If correct: a catastrophe in the Lorentz transformation

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  • #1
aawahab76
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The Lorentz transformation are given by (see the attachment)

x'=(x-vt)/√(1-v^2/c^2 )

t'=(t-vx/c^2)/√(1-v^2/c^2 )

Let us transform the event (10^100 m,1sec) in the x-frame to the x'-frame that is moving in the usual geometry with the speed v=10^(-10) c. Could you see that that t'≈-10^81 sec!! That is negative, right? Am I incorrect? Please help.
 

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  • #2
JesseM
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The Lorentz transformation are given by (see the attachment)

x'=(x-vt)/√(1-v^2/c^2 )

t'=(t-vx/c^2)/√(1-v^2/c^2 )

Let us transform the event (10^100 m,1sec) in the x-frame to the x'-frame that is moving in the usual geometry with the speed v=10^(-10) c. Could you see that that t'≈-10^81 sec!! That is negative, right? Am I incorrect? Please help.
Yes, what's the problem? A negative time-coordinate just means an event that occurred before the time arbitrarily labeled "0" in your coordinate system. For example, if an event at 11 A.M. is labeled with coordinate t=0 seconds, then an event at 10 A.M. must have a time-coordinate of t=-1 hour=-3600 seconds in the same coordinate system. In your example, t'≈-10^81 sec just means the event happened about 10^81 seconds before t'=0 in the x'-frame.
 
  • #3
aawahab76
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Not that simple. The usual geometry used in such problem is that the origins of the two frames concide with each other at t=t'=0 and as such t' is positive for positive t. on the other hand, notice that we are really in a nonrelativistic domain and such large time difference is not possible, right?
 
  • #4
Dale
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t' is positive for positive t
Not in general, as you have discovered. You are having trouble with the relativity of simultaneity, which is the most difficult concept in relativity for students to learn.
 
  • #5
JesseM
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Not that simple. The usual geometry used in such problem is that the origins of the two frames concide with each other at t=t'=0 and as such t' is positive for positive t.
No, as Dalespam said you need to look into the relativity of simultaneity. When two events have a spacelike interval between them, different frames can disagree about the order they occurred, so for example if event #1 is the one at t=t'=x=x'=0 and the event #2 is the one you wrote, then it can be true that in the unprimed frame event #2 happened after event #1, but in the primed frame event #2 happened before event #1.

If you learn to draw spacetime diagrams you can actually see how one frame's surface of simultaneity (a surface of constant time-coordinate in that frame) appears slanted when drawn from the perspective of another frame, and the farther along the x-axis you go, the greater the difference in judgment about the time-coordinate of a given event. This wikibook has a pretty good intro. to spacetime diagrams in the sections titled "Spacetime" and "Simultaneity, time dilation and length contraction".
on the other hand, notice that we are really in a nonrelativistic domain and such large time difference is not possible, right?
In classical Newtonian physics there is no relativity of simultaneity, so if the origins of two inertial frames in Newtonian physics coincide, then they can have zero disagreement about the time-coordinate of any event.
 
  • #6
aawahab76
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OK, it is not clear at all if the problem is in the simultaneity concept.

To make the problem clearer, I will try to reformulate it in the following different form: because v=10^(-10) c, then we can use the Galilean transformation rather than the Lorentz one,, that is
t'≈t and x'≈x-vt
and using the coordinates of the event given, that is (10^100 m, 1 sec), we find that t'≈t=1 sec and x'≈10^100 m
this is intuitively correct but is not the values given by the exact formula (the Lorentz transformation), right?!

In a different viewpoint, notice that the issue as I can see is much simpler to require the simultaneity subject. Simple as it is, the Lorentz transformation relates the coordinates of one event at two different inertial frames. The Lorentz transformation in the usual form used above, tacitly assumes that the event (0',0') in x' corresponds to (0,0) in the x-frame. Now do you think that an event at (10^100 m, 1 sec) in the x-frame will correspond to (≈10^100 m, -10^81 sec)?! Do not forget that v=10^(-10) c.
 
