Trouble with Lorentz transformations

[starthaus]

The correct formula is:
x2 = (x'2 + vt'2) - from above post by starthaus
to starthaus - isn't that what I wrote in this part (see the last line in the above quote?)
If so, then 15 ly is a correct calculation and so would be the 9 years for t₂ = 1.25(0 + 0.6*12) = 9 yr. However, I would have to adjust backwards the length at t = 0 (in other words -9*0.6 = -5.4. 15 - 5.4 = 9.6, so the length x₂ when t = 0 was actually 9.6 ly.

Guess what, uising the length contraction formula we get the same result. 12 * (1/\gamma) = 12/1.25 = 9.6 ly.

I can't follow all your numerical calculations, as a rule, you should be able to express your ideas in symbolic form. I repeat, you have no right to write what amounts to x_2=v*t since v is the speed between the frames.

Actually, using t₁ = t₂ = 0 as per starthaus gives these weird answers, but the "proper time" both ways comes out the same:

SQRT[(-7.2 - 0)² - (12 - 0)²] = \sqrt{-92.16

and

SQRT[(0 - 0)² - (9.6 - 0)² = \sqrt{-92.16

I never wrote such stuff, nor did I direct you to. Negative proper time is a clear indication of another error in your calculations.

 

[stevmg]

Let's put this to bed:

To wit, assume that there are two reference frames O (the "stationary" one) and O' (the "moving" one.) The moving frame (O') is going at 0.6c to the right.
The length of the rod is (in this example) 12 ly in the O' frame and is stationary with respect to the O' frame. This is a specific case of the x₂' - x₁' which pc2-brazil discusses. v = 0.6c and is the speed between the frames with O stationary and O' moving at the 0.6c to the right.
Thus, gamma is 1.25.
\gamma = 1/\sqrt{1 - v^2/c^2}
\gamma = 1/\sqrt{1 - (0.6c)^2/c^2}
\gamma = 1/\sqrt{1 - 0.36
\gamma = 1/\sqrt{0.64 = 1/0.8 = 1.25
By convention, we will assume that all measurements and times start from x = 0, x' = 0, t = 0 and t' will calculate to zero which we will see below.
By the specific example I gave, we were given a specific length in O' which I chose to be 12 ly. This would represent x₂' - x₁' as stated a few paragraphs above.
x₁ = 0, t₁ = 0
x₁' = \gamma*(x₁ - v*t₁) = 1.25(0 - 0.6*0) = 0 If I am not mistaken then this is the correct Lorentz transformation regarding distance.
t₁' = 1.25(t₁ - v*x₁) = 1.25(0 - 0.6*0) = 0 Likewise, this should be the correct Lorentz transformation with regards to time.
By the conditions imposed in the discourse between starthaus and pc2-brazil, t₁ = t₂. This was so because the L = (x₂ - x₁) was being measured or calculated in the stationary frame given the measure of L' = (x₂' - x₁') which is the length in the moving frame (or O').
t₁ = t₂ = 0. This is my adaptation of the requirement that t₁ = t₂.
Because of the relativity of simultaneity one cannot assume that t₁' = 0, so this must be calculated, if you need it. The question did not ask for t₂' but merely for the value of x₂ - x₁. We will use the given that t₁ = t₂ and with that we are to calculate x₂ - x₁. This means (and pc2-brazil states so) we are given x₂' - x₁' and which, in my particular case here, is 12 ly as stated above. Because of the values of x₁ = 0, t₁ = 0 and, x₁ = 0 and t₁' = 0 which are stated above, all we need to find is x₂ which will then be the value of x₂ - x₁. We can calculate t₂' which will give the value of t₂' - t₁'. Again, this was not asked for but doing this reveals some interesting intermediate results which only makes sense in the time-space Minkowski system.

We develop two equations in two unknowns which will give us answers for x₂
(which is the answer we are after) and t₂' which we can accompish at the same time.

Equation 1:
x₂ = \gamma*[(x₂' - x₁') + vt₂'] Now, I am positive that this is the correct form of the Lorentz transformation when going from O' back to O.
x₂ = 1.25*[12 ly + (0.6)t₂']
Equation 2:
t₂' = \gamma*[t₂ -vx₂] = 1.25*[0 - (0.6)x₂]
t₂' = -1.25*0.6x₂
t₂' = -0.75x₂
substituting equation 2 into equation 1:
x₂ = 1.25[12 ly + (0.6)(-0.75x₂]
1.5625x₂ = 15 ly
We divide both sides by 1.5625 and we get:
x₂ = 9.6 ly
This also represent (x₂ - x₁) = L from pc2-brazil. This is because x₁ = 0, so the value of x₂ is the L.
This is the right answer and does equal L'/\gamma or 12/1.25 which is the original pc2-brazil notation.

If we back substitute x₂ = 9.6 ly into t₂' = -0.75x₂ we get t₂' = -7.2 yr.

I also derived using the above equations but substituting for x₂ with t₂' into equation 1 and proceeded to solve the new equation for t₂'. This likewise came out with the same answer: -7.2 yr.

pc2-brazil's algebraic derivation "buried" the negative value of t₂' and it was not evident when you just use the differences of the x's and t's. When we approached it in two steps, we found it out again, if we looked for it. Now we are dealing with coordinates in space-time and because of the placing the origin of O' on top of the origin of O, we have a negative number for t₂' but it is all relative, hence, no harm no foul.

