# Lorentz Transformations

1. Sep 5, 2013

### smallgirl

1.Hey, I am rather stuck on this question which you can see in the attached PDF. Now I began by taylor expanding the Lorentz Gamma factor (γ), up to second order and inserting this into the equation wherever I saw the gamma function, then rearranging. But I can't seem to get a function for F, for me my F comes out to be 1/2

2. The relevant equations can be seen in the PDF

3. I'm new to this, so not sure how to write it all out, however I did a taylor expansion of gamma, then substituted and rearranged and ended up with F as 1/2. I know this is wrong as obviously this would act on the y and z components which it shouldn't, as they are not affected by the transformation.

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2. Sep 5, 2013

### Staff: Mentor

Hi smallgirl. Welcome to Physics Forums.

Try making use of equations 1 and 2, and getting $Δ'\phi '-Δ\phi$. This should lead to partials with respect to t and t' only.

3. Sep 5, 2013

### smallgirl

Hey,

I don't understand the reasoning in doing this...

4. Sep 5, 2013

### Staff: Mentor

$$Δ'\phi'=Δ\phi+\frac{1}{c^2}(\frac{\partial ^2\phi'}{\partial t'^2}-\frac{\partial ^2\phi}{\partial t^2})$$
The only thing you need to do is transform $\frac{\partial ^2\phi'}{\partial t'^2}$ to the non-primed coordinate system.

EDIT: I corrected a minor error in the earlier version of this response.

Last edited: Sep 6, 2013
5. Sep 6, 2013

### smallgirl

Would this give me the function F?

6. Sep 6, 2013

### smallgirl

Ok so I've been staring at it for a while, not sure how to get ∂2ϕ′/∂t′2 to the non-primed coordinate system

7. Sep 6, 2013

### Staff: Mentor

For an arbitrary function f,
$$df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx$$
$$\frac{\partial f}{\partial t'}=\frac{\partial f}{\partial t}\frac{\partial t}{\partial t'}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial t'}=γ(\frac{\partial f}{\partial t}+v\frac{\partial f}{\partial x})$$

8. Sep 6, 2013

### smallgirl

Hey,

Yes I had actually got that, but I didn't think what I was doing was right... I figured I would then need to differentiate again?

9. Sep 6, 2013

### Staff: Mentor

Yes. You do need to differentiate again. I only presented this result so that I could get you pointed in the right direction. So, differentiate again and show us what you got.

10. Sep 6, 2013

### smallgirl

I get zero...

Well ok I get this, but it seems wrong

gamma((d2f/dt'dt)+ V(d2f/dt'dx))

Last edited: Sep 6, 2013
11. Sep 6, 2013

### smallgirl

, Hey, see the attachment for what I got... Ignore the previous post...

I am really struggling with this :-(... I can't see what I am meant to be doing to get to the answer.. Like I can't see the path

12. Sep 6, 2013

### Staff: Mentor

You need a little practice doing coordinate transformations. Check this out:

$$\frac{\partial^2 f}{\partial t'^2}=γ(\frac{\partial^2 f}{\partial t\partial t'}+v\frac{\partial^2 f}{\partial x\partial t'})$$

13. Sep 6, 2013

### smallgirl

Ahh your kidding me right? I had that originally a you can see from my previous post

Anyways, given that I don't know what f is.. I'm not sure how to proceed

14. Sep 6, 2013

### Staff: Mentor

f is just an arbitrary function of the primed and/or the unprimed variables. Just substitute phi for f. Remember that, at a given location in 4D space time, phi prime is the same as phi.

15. Sep 6, 2013

### smallgirl

Hey,

Yeah I did that and substituted it in and got this

I'm confused as to where and when I'd do a Taylor expansion? Do I just Taylor expand the Lorentz factor? But that seems a bit wrong....

16. Sep 6, 2013

### Staff: Mentor

This is not what I got. Check the math. I got:

$$\frac{\partial^2 \phi}{\partial t'^2}=γ^2\left(\frac{\partial ^2 \phi}{\partial t^2}+2v\frac{\partial^2 \phi}{\partial x\partial t}+v^2\frac{\partial^2 \phi}{\partial x^2}\right)$$

17. Sep 6, 2013

### smallgirl

Hey so I figured the maths out and arrived at what you have..... Not sure what I'm meant to do now... I'm utterly lost....

Hmm maybe taylor expand gamma and then substitute in?

Last edited: Sep 6, 2013
18. Sep 6, 2013

### Staff: Mentor

OK. You're almost done. Now you substitute this back into the equation relating the Δ's, and combine the two terms containing the second partials with respect to t, by reducing them to a common denominator. It looks like your final result is going to have a linear term in v/c, which apparently the original problem statement omitted. At this point, you can approximate the γ's by 1. You will have 3 terms on the rhs, one with a factor of v/c and two with factors of (v/c)2. In any event, the presence of the linear term in v/c will not change the main conclusion from the problem statement.

Chet

19. Sep 6, 2013

### smallgirl

I don't understand what you mean by reducing them to a common denominator..

20. Sep 6, 2013

### Staff: Mentor

$$γ^2-1=\frac{1}{1-(\frac{v}{c})^2}-1=\frac{(\frac{v}{c})^2}{1-(\frac{v}{c})^2}=γ^2(\frac{v}{c})^2$$