A problem of congruency

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Introduction
Special relativity in the most simple of definitions holds that each inertial frame of reference can be considered stationary. As a consequence, calculations concerning an object in one frame of reference must be congruent with another, for which ever inertial frame produces a calculation that is fitting with experimental results can be considered zero. As all inertial frames should be able to be considered at rest, they must produce results that are equivalent to each other.

Example of congruency
An object with a clock on-board is observed passing by two stationary points, 1 light second apart with a relative velocity to the observer of 0.5C . This observer then calculates the elapsed time, taken to cover the distance between the two points, according to the object.

(1 light second/0.5C)x(1-(0.5)^2)^0.5 = 2(0.75)^0.5 = 1.732 seconds

A second observer observes this object with a velocity of 0.8C pass by two moving points, both with a velocity of 0.5C and a distance apart of (0.75)^0.5 light seconds. This observer then calculates the elapsed time, taken to cover the distance between the two points, according to the object.

((0.75)^0.5 light seconds/(0.8C – 0.5))x(1-(0.8)^2)^0.5 = 2(0.75)^0.5 = 1.732 seconds

As can be seen each observer calculated the same result, for the elapsed time according to the object.

Relativity of Simultaneity and Different Observers

A long rod that is (0.75)^0.5 long has a clock on each end that are synchronised according to an observer, that observes the rod to have a velocity of 0.5C. This observer then calculates the difference in clocks is for the rod.

(((0.75)^0.5light seconds/(1-(0.5)^2)^0.5))x(0.5C))/C^2 = leading clock 0.5 seconds ahead

A second observer observes the rod to have a length of 0.6 light seconds and to have a velocity of 0.8C, with the trailing clock ahead by 0.5 seconds in comparison to the leading. This observer then calculates the difference in clocks for the rod.

(((0.6 light seconds/(1-(0.8)^2)^0.5)x(0.8C))/C^2) – 0.5 seconds = leading clock 0.3 seconds ahead

As the observers have reached different answers there is something wrong.

Conclusion
After reading this an individual will arrive at 4 possible conclusions.

(1) congruency is not required, the relativity of simultaneity calculations are wrong, Special relativity is valid.
(2) congruency is not required, the relativity of simultaneity calculations are correct, Special relativity is valid
(3) congruency is required, the relativity of simultaneity calculations are wrong, Special relativity is valid
(4) congruency is required, the relativity of simultaneity calculations are correct, Special relativity is invalid.

If one can explain why they have arrived at a particular conclusion, then it may help me to find one also. Thank you
 

Answers and Replies

  • #2
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In the first example you calculate a frame invariant quantity and find that it is frame invariant. In the second example you calculate a frame variant quantity and find that it is frame variant.

So what is the problem?
 
  • #3
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can you link me to something that describes this in more detail?
 
  • #4
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  • #5
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With the relativity of simultaneity calculations, two different frames are considered to be at rest and calculations are done to determine the difference between two clocks according to the rod itself.

In the first example of congruency, the elapsed time that it takes the object to pass the two points is different for each observer. But when each observer is converted to the frame of the object, they arrive at the same answer.

In the simultaneity calculations the same thing occurs, one observer considers the clocks to be synchronised the second observer considers the trailing clock to be ahead by 0.5 seconds. When they are converted to the frame of the object, they arrive at different answers.

Both quantities vary depending on the frame they are measured, but the values for each frame should convert to the same value if the frame is converted to the same frame.

Can you see my problem yet?
 
  • #6
JesseM
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Relativity of Simultaneity and Different Observers

A long rod that is (0.75)^0.5 long has a clock on each end that are synchronised according to an observer, that observes the rod to have a velocity of 0.5C.
I take it you mean (0.75)^0.5 = 0.866 is the rod's length in the observer's frame, not its rest length, given that you later say that a different observer who sees it moving at 0.8c sees its length as 0.6.
striphe said:
This observer then calculates the difference in clocks is for the rod.

