# I think the special theory of relativity self-contradicts

• I
• Leepappas
Leepappas
Consider two rulers that have the same rest length. Denote it by L0.
Now let the two rulers be in relative motion. First consider things from the rest frame of the unprimed ruler illustrated below.

A'_____B'
............A_______________B. State 1

............A'_____B'
............A'_______________B. State 2

....................…..…..A'_____B'
............A_______________B. State 3

The primed ruler length contracts in the rest frame of the unprimed ruler. Let ∆t1 represent the amount of time from State one to state two as measured by a clock at rest in the unprimed frame. Let ∆t2 represent the amount of time between states two and three as measured by the same clock. The relative speed v of the rulers satisfies
v = L0/(∆t1+∆t2), because the point B" travels a distance L0 an amount of time ∆t1+∆t2. But using the same definition of speed the point B' travel a distance L in time ∆t1, so
v = L/∆t1, where L=L0√1-v^2/c^2.

And in going from state two to state three B' travels a distance L0 - L in amount of time ∆t2, so
v = (L0 - L)/∆t2.
Now let ∆t = ∆t1 + ∆t2, so that
∆t' = ∆t/√1 - v^2/c^2.
Now delta t prime is the time between states one and three as measured by a clock at rest in the primed frame. In that frame the unprimed ruler's length contracts as it moves from right to left.

A'_______________B'
.............................A_____B. State 1

A'_______________B'
..................A_____B state 3

Therefore v = L/∆t', since A travels a distance L in the primed frame in amount of time ∆t' as measured by a clock at rest in the primed frame.
Now since v ∆t2 = L0 - L it follows that
L/∆t' (∆t2) = L0 - L, so
L0-L=L0√1-v^2/c^2 (∆t2/∆t') which equals
L0√1-v^2/c^2 (∆t2√1-v^2/c^2 /∆t), which equals
L0(1-v^2/c^2)∆t2/∆t, which equals
L0(1-v^2/c^2)∆t2/(∆t1+∆t2).
But v = L0/(∆t1+∆t2), so
L0 - L = v (1-v^2/c^2) ∆t2, but v∆t2=L0-L, so
L0-L = (L0-L)(1-v2/c^2).
Since v isn't equal to 0, L0-L isn't zero so we can divide both sides by it to obtain:
1 = 1 - v^2/c^2. The only way the preceding equation can be true is if v= 0, but the rulers are in relative motion by hypothesis so v is not equal to 0. Therefore

v =0 and not( v = 0).
Thus the theory of special relativity self-contradicts.

What say you?

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davenn, weirdoguy, Motore and 1 other person
Leepappas said:
Thus the theory of special relativity self-contradicts.

What say you?
If you think you've found a self-contradiction in a theory with as much experimental confirmation as SR, your first idea should be that you've made a mistake somewhere, and if you need to ask for help, you should be asking for help in finding the mistake.

(If you want an excellent example of this, btw, consider the experimenters who got results from the OPERA experiments a few years ago that, at face value, seemed to be saying that neutrinos could travel faster than light. Did they claim they had found an error in SR? No, they asked for help in finding the mistake. And it took months to find--but yes, it turned out to be a mistake, a subtle error in the experimental setup, which, when corrected, made the results consistent with SR.)

davenn and russ_watters
@Leepappas rather than try to find a specific error in your analysis, I am going to give you a correct analysis for comparison with yours.

Your diagrams are very imprecise, which is at least partly due to the limitations of trying to use text to draw diagrams in a web forum. So before we can even properly analyze your scenario, we need to specify your three "States" in terms of something invariant.

From your diagrams, I think the appropriate specifications are:

State 1: Point B' is coincident with point A ("coincident" means "at the same point in spacetime", i.e., the worldlines of the two points intersect).

State 2: Point A' is coincident with point A.

State 3: Point B' is coincident with point B.

Now, a key point: the definitions of each of the three states pick out particular events in spacetime. But the events only involve one point on each ruler. And statements about "length" involve two points on each ruler: the two ends. So specifying the States above does not, by itself, specify the length of either ruler in any of the states, nor does it specify any events involving points on either ruler that are not named in the definitions of the states. A common error in analyzing relativity scenarios is to forget all of the above and make assumptions that seem innocuous, but in fact turn out to be false.

Now to the analysis.

The rest length of each ruler is ##L_0##. The relative speed of the rulers is ##v##. In each ruler's rest frame, the other ruler's length is therefore ##L_0 \sqrt{1 - v^2}## (I am using units in which ##c = 1##).

Now we can assign coordinates in the unprimed frame to each of the three defining events for the states above.

Event 1: Point B' is coincident with point A at coordinates ##(t, x) = (0, 0)##. We can simply assume this without loss of generality, because we are always free to choose the spacetime origin of a frame to suit our convenience.

