Invariance of the Lorentz transform

  • Thread starter tina21
  • Start date
  • #1
tina21
14
2
Homework Statement:
Prove the invariance of the electromagnetic wave equation by showing that the corresponding differential operator is an invariant.
Relevant Equations:
d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2 = d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2
of course y and z terms are invariant but for the x and t terms I am getting an additional factor of 1/1-v^2/c^2
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
24,067
15,765
Homework Statement: Prove the invariance of the electromagnetic wave equation by showing that the corresponding differential operator is an invariant.
Homework Equations: d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2 = d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2

of course y and z terms are invariant but for the x and t terms I am getting an additional factor of 1/1-v^2/c^2
Can you use Latex to show what you got?

https://www.physicsforums.com/help/latexhelp/
 
  • #3
tina21
14
2
I haven't used latex before. I hope these images are okay
 

Attachments

  • WhatsApp Image 2019-11-24 at 7.23.02 AM.jpeg
    WhatsApp Image 2019-11-24 at 7.23.02 AM.jpeg
    42.6 KB · Views: 68
  • WhatsApp Image 2019-11-24 at 7.23.16 AM.jpeg
    WhatsApp Image 2019-11-24 at 7.23.16 AM.jpeg
    24.8 KB · Views: 98
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
24,067
15,765
That doesn't look right. I prefer to differentiate a trial function. Imagine we have a function ##f(t', x')##, where ##t' = \gamma(t - vx/c^2), \ x' = \gamma(x - vt)##.

Now, we differentiate ##f## with respect to ##x## using the chain rule:

##\frac{\partial f}{\partial x} = \frac{\partial f}{\partial t'}\frac{\partial t'}{\partial x} + \frac{\partial f}{\partial x'}\frac{\partial x'}{\partial x} = \frac{\partial f}{\partial t'}(-\gamma v/c^2) + \frac{\partial f}{\partial x'}(\gamma)##

Now you need to take the second derivative of ##f## by repeating this process - and remember that you have to differentiate both terms using the chain rule, so you will get cross terms in ##\frac{\partial^2 f}{\partial t' \partial x'}##.

At the end, you can remove the trial function to leave, for example:

##\frac{\partial}{\partial x} = (-\gamma v/c^2)\frac{\partial }{\partial t'} + (\gamma)\frac{\partial}{\partial x'}##

That's what you get if you only wanted the first derivative.

Note that you can do the chain rule on differentials, but I prefer to have a function to differentiate, and then take the function away when I'm finished.
 
Last edited:
  • #5
tina21
14
2
yes, I tried doing it too. Made it a whole lot easier and I got the answer. Thanks so much !
 

Suggested for: Invariance of the Lorentz transform

Replies
2
Views
409
  • Last Post
Replies
24
Views
685
  • Last Post
Replies
4
Views
365
  • Last Post
Replies
1
Views
69
Replies
7
Views
642
Replies
35
Views
1K
Replies
3
Views
462
  • Last Post
Replies
2
Views
485
Top