Notation problem

  • Thread starter noamriemer
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  • #1
noamriemer
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Hi! I can't understand something in field theory and need your assistance:

I wish to understand why a particle of mass m^2<0 can't exist.
For a massive particle, in its reference frame, one would write:
[itex]p_\mu=(m,0,0,0)[/itex]. I understand that.
But for m=0, why is:
[itex]p_\mu=(p,0,0,p)[/itex]
And for [itex]m^2<0[/itex]
Why is:
[itex]p_\mu=(0,0,0,m)[/itex]
?
Thank you...!
 
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Answers and Replies

  • #2
Simon Bridge
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That would imply that the mass is imaginary ... what would that mean?
 
  • #3
noamriemer
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That would imply that the mass is imaginary ... what would that mean?

I don't know... an adjoint term maybe?
 
  • #4
Simon Bridge
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One of the things it may mean is that it is traveling faster than the speed of light - or that it is highly unstable... once you realize that m must be imaginary if m2 < 0 you'll have something to google for:
Wiki: http://en.wikipedia.org/wiki/Tachyonic_field
Here: https://www.physicsforums.com/showthread.php?t=107988
Arxiv: http://arxiv.org/pdf/physics/0604003.pdf (no date?!)
In contrast: http://www.quora.com/Quantum-Field-Theory/Can-real-particles-such-as-neutrinos-have-imaginary-mass

Some measurements of neutrinos mass have suggested that m2 < 0 is a possibility - however, experimental uncertainty tells us more about the measurement process than it does about the thing being measured. I'm guessing this is where you are coming from.

The second-to-last link above probably has the most accessible answer to your question.

When you think of things like this it is a good idea to try think through the consequences... try putting the imaginary mass into the momentum 4-vector for a simple problem and find an equation of motion or otherwise see what that does to the calculations ;) play around.
 
  • #5
noamriemer
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Thank you so much for a wonderful reply!
 
  • #6
Simon Bridge
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No worries. Have fun :)

I realized I didn't actually answer the whole question! You were talking about notation:
... a massless particle will be moving with momentum p and energy pc (think: photons) so the 4-vector is scatalogical for motion in the z direction if P0 = E/c ;)

The last one is because the imaginary mass gives it a space-like 4-momentum.
Find the inner product of that vector with itself and you'll get a negative mass-squared out.
If you try a naive construction with E=(-m)c2 you won't get negative mass-squared out.
http://en.wikipedia.org/wiki/Minkowski_space#Causal_structure
 
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  • #7
noamriemer
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I will take advantage of your kindness and ask another thing:

For m^2<0:
[itex]W^\mu=\frac {1} {2} \varepsilon^{\alpha\beta\gamma\delta}M_{\beta\alpha} p[/itex]
out of it we will get both L's and K's, meaning the algebra is not close in su2. so, there is no finite way to write the states you can get. That is not physically, and therefore, there exist no such particle.
But if we look at a wider range, and include both K's and L's, the algebra does form a complete "basis". Why is this not enough for such particle to exist?

I hope my intentions are possible to understand... :)
Thanks!
 
  • #8
Simon Bridge
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Well what you've shown is that you have a consistent mathematics (I didn't check - don't take my word for it) to allow imaginary-mass particles. You'll see the idea discussed in the literature ... so where did you get the idea that these "cannot exist" as a consequence of the math?

Your next step is to propose an imaginary math particle and work out the consequences.
 

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