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michel123456's relativity thread (from Time dilation dependence on direction)


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1 hour ago, swansont said:

In the FOR of B:

  • E and X travel a distance of 0.6c
  • It takes them 45 minutes

(B is at rest in its frame)

I stand corrected.

But I made another error, so again:

In the FOR of B:

  • E and X travel a distance of 0.6Lh
  • It takes them 45 minutes
58 minutes ago, michel123456 said:

You have evaded the question: what is the speed of B in the FOR of the virtual observer?

I already answered that:

2 hours ago, Eise said:

The speed of B in the FOR of E and X is 0.8c

And the observer is in the same FOR as E and X. Don't you see that E and X are standing still in Janus' first animation? That means the virtual observer is in the same FOR as E and X.

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2 hours ago, michel123456 said:

You have evaded the question: what is the speed of B in the FOR of the virtual observer?

No, I simply referred you to the answer that’s been provided perhaps 20 times in this thread. I chose to answer in a way that referenced physics, to make it marginally harder for you to engage in denial. (and to forestall the followup of asking what the distance and time are)

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1 hour ago, Eise said:
2 hours ago, michel123456 said:

You have evaded the question: what is the speed of B in the FOR of the virtual observer?

I already answered that:

3 hours ago, Eise said:

The speed of B in the FOR of E and X is 0.8c

And the observer is in the same FOR as E and X. Don't you see that E and X are standing still in Janus' first animation? That means the virtual observer is in the same FOR as E and X.

Do you say that the speed of B related to Virtual Observer is 0.8c?

But the distance between B and V.O. do not change. How do you explain that?

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23 minutes ago, michel123456 said:

Do you say that the speed of B related to Virtual Observer is 0.8c?

But the distance between B and V.O. do not change. How do you explain that?

B is moving essentially tangent to the observer. The radial velocity is zero. The tangential velocity is not. 

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12 hours ago, michel123456 said:

Do you say that the speed of B related to Virtual Observer is 0.8c?

Yes, of course. Let's say you want to do calculate the impact of a collision between a flying drone and a wall, i.e. you want to calculate the impulse of the drone, and its energy. You and the wall are in the same FOR. In both you need the velocity of the drone (mv, ½mv2 resp.). So damage to the drone (and wall?) will be dependent on the velocity. Does it make a difference from which distance you are looking at it? Or does it solely depend on the relative velocity of drone and wall?

Notice that in Janus' first animation, B's clock is contracted in the direction of its own flight. It definitely does not depend on how big the component of the velocity in the observer's direction is, just as in the example of the drone and the wall.

And another viewpoint that might help: the simplest explanation of time dilation that I know of is the light clock. To be sure that we make a clean derivation, we must exclude all effects of a velocity in the radial component, because it might have some effect. In other words, the explanation works if we look at a velocity that is perpendicular to the line of sight. So just imagine that the traveling B has such a light clock aboard. The farther away you are, the closer the light clock shows the time dilation directly.

Again, you seem to think that relativistic phenomena have something to do with what an observer sees, and with signal delay. That is just not true. It has only to do with the relative velocity of two inertial systems. The effects of a relativistic drone flying into a wall in your FOR (e.g. the damage caused by the collision) is of course independent of your viewpoint, if your are close by, or far away. Even better, if we assume we live in a 'consistent universe', the effects must be the same for observers in all FORs. Careful considerations of what the latter means lead you quickly to E = mc2.

 

Edited by Eise
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On 10/3/2020 at 9:48 PM, Janus said:

Animatione2.gif.4d0353946ab16a1bac4aa5f67e55c157.gif

B is at rest while the Earth and planet X move to the left at 0.8c.

the Earth, planet X, their clocks, and the distance between them is length contracted. B is not.

It take 45 min by B's clock for planet X to travel from being 0.6 lh away to B at 0.8c

Earth and planet X's clocks are time dilated and only accumulate 27 min during this time.

Why are the clocks E and X not synchronized?

 

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4 hours ago, michel123456 said:

Let's admit that there is a resultant velocity. Is this velocity equal to 0.8c?

 

Yes. The observers are all in the same frame and measure the same thing. E and X are separated by 1 LH and the trip takes 1:15. Thus the measured speed is 0.8c

 

30 minutes ago, michel123456 said:

Why are the clocks E and X not synchronized?

