Galilean and Lorentz invariant

In summary: I think I need more coffee this morning.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... I think I need more coffee this morning.Haha, yeah, I'm definitely feeling the need for more coffee too. I think I've got too many things going on this morning and I'm not thinking very clearly.
  • #1
Vrbic
407
18

Homework Statement


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

Homework Equations


Galilean transformation:
[itex]x'=x-vt, t'=t[/itex]
Lorentz transformation:
[itex]'x=\frac{x-vt}{\sqrt{1-(\frac{v}{c})^2}}, t'=\frac{t-\frac{tv}{c}}{\sqrt{1-(\frac{v}{c})^2}}[/itex]

The Attempt at a Solution


(A) I suppose that part with [itex]t_2,t_1[/itex] is not important for Galilean transformation it is same. And if I transform [itex] r_1, r_2[/itex] to [itex]r'_1=r_1-vt, r'_2=r_2-vt[/itex] 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?
 
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  • #2
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
kuruman said:
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.
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?
 

Attachments

  • quiz1_2006_sol.pdf
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  • #4
I see no mistake in the file that you attached.
Vrbic said:
In Lorentz case I have problem with time
Can you post exactly what you did and what kind of problem you have with time?
 
  • #5
kuruman said:
I see no mistake in the file that you attached.
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
kuruman said:
Can you post exactly what you did and what kind of problem you have with time?
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.
[tex] r'_1=\frac{r_1-vT_1}{\sqrt{1-(\frac{v}{c}})^2} [/tex]
[tex] r'_2=\frac{r_2-vT_2}{\sqrt{1-(\frac{v}{c}})^2} [/tex]
My question is if [itex]T_1==T_2[/itex]? But if I think about that now...[itex]r_1[/itex] and [itex]r_2[/itex] 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
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:
  • #8
kuruman said:
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?
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
Vrbic said:
My question is if [itex]T_1==T_2[/itex]? But if I think about that now...[itex]r_1[/itex] and [itex]r_2[/itex] 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?

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.
 
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  • #10
vela said:
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.
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
vela said:
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.
I understand, Even if in S system ##t_1=t_2## in S' ##t'_1=t'_2## may not be true.
 
  • #12
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
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
kuruman said:
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)?
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.
 
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  • #15
Vrbic said:
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?
Think about what value of ##t## you're referring to when you say ##r \to r' = r-vt##.
 
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  • #16
vela said:
Think about what value of ##t## you're referring to when you say ##r \to r' = r-vt##.
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
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
vela said:
... which makes makes me feel better now.
Me too.
 
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  • #19
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
Vrbic said:
So Galilean (again):
##(r_1−r_2)=(r′_1+v dt_1−r′_2−v dt_2)## , ##dt1≠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?
That's not what you are asked to find if it is Galilean invariant. Start with
## \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
kuruman said:
That's not what you are asked to find if it is Galilean invariant. Start with
## \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.
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?
 
  • #23
kuruman said:
You are making it more complicated than it really is. Check this out
https://en.wikipedia.org/wiki/Postulates_of_special_relativity
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
Vrbic said:
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?
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
vela said:
Hint: You can simplify the algebra a lot by noting, for example,
$$\Delta t' = t'_2 - t'_1 = \gamma(t_2 - \beta x_2) - \gamma(t_1 - \beta x_1) = \gamma[(t_2-t_1) - \beta(x_2-x_1)] = \gamma(\Delta t -\beta \Delta x).$$
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.
Thank you for your patience.
 
  • #26
Showing the function is Lorentz invariant is literally a few lines of algebra. I don't think we can actually give you more hints without essentially doing the problem for you.
 

Related to Galilean and Lorentz invariant

1. What is the difference between Galilean and Lorentz invariance?

Galilean invariance is a principle of classical mechanics that states that the laws of motion are the same in all inertial reference frames. Lorentz invariance, on the other hand, is a principle of special relativity that states that the laws of physics are the same in all inertial reference frames and that the speed of light is constant.

2. Why is Lorentz invariance considered more accurate than Galilean invariance?

Lorentz invariance takes into account the effects of time dilation and length contraction at high speeds, while Galilean invariance does not. This makes it a more accurate and comprehensive description of the laws of physics.

3. Can the Galilean and Lorentz invariance principles coexist?

No, the Galilean and Lorentz invariance principles are incompatible. Galilean invariance only holds true at low speeds, while Lorentz invariance applies to all speeds, including the speed of light.

4. How do the Galilean and Lorentz transformations differ?

The Galilean transformation is a set of equations that describe how measurements of time and space in one inertial reference frame relate to measurements in another inertial reference frame. The Lorentz transformation takes into account the effects of time dilation and length contraction at high speeds, making it more complex than the Galilean transformation.

5. What is the significance of the Galilean and Lorentz invariance principles?

These principles are important in understanding the fundamental laws of physics and how they apply in different reference frames. They also play a crucial role in special relativity and the development of modern physics theories, such as Einstein's theory of general relativity.

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