Event 2: Point A' is coincident with point A at coordinates ##(t, x) = (L_0 \sqrt{1 - v^2} / v, 0)##. The ##t## coordinate is the time it takes for the (length contracted) primed ruler to pass point A in the unprimed frame, which is simply the length contracted length divided by the speed.

Event 3: Point B' is coincident with point B at coordinates ##(t, x) = (L_0 / v, L_0)##. Here the ##t## coordinate is the time it takes for point B' to travel the length of the unprimed ruler, which is just its length divided by the speed; and the ##x## coordinate is just the length of the unprimed ruler, since that is the distance of the point B end from the point A end, and the point A end is at the origin.

Now, to see what things look like in the primed frame, we just use the Lorentz transformation: ##t' = \gamma \left( t - v x \right)##, ##x' = \gamma \left( x - v t \right)##. That gives:

Event 1: Point B' is coincident with point A at coordinates ##(t', x') = (0, 0)##. The spacetime origin's coordinates are the same in both frames.

Event 2: Point A' is coincident with point A at coordinates ##(t', x') = (L_0 / v, - L_0)##. This makes sense because point A takes time ##L_0 / v## to move the length of the primed ruler (which is ##L_0## in this frame), and point A' is a distance ##L_0## to the left (minus ##x'## direction) of point B'.

Event 3: Point B' is coincident with point B at coordinates ##(t', x') = (L_0 \sqrt{1 - v^2} / v, 0)##. Notice that this event is at the spatial origin of the primed frame, ##x' = 0##. That is because point B' is the spatial origin of that frame. The time here makes sense because it is the time it takes the unprimed ruler, which is length contracted in this frame, to move past point B'.

So everything here makes sense--but you will notice a key fact, that in the unprimed frame, Event 2 happens before Event 3, whereas in the primed frame, Event 2 happens after Event 3. This is an illustration of the fact that the time ordering of spacelike separated events is frame dependent. We can verify that these events are spacelike separated by computing the squared coordinate deltas and seeing that ##\Delta x > \Delta t## and ##\Delta x' > \Delta t'## (in fact the coordinate deltas are the same in both frames), which means the events are spacelike separated.

In your analysis, it looks like you are trying to capture the first and last events in the sequence in each frame, so you can compare the total time deltas. But to do that correctly, you need to compare, not the times between Event 1 and Event 3 in both frames, but the time between Event 1 and Event 3 in the unprimed frame, with the time between Event 1 and Event 2 in the primed frame.

Leepappas, russ_watters, topsquark and 1 other person
Leepappas said:
What say you?
As in almost every case where someone "finds a contradiction" in special relativity, you are only using length contraction and time dilation and have forgotten to take into account the relativity of simultaneity. In particular, that means that your two diagrams labelled "State 1" and the two labelled "State 3" actually represent four different states because the two frames have different notions of simultaneity.

So it's your application of the simpler special case length contraction and time dilation equations to this case in an invalid way that is the self-contradiction. That is, it's in your description of your experiment, not in relativity. Repeat your calculations using the full Lorentz transforms and your contradiction will vanish.

russ_watters, Vanadium 50, topsquark and 3 others
Leepappas said:
Thus the theory of special relativity self-contradicts.

What say you?
Thus, you have an erroneous understanding of SR. If mathematics was claimed to be self-contradictory every time a student got a calculation wrong, where would we be?

FactChecker, russ_watters, berkeman and 1 other person
PeterDonis said:
@Leepappas rather than try to find a specific error in your analysis, I am going to give you a correct analysis for comparison with yours.

Your diagrams are very imprecise, which is at least partly due to the limitations of trying to use text to draw diagrams in a web forum. So before we can even properly analyze your scenario, we need to specify your three "States" in terms of something invariant.

From your diagrams, I think the appropriate specifications are:

State 1: Point B' is coincident with point A ("coincident" means "at the same point in spacetime", i.e., the worldlines of the two points intersect).

State 2: Point A' is coincident with point A.

State 3: Point B' is coincident with point B.

Now, a key point: the definitions of each of the three states pick out particular events in spacetime. But the events only involve one point on each ruler. And statements about "length" involve two points on each ruler: the two ends. So specifying the States above does not, by itself, specify the length of either ruler in any of the states, nor does it specify any events involving points on either ruler that are not named in the definitions of the states. A common error in analyzing relativity scenarios is to forget all of the above and make assumptions that seem innocuous, but in fact turn out to be false.

Now to the analysis.

The rest length of each ruler is ##L_0##. The relative speed of the rulers is ##v##. In each ruler's rest frame, the other ruler's length is therefore ##L_0 \sqrt{1 - v^2}## (I am using units in which ##c = 1##).

Now we can assign coordinates in the unprimed frame to each of the three defining events for the states above.

Event 1: Point B' is coincident with point A at coordinates ##(t, x) = (0, 0)##. We can simply assume this without loss of generality, because we are always free to choose the spacetime origin of a frame to suit our convenience.

