Trouble with Lorentz transformations

[AdVen]

I think the point is that if you are starting from the assumption that you start out knowing the separation in the S' frame dx'=0, then it makes more sense if you also assume you start out knowing the time interval in the S' frame and are using it to find the time interval in the S frame, rather than starting from the time interval in the S frame and using it to find the time interval in the S' frame (usually in a basic SR problem, you'll be given all the information about coordinates in one frame and then you have to find the coordinates in the other frame, rather than initially being given the distance in the S' frame but the time in the S frame). In other words it would make more sense for your final equation to give dt as a function of dt', not dt' as a function of dt, so the final equation should be dt = dt'*gamma rather than dt' = dt/gamma.

In my opinion I think you are quite right, if you are aiming at a conclusion such as:

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

However, a conclusion which is consistent with this conclusion is:

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

It seems to me, that, if you do not have any preference for either of these two conclusions, and why should you, it does not matter what approach you choose.

Anyway, you have to specify what exactly you mean with:

then it makes more sense.

 

[AdVen]

I think the point is that if you are starting from the assumption that you start out knowing the separation in the S' frame dx'=0, then it makes more sense if you also assume you start out knowing the time interval in the S' frame and are using it to find the time interval in the S frame, rather than starting from the time interval in the S frame and using it to find the time interval in the S' frame (usually in a basic SR problem, you'll be given all the information about coordinates in one frame and then you have to find the coordinates in the other frame, rather than initially being given the distance in the S' frame but the time in the S frame). In other words it would make more sense for your final equation to give dt as a function of dt', not dt' as a function of dt, so the final equation should be dt = dt'*gamma rather than dt' = dt/gamma.

You prefer:

starting from the time interval in the S frame and using it to find the time interval in the S' frame.

I believe I did so on:

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

If I made some errors, please, let me know. If you would like to rephrase some of the sentences, please let me know too. Thanks in advance.

In connection with the derivation of the length contraction formula you could have argued similarly:

starting from the length interval in the S frame and using it to find the lentgh interval in the S' frame.

I believe I did so for the length contraction on:

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

 

[stevmg]

In my opinion I think you are quite right, if you are aiming at a conclusion such as:

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

However, a conclusion which is consistent with this conclusion is:

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

It seems to me, that, if you do not have any preference for either of these two conclusions, and why should you, it does not matter what approach you choose.

Anyway, you have to specify what exactly you mean with:

Something got lost in translation here.

That does make sense when we speak of the twin or clock paradox. The time elapsed in the moving frames (the spaceship twin and the clock is in the spaceship) is shorter than the eartbound twin. If $$\gamma$$ were, say, 0.8, then an eight year trip on the spaceship (discounting acceleration, deceleration) would be a 10 year lapse at home on earth.

Hell, it's even possible to have a virtual no elapsing of time for the spaceship twin if he/she moved near the speed of light out and back (virtual light-like) while the earthbound partner aged.

In that .pdf I think something is wrong but I don't know what.

 

[AdVen]

Hell, it's even possible to have a virtual no elapsing of time for the spaceship twin if he/she moved near the speed of light out and back (virtual light-like) while the earthbound partner aged.

I would say that one has even NO elapsing of time for a foton moving with the speed of light.

 

[jtbell]

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.

If you have LaTex source code, you can post it here. Enclose it in [ tex ] and [ /tex ] tags (remove the spaces inside the brackets) and the forum's software will convert the LaTeX code and display the equations. (I had to add spaces to the tags in this example, otherwise the forum software would interpret them as actual tags and try to render the text in between as LaTeX code.

Or click on this equation and you can see the LaTeX code and the tags in a popup window:

$$x^{\prime} = \gamma (x - vt)$$

 

[jtbell]

Also, if you use the QUOTE button to respond to a post, it includes the original LaTeX code in the quote, and you can edit it just like the quoted plain text.

$$x^{\prime} = \gamma (x - vt)$$

 

[AdVen]

If you have LaTex source code, you can post it here. Enclose it in [ tex ] and [ /tex ] tags (remove the spaces inside the brackets) and the forum's software will convert the LaTeX code and display the equations. (I had to add spaces to the tags in this example, otherwise the forum software would interpret them as actual tags and try to render the text in between as LaTeX code.

Or click on this equation and you can see the LaTeX code and the tags in a popup window:

$$x^{\prime} = \gamma (x - vt)$$

Thanks a lot. I will try soon.

 

[AdVen]

It did not work with the original Latex source file enclosed as an attachement text file.

