Lorentz Transformations and their Inverse

[NanakiXIII]

Ah, I see. Well, though it hasn't come through fully just yet, I do believe my mistaken thoughts lie with the relativity of simultaneity, as you pointed out in post #26. Some pondering and going over what new knowledge I acquired is probably in place for now. Thanks to everyone who replied.

 

[NanakiXIII]

Well, this next question is about a somewhat different topic, but still about the Lorentz transformations. A little bit down the page in the post I linked to earlier, there are also used these "generalized" Lorentz transformations. They look to me like two Lorentz transformations simply subtracted, but why would you do that? How does that generalize the transformation?

 

[Hootenanny]

Well, this next question is about a somewhat different topic, but still about the Lorentz transformations. A little bit down the page in the post I linked to earlier, there are also used these "generalized" Lorentz transformations. They look to me like two Lorentz transformations simply subtracted, but why would you do that? How does that generalize the transformation?

In the Lorentz Transformations (LT) you have met so far, we have assumed that at t=t'=0 then x=x'=0 , in other words the origins of both reference frames (spatial and temporal coordinates) coincide at the start. The more general version jtbell quoted allows us to consider situations where the origins of the two reference frames do not coincide at t = 0. It may be more intuitive if you consider time and position intervals or changes in time and displacement (as Berhard mentioned previously). If we take jtbell's general LT for position;

$$x_1^{\prime \prime} - x_0^{\prime \prime} = \gamma [(x_1 - x_0) - v (t_1 - t_0)]$$

$$\text{Define:}\hspace{1cm}\Delta x^{\prime \prime}:= x_1^{\prime \prime} - x_0^{\prime \prime}\hspace{1cm}\Delta x := x_1 - x_0 \hspace{1cm}\Delta t := t_1 - t_0$$

Then we have;

$$\Delta x^{\prime \prime} = \gamma [\Delta x - v \Delta t]$$

Hence, we obtain an expression for a change in coordinates, i.e. if we travel some distance $\Delta x$ in the S frame, which takes $\Delta t$ seconds. Then in the S'' frame traveling at constant velocity v in the positive x direction, the measured distance will be $\Delta x^{\prime\prime}$. The expression is just a subtraction of two LT's at different coordinates, hence we obtain intervals. Does that make sense?

 

[JesseM]

Well, this next question is about a somewhat different topic, but still about the Lorentz transformations. A little bit down the page in the post I linked to earlier, there are also used these "generalized" Lorentz transformations. They look to me like two Lorentz transformations simply subtracted, but why would you do that? How does that generalize the transformation?

Those equations just more general in the sense that you're no longer assuming the spatial origins of the two coordinate systems (in one dimension, the spatial origins are just x=0 and x''=0) coincide at the time t=t''=0 (which was one of the conditions I mentioned in my first post on this thread), but instead just assuming you know the coordinates of some "reference event" in both coordinate systems (with the reference event having some coordinates $$x_0$$, $$t_0$$ in one system and some other coordinates $$x_0^{\prime \prime}$$ and $$t_0^{\prime \prime}$$ in the other...they could be anything, like $$x_0$$ = 3 light years and $$t_0$$ = 27 years).

 

[NanakiXIII]

Thanks, Hootenanny and JesseM, I believe I understand these generalized transformations now. I'll be testing this understanding soon, but in the mean time I have another question. When substituting the values into the generalized transformation, jtbell comes to the conclusion that x" = -2.4. I figured this was the position of Earth in S''. However, a few lines down:

just after the turnaround, you are 8.2 light years in front of him

Where did this 8.2 come from? Also, if the Earth is now 8.2 ly away, how can the traveling twin bridge this gap in 1.8 years?EDIT: No, I do have another question about those generalized transformations. If we can't use the Lorentz transformations between co-ordinates of S and S'', then why can we use them between intervals on S and S''? What's the difference?

 

[jtbell]

Where did this 8.2 come from?

A typographical error. Thanks for catching it! I've just now corrected it to 2.4.

 

[NanakiXIII]

You're welcome, jtbell, I wouldn't have guessed I'd be right about it. Anyway, if I was right in my previous post, then isn't the following statement in your explanation also false:

Apparently your position has shifted by 5.8 light-years

 

[Hootenanny]

No, I do have another question about those generalized transformations. If we can't use the Lorentz transformations between co-ordinates of S and S'', then why can we use them between intervals on S and S''? What's the difference?

Because you are looking at changes in temporal and spatial coordinates, not absolute coordinates. In the general case, we are looking at the difference between an initial position and time and a final position and time. However, in the case of the other ('non general') LT's, it is a condition that the origins of the two frames coincide at t=t''=0. This does not have to be the case in the general LT's since we are only considering the difference between two defined sets of coordinates. For example, take jtbell's general LT;

$$x_1^{\prime \prime} - x_0^{\prime \prime} = \gamma [(x_1 - x_0) - v (t_1 - t_0)]$$

Now, if we assume that the origins of the two frames (S and S'') coincide at t=t''=0 then this implies that x₀''=0 ; x₀=0 and t₀ = 0 i.e. the origin of the S'' frame lies on the origin of the S frame, both frames start from the same position at zero displacement. Does that make sense?

