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I am currently learning electromagnetism and I am getting really annoyed using a physical hand to accomplish the right hand rule. I get it, i've seen diagrams, and i've solved problems with it but still somehow I mess up because I bend a finger or twist my wrist a strange way. I am tired of using my hand.

So my question is can I find an entirely mathematical basis for this?

Lets say for any given diagram I define the positive $x$ axis as going to the right, the positive $y$ axis as going up and the positive $z$ axis as going into the page. Then I just did unit vectors for everything to determine the missing component?

As I understand it $\vec{F} = q (\vec{V}\times\vec{B})$ and with that definition I am confident I could find out the direction of the electric field given the velocity of a particle of charge and the magnetic field but how can I re-arrange this so I can solve for other variables? For example what does $\vec{B}$ equal in terms of a cross product involving $\vec{V}$ and $\vec{F}$?

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    $\begingroup$ If the question is just "Given $\vec{F}=\vec{V}\times\vec{B}$ find $\vec{B}$" then unfortunately the answer is that there are multiple such $\vec{B}$, because in doing the cross product you throw away any information about the component of $\vec{B}$ along $\vec{V}$. $\endgroup$ – jacob1729 May 15 at 23:26
  • $\begingroup$ On the other hand if you want to know if you can compute cross products algebraically, then yes of course you can. But wikipedia already explains this well. $\endgroup$ – jacob1729 May 15 at 23:27
  • $\begingroup$ Cross-product simply encodes the chosen order of the coordinates. The even order is XYZ, YZX, or ZXY, i.e. you always read XYZ from left to right and 'loop back' when you reach the end. If you read XYZ from right to left, the order is odd (i.e. it is an odd permutation of XYZ). So let's say you have $\mathbf{F}=\mathbf{a}\times\mathbf{b}$ and you want $F_y$. The relevant triads are those that start with Y, so YZX is even YXZ is odd, so: $F_y = (a_z b_x )+ (-a_x b_z)$ $\endgroup$ – Cryo May 15 at 23:38
  • $\begingroup$ Also works the other way round. Lets say $\mathbf{a}\times\mathbf{b}=\mathbf{F}$ and you want $F_y$, the relevant triads are the ones that end with Y, so: ZXY is even and XZY is odd, thus $(a_z b_x)+ (-a_x b_z)= F_y$, which is the same as my comment above $\endgroup$ – Cryo May 15 at 23:45
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    $\begingroup$ "I define the positive x-axis as going to the right, the positive y-axis as going up and the positive z-axis as going into the page." I'm pretty sure you just defined a left-handed coordinate system. $\endgroup$ – Michael Seifert May 16 at 13:02
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So my question is can I find an entirely mathematical basis for this?

Lets say for any given diagram I define the positive x axis as going to the right, the positive y axis as going up and the positive z axis as going into the page. Then I just did unit vectors for everything to determine the missing component?

The "Right hand rule" is a convention both for cross-products and for x-y-z coordinate systems. In pre-war Germany it was common to use a LEFT-HANDED coordinate system where if x increased "to the right" and y increased "up the page" then z INCREASED "into the page." However, nowadays this is very uncommon and we use right-handed coordinate systems where if x increases "to the right" and y increases "up the page" then z increases "out of the page."

Clearly (?) there is no one intrinsically "correct" convention.

But, as stated above, nowadays we almost exclusively use the right-handed coordinate convention where $$ \hat x \times \hat y = \hat z\;. $$

Given the above convention you can "mathematically" compute cross products.

For example, if $$ \vec V = |V|\hat x $$ and $$ \vec B = |B|\hat y $$ then $$ F = q|V||B|\hat z $$ because, by convention: $$ \hat x \times \hat y = \hat z\;. $$


Update (per the comment):

The example above makes use of the right-hand rule for $\hat x \times \hat y$. But, in general, you will also find these two other rules useful: $$ \hat z \times \hat x = \hat y $$ $$ \hat y \times \hat z = \hat x $$

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You can also use determinants: $$ \begin{vmatrix} i&j&k\\ v_x&v_y&v_z\\ B_x&B_y&B_z \end{vmatrix}$$ plug them in and do a determinant and you will get the correct cross product.

