How Does Relativity Affect the Synchronization and Position of Moving Rods?

[alvaros]

Homework Statement

Referring to the drawing attached:
Two rods of the same length ( say 3 m ).
Rod A has a velocity of v = 0.866 c in x.
Rods are very close in y.
The drawing is taken from the point of view of rod B.
1 - Is the drawing correct ?
2 - Could we say that the right ends of the rods are at the same point and at the same time ?
3 - Could we say that the left end of rod A is at the same point and at the same time that the middle ( 1.5 m ) point of rod B ?

Homework Equations

l' = l . [tex]\gamma[/tex]

The Attempt at a Solution

The attempt at a solution is the drawing.

 

[alvaros]

No answer ?

The next question would be: what happens from the point of view of rod A ? In this case:
- rod B would be shorter than rod A
- the right end of rod B would be at the same point and at the same time than the middle point of rod A.

 

[nrqed]

Homework Statement

Referring to the drawing attached:
Two rods of the same length ( say 3 m ).
Rod A has a velocity of v = 0.866 c in x.
Rods are very close in y.
The drawing is taken from the point of view of rod B.
1 - Is the drawing correct ?
2 - Could we say that the right ends of the rods are at the same point and at the same time ?
3 - Could we say that the left end of rod A is at the same point and at the same time that the middle ( 1.5 m ) point of rod B ?

Yes to al your questions at the condition that by "at the same time" you mean at the same time as measured in the frame of B.

 

[nrqed]

No answer ?

The next question would be: what happens from the point of view of rod A ? In this case:
- rod B would be shorter than rod A
- the right end of rod B would be at the same point and at the same time than the middle point of rod A.

From the point of view of A, rod B is shorter, yes. I can't quite make sense of your second sentence, it's incomplete.

 

[alvaros]

The next question would be: what happens from the point of view of rod A ? In this case:
- rod B would be shorter than rod A
- the right end of rod B would be at the same point and at the same time than the middle point of rod A.

Ive made another drawing from the point of view of rod A.

Then:

At the same readings on clocks A and B the right ends of the rods are at the same point.
( suppose both clocks give "0" )
But from the point of view of B its middle point is at the same point than the left point of rod A. ( first drawing, "dibujo1.bmp" ) and
From the point of view of A its middle point is at the same point than the left point of rod B. ( second drawing, "dibujo2.bmp" )

 

[nrqed]

Ive made another drawing from the point of view of rod A.

Then:

At the same readings on clocks A and B the right ends of the rods are at the same point.
( suppose both clocks give "0" )
But from the point of view of B its middle point is at the same point than the left point of rod A. ( first drawing, "dibujo1.bmp" ) and
From the point of view of A its middle point is at the same point than the left point of rod B. ( second drawing, "dibujo2.bmp" )

Again your sentences are not complete (sorry, I am not trying to be difficult, it's just that most of the apparent paradoxes in SR arise from a careless use of words which lead to misunderstandings so I am just trying to be careful with the wording).

here would be two correct statements:

But from the point of view of B and at t=0 as measured by B its middle point is at the same point than the left point of rod A.

From the point of view of A abd at t=0 as measured by A its middle point is at the same point than the left point of rod B.


The reason those two statements don't contradict each other is that two events which are simultaneous in one frame are not simultaneous in another frame.

 

[Doc Al]

Ive made another drawing from the point of view of rod A.

Then:

At the same readings on clocks A and B the right ends of the rods are at the same point.
( suppose both clocks give "0" )

OK. So when the right ends of A and B overlap, the clocks at those locations read the same time t = 0. No problem.

Just to be clear, let's imagine that each rod has three synchronized clocks located at the left end, middle, and right end. Let's call the clocks AL, AM, AR and BL, BM, BR.

So, to restate what we already stated: When clocks AR and BR overlap, they read the same reading, t = 0.

But from the point of view of B its middle point is at the same point than the left point of rod A. ( first drawing, "dibujo1.bmp" )

No problem. The first drawing is from B's point of view: According to B, at the instant shown all three of B's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod A is aligned with the middle of B. Note that clock AL does not read t = 0. (According to B, clock AL is behind clock AR.)

From the point of view of A its middle point is at the same point than the left point of rod B. ( second drawing, "dibujo2.bmp" )

No problem. This drawing is from A's point of view: According to A, at the instant shown all three of A's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod B is aligned with the middle of A. Note that clock BL does not read t = 0. (According to A, clock BL is ahead of clock BR.)

So both diagrams are perfectly consistent.

(nrqed beat me to it!)

