (and not 1h 25 min.). Correct?

Generally I am confused with the mix of 1.25 & 1.15 in the text above.

They are not mixed.

The text:
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.)

1.15, for example, is not used. All of the time-of-day readings (as one might see on a digital clock) or elapsed times denoting hours and minutes use a colon, per the ISO 8601 standard (further, all clock readings are denoted as such)

The decimals are used for elapsed time calculations, as one might expect when the values used don't result in a whole number

The two notations are used consistently. Decimal point used for decimal numbers, colon used for hours:minutes

##### Share on other sites
45 minutes ago, swansont said:

They are not mixed.

The text:
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.)

1.15, for example, is not used. All of the time-of-day readings (as one might see on a digital clock) or elapsed times denoting hours and minutes use a colon, per the ISO 8601 standard (further, all clock readings are denoted as such)

The decimals are used for elapsed time calculations, as one might expect when the values used don't result in a whole number

The two notations are used consistently. Decimal point used for decimal numbers, colon used for hours:minutes

Well understood now. Thx.

1 hour ago, Janus said:

How is using 1.25 hrs any different than saying 1 and a quarter hrs?  or 1 1/4 hrs.  The fact that the number is followed by "hrs" indicates that the numerical value is applies to the one unit, hours, while with 1:15  the ":" denotes a marker between units, or 1 hr, 15 min.

I'm sorry if this confuses you, but when working with such problems it is easier to the work the math out in decimal hrs and then convert back to Hr and min afterwards if you need to.

If the clock left you moving at 0.8c to a point 1 light hr away ( as measured by you), while both it and your clock both read 12:00, then it will take, by your determination, 1 hr 15 min to reach that point. You, however, will not see him arrive when your clock reads 1:15, as It will take an another hr for the light from his arrival to reach you, so you will see the image of his arrival 2 hr and 15 min after he left, when your clock reads 2:15. In other words, you will "see" his entire trip as being stretched out over 2 hrs 15 min.

As you watched him recede from you, you will see him via relativistic Doppler shift. This does not only effects the measured frequency of the light you get from him, but also the rate at which you would "see" his clock tick.  The Relativistic Doppler shift rate for 0.8c is 1/3.   Since you see this happening for the entire 2 hr and 15 min, you will see that clock ticking off only 1/3 of that, or 45 min.  Thus when you see the clock arrives at that point 1 light hr away, you will see it as reading 12:45.

Thx, that is now clear as crystal water.

now the return trip:

On 4/26/2017 at 6:43 PM, Janus said:

Please explain the bold part, does it come from the dilation formula you posted above?

##### Share on other sites
26 minutes ago, michel123456 said:

Well understood now. Thx.

Thx, that is now clear as crystal water.

now the return trip:

Please explain the bold part, does it come from the dilation formula you posted above?

Again. Relativistic Doppler effect. (which is what you get by combining both light propagation and time dilation)   If the clock is receding at 0.8 c, you see it ticking 1/3 as fast as your own.   If it is approaching at 0.8 c, you see it running 3 times faster.

One way to look at it is that by the time you visually see that clock begin its return trip, the clock has already been on its return leg for an hr, and only 15 min remain until it arrives. All the light emitted from it during the return leg is squeezed into that 15 min by your clock.  If you see the returning clock tick 3 times faster, 3*15 min = 45 min is the time you see pass on that clock during its return leg.

If you factor out the light propagation delay you will conclude that the clock left reading 12:45, when your clock read 1:15.  It took 1 hr 15 min to cross the 1 light hr at 0.8c,  and arrives when your clock reads 2:30.   The clock read 12:45 when it left, and reads 1:30 when it arrives, thus ticked off 45 min in that 1 hr 15 min by your clock, or  6/10 of  1 hr 15 min(75 min)

The time dilation factor for 0.8c is 0.6.

##### Share on other sites
7 minutes ago, Janus said:

Again. Relativistic Doppler effect. (which is what you get by combining both light propagation and time dilation)   If the clock is receding at 0.8 c, you see it ticking 1/3 as fast as your own.  If it is approaching at 0.8 c, you see it running 3 times faster.

So it depends on direction after all.

##### Share on other sites
35 minutes ago, michel123456 said:

So it depends on direction after all.

The Doppler effect does, but the time dilation does not.

