Special Relativity and radioactive nucleus

  • #1

Main Question or Discussion Point

I was wondering if I could get some guidance regarding the following question:

"A radioactive nucleus, moving in a straight line, decays and emits an alpha particle. In the rest frame of the nucleus, the emitted particle moves with a velocity 6c/7, at an angle of 60 degrees to the direction of motion of the nucleus. If a laboratory experimenter observes the particle to be emitted at an angle of 30 degrees to the original velocity of the nucleus, calculate the speed of the radioactive nucleus in the lab frame."

A nudge in the right direction would be brilliant.

Thanks much,

Pyro...
 

Answers and Replies

  • #2
jcsd
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Think about:

x'= (x - ut)/√(1 - u2/c2)

y' = y
 
  • #3
jcsd
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and of course the transformations for t and vx (this is assuming you don't already have the transformation for vy
 
  • #4
Cheers,

I shall now make an attempt to solve it.
 
  • #5
I'm still slightly baffled, any further help would be brilliant!

Cheers.
 
  • #6
jcsd
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Think about the problem in terms of the components of velocity (and as you need only consider two dimensions they are vx, which you should already have the formula for, and vy which you may not have the formula for, but should be able to derive) of the emitted particle in the rest frame of the nucleus and the rest frame of the lab and how they are related.
 
  • #7
Okay, if u(x) = 6c/7cos(theta) then I think I'm on the right track.

And u(y) = u(y)'.

Thanx.
 
  • #8
jcsd
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I'm not sure exactly what you want your notaion to mean but I think you may of made a mistake as vy is not equal to vy'.
 
  • #9
Yes I see that now I was just being slow of brain, however I have an answer now for v using just the values of u(x) and u(x)', and the transformation velocity equation for u(x). Is that the velocity of the S' frame and is it needed?
 
  • #10
jcsd
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I'm trying to work out exactl;y what you've claculated, but get the answer you need to work out an equation relating uy to uy'
 
  • #11
jcsd
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To make it easier for you (though you will probably still need to derive this, if it's not given):

uy' = uy√(1 - vx2/c2)/(1 + uxvx/c2)

In this case u is the velocity of the alpha particle in the rest frame of the lab, u' is the velocity of the alpha particle in the rest frame of the nucleus and v is the velocity of the nucleus in the rest frmae of the lab.
 
Last edited:
  • #12
I have an equation:

u(y)' = u(y)/(gamma)(1 - u(x)v/c^2).

Is this the equation?
 
  • #13
jcsd
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Yes the two equations are equivalent,
 
  • #14
Sorry, didnt see your last reply. Our equations look similar (apart from a difference in sign, mine is probably wrong then) and obviously v is the speed of the nucleus which is the moving frame.
 
  • #15
jcsd
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I've just realized, I misread your first post, I thought 6c/7 was the velocity of the nucleus not the alpha particle, anyway it shouldn't matter too much it just means that the answer that I worked out for myself is wrong.
 
  • #16
Thanx for all your help. Should have seen it all along!
 

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