Why Are Maxwell's Equations Invariant Across Reference Frames?

[universal2013]

I am trying to understand why maxwell equations are correct in any reference frames? While i started to understand of his laws of physics a bit i could not imagine why he uses hyperbolic functions such as coshw instead of spherical ones in position and time relation between moving frames. Velocity of a frame = sinhw/coshw .. ?

 

[haushofer]

Basically, it's because th Maxwell eqns are second order in both space and time derivatives and Poincare transformations treat both space and time equally.

By the way, they don't have the same form in ANY reference frame, but only in any INERTIAL frame. You can check that by e.g. transforming to accelerating coordinates.

If you want people to help you, it helps to say what book you're using.

 

[Nugatory]

I am trying to understand why maxwell equations are correct in any reference frames?

It is an experimental fact that the speed of light is the same in all inertial frames, so we expect that Maxwell's equations have to be correct in any inertial frame - otherwise they would disagree with the observed behavior of the universe.

 

[Matterwave]

It is an experimental fact that the speed of light is the same in all inertial frames, so we expect that Maxwell's equations have to be correct in any inertial frame - otherwise they would disagree with the observed behavior of the universe.

Although I agree with your conclusion, I don't think this answers OP's question. Unless I'm misreading, this statement simply assumes Maxwell's equations agree with experimental fact, which they do, but the OP is asking why they do.

 

[Nugatory]

I don't think this answers OP's question. Unless I'm misreading...the OP is asking why they do.

That may be what OP is asking, but if so the answer is the same as for most other "why?" questions: That's how the universe we live in behaves. Empirical science is very good at telling us how our universe behaves, but no amount of measurement and observation on the universe we have will ever tell us why we have that one and not some other.

For example, there is a mathematically consistent physics that works if the speed of light is not constant in all inertial frames. The only reason that we reject that one and accept relativity instead is that observation and experiment tell us that the universe doesn't work that way. It could, but it doesn't.

 

[m4r35n357]

I am trying to understand why maxwell equations are correct in any reference frames?

The equations do not specify a reference frame so how can they depend on one? That is a big enough clue for me!

 

[pervect]

I am trying to understand why maxwell equations are correct in any reference frames? While i started to understand of his laws of physics a bit i could not imagine why he uses hyperbolic functions such as coshw instead of spherical ones in position and time relation between moving frames. Velocity of a frame = sinhw/coshw .. ?

It appears that the question may be about rapidity, https://en.wikipedia.org/w/index.php?title=Rapidity&oldid=879813333

I say this because Maxwell's equations don't have any cosh or sinh in them, but rapidity does, for instance in the above article we have:

$$\frac{v}{c} = \frac{\sinh w}{\cosh w}$$

Maxwell's equtions can be written in a differential form or an integral form, but neither of these forms has hyperbolic functions in it.

But while I can guess that the question is probably involves rapidity (or perhaps it's about the Lorentz transform?), I'm not sure exactly what the question really is.

Maxwell's equations lead to the wave equation (the vacuum solution of maxwell's equations, the homogeneous solution, is the wave equation). The wave equation lead to the constancy of the speed of light in a vacuum in any frame.

The constancy of the speed of light also leads to special relativity and the Lorentz transform. The hyperbolic functions of rapidity are one way of writing the Lorentz transform, but there are other formulations that don't use hyperbolic functions. The link there is the link between the Lorentz transform and hyperbolic rotations.

But it's unclear to me exactly which of these points the OP needs clarification on.

 

[Matterwave]

That may be what OP is asking, but if so the answer is the same as for most other "why?" questions: That's how the universe we live in behaves. Empirical science is very good at telling us how our universe behaves, but no amount of measurement and observation on the universe we have will ever tell us why we have that one and not some other.

For example, there is a mathematically consistent physics that works if the speed of light is not constant in all inertial frames. The only reason that we reject that one and accept relativity instead is that observation and experiment tell us that the universe doesn't work that way. It could, but it doesn't.

