The Lorentz Transformations and vector components

  • #1

Main Question or Discussion Point

If a rod is traveling with a velocity 'v' and its proper length is [tex]L_0[/tex]

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

[tex] L_0 = \frac {L} { \sqrt {1 - \frac {{v_x}^2} {c^2}}} [/tex]

[tex] L_0 = \frac {L} { \sqrt {1 - \frac {{v_y}^2} {c^2}}} [/tex]
 
Last edited by a moderator:

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
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 [itex] L_0 [/itex] as a function of v ? Or did you mean to write:

[tex]L =\frac {L_0} \sqrt {1 - {( \frac v c )^2}}[/tex]
 
Last edited:
  • #3
Err...i meant to write L to be the Length as observed by an observer who is observing the moving rod
 
  • #4
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
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
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
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
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 [itex] L_0 [/itex] as a function of v ? Or did you mean to write:

[tex]L =\frac {L_0} \sqrt {1 - {( \frac v c )^2}}[/tex]
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)

[tex]L = {L_0} \sqrt {1 - {( \frac v c )^2}}[/tex]

There, now we have contraction!
 
  • #7
ok.But what of we hav a rod travelling at an angle?How would we calculate the length conrtraction without taking its components?
 
  • #8
2,063
2
In special relativity the motion is assumed to be aligned with the x axis. The Lorentz transform of the y axis is

y= y'
That is something we use for simplicity, isn't it? It makes deriving formulae easier.
 
  • #9
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
Since in Special Relativity, motion is linear, you can always align the axis with the motion. Why wouldn't you?
 
  • #10
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
[tex] L_0_y = \frac {L_y} { \sqrt {1 - \frac {{0}^2} {c^2}}} [/tex]

would giv us [tex]L_0_y=L_y[/tex]

therefore there isnt any contraction in the direction of the 'y' axis!
 
Last edited:
  • #11
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
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
[tex] L_0 = \frac {L} { \sqrt {1 - \frac {{0}^2} {c^2}}} [/tex]

would giv us [tex]L_0_y=L_y[/tex]

therefore there isnt any contraction in the direction of the 'y' axis!
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
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
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
Once again, align your coordinate system with the direction of linear motion.
 
  • #14
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
1,997
5
Since in Special Relativity, motion is linear, you can always align the axis with the motion. Why wouldn't you?
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:
  • #16
a simpler problem but instructive

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)
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
Umm...according to me,the radius of the rod(cylindrical) should change...correct?Are u a teacher?
 
  • #18
2,063
2
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.
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?

[tex]L_{x} = \frac{L_{0x}}{\gamma\left(v_{cm}\right)} = \frac{L_{0}\cos{\theta}}{\gamma\left(v_{cm}\right)}
[/tex]

[tex]L_{y} = \frac{L_{0y}}{\gamma\left(v_{cm}\right)} = \frac{L_{0}\sin{\theta}}{\gamma\left(v_{cm}\right)}
[/tex]

[tex]\therefore L = \sqrt{L_{x}^{2} + L_{y}^{2}} = \frac{L_0}{\gamma}[/tex]

Edit: After reading this post by jtbell, I think I might've made a mistake.:blushing:
 
Last edited:
  • #19
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
Doc Al
Mentor
44,892
1,143
neutrino's analysis is in error; as jtbell explains, only the component of length parallel to the direction of motion will contract.
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
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
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
neutrino's analysis is in error; as jtbell explains,
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.

Please guide me

Thank you
 
  • #23
the oblique rod poblem

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.

Please guide me

Thank you
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
Doc Al
Mentor
44,892
1,143
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.

Please guide me
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:

Related Threads on The Lorentz Transformations and vector components

Replies
10
Views
2K
Replies
7
Views
563
Replies
19
Views
6K
Replies
1
Views
5K
Replies
6
Views
901
Replies
19
Views
2K
Replies
15
Views
1K
Replies
47
Views
10K
Replies
5
Views
4K
Top