Since this not a trip but a rod, we can have two different but simultaneous events, and they are simultaneous because they are being done at the same time in frame O. If one were himself or herself in O' we would "see" the measurement at the front of of the rod being done 7.2 years (-7.2, 12, 0, 0) ahead of the measurement at the back end of the rod (0, 0, 0, 0).

In most texts I have seen, they gloss over all this middle men part of the calculations and just state (using the terminology of pc2-brazil and starthaus) L = L'/\gamma which yields the same results and does not require delving into Minkowski space.

 

[starthaus]

In most texts I have seen, they gloss over all this middle men part of the calculations and just state (using the terminology of pc2-brazil and starthaus) L = L'/\gamma which yields the same results and does not require delving into Minkowski space.

The standard texts are very simple, they start with the Lorentz transform:

x'=\gamma(x-vt)
t'=\gamma(t-vx/c^2

then, they calculate:dx'=\gamma(dx-vdt)
dt'=\gamma(dt-vdx/c^2

In order to calculate the length L'=dx' as a function of the length L=dx
the texts make use of the fact that dt'=0 (you must mark both ends of L' simultaneously in frame F'). This gives:

dt=vdx/c^2

Substitute in the first equation and you will get:

L'=L/\gamma

None of your the impossible to follow manipulations. For example:

Equation 2:
t₂' = \gamma*[t₂ -vx₂] = 1.25*[0 - (0.6)x₂]
t₂' = -1.25*0.6x₂
t₂' = -0.75x₂

is not usable since t'_1=0 is DIFFERENT from your t'_2 above. The theory of measurement REQUIRES that you mark both ends of the rod simultaneously, i.e. t'_1=t'_2. So, the fact that you obtained the correct result for L' is purely accidental.

 

[stevmg]

Posts 1 - 5 of this topic says the opposite of what you just stated above in post 53. In those earlier posts you stated that t₁' does not equal t₂' and you wanted t₁ = t₂ because we were measuring the result in the unprimed frame.

The answer of 9.6 ly is not accidental. It follows from the already proven L = L'/\gamma.

Remember, as per pc2-brazil, we were given (x₂' - x₁'. That's what pc2-brazil stated and you went along with. He wanted to use t₁' = t₂' but you changed that to t₁ = t₂ and stated that t₁' could not equal t₂' because of the relativity of simultaneity. Just reread those posts. I didn't write them, you and pc2-brazil did.

 

[starthaus]

Posts 1 - 5 of this topic says the opposite of what you just stated above in post 53. In those earlier posts you stated that t₁' does not equal t₂' and you wanted t₁ = t₂ because we were measuring the result in the unprimed frame.

If you want to measure x'_2-x'_1 you need to make t'_2=t'_1.
If you want to measure x_2-x_1 you need to make t_2=t_1.
I think I already told you this a few times.

Remember, as per pc2-brazil, we were given (x₂' - x₁'. That's what pc2-brazil stated and you went along with. He wanted to use t₁' = t₂' but you changed that to t₁ = t₂ and stated that t₁' could not equal t₂' because of the relativity of simultaneity. Just reread those posts. I didn't write them, you and pc2-brazil did.

At this rate , you'll never learn anything.

 

[stevmg]

Well, actually I have learned a lot.

Too bad that we are not in a classroom with a blackboard, where the give and take is quicker and sometimes the light bulb goes off sooner. Right now, you're right. I am stuck in neutral. But I can tell you this:

If you have a rod of length L' as measured in O' and with O' moving at v relative to O, by length contraction when you transform this to what an observer in O (unprimed frame) sees, he/she will see a rod of length L = L'/\gamma.

JesseM and DaleSpam went over that with me (as well as with many others) in the Einstein train thread. That isn't hard to understand. Dr. Einstein demonstrated that in a simplistic way (he was great at breaking things down to simple elements) in Section XII. "Rods and Clocks in motion" in Relativity.

If you want to have one more pass before you say the hell with me, consider below

"If you want to measure x₂' - x₁' you need to make t₁' = t₂'
If you want to measure x₂ - x₁ you need to make t₁ = t₂"
This is quoting what you have stated above several different times in the thread.

It appears that in the case that I presented, we are to apply the 12 ly (x₂' - x₁') to O' (because that is where it is "measured") and use t₁' = t₂' which would be zero. We would then transform this to O via Einstein-Lorentz and would wind up with an x₁ and x₂ which would be the length (L) in O. Are we on the same sheet of music? I know it seems redundant and beyond elemental but just answer the question. In other words, we "measured" L' and from that will calculate L.

 

[starthaus]

It appears that in the case that I presented, we are to apply the 12 ly (x₂' - x₁') to O' (because that is where it is "measured") and use t₁' = t₂' which would be zero. We would then transform this to O via Einstein-Lorentz and would wind up with an x₁ and x₂ which would be the length (L) in O. Are we on the same sheet of music?

Nope. This is wrong. Remember the rule, in order to get a valid answer in O you need to measure x_2-x_1 for t_2=t_1. When you "Lorentz transform" t'_2 and t'_1 into O, you will NOT get t_2=t_1, so you will not get a valid measurement. Why is this so, Steve?

 

[stevmg]

That is because of the relativity of simultaneity, i.e., when t₂'=t₁' in O', when these are transformed, the "corresponding t's" are not equal any longer. They are only equal in the original frame (O') because they were "set" that way by virtue of a measurement of the front and back of the rod being done together. The measurement of the rod in O' means that all elements from front to back are measured at the same time.