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  • #7
JesseM
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To make the problem clearer, I will try to reformulate it in the following different form: because v=10^(-10) c, then we can use the Galilean transformation rather than the Lorentz one,
No, you can't. Whether the Galilei transformations give you approximately the correct answer or not depends not only on the absolute magnitude of the velocity but also on the distance and the time between the events involved, as your example shows.
aawahab76 said:
In a different viewpoint, notice that the issue as I can see is much simpler to require the simultaneity subject. Simple as it is, the Lorentz transformation relates the coordinates of one event at two different inertial frames. The Lorentz transformation in the usual form used above, tacitly assumes that the event (0',0') in x' corresponds to (0,0) in the x-frame. Now do you think that an event at (10^100 m, 1 sec) in the x-frame will correspond to (≈10^100 m, -10^81 sec)?!
Yes, of course it would, why do you think this is a problem? Suppose you had an two enormous rulers over 10^100 meters long moving alongside one another at a relative velocity of 10^(-10) c, and along each ruler were fixed clocks at regular intervals that had been synchronized using the Einstein clock synchronization convention, and at the moment the 0-meter marks of the two rulers lined up the clocks affixed to each ruler read t=t'=0. (So the setup of this experiment would be similar to the one I depicted in the diagrams on this thread, although the velocity in this case is much smaller.) Then it would also be true that at the moment the 10^100 meter mark on the unprimed ruler lined up with the gamma*(10^100 - (10^(-10)*c)) ≈10^100 m mark on the primed ruler, the unprimed clock at that position would read 1 second while the unprimed clock at that position would read ≈10^-81 seconds. This is a real physical fact about what would be true in a certain type of experiment in relativity, not merely a fact about coordinate systems.
 
  • #8
aawahab76
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Dear JesseM,

What you say is mathematically correct, but this is not my real concern. Of course it might be that such subjects are to be only treated with mathematics but still we can discuss the physics. So does any body think that the result we stated is physically reasonable! Does such huge difference between the measurements in the two inertial frame resulting from such small relative speed make sense?

On the other hand, when you say "No, you can't. Whether the Galilei transformations give you approximately the correct answer or not depends not only on the absolute magnitude of the velocity but also on the distance and the time between the events involved, as your example shows." then does this means that in general the Lorentz transformation does not reduce to the Galilean transformation?

I will try to reformulate the whole problem in a different mathematically sound form in a subsequent replay, if I could find such.
 
  • #9
JesseM
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What you say is mathematically correct, but this is not my real concern. Of course it might be that such subjects are to be only treated with mathematics but still we can discuss the physics.
I did discuss the physics, that's why I brought up the scenario of two physical rulers with physical clocks along them synchronized using the Einstein synchronization convention (this is how Einstein originally defined the coordinates of a relativistic inertial frame in physical terms), and what different pairs of clocks would read at the moment they passed one another.
On the other hand, when you say "No, you can't. Whether the Galilei transformations give you approximately the correct answer or not depends not only on the absolute magnitude of the velocity but also on the distance and the time between the events involved, as your example shows." then does this means that in general the Lorentz transformation does not reduce to the Galilean transformation?
If by "in general" you mean "for arbitrarily large separations between the events", then no, it doesn't reduce to the Galilean transform in general. But I think you can say it reduces to the Galilei transform in some appropriate limit which involves both the spacetime interval between the event you're interested in and the origin and the relative velocity between the frames--I haven't thought about what this limit would look like though, might be an interesting problem to consider.
 
  • #10
bobc2
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...If by "in general" you mean "for arbitrarily large separations between the events", then no, it doesn't reduce to the Galilean transform in general. But I think you can say it reduces to the Galilei transform in some appropriate limit which involves both the spacetime interval between the event you're interested in and the origin and the relative velocity between the frames--I haven't thought about what this limit would look like though, might be an interesting problem to consider.

aawahab76, here is a sketch of the so-called Penrose Andromeda Paradox. The sketch illustrates just what JesseM has been explaining to us. Two people walk past each other very slowly (your example of relative velocities so small that you might think you could approximate the situation with the Galilean concept without any problems). Way out in the Andromeda galaxy, in one person's "NOW" the aliens are having a meeting to decide whether to attack earth. But, in the other person's "NOW" the decision has already been made and the alien fleet is already on its way for the attack.