 

[alvaros]

nrqed:

here would be two correct statements:

But from the point of view of B and at t=0 as measured by B its middle point is at the same point than the left point of rod A.

From the point of view of A abd at t=0 as measured by A its middle point is at the same point than the left point of rod B.

Yes, this is what I meant.

Doc Al:

No problem. The first drawing is from B's point of view: According to B, at the instant shown all three of B's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod A is aligned with the middle of B. Note that clock AL does not read t = 0. (According to B, clock AL is behind clock AR.)
No problem. This drawing is from A's point of view: According to A, at the instant shown all three of A's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod B is aligned with the middle of A. Note that clock BL does not read t = 0. (According to A, clock BL is ahead of clock BR.)

I suppose you are using t' = γ(t-vx/c²)
"According to B, clock AL is behind clock AR"
"According to A, clock BL is ahead of clock BR"

1 - The change in sign comes from the change in the direction of velocity "v" ?
2 - Anyway, its very difficult to understand that movements from left to right have different results than right to left.
Why the left clock ( A ) that moves to the left is behind and
the left clock ( B ) that moves to the right is ahead ?

 

[Doc Al]

clock desynchronization

I suppose you are using t' = γ(t-vx/c²)
"According to B, clock AL is behind clock AR"
"According to A, clock BL is ahead of clock BR"

This is the relativity of simultaneity. The rule of thumb for moving clocks is: Moving clocks, synchronized in their rest frame but separated by a distance D along their direction of motion, are not synchronized in the stationary frame; The front clock lags the rear clock by an amount:
[tex]T = Dv/c^2[/tex].

I think this is easier to see using:
[tex]t = \gamma(t' + vx'/c^2)[/tex]

From A's point of view: What do B's clocks read when the time is t = 0 on A's clocks? The above transformation gives us:
[tex]0 = \gamma(t' + vx'/c^2)[/tex]
Or:
[tex]t' = -vx'/c^2[/tex]

Clock BR is at x'=0, so clock BL is at some negative value of x' making it have a time t' > 0.

1 - The change in sign comes from the change in the direction of velocity "v" ?

Sure.

2 - Anyway, its very difficult to understand that movements from left to right have different results than right to left.

They don't! The physics works the exact same way. Of course the velocity (and coordinate system) is different depending upon which direction you are moving, but the effect is completely symmetric. To see this, just rotate both of your diagrams by 180 degrees. Do you think the results would be different?

Why the left clock ( A ) that moves to the left is behind and
the left clock ( B ) that moves to the right is ahead ?

See my explanation above.

 

[alvaros]

The rule of thumb for moving clocks is: Moving clocks, synchronized in their rest frame but separated by a distance D along their direction of motion, are not synchronized in the stationary frame; The front clock lags the rear clock by an amount:
T = Dv/c^2

This is a very simple and clear explanation ( at least to me )

They don't! The physics works the exact same way. Of course the velocity (and coordinate system) is different depending upon which direction you are moving, but the effect is completely symmetric. To see this, just rotate both of your diagrams by 180 degrees. Do you think the results would be different?

Ive rotated the diagrams 180 degrees and the results are OK.

Thanks.

But one question that I see now is: the same ( real ) clock, at the same point, is seen different if you are moving or not. So, at the same point, you see different events, depending on your velocity.

 

[Doc Al]

But one question that I see now is: the same ( real ) clock, at the same point, is seen different if you are moving or not. So, at the same point, you see different events, depending on your velocity.

I don't quite know what you mean by seeing different events at the same point. An "event" is something that happens at a particular place and time, but the measurement of the position and time of that event depends on who's doing the observing.

In this example, what the the two observers disagree about is whether the "events" (say the left-end and right-end clocks showing 2 pm) happen at the same time or not. A says B's clocks strike 2pm at different times (according to A's clocks). Of course B says the same thing about A's clocks.

 

[alvaros]

I don't quite know what you mean by seeing different events at the same point.

I call one event "A real clock whit its hand pointing to 10 s" ( Observer 1 )
Another event "The same real clock whit its hand pointing to 0 s" ( Observer 2 )

If the clock is programmed to explode at +10s Observer 1 is seeing the explosion and Observer2 will see the explosion ( at another point ) ...

unless Observer 2 destroys the clock ( they are at the same point, can he destroy the clock ? )

 

[nrqed]

I call one event "A real clock whit its hand pointing to 10 s" ( Observer 1 )
Another event "The same real clock whit its hand pointing to 0 s" ( Observer 2 )

If the clock is programmed to explode at +10s Observer 1 is seeing the explosion and Observer2 will see the explosion ( at another point ) ...

unless Observer 2 destroys the clock ( they are at the same point, can he destroy the clock ? )

The point is that an event is not only something that oocurs at one specific time but also at one specific location. two observers in different frames will assign different coordinates (values of x and t) to a single event but they both agree that the event occurred at a specific time and location.