Event 2: Point A' is coincident with point A at coordinates ##(t, x) = (L_0 \sqrt{1 - v^2} / v, 0)##. The ##t## coordinate is the time it takes for the (length contracted) primed ruler to pass point A in the unprimed frame, which is simply the length contracted length divided by the speed.

Event 3: Point B' is coincident with point B at coordinates ##(t, x) = (L_0 / v, L_0)##. Here the ##t## coordinate is the time it takes for point B' to travel the length of the unprimed ruler, which is just its length divided by the speed; and the ##x## coordinate is just the length of the unprimed ruler, since that is the distance of the point B end from the point A end, and the point A end is at the origin.

Now, to see what things look like in the primed frame, we just use the Lorentz transformation: ##t' = \gamma \left( t - v x \right)##, ##x' = \gamma \left( x - v t \right)##. That gives:

Event 1: Point B' is coincident with point A at coordinates ##(t', x') = (0, 0)##. The spacetime origin's coordinates are the same in both frames.

Event 2: Point A' is coincident with point A at coordinates ##(t', x') = (L_0 / v, - L_0)##. This makes sense because point A takes time ##L_0 / v## to move the length of the primed ruler (which is ##L_0## in this frame), and point A' is a distance ##L_0## to the left (minus ##x'## direction) of point B'.

Event 3: Point B' is coincident with point B at coordinates ##(t', x') = (L_0 \sqrt{1 - v^2} / v, 0)##. Notice that this event is at the spatial origin of the primed frame, ##x' = 0##. That is because point B' is the spatial origin of that frame. The time here makes sense because it is the time it takes the unprimed ruler, which is length contracted in this frame, to move past point B'.

So everything here makes sense--but you will notice a key fact, that in the unprimed frame, Event 2 happens before Event 3, whereas in the primed frame, Event 2 happens after Event 3. This is an illustration of the fact that the time ordering of spacelike separated events is frame dependent. We can verify that these events are spacelike separated by computing the squared coordinate deltas and seeing that ##\Delta x > \Delta t## and ##\Delta x' > \Delta t'## (in fact the coordinate deltas are the same in both frames), which means the events are spacelike separated.

In your analysis, it looks like you are trying to capture the first and last events in the sequence in each frame, so you can compare the total time deltas. But to do that correctly, you need to compare, not the times between Event 1 and Event 3 in both frames, but the ti

PeterDonis said:
@Leepappas rather than try to find a specific error in your analysis, I am going to give you a correct analysis for comparison with yours.

Your diagrams are very imprecise, which is at least partly due to the limitations of trying to use text to draw diagrams in a web forum. So before we can even properly analyze your scenario, we need to specify your three "States" in terms of something invariant.

From your diagrams, I think the appropriate specifications are:

State 1: Point B' is coincident with point A ("coincident" means "at the same point in spacetime", i.e., the worldlines of the two points intersect).

State 2: Point A' is coincident with point A.

State 3: Point B' is coincident with point B.

Now, a key point: the definitions of each of the three states pick out particular events in spacetime. But the events only involve one point on each ruler. And statements about "length" involve two points on each ruler: the two ends. So specifying the States above does not, by itself, specify the length of either ruler in any of the states, nor does it specify any events involving points on either ruler that are not named in the definitions of the states. A common error in analyzing relativity scenarios is to forget all of the above and make assumptions that seem innocuous, but in fact turn out to be false.

Now to the analysis.

The rest length of each ruler is ##L_0##. The relative speed of the rulers is ##v##. In each ruler's rest frame, the other ruler's length is therefore ##L_0 \sqrt{1 - v^2}## (I am using units in which ##c = 1##).

Now we can assign coordinates in the unprimed frame to each of the three defining events for the states above.

Event 1: Point B' is coincident with point A at coordinates ##(t, x) = (0, 0)##. We can simply assume this without loss of generality, because we are always free to choose the spacetime origin of a frame to suit our convenience.

Event 2: Point A' is coincident with point A at coordinates ##(t, x) = (L_0 \sqrt{1 - v^2} / v, 0)##. The ##t## coordinate is the time it takes for the (length contracted) primed ruler to pass point A in the unprimed frame, which is simply the length contracted length divided by the speed.

Event 3: Point B' is coincident with point B at coordinates ##(t, x) = (L_0 / v, L_0)##. Here the ##t## coordinate is the time it takes for point B' to travel the length of the unprimed ruler, which is just its length divided by the speed; and the ##x## coordinate is just the length of the unprimed ruler, since that is the distance of the point B end from the point A end, and the point A end is at the origin.

Now, to see what things look like in the primed frame, we just use the Lorentz transformation: ##t' = \gamma \left( t - v x \right)##, ##x' = \gamma \left( x - v t \right)##. That gives:

Event 1: Point B' is coincident with point A at coordinates ##(t', x') = (0, 0)##. The spacetime origin's coordinates are the same in both frames.

