How Do Lorentz Transformations Work with Y and Z Components of Speed?

[quasar987]

In all the textbooks I read on SR, they always list the LT assuming y'=y and z'=z. But how does the time coordinate transform if the speed has a y and a z component?

I'm guessing

$$t' = \frac{t-(v_x x + v_y y +v_z z )/c^2}{\sqrt{1-v^2/c^2}}$$

 

[da_willem]

I think so too.

 

[pmb_phy]

In all the textbooks I read on SR, they always list the LT assuming y'=y and z'=z. But how does the time coordinate transform if the speed has a y and a z component?

I'm guessing

$$t' = \frac{t-(v_x x + v_y y +v_z z )/c^2}{\sqrt{1-v^2/c^2}}$$

Yes. Absolutely.

Pete

 

[JesseM]

Yes. Absolutely.

Pete

Did you actually verify this using the Poincare transformation? I tried starting to figure this out by imagining a ruler moving diagonally with clocks mounted on it that were synchronized by light-signals in the ruler's own frame, and things were coming out pretty complicated, although I haven't finished the calculation yet.

 

[pmb_phy]

Did you actually verify this using the Poincare transformation?

This is the Lorentz transformation we're talking about. Not the Poincare transformation.

I tried starting to figure this out by imagining a ruler moving diagonally with clocks mounted on it that were synchronized by light-signals in the ruler's own frame, and things were coming out pretty complicated, although I haven't finished the calculation yet.

Its a standard equation which can be found it many relativity texts. However I did doublecheck it in Classical Electrodynamics - 2nd Ed., J.D. Jackson. I also recall verifying it myself in the past.

What do you mean by a ruler moving diagonally? Remember that this equation still relates to axes between two systems for which the axes are parallel. Only thing that's changed is the direction of motion of the origin.

Note that $v_x x + v_y y +v_z z$ can can be replaced by [itex]\beta r[/tex] where $\beta r$ is the dot product of $\beta$ = v/c and r = (x, y, z). Then we simply replace

$$t' = \frac{t - (v_x x + v_y y +v_z z )/c^2}{\sqrt{1-v^2/c^2}}$$

with

$$t' = \frac{t - \beta r/c^2}{\sqrt{1-v^2/c^2}}$$

I.e. all you do is rotate the axes. TGive it a try. Rotations are part of the Lorentz group. Use matrices for this transformation. It should make it simple. I think that will give the correct result. Give it a whirl and let me know how it works out.

Pete

 

[JesseM]

What do you mean by a ruler moving diagonally? Remember that this equation still relates to axes between two systems for which the axes are parallel. Only thing that's changed is the direction of motion of the origin.

Ah, that's what I got confused about. If you actually rotate the axes, would you have to use the Poincare transformation then? I was under the impression that the Lorentz transform could only be used when the axes are parallel and the origins of the two systems coincide, and the Poincare transformation is more general and can deal with systems not covered by the Lorentz transform, but maybe I've got it wrong, I don't think I ever actually studied the Poincare transformation in my college SR class.

 

[Fredrik]

If you actually rotate the axes, would you have to use the Poincare transformation then?

Nope.

Rotations = Those Lorentz transformations that leave the t coordinate unchanged

Lorentz transformations = Poincaré transformations with zero translation

 

[pmb_phy]

Ah, that's what I got confused about. If you actually rotate the axes, would you have to use the Poincare transformation then? I was under the impression that the Lorentz transform could only be used when the axes are parallel and the origins of the two systems coincide, and the Poincare transformation is more general and can deal with systems not covered by the Lorentz transform, but maybe I've got it wrong, I don't think I ever actually studied the Poincare transformation in my college SR class.

Lorentz transformations are those transformations of the form x' = Lx where x',x are column vectors and L is a matrix. Poincare transformations are those transformations of the form x' = Lx + a where a is a point in spacetime.

No. Lorentz transformations can represent transformations such as rotating axes. give it a try. Employ the usual coordinate transformation for a rotation of axes. You'll see what I mean.

Pete

 

[jcsd]

Within each spatial slice the chocie of the x coordiante is arbiatry and is usually choosen for convenience, so the product of a spatial rotation and a Lorentz boost (a Lorentz boost is a change in velocity) is also a Lorentz transformation (as has really already been pointed out). The Lorentz transformation is best viewed as a generalized 'rotation' in spacetime (or at least that's how I like to view it at an intuitve level YMMV)

 

[quasar987]

In all the textbooks I read on SR, they always list the LT assuming y'=y and z'=z. But how does the time coordinate transform if the speed has a y and a z component?

I'm guessing

$$t' = \frac{t-(v_x x + v_y y +v_z z )/c^2}{\sqrt{1-v^2/c^2}}$$

Mmmh, and also in this case, the transformations of position have different values of gamma, right? For exemple

$$y' = \gamma_y (y - v_y t)$$

where

$$\gamma_y = \frac{1}{\sqrt{1-v_y^2/c^2}}$$

 

[dextercioby]

Nope,IIRC,for all possible cases of LT,there's only one "gamma",namely the one with the "beta" squared in the denominator.

Daniel.

 

[quasar987]

Are you certain? It seem very weird that if we have a square, moving in the xy plane at speed near c with all of its speed in the x direction, its height suffers no contraction at all. But if most of the component of speed is in the x direction and very little is in the y direction. Then the height is contracted just as much as the width, and just as much as if most of the speed were in the y direction.