Now, a clock indicating 10 seconds is a valid event. Everybody (in all frames!) will agree that this event occurred and that they saw that clock indicating 10seconds when it was at some location in space. Nobody can disagree on the event itself . when all observers assign a time and a value of x to this particular event, they will all have different results (and values of time which may have nothing to do with the value indicated by the clock).

 

[alvaros]

nrqed:

The point is that an event is not only something that oocurs at one specific time but also at one specific location. two observers in different frames will assign different coordinates (values of x and t) to a single event but they both agree that the event occurred at a specific time and location.

I don't understand what you mean, but in the given example the two events occurs at the same point ( I don't know why you talk about x coordinates ),
and the two events are: a real clock with its hand at one position ( 0 s ) and the same clock with its hand at another position ( 10 s ).

(and values of time which may have nothing to do with the value indicated by the clock).

Are you saying that there is something called "time" that has nothing to do with clocks ?
How do you measure this "time" ?

 

[nrqed]

nrqed:


I don't understand what you mean, but in the given example the two events occurs at the same point ( I don't know why you talk about x coordinates ),
and the two events are: a real clock with its hand at one position ( 0 s ) and the same clock with its hand at another position ( 10 s ).

whn you say that the two events occur at the same "point" , you are making a statement which depends on the frame. There is only one frame of reference for which the two events occur at the same point (i.e. at the same value of x...I am considering only frames moving in one dimension, along the x axis) and that's the frame attached to the clock. Look, if I am moving at some speed relative to the clock, the two events will NOT occur at the same value of x in my frame! Between the instant the clock will indicate 0 and 10 seconds, the clock will have moved with respect to me! So the two events will occur at different locations fo rme. You were restricting yourself to the frame of the clock!

Are you saying that there is something called "time" that has nothing to do with clocks ?
How do you measure this "time" ?

Yes, you measure time using clocks, but you must use clocks in yoru frame of reference, that is at rest with respect to you! . If I see a flying clock zooming by me, I cannot use this to measure time in my frame

 

[alvaros]

I don't know if you don't understand or you don't want to understand.

Doc Al:

the right sides of both rods are aligned,

What do you think "are aligned" means ?

 

[nrqed]

I don't know if you don't understand or you don't want to understand.

I guess I don't understand special relativity.

Goodbye

 

[Doc Al]

I don't know if you don't understand or you don't want to understand.

I beg your pardon? You might wish to rephrase that. You are the one looking for help, not us.

Doc Al:

What do you think "are aligned" means ?

Two points on the the moving rods "are aligned" when they are momentarily passing each other.

For example: When the right ends of each rod pass each other ("are aligned"), the clocks at those points (AR & BR) both read t = 0. Everyone agrees that that is true: that those clocks really read those times as they passed each other. What they disagree about is the time that the other clocks read at that moment.

 

[alvaros]

Quote:
Originally Posted by alvaros
I don't know if you don't understand or you don't want to understand.

I beg your pardon? You might wish to rephrase that. You are the one looking for help, not us.

You seem to be a lawyer, not a physicist.
Yes, I am looking for help, so, please, help me, and tell me if the following sentences are true or false.

What do you think "are aligned" means ?
1 - It means that they are at the same point. ( True / False )
2 - Time is measured by clocks, so time = reading of a clock ( T / F )
3 - One clock, at rest, seeing by an observer at rest, gives a time, but the same clock seeing by a moving observer, gives a different time ( both observers at the same point ) ( T / F )

Last, Doc Al:

What they disagree about is the time that the other clocks read at that moment

.
You introduce a new concept ( to me ) in SR: the moment

 

[Doc Al]

You seem to be a lawyer, not a physicist.

Since, as is evident from this thread, SR is a subtle and tricky subject one must be precise. If you are serious about learning relativity (or physics in general), you'd better get used to being precise.

Yes, I am looking for help, so, please, help me, and tell me if the following sentences are true or false.

What do you think "are aligned" means ?

I already told you how I used the term. Go back and reread what I wrote earlier.

1 - It means that they are at the same point. ( True / False )

The way I used the term, the ends of the rod are momentarily "aligned" when they pass each other. When the ends are so "aligned" they are at the same point.

