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SR Velocity Transformation

  • Thread starter rbn251
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  • #1
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Homework Statement


Two rockets A and B are moving away from the earth in opposite directions at 0.85c and -0.75c respectively.
How fast does A measure B to be travelling?

Now I have worked out v = -0.85-0.75/(1- -0.85*-0.75) = -0.997. This is correct.

Now I would like to work it out backwards to check my understanding so:
According to A:
B travels at 0.997c
The earth frame travels at 0.75c

Q - How fast does the earth measure B to be travelling?

Homework Equations


w=u-v/(1-uv)

The Attempt at a Solution



I expected the answer to be 0.85c but:
v=0.997-0.75/(1- 0.997*0.75) == 0.979 and not 0.85.

Thanks for any help
 

Answers and Replies

  • #2
PeroK
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Homework Statement


Two rockets A and B are moving away from the earth in opposite directions at 0.85c and -0.75c respectively.
How fast does A measure B to be travelling?

Now I have worked out v = -0.85-0.75/(1- -0.85*-0.75) = -0.997. This is correct.

Now I would like to work it out backwards to check my understanding so:
According to A:
B travels at 0.997c
The earth frame travels at 0.75c

Q - How fast does the earth measure B to be travelling?

Homework Equations


w=u-v/(1-uv)

The Attempt at a Solution



I expected the answer to be 0.85c but:
v=0.997-0.75/(1- 0.997*0.75) == 0.979 and not 0.85.

Thanks for any help
Are you sure?

Technically you're missing some brackets in that equation.
 
  • #3
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As far are I know everything is correct 'mathematically'. You are right about the brackets but I have used the equation correctly in both cases as such.
 
  • #4
stevendaryl
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Homework Statement


Two rockets A and B are moving away from the earth in opposite directions at 0.85c and -0.75c respectively.
How fast does A measure B to be travelling?

Now I have worked out v = -0.85-0.75/(1- -0.85*-0.75) = -0.997. This is correct.

Now I would like to work it out backwards to check my understanding so:
According to A:
B travels at 0.997c
The earth frame travels at 0.75c
Your problem is just keeping your variables straight. In my opinion, it's always good to make up variables for all the values in a problem, and do as much of the derivation as possible symbolically, and only substitute numbers into the equations at the last step.

So, let's make up some variable names:
[itex]v_{AE} = [/itex] velocity of [itex]A[/itex] as measured in the Earth's rest frame
[itex]v_{BE} = [/itex] velocity of [itex]B[/itex] as measured in Earth's rest frame
[itex]v_{BA} = [/itex] velocity of [itex]B[/itex] as measured in A's rest frame

The velocity addition formula tells you that:

[itex]v_{BE} = \dfrac{v_{BA} + v_{AE}}{1 + \frac{v_{BA} v_{AE}}{c^2}}[/itex]

You have: [itex]v_{BA} = -0.997c[/itex], [itex]v_{AE} = 0.85c[/itex].

[edit]: [itex]v_{BA}[/itex] is negative, so it should be [itex]-0.997c[/itex].
[second edit]: As PeroK points out, it should actually be [itex]-0.977c[/itex]
 
  • #5
PeroK
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I think it should be 0.977 not 0.997. That's the problem.
 
  • #6
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haha I think you're right - typo in the text book!
 
  • #7
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yup f*s sry for bothering!
 
  • #8
PeroK
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I prefer:

##v' = \frac{u+v}{1+uv}##

##v'' = \frac{v'-u}{1-v'u} = \frac{v(1-u^2)}{1-u^2} = v##

Then you can relax!
 
  • #9
stevendaryl
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I prefer:

##v' = \frac{u+v}{1+uv}##

##v'' = \frac{v'-u}{1-v'u} = \frac{v(1-u^2)}{1-u^2} = v##

Then you can relax!
But my point is that for each of the variables [itex]u, v, v', v''[/itex] you need to get clear in your mind: Whose velocity is it, and relative to which frame? And you also have to keep in mind that velocity has a direction, as well as a magnitude. (In these 1-D problems, "direction" means "sign")
 

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