There's the "jump" where I am hung up. I comprehend the measurement in O' as pc2-brazil presented (x₂' - x₁' = L')

But I want to stop here and have you "OK" what I have said so far. If so, I will then proceed.

 

[starthaus]

That is because of the relativity of simultaneity, i.e., when t₂'=t₁' in O', when these are transformed, the "corresponding t's" are not equal any longer. They are only equal in the original frame (O') because they were "set" that way by virtue of a measurement of the front and back of the rod being done together. The measurement of the rod in O' means that all elements from front to back are measured at the same time.

Yes, this is correct. Now you understand why I objected to your method?

 

[stevmg]

I will understand it more when I successfully complete the problem.

As I read pc2-brazil's opening statement (post 1, this thread) it appears that he knows L' (or x₂' - x₁') and, by consequence, t₁' = t₂' because he measured the position of the front of the rod (x₂') at the same time (t₂' = t₁') he measured the position of the back of the rod (x₁').

That is the core of my confusion: when you used the word "measure" it would seem that you mean "calculate." We are given x₁', x₂', t₁' = t₂' so we would not need to measure x₂' - x₁' therefore we would want to calculate x₂ - x₁ or L.

We would not get a "good" calculation of x₂ - x₁ unless t₁ = t₂. If I calculate t₁ and t₂ from the data given on x₁', x₂', t₁' and t₂', I will NOT get t₁ = t₂ because of this relativity of simultaneity.

At this point I will stop again, to wait for the "OK" on what I have just stated. So, let's make sure I have this correct so far.

 

[starthaus]

We would not get a "good" calculation of x₂ - x₁ unless t₁ = t₂. If I calculate t₁ and t₂ from the data given on x₁', x₂', t₁' and t₂', I will NOT get t₁ = t₂ because of this relativity of simultaneity.

Right. So you can EITHER calculate L as L'/gamma OR L' as L/\gamma.

 

[pc2-brazil]

starthaus: Let's say I make the measurement of L from O and want to measure by which factor L is smaller than L', the proper length (length measured by O', according to which the measured object is at rest). Why can't I obtain this relation (which is L = L'/γ) by writing x'₂ - x'₁ with t₂ = t₁?
I still don't understand why this is wrong. This seems to be calculated like this very commonly. For example:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

 

[starthaus]

starthaus: Let's say I make the measurement of L from O and want to measure by which factor L is smaller than L', the proper length (length measured by O', according to which the measured object is at rest). Why can't I obtain this relation (which is L = L'/γ) by writing x'₂ - x'₁ with t₂ = t₁?
I still don't understand why this is wrong.

Because the theory of measurement says that , in order to calculate the length of an objects in a frame, you need to mark its endpoints simultaneously in THAT particular frame.

This seems to be calculated like this very commonly. For example:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

It says quite clearly, that it calculates the length x_2-x_1 for t_2=t_1, exactly what I have been saying several times in this thread.

 

[pc2-brazil]

It says quite clearly, that it calculates the length x_2-x_1 for t_2=t_1, exactly what I have been saying several times in this thread.

And also what I was trying to say.
If that in the site is written correctly, then what I wrote in the last phrase of post #14 is correct.
Unless I expressed myself badly. I want to measure x₂ - x₁ with t₂ = t₁, but start the resolution of the problem by writing x'₂ - x'₁ and substituting it for the Lorentz transformations instead of doing it the other way: writing x₂ - x₁. Then, I obtain L = L'/γ.

 

[starthaus]

And also what I was trying to say.
If that in the site is written correctly, then what I wrote in the last phrase of post #14 is correct.
Unless I expressed myself badly. I want to measure x₂ - x₁ with t₂ = t₁, but start the resolution of the problem by writing x'₂ - x'₁ and substituting it for the Lorentz transformations instead of doing it the other way: writing x₂ - x₁. Then, I obtain L = L'/γ.

ok, then

 

[stevmg]

Dr. Einstein, with very simple algebra and calculations proves what both of you (pc2-brazil and starthaus) are saying: L = L'/\gamma. In his book, Relativity, Section XII "The Behaviour of Measuring Rods and Clocks in Motion" he puts his derivation in this format as opposed to L' = \gammaL. If you assume that x₁' and x₂' are given in O' and x₁' and x₂' are fixed in O' and you "Lorentz" them to x₁ and x₂ in O and you do it in a way that pc2-brazil algebraically defines, you will wind up with t₁ = t₂ (because that's the way he set it up - to come out that way.) Thus, you are measuring x₁ and x₂ at he same time (simultaneously) in O. Because simultaneity is not preserved from different frames of reference, in O', the "corresponding t' s" are not equal to each other as they were in O. This is algebraically established in the second to last paragraph of this post. But that doesn't make any difference. Since x₁' and x₂' are fixed in O' as this was a given, the actual times of measurement of x₁' and x₂' in O' are immaterial to their measurements because these measurements never change in that frame of reference.

Final point, to establish that I do understand the derivation of L = L'/\gamma or L' = \gammaL:

x₂'= \gamma[x₂ - vt₂]
x₁'= \gamma[x₁ - vt₁]
x₂' - x₁' = L' = \gamma[x₂ - x₁ - (vt₂ - vt₁)]
but t₂ = t₁, therefore
x₂' - x₁' = L' = \gamma[x₂ - x₁ - (vt₂ - vt₂)] = \gamma(x₂ - x₁) = L
L' = \gammaL
Q.E.D.