2 - Time is measured by clocks, so time = reading of a clock ( T / F )

Not sure I understand the second part of that statement. Every frame uses their own clocks to measure time.

3 - One clock, at rest, seeing by an observer at rest, gives a time, but the same clock seeing by a moving observer, gives a different time ( both observers at the same point ) ( T / F )

I don't understand your question; If I misinterpret your question, please restate with a clear, specific example. I don't know what you mean by "both observers at the same point". Do you mean that the observations take place as the two observers--each with their own clock--pass each other and look at each other's clock? If that's what you mean, then the answer is clear: If observer A's clock reads a certain time--as observer B passes it--then moving observer B will also see A's clock read that same time. How could it be otherwise?

You introduce a new concept ( to me ) in SR: the moment

SR deals with time and simultaneity, so the idea of "now" should not be new.

 

[alvaros]

Quote:
2 - Time is measured by clocks, so time = reading of a clock ( T / F )

Not sure I understand the second part of that statement. Every frame uses their own clocks to measure time.

t is the reading of clocks at rest, t' is the reading of moving clocks, and there is no any other "time".

I don't understand your question; If I misinterpret your question, please restate with a clear, specific example. I don't know what you mean by "both observers at the same point". Do you mean that the observations take place as the two observers--each with their own clock--pass each other and look at each other's clock?

Yes !

If that's what you mean, then the answer is clear: If observer A's clock reads a certain time--as observer B passes it--then moving observer B will also see A's clock read that same time. How could it be otherwise?

! Time dilation can't be observed ! The formula you gave me T = Dv/c^2 is useless because you can't observe T !

 

[Doc Al]

! Time dilation can't be observed ! The formula you gave me T = Dv/c^2 is useless because you can't observe T !

How did you draw that bizarre (and quite silly) conclusion? Do you even know what that formula means?

 

[alvaros]

All quotes from Doc Al:

Referring to the first drawing.

No problem. The first drawing is from B's point of view: According to B, at the instant shown all three of B's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod A is aligned with the middle of B. Note that clock AL does not read t = 0. (According to B, clock AL is behind clock AR.)

Referring to the second drawing ( first and second drawing refer to the same moment )

No problem. This drawing is from A's point of view: According to A, at the instant shown all three of A's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod B is aligned with the middle of A. Note that clock BL does not read t = 0. (According to A, clock BL is ahead of clock BR.

)

This is what you thought.
And now:

If that's what you mean, then the answer is clear: If observer A's clock reads a certain time--as observer B passes it--then moving observer B will also see A's clock read that same time. How could it be otherwise?

And because I am

the one looking for help

please, help me and tell me where were you wrong or what am I missing.

 

[Doc Al]

Referring to the second drawing ( first and second drawing refer to the same moment )

This is where you are wrong: The first and second drawings DO NOT refer to the same moment.

Your first diagram shows things from B's point of view. According to B, the following two events take place at the same time (when B's clocks all read t = 0):
(Event #1) The left side of rod A passes the midpoint of rod B
(Event #2) The right side of rod A passes the right side of rod B

(Observers on rod A do not agree! According to A, those two events took place at different times: Event #1 happened before Event #2.)

Your second diagram shows things from A's point of view. According to A, the following two events take place at the same time (when A's clocks all read t = 0):
(Event #3) The left side of rod B passes the midpoint of rod A
(Event #2) The right side of rod A passes the right side of rod B

(Observers on rod B do not agree! According to B, those two events took place at different times: Event #2 happened before Event #3.)

Note that the second diagram does not show Event #1 and the first diagram does not show Event #3. The only thing that is "the same" about the two diagrams is that they both show Event #2: the overlapping of the right ends of each rod. (Everybody agrees that Event #2 takes place at t = t' = 0.)

Questions?

 

[alvaros]

Your post #7:

No problem. The first drawing is from B's point of view: According to B, at the instant shown all three of B's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod A is aligned with the middle of B. Note that clock AL does not read t = 0. (According to B, clock AL is behind clock AR.)

Your post #24:

Your first diagram shows things from B's point of view. According to B, the following two events take place at the same time (when B's clocks all read t = 0):
(Event #1) The left side of rod A passes the midpoint of rod B
(Event #2) The right side of rod A passes the right side of rod B
(Observers on rod A do not agree! According to A, those two events took place at different times: Event #1 happened before Event #2.)

Your post #7:

No problem. This drawing is from A's point of view: According to A, at the instant shown all three of A's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod B is aligned with the middle of A. Note that clock BL does not read t = 0. (According to A, clock BL is ahead of clock BR.)

