Are the Lorentz transformation formulas on wikipedia correct?

[robphy]

Sorry for jumping in late and being too lazy to read everything up to this point...
But isn't the essence of the issue already seen in the Galilean Transformation?
$\left(\begin{array}{rr} 1 & 0 \\ -v & 1 \end{array}\right)^{-1} = \left(\begin{array}{rr} 1 & 0 \\ v & 1 \end{array}\right)$
(Trying to resolve it with Lorentz Transformations adds unnecessary complication to the issue.)
Implicitly, the observers face the "same x"-direction and agree that that direction is positive.

The issue also arises with rotations.

Maybe $v$ needs some extra notation to clarify its meaning and who is doing the measuring.
$\left(\begin{array}{r} t^{Bob} \\ x^{Bob} \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ -v_{Bob}{}^{Alice} & 1 \end{array}\right) \left(\begin{array}{r} t^{Alice} \\ x^{Alice} \end{array}\right)$

Maybe I'm missing some aspect of the issue at hand...

 

[Nugatory]

How would an Observer who has not communicated with the other to decide who is A and who is A' know to use the inverse or non inverse LT formula when transforming an event? He couldn't, so you must be wrong.

You are still misunderstanding what a transformation does, perhaps because so many presentations introduce an observer into every frame. Those observers are just there to make the implications of the math a bit easier to visualize - neither observers nor communication between observers is necessary.

Here's the observer-free language:
We have a coordinate system that labels every point in flat spacetime with coordinates x, y, z, and t. We have a second coordinate system that labels every point in the same spacetime with coordinates x', y', z', and t'. How do we write x', y', z', and t' as functions of x, y, z, and t and vice versa? The Lorentz transformations in the wikipedia article are the answer to that question for the case in which the origin of the second coordinate system is moving with speed v along the x-axis (and some other conditions are met).

 

[stevendaryl]

Of course something as energy and lengths does not require any communication because a length is always positive and so is energy, but you are wrong in that there is no communication required between two observers traveling at vrel when handling it with the LTs given by wikipedia. How would an Observer who has not communicated with the other to decide who is A and who is A' know to use the inverse or non inverse LT formula when transforming an event? He couldn't, so you must be wrong.

I'm not exactly sure what point you are making. But you are right--If I see an alien creature hurtling through space at some velocity \vec{v}, there is no way for me to write down a transformation relating my coordinate system to the coordinate system of the alien. To figure out the coordinate system of the alien, we need to know:

1. Is he using isotropic inertial coordinates? (Isotropic for our purposes just means for these purposes that the coordinates are such that the speed of light is the same in every direction)
2. What is the origin of his coordinate system?
3. What directions does he consider his x-axis, y-axis and z-axis?
4. Does he consider himself to be at rest? (That's the usual assumption)
5. What units is he using for distance and time?

You probably can't answer those questions without talking to the alien. But given the answers to those questions, you can uniquely determine a transformation between his coordinates and yours. And given the same information about you, he can uniquely determine a transformation between your coordinates and his. What's not necessary is for the two of you to decide whose coordinates get a "prime" on them.

 

[stevendaryl]

Sorry for jumping in late and being too lazy to read everything up to this point...
But isn't the essence of the issue already seen in the Galilean Transformation?

[stuff deleted]

The issue also arises with rotations.

[stuff deleted]

Maybe I'm missing some aspect of the issue at hand...

No, you're not. The issues brought up by Jeronimus are just as relevant (and as trivial) for Galilean transformations and rotations.

 

[PeterDonis]

The involution formulas i am imagining will be universal and will only require to be fed with the proper vrel between the TWO systems you want to transform.

Suppose this is true (again, I haven't tried to verify any of your proofs). That still doesn't change anything I said, or address any of the issues I raised. Your transformations are not Lorentz transformations; they are Lorentz transformations + parity reversals. Those are perfectly good transformations mathematically speaking, but they do not have certain properties that the group of Lorentz transformations alone has. Most physicists want to use a group of transformations that has those properties; that's why they use the standard Lorentz transformations instead of yours.

 

[PeterDonis]

i would like to see those involution formulas (for x,x' and t,t') i am imagining exist

Compose the formulas for standard Lorentz transformations with the formulas for a parity reversal. The latter is simply $x' = -x$, with all other coordinates unchanged.

 

[Ibix]

I think Peter is saying the same thing I said in #22. You just tag a minus sign on to the forward x transform and you need to be careful about what v means in the inverse transform.