 

They are in a moving frame, according to B. This falls under the relativity of simultaneity: Things that are simultaneous in one frame are not simultaneous in another.  

You should be able to figure out why: you synchronize clocks by sending a light signal back and forth between them. In B's frame, this light travel is not symmetric - when X sends out a signal, it takes a certain amount of time to reach E, but E is moving away from the light, so the light has to travel a greater distance than their separation. When E sends out the signal, it travels a shorter path, because X is moving toward the signal.

In the earth frame, the path is symmetric and the clocks are synchronized. Clocks striking <whatever hour> will be measured as simultaneous in that frame. But not in B's frame, because of the relative motion.

 

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55 minutes ago, michel123456 said:

Why are the clocks E and X not synchronized?

This seems typical for the way how you cope with relativity, or better, with the reactions you get. After you got 2 reactions that also the 'virtual observer' measures a velocity of 0.8c for the traveling B, you do not react anymore on it. I would expect an 'ah, now I understand', or another argument why the velocity would not be 0.8c. Instead you just jump to another point. It looks very much like "Damn't, I cannot argue against that, so let's go to another point. These explanations must be wrong, so let's try another argument".

Janus already said it: two synchronised clocks in a FOR at a certain distance from each other, are not synchronised anymore in another FOR (i.e. a FOR that moves relative to the FOR of the clocks). It is the relativity of simultaneity. See e.g. here.

Relativity_of_Simultaneity_Animation.gif

The white lines are lines of simultaneity in the 3 different FORs.

Edited by Eise
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6 hours ago, michel123456 said:

Why are the clocks E and X not synchronized?

 

Relativity of simultaneity. Since the clocks are synchronized in the Rest frame of Earth and X, they can't be in the rest frame of B.

Below we have three clocks all a t rest in the same frame.  The clocks are synchronized in this frame.  The central clock emits a pulse of light which expands outward at c, and when it reaches the other clocks, they emit each emit a light pulse.  These pulses meet at the center clock.   Note that int this frame, all three clocks read the same when the pulses are emitted and when they meet at the central clock.

649682976_CLOCK_SYNC1.gif.4cd7e33845af1f852befa3d25af96b61.gif

Now we consider the exact same scenario,  except now we consider events according to a frame of reference that is moving relative to the clocks.

 

clock_sync2.gif.d7550c349bf2a5b74d0e94a4ee053a8f.gif

Unlike the first animation, the clocks do not remain at the center of the light pulses they emit.  This is because light must move at c relative to this frame as measured from this frame. Thus each pulse expands at c from a point that remains at rest with respect to this frame of reference, and the clocks move away from that emission point after emitting the pulse.

In this frame, the clocks run slower, they and the distance between them is length contracted.    What does remain consistent is what time each clock reads when a light pulse reaches it or it emits a pulse.

Now, when the center clock emits its pulse, the clock at moving relative to the center emission point from which the pulse is expanding at c.  as a result, the leftmost clock runs into the first pulse and emits its pulse before the rightmost clock does.  Each of these pulses expand out at c from their emission points and meet up at the center of the central clock.

In order for the pulses to arrive at each clock when that clock is reading the same time as it was in the first animation, the clocks cannot be in sync in this frame. 

This the reality of Relativity.  You can complain all you want about how "this doesn't make sense" or that you can't accept this as being true, but that will not change the reality of it.

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So i have to rephrase my question:

In animation 1 the clocks are synchronized and show the same time. If they show the same time, it means that it cannot be the situation in the FOR of the Earth. it is not the situation in the FOR of X either (there is an hour of difference: for the Earth departure time is 12.00 & it is 11.00 on X, for X departure time is 13.00 & it is 12.00 on Earth). The animation does not show the 2h15min that were needed for Earth to receive the signal of the arrival at X. IOW animation 1 is that of a virtual observer.

Animation 2 has a different concept: it is from the FOR of B. 

So the question is; why is animation 2 not from a virtual observer? that would show the clocks synchronized I suppose, like animation 1.

 

7 hours ago, Eise said:

It looks very much like "Damn't, I cannot argue against that, so let's go to another point. These explanations must be wrong, so let's try another argument".

