Lorentz boost, electric field along x-axis, maths confusion?

In summary, the problem asks to show that the electric field component Ex transforms according to Ex' = Ex under a Lorentz boost along the x-axis, given that (φ/c, A) is a 4-vector. The solution involves using the chain rule and Lorentz coordinate transformation to find the partial derivatives w.r.t. x' in terms of x and t.
  • #1
JesseC
251
2

Homework Statement



Given that (φ/c,A) is a 4-vector, show that the electric field component Ex for a
Lorentz boost along the x-axis transforms according to Ex' = Ex.

Homework Equations



[tex]E_x = -\frac{\partial \phi}{\partial x} - \frac{\partial A_x}{\partial t}[/tex]

[tex]A_x[/tex] being the x component of the vector potential

The Attempt at a Solution



So I don't have a problem getting φ'/c or Ax'

Obviously:
[tex]E_x^{\prime} = -\frac{\partial \phi^{\prime}}{\partial x^{\prime}} - \frac{\partial A_x^{\prime}}{\partial t^{\prime}}[/tex]

But I don't understand how to get the partial derivative w.r.t. x' in terms of x and t. Likewise for the partial derivative w.r.t. t'.

In the solutions it seems:

[tex]\frac{\partial}{\partial x^{\prime}} = - \frac{\gamma \beta}{c}\frac{\partial}{\partial t} - \gamma \frac{\partial}{\partial x}[/tex]

?? Does it come from the lorentz co-ord transformation?
 
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  • #2
Hi JesseC! :smile:
JesseC said:
But I don't understand how to get the partial derivative w.r.t. x' in terms of x and t. Likewise for the partial derivative w.r.t. t'.

Chain rule: ∂/∂x' = ∂x/∂x' ∂/∂x + ∂t/∂x' ∂/∂t :wink:
 

1. What is a Lorentz boost?

A Lorentz boost is a transformation used in special relativity to describe how the coordinates of an event in one frame of reference are related to the coordinates of that same event in another frame of reference that is moving at a constant velocity relative to the first frame.

2. How is an electric field along the x-axis calculated?

The electric field along the x-axis is calculated using the equation E = F/q, where E is the electric field in newtons per coulomb, F is the force in newtons, and q is the charge in coulombs. This equation only applies for a point charge along the x-axis. For more complex situations, the electric field can be found by taking the derivative of the electric potential with respect to x.

3. What are some common sources of confusion in the math involved with Lorentz boosts?

One common source of confusion is the use of different units for time and distance in different frames of reference. Another is the use of the Lorentz transformation equations, which can be difficult to understand and apply without a strong understanding of special relativity and vector calculus.

4. How is the direction of the electric field related to the direction of the Lorentz boost?

The direction of the electric field is always perpendicular to the direction of motion in a Lorentz boost. This means that if the boost is along the x-axis, the electric field will be in the y-z plane.

5. Is there a simple way to visualize the effects of a Lorentz boost on an electric field?

One way to visualize the effects of a Lorentz boost on an electric field is to imagine a magnetic field running perpendicular to the direction of motion. This magnetic field will cause the electric field to rotate and become stronger in the direction of motion.

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