Credit: "The Emperor's New Mind" by Roger Penrose (OXFORD) - page 260 paper back edition.



PenroseSpaceTime-1.jpg
 
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  • #11
Dale
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What you say is mathematically correct, but this is not my real concern. Of course it might be that such subjects are to be only treated with mathematics but still we can discuss the physics.
The math correctly describes the physics. This has been tested to enormous precision.

So does any body think that the result we stated is physically reasonable! Does such huge difference between the measurements in the two inertial frame resulting from such small relative speed make sense?
At such a large distance, yes. You are trying to synchronize two events that are a google meters apart. Why are you surprised that it is sensitive to very small speeds? The only thing that is not physically reasonable is the distance. Once you stipulate that, the rest follows.
 
  • #12
aawahab76
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At such a large distance, yes. You are trying to synchronize two events that are a google meters apart. Why are you surprised that it is sensitive to very small speeds? The only thing that is not physically reasonable is the distance. Once you stipulate that, the rest follows.[/QUOTE]

Before going more on the subject, we are here talking about one and only one event, right? this event is named in the x-frame (x=10^100 m, t=1 sec) or the corresponding (x',t') in the x'-frame. I do not see "simultaneity" issue here except when used to build the two frames of reference using the rulers and synchronized clocks as explained in relativity books.

Now let us get a little deeper into the whole physical process we described. At first, we have two inertial frames with their corresponding coordinates and satisfying the condition that (0,0) corresponds to (0',0'). From this moment to before t=1 sec in the x-frame (or from 0' to t' that corresponds to (x=10^100 m, t=1 sec) in the x'-frame), no things happened. Suddenly a flash of light appeared at a point (we call it event P) and was registered by the x-frame to be located at (10^100 m, 1sec). This is a real physical event which is not related to any coordinate or frame of reference . Are not we sure, physically and mentally, at this moment that the x'-frame will register this event to happen some where after (0',0')? Notice that (x',t') corresponding to (10^100 m, 1sec) will be registered by the already synchronized clock located at the event P. Will not this be a positive time, that is after the (0'0') event?

Notice that when I speak about mathematical versus physical I do not intend to differentiate between them such that one is true and the other is not, but my whole problem in this thread is to find why my physical intuition seems (wright or wrong) to contradict the mathematical structure that we believe in.
 
  • #13
Dale
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Before going more on the subject, we are here talking about one and only one event, right? this event is named in the x-frame (x=10^100 m, t=1 sec) or the corresponding (x',t') in the x'-frame. I do not see "simultaneity" issue here
The second event is obviously the origin (0,0). You are surprised that one event may be later than the other in one frame and that same event may be earlier than the other in another frame. That is a clear example of the relativity of simultaneity.

Btw, by convention we put the time coordinate first. Not a big deal, but you will probably see it the other way more often than the way you wrote it.
 
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  • #14
aawahab76
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(0,0) is the first event is true and equivalent to what I said in my original post that simultaneity appears when building the frame of reference.
However, still the event P happened after (0,0) and (0',0') (so we are not speaking about simultaneous events) for the x and x'-frames respectively so why the Lorentz transformation gives a negative t' value? It is possible here that I am missing some thing. Keep attention that physically when the two observers in the two frame of references meet, event P already happened with respect to x'-frame but not yet for the x-frame, right?

And for the (x, t) or (t,x), yes you are right and I am sorry, every event should read (t,x) or even (ct,x).
 
  • #15
Dale
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However, still the event P happened after (0,0) and (0',0') (so we are not speaking about simultaneous events) for the x and x'-frames respectively so why the Lorentz transformation gives a negative t' value?
No, the event P happened after (0,0) in the x frame and before (0',0') in the x' frame. That is what is meant by the relativity of simultaneity: that whether or not one event is in the future, same time, or past of a second event depends on the frame of reference.
 
