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A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' axis.

Required:
a. Show that the length of the rod as measured by a stationary observer is L=L0[1-v^2/c^2 cos^2(θ)0].
b. Show that the angle the rod makes iwth the x-axis is given by the expression tan(theta)=tan(θ)0/(1-v^2/c^2)^.5

Answer :

Answer:

From the question we are told that

  The length of the rod is  [tex]L_o[/tex]

    The  speed is  v  

     The angle made by the rod is  [tex]\theta[/tex]

     

Generally the x-component of the rod's length is  

     [tex]L_x =  L_o cos (\theta )[/tex]

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       [tex]L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }[/tex]

=>     [tex]L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }[/tex]

Generally the y-component of the rods length  is mathematically represented as

      [tex]L_y  =  L_o  sin (\theta)[/tex]

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e [tex]L_y [/tex]

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     [tex]L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}[/tex]

=>  [tex]L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}[/tex]

=>  [tex]L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}[/tex]

=>   [tex]L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}[/tex]

=> [tex]L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}[/tex]

=> [tex]L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }[/tex]

=> [tex]L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }[/tex]

Hence the length of the rod as measured by a stationary observer is

       [tex] L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }[/tex]

   Generally the angle made is mathematically represented

[tex]tan(\theta) =  \frac{L_y}{L_x}[/tex]

=>  [tex]tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }[/tex]

=> [tex]tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }[/tex]

Explanation:

     

     

       

The special relativity relations allow to find the results for the questions about the measurements made by an observed at rest on the rod are:

     a) The length of the rod is: [tex]L = L_o \sqrt{1 - \frac{v^2}{c^2} \ cos^2\theta_o }[/tex]  

     b) The angle with respect to the x axis is:   [tex]tan \theta = \frac{tan \theta_o}{\sqrt{1- \frac{v^2}{c^2} } }[/tex]

Special relativity studies the motion of bodies with speeds close to the speed of light, with two fundamental assumptions.

  • The laws of physics are the same in all inertial systems.
  • The speed of light in vacuum has the same value for all inertial systems.

If we assume that the two systems move in the x-axis, the relationship between the components of the length are:

        [tex]L_x = L_{ox} \ \sqrt{1- \frac{v^2}{c^2} }[/tex]  

        [tex]L_y = L_o_y \\L_z = L_{oz}[/tex]  

           

Where the subscript "o" is used for the fixed observed on the rod, that is, it is at rest with respect to the body, v and c are the speed of the system and light, respectively.

a) They indicate that the length of the rod is L₀ and it forms an angle θ with the horizontal.

Let's use trigonometry to find the components of the length of the rod in the system at rest, with respect to it.

         sin θ = [tex]\frac{L_{oy}}{L_o}[/tex]  

         cos θ = [tex]\frac{L_{ox}}{L_o}[/tex]  

         [tex]L_{oy}[/tex] = L₀ sin θ

         L₀ₓ = L₀ cos θ

Let us use the transformation relations of the length of the special relativity rod.

x-axis

      [tex]L_x = (L_o cos \theta_o) \ \sqrt{1- \frac{v^2}{c^2} }[/tex]  

       

y-axis  

      [tex]L_y = L_{o} sin \theta_o[/tex]  

The length of the rod with respect to the observer using the Pythagorean theorem is:

      L² = [tex]L_x^2 + L_y^2[/tex]  

      [tex]L^2 = (L_o cos \theta_o\sqrt{1- \frac{v^2}{c^2} })^2 + (L_o sin \theta_o)^2[/tex]

      [tex]L_2 = L_o^2 ( cos^2 \theta_o - cos^2 \theta_o \frac{v^2}{c^2} + sin^2\theta_o)[/tex]  

      [tex]L^2 = L_o^2 ( 1 - \frac{v^2}{c^2} \ cos^2 \theta_o)[/tex]  

      [tex]L= Lo \sqrt{1- \frac{v^2}{c^2} cos^2 \theta_o}[/tex]  

b) the angle with the x-axis measured by the stationary observer is:

     [tex]tna \theta = \frac{L_y}{L_x}[/tex]

    [tex]tan \ theta = \frac{L_o sin \theta_o}{L_o cos \theta_o \sqrt{1- \frac{v^2}{c^2} } }[/tex]  

    [tex]tan \theta = \frac{tan \theta_o}{\sqrt{1-\frac{v^2}{c^2} } }[/tex]  

In conclusion, using the special relativity relations we can find the results for the questions about the measurements made by an observed at rest on the rod are:

      a) The length of the rod is:  [tex]L = L_o \sqrt{1- \frac{v^2}{c^2} \ cos^2\theta_o }[/tex]

      b) The angle to the x axis is:  [tex]tan \theta = \frac{tan \theta_o}{\sqrt{1- \frac{v^2}{c^2} } }[/tex]

Learn more about special relativity here: brainly.com/question/9820962

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