Event 2: Point A' is coincident with point A at coordinates ##(t', x') = (L_0 / v, - L_0)##. This makes sense because point A takes time ##L_0 / v## to move the length of the primed ruler (which is ##L_0## in this frame), and point A' is a distance ##L_0## to the left (minus ##x'## direction) of point B'.

Event 3: Point B' is coincident with point B at coordinates ##(t', x') = (L_0 \sqrt{1 - v^2} / v, 0)##. Notice that this event is at the spatial origin of the primed frame, ##x' = 0##. That is because point B' is the spatial origin of that frame. The time here makes sense because it is the time it takes the unprimed ruler, which is length contracted in this frame, to move past point B'.

So everything here makes sense--but you will notice a key fact, that in the unprimed frame, Event 2 happens before Event 3, whereas in the primed frame, Event 2 happens after Event 3. This is an illustration of the fact that the time ordering of spacelike separated events is frame dependent. We can verify that these events are spacelike separated by computing the squared coordinate deltas and seeing that ##\Delta x > \Delta t## and ##\Delta x' > \Delta t'## (in fact the coordinate deltas are the same in both frames), which means the events are spacelike separated.

In your analysis, it looks like you are trying to capture the first and last events in the sequence in each frame, so you can compare the total time deltas. But to do that correctly, you need to compare, not the times between Event 1 and Event 3 in both frames, but the time between Event 1 and Event 3 in the unprimed frame, with the time between Event 1 and Event 2 in the primed frame.
In my analysis I am not trying to compare the first and last events in the sequence in each frame so I can compare the total time deltas that's not what I'm doing. There are two moments in time or states of the universe that I am focusing on. One moment in time is state one and the other moment in time is state three and I did it right and I got the contradiction what say you?

davenn and PeroK
PeterDonis said:
If you think you've found a self-contradiction in a theory with as much experimental confirmation as SR, your first idea should be that you've made a mistake somewhere, and if you need to ask for help, you should be asking for help in finding the mistake.

(If you want an excellent example of this, btw, consider the experimenters who got results from the OPERA experiments a few years ago that, at face value, seemed to be saying that neutrinos could travel faster than light. Did they claim they had found an error in SR? No, they asked for help in finding the mistake. And it took months to find--but yes, it turned out to be a mistake, a subtle error in the experimental setup, which, when corrected, made the results consistent with SR.)
Have you actually done any experiments measuring the speed of light or are you just quoting other people's results?

davenn, russ_watters, phinds and 2 others
Leepappas said:
There are two moments in time or states of the universe that I am focusing on.
No, there are four states. Thinking that there are only two is your underlying problem. The details of the correct analysis are in Peter's post.

Dale
Ibix said:
No, there are four states. Thinking that there are only two is your underlying problem. The details of the correct analysis are in Peter's post.
Peter did the analysis correct but his interpretation of what I was doing was incorrect. Furthermore he only used the three states State one state two and state three he didn't use four. Where is this fourth state that you're concerned with I don't see it.

davenn
Leepappas said:
Peter did the analysis correct but his interpretation of what I was doing was incorrect. Furthermore he only used the three states State one state two and state three he didn't use four. Where is this fourth state that you're concerned with I don't see it.
As I said, your two diagrams labelled "State 1" are two different states, as are your two diagrams labelled "State 3". That's four states.

I have something to do right now, but I can probably post a Minkowski diagram in an hour or so which will show you what I mean. In the meantime, why don't you go and google "relativity of simultaneity" and see if you can work out why your two states 1 and two states 3 are different.

Dale
Ibix said:
As I said, your two diagrams labelled "State 1" are two different states, as are your two diagrams labelled "State 3". That's four states.

I have something to do right now, but I can probably post a Minkowski diagram in an hour or so which will show you what I mean. In the meantime, why don't you go and google "relativity of simultaneity" and see if you can work out why your two states 1 and two states 3 are different.
I understand the Crux of your argument is that there are two states labeled State One, but there is only one initial state when B prime coincides with A at t=t'=0. How can there be two initial States when there is only one. It merely looks different from the two different frames but it's still just one state. I can wait an hour for your response.

Thank you

davenn, PeroK and Dale
Leepappas said:
What say you?
You forgot the relativity of simultaneity.

Solution: use the full Lorentz transform formula instead of the length contraction formula.

Leepappas said:
I did it right
No. You really didn’t. You made an obvious mistake neglecting the relativity of simultaneity, which is the usual mistake most students make. Use the Lorentz transform, not the length contraction and time dilation formulas.

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Demystifier, davenn, PeterDonis and 2 others
Leepappas said:
Thank you
I've been a member of this site for nearly 10 years. In that time, I would say that at least 20-30 students have posted the same or similar "contradiction" in SR by considering only time dilation and length contraction and neglecting the relativitity of simultaneity.