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  • $\begingroup$ Thanks for the edit. I need to learn mathjax $\endgroup$ – ggcg May 16 at 12:55
  • $\begingroup$ No problem, glad I could help. $\endgroup$ – Manvendra Somvanshi May 16 at 13:13
  • $\begingroup$ But what if it's a left-handed determinant? :-) $\endgroup$ – Carl Witthoft May 16 at 13:15
  • $\begingroup$ @CarlWitthoft, :-(. $\endgroup$ – ggcg May 16 at 13:31
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It actually is possible(*) to find a purely algebraic way to analyze such problems mathematically, without using your hands or equivalent devices. The method is actually quite closely related to ggcg's answer, though it will look quite different. And it goes beyond that by actually dispensing with the cross product entirely, and replacing it with what we call the "wedge" or "exterior" product. Rather than taking two vectors and getting one other vector out of it, as in $\vec{a} \times \vec{b}$, when you take the wedge product you get a plane out: the wedge product $\vec{a} \wedge \vec{b}$ represents the plane spanned by $\vec{a}$ and $\vec{b}$, which is orthogonal to the vector $\vec{a} \times \vec{b}$. Moreover, there is also a "sense" of this plane, which substitutes for the right-hand rule, and is related to the order in which you take the product: $\vec{b} \wedge \vec{a}$ represents the same exact plane, except with the opposite "sense", just as $\vec{b} \times \vec{a}$ represents the same vector in the opposite direction. You can rewrite expressions for things like the Lorentz force law and rotations (and everything else in physics that uses a cross product) with the wedge product, and it all just works out beautifully. But in this case, rather than using the right-hand rule, you just need to decide on an order for your basis vectors: do you order them as $(\vec{x}, \vec{y}, \vec{z})$, or as $(\vec{x}, \vec{z}, \vec{y})$, for example? Making a choice for that ordering allows you to completely specify everything you need, expand in the basis, and compute the wedge product correctly without ever referring to handedness.(**)

This approach is just one small part of something called "geometric algebra". One nice feature of this approach is that it actually generalizes to other dimensions. In two dimensions, you get a nicer understanding of complex algebra. In four dimensions, you can use the same exact techniques to do special relativity more easily. It's actually just a coincidence that the cross product even works in three dimensions; if you want something that will work in three dimensions and any other dimension, you actually have to go to the wedge product.

Geometric algebra is actually a really great pedagogical approach, and we would all be better off if everyone would just use it, and ditch cross products forever. Unfortunately, that's not going to happen while you're a student. So while I encourage you to learn geometric algebra, and I would bet that you'd really get better at physics if you do, remember that you'll still probably be taught and asked to use cross products.


(*) I just need to dispense some practical advice here. It sounds like you're a relatively new (and probably talented) student of physics. Physicists (including teachers) are all very familiar with the various problems with the right-hand rule. And it is unfortunate. I wish that it weren't so, but as a purely practical matter, if you stick with physics you'll need to keep interacting with the cross product for at least another couple years. So my advice is to stick with it, and become really good at using it correctly. It's not that hard, and maybe it'll exercise your brain in ways that come in handy.

(**) It's still true that you might eventually need to relate your particular choice of directions to someone else's idea of the "correct" orientation. Basically, you would need to make your choice of ordering consistent with their choice, which would probably be based on the right-hand rule. However, unless a question specifically asks for that orientation, you could very well use a left-handed orientation and still get correct answers (e.g., force going in or out of the page, etc.) using the wedge product.

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  • $\begingroup$ Specifically, if you wanted the cross product (for 3D GA) for some bizarre reason, you'd have a formula like $u\times v = (u\wedge v)I^{-1}$ where $I$ is the unit pseudoscalar, except that there're two unit pseudoscalars, $\vec x\wedge\vec y\wedge\vec z$ and $\vec x\wedge\vec z\wedge\vec y$. This is where the handedness of the cross product comes in. $\vec x\times\vec y=\pm\vec z$ becomes revealed as $\vec x\wedge\vec y=I\vec z$ which, by having the pseudoscalar explicit, is independent of handedness. What is dependent is whether $I\vec z$ is positive $\vec x\wedge\vec y$ or negative. $\endgroup$ – Derek Elkins May 16 at 17:52
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Indeed, the definition of the cross-product is inextricably linked to the choice of orientation for the three-dimensional vector space. Locally, we communicate the difference between right and left-handed orientations with visual cues. Most humans are "right-handed," or we can reference the rotation sense of the moon, earth, or celestial sphere, etc. By such visual cues and conventions, we communicate the "preferred" definition of the cross-product.

Now imagine you could have a conversation with someone from a very distant part of the universe, with the rule that all the communication had to be by texting with a stream of 0s and 1s--so no emojis, no pictures, etc. Could you actually communicate our local preferred convention? It's not so clear. Perhaps a text discussion about the weak force could accomplish this. But mathematically, there is no mechanism to universally distinguish one of the two orientations.

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I prefer the linear algebra approach to cross products.

$$ \begin{aligned} \boldsymbol{F} & = q \left( \boldsymbol{v} \times \boldsymbol{B} \right) \\ & = q \left( [\boldsymbol{v} \times] \boldsymbol{B} \right) \\ & = q \begin{bmatrix}0 & -v_z & v_y \\ v_z & 0 & -v_x \\ -v_y & v_x & 0 \end{bmatrix} \begin{bmatrix} B_x \\ B_y \\ B_z \end{bmatrix} \\ & = \begin{bmatrix} q(v_y B_z - v_z B_y) \\ q(v_z B_x - v_x B_z) \\ q(v_x B_y - vy B_x) \end{bmatrix} \end{aligned} $$

where $[\boldsymbol{r}\times] $ is the 3×3 skew-symmetric cross product matrix operator. It is defined as:

$$ \pmatrix{x \\ y \\ z} \times = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$$

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