Just for the record:
t₂' = \gamma(t₂ - vx₂)
t₁' = \gamma(t₁ - vx₁)
but t₂ = t₁, therefore
t₂' = \gamma(t₂ - vx₂)
t₁' = \gamma(t₂ - vx₁)
t₂' - t₁' = \gamma[v(x₁ - x₂)] which is not zero.
This shows that t₂' does not = t₁'. Again, this is irrelevant as x₁' and x₂' never change with time in O' because this was stipulated as a given so the L' = \gammaL relationship holds.

If I am wrong, please, please, please do NOT call me a dolt. I was born during the day, but not yesterday. Pardon any typos. I tried to clean them up but this is so damn cumbersome I may have missed a few.

 

[stevmg]

To starthaus:

These two quotes answers the "why" one can use x₁' and x₂' in which t₁' \neq t₂' (i.e, differing times when the rod is measured between front and back) for the calculation which t₁ = t₂ in S (or frame O in the pc2-brazil question.) Up until that point I couldn't make the "jump" and figure how you were still able to use the length of the rod in S' or L' as pc2-brazil stated it with its differing times of measurement of the front and back to calculate the length (L) in frame O in which the times of measurement of the front and back were the same.

Back to Frame-Dragger:

If I do place a rod of certain length l₀ in the moving frame S' which moves at a velocity of v to the right with respect to the stationary frame S and measure it in the stationary frame S at x₁ and x₂ simultaneously at t₁ = t₂, we get the length l₀ = x₂ - x₁ according to the length contraction formula l₀ = l/gamma. However, when we solve the simultaneous Lorentz equations for t_1' and t₂' in in S' which are the corresponding times for the t₁ = t₂ in S we note that t₁' does not = t₂\neq' which means we measured the front and back of the rod at different times in S'.

Since the rod was stationary in S' it does not make a difference for l = x₂' - x₁' as both x₂' and x₁' are fixed in S' so that l will always be the same no matter when you measure the rod even if you measure the front and back at different times in that moving frame.

Is that logic correct? If so, this would justify the equation l₀ = l/gamma

l = length in moving frame S'
l₀ = length in resting frame S

Yes, exactly so. Going the other way, if we measure the ends of the rod simultaneously in S' at t₂' = t₁', we similarly find that in S, t₂ ≠ t₁. This matters because the rod is not stationary in S.

 

[AdVen]

One end of the object is at x'₁ x'₂. Then, the length of the object in the primed reference frame is L' = x'₂ - x'₁.

Two questions:

Question 1:

What do you mean exactly by the sentence: "One end of the object is at x'₁ x'₂."

Question 2:

Should it not be the case, that x'₂ > x'₁?

PS. I followed the same approach as you did in the thread: Derivation of Proper Time and Proper Space.

 

[stevmg]

Two questions:

Question 1:

What do you mean exactly by the sentence: "One end of the object is at x'₁ x'₂."

Should read one end of the object is at x'₁ and the further end is at x'₂

Question 2:

Should it not be the case, that x'₂ > x'₁?

Yes, that is correct.

PS. I followed the same approach as you did in the thread: Derivation of Proper Time and Proper Space.

Where have you been? I wrote that weeks ago! Not harassing you - just asking.

 

[AdVen]

The math that you wrote below contradicts your above statement.





When you write x'₂ - x'₁ it means that you are measuring in the primed frame.



One more time, L' = γL is wrong. If you did things correctly you should have gotten L' = L/γ

If L' = L/γ then one would expect L = γ L'.

 

[JesseM]

If L' = L/γ then one would expect L = γ L'.

How are you defining L' and L? In post #62 pc2-brazil defined L' as the length of the object in its own rest frame and L as its length in the frame that sees it moving, and likewise in post #1 stevmg said "Suppose I have a thin rod moving along with the primed reference frame" indicating that L' was the rest length of the rod. If L' is the rest length and L is the length in the frame where the rod is moving, the correct equations are L = L'/γ and L' = γL, not the equations you write above. I don't know why starthaus said "One more time, L' = γL is wrong" in post #12, perhaps he/she was confused and thought L was supposed to be the rest length (this is the more common convention when the length contraction equation is written in textbooks and such).

 

[stevmg]

If L' = L/γ then one would expect L = γ L'.

It's been a long time since this topic was started so I'm turned around.

As I understand it, the original "measurement" was L' in the O' (the so-called "moving" frame of reference) and we are after L in O the so-called "stationary" frame of reference.

As starthaus has pointed out, we were to measure L in O such that t₁ = t₂ so that we would get a meaningful length L = x₂ -x₁ where both ends were measured at the same time in O.

In O', where L' was originally situated, since the so-called rod of length L' = x'₂ - x'₁ and since the rod is stationary with respect to O' it doesn't make any difference when t'₁ or t'₂ are because the measurements of x'₁ and x'₂ never change no matter when you measure them in O'

Thus L = L'/\gamma (the rod measures "shorter" in O than in O')

Use these assumptions that I just described to come up with the L = L'/\gamma.

