If correct: a catastrophe in the Lorentz transformation

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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
 

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