Sorry, I have to digest the answers.

 

 

7 hours ago, Eise said:

This seems typical for the way how you cope with relativity, or better, with the reactions you get. After you got 2 reactions that also the 'virtual observer' measures a velocity of 0.8c for the traveling B, you do not react anymore on it. I would expect an 'ah, now I understand', or another argument why the velocity would not be 0.8c.

Yes i am baffled. To me, geometry shows evidently that the relative velocity is almost null since it is perpendicular to the line of sight. The fact that the virtual observer is in the same FOR of earth & planet X has nothing to do with that. Geometry still counts.

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21 minutes ago, michel123456 said:

So i have to rephrase my question:

In animation 1 the clocks are synchronized and show the same time. If they show the same time, it means that it cannot be the situation in the FOR of the Earth. it is not the situation in the FOR of X either (there is an hour of difference: for the Earth departure time is 12.00 & it is 11.00 on X, for X departure time is 13.00 & it is 12.00 on Earth). The animation does not show the 2h15min that were needed for Earth to receive the signal of the arrival at X. IOW animation 1 is that of a virtual observer.

Animation 2 has a different concept: it is from the FOR of B. 

So the question is; why is animation 2 not from a virtual observer? that would show the clocks synchronized I suppose, like animation 1.

 

Sorry, I have to digest the answers.

 

 

I don't think you understand what "frame of reference" means in Relativity.  It is not the same as a "point" of reference.  A frame of reference includes all points regardless of their relative position that are at rest with respect to each other.   So as long as Earth and X are motionless with respect to each other, they share the same frame of reference, and there is not an "Earth frame of reference" and a "Planet X frame of reference", but a single reference frame which Earth and Planet X are at rest with respect to.  And in that reference frame, the Earth and Planet X clocks are synchronized.  The fact that different observers placed at different positions in that reference frame would visually see different times on those clocks does not change the fact that all these observers would agree that these clocks are "in fact" synchronized with each other. 

Differences in position has no effect on the determination of simultaneity, only differences in velocity.

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3 minutes ago, Janus said:

I don't think you understand what "frame of reference" means in Relativity.  It is not the same as a "point" of reference.  A frame of reference includes all points regardless of their relative position that are at rest with respect to each other.   So as long as Earth and X are motionless with respect to each other, they share the same frame of reference, and there is not an "Earth frame of reference" and a "Planet X frame of reference", but a single reference frame which Earth and Planet X are at rest with respect to.  And in that reference frame, the Earth and Planet X clocks are synchronized.  The fact that different observers placed at different positions in that reference frame would visually see different times on those clocks does not change the fact that all these observers would agree that these clocks are "in fact" synchronized with each other. 

Differences in position has no effect on the determination of simultaneity, only differences in velocity.

So you are saying that the 1h delay between Earth & X does not count for relativity?

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56 minutes ago, michel123456 said:

To me, geometry shows evidently that the relative velocity is almost null since it is perpendicular to the line of sight.

Thought experiment: Let's assume you stretch one arm out far enough so your hand gets in the way of the object moving perpendicular to you. Is the impact at "almost null velocity"? No it is not.

 

(edit: x-post Janus explanation better than my thought experiment)

Edited by Ghideon
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4 minutes ago, michel123456 said:

So you are saying that the 1h delay between Earth & X does not count for relativity?

No, not really.   In most discussions dealing with Relativity, we factor out the light signal delay effects.

Thus if the Earth "sees" 11:00 on Planet X's clock when the Earth clock read 12:00, The Earth knows that the light carrying that image left planet X 1 hr ago when the Earth clock read 11:00 and thus both clocks read 11:00 at that moment.  And assuming that Planet X remained at rest with respect to the Earth, that in the time it took the light to reach the Earth, the Planet x clock ticked off 1 hr and thus reads 12:00 when that light arrives at Earth and the Earth clock reads 12:00.  Thus the Earth would conclude that the the two clocks are in sync. Planet X would conclude the same from its observations of Earth's clock from a distance of 1 light hr.

Relativity is what is "left over" after you account for light signal delay.

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michel123456, consider John in the 1700's rides his horse from London to Glasgow for business.

