Trouble with Lorentz transformations

[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'.

 

[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.