The relativistic Doppler shift is due to two compounding effects:

Changing light propagation delay due to changing distance, the effect of which is determined whether the propagation delay is increasing or decreasing.

Time dilation, which only depends on relative speed and is independent of direction.

So with the above example, Doppler shift gives a value of 1/3 on the for the receding leg and 3 for the return leg, but time dilation has a factor of 0.6 for both legs.

##### Share on other sites
22 minutes ago, Janus said:

The Doppler effect does, but the time dilation does not.

The relativistic Doppler shift is due to two compounding effects:

Changing light propagation delay due to changing distance, the effect of which is determined whether the propagation delay is increasing or decreasing.

Time dilation, which only depends on relative speed and is independent of direction.

So with the above example, Doppler shift gives a value of 1/3 on the for the receding leg and 3 for the return leg, but time dilation has a factor of 0.6 for both legs.

So, from your example I can conclude that:

1. the 3rd clock traveled 1 LightHour in 45 minutes, but that was not faster than SOL

2. as seen from A, the same clock popped in 15 minutes from 1 LH away, but that is not considered faster than SOL.

##### Share on other sites
21 minutes ago, michel123456 said:

So, from your example I can conclude that:

1. the 3rd clock traveled 1 LightHour in 45 minutes, but that was not faster than SOL

2. as seen from A, the same clock popped in 15 minutes from 1 LH away, but that is not considered faster than SOL.

That error has already been pointed out.

The example shows clock 3 traveling the light hour in 5/4 hours. That's what it means to be moving at 0.8c. Due to time dilation, clock 3 just happens to have logged only 3/4 hours during that trip.

Relative to the clock's own frame, only 45 undilated minutes pass, but it is stationary (by definition), which is hardly traveling faster than SoL. In that frame, clock 2 comes to it at 0.8c, and moves 36 light-minutes during those 45 minutes, which is also less than SoL.

2: As seen some observer A, the incoming clock is seen to go from a light hour away to 'here' in 15 minutes, yes. That's the difference between its actual speed in that frame and the speed of the light the observer is observing.

Edited by Halc

##### Share on other sites
1 hour ago, michel123456 said:

So, from your example I can conclude that:

1. the 3rd clock traveled 1 LightHour in 45 minutes, but that was not faster than SOL

2. as seen from A, the same clock popped in 15 minutes from 1 LH away, but that is not considered faster than SOL.

1. No, the 3rd clock traveled 1 light hr in 1hr 15 min as measured by you, it just also ran slow by a factor of 0.6 and thus only accumulated 45 min , as measured by you.   Visually, you saw it take 2 hr and 15 min to reach that 1 light hr distance.

2. Again, it took the clock 1hr 15 min to travel the distance.  The fact that it arrived only 15 min after you saw it start its trip is just because it was following closely behind it's own light.

Two cars leave a point 100 miles from you, driving towards you.   One car, traveling at 100 mph carries a message that says the other car is also just leaving. The second car travels at 80 mph.

The faster car arrives at 1:00, while the second car arrives 15 min later at 1:15.   Now just because the message saying that the second car was leaving arrived just 15 min before the second car arrives, this does not mean that you will conclude that the second car only took 15 min to cross the distance and had to be traveling at 400 mph.  The second car took 1 hr 15 min to make the trip, it was just following not too far behind the faster car.( By the time the first car arrives, the second car is only 20 miles away)

Again, you are conflating "what someone visually sees", with what they say is actually happening.

If I am 1 light hour from a clock stationary with respect to me, and see it reading 11:00, I don't think that the clock reads 11:00 at that moment, but that, assuming the clock continued to run, it ticked off an additional hr since the image I am now seeing left it, thus I will say that the clock now reads 12:00.  Conversely, if your clock reads 12:00, you know that the light left when your clock read 11:00 and thus both clocks read 11:00 at that moment.

This works if even if the other clock is moving relative to you.   Light that left it when it was 1 light hr away will carry the image of what that clock read 1 hr ago by your clock.

So if your clock reads 2:15 when you see the image of the 3rd clock carried by light that left it when it was 1hr away,  You know that image left when your clock read 1:15. and if that image is of the 3rd clock reading 12:45, you know that the 3rd clock read 12:45 at the same moment your clock read 1:15.

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account