(((0.75)^0.5light seconds/(1-(0.5)^2)^0.5))x(0.5C))/C^2 = leading clock 0.5 seconds ahead
OK, suppose at t=0 in the observer's frame the trailing end is at x=0 and the clock there reads T=0, while the leading end is at x=0.866 and the clock there also reads T=0. In this case, in the rod's own frame the event of the clock at the trailing end reading T=0 occurs at t'=0, while the event of the clock at the leading end reading T=0 occurs at:

t' = (1/0.866) * (0 - 0.5*0.866) = -0.5 seconds. So, 0.5 seconds later at t'=0 when the trailing clock reads T=0, the clock at the leading end reads T=0.5, so I agree with your calculation.
striphe said:
A second observer observes the rod to have a length of 0.6 light seconds and to have a velocity of 0.8C, with the trailing clock ahead by 0.5 seconds in comparison to the leading.
But this is a physically different scenario! There's no way it can both be true that the observer who sees the rod moving at 0.5c sees the clocks synchronized and that the observer who sees the rod moving at 0.8c sees the trailing clock ahead by 0.5 seconds. Two simultaneous events on either end of the rod in the frame of the observer who sees the rod moving at 0.8c must have been separated by the following amount of time in the rod's own frame:

(1/0.6)*(0 + 0.8*0.6) = 0.8 seconds

And in the rod's own frame, two events on either end which happen 0.8 seconds apart have clock readings which are separated by a time of 0.8 + 0.5 = 1.3 seconds or 0.8 - 0.5 = 0.3 seconds, depending on whether the earlier event happens on the end where the clock is 0.5 seconds ahead or whether it happens on the end where the clock is 0.5 seconds behind (which depends on whether this second observer sees the rod moving in the same direction as the first observer or the opposite direction). So, the observer who sees the rod moving at 0.8c must see the two clocks out-of-sync by either 0.3 seconds or 1.3 seconds, to be compatible with the earlier statement that the observer who sees the rod moving at 0.5 sees the clocks synchronized.

You can't just make up random numbers for how much the clocks are out-of-sync in each observer's frame and expect them to both be compatible with a single physical situation!
 
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  • #7
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When they are converted to the frame of the object, they arrive at different answers.
Then you did something wrong. The forward and inverse transforms gives you the identity transform.

If you can elaborate your calculations I may be able to help, but I can't decipher them as they are.
 
  • #8
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I am with Dalespam on this one JesseM, I really do think that the calculations must arrive at the same answer in special relativity. Also I have made note on which clock is ahead of the other in each frame, so there is no confusion over if the answer should be 1.3 seconds or 0.3 seconds when converting from the second observer to the rod.

I don't have a whole lot of time atm, so ill come back later and explain my calculations better.
 
  • #9
JesseM
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I am with Dalespam on this one JesseM, I really do think that the calculations must arrive at the same answer in special relativity.
You think the calculations must arrive at the same answer when you just make up off the top of your head how out-of-sync the clocks are in the two frames without bothering to check if they are physically consistent? How about if I said the following:

"suppose that in the frame of an observer who sees the rod moving at 0.6c, the clocks on either end of the rod are out-of-sync by 1 second. And suppose that in the frame of a different observer who sees the rod moving at 0.6000001c, the clocks on either end of the rod are out-of-sync by 57832592478924 trillion years. Yet when we transform both these claims into the rod rest frame, we get incongruent answers to how much the clocks are out-of-sync by in the rod rest frame!"

Do you think the above represents a "problem" for relativity, or do you agree that the problem above is that I just made up completely random numbers for how much the clocks are out-of-sync in each frame, so there's no reason to expect that both descriptions could be consistent with the selfsame physical scenario?

striphe said:
Also I have made note on which clock is ahead of the other in each frame, so there is no confusion over if the answer should be 1.3 seconds or 0.3 seconds when converting from the second observer to the rod.
No you didn't, you just referred to the "trailing" clock being ahead of the "leading" clock, but that doesn't rule out the possibility that the different observers see the rod moving in different directions and thus disagree in their labels of which clock is "leading" and which is "trailing". But from your response I think you probably meant them to use the same labels (so they both see the rod moving in the same direction relative to themselves), in which case the observer who sees the rod moving at 0.8c should see the trailing clock ahead of the leading clock by 0.3 seconds, not 0.5 seconds as you suggested. If you use this number you will see that everything is "congruent". For example, in the frame of this observer we can say that the event of the trailing clock reading T=0 happens at x=0, t=0 and the event of the leading clock reading T=-0.3 happens at x=0.6, t=0 (so at the time t=0, the leading clock is 0.3 seconds behind the trailing clock). If we transform the first event into the rod frame it happens at x'=0, t'=0 and if we transform the second event into the rod frame it happens at a time of:

t' = gamma*(t - vx) = (1/0.6)*(0 - 0.8*0.6) = -0.8

So in the rod frame, at t'=-0.8 the leading clock reads T=-0.3, and the trailing clock is synced to coordinate time in this frame so at t'=-0.8 it must read T=-0.8, which means the leading clock is ahead by 0.5 seconds, same as what you concluded from the fact that the clocks were synchronized in the frame of the observer who saw the rod moving at 0.5c.