The string is too damn long to go through to find all the mistakes (albeit mine) that were made. Just look at it from scratch:

O' moving at v with respect to O. L' (i.e., x'₁ and x'₂)measured in O' as a given. L' is stationary with respect to O' (thus x'₁ and and x'₂ never, never change). From that we are to calculate L = x₁ - x₂ in O where x₁ and x₂ are measured at the same time (t₁ = t₂).

Again, don't try to fix any prior entries from me. Just go from today post #72 as this has been too long a post and too convoluted (again, probably my fault) to try to untangle.

 

[pc2-brazil]

AdVen: I confirm stevmg's correction to my phrasing. I wrote that phrase in a hurry and ended up forgetting to complete it.
Everyone:
When starthaus said that L' = γL is wrong, I believe he was referring to the case in which L is the length of an object at rest in frame O (posts #62 to #65 conclude the divergence, since I clear up the fact that the situation I was referring to is one in which L' is the rest length).

 

[AdVen]

I am very, very delighted with your response. I can tell you, that my problems with relativistic theory are not primarily the mathematics of it, but much more the 'what is what' of things or 'what is relative to what' or 'which observer in which reference system sees what in another or the same reference sysyem'. I do not know how to tell it differently. I am not a native speaker. You are very clear in this respect for which I am very thankful. You wrote:

I don't know why starthaus said "One more time, L' = γL is wrong" in post #12, perhaps he/she was confused and thought L was supposed to be the rest length (this is the more common convention when the length contraction equation is written in textbooks and such).

I prefer the more common convention and am going to write the derivations and will inform you when I am ready.

 

[AdVen]

How are you defining L' and L? In post #62 pc2-brazil defined L' as the length of the object in its own rest frame and L as its length in the frame that sees it moving, and likewise in post #1 stevmg said "Suppose I have a thin rod moving along with the primed reference frame" indicating that L' was the rest length of the rod. If L' is the rest length and L is the length in the frame where the rod is moving, the correct equations are L = L'/γ and L' = γL, not the equations you write above. I don't know why starthaus said "One more time, L' = γL is wrong" in post #12, perhaps he/she was confused and thought L was supposed to be the rest length (this is the more common convention when the length contraction equation is written in textbooks and such).

I hope I do understand your words correctly if I conclude as follows:

If L is the rest length and L' is the length in the frame where the rod is moving, the correct equations are L' = L/γ and L = γL'.

 

[AdVen]

Hi JesseM,

I have made a mathematical derivation of the Length Contraction (Proper Length) now by following the more common convention. Please, would you check whether it is finally oké now. You can find the derivation by clicking underneath:

http://www.socsci.ru.nl/~advdv/ProperLengthFinal.pdf

I really hope it is oké now and I would appreciate it very much if you would look into it.

Yours, AdVen.

 

[stevmg]

Hi JesseM,

I have made a mathematical derivation of the Length Contraction (Proper Length) now by following the more common convention. Please, would you check whether it is finally oké now. You can find the derivation by clicking underneath:

http://www.socsci.ru.nl/~advdv/ProperLengthFinal.pdf

I really hope it is oké now and I would appreciate it very much if you would look into it.

Yours, AdVen.

Nice article AdVen. In my opinion, which means nothing on this forum as I am, too, a complete novice, if the given length L is in the rest frame O, the "measured" length L' in S' (the moving frame) at t'₁ = t'₂ is L' = L/\gamma.

I hope JesseM confirms this which will make my understanding likewise correct.

 

[JesseM]

Hi JesseM,

I have made a mathematical derivation of the Length Contraction (Proper Length) now by following the more common convention. Please, would you check whether it is finally oké now. You can find the derivation by clicking underneath:

http://www.socsci.ru.nl/~advdv/ProperLengthFinal.pdf

I really hope it is oké now and I would appreciate it very much if you would look into it.

Yours, AdVen.

Yup, the derivation looks good to me.

I hope I do understand your words correctly if I conclude as follows:

If L is the rest length and L' is the length in the frame where the rod is moving, the correct equations are L' = L/γ and L = γL'.

Right.

 

[AdVen]

Yup, the derivation looks good to me.

Right.

I suppose that 'yup' means 'yes'. Anyhow, I am very grateful that you have looked into it.

I am now trying to derive in a similar way the Time Dilatation Proper Time formula. I hope you will soon hear from me. If you might have any suggestions, please, let me hear.

 

[starthaus]

I suppose that 'yup' means 'yes'. Anyhow, I am very grateful that you have looked into it.

You can shorten your proof considerably this way:

x'=\gamma(x-vt)
t'=\gamma(t-xv/c^2

Differentiating the above you obtain:dx'=\gamma(dx-vdt)
dt'=\gamma(dt-vdx/c^2)

Now:

L'=dx' is the length you measure in the moving frame S' and L=dx is the length of the rod measured in the co-moving frame S (proper length). In order for your measurement in S' to be valid, you must mark both ends simultaneously in S':

dt'=0

so:

0=\gamma(dt-vdx/c^2)

meaning that :

dt=vdx/c^2=vL/c^2

Therefore:

L'=dx'=\gamma(dx-vdt)=\gamma(L-v^2L/c^2)=L/\gamma

The derivation for time dilation follows a similar pattern:

dx=0 (the events a happen at the same location in frame S)

dt'=\gamma dt

That's it.

 

[JesseM]

I am now trying to derive in a similar way the Time Dilatation Proper Time formula. I hope you will soon hear from me. If you might have any suggestions, please, let me hear.