 

[dextercioby]

I'm absolutely sure.I've checked my EM course and,for arbitrary LT,it's still the "old" gamma involved.
You may want to check Jackson,if u don't believe me...

Daniel.

 

[quasar987]

Ok, I believe you.

Another weird effect of this is that if we look at the square in a way such that its speed only has an x component, it is a rectangle. But if we tilt our head/coordinate system just a tiny winy bit (or by any amount in $]0,\pi[$ for that matter), it instantaneously becomes a tiny square.

There is a discontinuity in the process of transformation of the figure from rectangle to tiny square.

 

[jcsd]

Are you certain? It seem very weird that if we have a square, moving in the xy plane at speed near c with all of its speed in the x direction, its height suffers no contraction at all. But if most of the component of speed is in the x direction and very little is in the y direction. Then the height is contracted just as much as the width, and just as much as if most of the speed were in the y direction.

It is contracted in the direction of motion, so no height and width are not contracted equally as much as that suggests that the square shrinks, infact in the case where some of it's motion is not parallel to one of the square's sides it will no longer be a square in that frame.

 

[quasar987]

It is contracted in the direction of motion, so no height and width are not contracted equally as much

But aren't the contraction governed by

$$\Delta x = \frac{\Delta x'}{\gamma}$$

$$\Delta y = \frac{\Delta y'}{\gamma}$$?

For equal gammas and equal sides of the square in its rest frame (S'), $\Delta x = \Delta y$

 

[jcsd]

This would suggest that Lorentz-FitzGerald contraction is not just depednent on the velocity, but on spatial orienation also which is not the case. If the square shrinks then there MUST be contraction in the direction perpendicular to the motion of the object (which is actually can be shown to be in violation the postulates of relativity). What happens when the motion is not parallel to a set of the square side is thay the square appears to be a parallelogram.

see the attachment for illustration of this (it's just a quick sketch done in paint so obviously it's not 100% accuarte).

 

[quasar987]

I see, thanks for the sketch it helps a lot. But are the equations I wrote in the last post wrong?

 

[selfAdjoint]

I see, thanks for the sketch it helps a lot. But are the equations I wrote in the last post wrong?

Since the motion you describe is along a diagonal you should put in a factor of $$\sqrt{2}/2$$ in each one

 

[jcsd]

You just need to use a bit of trig.

$$\Delta x'=\sqrt{\frac{1}{\gamma^2}\Delta x^2 \cos^2 \theta + \Delta x^2 \sin^2\theta}$$

Where theta is the angle that the length Delta x makes with the direction of motion (I've not simplifed the formula in odrer that you can see where all the bits come from).

edited to correct equation.

 

[jcsd]

As that square could be a square on a coordinate grid, you should also see that when you perform a Lorentz boost on an orthogonal coordinate system where none of the basis vectors are parallel to v you get a skew (non-orthgonal) coordinate system.

 

[quasar987]

Why did you change the expression of $\Delta x'$ in post #20 ? I understood it before you changed it but not anymore.

 

[quasar987]

You may want to check Jackson

Do you know what page and what's the entire title of his book?

 

[jcsd]

Because Delta x is the proper length (and I tidied it up a little so that it's easier to see how it could be simplified) and Delta x' is the length in the frame where the length is moving so when theta = 0 Delta x' = gamma*Delta x.

 

[jcsd]

oops yes you're right I didn't really think about it did I.

 

[quasar987]

Alright, now I understand this thing. I understand it in terms of proper lenghts and lenghts contraction.. but how does it follow from Lorentz transformations?! My first relativity teacher in high school told us the complete set of Lorentz transformations was, if I remember correctly,

$$x' = \gamma (x - v_x t)$$
$$y' = \gamma (y - v_y t)$$
$$z' = \gamma (z - v_z t)$$

Plus the time coordinate which I wrote in firts post. Are these right? If so, how does the contraction formula you state follows from these?

 

[reilly]

This issue is very basic. As pmb_phy mentions, rotation is the key. That is, prior to the LT, rotate so that the velocity is along a coordinate axis -- does not matter which one. Then do the LT. This type of manipulation is fundamental to much of particle physics. Whenever possible, great simplicity obtains when one works with the Center of Momentum frame for the system of interest. Relativistic kinematics as used in much of physics, is based on both LTs and rotations. (Sometimes LTs are referred to as boosts; stemming from the nomenclature of the Lorentz and Poincare groups.) Also note that in many treatments, vectors rather than components are used.
Regards,
Reilly Atkinson

 

[dextercioby]

Do you know what page and what's the entire title of his book?

J.D.Jackson:"Classical electrodynamics" (any edition),second volume,i can't remember the page,but it has to be there...

Daniel.

 

[quasar987]

This issue is very basic. As pmb_phy mentions, rotation is the key. That is, prior to the LT, rotate so that the velocity is along a coordinate axis -- does not matter which one. Then do the LT.

That's fine is you want to compute the contracted length of just one object. but if you have many object with many different orientations and you want to find their lenghts withouth switching coordinate system (i.e. without using jcsd's pythagore trick). What are the equations that tell you this?

Are the ones I have written down correct or incorrect, that's what I'm asking.

 

[quasar987]

They are incorect. Only the time transformation is ok. The transformations for x', y' and z' are super long and complicated.