# Homework Help: Galilean and Lorentz invariant

1. Oct 19, 2016

### Vrbic

1. The problem statement, all variables and given/known data
Professor C. Rank claims that a charge at $(r_1, t_1)$ will contribute to the air pressure
at $(r_2, t_2)$ by an amount $B \sin[C(|r_2 − r_1|^2− c^2|t_2 − t_1|^2)]$, where B and C are constants.
(A) Is this effect Galilean invariant?
(B) Is this effect Lorentz invariant?

2. Relevant equations
Galilean transformation:
$x'=x-vt, t'=t$
Lorentz transformation:
$'x=\frac{x-vt}{\sqrt{1-(\frac{v}{c})^2}}, t'=\frac{t-\frac{tv}{c}}{\sqrt{1-(\frac{v}{c})^2}}$

3. The attempt at a solution
(A) I suppose that part with $t_2,t_1$ is not important for Galilean transformation it is same. And if I transform $r_1, r_2$ to $r'_1=r_1-vt, r'_2=r_2-vt$ the extra terms are deducated.
So it is Galilean invariant? True?
(B) In Lorentz case I have problem with times. Could you suggest some point of view?

2. Oct 20, 2016

### kuruman

The Galilean part is correct.
Your problem with times in the Lorentz case may have something to do with the incorrect expression for $t'$ that you have posted.

3. Oct 20, 2016

### Vrbic

Ehm no :) Sorry it just typo :) Ok the reason why I'm dealing with such easy things is that I have results that claim opposite than me. Check attached file. Is there mistake or I'm wrong?

#### Attached Files:

• ###### quiz1_2006_sol.pdf
File size:
148.6 KB
Views:
50
4. Oct 20, 2016

### kuruman

I see no mistake in the file that you attached.
Can you post exactly what you did and what kind of problem you have with time?

5. Oct 21, 2016

### Vrbic

You didn't? I claimed that this is Galilean invariant and you agree. In this file is an answer NO. So what I'm missing? And for Lorentzian there is an answer YES. I mean it is reversely.

6. Oct 21, 2016

### Vrbic

I intuitively mean, that it is not Lorentz invariant. But if I would like to do it carefully I would I put there the transformed terms. And I'm not sure if the time which appears in r_1 and r_2 transformation is same or what should be there.
$$r'_1=\frac{r_1-vT_1}{\sqrt{1-(\frac{v}{c}})^2}$$
$$r'_2=\frac{r_2-vT_2}{\sqrt{1-(\frac{v}{c}})^2}$$
My question is if $T_1==T_2$? But if I think about that now...$r_1$ and $r_2$ are in same system so if something move with respect to one it moves same in respect to second so times should be same. Am I right? And in this file it is wrong. Do you agree?

7. Oct 21, 2016

### kuruman

Sorry for confusing you. I read the question rather hastily. No, the expression is not Galilean invariant. Did you actually calculate $|{r'}_2^2-{r'}_1^2|$ under the Galilean transformation? If so, what did you get?
For the Lorentz part you don't even need to do that. The sine function can be thought of representing a traveling wave. What is the speed of that wave?

On edit: In view of @vela's comment below, I meant $\lvert \vec{r}'_2 - \vec{r}'_1 \rvert^2$.

Last edited: Oct 21, 2016
8. Oct 21, 2016

### vela

Staff Emeritus
The problem involves $\lvert \vec{r}_2 - \vec{r}_1 \rvert^2$, not $|{r'}_2^2-{r'}_1^2|$. It looks invariant under a Galilean transformation to me since it depends on the spatial distance between the two points.

9. Oct 21, 2016

### vela

Staff Emeritus

No, you can't assume $T_1 = T_2$. As measured in the S frame, event 1 occurs in space at $\vec{r}_1$ at time $T_1$, and event 2 occurs in space at $\vec{r}_2$ at time $T_2$. There's no reason to assume the events are simultaneous.

By the way, you should find that the expression is Lorentz invariant.

10. Oct 21, 2016

### kuruman

I interpreted "Galilean invariant" to mean that
$\sin \left[ C \left( | \vec{r}'_2-{\vec{r}'}_1|^2-c^2|{t'}_2-{t'}_1|^2\right) \right]=\sin \left[ C \left( | (\vec{r}_2-\vec{r}_1|^2-c^2|{t}_2-{t}_1|^2 \right) \right]$
under the Galilean transformation, $\vec{r}'=\vec{r}-\vec{v}~t;~~t'=t$. Was I wrong (again)?

11. Oct 21, 2016

### Vrbic

I understand, Even if in S system $t_1=t_2$ in S' $t'_1=t'_2$ may not be true.

