# Lorentz transormation for electomagnetic field

• Maximtopsecret
In summary, in order to find the frame where the vectors E and H are parallel, a formula for the angle between the fields in the new frame must be obtained. This formula will involve the velocity and the given angle in the old frame. By setting the new angle to zero and solving for the velocity, a solution can be found. It is important to choose the coordinates that give the simplest description of the data, such as the x-axis along E and the y-axis so that the x,y plane contains B. A boost in the z-direction will not help in this case. Additionally, when boosting perpendicular to the E-H plane, the vector [v, H] will lie along the z-axis and E' and H' will not
Maximtopsecret

## Homework Statement

In laboratory's frame the angle between vectors E and H is given φ. Find the frame where these vectors will be parallel.

## 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?

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.

MarcusAgrippa said:
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.

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...

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.

TSny said:
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.
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.

Maximtopsecret said:
1) [v H] will lie along z-axis.
2)Sure
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.
3)Yes
OK. So, H' will lie in the xy plane while E' will not lie in the xy plane. So, they can't be parallel.
4)I am not sure, but following a given advice I made calculations above and some adequate result was obtained...
It looks like you got the result v = c/ cosΦ. That makes v greater than the speed of light.
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.
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'?

Maximtopsecret
TSny said:
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'?
Cross product will be =|E'|*|H'|.

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##.

## What is the Lorentz transformation for electromagnetic field?

The Lorentz transformation is a mathematical formula used to describe how the electric and magnetic fields of an object change when viewed from different reference frames, specifically when transforming between the stationary frame of the observer and the moving frame of the observed object.

## Why is the Lorentz transformation important?

The Lorentz transformation is important because it allows us to understand and predict how electromagnetic fields behave under different conditions, such as when an object is moving at high speeds. It is a fundamental concept in the theory of special relativity and is essential for many modern technologies, including GPS and particle accelerators.

## How is the Lorentz transformation derived?

The Lorentz transformation is derived from the principles of special relativity, which state that the laws of physics should be the same for all observers moving at a constant velocity. It is also derived from the equations of electromagnetism, specifically Maxwell's equations, which describe the behavior of electric and magnetic fields.

## What are the components of the Lorentz transformation?

The Lorentz transformation has four components: time dilation, length contraction, velocity addition, and the transformation of electric and magnetic fields. These components are all interconnected and work together to describe how the electromagnetic fields of an object change when viewed from different reference frames.

## How is the Lorentz transformation applied in real-world situations?

The Lorentz transformation is applied in many real-world situations, including GPS technology, particle accelerators, and high-speed spacecraft. It is also used in the development of theoretical models, such as the Standard Model of particle physics, and in the study of cosmology and the behavior of objects in the universe.

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