When he gets to Glasgow he mails a letter home to say he got there safely.

Back home in London his family get the letter 2 weeks after he left.

Question: do they think it took him 2 weeks to get to Glasgow?

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12 hours ago, michel123456 said:

If they show the same time, it means that it cannot be the situation in the FOR of the Earth. it is not the situation in the FOR of X either (there is an hour of difference: for the Earth departure time is 12.00 & it is 11.00 on X, for X departure time is 13.00 & it is 12.00 on Earth). The animation does not show the 2h15min that were needed for Earth to receive the signal of the arrival at X. IOW animation 1 is that of a virtual observer.

I can only second Janus here. Every object, how far or how near, that has no relative velocity in any direction to other objects in that FOR, belongs to the same inertial frame. All observers in that FOR agree on relative velocities of other objects, of the directions of their movements, etc. And they also  can agree on time. They can synchronise their clocks. Even if their distance is several light years. But of course they will see other clocks running behind, but not running slower. All clocks in their common FOR tick at the same rate. Only clocks 'ticking' in another FOR, i.e. move relative to the first FOR (Earth and X in your example), tick slower. And that is what time dilation is about.

Now the 'virtual observer' in the first animation is 'just some observer' in the same FOR as Earth and X. He knows what time it is in the FOR, independently of what he actually sees on other clocks in the same FOR. He agrees on how fast other objects, not in that FOR (e.g. B in the animations) move. And that is the velocity that counts in relativity, not the direction from the viewpoint of the observer.

12 hours ago, michel123456 said:

So the question is; why is animation 2 not from a virtual observer? that would show the clocks synchronized I suppose, like animation 1.

The answer should be simple by now: animation 1 is for an observer in the same FOR as Earth and X, animation 2 is for an observer in the same FOR as B. That is the only difference. Time differences due to signal delay in animation 1 are 'calculated out'. Or simpler said: the observer knows what time it is on Earth and X, because he is in the same FOR as them.

12 hours ago, michel123456 said:

Yes i am baffled. To me, geometry shows evidently that the relative velocity is almost null since it is perpendicular to the line of sight. The fact that the virtual observer is in the same FOR of earth & planet X has nothing to do with that. Geometry still counts.

No, as Swansont also repeated many times: there is only one velocity of B relative to the FOR of Earth and X, independent from how far an observer is in that frame, independent of the direction of the moving object.

In the case at hand geometry only counts if you want to derive what observers in the FOR of Earth and X actually see. Imagine it as a two step process: with relativity one calculates what actually happens; and then, using geometrical considerations. one can derive what observers actually see. To give just one example: when B flies back to Earth, its clock is time dilated for an observer on Earth. But he may see it actually run faster, because B is moving to Earth. (OTOH, an observer on X, in the same FOR as Earth will see B's clock run even slower than relativity predicts. But Earth and X fully agree about the 'real time dilation' of B, because they agree, per definition of a FOR, what the velocity of B is.) 

12 hours ago, michel123456 said:

Sorry, I have to digest the answers.

Then digest the answers, before firing off another argument!

 

Edited by Eise
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On 9/19/2020 at 6:52 PM, michel123456 said:

image.png.a46a8254b5667171cef45e11518d95b5.png

In the example, the ticking rate for the outbound travel is 1/3.

The ticking rate for the return travel is 3.

What is about the formula above? Is it about what someone sees, or about what actually happens?

On 10/3/2020 at 9:48 PM, Janus said:

If we now consider things from B's inertial frame of reference, you get this.

Animatione2.gif.4d0353946ab16a1bac4aa5f67e55c157.gif

B is at rest while the Earth and planet X move to the left at 0.8c.

the Earth, planet X, their clocks, and the distance between them is length contracted. B is not.

It take 45 min by B's clock for planet X to travel from being 0.6 lh away to B at 0.8c

Earth and planet X's clocks are time dilated and only accumulate 27 min during this time.

However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B.  The times on Earth's clock and B's clock when they are next to each other agree with the first animation, and the Times shown on B's clock and Planet X's clock when they are next to each other also agree with the first animation.

Again: Please explain the bold part:

Please correct me: the planet X clock stays in X's FOR. Which means (by me) that X clock reads 13.00 when B passes next to E (for clock X, the distance is not contracted). Where do the 48 min. come from?