My impression, and I could certainly be totally wrong on this, is that the OP was asking for a mathematical derivation that says that Maxwell is consistent with relativity (i.e. starting from Maxwell's equations, derive that they give a constant speed $c$ for EM waves in all inertial reference frames)... at least that was my initial reaction. Now re-reading his question, I'm not so sure.

 

[sweet springs]

I am trying to understand why maxwell equations are correct in any reference frames?

The fact that there exists tensor form of Maxwell equation,
F^{\mu\nu}_{:\mu}=\mu_0j^{\nu}
shows us that Maxwell equation stands in any reference frames.

Why correct is a never ending, multi layer lasting tough question.

 

[universal2013]

Let me clarify my question in this way, when i use a coordinate transformation lastly i put it in this way x' = xcos(theta) - tsin(theta) and also for t' = tcos(theta) - xsin(theta)
but in the beginning of this equation i could not see how the light has the same velocity in every inertial referance frame unless i used the proper way of interpreting x' = xcosh(theta) - tsinh(theta) and also for t' = tcosh(theta) - xsinh(theta) then i got the equation in this form x'^2 - t'^2 = x^2 - t^2
Thanks for your answers !

 

[robphy]

So, it seems you are looking to see how
the fields and sources transform under a boost,
the Maxwell Equations are preserved in form,
[and how the resulting wave equation are preserved in form with the same speed of propagation].
Is that what you seek? (as @Matterwave had envisioned)

http://www.feynmanlectures.caltech.edu/II_26.html#Ch26-T2 (fields under boost transformations... which can be written in terms of rapidities, instead of \gamma and v)
http://www.feynmanlectures.caltech.edu/II_18.html#mjx-eqn-EqII1826 (Maxwell implies the Wave Equation)
http://ricardoheras.com/wp-content/uploads/2013/10/LTDerivation.pdf (Wave Equation invariant under boosts)
https://iopscience.iop.org/article/10.1088/0143-0807/38/1/019401 (addendum to previous article)

 

[pervect]

Let me clarify my question in this way, when i use a coordinate transformation lastly i put it in this way x' = xcos(theta) - tsin(theta) and also for t' = tcos(theta) - xsin(theta)
but in the beginning of this equation i could not see how the light has the same velocity in every inertial referance frame unless i used the proper way of interpreting x' = xcosh(theta) - tsinh(theta) and also for t' = tcosh(theta) - xsinh(theta) then i got the equation in this form x'^2 - t'^2 = x^2 - t^2
Thanks for your answers !

Your observation that the first coordinate transformation with cos and sin doesn't lead to a constant velocity of light, while the second coordinate transformation with cosh and sinh does is essentially correct, though you've skipped over all the details of how you write down and perform the necessary coordinate transformation.

There is a sudden leap here where you are talking about coordinates, then you are talking about light. But you never talk in detail about the equations that govern the propagation of light.

If you write down the equations that govern the propagation of light , you can discuss how the equations transform when you change the coordinates. And you can show that under the second coordinate transformation, the equations remain the same.

It is a lot to write down all the equations, I suggest approaching it from the point of view of writing down the wave equation in one set of coordinates

i.e. in the Lorenz gauge in a vacuum with $\rho = J = 0$ you can write down the solution to Maxwell's equations in terms of the potentials as

$$\left(-\frac{\partial ^2}{\partial x^2} + \frac{1}{c^2} \frac{\partial ^2}{\partial t^2} \right) \varphi = 0$$
$$\left(-\frac{\partial ^2}{\partial x^2} + \frac{1}{c^2} \frac{\partial ^2}{\partial t^2} \right) A= 0$$

see https://en.wikipedia.org/w/index.php?title=Maxwell's_equations&oldid=879554402

Then you can concentrate on how the wave equation

$$-\frac{\partial ^2}{\partial x^2} + \frac{1}{c^2} \frac{\partial ^2}{\partial t^2} $$

transforms under a change of variables. It's just a mathematical fact that the wave equation is invariant under the second transformation you write with cosh and sinh, and not the first you write with cos and sin.

Perhaps you still need to know how one goes about doing the change of variables to change coordinates? It's basically done via the chain rule.