How Does Particle Velocity Change in Different Reference Frames?

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SUMMARY

The discussion focuses on calculating the velocity of a particle in different reference frames using both Galilean and Lorentz transformations. The key equation used is u' = (u + v) / (1 + (uv/c²)), where u represents the particle's velocity and v represents the observer's velocity. Participants are tasked with calculating and graphing the velocities u_{x}'_{Gal} and u_{x}'_{SR} for various values of β ranging from -0.95c to +0.95c, as well as for particles moving at speeds of +0.5c, +0.8c, +0.99c, and +0.9999c. The correct interpretation of β as a fraction of the speed of light is also clarified.

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Homework Statement



A particle moves at speed [tex]u_{x}= +0.0c[/tex] according to an observer in the frame S.

a)Calculate the velocity of the particle, [tex]u_{x}'_{Gal}[/tex] and [tex]u_{x}'_{SR}[/tex] as measured by an observer in S'. Let [tex]\beta[/tex] range from -0.95c to +0.95c in increments of 0.05. Put your results in a table.

b)Graph [tex]u_{x}'_{Gal}[/tex] and [tex]u_{x}'_{SR}[/tex] as a function of [tex]\beta[/tex].

c) Repeat steps a and b for particles with [tex]u_{x}=+0.5c[/tex], [tex]u_{x}=+0.8c[/tex], [tex]u_{x}=+0.99c[/tex] and [tex]u_{x}=+0.9999c[/tex].

*Note* [tex]u_{x}'_{Gal}[/tex] and [tex]u_{x}'_{SR}[/tex] mean the velocities with the Galilean and the lorentz trans., respectfully.

Homework Equations



[tex]u'=\frac{u+v}{1+\frac{(u)(v)}{c^{2}}}[/tex]

The Attempt at a Solution



These are kind of dumb questions but here goes:

First of all the equation with beta in it looks like:

[tex]u'=\frac{u+v}{1+\frac{(u)(\beta)}{c}}[/tex]

Right?

Secondly, when ux=+0.0c the equation just reduces to u'x=v correct?

Thirdly, when it says graph u'Gal and u'SR as functions of beta that means beta on the horizonal and the velocities on the vertical right?
 
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CINA said:

Homework Statement



A particle moves at speed [tex]u_{x}= +0.0c[/tex] according to an observer in the frame S.

a)Calculate the velocity of the particle, [tex]u_{x}'_{Gal}[/tex] and [tex]u_{x}'_{SR}[/tex] as measured by an observer in S'. Let [tex]\beta[/tex] range from -0.95c to +0.95c in increments of 0.05. Put your results in a table.

b)Graph [tex]u_{x}'_{Gal}[/tex] and [tex]u_{x}'_{SR}[/tex] as a function of [tex]\beta[/tex].

c) Repeat steps a and b for particles with [tex]u_{x}=+0.5c[/tex], [tex]u_{x}=+0.8c[/tex], [tex]u_{x}=+0.99c[/tex] and [tex]u_{x}=+0.9999c[/tex].

*Note* [tex]u_{x}'_{Gal}[/tex] and [tex]u_{x}'_{SR}[/tex] mean the velocities with the Galilean and the lorentz trans., respectfully.

Homework Equations



[tex]u'=\frac{u+v}{1+\frac{(u)(v)}{c^{2}}}[/tex]

The Attempt at a Solution



These are kind of dumb questions but here goes:

First of all the equation with beta in it looks like:

[tex]u'=\frac{u+v}{1+\frac{(u)(\beta)}{c}}[/tex]

Right?
First, what is β defined as? The velocity of what relative to what? Second, in the problem statement you gave, β is proportional to c, so it's a regular velocity. Usually β denotes a velocity as a fraction of the speed of light, i.e. β=0.5 not β=0.5c. If you're using this convention, then yes, what you wrote above is correct because one of the factors of c in the formula was pulled into β.

Secondly, when ux=+0.0c the equation just reduces to u'x=v correct?
Yes.

Thirdly, when it says graph u'Gal and u'SR as functions of beta that means beta on the horizonal and the velocities on the vertical right?
Yes.
 

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