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Evaluating Tensor in Special Relativity

  • #1
206
6

Homework Statement



Using the dyadic form of the velocity dependent mass, find the acceleration vector a of an electron whose initial velocity is v =0.9c i at the instant a force is applied given by:

F = F0 (i + j) / [tex]\sqrt{}\pi[/tex]

F0 = 106 N

2. The attempt at a solution

I feel like I've got this to a point, and then I'm not sure how to evaluate the dyadic. So here it is:

After a good bit of math I arrive at:

dv/dt = F c2 /E - v / E (F * v)

which in dyadic form is dv/dt = (c2 / E) F (I - vv/c2 )

That form is the acceleration and is where I draw a blank. The first step would seem to be to plug in the given values for velocity and F , but it still leaves the identity dyad which is what I'm not totally sure how to handle. So; am I going at this correctly, and how do I evaluate the dyad?
 

Answers and Replies

  • #2
169
0
Dyads are ugly old names for tensors of second rank, using index calculus makes everything much simpler. But well. This "identity dyad" is just a unit matrix, and the dyad "[tex]\mathbf{v} \mathbf{v}[/tex]" is better written [tex]\vec{v} \otimes \vec{v}[/tex] -- it is the tensorial or outer product, which in 3D reads
[tex](v1, v2, v3) \otimes (v1,v2,v3) = \begin{pmatrix} v1 v1 & v1 v2 & v1 v3 \\ v2 v1 & v2 v2 & v2 v3 \\ v3 v1 & v3 v2 & v3 v3 \end{pmatrix}[/tex]. The one important rule for that product is [tex] \vec{F} ( \vec{v} \otimes \vec{v} ) = \vec{v} ( \vec{F} \cdot \vec{v} )[/tex], where the central dot is the usual scalar product of two vectors. I think you can go on now yourself :)
 
  • #3
206
6
I hope you can excuse my ignorance on tensors... one of these days it will click, I hope.

Do I need to do anything with the subtraction of the identity dyad and the velocity dyad, or can I just jump right to the dot product as usual and get the very straightforward result?
 
  • #4
169
0
Of course you have the linearity rule, so F(1-vv) = F1 - Fvv. Straightforward if that's what you meant.
 
  • #5
206
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As straightforward as tensors get I suppose :)
 

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