  • #16
Mentz114
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The diagrams show the worldlines of the ends of a green and blue ruler of equal length that pass each other on the x-axis. Event 0 is the coincidence of one end of the rulers, event 1 is the coincidence of the other (leading) ends. In the blue frame event 0 happens before event 1, while in the green frame it is the other way round. The diagrams are accurately scaled, the relative velocity is 0.197c and the rest length of the rulers 6 light years.

If the clocks were synchronised at event 0 so t = t'= 0, the the time of event 2 on the green clock is negative.

It seems clear that if we increase the length of the rulers the time difference between the two clock readings of event 1 will grow.

There's nothing weird happening, it is consistent.
 

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  • #17
aawahab76
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No, the event P happened after (0,0) in the x frame and before (0',0') in the x' frame. That is what is meant by the relativity of simultaneity: that whether or not one event is in the future, same time, or past of a second event depends on the frame of reference.
the event P never happened before the two observers meet so how it happened for the x'-frame before that?
 
  • #18
aawahab76
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The diagrams show the worldlines of the ends of a green and blue ruler of equal length that pass each other on the x-axis. Event 0 is the coincidence of one end of the rulers, event 1 is the coincidence of the other (leading) ends. In the blue frame event 0 happens before event 1, while in the green frame it is the other way round. The diagrams are accurately scaled, the relative velocity is 0.197c and the rest length of the rulers 6 light years.

If the clocks were synchronised at event 0 so t = t'= 0, the the time of event 2 on the green clock is negative.

It seems clear that if we increase the length of the rulers the time difference between the two clock readings of event 1 will grow.

There's nothing weird happening, it is consistent.
thanks but I do not have an objection on the consistency of the Lorentz transformation or the graphical construction, my concern is the physical picture, that is all.
 
  • #19
bobc2
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thanks but I do not have an objection on the consistency of the Lorentz transformation or the graphical construction, my concern is the physical picture, that is all.

aawahab, I hope I am not overly belaboring the point, but you seem to not accept that the Lorentz transformation and the graphical construction is the actual physical picture. And the Penrose sketch presented above is both consistent with the Lorentz transformation and is a very good representation of the physical picture. At this point in our understanding of the world, that is exactly the way nature is working.

You might want to go back and reflect on JesseM's posts, nos. 7 & 9.
 
  • #20
JesseM
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the event P never happened before the two observers meet so how it happened for the x'-frame before that?
Why do you believe it "never happened before the two observers meet"? Do you believe there is some real, frame-independent truth about whether one event A happened before or after another event B with a spacelike separation from A? If so, you are rejecting the relativity of simultaneity! (Unless you believe that the truth of which happened first is of a purely metaphysical sort that could be known only by God...as long as a person accepts that there is no empirical way to show that one frame's definition of simultaneity is 'more correct' than any others, there views are compatible with relativity).
 
  • #21
Dale
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the event P never happened before the two observers meet so how it happened for the x'-frame before that?
It did happen before the observers meet in the x' frame. It only happened afterwards in the x frame. Whether it happened before or after depends on the frame.

Please Google "relativity of simultaneity" and read up a bit and come back with questions. This is the most challenging concept in special relativity.
 
  • #22
Dale
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Now let us get a little deeper into the whole physical process we described. At first, we have two inertial frames with their corresponding coordinates and satisfying the condition that (0,0) corresponds to (0',0'). From this moment to before t=1 sec in the x-frame (or from 0' to t' that corresponds to (x=10^100 m, t=1 sec) in the x'-frame), no things happened. Suddenly a flash of light appeared at a point (we call it event P) and was registered by the x-frame to be located at (10^100 m, 1sec). This is a real physical event which is not related to any coordinate or frame of reference . Are not we sure, physically and mentally, at this moment that the x'-frame will register this event to happen some where after (0',0')? Notice that (x',t') corresponding to (10^100 m, 1sec) will be registered by the already synchronized clock located at the event P. Will not this be a positive time, that is after the (0'0') event?
I think you are forgetting that the speed of light is finite.