We can also show that the clocks should be out-of-sync by 0.3 seconds in the frame of the second observer just by transforming from the second observer's frame to the first observer's frame. First we must figure out how fast the first observer is moving in the second observer's frame--since the first observer is moving at -0.5c in the frame of the rod, and the rod is moving at 0.8c in the frame of the second observer, by the velocity addition formula the first observer's velocity in the second observer's frame must be (0.8c - 0.5c)/(1 - 0.8*0.5) = 0.3c/0.6 = 0.5c.

So, let's again say that in the frame of the second observer, the event of the trailing clock reading T=0 happens at x=0, t=0 and the event of the leading clock reading T=-0.3 happens at x=0.6, t=0. Since the leading clock is slowed down by a factor of 0.6 in the second observer's frame, it will take the leading clock a coordinate time of 0.3/0.6 = 0.5 seconds to tick forward by 0.3 seconds of its own time, so the leading clock will read T=0 at t=0.5, and during that time it has moved forward in space by a distance of 0.5*0.8 = 0.4 light-seconds, so if it read T=-0.3 when it was at x=0.6, it will read T=0 at x=0.6+0.4=1 light-second. So, in the second observer's frame the leading clock reads T=0 at x=1, t=0.5.

Now if we transform these events of the clocks at either end reading T=0 into the first observer's frame (with the first observer moving at 0.5c in the frame of the second observer), the event of the trailing clock reading T=0 happens at x'=0, t'=0, while the event of the leading clock reading T=0 happens at:

t' = gamma*(t - vx) = (1/(0.75)^0.5)*(0.5 - 0.5*1) = 0

So, in the first observer's frame the event of the leading clock reading T=0 also happens at a coordinate time t'=0 just like the event of the trailing clock reading T=0, consistent with the original notion that the two clocks were synchronized in the first observer's frame.
 
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  • #10
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I am with Dalespam on this one JesseM
JesseM and I are not disagreeing here. Let me explain what I am saying and what JesseM are saying in terms of the Lorentz group.

You have three reference frames, P, Q, R, so lets denote the Lorentz transform from P to Q as QP, and similarly with the other frames. Then since the set of all Lorentz transforms forms a group we have:
I = PQ.QR.RP
QP.I = QP.PQ.QR.RP = (QP.PQ).QR.RP = I.QR.RP
QP = QR.RP

So I am saying that it doesn't matter if you transform from the first frame to the rod's frame directly or from the first frame to the second frame and then from the second frame to the rod's frame, they are identical. Since you did not get identity then you must have made a mistake somewhere.

JesseM is saying that you made a mistake in your transformation from the first frame to the second frame, which is consistent with what I said.
 
  • #11
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this is the formulation for the calculations, tell me where it is wrong.

(((observed length/factor of contraction due to velocity) X Velocity)/C^2) - trailing clocks lead on leading clock according to observations

My imputs:

I began with a rest length for the rod of one light second, which is observed by the first observer as root0.75 light seconds due to the relative velocity of 0.5C and the second observer as 0.6 light seconds due to the relative velocity of 0.8C.

Observer one sees the sees the two clocks as being synchronised. If an object with a rest length of one light second is synchronised then observer two with a relative velocity to observer one is 0.5C sees the objects clocks as being out from one another by 0.5 seconds.
 
  • #12
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I don't know that formula. I would recommend using the Lorentz transform directly instead of using weird formulas of potentially limited applicability.
 
  • #13
JesseM
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(((observed length/factor of contraction due to velocity) X Velocity)/C^2) - trailing clocks lead on leading clock according to observations
This formula is only correct in the case where the two clocks are synchronized in their own rest frame, which is not true in your scenario; as DaleSpam said it's safer to use the Lorentz transform if you aren't 100% clear on the assumptions used in a given formula. Also, it's simpler to write that formula as (rest length)*v/c^2 = trailing clock lead on leading clock in frame where clocks moving at velocity v.
 