Even if you can't accept that SR is correct, you should accept that many thousands of students before you have made the same mistake/found the same fundamental flaw in modern physics. The fact that modern physics is still standing suggests that the students are mistaken and that the theory is valid. All particle physics uses SR kinematics, for example.

The idea that you have blown apart modern physics with a few simple calculations must strike you as odd? Were all the physicists of the last hundred years, from Einstein to Feynman to Hawking really that dumb?

davenn, russ_watters, Vanadium 50 and 1 other person
PeroK said:
If mathematics was claimed to be self-contradictory every time a student got a calculation wrong, where would we be?

Integration table:
##\int \frac{1}{x}dx = \ln |x|##
##\int u \hspace{2pt} dv = uv - \int v du##
Set ##u:=x## and ##v:=\frac{1}{x}##

##\int u \hspace{2pt} dv = uv - \int v du = 1 - \int \frac{1}{x}dx = 1 - \ln |x| \ \ \ \ \ (1)##
##\int u \hspace{2pt} dv = \int \frac{1}{v}dv = \ln |v| = \ln |\frac{1}{x}| \ \ \ \ \ (2)##

Set equations ##(1) = (2) \Rightarrow##

##1 - \ln |x| = \ln |\frac{1}{x}|##
##\frac{e^1}{|x|} = |\frac{1}{x}|##
##\Rightarrow##
##e=1##.

Leepappas said:
I understand the Crux of your argument is that there are two states labeled State One, but there is only one initial state when B prime coincides with A at t=t'=0. How can there be two initial States when there is only one. It merely looks different from the two different frames but it's still just one state. I can wait an hour for your response.
This is a Minkowski diagram showing the worldlines of the front and rear of each of your two rods. Stripped of jargon, this is a simple displacement-vs-time graph that you should have encountered at school, except that time is drawn up the page and displacement across the page by convention.

This is drawn in your unprimed frame. The two reddish lines represent the front and rear of the stationary rod and the blue lines represent the front and rear of the rod moving to the right. Due to length contraction, this blue rod appear slightly shorter in this frame. Now let's mark on the times of your State 1, 2 and 3 diagrams in this frame with fine red lines:

So far so good, I hope. The first three diagrams in your OP can be generated just by taking horizontal slices through the diagram along the three fine lines.

Now here's where you went wrong. The Lorentz transform for time says that ##t'=\gamma(t−\frac{v}{c^2}x)##, which we can rearrange to ##t=\frac1\gamma t'+\frac{v}{c^2}x##. That means that a line of fixed time in the primed frame (##t'=\mathrm{const}##) is not a horizontal line in this frame. Let's draw the three lines of fixed ##t'## through the three key events as fine blue lines:

The second set of diagrams you labelled "State 1" and "State 3" can be drawn by taking sections along those blue lines (an scaling down by a factor of ##\gamma## because you can't draw a Minkowski geometry completely faithfully on a Euclidean plane). Note that they are not the same as the reddish lines! That's what I mean about there being four states here - the two you drew by looking along the top and bottom horizontal lines, and the two you drew along the top and bottom diagonal lines.

What are the implications of that? Well, first, let's draw that last diagram in the primed frame:

Hopefully you can see that this is a mirror image of the second image, because this time it's the red rod that is moving and length contracted. Hopefully that makes sense. Now let's go back to the unprimed frame and add one green line:

The vertical distance between the middle and upper lines is what you called ##\Delta t_2##. The green line marks the time interval ##\Delta t_2/\gamma##, which is the time ticked off by a clock attached to the left hand end of the blue rod in what the unprimed frame calls that same time interval. But now look at it when it's transformed into the primed frame:

It is not the time interval between the right hand end of the red rod meeting the right and left ends of the blue rod. And your "proof" relies on the assumption that it is, if I read your mess of badly typeset maths correctly (we have ##\LaTeX## enabled on the forum for a reason).

As has been said several times already, you did not account for the relativity of simultaneity.

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topsquark, Dale and PeroK
Dale said:
You forgot the relativity of simultaneity.

Solution: use the full Lorentz transform formula instead of the length contraction formula.

No. You really didn’t. You made an obvious mistake neglecting the relativity of simultaneity, which is the usual mistake most students make. Use the Lorentz transform, not the length contraction and time dilation formulas.
I didn't forget the relativity of simultaneity. State three in the unprimed frame has Lorentz coordinates
## (L, \Delta t) ##

State three in the prime frame has Lorentz coordinates
##(0, \Delta t' )##
The relationship between delta t and delta t prime is

## \gamma \Delta t = \Delta t' ##

As they are the same moment in time gamma equals one. That can only be true if the speed equals zero but the rulers are in relative motion by hypothesis hence the contradiction. You can use delta t and delta t prime because of the following
## \Delta t = t - 0 = t ##
And
## \Delta t' = t' - 0 = t' ##

The point B prime and the point B will pass each other only once in the history of the universe there is a unique moment in Time that happens. Therefore gamma equals one, v equals 0, and my proof stands.