 

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2 hours ago, michel123456 said:

What is about the formula above? Is it about what someone sees, or about what actually happens?

That is the formula for the Doppler effect for light. So it is about what you see. And this formula is only valid if the movement of sender and receiver is along the same line as the straight line between source and receiver. So here you have definitely dependency on the direction of the movement.

2 hours ago, michel123456 said:

Where do the 48 min. come from?

From the facts that the clocks are synchronised in their FOR, but therefore not in other FORs. No doubt, @Janus (or @swansont) can give the exact calculation. I do not have the complete in-depth knowledge (and experience) with relativity. I assume one can derive this from the Lorentz transformations. Maybe I'll try if my databases let me...

Edited by Eise
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40 minutes ago, Eise said:
2 hours ago, michel123456 said:

What is about the formula above? Is it about what someone sees, or about what actually happens?

That is the formula for the Doppler effect for light. So it is about what you see. And this formula is only valid if the movement of sender and receiver is along the same line as the straight line between source and receiver. So here you have definitely dependency on the direction of the movement.

I was just asking because from this formula was extracted what is happening (according to all other members here).

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2 minutes ago, michel123456 said:

I was just asking because from this formula was extracted what is happening (according to all other members here).

Of course you can. Subtract the effect of the signal delay, and you should get the relativistic time dilation. But I am not so versed in such calculations anymore.

58 minutes ago, Eise said:

I assume one can derive this from the Lorentz transformations. Maybe I'll try if my databases let me...

OK, it really was so simple:

Take the inverse  Lorentz transformation for time (we are looking at how B sees X at the moment of passing Earth, not how Earth sees B's time):

image.png.82c66f5d02d76aef1da067ae1552fa82.png

Whereby gamma:

image.png.31071062a5c8ccd7b7946692cdecddab.png

We take t = 0 for ease, so in our example we get:

vx'/gamma =(0.8 x 0.6)/0.6 = 0.8.

1 Lh = 60 LMinutes, so 0.8 x 60 LMinutes makes 48 minutes.

 

Now back to my databases...

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7 hours ago, michel123456 said:

What is about the formula above? Is it about what someone sees, or about what actually happens?

Again: Please explain the bold part:

Please correct me: the planet X clock stays in X's FOR. Which means (by me) that X clock reads 13.00 when B passes next to E (for clock X, the distance is not contracted). Where do the 48 min. come from?

 

Again, You are not getting the concept of inertial frames of reference. Planet X and Earth ( and their associated clocks) are at rest with respect to a single inertial reference frame. According to that reference frame the Earth clock and Planet X clock are in sync.  B is also "in" that inertial frame, but not at rest with respect to it.

Likewise,  there is a reference frame where B is at rest, while Earth and planet X are in motion ( in fact, there are an infinite number of inertial frame two choose from, It is just that these two are the most convenient to work with.)

So in the second animation, we are making our determination from  the later of these two frames. In that frame, the distance between Earth and Planet X is 0.6 light hr.  And is as measured from that frame that planet X's clock is 48 min ahead of Earth's clock.   In the "rest frame" of Planet X and the Earth, the distance is 1 light hr, and the clocks are in sync ( planet X's clock reads 12:00 when B passes Earth)

What you are trying to do is called "mixing frames" ( trying to use measurements from two different frames at the same time.). This only leads to confusion. 

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6 hours ago, Eise said:

Of course you can. Subtract the effect of the signal delay, and you should get the relativistic time dilation. But I am not so versed in such calculations anymore.

OK, it really was so simple:

Take the inverse  Lorentz transformation for time (we are looking at how B sees X at the moment of passing Earth, not how Earth sees B's time):

image.png.82c66f5d02d76aef1da067ae1552fa82.png

Whereby gamma:

image.png.31071062a5c8ccd7b7946692cdecddab.png

We take t = 0 for ease, so in our example we get:

vx'/gamma =(0.8 x 0.6)/0.6 = 0.8.

1 Lh = 60 LMinutes, so 0.8 x 60 LMinutes makes 48 minutes.

 

Now back to my databases...

gamma is 1.66666 in our example.

1/gamma is 0.6

Edited by michel123456
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