 

[JesseM]

In my opinion I think you are quite right, if you are aiming at a conclusion such as:

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

However, a conclusion which is consistent with this conclusion is:

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

It seems to me, that, if you do not have any preference for either of these two conclusions, and why should you, it does not matter what approach you choose.

Any physics equation involves giving you some unknown variable as a function of some known variable. dt = dt'*gamma and dt' = dt/gamma are mathematically equivalent equations, but dt = dt'*gamma would be used in a scenario where you know the value of dt' and want to find the value of dt, while dt' = dt/gamma would be used in a scenario where you know the value of dt and want to find the value of dt'. My point was that in a basic relativity problem, the usual situation is that you know all the coordinates in one frame, and want to find the (unknown) coordinates in a different frame as a function of the coordinates in the first frame. Since you start out assuming the value of dx' is known (it's 0), it makes more sense to also assume dt' is known and dt is unknown. This is just an issue of presentation, there is nothing physically or mathematically incorrect about deriving the equation dt' = dt/gamma instead of dt = dt'*gamma.

 

[AdVen]

Something got lost in translation here.

That does make sense when we speak of the twin or clock paradox. The time elapsed in the moving frames (the spaceship twin and the clock is in the spaceship) is shorter than the eartbound twin. If $$\gamma$$ were, say, 0.8, then an eight year trip on the spaceship (discounting acceleration, deceleration) would be a 10 year lapse at home on earth.

Do you know any text available on Internet where the twin paradox is solved using SR only?

 

[stevmg]

Do you know any text available on Internet where the twin paradox is solved using SR only?

I am posting an attachment which should help you out. It was and still is on the Internet.

 

[Fredrik]

Do you know any text available on Internet where the twin paradox is solved using SR only?

Did you read any of the replies you got in the thread you started about the twin paradox?

https://www.physicsforums.com/showthread.php?t=399741

Every solution is using SR only (because the problem is specified using SR only). That includes the two solutions you got from me, and the one you linked to yourself.

 

[stevmg]

Did you read any of the replies you got in the thread you started about the twin paradox?

https://www.physicsforums.com/showthread.php?t=399741

Every solution is using SR only (because the problem is specified using SR only). That includes the two solutions you got from me, and the one you linked to yourself.

Fredrik -

I just gave Adven an online source for a lot of things about SR including the twin paradox. There are on this forum alone about ten zillion solutions to the twin paradox which involve SR alone. The .pdf "book" I cited has an explanation which is about three hundred times more complicated than is needed to show that there is no paradox.

Adven has the solution - which he wrote but I guess he wanted some online text to refer to, so that is what I responded to.

 

[starthaus]

I am posting an attachment which should help you out. It was and still is on the Internet.

This is a very good book.

 

[AdVen]

I am posting an attachment which should help you out. It was and still is on the Internet.

Dear stevmg,

Thanks a lot for the attachement. I am certainly going to read the text. It is my intention to derive the well-known formulas of SR, such as the formulas for proper length and proper time directly from the two formulas of the Lorentz transformation. I am planning to do something similar with the solution of the twin paradox. In the mean time I would appreciate it very much if you would read the two attached files just to check whether they do not contain any nonsense.

Thanks a lot, Ad.

PS. I have asked the same question to some other guy of PhysicsForum.

 

[AdVen]

Did you read any of the replies you got in the thread you started about the twin paradox?

https://www.physicsforums.com/showthread.php?t=399741

Every solution is using SR only (because the problem is specified using SR only). That includes the two solutions you got from me, and the one you linked to yourself.

Dear Fredrik,

I was not aware that I had linked a solution to myself nor was I aware of the two solutions I got from you. Would you be so kind to tell me were exactly I can find these solutions?

It is my intention to derive the well-known formulas of SR, such as the formulas for proper length and proper time directly from the two formulas of the Lorentz transformation (see the two attached PDF files). I am planning to do something similar with the solution of the twin paradox, something like what Professor Park did at

http://kspark.kaist.ac.kr/Twin Paradox/Twin-Paradox Events and Transformations.htm

However, I am still not quite sure whether this solution is correct.

In the mean time I would appreciate it very much if you would read the two attached files just to check whether they do not contain any nonsense.

Thanks a lot for your information, Ad.

PS. I have asked the same question to some other guys of PhysicsForum.

 

[AdVen]

Adven has the solution - which he wrote but I guess he wanted some online text to refer to, so that is what I responded to.

I was not aware that I had a solution. Please, can you tell me where I can find the text?

Ad.

 

[AdVen]

Hi JesseM,

Could you check whether the following texts are correct:

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

and

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

Thanks a lot for your time and effort, Ad.