davenn and PeroK
Leepappas said:
I didn't forget the relativity of simultaneity. …
The relationship between delta t and delta t prime is

## \gamma \Delta t = \Delta t' ##
And right there you are showing that you indeed forgot the relativity of simultaneity. The equation that you posted is the time dilation equation, not the Lorentz transform which accounts for the relativity of simultaneity. The correct equations to use are:
$$t’=\gamma \left( t-\frac{xv}{c^2}\right)$$$$x’=\gamma \left(x-vt\right)$$

Leepappas said:
my proof stands
No, it doesn’t. In fact, it shows that you are unequipped to make proofs in relativity since you don’t even understand the relevant concepts and don’t know the correct formulas.

If you think your proof does stand, then I challenge you to find a professional scientific reference that uses your proof. That is the standard used here in this forum

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Leepappas said:
Have you actually done any experiments measuring the speed of light or are you just quoting other people's results?
Why yes. Yes I have. And yes, I have measured time dilation as well. (Done more of that, in fact)

Have you? If you did and got a result that differs from the predictions of relativity, publish that. That is better evidence that relativity is wrong than any calculation. If you haven't, why are you demanding it of others?

ersmith, PeterDonis, russ_watters and 2 others
Why yes. Yes I have. And yes, I have measured time dilation as well. (Done more of that, in fact)

Have you? If you did and got a result that differs from the predictions of relativity, publish that. That is better evidence that relativity is wrong than any calculation. If you haven't, why are you demanding it of others?
No I have not but if my proof is right then I don't need to.

davenn, russ_watters and Borg
Ibix said:
This is a Minkowski diagram showing the worldlines of the front and rear of each of your two rods. Stripped of jargon, this is a simple displacement-vs-time graph that you should have encountered at school, except that time is drawn up the page and displacement across the page by convention.
View attachment 332483
This is drawn in your unprimed frame. The two reddish lines represent the front and rear of the stationary rod and the blue lines represent the front and rear of the rod moving to the right. Due to length contraction, this blue rod appear slightly shorter in this frame. Now let's mark on the times of your State 1, 2 and 3 diagrams in this frame with fine red lines:
View attachment 332484
So far so good, I hope. The first three diagrams in your OP can be generated just by taking horizontal slices through the diagram along the three fine lines.

Now here's where you went wrong. The Lorentz transform for time says that ##t'=\gamma(t−\frac{v}{c^2}x)##, which we can rearrange to ##t=\frac1\gamma t'+\frac{v}{c^2}x##. That means that a line of fixed time in the primed frame (##t'=\mathrm{const}##) is not a horizontal line in this frame. Let's draw the three lines of fixed ##t'## through the three key events as fine blue lines:
View attachment 332485
The second set of diagrams you labelled "State 1" and "State 3" can be drawn by taking sections along those blue lines (an scaling down by a factor of ##\gamma## because you can't draw a Minkowski geometry completely faithfully on a Euclidean plane). Note that they are not the same as the reddish lines! That's what I mean about there being four states here - the two you drew by looking along the top and bottom horizontal lines, and the two you drew along the top and bottom diagonal lines.

What are the implications of that? Well, first, let's draw that last diagram in the primed frame:
View attachment 332486
Hopefully you can see that this is a mirror image of the second image, because this time it's the red rod that is moving and length contracted. Hopefully that makes sense. Now let's go back to the unprimed frame and add one green line:
View attachment 332487
The vertical distance between the middle and upper lines is what you called ##\Delta t_2##. The green line marks the time interval ##\Delta t_2/\gamma##, which is the time ticked off by a clock attached to the left hand end of the blue rod in what the unprimed frame calls that same time interval. But now look at it when it's transformed into the primed frame:
View attachment 332488
It is not the time interval between the right hand end of the red rod meeting the right and left ends of the blue rod. And your "proof" relies on the assumption that it is, if I read your mess of badly typeset maths correctly (we have ##\LaTeX## enabled on the forum for a reason).

As has been said several times already, you did not account for the relativity of simultaneity.
I followed you and agree up to the point where you say now here's where you went wrong. Then for some reason you assumed that t prime is a constant, when it's clearly not.
## t' = \gamma (t- \frac{vx}{c^2})##
On the right hand side of the equation t is a variable

Leepappas said:
Then for some reason you assumed that t prime is a constant,
It's a constant across space at one instant of time. Each of your diagrams is drawn at one instant of time is it not?

Ibix said:
It's a constant across space at one instant of time. Each of your diagrams is drawn at one instant of time is it not?
Yes I reread it now I understand what he did.