 

[Jeronimus]

Compose the formulas for standard Lorentz transformations with the formulas for a parity reversal. The latter is simply $x' = -x$, with all other coordinates unchanged.

Before i end up typing more garbage, can you give us an example how this would work for x=10ls(lightseconds) t=3s v=0,5c for example, and how this would be an involution where going from x to x' and back from x' to x would work with just one involution formula?

 

[pervect]

One might also consider the question:

Are the Gallilean transforms and their inverses correct in the Newtonian limit, i.e.

$$x' = x - vt \quad t' = t \quad \quad x = x' + vt' \quad t = t'$$

The answer is yes, they are. To illustrate, it's basic algebra that if x' = x - vt then x = x' + vt (add vt to both sides). By substituting t=t', we then get x = x'+vt'

 

[DrGreg]

Post #22 tells you what to do, but to spell it out more explicitly, take these equations:$$
\begin{align*}
t' &= \gamma \left( t - \frac{vx}{c^2} \right) \\
x' &= \gamma \left( x - vt \right) \\
t'' &= t' \\
x'' &= -x' \\
\end{align*}
$$and get expressions for $t''$ and $x''$ in terms of $t$ and $x$. Then rearrange the equations to get $t$ and $x$ in terms of $t''$ and $x''$. You will need to make use of the fact that $\gamma^2 ( 1 - v^2 / c^2 ) = 1$. All of the above is just algebra; no physics required.

 

[Herman Trivilino]

For example, what you said about velocities. If one astronaut sees the other approaching from the front, hence negative velocity, the other will see the same from his perspective. Hence, also negative velocity.

Yes, but note that the $x$-axis and the $x'$-axis point in opposite directions. The Lorentz transformation equations in the form you see them in Wikipedia and in virtually every textbook have those two axes pointing in the same direction. The reason for this is because of the purpose of the equations. Their purpose is to describe the motion of objects. In every example you've given the object of interest is at rest in one of the frames. More generally, that's not true. The object is in motion in both frames. Using the common form of the Lorentz transformations, the sign of that object's velocity is the same in both frames. Thus, if its velocity is positive in the unprimed frame, it's also positive in the primed frame.

As Ibix said, you can use a different form of the transformation equations if you like, but it will introduce complexities that aren't there when you use the common form. In other words, it's a matter of convention. When things are a matter of convention we tend to choose a convention that makes things less complicated, although this often backfires. For example, the two forms of the First Law of Thermodynamics. One of them is more convenient in some situations, the other in other situations, so you find some authors using one convention, and other authors using the other. One is sometimes called the European convention (the convention used in most chemistry textbooks) where $\Delta U=Q+W$. The so-called American convention (used in most physics textbooks) is $\Delta U=Q-W$.

If astronaut A measures a ruler at length L but astronaut A' measures that ruler which is moving at Vrel to be shorter by a factor 1/γ, hence L' = L*(1/γ)

Let's say, for purposes of keeping the discussion uncluttered, that ruler is at rest in the unprimed frame.

then a ruler which A' measures to be of the length L₂' = L will be measured by astronaut A to be shorter by the same factor 1/γ.

Such would be the case, for example, if the ruler were at rest in the primed frame. Thus the two rulers are identical. Each observer concludes his ruler has a length $L$ and his partner's has a length $\frac{L}{\gamma}$.

Continuing this line of reasoning, it follows (unless my mind plays tricks on me) that

If A measures an event e1 at x=5 and t=10 and A' measures this event to be "happening" at let's say e1' at x'=3 t'=11 (no calculations made, just fictive numbers to make the point), then it would follow that if A' was to measure an event e2' at x'=5 and t'=10, A would measure this event to be happening at e2 with coordinates x=3 t=11.

I not only don't see how it follows, I don't see how it's even possible. The Lorentz transformations tell you what's possible. The fact that they don't work in your scenario tells us that your scenario is not possible.

A much simpler way of writing the coordinates of these events is ...

$x_1=5$, $t_1=0$; $x'_1=3$, $t'_1=11$.

$x_2=3$, $t_2=11$; $x'_2=5$, $t'_2=10$.

You could use the values given in the first sentence (corresponding to the first event) to solve one of the transformation equations for $v$. If you do the same with the other transformation equation, you get a different value for $v$. You can remedy this by writing a different pair of transformation equations that follow a different convention.