12. Oct 21, 2016

### Vrbic

So for Galilean transformation:
$r'_1=r_1-vt, r'_2=r_2-vt, t'_1=t_1, t'_2=t_2$ => $(r_1-r_2)=(r'_1+vt -r'_2-vt)=(r'_1-r'_2)$ and $(t_1-t_2)=(t'_1-t'_2)$. Am I right? And from this it has to be Galilean invariant. Do you agree?

13. Oct 21, 2016

### Vrbic

For Lorentz transformation:
I'm not rewriting the rules again.
$(r_1-r_2)=\gamma (r'_1+v t'_1-r'_2-v t'_2)$ and $(t_1-t_2)=\gamma (t'_1+\frac{r'_1 v}{c}-t'_2 - \frac{r'_2 v}{c})$ where $\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}$ and $v$ is velocity between S and S'. From this I suppose it isn't Lorentz invariant.

14. Oct 21, 2016

### vela

Staff Emeritus
No, that's what I took it to mean as well, but I see the mistake I made. I agree with the provided answer it's not invariant under a Galilean transformation, which makes makes me feel better now.

15. Oct 21, 2016

### vela

Staff Emeritus
Think about what value of $t$ you're referring to when you say $r \to r' = r-vt$.

16. Oct 21, 2016

### Vrbic

Ha :) That's the reason why I wrote I have problem with meaning of times. Now I probably understand...now recompute it. $t$ is in such transformation difference between the time when the systems met each other and time of measurement. True?

17. Oct 21, 2016

### Vrbic

So Galilean (again):
$(r_1−r_2)=(r′_1+v dt_1−r′_2−v dt_2)$ , $dt_1\neq dt_2$ because $r_1$ and $r_2$ has generaly different position so met S' system in diferent time. So it is not Galilean invarinat. Do you agree with my explanation?

18. Oct 21, 2016

### kuruman

Me too.

19. Oct 22, 2016

### Vrbic

Is my $dt$ definition right? I'm struggling with Lorentz invariant. There is many terms but I can't see any connection among them.

20. Oct 22, 2016

### kuruman

$\vec{r}'_1=\vec{r}_1-\vec{v}~t_1$, $\vec{r}'_2=\vec{r}_2-\vec{v}~t_2$, $t'_1=t_1$ and $t'_2=t_2$.
Then note that $| \vec{r}'_2-{\vec{r}'}_1|^2-c^2|{t'}_2-{t'}_1|^2= (\vec{r}'_2-\vec{r}'_1) \cdot (\vec{r}'_2-\vec{r}'_1)-c^2|{t'}_2-{t'}_1|^2$
Then replace the primed quantities on the right hand side with unprimed ones as given by the Galilean transformation and see what you get.

21. Oct 22, 2016

### Vrbic

Ok, understand I done it. It is for me an esyier part. But if I do it in Lorentz case, there arise this:
$|r_2-r_1|^2-c^2|t_2-t_1|^2=\gamma ^2 [r' ^2_2+r' ^2_1+vt'^2_2+vt'^2_1+2vr'_2t'_2-2r'_1r'_2-2vr'_2t'_1-2vr'_1t'_1-2v^2t'_1t'_2+2vr'_1t'_1-((ct'_2)^2+(ct'_1)^2+(vr'_2)^2/c^2+(vr'_1)^2/c^2+2vt'_2r'_2-2t'_1t'_2-2vt'_2r'_1-2vt'_1r'_2-2v^2r'_1r'_2/c^2+2vt'_1r'_1)]$ (I hope I didn't do any mistake :) )
And I'm not sure what should I use to reduce it... any advice?

22. Oct 22, 2016

### kuruman

23. Oct 22, 2016

### Vrbic

Ok, I have seen it, the problem is that I'm not able to apply it on this particular problem. As you said, I'm doing it more complicated. On the other hand, way which I used, should lead to solution. It is complicated way but it should. No? If not, what is wrong on that?

24. Oct 22, 2016

### vela

Staff Emeritus
Hint: You can simplify the algebra a lot by noting, for example,
$$\Delta t' = t'_2 - t'_1 = \gamma(t_2 - \beta x_2) - \gamma(t_1 - \beta x_1) = \gamma[(t_2-t_1) - \beta(x_2-x_1)] = \gamma(\Delta t -\beta \Delta x).$$

25. Oct 23, 2016

### Vrbic

Ok, but this is just shorthand. I will have only half of terms, but nothing changes. Could you advise me more. I try to think but I stuck in this and I don't believe I can get out alone.