Trouble with Lorentz transformations

[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'.

 

[starthaus]

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'.

$$dt'$$ is a shorthand for your $$t'_1-t'_2$$
$$dx'$$ is a shorthand for your $$x'_1-x'_2$$

 

[AdVen]

$$dt'$$ is a shorthand for your $$t'_1-t'_2$$
$$dx'$$ is a shorthand for your $$x'_1-x'_2$$

Thanks a lot for your explanation. However, I am very sorry to say, that I still do not understand what it is, that you are trying to say. I have understood, that the condition $$x_1=x_2$$ should be satisfied if the clock is at rest in frame S

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

and the condition $$x'_1=x'_2$$ should be satisfied if the clock is at rest in frame S'

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

 

[JesseM]

Thanks a lot for your explanation. However, I am very sorry to say, that I still do not understand what it is, that you are trying to say. I have understood, that the condition $$x_1=x_2$$ should be satisfied if the clock is at rest in frame S

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

and the condition $$x'_1=x'_2$$ should be satisfied if the clock is at rest in frame S'

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

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.

 

[starthaus]

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.

Precisely.

 

[stevmg]

For me, the novice, the dt = dt'gamma makes intuitive sense in that the "moving" frame (that's where dt' is) measures (in time) shorter than the "static" fame (where dt is.)

The algebraic (JesseM, Adven) or calculus (starthaus) derivations speak for themselves and they do not need to be rehashed here.

 

[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.