# Lorentz transormation for electomagnetic field

1. Jun 2, 2015

### Maximtopsecret

1. The problem statement, all variables and given/known data
In laboratory's frame the angle between vectors E and H is given φ. Find the frame where these vectors will be parallel.
2. Relevant equations

3. The attempt at a solution
I suppose I should require equal angle between each of the E' and H' and axes in the ''new'' frame.
Means this: E'||/|E'|=H'||/|H'|
After this I lose the key. How can I implement the given angle φ?

2. Jun 2, 2015

### MarcusAgrippa

Get a formula for the angle between the fields in the new frame. It will involve v and the given angle in the old frame. Set the new angle to zero and solve for v.

The most obvious way to proceed is by adapting a coordinate system to your data - we can choose whatever coordinates we please, so choose the coordinates that give the simplest description of your data. I would choose x-axis along E, and then y-axis so that the x,y plane contains B. A boost in the z-direction won't help you (why not?), so boost in the x,y plane. The simplest is to boost in the direction of E. See if you can make this work.

3. Jun 2, 2015

### Maximtopsecret

Here is my solution. I also tried another way using scalar product which should be equal to 1 in the new frame. I obtained the same result.

If this is right, this solution means the new frame moves with V along x-axis. I am curious if this solution is not unique.
We started with a condition E to be parallel to x-axis to simplify the process. There might be lots of other solutions...

4. Jun 3, 2015

### TSny

If you boost parallel to $\vec{E}$, what would be the direction of $\vec{v} \times \vec{H}$?

So, will $\vec{E}'$ lie in the xy plane?

Will $\vec{H}'$ lie in the xy plane?

Will a boost in a direction parallel to $\vec{E}$ work?

You will want to consider a boost perpendicular to the E-H plane.

5. Jun 4, 2015

### Maximtopsecret

1) [v H] will lie along z-axis.
2)Sure
3)Yes
4)I am not sure, but following a given advice I made calculations above and some adequate result was obtained...
5)When boosting perpendicular to x-y (E-H) plane the only case with some common sense is when x-axis divides angle φ in 2 halves φ/2.

6. Jun 4, 2015

### TSny

Note that the vector [v H] gets added to the vector E when determining the vector E'. So, E' will pick up a z component. That pushes it out of the xy plane.
OK. So, H' will lie in the xy plane while E' will not lie in the xy plane. So, they can't be parallel.
It looks like you got the result v = c/ cosΦ. That makes v greater than the speed of light.
When boosting along the z axis, it will not matter how you choose the orientation of the x and y axes.

When boosting along z, then E and H are the same as E and H. You have general expressions for how E and H transform. If the boost makes E' and H' parallel, then what can you say about the cross product of E' and H'?

7. Jun 4, 2015

### Maximtopsecret

Cross product will be =|E'|*|H'|.

8. Jun 4, 2015

### TSny

But what will be the value of $\vec{E}\,' \times \vec{H}\,'$ if $\vec{E}\, '$ and $\vec{H}\,'$ are parallel? If you see what it is, then you can set up an equation for the boost velocity $v$.