The Lorentz Transformations and vector components

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Discussion Overview

The discussion revolves around the Lorentz transformations and their application to length contraction, particularly in the context of a rod moving at an angle relative to the x-axis. Participants explore the implications of special relativity on the measurement of lengths and angles, including the effects of motion on these measurements.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the Lorentz transformations for length contraction can be expressed as \( L_0 = \frac{L}{\sqrt{1 - \frac{v_x^2}{c^2}}} \) and \( L_0 = \frac{L}{\sqrt{1 - \frac{v_y^2}{c^2}}} \), while others challenge this formulation.
  • One participant argues that the formula does not correctly represent contraction, suggesting it expands instead, and questions the intent behind expressing \( L_0 \) as a function of \( v \).
  • There is a discussion about the necessity of calculating the contraction of the x-component of a rod's length before determining the overall length when the rod is at an angle.
  • Some participants assert that in special relativity, motion is typically aligned with the x-axis, simplifying the application of Lorentz transformations.
  • One participant states that an object only contracts in the direction of motion, implying that there is no contraction in the direction perpendicular to motion.
  • There is a question raised about how to calculate length contraction for a rod moving at an angle, prompting further exploration of the topic.
  • Participants discuss the implications of motion on the angles of shapes, such as an equilateral triangle, and whether they would appear different to an observer in motion.
  • Some participants express uncertainty about the effects of motion on the radius of a cylindrical rod and whether it changes due to the angle of motion.
  • There is a contention regarding the interpretation of the rod's motion and whether it can be treated as a vector, with some participants seeking clarification on this point.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct application of the Lorentz transformations in the context of length contraction, particularly when considering components and angles. Multiple competing views remain regarding the treatment of motion and contraction in different directions.

Contextual Notes

Limitations include unresolved assumptions about the alignment of motion with coordinate axes, the treatment of angles in moving frames, and the implications of different interpretations of the rod's geometry.

anantchowdhary
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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}}}
 
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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}}
 
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Err...i meant to write L to be the Length as observed by an observer who is observing the moving rod
 
and if the rod is going at an angle relative to the x-axis or to the y axis,then wouldn't we first calculate the contraction of its 'x' component and then see the new length?
 
In special relativity the motion is assumed to be aligned with the x axis. The Lorentz transform of the y-axis is

y= y'
 
Integral said:
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}}

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!
 
ok.But what of we have a rod traveling at an angle?How would we calculate the length conrtraction without taking its components?
 
Integral said:
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.
 
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 isn't 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 isn't any contraction in the direction of the 'y' axis!
 
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  • #11
anantchowdhary said:
i agree with neutrino.But why isn't 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 = \frac {L} { \sqrt {1 - \frac {{0}^2} {c^2}}}

would giv us L_0_y=L_y

therefore there isn't 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
Once again, align your coordinate system with the direction of linear motion.
 
  • #14
ok,ill ask a question related to the topic:

If we have say an equilateral trinagle(a cardboard cutout) and its traveling 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
Integral said:
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!
 
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  • #16
a simpler problem but instructive

anantchowdhary said:
ok,ill ask a question related to the topic:

If we have say an equilateral trinagle(a cardboard cutout) and its traveling 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
bernhard.rothenstein said:
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?

L_{x} = \frac{L_{0x}}{\gamma\left(v_{cm}\right)} = \frac{L_{0}\cos{\theta}}{\gamma\left(v_{cm}\right)}<br />

L_{y} = \frac{L_{0y}}{\gamma\left(v_{cm}\right)} = \frac{L_{0}\sin{\theta}}{\gamma\left(v_{cm}\right)}<br />

\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.:blushing:
 
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  • #19
I don't 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 traveling at the same alignment
 
  • #20
neutrino's analysis is in error; as jtbell explains, only the component of length parallel to the direction of motion will contract.
anantchowdhary said:
I don't 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 traveling 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 can't get this.Why would the angle and length be different.The rod is still traveling parallel to the x axis.Even though it is at an angle,its motion is not at an angle with the x-axis isn't it?
 
  • #22
Doc Al said:
neutrino's analysis is in error; as jtbell explains,
Why is there an error?Also,if I've understood the question correctly, the rod isn't 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

anantchowdhary said:
Why is there an error?Also,if I've understood the question correctly, the rod isn't 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
anantchowdhary said:
Why is there an error?Also,if I've understood the question correctly, the rod isn't 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.)
 
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