Leepappas said:
What say you?
As I have said here before
When you come up with an idea that seems to completely be against established science, it is not a good idea to start off reaching different conclusions and stating them as correct but rather to start off with the assumption that you have made a mistake somewhere and try to find out where it is. If you have NOT made a mistake you will find the flaw in the established science, but that is extraordinarily unlikely to happen. If you start off thinking that you have overturned established science you are likely to just end up embarrassed.
As has been pointed out, you clearly do not fully understand SR.

davenn
phinds said:
As I have said here before

As has been pointed out, you clearly do not fully understand SR.
Oh no I can win this argument there's an error in his drawing. It would help if I knew how to make the same kind of graph as he did to show everyone his mistake.

davenn and phinds
The first rule of holes is, when you find yourself in one, STOP DIGGING.

You need to step back and take a deep breath. Do you REALLY think that you have discovered a flaw in a theory that is over 100 years old and has been shown conclusively to be correct by thousands of experiments?

Really? Seriously?

davenn
phinds said:
The first rule of holes is, when you find yourself in one, STOP DIGGING.

You need to step back and take a deep breath. Do you REALLY think that you have discovered a flaw in a theory that is over 100 years old and has been shown conclusively to be correct by thousands of experiments?

Really? Seriously?
It's it's child's Play for me so yes.

davenn, PeroK and phinds
Ibix said:
This is a Minkowski diagram showing the worldlines of the front and rear of each of your two rods. Stripped of jargon, this is a simple displacement-vs-time graph that you should have encountered at school, except that time is drawn up the page and displacement across the page by convention.
View attachment 332483
This is drawn in your unprimed frame. The two reddish lines represent the front and rear of the stationary rod and the blue lines represent the front and rear of the rod moving to the right. Due to length contraction, this blue rod appear slightly shorter in this frame. Now let's mark on the times of your State 1, 2 and 3 diagrams in this frame with fine red lines:
View attachment 332484
So far so good, I hope. The first three diagrams in your OP can be generated just by taking horizontal slices through the diagram along the three fine lines.

Now here's where you went wrong. The Lorentz transform for time says that ##t'=\gamma(t−\frac{v}{c^2}x)##, which we can rearrange to ##t=\frac1\gamma t'+\frac{v}{c^2}x##. That means that a line of fixed time in the primed frame (##t'=\mathrm{const}##) is not a horizontal line in this frame. Let's draw the three lines of fixed ##t'## through the three key events as fine blue lines:
View attachment 332485
The second set of diagrams you labelled "State 1" and "State 3" can be drawn by taking sections along those blue lines (an scaling down by a factor of ##\gamma## because you can't draw a Minkowski geometry completely faithfully on a Euclidean plane). Note that they are not the same as the reddish lines! That's what I mean about there being four states here - the two you drew by looking along the top and bottom horizontal lines, and the two you drew along the top and bottom diagonal lines.

What are the implications of that? Well, first, let's draw that last diagram in the primed frame:
View attachment 332486
Hopefully you can see that this is a mirror image of the second image, because this time it's the red rod that is moving and length contracted. Hopefully that makes sense. Now let's go back to the unprimed frame and add one green line:
View attachment 332487
The vertical distance between the middle and upper lines is what you called ##\Delta t_2##. The green line marks the time interval ##\Delta t_2/\gamma##, which is the time ticked off by a clock attached to the left hand end of the blue rod in what the unprimed frame calls that same time interval. But now look at it when it's transformed into the primed frame:
View attachment 332488
It is not the time interval between the right hand end of the red rod meeting the right and left ends of the blue rod. And your "proof" relies on the assumption that it is, if I read your mess of badly typeset maths correctly (we have ##\LaTeX## enabled on the forum for a reason).

As has been said several times already, you did not account for the relativity of simultaneity.
In the second dstance versus time graph the horizontal line representing state three is drawn too high. It should intersect the lower Blue line where the lower Blue line intersects the rightmost vertical Red line. That's the moment B prime coincides with B. The way you have state three drawn is the moment when A prime coincides with B.

Leepappas said:
No I have not but if my proof is right then I don't need to.
Publish and claim your Nobel Prize!

Demystifier, Doc Al and Dale
I will re-analyze your scenario using the correct formulas (the Lorentz transform). I will make one notational change that I feel is important and worth explaining.

I will use ##'## to indicate a quantity in the "moving" frame and the absence of ##'## to indicate the same quantity in the "stationary" frame where as usual "moving" and "stationary" are just verbal labels. The two rods are physically distinct rods, so what you have labeled as A and A' are not the same thing seen in different frames, but different things. I will indicate the rod that is at rest in the stationary frame with capital letters ##A## for the left end and ##B## for the right end. I will indicate the rod that is at rest in the moving frame with lower case letters ##a## for the left end and ##b## for the right end. Thus ##A'## is the left end of the rod which is at rest in the stationary frame expressed in the coordinates of the moving frame.

For conciseness I will use units where ##c=1## and I will express coordinates in standard four vector notation with a single spatial dimension, i.e. ##(t,x)##.