 

[NanakiXIII]

Yes, but to transform between S and S'', you're still just using the Lorentz transformation. You're saying:

$$\Delta x^{\prime \prime} = \gamma [\Delta x - v \Delta t]$$

And thus:

$$x''=\gamma (x-vt)$$ and $$x''_0=\gamma (x_0-vt_0)$$

Where is my error in this?

 

[Hootenanny]

Yes, but to transform between S and S'', you're still just using the Lorentz transformation. You're saying:

$$\Delta x^{\prime \prime} = \gamma [\Delta x - v \Delta t]$$

And thus:

$$x''=\gamma (x-vt)$$ and $$x''_0=\gamma (x_0-vt_0)$$

Where is my error in this?

There is no error. I never said there was.

 

[NanakiXIII]

If there is no error in what I said, then you're using the non-generalized Lorentz transformation to transform between S and S''.

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

And if that is true, the generalized transformations are redundant. So there must be an error in my previous post, mustn't there?

 

[Hootenanny]

This has already been explained, please reread the following posts

https://www.physicsforums.com/showpost.php?p=1220390&postcount=33"
https://www.physicsforums.com/showpost.php?p=1220390&postcount=34"
https://www.physicsforums.com/showpost.php?p=1220390&postcount=38"

Note what happens to the following LT when the origins coincide, that is when x₀''=0 ; x₀=0 and t₀ = 0

$$x_1^{\prime \prime} - x_0^{\prime \prime} = \gamma [(x_1 - x_0) - v (t_1 - t_0)]$$

 

[bernhard.rothenstein]

inverse LT

So, what I gather is that one way of looking at inverse Lorentz transformation is that there's just a plus sign because the velocity is negative. Am I in any way correct here? (I'm aware that the inverse transformations can be derived from the normal transformations, I'm just looking to know whether the thought described above is correct.)



Is that a Socratic question? I haven't a clue.

Socratic or not, start with
x=g(x'+Vt')
t=g(t'+Vx'/cc)[/
and with some simple algebra solve them for x' and t' in order to obtain the inverse transformaions!B]

 

[NanakiXIII]

Those links don't seem to work for me, but I found the posts anyway. I think I understand if the following is true:

The Lorentz transformation apply per definition to intervals rather than to co-ordinates.

If that is the case, however, how does that show from the derivation of the Lorentz transformations? In this case, I'll have to refer to the only derivation I know, which is Einstein's derivation as described in Relativity, a book he wrote. The derivation can also be found at http://www.bartleby.com/173/a1.html.

Or, if easier, how else could this be proven or at least be made plausible?In reply to bernhard.rothenstein: I am aware of that, I just didn't know what you meant by asking what it was called. Thanks for pointing it out in the first place, though.

 

[MeJennifer]

The Lorentz transformation apply per definition to intervals rather than to co-ordinates.

Be aware that the x and y coordinates in the presented Euclidean-Cartesian frames are not distances and time in the way the world works.

In reality, the physical distance between two objects is the amount of proper travel time taken for an object to go from one to the other.

You cannot take time and space in isolation.

"Henceforth Space by itself, and Time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality."
- Minkowski

Henceforth? Well apparently not by everyone, after more than 100 years of SR many still consider, lengths contracting and time slowing down, in isolation rather than in relation.

 

[robphy]

In reality, the physical distance between two objects is the amount of proper travel time taken for an object to go from one to the other.

Can you clarify? That statement doesn't make much sense to me.

 

[NanakiXIII]

I see your point, MeJennifer, but I'm not quite sure how to relate it to my problem.

 

[JesseM]

Yes, but to transform between S and S'', you're still just using the Lorentz transformation. You're saying:

$$\Delta x^{\prime \prime} = \gamma [\Delta x - v \Delta t]$$

And thus:

$$x''=\gamma (x-vt)$$ and $$x''_0=\gamma (x_0-vt_0)$$

Where is my error in this?

The second doesn't follow from the first--if S and S'' do not share a common origin, the first is true while the second is false. As a simple analogy, consider two cartesian coordinate systems in 2D space, with the x' axis parallel to the x-axis and the y' axis parallel to the y axis, but with the origin of S' (x' = 0, y'=0) located at coordinates x=5, y=8 in the S system. In this case, the coordinate transform is just:

x' = x - 5
y' = x - 8

However, the displacement between a given pair of points is the same in both coordinate systems:

$$\Delta x' = \Delta x$$

The Lorentz transformation apply per definition to intervals rather than to co-ordinates.

No, quite the opposite in fact.