With the time dilation equation, just remember that the equation deals specifically with the case of two events that happen at the same spatial location in one of the two frames, but different times (like successive readings on a clock that's at rest in the unprimed frame). This makes it pretty simple, since in the equation dt' = gamma*(dt - vdx/c^2) [or if you prefer, the equation (t'₂ - t'₁) = gamma*((t₂ - t₁) - v*(x₂ - x₁)/c^2)], you know dx = 0.

 

[AdVen]

Thanks a lot. I am very glad with your suggestion. I am going to look into it tomorrow. I hope to reply tomorrow.

 

[AdVen]

With the time dilation equation, just remember that the equation deals specifically with the case of two events that happen at the same spatial location in one of the two frames, but different times (like successive readings on a clock that's at rest in the unprimed frame). This makes it pretty simple, since in the equation dt' = gamma*(dt - vdx/c^2) [or if you prefer, the equation (t'₂ - t'₁) = gamma*((t₂ - t₁) - v*(x₂ - x₁)/c^2)], you know dx = 0.

I already knew what you are telling in the first sentence. However, thanks a lot for this idea too. I am also very glad with this suggestion. I am going to look into it tomorrow also. I hope to reply tomorrow.

 

[AdVen]

You need to calculate x₂ - x₁ for the condition: t₂ = t₁. You are inadvertently calculating it for t'₂ = t'₁, this is why you get the error.

I am trying to derive the formula for the time dilatation. I consider the clock is situated in the rest frame S and the observer is situated in the moving frame S'. I want to derive the length of time on the clock as seen by the observer in the moving frame. Therefore I think I need to calculate t'₂-t'₁ for the condition: x'₂ = x'₁. This leads to x₂-x₁ = (t₂-t₁)*v. As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma, where gamma is the Lorentz factor.

Is this correct? I am afraid it is not. At least in Wikipedia I get a different result: http://en.wikipedia.org/wiki/Time_dilation

 

[starthaus]

I am trying to derive the formula for the time dilatation. I consider the clock is situated in the rest frame S and the observer is situated in the moving frame S'. I want to derive the length of time on the clock as seen by the observer in the moving frame. Therefore I think I need to calculate t'₂-t'₁ for the condition: x'₂ = x'₁. This leads to x₂-x₁ = (t₂-t₁)*v. As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma, where gamma is the Lorentz factor.

Is this correct? I am afraid it is not. At least in Wikipedia I get a different result: http://en.wikipedia.org/wiki/Time_dilation

It's correct, you are doing fine. For a simpler derivation, see the last two lines in post 80.

 

[JesseM]

I am trying to derive the formula for the time dilatation. I consider the clock is situated in the rest frame S and the observer is situated in the moving frame S'. I want to derive the length of time on the clock as seen by the observer in the moving frame. Therefore I think I need to calculate t'₂-t'₁ for the condition: x'₂ = x'₁.

If the clock is at rest in the unprimed frame, then the condition should be x₂ = x₁ (the clock's unprimed position coordinate remains unchanged), not x'₂ = x'₁.

This leads to x₂-x₁ = (t₂-t₁)*v.

If you want to start from the part of the Lorentz transformation that deals with x-coordinates, you'd use (x₂-x₁) = gamma*((x'₂-x'₁) + v*(t'₂-t'₁)), and then with x₂-x₁ = 0 you're left with x'₂-x'₁ = -v*(t'₂-t'₁). However, if you're trying to derive the time dilation equation it's much better to start with the part of the Lorentz transformation that deals with t-coordinates, namely (t'₂ - t'₁) = gamma*((t₂ - t₁) - v*(x₂ - x₁)/c^2), then if you substitute in (x₂ - x₁) = 0 you're left with (t'₂ - t'₁) = gamma*(t₂ - t₁), which is the correct time dilation equation.

As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma, where gamma is the Lorentz factor.

Oddly this equation is correct even though your earlier step x'₂ = x'₁ was incorrect! Maybe you made an algebra error too?

Is this correct? I am afraid it is not. At least in Wikipedia I get a different result: http://en.wikipedia.org/wiki/Time_dilation

Why do you think wikipedia gives a different result? The time dilation formula in this section is the same as the one you just wrote.

 

[AdVen]

Oddly this equation is correct even though your earlier step x'₂ = x'₁ was incorrect! Maybe you made an algebra error too?

Why do you think wikipedia gives a different result? The time dilation formula in this section is the same as the one you just wrote.

You can find the mathematics (without text) on:

http://www.socsci.ru.nl/~advdv/TimeDilatationFinal.pdf

Hopefully, you can understand. Otherwise I will add explaining text to it.

 

[JesseM]

You can find the mathematics (without text) on:

http://www.socsci.ru.nl/~advdv/TimeDilatationFinal.pdf

Hopefully, you can understand. Otherwise I will add explaining text to it.

OK, the steps of going from lines 2 and 3 to line 4 indicate you are assuming x'₁ = x'₂, which as I said is the wrong physical assumption if you want the coordinates to be those of a clock at rest in the unprimed frame. Still, the assumption would work if they were the coordinates of a clock at rest in the primed frame, in which case the correct final time dilation equation should be t'₂ - t'₁ = (t₂ - t₁)/gamma, which is what you got in the pdf. In the earlier post you had said "As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma" which is why I was confused (since here you are multiplying the unprimed time difference by gamma rather than dividing it by gamma), but that's not the equation that appears in the pdf.

