The Lorentz Transformations and vector components

1. Feb 27, 2007

anantchowdhary

If a rod is traveling with a velocity 'v' and its proper length is $$L_0$$

will the lorentz transformations given below hold true for the length contraction

$$L_0 = \frac {L} { \sqrt {1 - \frac {{v_x}^2} {c^2}}}$$

$$L_0 = \frac {L} { \sqrt {1 - \frac {{v_y}^2} {c^2}}}$$

Last edited by a moderator: Feb 27, 2007
2. Feb 27, 2007

Integral

Staff Emeritus
No, that is incorrect, first you apply the Gamma transform in one dimension. Second your formula (did you try it?) does not contract, it expands.

Let v= .9c what does your formula predict for the length?

OH! pardon me, did you really want to express $L_0$ as a function of v ? Or did you mean to write:

$$L =\frac {L_0} \sqrt {1 - {( \frac v c )^2}}$$

Last edited: Feb 27, 2007
3. Feb 27, 2007

anantchowdhary

Err...i meant to write L to be the Length as observed by an observer who is observing the moving rod

4. Feb 27, 2007

anantchowdhary

and if the rod is going at an angle relative to the x axis or to the y axis,then wouldnt we first calculate the contraction of its 'x' component and then see the new length?

5. Feb 27, 2007

Integral

Staff Emeritus
In special relativity the motion is assumed to be aligned with the x axis. The Lorentz transform of the y axis is

y= y'

6. Feb 27, 2007

Integral

Staff Emeritus
Opps! I kinda messed that up, this why I do not like to work off the top of my head. (I do not have a reference handy)

$$L = {L_0} \sqrt {1 - {( \frac v c )^2}}$$

There, now we have contraction!

7. Feb 27, 2007

anantchowdhary

ok.But what of we hav a rod travelling at an angle?How would we calculate the length conrtraction without taking its components?

8. Feb 27, 2007

neutrino

That is something we use for simplicity, isn't it? It makes deriving formulae easier.

9. Feb 27, 2007

Integral

Staff Emeritus
Since in Special Relativity, motion is linear, you can always align the axis with the motion. Why wouldn't you?

10. Feb 27, 2007

anantchowdhary

i agree with neutrino.But why isnt the calculation of the contraction taking components correct?

for example,as u said if we assume the motion to be in the direction of the x axis,then we can simply say the relative velocity along the y axis is zero

Therefore
$$L_0_y = \frac {L_y} { \sqrt {1 - \frac {{0}^2} {c^2}}}$$

would giv us $$L_0_y=L_y$$

therefore there isnt any contraction in the direction of the 'y' axis!

Last edited: Feb 27, 2007
11. Feb 27, 2007

Integral

Staff Emeritus
Your equation, as written, is pretty useless. It seems to imply that L 0 is a function of L. Do you really want to say that L 0 is changing?

An object only contracts in the direction of motion. That is a simple fact, perhaps you need to look a bit more carefully at the FULL set of Lorentz transforms.

12. Feb 27, 2007

anantchowdhary

PLease tell me then,how would we calculate the length contraction if the object is moving with an angle with respect to the horizontal?

13. Feb 27, 2007

Integral

Staff Emeritus
Once again, align your coordinate system with the direction of linear motion.

14. Feb 27, 2007

anantchowdhary

ok,ill ask a question related to the topic:

If we hav say an equilateral trinagle(a cardboard cutout) and its travelling with a certain velocity with respect to a person A.Would he observer the angles to be any different from 60 degrees?(when compared to the proper angles)

15. Feb 27, 2007

MeJennifer

Obviously for three or more objects with different velocities, except for trivial cases, that would not be the case.

In such cases you can use the velocity addition formula.
Note, however, that this formula is neither commutative or associative except for trivial cases!

Last edited: Feb 27, 2007
16. Feb 27, 2007

bernhard.rothenstein

a simpler problem but instructive

I used to solve with myh students the follolwing problem. Consider a rod of given proper length at rest in I. It makes a given angle (theta') with the positive direction of the O'X' axis. Find out its length measured by observers of the I frame relative to which it moves with V.

17. Feb 27, 2007

anantchowdhary

Umm...according to me,the radius of the rod(cylindrical) should change...correct?Are u a teacher?

18. Feb 27, 2007

neutrino

I thought it was at rest w.r.t I?

Or did you mean to say "length measured by observers of the I' frame relative to which it moves with V." If that's the case, will you accept this soltuion?

$$L_{x} = \frac{L_{0x}}{\gamma\left(v_{cm}\right)} = \frac{L_{0}\cos{\theta}}{\gamma\left(v_{cm}\right)}$$

$$L_{y} = \frac{L_{0y}}{\gamma\left(v_{cm}\right)} = \frac{L_{0}\sin{\theta}}{\gamma\left(v_{cm}\right)}$$

$$\therefore L = \sqrt{L_{x}^{2} + L_{y}^{2}} = \frac{L_0}{\gamma}$$

Edit: After reading this post by jtbell, I think I might've made a mistake.

Last edited: Feb 27, 2007
19. Feb 27, 2007

anantchowdhary

I dont think uve made a mistake!I think only the radius of the cylinder should change,as the direction of motion of the rod is in line with the x axis only the rod is at an angle but travelling at the same alignment

20. Feb 27, 2007

Staff: Mentor

neutrino's analysis is in error; as jtbell explains, only the component of length parallel to the direction of motion will contract.
What are you talking about with the radius of the cylinder? Assume it's a thin rod. A moving observer will measure both the length and the angle of the rod to be different, compared to observations made in the stick's proper frame.

21. Feb 27, 2007

anantchowdhary

err...i still cant get this.Why would the angle and length be different.The rod is still travelling parallel to the x axis.Even though it is at an angle,its motion is not at an angle with the x-axis isnt it?

22. Feb 27, 2007

anantchowdhary

Why is there an error?Also,if ive understood the question correctly, the rod isnt a vector!Its velocity is a vector which is just parallel to the x axis.

Thank you

23. Feb 27, 2007

bernhard.rothenstein

the oblique rod poblem

In my statement of the problem consider please that the rod (not a cylinder) is at rest in I'. Take into account that distances measured perpendicular to the direction of relative motion are invariants and that the OX(O'X') components contract
L(x)=L(0)cosw'/g(V)
L(y)=L(0)sinw'
where L(0) stands for the proper length, w' for the angle made by the rod with the positive direction of the overlaped axes and g(V) is the gamma factor. In order to simplify the problem consider that one of the ends of the rod is located at the origin O' of the I' frame. The fact that you consider the rod as a vector does not change the problem.
In order to finish apply Pythagoras' theorem. As an interesting fact notice that when detected from I the rod is rotated and you can find out the angle w measured by observers from I
tgw=L(y)/L(x).
Expressing the right side of that equation as a function of physical quantities measured in I' we obtain the relationship between w and w'.
I used to ask my students capisci?

24. Feb 28, 2007

Staff: Mentor

You will agree that any object has dimensions parallel and perpendicular to its relative velocity vector, right? In this case, the velocity is parallel to the x-axis, so all length measurements parallel to the x-axis will be contracted. (And length measurements perpendicular to the x-axis will remain unchanged.)

Rather than discuss this same issue in multiple threads, followups should go here: https://www.physicsforums.com/showthread.php?t=158375". (Especially since jtbell has given the explicit calculation in that thread.)

Last edited by a moderator: Apr 22, 2017