Electromagnetic tensor

  • Thread starter shadi_s10
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  • #1
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dear all,

I have a question about Electromagnetic field tensor.
As I was reading introduction to relativity by Hobson, I saw this sentence:
"in order that the rest mass of a particle is not altered by the action of the erforce we require the latter to be a pure force, so:
u.f=0
1) what does it mean?

The other question is about the rest of this text:

in the component form this reads:

f_μ u^μ=q F_μσ u^μ u^σ=0
so that F must be an anti symmetric tensor. 2) why?!!!!!
 

Answers and Replies

  • #2
5,428
291
If this holds

Fμσuμ uσ = 0

then Fμμ = 0, and Fμσ = -Fσμ

It seems obvious to me if the full expansion is written out, but I'm not sure if what I've written is rigorous or helpful.
 
  • #3
89
0
If this holds

Fμσuμ uσ = 0

then Fμμ = 0, and Fμσ = -Fσμ

It seems obvious to me if the full expansion is written out, but I'm not sure if what I've written is rigorous or helpful.
Dear friend,

I agree with he first one but could you please explain the second one more?
Fμσ = -Fσμ

I don't know why it seems so odd to me!
 
  • #4
5,428
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F12u1 u2 + F21 u2 u1 = 0 => F12 = -F21 ?
 
  • #6
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does anyone know the answer to the first question?
why should it be like this:
u.f=0?!!
what does hobson mean by that?
 
  • #7
Bill_K
Science Advisor
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Relativistic mechanics can be expressed in terms of 4-vectors f, a, v, related to each other in the same way that we're used to seeing in three dimensions except for the use of proper time τ in place of coordinate time, f = ma = m dv/dτ.

Consequently f·v = m v·dv/dτ = ½ m d(v·v)/dτ. But the 4-velocity has constant length c, so this requires that f·v = 0. A realistic 4-force must always be orthogonal to the 4-velocity.

The only way to have f·v ≠ 0 is to consider a system with variable mass.
 
  • #8
89
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Thank you. I understand it now.
 

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