 

[AdVen]

OK, the steps of going from lines 2 and 3 to line 4 indicate you are assuming x'₁ = x'₂, which as I said is the wrong physical assumption if you want the coordinates to be those of a clock at rest in the unprimed frame. Still, the assumption would work if they were the coordinates of a clock at rest in the primed frame, in which case the correct final time dilation equation should be t'₂ - t'₁ = (t₂ - t₁)/gamma, which is what you got in the pdf. In the earlier post you had said "As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma" which is why I was confused (since here you are multiplying the unprimed time difference by gamma rather than dividing it by gamma), but that's not the equation that appears in the pdf.

I am very, very grateful for your comment. I am going to read this very carefully and will reply as soon as possible. I think my major problem is understanding what is what (I do not know how to say it differently, as I am not a native speaker). You have many possibilities:

the clock at rest in the unprimed frame,
the clock at rest in the primed frame,
the observer at rest in the unprimed frame,
the observer at rest in the primed frame

and so on

and how these are related to the different assumptions:

x₁ = x₂
x'₁ = x'₂

Since the time I am studying special relativity I have great diffculties with what I call above 'what is what'.

Among other things the expression the 'moving observer' gives me difficulties. I know he/she is moving with respect to the clock. But the usual Lorentz transformation is about a rest frame S and a moving frame S'. Now, if the clock is located in S' then the so called 'moving observer' is located in the rest frame, which is not moving. I hope you can understand my confusion.

 

[AdVen]

Still, the assumption would work if they were the coordinates of a clock at rest in the primed frame, in which case the correct final time dilation equation should be t'₂ - t'₁ = (t₂ - t₁)/gamma, which is what you got in the pdf.

Yes, I imagine the following situation:

S: frame at rest (x and t are coordinates in S).

S': moving frame with respect to S (x' and t' are coordinates in S').

Lorentz transformations::

x' = γ(x - vt)

t' = γ(t - vx/c²)

Where γ is the Lorentz factor and c is the speed of light in vacuum.

Clock: clock is at rest in S'.

t₂ - t₁: time difference observed by the observer, which is located in S, when looking at the clock in S'. This observer is the so-called moving observer. Although he is NOT moving with respect to S, he IS moving relative to S' and, therefore also relative to the clock in S', which is at rest in S'.

t'₂ - t'₁: time difference observed by the observer at rest in S' when looking at the clock in S' . This observer is located in S'.

The expression t'₂ - t'₁ = (t₂ - t₁)/γ expresses the fact that for the resting observer the period of the clock is shorter then for the moving observer.

Do you think, that this correct now?

Thanks a lot for your concern and for your time and effort.

 

[JesseM]

I am very, very grateful for your comment. I am going to read this very carefully and will reply as soon as possible. I think my major problem is understanding what is what (I do not know how to say it differently, as I am not a native speaker). You have many possibilities:

the clock at rest in the unprimed frame,
the clock at rest in the primed frame,
the observer at rest in the unprimed frame,
the observer at rest in the primed frame

and so on

and how these are related to the different assumptions:

x₁ = x₂
x'₁ = x'₂

Since the time I am studying special relativity I have great diffculties with what I call above 'what is what'.

Make sure you keep in mind what specific events are being assigned coordinates! In the case of the time dilation equation, you're always picking two events on the worldline of a clock, like having x₁, t₁ being the coordinates (in the unprimed frame) of the clock reading 10 AM, and x₂, t₂ being the coordinates of the same clock reading 11 AM. So if x₁ represents the position of the clock at one time (when it shows 10 AM) and x₂ represents the position of the clock at another time (when it shows 11 AM), that tells you that if the clock is at rest in the unprimed frame, its position coordinate in the unprimed frame shouldn't change from one moment to another (that's what it means to be at rest in a given frame), so x₁ = x₂

Among other things the expression the 'moving observer' gives me difficulties. I know he/she is moving with respect to the clock.

I would prefer not to use language like "moving observer" without referring to a specific frame, since in relativity all motion is relative. Better to say something like "moving relative to the clock" or "moving relative to the unprimed frame" to make clear that all motion is relative to something, that there is no absolute motion.

But the usual Lorentz transformation is about a rest frame S and a moving frame S'.

In most textbooks I've seen they don't label one frame "the rest frame" and the other "the moving frame", the two frames S and S' are just two frames on equal footing. You might say that S is one particular object's rest frame, or that it's one particular observer's rest frame, but you wouldn't just call it "rest frame" without naming something specific that it's the rest frame for.

 

[JesseM]

Yes, I imagine the following situation:

S: frame at rest (x and t are coordinates in S).

S': moving frame with respect to S (x' and t' are coordinates in S').

Like I said in the last post, I think it's better not to use the language of a particular frame being "at rest", especially if this has been causing you confusion; maybe better to just say S is the rest frame of a clock, and S' is the rest frame of an observer who the clock is moving relative to? (or vice versa if you prefer, but as I said before, the more common convention is to have the unprimed frame as the clock's rest frame when writing the time dilation equation)

Lorentz transformations::

x' = γ(x - vt)

t' = γ(t - vx/c²)

Where γ is the Lorentz factor and c is the speed of light in vacuum.

Clock: clock is at rest in S'.