But you cannot then use those same transformation equations and in the second sentence (corresponding to the second event). You will get another pair of values for #v## and there's no freedom left to get them to match each other. The fact that they don't match what you got from the first event is the issue you seem to be illustrating, but the problem is worse than that. It won't work because it's not physically possible for two different frames of reference to measure those coordinate values for any pair of events. For example, it would require the two frames to have a faster-than-light relative velocity.

 

[robphy]

Continuing on the "parity" notion introduced by @PeterDonis in #24...

In the boosts, we have (in the simple case) that $y'=y$ and $z'=z$...that is, their corresponding y and z axes coincide.*
In the "symmetric transformation sought by the OP", the parity flip of the x'-axis (noted by @PeterDonis)
would also flip the handedness of the spatial-axes from right-handed to left.

(Off-topic but related...This whole discussion has some similarities to "trying to hide minus signs (or negative numbers)" in analyzing circuits [or free-body diagrams or optics ray diagrams] by talking about magnitudes and directions, rather than just let the quantity be a signed-quantity with reference to an "arbitrary but chosen once and for all" direction.)

[*As vectors in spacetime, the observer's y and y' axes do coincide, and similarly for z and z'.
However, (as vectors in a spacetime diagram) the observer's x and x' axes [in standard form] don't coincide, although they "point" in the same forward direction.]

 

[Dale]

For this to be the case, we could not possibly use the two _different_ formulas we get on wikipedia.

Of course not. As I said before, the wikipedia formula very clearly specifies the assumptions. Specifically, the axes are assumed to be in the standard configuration. I.e. they are facing the same way.

a symmetrical scenario, both astronauts facing each other (or both having their backs turned towards each other)

You have the axes rotated by 180 degrees, so you would have to use a different formula.

Again, the wikipedia formula is correct. It makes explicit assumptions and if you choose to make different assumptions then it won't apply. Not because the formula is wrong, but because it doesn't apply to your assumed scenario.

 

[Jeronimus]

Of course not. As I said before, the wikipedia formula very clearly specifies the assumptions. Specifically, the axes are assumed to be in the standard configuration. I.e. they are facing the same way..

Yes, and as i said before, there is something terribly wrong with the reasoning of the wikipedia article. Of course the results are "right" using the non-inverse and inverse LTs, but the reasoning is wrong.

I am not questioning arriving at proper results. The reasoning is false as i stated before.

"This "trick" of simply reversing the direction of relative velocity while preserving its magnitude..." - Wikipedia *snap*" is simply a wrong statement. There is no direction for relative velocity.

Also, i believe that handling the mapping of x to x' and back from x' to x in the manner it is done, without considering the two symmetrical cases as i described them:...

Unfortunately i was sloppy in explaining the "facing" part.

The derivation for one of the symmetric cases will be astronauts facing each other as they approach each other and once they pass each other, they will of course have their backs turned against each other.

The other symmetric case would be astronauts having their backs turned against each other as they measure each other to be approaching from the negative x-asis and then when they just pass each other they will be facing each other, measuring each other to be moving away towards the positive axis.

...and their importance, without using "tricks", is clouding the understanding of SR.

 

[Herman Trivilino]

"This "trick" of simply reversing the direction of relative velocity while preserving its magnitude..." - Wikipedia *snap*" is simply a wrong statement.

What one is actually doing when changing the sign is reversing the direction of a component of the velocity. This is a liberty that virtually all textbook authors take, and as has been mentioned already in this thread, not unique to a Lorentz transformation. Velocity is a vector, expressed using the notation $\vec{v}$ but authors get sloppy when they use the notation $v$. Sometimes it means the magnitude of the velocity, in which case it's never negative, and sometimes it means the component of the velocity along an axis that's parallel to the direction of motion, in which case it can be either positive or negative, depending on whether the position numbers on that axis are increasing or decreasing..

There is no direction for relative velocity.

Of course there is. It can be towards or away, up or down, left or right. When I'm navigating my vehicle through traffic I pay very close attention to the direction of my vehicle's motion relative to other vehicles, the roadway, people walking, and objects anchored to the ground, such as buildings, fences, and guard rails.

 

[PeterDonis]

there is something terribly wrong with the reasoning of the wikipedia article

No, there isn't. The difference between what the Wikipedia article is describing--the standard Lorentz transformation--and the different kind of transformation you have in mind has been explained repeatedly in this thread. At this point you are simply repeating your erroneous opinion without responding to what others have been telling you, so there is no point in continuing the discussion. Thread closed.