So, at t = 1 s a flash is emitted at x = 10^100 m. Then at t = 3.33564095198 10^91 s the flash is received by the observer at x = 0 m. The observer divides the distance by c to determine how long it took the flash of light to arrive and thereby determines that the flash occured at t = 1 s.

Because of the relative motion, that same flash of light was received at t' = 3.33564095165 10^91 s by the primed observer at x' = 0. The observer also divides the distance by c to determine how long it took the flash of light to arrive and thereby determines that the flash occured at t' = -3.33564095198 10^81 s.
 
  • #23
aawahab76
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It did happen before the observers meet in the x' frame. It only happened afterwards in the x frame. Whether it happened before or after depends on the frame.

Please Google "relativity of simultaneity" and read up a bit and come back with questions. This is the most challenging concept in special relativity.
I am not sure if that the way to register the event coordinates. You see that the x and x'-frames already defined their global time coordinates and so any event will be registerd by the clock located at the event's location, am I wrong?

I will come back soon
 
  • #24
aawahab76
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BY the way, I am still not able to understand the effect of the very long distance in the whole process in the x'-frame but not on the x-frame such that the small relative speed leads to such a huge difference in the time measurements!!
 
  • #25
Dale
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I am not sure if that the way to register the event coordinates. You see that the x and x'-frames already defined their global time coordinates and so any event will be registerd by the clock located at the event's location, am I wrong?
You are correct, but you don't understand the relativity of simultaneity. The x clock at the event's location reads 1 s and the x' clock at the event's location reads -3.3 10^81 s. The clocks read differently because of the relativity of simultaneity.
 
  • #26
JesseM
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I am not sure if that the way to register the event coordinates. You see that the x and x'-frames already defined their global time coordinates and so any event will be registerd by the clock located at the event's location, am I wrong?
Yes, but why do you say this in response to DaleSpam's comment? Nothing he said contradicted the idea that each frame defines their time coordinates using local readings on clocks! But remember, each frame synchronizes their clocks using the Einstein synchronization convention (do you know what that is?), which is based on the assumption that light travels at a constant speed in that frame--so, if observer B wants to synchronize a pair of his clocks attached to different points on his ruler, he can set off a flash at the exact midpoint between these clocks, and then since in his frame the light has the same distance to travel from the flash to each clock, if he assumes the light travels at the same speed in both directions in his frame he should conclude the clocks are "synchronized" if they both show the same time at the moment the light from the flash reaches them. But supposed observer A is watching observer B synchronize his clocks this way. In observer A's frame, the two clocks are moving, so the rear clock is moving towards the position (on A's ruler) where the flash was set off, while the lead clock is moving away from that position. So if A assumes both light beams travel at the same speed in his own frame, he should conclude that B's rear clock will catch up with the light beam before the light hits B's lead clock, thus if B sets both clocks so they read the same time when the light hits them, in A's frame the two clocks are out-of-sync.
BY the way, I am still not able to understand the effect of the very long distance in the whole process in the x'-frame but not on the x-frame such that the small relative speed leads to such a huge difference in the time measurements!!
If you think about the above argument involving the flash used to synchronize B's two clocks, you can see that the greater the distance between the two clocks, the greater the difference in time in A's frame between the light hitting the two clocks, so the farther out-of-sync those two clocks are. For example, suppose in A's frame the flash is set off at position x=0 light-seconds at t=0 seconds, and at t=0 B's clock #1 is at position x=-60 while B's clock #2 is at position +60, and both are moving in the +x direction at 0.2c. Then at time t=50, the clock that was at x=-60 will now be at x=-60 + 0.2*50 = -50, and the light beam that was emitted in the -x direction from x=0 at t=0 will now be at x=-50, so t=50 is the time in A's frame the light will hit the rear clock, at which point B sets the rear clock can be set to some time T. And at t=75, the clock that was at x=+60 will now be at x=+60 + 0.2*75 = 75, and the light that was emitted in the +x direction from x=0 at t=0 will now be at x=75, so t=75 is the time in A's frame when the light will hit the lead clock, at which point B sets the lead clock to the same time T. So although the clocks are synchronized in B's frame, in A's frame there is a 25-second gap between when the rear clock shows a time of T and when the lead clock shows a time of T.