OK, so you're using the convention that the primed frame is the clock's rest frame. As I said it's more common to see the time dilation equation written with the unprimed frame as the clock's rest frame, but as long as you keep things consistent this is fine.

t₂ - t₁: time difference observed by the observer, which is located in S, when looking at the clock in S'. This observer is the so-called moving observer.

Again, I'd prefer to only use terms like "rest" and "moving" in a relative sense, like "moving relative to the clock" or "moving relative to S'." I don't think most textbooks discussing the time dilation equation would use a phrase like "the moving observer".

t'₂ - t'₁: time difference observed by the observer at rest in S' when looking at the clock in S' . This observer is located in S'.

Yes, although you don't need to have two (or even one) "observers", you can also just talk about the time between the events in the clock's rest frame (talking about 'observers' is basically just a shorthand way of talking about inertial frames, anyway).

The expression t'₂ - t'₁ = (t₂ - t₁)/γ expresses the fact that for the resting observer the period of the clock is shorter then for the moving observer.

Yes, the time between two events on the clock's worldline is shorter for an observer at rest relative to the clock (in the clock's rest frame) than the time between those same two events in a frame that's moving relative to the clock. For example, the time between the event of the clock reading "0 seconds" and the clock reading "100 seconds" would be 100 seconds in the clock's own rest frame (the primed frame according to your convention), but in an unprimed frame moving at 0.6c relative to the clock, 125 seconds would elapse between these same two events.

 

[AdVen]

I am again very grateful for your comments. I think I have understood now.
Expressions like 'rest frame' and 'moving frame' should be avoided.
The use of an 'observer' is also not really necessary. However, it is necessary
to say in which frame the clock is located or with respect to which frame the clock is at rest.
The same holds for a rod in the case of length contraction.

S: unprimed frame (x and t are coordinates in S).

S': primed frame (x' and t' are coordinates in S').

The primed frame S' has a velocity v with respect to the unprimed frame S.

Lorentz transformations::

x' = γ(x - vt)

t' = γ(t - vx/c²)

Where γ is the Lorentz factor and c is the speed of light in vacuum.

Clock: clock is at rest in the primed frame S'.

t₂ - t₁: time difference measured in S.

t'₂ - t'₁: time difference measured in S'.

The expression t'₂ - t'₁ = (t₂ - t₁)/γ expresses the fact that the time difference measured in S' is shorter then the time difference measured in S.

I hope it is correct now.

Perhaps, you can still elaborate on the way you are saying this:

Yes, the time between two events on the clock's worldline is shorter for an observer at rest relative to the clock (in the clock's rest frame) than the time between those same two events in a frame that's moving relative to the clock.

Finally, I am planning to make the derivations for the more common convention in which one has the unprimed frame as the clock's rest frame when writing the time dilation equation. I let you know when I am ready.

 

[AdVen]

Yes, the time between two events on the clock's worldline is shorter for an observer at rest relative to the clock (in the clock's rest frame) than the time between those same two events in a frame that's moving relative to the clock. For example, the time between the event of the clock reading "0 seconds" and the clock reading "100 seconds" would be 100 seconds in the clock's own rest frame (the primed frame according to your convention), but in an unprimed frame moving at 0.6c relative to the clock, 125 seconds would elapse between these same two events.

Hi JesseM,

As a result of all your very valuable suggestions I have now made the final derivation according to 'my convention'. Go to:

http://www.socsci.ru.nl/~advdv/TimeDilatationFinal.pdf

I am very curious to your opinion. Anyway, I hope it is correct now at last. I will also derive the equation according to the 'normal' convention. As soon as I have done this I will inform you.

You are a wonderful guy (or girl, I do not know the gender of Jesse), Ad.

 

[starthaus]

Hi JesseM,

As a result of all your very valuable suggestions I have now made the final derivation according to 'my convention'. Go to:

http://www.socsci.ru.nl/~advdv/TimeDilatationFinal.pdf

I am very curious to your opinion. Anyway, I hope it is correct now at last. I will also derive the equation according to the 'normal' convention. As soon as I have done this I will inform you.

You are a wonderful guy (or girl, I do not know the gender of Jesse), Ad.

There is a small correction of style (the math is correct):

-you started with dx'=0
-you should present your result as dt=\gamma dt' , not the other way around

 

[AdVen]

If I quote your comments I get:

-you started with dx'=0
-you should present your result as dt=\gamma dt' , not the other way around

Is this correct?

 

[AdVen]

Are you familiar with Latex. I could send you the source file. You could make the changes your self. It seems to me that it is not much work.

 

[AdVen]

-you started with dx'=0

Why and should it not be Delta x' = 0 (with capital delta)?

-you should present your result as dt=\gamma dt' , not the other way around

Is this a matter of convention and should it not be Delta t = gamma Delta t' (with capital delta)?

 

[starthaus]

Why and should it not be Delta x' = 0 (with capital delta)?Is this a matter of convention and should it not be Delta t = gamma Delta t' (with capital delta)?

That wasn't the point, you start with the time separation dt' in frame F' where dx'=0 and you try to figure out dt as a function of dt'

 

[AdVen]

That wasn't the point, you start with the time separation dt' in frame F' where dx'=0 and you try to figure out dt as a function of dt'

I am sorry, but these expressions do not occur on my derivation at:

http://www.socsci.ru.nl/~advdv/TimeDilatationFinal.pdf

Neither do I hade F'. I use S'.