Electromagnetic Tensor: Questions & Answers

[shadi_s10]

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?!

 

[Mentz114]

If this holds

F_μσu^μ u^σ = 0

then F_μμ = 0, and F_μσ = -F_σμ

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.

 

[shadi_s10]

If this holds

F_μσu^μ u^σ = 0

then F_μμ = 0, and F_μσ = -F_σμ

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?
F_μσ = -F_σμ

I don't know why it seems so odd to me!

 

[Mentz114]

F₁₂u¹ u² + F₂₁ u² u¹ = 0 => F₁₂ = -F₂₁ ?

 

[shadi_s10]

F₁₂u¹ u² + F₂₁ u² u¹ = 0 => F₁₂ = -F₂₁ ?

U r right!
Thank you!

 

[shadi_s10]

does anyone know the answer to the first question?
why should it be like this:
u.f=0?!
what does hobson mean by that?

 

[Bill_K]

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.

 

[shadi_s10]

Thank you. I understand it now.