Now consider what happens if we keep the speed of B relative to A the same (0.2c) but increase the distances by a factor of 10, so that the rear clock starts at x=-600 and the lead clock starts at x=+600. In this case it's not hard to show that all the other numbers increase by a factor of 10 too, so in A's frame the light will catch up with the rear clock at t=500 and the light will catch up with the lead clock at t=750, so here there is a 250-second gap between when the rear clock shows a time of T and when the lead clock shows a time of T. Similarly if you increased the distances from ±60 to ±6*10^100, then the difference in A's frame between each clock showing a time of T would now by 2.5*10^100.

Also, I again recommend checking out the diagrams I drew of two ruler-clock systems moving alongside each other on this thread, it shows visually how in A's frame each clock on B's ruler is out-of-sync with the nearest one to the right by the same amount, so clocks with a larger and larger separation on B's ruler are more and more out-of-sync in A's frame (and vice versa).
 
  • #27
Dale
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BY the way, I am still not able to understand the effect of the very long distance in the whole process in the x'-frame but not on the x-frame such that the small relative speed leads to such a huge difference in the time measurements!!
The difference in time is not big, it is small. In natural units it is 10 orders of magnitude smaller than the distance, as you would expect for a difference in speed 10 orders of magnitude smaller than c.
 
  • #28
aawahab76
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I really thank every body contributing for our understanding but before proceeding further I would like to stress that I understand very well the meaning of simultaneity in classical and relativistic physics and all related subjects. However, I cannot, and I believe many others, accept this easily that our physical intuition is so remote, or as we think, from the mathematical structure of the theory. Here, I am trying very hard to take the matter to its most even if my questions are or seems to be stupid. I believe that physics and mathematics are two different structures and the aim of physics is to build the relation between them guided by physics and not the other way round.
 
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  • #29
JesseM
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I really thank every body contributing for our understanding but before proceeding further I would like to stress that I understand very well the meaning of simultaneity in classical and relativistic physics and all related subjects.
Do you also understand the meaning of Lorentz-invariance? As long as the equations of all the fundamental laws of physics (quantum field theory, for example), are Lorentz-invariant, that implies that it should be impossible in principle for any experiment to pick out a preferred inertial frame, the equations of the laws of physics will look the same when expressed in the coordinates of any inertial frame. This implies that no possible experiment could pick out a preferred definition of simultaneity, although as I said earlier you are free to adopt some sort of metaphysical belief that there is a "real truth" about which of a given pair of spacelike-separated events happened earlier (or if they 'really' happened simultaneously), as long as you acknowledge that this truth couldn't be discovered by any possible experiment your view won't conflict with relativity (but metaphysically I prefer eternalism to presentism, so I don't see the need for there to be any objective truth about which events are simultaneous and which aren't).
 
  • #30
aawahab76
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Do you also understand the meaning of Lorentz-invariance? As long as the equations of all the fundamental laws of physics (quantum field theory, for example), are Lorentz-invariant, that implies that it should be impossible in principle for any experiment to pick out a preferred inertial frame, the equations of the laws of physics will look the same when expressed in the coordinates of any inertial frame. This implies that no possible experiment could pick out a preferred definition of simultaneity, although as I said earlier you are free to adopt some sort of metaphysical belief that there is a "real truth" about which of a given pair of spacelike-separated events happened earlier (or if they 'really' happened simultaneously), as long as you acknowledge that this truth couldn't be discovered by any possible experiment your view won't conflict with relativity (but metaphysically I prefer eternalism to presentism, so I don't see the need for there to be any objective truth about which events are simultaneous and which aren't).
It is obvious by now that the subject is getting broader and broader so let me collect my mind and ask the following questions. Is it physically, that is according to relativity, possible that event A happened before event B in the x-frame while event B happened before A in the x'-frame? Under which conditions is that possible?
 
