Evaluating Tensor in Special Relativity

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Homework Help Overview

The problem involves evaluating the acceleration vector of an electron in the context of special relativity, using the dyadic form of velocity-dependent mass and a given force. The initial velocity is specified as v = 0.9c i, and the force is expressed in a specific form with a constant F0.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to derive the acceleration vector using the dyadic form but expresses uncertainty about evaluating the identity dyad and the outer product. Some participants suggest using index calculus for simplification and clarify the nature of the dyads involved.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of dyads and tensors. There is an exploration of the operations involving the identity dyad and the velocity dyad, with some guidance offered on how to proceed with the calculations.

Contextual Notes

Participants note the complexity of working with tensors and express varying levels of familiarity with the concepts involved. The original poster indicates a desire to understand the evaluation process better.

atomicpedals
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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?
 
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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 :)
 
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?
 
Of course you have the linearity rule, so F(1-vv) = F1 - Fvv. Straightforward if that's what you meant.
 
As straightforward as tensors get I suppose :)
 

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