# Einstein's applications of tensor calculus

1. Nov 8, 2013

### wavepart7cle

Hey everyone, I recently learned that my certified genius weird-uncle-who-lives-at-home (IQ over 200 something, legitimate 'genius') or WULAH for short, passes his spare time by lounging around his place and doing tensor calculus. I've done some calc in 3d in college and I know that's commonly called vector calculus. I also know that tensors are vectors that also change in time. Therefore tensor calculus is calc in 4d if I am not mistaken.

My question is this: what real applications does tensor calculus have as far as quantum physics is concerned? What valuable answers can one use it to uncover? Or in other words, what types of things can you investigate further once you have the knowledge of tensor calc?

I plan to research this a little more, so if anyone has some insightful comments or useful links, they would be much appreciated!

Peace and love,

~`wavepart7cle

Last edited: Nov 8, 2013
2. Nov 8, 2013

### WannabeNewton

Tensor calculus is pretty much non-existent in QM when compared with its ubiquity in GR. Also, it is valid in any number of dimensions, it's just applied to 4 dimensions in GR. Tensor calculus does come into play once you get into QFT though.

3. Nov 8, 2013

4. Nov 8, 2013

### HallsofIvy

That is incorrect. "Vectors that change in time" are just vector functions in which one of the variables is time. They are exactly the kind of vectors you deal with in Calc III.

Nor do you have to have "time" in order to have tensors. One common use of tensors, outside of relativity, is in "Elasticity", the mathematical study of deformable bodies. At each point in a deformable body, we can imagine three "planes", the xy, xz, and yz planes of a Cartesian Coordinate system. And the forces applied to each of those plane can have a three dimensional "stress" vector applied to it, giving 9 numerical values. Those 9 values form the "stress tensor" at that point.

Just as a three-dimensional vector can be represented by three numbers, in a given coordinate system, so a three dimensional tensor (strictly speaking a "tensor or order 2". A number is a "tensor of order 0" and a vector is a "tensor of order 1") can be represented by a 3 by 3 matrix. I emphasise "represented". Just as a vector is NOT just those three numbers, so a tensor is not just a matrix. They have an existence independent of the coordinate system and we have to give a method for determing how components change from one coordinate system to another.

Just as for vectors, the transformation rules for tensors are "linear" and "homogeneous". If we have a four dimensional vector, with components $v_1$, $v_2$, $v_3$, and $v_4$ given in a coordinate system $x_1$, $x_2$, $x_3$, and $x_4$ , then that same vector will have components $v_1'$, $v_2'$, $v_3'$, and $v_4'$ given in a coordinate system $x_1'$, $x_2'$, $x_3'$, and $x_4'$ by
$$v_i'= \sum_{j=1}^4 \frac{\partial x_i'}{\partial x_j}v_j$$
(That could be conveniently written as a matrix of the partial derivatives times the vector.)

For a three dimensional second order tensor, with components given by $A_{ij}$ in the first coordinate system and $A'_{pq}$ in the second, then we have
$$A'_{mn}= \sum_{i= 1}^{4}\sum_{j= 1}^4 \frac{\partial x_m'}{\partial x_i}\frac{\partial x_n'}{\partial x_j}A_{ij}$$.

The fact that the formula is "homogeneous" means that if a tensor has entries all "0" in one coordinate system then it has all "0" entries in any coordinates system. That is crucially important because it means if a tensor equation is true in one coordinate system then it is true in any coordinate system!

For example, if A= B, where A and B are tensors, in one coordinate system, then, in that coordinate system, A- B= 0- the tensor having all 0 entries. But then, in any other coordinate sytem (A- B)'= 0 also (the ' indicating the components in the new coordinate system. Now, because or "linearity", (A- B)'= A'- B' so A'- B'= 0 or A'= B'. So to prove that a tensor equation is true in any coordinate system, it is sufficient to choose a convenient coordinate system and prove it true in that.

You are mistaken. You can do ordinary vector calculus in any number of dimensions.

I don't know of any applications of tensors to quantum physics. They are of greatest strength in dealing with non-Euclidean geometries and geometry is of importance in general relativity, not quantum physics.

Last edited by a moderator: Nov 8, 2013
5. Nov 8, 2013

### Mandelbroth

Well, if you think that vector calculus is limited to three dimensional Euclidean space, I'm thinking you had a pretty bad course in vector calculus! Vectors are actually an area of study in a field of mathematics known as "algebra." Essentially, a vector is any element of a vector space. Typically, we talk about doing "vector calculus" on $\mathbb{R}^n$.

Tensor calculus is basically an extension of this idea to more arbitrary tensors. Awhile back, I tried to imagine tensors geometrically. This is hard without some intuition. Instead, we might think of them as multilinear maps. In that case, a simple familiarity with vector spaces will get you pretty far.

Any number of dimensions? How about 0? :tongue:

I jest, of course. This is a wonderful post. You should thoroughly read it, wavepart7cle, if you want to understand.

6. Nov 11, 2013

### wavepart7cle

Sorry for the hasty assumptions in my first post, seems I don't get this quite as well as I thought I did. My bad!

Halls of Ivy, thanks for your explanation! It's a little bit clearer now, I think.
So are tensors applicable to the interaction of particles with the Higgs-Boson field then? whereas the particle exists in a 3-dimensional Euclidean space but then has the interaction with the Higgs-Boson field to be considered as well? Thanks again