  • #31
JesseM
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It is obvious by now that the subject is getting broader and broader so let me collect my mind and ask the following questions. Is it physically, that is according to relativity, possible that event A happened before event B in the x-frame while event B happened before A in the x'-frame? Under which conditions is that possible?
Yes, it is definitely possible for two frames to disagree on the order of two events, the condition where this can occur is when the spacetime interval between the events is "space-like", which among other things means that neither event is in the past or future light cone of the other event (so there can be no causal relation between the two events). See here for more discussion of the meaning of time-like, light-like, and space-like intervals.
 
  • #32
aawahab76
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Yes, it is definitely possible for two frames to disagree on the order of two events, the condition where this can occur is when the spacetime interval between the events is "space-like", which among other things means that neither event is in the past or future light cone of the other event (so there can be no causal relation between the two events). See here for more discussion of the meaning of time-like, light-like, and space-like intervals.
I completely agree with this, so any two events A and B that are causally unrelated (that is cannot be joined by a light ray, we may return to the causality concept in the future) can have A before B or B before A or A simultaneous with B depending on the two events and on the states of the two frames, right? In addition if A and have time-like intervals between them, A is before B or B before A for both frames, right? Now apply this to our initial problem with the event (10^100 m, 1 sec) in the x=frame, what are the two events here? Are they space-like separated? Are the state of the two frames capable to make A before B in the x-frame and the other way round for the x'-frame?
 
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  • #33
JesseM
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I completely agree with this, so any two events A and B that are causally unrelated (that is cannot be joined by a light ray, we may return to the causality concept in the future) can have A before B or B before A or A simultaneous with B depending on the two events and on the states of the two frames, right?
That's right, in fact for any pair of events that are space-like separated it is always possible to find three frames where each of these is true.
In addition if A and have time-like intervals between them, A is before B or B before A for both frames, right?
Yes, for either time-like or light-like intervals, all frames will agree on the order.
Now apply this to our initial problem with the event (10^100 m, 1 sec) in the x=frame, what are the two events here? Are they space-like separated? Are the state of the two frames capable to make A before B in the x-frame and the other way round for the x'-frame?
Well, you were before expressing puzzlement that in one frame that event happened after the event of the origins meeting, and in another it happened before, so presumably the two events would be the event you mentioned above and the event of the origins meeting at (0 m, 0 sec). The interval between these events is space-like, because the definition of space-like is that [tex]\Delta x^2 > c^2 \Delta t^2[/tex], and here we have [tex]\Delta x^2[/tex] = (10^100 meters)^2 = 10^200 m^2, and [tex]c^2 \Delta t^2[/tex] = (299792458 m/s)^2 * (1 second)^2 = 8.98755179 * 10^16 m^2. And we've already shown, just by using the Lorentz transformation, that the relative velocity of the two frames is indeed large enough so that the order of the events is different in each frame.
 
  • #34
aawahab76
44
0
That's right, in fact for any pair of events that are space-like separated it is always possible to find three frames where each of these is true.

Yes, for either time-like or light-like intervals, all frames will agree on the order.

Well, you were before expressing puzzlement that in one frame that event happened after the event of the origins meeting, and in another it happened before, so presumably the two events would be the event you mentioned above and the event of the origins meeting at (0 m, 0 sec). The interval between these events is space-like, because the definition of space-like is that [tex]\Delta x^2 > c^2 \Delta t^2[/tex], and here we have [tex]\Delta x^2[/tex] = (10^100 meters)^2 = 10^200 m^2, and [tex]c^2 \Delta t^2[/tex] = (299792458 m/s)^2 * (1 second)^2 = 8.98755179 * 10^16 m^2. And we've already shown, just by using the Lorentz transformation, that the relative velocity of the two frames is indeed large enough so that the order of the events is different in each frame.

ok, accepting your calculation, still how do you decide the order of the two events in each of the two given frames?
 
  • #35
JesseM
Science Advisor
8,518
15
ok, accepting your calculation, still how do you decide the order of the two events in each of the two given frames?
Just figure out the time coordinate of both events in each frame (if you know the position and time coordinates of each event in one frame, you can use the Lorentz transformation to find their time coordinates in the other). Whichever has the larger time-coordinate in a particular frame is by definition the later one in that frame.
 

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