Special Relativity and radioactive nucleus

  • #1
Pyrokenesis
19
0
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
Science Advisor
Gold Member
2,101
12
Think about:

x'= (x - ut)/√(1 - u2/c2)

y' = y
 
  • #3
jcsd
Science Advisor
Gold Member
2,101
12
and of course the transformations for t and vx (this is assuming you don't already have the transformation for vy
 
  • #4
Pyrokenesis
19
0
Cheers,

I shall now make an attempt to solve it.
 
  • #5
Pyrokenesis
19
0
I'm still slightly baffled, any further help would be brilliant!

Cheers.
 
  • #6
jcsd
Science Advisor
Gold Member
2,101
12
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
Pyrokenesis
19
0
Okay, if u(x) = 6c/7cos(theta) then I think I'm on the right track.

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

Thanx.
 
  • #8
jcsd
Science Advisor
Gold Member
2,101
12
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
Pyrokenesis
19
0
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
Science Advisor
Gold Member
2,101
12
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
Science Advisor
Gold Member
2,101
12
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
Pyrokenesis
19
0
I have an equation:

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

Is this the equation?
 
  • #13
jcsd
Science Advisor
Gold Member
2,101
12
Yes the two equations are equivalent,
 
  • #14
Pyrokenesis
19
0
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
Science Advisor
Gold Member
2,101
12
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
Pyrokenesis
19
0
Thanx for all your help. Should have seen it all along!
 

Suggested for: Special Relativity and radioactive nucleus

  • Last Post
Replies
10
Views
597
  • Last Post
Replies
6
Views
360
Replies
3
Views
193
Replies
4
Views
316
  • Last Post
Replies
8
Views
385
  • Last Post
Replies
1
Views
246
Replies
4
Views
506
  • Last Post
Replies
6
Views
381
  • Last Post
Replies
7
Views
487
Replies
12
Views
475
Top