Your post #24:

Your second diagram shows things from A's point of view. According to A, the following two events take place at the same time (when A's clocks all read t = 0):
(Event #3) The left side of rod B passes the midpoint of rod A
(Event #2) The right side of rod A passes the right side of rod B
(Observers on rod B do not agree! According to B, those two events took place at different times: Event #2 happened before Event #3.)

You have come back to post #7 ( you are saying the same in posts 7 and 24 )

And because this leads to a paradox: ( My post #12 )

If the clock is programmed to explode at +10s Observer 1 is seeing the explosion and Observer2 will see the explosion ( at another point ) ...
unless Observer 2 destroys the clock ( they are at the same point, can he destroy the clock ? )

you change your explanation: ( from your post #20 )

If that's what you mean, then the answer is clear: If observer A's clock reads a certain time--as observer B passes it--then moving observer B will also see A's clock read that same time. How could it be otherwise?

Are you going now to repeat post #20 ?

You keep that If observer A's clock reads a certain time--as observer B passes it--then moving observer B will also see A's clock read that same time. !

 

[Doc Al]

You keep that If observer A's clock reads a certain time--as observer B passes it--then moving observer B will also see A's clock read that same time. !

Well, it's a true statement--as were all the other statements I (and nrqed) have been making.

Let's take the example of Event #1: The left side of rod A passes the midpoint of rod B. Assume that the A frame has clocks located at each end and the middle of rod A and that the B frame has its own clocks at each end and the middle of rod B. At the time of this event, B's clock (at the midpoint of rod B) reads time t = 0, but A's clock (at the left end of rod A) reads a different time, let's say t' = -10 seconds. [The rods are assumed to be extremely long in order to have observable time differences.] Of course B agrees that A's clock reads whatever it says it reads, but B does not agree that A's clocks (at left and right ends of rod A) are synchronized.

If clock A (say the one at the right end of rod A) is set to explode when it reads t' = +10 seconds, I assure you ALL OBSERVERS will agree that it blows up when clock A reads t' = +10 seconds. And all observers will be able to calculate what time that explosion will take place according to B's clocks and exactly where along rod B that clock will be when it explodes. (Of course, to correctly predict the time and place, one must use relativity.)

This is basic stuff. Until you understand this much, further discussion is pointless. So I'll stop repeating myself!

 

[alvaros]

Let's take the example of Event #1: The left side of rod A passes the midpoint of rod B. Assume that the A frame has clocks located at each end and the middle of rod A and that the B frame has its own clocks at each end and the middle of rod B. At the time of this event, B's clock (at the midpoint of rod B) reads time t = 0, but A's clock (at the left end of rod A) reads a different time, let's say t' = -10 seconds. [The rods are assumed to be extremely long in order to have observable time differences.] Of course B agrees that A's clock reads whatever it says it reads, but B does not agree that A's clocks (at left and right ends of rod A) are synchronized.

"At the time of this event" what do you mean by time ? Ah, you are talking about the moment !

Of course B agrees that A's clock reads whatever it says it reads ( you, talking about AL clock ) Following this reasoning:

Of course B agrees that A's clock reads whatever it says it reads ( me,talking about AR clock )

So AL = AR from the point of view of B -> A's clocks are synchronized from the point of view of B !

If clock A (say the one at the right end of rod A) is set to explode when it reads t' = +10 seconds

I said AL clock, don't make the discussion more confusing.

So I'll stop repeating myself!

I suppose that being you a PF MENTOR nobody will dare to continue discussing, at least from the PF staff. What a pity !
Anyway, Thank you by your time.

 

[Doc Al]

"At the time of this event" what do you mean by time ? Ah, you are talking about the moment !

I'm talking about the time of one specific event, Event #1.

Of course B agrees that A's clock reads whatever it says it reads ( you, talking about AL clock )

Right! ALL observers agree on what clock AL and clock BM read at the moment they pass each other, which is Event #1.

Following this reasoning:

Of course B agrees that A's clock reads whatever it says it reads ( me,talking about AR clock )

If you wish to talk about some other clock, not present at Event #1 but some distance away, you have to define when you are talking about it. And since A and B disagree on whether clocks AL and AR read the same time, you need to define "when" according to who.

But the simple fact remains: Whenever two clocks pass each other, everyone agrees on the time that each clock reads.

So AL = AR from the point of view of B -> A's clocks are synchronized from the point of view of B !

Uh... nope!

Again, to get anywhere, you must be precise.

I said AL clock, don't make the discussion more confusing.

It's only your half of the discussion that's confusing!