What Does q*a Dot Product Mean?

[Gogsey]

If someone says that you have to take the dot product of q*a and a is a length, and q is the 111 direction, or alonfg the 111 direction. What does that mean.

Is it simply 111 dot length a.

We have a exponential function to the power q dot a and want to know what to do.

 

[Gogsey]

So, if we have an atom at 0,0,0 and one at 1/2,1/2,1/2, and with bond length sqrt(3)/2 and are looking along the 111 direction. If we wanted to cmpute the dot product, we would do

1/2*sqrt(3)/2 + 1/2*sqrt(3)/2 + 1/2*sqrt(3)/2 = 3sqrt(3)/4?

Its q dot a, and q along 111 direction. a = distance bewtween these 2 atoms.

 

[gneill]

The dot product a dot b (for vectors a and b) produces the value |a||b|cos(Θ), where Θ is the angle between the vectors. If a and b are collinear, then cos(Θ) = 1, and the value of the dot product is simply |a||b|.

 

[Gogsey]

Its a vector dotted with a length sqrt(3)/2. Vector 1/2,1/2,1/2,

 

[gneill]

Its a vector dotted with a length sqrt(3)/2. Vector 1/2,1/2,1/2,

It can't be just a "naked" length. It must have a direction associated with it (perhaps it's along the line joining the two atoms). The dot product is defined on vectors; There must be two vectors.

Suppose v is your vector <1/2,1/2,1/2>. Then the magnitude of v is |v| = sqrt(3)/2.

Now, what is your a? You say it's a length, but what length? Does it have a direction?

 

[Gogsey]

The distance between the 2 atoms is sqrt(3)/2 and the vector is from 0,0,0, to 1/2,1/2,1/2.

So we have q dot a(the vecor given above) and q is the 111 direction.

 

[gneill]

The distance between the 2 atoms is sqrt(3)/2 and the vector is from 0,0,0, to 1/2,1/2,1/2.

So we have q dot a(the vecor given above) and q is the 111 direction.

You are saying that a is the vector <1/2,1/2,1/2>, and that another vector q lies in the direction <1,1,1>, and you want to know about the dot product a.q?

 

[Gogsey]

I guess that's it. Would it be (3/4)*sqrt(3)?

There are also vectors (a) in all 8 combinations of +/-1/2,+/-1/2,+/-1/2 as well. Does it mean that we have to compute all the dot product for q and a with all 8 combinations? Or since we only care about the 111 direction, do we only do this for the 1/2,1/2,1/2 case?

 

[gneill]

I guess that's it. Would it be (3/4)*sqrt(3)?

You should be absolutely specific about what vectors you are talking about! What exactly are the components of a. What exactly are the components of q. Otherwise you are asking for an answer to a question that is not well defined. The best I can offer in that case is to say that a dot q = |a||q|cos(Θ), where Θ is the angle between the vectors. Or, if a = <a1,a2,a3> and q = <q1,q2,q3>, then a dot q = a1q1 + a2q2 + a3q3. If the vectors are collinear, then a dot q = |a||b| multiplied by + or - 1 depending upon whether they are parallel or anti-parallel (in the same or opposite directions).

There are also vectors (a) in all 8 combinations of +/-1/2,+/-1/2,+/-1/2 as well. Does it mean that we have to compute all the dot product for q and a with all 8 combinations? Or since we only care about the 111 direction, do we only do this for the 1/2,1/2,1/2 case?

Yes, since the vectors will point in various directions that are not aligned with the <1,1,1> direction (which is the same direction as <1/2,1/2,1/2>).

 

[Gogsey]

I apologize. I'm not sure myself what q is. It just says q in the 111 direction, so I'm assumeing q is 111.

Sorry, was what yes we need to do all 8 combinations, or yes yes we care only about the 111 direction? The question does say that we need to sketch the dispersion relaions for the phonon modes.

 

[gneill]

I don't know what the specific problem is, or how your a and q vectors relate to it, so I don't know how to tell you what is relevant and what is not. All I can say is that if you have different q vectors pointing in various directions, then in general you need to perform each dot product separately.

If the results of each dot product are being summed in some fashion, then there may be symmetries that you can exploit that will make some contributions cancel with others, so be on the lookout for those.