  • #14
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As far as I am aware the formula (rest length)*v/c^2 works vise versa. If two events are simultaneous on board then it can be used to determine how an external observer witnesses them and if an external observer considers two events on board to be simultaneous it can be used to determine what an on board observer considers them to be.

The legitimacy of my calculations does not rest on the calculations themselves but rather an input "trailing clock ahead by 0.5 seconds in comparison to the leading" (control F to find the spot in the op) if this value were 0.3 seconds there would be no problem. I don't understand how one is supposed to arrive at 0.3 seconds instead of 0.5 seconds for this value.
 
  • #15
JesseM
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As far as I am aware the formula (rest length)*v/c^2 works vise versa. If two events are simultaneous on board then it can be used to determine how an external observer witnesses them and if an external observer considers two events on board to be simultaneous it can be used to determine what an on board observer considers them to be.
Yes, I suppose it would also work for the case of clocks synchronized in the frame of the observer who sees the clocks in motion, if we wanted to know how out-of-sync they are in their own rest frame. If a rod of rest length L is moving at v in my frame, and at t=0 in my frame the trailing clock is at x=0 and reads T=0 while the leading clock is at x=L/gamma and reads T=0, then in the rod's own frame the event of the trailing clock reading T=0 occurs at coordinate time t'=0, while the event of the leading clock reading T=0 occurs at:

t' = gamma*(t - vx/c^2) = gamma*(0 - v(L/gamma)/c^2) = -vL/c^2.

And since the leading clock is running at the same rate as coordinate time in this frame, at t'=0 a time-interval of vL/c^2 has passed, so the leading clock reads T=vL/c^2 at t'=0.

Still, neither of these cases tell you how out-of-sync the clocks should be in the frame of the observer who sees the rod moving at 0.8c in your example, since in that case the clocks aren't synchronized in either that observer's frame or the rod's own rest frame. So, you can't use the formula above in this situation, do you understand that?
striphe said:
The legitimacy of my calculations does not rest on the calculations themselves but rather an input "trailing clock ahead by 0.5 seconds in comparison to the leading" (control F to find the spot in the op) if this value were 0.3 seconds there would be no problem. I don't understand how one is supposed to arrive at 0.3 seconds instead of 0.5 seconds for this value.
Well, in post #6 I gave one argument--if we consider two events on either end of the rod which are simultaneous in the frame of the observer who sees the rod moving at 0.8c, with one event occuring at x=0, t=0 and the other at x=0.6, t=0 in his frame, then using the Lorentz transformation we can show that in the rod's rest frame the first event happened at t'=0, while the second event happened at:

t' = gamma*(t - vx/c^2) = (1/0.6)*(0 - 0.8*0.6) = -0.8

So, these two events which are simultaneous in the frame of the observer must have happened 0.8 seconds apart in the frame of the rod, with the event on the trailing end happening 0.8 seconds later than the one on the leading end. But we know that in the rod's frame the clock on the trailing end is behind by 0.5 seconds, so the first event happens on the leading end when it reads T while the trailing end reads T-0.5, and the second event happens on the trailing end 0.8 seconds later in this frame, then at the time of the second event the trailing end will read T-0.5+0.8=T+0.3.

In post #9 I gave a different derivation which doesn't consider how things look in the rod's own frame at all, but just considers how out-of-sync the clocks must be in the second observer's frame (the one who sees the rod moving at 0.8c) given that they are synchronized in the first observer's frame (the one who sees the rod moving at 0.5c), and given that the first observer is moving at 0.5c in the frame of the second observer (which follows from the velocity addition formula). There I showed that if we started out by assuming the clocks were out-of-sync by 0.3 seconds in the second observer's frame, you could use this to derive the fact that they are synchronized in the first observer's frame, again using the Lorentz transformation. And of course you also start from the assumption that the clocks are synchronized in the first observer's frame and use that to derive the fact that they are out-of-sync by 0.3 seconds in the second observer's frame. Suppose at t'=0 in the first observer's frame, the trailing clock is at position x'=0 reading a time of T=0, and the leading clock is at position x'=(0.75^0.5) also reading at time of T=0. Using the reverse Lorentz transformation equation t=gamma*(t' + vx'/c^2), this means that in the second observer's frame the trailing clock reads T=0 at t=0, while the leading clock reads T=0 at:

t=(1/0.75^0.5)*(0 + 0.5*(0.75^0.5)) = 0.5

So if the leading clock read T=0 at t=0.5 in this frame, what did it read 0.5 seconds earlier at t=0 in this frame? Well, since the clock is moving at 0.8c it's running slow by a factor of 0.6, so in 0.5 seconds it only ticks forward by 0.5*0.6 = 0.3 seconds, so 0.5 seconds before it read T=0 it must have read T=-0.3. So, at t=0 in the frame of the second observer, the trailing clock read T=0 and the leading clock read T=-0.3, showing that in the second observer's frame the trailing clock is ahead by 0.3 seconds.
 
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  • #16
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I don't understand how one is supposed to arrive at 0.3 seconds instead of 0.5 seconds for this value.
Just use the Lorentz transform, not some shortcut formula.
 
  • #17
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Just use the Lorentz transform, not some shortcut formula.
Just to do the example quickly, using four-vector notation (ct,x,y,z), setting the origin at the event, a, where the trailing clock reads 0, letting v be the four-velocity of the rod, and letting b be the event where the leading clock reads 0 then.
In Frame 1:
a=(0,0,0,0)
b=(0,.866,0,0) -> synchronized
v=(1.15,.577,0,0) -> speed is .5c
boosting to rod frame (Lorentz transform with u=.5c)
a'=(0,0,0,0)
b'=(-.5,1,0,0) -> lead clock ahead by .5
v'=(1,0,0,0) -> speed is 0

In Frame 2: (obtained by Lorentz transform from Frame 1 with u=-.5c)
a''=(0,0,0,0)
b''=(.5,1,0,0) -> lead clock behind by .5
v''=(1.66,1.33,0,0) -> speed is .8c
boosting to rod frame (Lorentz transform with u=.8c)
a'=(0,0,0,0)
b'=(-.5,1,0,0) -> lead clock ahead by .5
v'=(1,0,0,0) -> speed is 0

All of that is simply obtained using the standard Lorentz transform with the indicated speeds. Just stick with the Lorentz transform and be careful in your work.
 
  • #18
JesseM
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In Frame 2: (obtained by Lorentz transform from Frame 1 with u=-.5c)
a''=(0,0,0,0)
b''=(.5,1,0,0) -> lead clock behind by .5
The bolded part is worded in a slightly confusing way (and perhaps this sort of word-confusion was the original source of striphe's error, since the OP said 'the trailing clock ahead by 0.5 seconds in comparison to the leading'). At any given instant in frame 2 the lead clock is only behind by 0.3 seconds, so at t''=0 when the trailing clock reads 0, the lead clock will read -0.3. But because of time dilation, it will take 0.5 seconds in this frame for the lead clock to go from reading -0.3 to reading 0, and you said earlier that b'' represented the event of the lead clock reading 0, so that's why you have b'' happening at t''=0.5 even though the lead clock is actually behind by 0.3, not 0.5. Of course, here I have used time dilation in my argument, when you wanted to use only the Lorentz transform! But you could have just included an additional step where you picked the event c''=(0,.6,0,0) on the worldline of the lead clock which occurred at t''=0 (simultaneously in frame 2 with the trailing clock reading 0), and showed that in the rod frame it occurred at c'=(-.8,1,0,0), and since you already knew from b' that the lead clock read 0 at a time of -.5 in the rod frame, that suggests that at a time of -.8 in the rod frame the lead clock must have read -.3 (just based on the idea that both clocks are ticking at normal speed in their own rest frame). And we picked this event to be simultaneous in frame 2 with the event of the trailing clock reading 0, so this shows that in frame 2 the trailing clock is .3 seconds ahead of the lead clock.
 
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  • #19
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You have a point. The wording is confusing. Even though the lead clock reads -.3 at t''=0 I still think of it as being behind by .5 s, not .3 s. Regardless of what it reads at t''=0 it will not read the same as the trailing clock for another .5 s in the second frame.

I don't know if there is standard terminology for the distinction. In any case, the math is clear and correct, regardless of the ambiguities in the English.
 

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