 

[starthaus]

Hi JesseM,

Could you check whether the following texts are correct:

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

and

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

Thanks a lot for your time and effort, Ad.

both files are correct now.

 

[starthaus]

I was not aware that I had a solution. Please, can you tell me where I can find the text?

Ad.

Here

 

[Fredrik]

I just gave Adven an online source for a lot of things about SR including the twin paradox.
...
I guess he wanted some online text to refer to, so that is what I responded to.

You don't have to explain yourself to me. The things I said were for AdVen, not you. (And so are the things below this).

I was not aware that I had linked a solution to myself nor was I aware of the two solutions I got from you. Would you be so kind to tell me were exactly I can find these solutions?

Now you're really confusing me. I included a link to the thread you started, so it must have been obvious that I was referring to my posts in that thread, and to your posts in that thread.

...something like what Professor Park did at

http://kspark.kaist.ac.kr/Twin Paradox/Twin-Paradox Events and Transformations.htm

I don't know how you can say "I was not aware that I had linked a solution" and then immediately link to it again. It must have been clear that that's the link I meant, since I included a link to the thread where you posted it and asked for comments. One of the comments you got was from me:

It looks good. I don't like that he says "is in S" when he should be saying something like "has velocity 0 in S", but the explanation is fine.

 

[AdVen]

You don't have to explain yourself to me. The things I said were for AdVen, not you. (And so are the things below this).


Now you're really confusing me. I included a link to the thread you started, so it must have been obvious that I was referring to my posts in that thread, and to your posts in that thread.

I don't know how you can say "I was not aware that I had linked a solution" and then immediately link to it again. It must have been clear that that's the link I meant, since I included a link to the thread where you posted it and asked for comments. One of the comments you got was from me:

I am very sorry for the confusion. Now it is clear to me. We both were referring to Park's solution. If I understand you well this solution is oké except for some minor phrasing, such as:

"instead of "is in S" he should be saying something like "has velocity 0 in S":

Thanks again, Ad.

 

[AdVen]

Here

I am going to look into it very carefully. Thanks for the information.

 

[AdVen]

Here

In this text they use acceleration and deceleration. SR is only about inertial systems. It looks as if GR is used here or am I wrong?

 

[starthaus]

In this text they use acceleration and deceleration. SR is only about inertial systems. It looks as if GR is used here or am I wrong?

You are wrong, this is sadly a very common misconception. SR deals with accelerated frames. See the many attachments in my blog on this very subject.

 

[JesseM]

In this text they use acceleration and deceleration. SR is only about inertial systems. It looks as if GR is used here or am I wrong?

Accelerated motion can be analyzed just fine from the perspective of an inertial frame (and technically a modern physicist would generally say that even if you use a non-inertial frame with a pseudo-gravitational field as discussed here, this is still 'SR' if spacetime is not curved). Any smoothly-curve path in spacetime can be approximated as a polygonal path made up of a series of straight segments of constant velocity joined by instantaneous accelerations, and if you take the limit (in the calculus sense) as the segments become shorter and shorter (and the number of segments becomes greater and greater), this should approach perfect agreement with the original smooth path. If a given straight segment has a velocity v as seen in the inertial observer's frame, and the time between the beginning and end of the segment in the inertial observer's frame is dt, then the proper time along that segment should be $$\sqrt{1 - v^2/c^2} dt$$. So, if you have a smoothly-curved path where the velocity as a function of time in the inertial observer's frame is some function v(t), the proper time along this path between two moments t₀ and t₁ (which could be the moments of departing from and reuniting with a twin, for example) would be given by the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2}\,dt$$ (because in calculus an integral is just the limiting case of a sum in which the size of each segment dt becomes arbitrarily short)

 

[AdVen]

You are wrong, this is sadly a very common misconception. SR deals with accelerated frames. See the many attachments in my blog on this very subject.

Where can I find your blog?

 

[AdVen]

Accelerated motion can be analyzed just fine from the perspective of an inertial frame (and technically a modern physicist would generally say that even if you use a non-inertial frame with a pseudo-gravitational field as discussed here, this is still 'SR' if spacetime is not curved). Any smoothly-curve path in spacetime can be approximated as a polygonal path made up of a series of straight segments of constant velocity joined by instantaneous accelerations, and if you take the limit (in the calculus sense) as the segments become shorter and shorter (and the number of segments becomes greater and greater), this should approach perfect agreement with the original smooth path. If a given straight segment has a velocity v as seen in the inertial observer's frame, and the time between the beginning and end of the segment in the inertial observer's frame is dt, then the proper time along that segment should be $$\sqrt{1 - v^2/c^2} dt$$. So, if you have a smoothly-curved path where the velocity as a function of time in the inertial observer's frame is some function v(t), the proper time along this path between two moments t₀ and t₁ (which could be the moments of departing from and reuniting with a twin, for example) would be given by the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2}\,dt$$ (because in calculus an integral is just the limiting case of a sum in which the size of each segment dt becomes arbitrarily short)

Thank you very much for your clear explanation.