It is easiest to state each the coordinates of each rod end in the corresponding rest frame. So $$A=(t,0)$$$$B=(t,L_0)$$$$a'=(t',-L_0)$$$$b'=(t',0)$$

To find the coordinates in the other frame we simply use the Lorentz transform posted above, and simplify to get $$A'=(t',-vt')$$$$B'=\left(t',-vt'+\frac{L_0}{\gamma}\right)$$$$a=\left(t,vt-\frac{L_0}{\gamma}\right)$$$$b=(t,vt)$$

Here we are actually essentially done. We can determine the velocity simply by taking the derivative of the ##x## coordinate. $$\frac{dx_{a}}{dt}=v=\frac{dx_{b}}{dt}$$$$\frac{dx'_{A'}}{dt'}=-v=\frac{dx'_{B'}}{dt'}$$All velocities must come back to those derivatives, regardless of which pairs of events we choose. So it is redundant, but we can go ahead and explicitly calculate according to your designations of events.

Your event 1 is where ##A## coincides with ##b## so we can find the time in the unprimed frame by setting ##A|_{t_1}=b|_{t_1}## and solving for ##t_1##. This gives us ##t_1=0##. So event 1 is $$A|_{t_1}=b|_{t_1}=(0,0)$$

Your event 3 is where ##B## coincides with ##b## so we can find the time in the unprimed frame by setting ##B|_{t_3}=b|_{t_3}## and solving for ##t_3##. This gives us ##t_3=L_0/v##. So event 3 is $$B|_{t_3}=b|_{t_3}=\left( \frac{L_0}{v},L_0\right)$$

So in the stationary frame ##b## travels a displacement ##L_0## in a time ##t_3=L_0/v## so the velocity as determined by the stationary frame is $$\frac{L_0}{L_0/v}=v$$

Now, in the moving frame event 1 is where ##A'## coincides with ##b'##. So similarly we get ##t'_1=0## and event 1 is $$A'|_{t'_1}=b'|_{t'_1}=(0,0)$$

In the moving frame event 2 is where ##A'## coincides with ##a'##. So similarly we get ##t'_2=L_0/v## and event 2 is $$A'|_{t'_2}=a'|_{t'_2}=\left(\frac{L_0}{v},-L_0\right)$$

So in the moving frame ##A'## travels a displacement ##-L_0## in a time ##t'_2=L_0/v## so the velocity as determined by the moving frame is again $$\frac{-L_0}{L_0/v}=-v$$

Last edited:
Ibix, russ_watters, Motore and 4 others
@Leepappas is correct in posts #25 and #28 that I made a mistake in my diagrams. It changes nothing about the conclusions.

The first diagram I posted is correct:

I've added labels for the ends A and B of the red rod and A' and B' of the blue rod. The next diagram was where the error began - "State 1" and "State2" are correct, but "State 3" is when the B and B' lines cross. The corrected diagram is:

As before, the definition of simultaneity in the primed frame yields sloped lines in this frame:

Note that, in the primed frame, "State 2" is after "State 3"! This is the relativity of simultaneity striking - the definitions of the states are anchored to spacelike separated events, and the ordering of spacelike separated events is frame-dependent.

Again we can transform this last diagram into the primed frame:

And again, we can switch back to the original unprimed frame and add a green line showing the time ##\Delta t_2/\gamma## that corresponds to what that frame calls "during" the gap between "State 2" and "State 3":

And again we can show that in the primed frame:

And again, we can see that the green line does not correspond to the gap between any pair of fine blue lines. The relativity of simultaneity cannot be ignored.

Does the green line still not show the interval you were thinking of? It doesn't matter. None of the intervals match up once you take into account the relativity of simultaneity, and this "proof" needs some of them to match up.

This is a rookie mistake. Most of us have made it at some point early in our learning, and understanding why it's a mistake is one of the big steps in developing insights into relativity.

Dale
Leepappas said:
In my analysis I am not trying to compare the first and last events in the sequence in each frame so I can compare the total time deltas that's not what I'm doing. There are two moments in time or states of the universe that I am focusing on. One moment in time is state one and the other moment in time is state three and I did it right and I got the contradiction what say you?
How can that be only two times for all those length measurements? Even measuring a single length of an object involves the location of the beginning and the end at "the same time" defined by the IRF, IRF1, that the length is being measured in. But the "same time" in IRF1 is two different times in any other IRF. Have you taken that into account?

Leepappas said:
It's it's child's Play for me so yes.
It is really not. As mentioned above, you clearly don’t understand the relevant concepts and don’t recognize which formulas to use. You simply are not qualified to make the claims you are making.

I challenged you above to post a professional scientific reference that makes the same proof you are claiming here. You have failed to do so. That is the standard used on this forum.

As such, this thread is closed until you find such a proof. Once (if) you do so, just PM me and I can reopen it. Until then the two disproofs you have received and the many recommendations to study the relativity of simultaneity and the Lorentz transform will suffice.

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