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

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SUMMARY

The discussion focuses on the transformation of the electric field component Ex under a Lorentz boost along the x-axis, specifically demonstrating that Ex' = Ex. The relevant equations include the electric field expression E_x = -∂φ/∂x - ∂A_x/∂t and the transformation of derivatives, which is derived from the Lorentz coordinate transformation. The participants clarify the use of the chain rule to express partial derivatives with respect to the transformed coordinates x' and t'.

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Students and professionals in physics, particularly those studying electromagnetism and special relativity, as well as educators looking to clarify concepts related to Lorentz boosts and electric field transformations.

JesseC
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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



E_x = -\frac{\partial \phi}{\partial x} - \frac{\partial A_x}{\partial t}

A_x 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:
E_x^{\prime} = -\frac{\partial \phi^{\prime}}{\partial x^{\prime}} - \frac{\partial A_x^{\prime}}{\partial t^{\prime}}

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:

\frac{\partial}{\partial x^{\prime}} = - \frac{\gamma \beta}{c}\frac{\partial}{\partial t} - \gamma \frac{\partial}{\partial x}

?? Does it come from the lorentz co-ord transformation?
 
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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:
 

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