 

[AdVen]

Here

Hello starthaus,

I copied from the Twin Paradox in Wikipedia the following part:

Consider a spaceship traveling from Earth to the nearest star system outside of our solar system: a distance $$d = 4.45$$ light years away, at a speed $$v = 0.866c$$ (i.e., 86.6 percent of the speed of light). The Earth-based mission control reasons about the journey this way (for convenience in this thought experiment the ship is assumed to immediately attain its full speed upon departure): the round trip will take $$t = 2d/v = 10.28$$ years in Earth time (''i.e.'' everybody on Earth will be 10.28 years older when the ship returns). The amount of time as measured on the ship's clocks and the aging of the travelers during their trip will be reduced by the factor $$\epsilon = \sqrt{1 - v^2/c^2}$$, the reciprocal of the (Lorentz factor). In this case $$\epsilon = 0.500 \,$$ and the travelers will have aged only 0.500*10.28 = 5.14 years when they return.

The ship's crew members also calculate the particulars of their trip from their perspective. They know that the distant star system and the Earth are moving relative to the ship at speed $$v$$ during the trip. In their rest frame the distance between the Earth and the star system is $$\epsilon d = 0.5d$$ = 2.23 light years ((length contraction)), for both the outward and return journeys. Each half of the journey takes $$2.23/v$$ = 2.57 years, and the round trip takes 2*2.57 = 5.14 years. Their calculations show that they will arrive home having aged 5.14 years. The travelers' final calculation is in complete agreement with the calculations of those on Earth, though they experience the trip quite differently.

If a pair of twins are born on the day the ship leaves, and one goes on the journey while the other stays on Earth, they will meet again when the traveler is 5.14 years old and the stay-at-home twin is 10.28 years old. The calculation illustrates the usage of the phenomenon of length contraction and time dilation to describe and calculate consequences and predictions of Einstein's (special theory of relativity).

and I finally understood the solution of the twin paradox. I am going to work this out with reference to the length contraction and time dilation formulas and will show you the result as soon as I am finished.

I am very grateful to you.

 

[starthaus]

Hello starthaus,

I copied from the Twin Paradox in Wikipedia the following part:



and I finally understood the solution of the twin paradox. I am going to work this out with reference to the length contraction and time dilation formulas and will show you the result as soon as I am finished.

I am very grateful to you.

You are welcome.

 

[starthaus]

Where can I find your blog?

Here : https://www.physicsforums.com/blog.php?u=241315

 

[stevmg]

starthaus -

I have read Taylor/Wheeler Spacetime Physics the First edition (1962 version) which is easier for a novice to comprehend than the later 2002 2nd edition as the later edition has too many bells and whistles. The First Edition is more to the point. Remember I have no one else around me to even discusss this subject matter with other than this Forum and so digesting the material can get slow.

The third chapter goes into GR. It explains the curvature of spacetime as the source of gravity. What is not explained in any text that I ever read or any comment anywhere is that what would make an object at rest (I know, there is no such thing as "at rest") appear to be pushed (curved worldline)?

Then it dawned on me - every object anywhere has a worldline that is forever growing and hence, other than the old general saying that "everything's in motion" and all frames of reference are relative to each other with no central one favored, there is the motion - presumably all the worldines would be traveling along a geodesic (I guess that's the right term) and there would always appear to be a "force" acting on them as all geodesics are curved.

I didn't have anybody here to tell me that and I never saw it anywhere else. All the 2D analogies which showed by bending a 2D world in a third dimension, objects would appear to be pushed together as they moved (page 184 of this 1st Edition), never showed why they would move in the first place.

In a 3D world, the unseen 4th dimension, time, makes them move (i.e. - the worldline.)

I hope I am correct.

 

[AdVen]

starthaus -
Remember I have no one else around me to even discusss this subject matter with other than this Forum and so digesting the material can get slow.

I am very curious about your situation:

starthaus -
I have no one else around me to even discusss this subject matter with other than this Forum.

Could you tell me more about it? Just for the record.

 

[stevmg]

I am a retired AF officer and delving into matters I never had time to do before. Have a math (not physics) background and my math degree is from 1963.

 

[stevmg]

I am very curious about your situation:


Could you tell me more about it? Just for the record.

I gave you the answer, now tell me the same about you...