Relavistic velocity lorentz transformation problem.

AI Thread Summary
The discussion focuses on calculating the relative speed of two spaceships approaching each other at high velocities. For part (a), when each ship travels at 0.9c relative to Earth, the appropriate Lorentz transformation equation is used to find the relative speed. The participants clarify that since both ships are moving along a common line, calculations can be simplified to one dimension, specifically the x-axis. The key formula for relative velocity is emphasized, allowing for the determination of one ship's speed from the other's frame of reference. This approach streamlines the problem-solving process for relativistic velocity transformations.
Benzoate
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Homework Statement


Two space ships are approaching each other

a)if the speed of each is .9 *c relative to Earth , what is the speed of one relative to the other

b)if the speed each relative to Earth is 30000m/s, what is the speed of one relative to the other


Homework Equations


The problem does not state if the two rockets are headed toward each other on the x-axis , y-axis or z-axis , so I'm not really sure what velocity equations I should used and so I will list them all.
u'(z)= u(z)/(gamma*(1-(v*u(z))/c^2)
u'(y)= u(y)/(gamma*(1-(v*u(x))/c^2)
u'(x)=u(x)-v/(1-v*u(x)/c^2)


/

The Attempt at a Solution



Here is my attempted solution for the first part part of the problem:
a) u'(z)=u(z)-.9c/(2.29*((1)- u(z)*(.9c)/c^2))
u'(y)= u(y)/(2.29*((1)-u(y)*(.9c)/c^2))
u'(x)=u(x)-.9c/(1-.9c*u(x)/c^2)

not sure how to calculate u(z),u(x), or u(y).
 
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Benzoate said:

Homework Statement


Two space ships are approaching each other

a)if the speed of each is .9 *c relative to Earth , what is the speed of one relative to the other

b)if the speed each relative to Earth is 30000m/s, what is the speed of one relative to the other


Homework Equations


The problem does not state if the two rockets are headed toward each other on the x-axis , y-axis or z-axis , so I'm not really sure what velocity equations I should used and so I will list them all.
u'(z)= u(z)/(gamma*(1-(v*u(z))/c^2)
u'(y)= u(y)/(gamma*(1-(v*u(x))/c^2)
u'(x)=u(x)-v/(1-v*u(x)/c^2)


/

The Attempt at a Solution



Here is my attempted solution for the first part part of the problem:
a) u'(z)=u(z)-.9c/(2.29*((1)- u(z)*(.9c)/c^2))
u'(y)= u(y)/(2.29*((1)-u(y)*(.9c)/c^2))
u'(x)=u(x)-.9c/(1-.9c*u(x)/c^2)

not sure how to calculate u(z),u(x), or u(y).


You don't have to do all three! They are moving stratight toward one another, so both motions are along a comon straight line. You may call this line the x-axis and just do th ecalculation along x. "v" will be the speed of one of the spaceship as measured from Earth, u(x) will be the velocity of the second spaceship as seen from Earth and u'(x) will be the answer you are looking for.
 
nrqed said:
You don't have to do all three! They are moving stratight toward one another, so both motions are along a comon straight line. You may call this line the x-axis and just do th ecalculation along x. "v" will be the speed of one of the spaceship as measured from Earth, u(x) will be the velocity of the second spaceship as seen from Earth and u'(x) will be the answer you are looking for.

so for one rocket ship , v=.9*c ,u(x)= u'(x) +v/((1+vu'(x)/c^2);

for the other rocket ship , v=.9c , u'(x)=(u(x)-v)/(1-vu'(x)/c^2)

not sure how to calculate u'(x) for the first rocket nor how to calculate u'(x) for the second rocket.
 
u'(x)=u(x)-v/(1-v*u(x)/c^2)

This is the only formula you need... you want to see what the second rocket's speed is, in the first rocket's frame of reference... so if the first rocket is going at 0.9c relative to the earth, v=0.9c.

If the second rocket is going at -0.9c relative to the earth, then u(x)=-0.9c...

So just plug in u(x) and v into the equation for u'(x)... that gives you the speed of the second rocket as seen in the first rocket's frame of reference.
 
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