Particle moving at speed ##bc## relative to laboratory frame

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

My working is

(a) $\Delta t = \frac{a}{\sqrt{1 - b^2}}~ns$
(b) $\frac{abc}{(1 \times 10^9)\sqrt{1 - b^2}}~m= d_1$
(c) By symmetry, $d_2 = -d_1$

$-\frac{abc}{(1 \times 10^9) \sqrt{1 - b^2}}~m= d_2$, however, I'm not sure whether my explanation for (c) is enought, does someone please know how to explain it better?

Thanks!

 

[Orodruin]

It is not enough as it is incorrect. The particle lifetime in its rest frame is not the same as in the lab frame.

 

[Ibix]

You don't seem to have used $a$ at all.

For (c), think about what the particle frame says about the lab's velocity. Is it consistent with your answer?

 

[member 731016]

It is not enough as it is incorrect. The particle lifetime in its rest frame is not the same as in the lab frame.

You don't seem to have used $a$ at all.

For (c), think about what the particle frame says about the lab's velocity. Is it consistent with your answer?

Thank you for your replies @Orodruin and @Ibix!

Sorry there was a typo in my working, what I was intending was this:

(a) $\Delta t = \frac{a}{\sqrt{1 - b^2}}~ns$
(b) $\frac{abc}{(1 \times 10^9)\sqrt{1 - b^2}}~m= d_1$
(c) By symmetry, $d_2 = -d_1$

$-\frac{abc}{(1 \times 10^9) \sqrt{1 - b^2}}~m= d_2$

Thanks!

 

[Orodruin]

(c) By symmetry, $d_2 = -d_1$

This argument is still wrong.

 

[member 731016]

This argument is still wrong.

Thank you for your reply @Orodruin!

Do you please know why? Going from the spaceship frame and the earth frame should be symmetrical should it not?

Thanks!

 

[Hill]

Using the number $(1 \times 10^9)$ in (b) and (c) is also wrong.

 

[member 731016]

Using the number $(1 \times 10^9)$ in (b) and (c) is also wrong.

Thank you for your reply @Hill!

Do you please know why it is wrong? The $(1 \times 10^9)$ factor comes from converting the time in nanoseconds to the time in seconds

Thanks!

 

[Hill]

Thank you for your reply @Hill!

Do you please know why it is wrong? The $(1 \times 10^9)$ factor comes from converting the time in nanoseconds to the time in seconds

Thanks!

Why do you convert?

 

[member 731016]

Why do you convert?

Thank you for your reply @Hill!

So that I can I get the distance in $m$.

Thanks!

 

[Hill]

Thank you for your reply @Hill!

So that I can I get the distance in $m$.

Thanks!

1. There is no such requirement in the problem.
2. It does not ensure that you will get the distance in $m$, anyway.

 

[member 731016]

Thank you for your reply @Hill! So removing the factor $1 \times 10^9$ gives the correct solution thought?


(a) $\Delta t = \frac{a}{\sqrt{1 - b^2}}$
(b) $\frac{abc}{\sqrt{1 - b^2}} = d_1$
(c) By symmetry, $d_2 = -d_1$

$-\frac{abc}{\sqrt{1 - b^2}} = d_2$

Thanks!

 

[Hill]

Thank you for your reply @Hill! So removing the factor $1 \times 10^9$ gives the correct solution thought?


(a) $\Delta t = \frac{a}{\sqrt{1 - b^2}}$
(b) $\frac{abc}{\sqrt{1 - b^2}} = d_1$
(c) By symmetry, $d_2 = -d_1$

$-\frac{abc}{\sqrt{1 - b^2}} = d_2$

Thanks!

The answer to (c) is still wrong, as is said in #5.

If it were correct, then the laboratory has moved in the particle's reference frame during the lifetime of the particle the distance $\frac{abc}{\sqrt{1 - b^2}}$. It moved with the speed $bc$. Thus, the lifetime of the particle in the particle's reference frame was $\frac{abc}{(\sqrt{1 - b^2})(bc)} = \frac{a}{\sqrt{1 - b^2}}$. But it should be $a$. You get a contradiction.

 

[member 731016]

The answer to (c) is still wrong, as is said in #5.

If it were correct, then the laboratory has moved in the particle's reference frame during the lifetime of the particle the distance $\frac{abc}{\sqrt{1 - b^2}}$. It moved with the speed $bc$. Thus, the lifetime of the particle in the particle's reference frame was $\frac{abc}{(\sqrt{1 - b^2})(bc)} = \frac{a}{\sqrt{1 - b^2}}$. But it should be $a$. You get a contradiction.

Thank you for your reply @Hill!

Sorry do you please know how to what the asymmetry is between the results? I don't understand how to find the asymmetry between them.

Thanks!

 

[Hill]

Thank you for your reply @Hill!

Sorry do you please know how to what the asymmetry is between the results? I don't understand how to find the asymmetry between them.

Thanks!

There is asymmetry in the distances like there is asymmetry in the lifetimes.

 

[Ibix]

Sorry do you please know how to what the asymmetry is between the results? I don't understand how to find the asymmetry between them.

See #3. What is the lab's velocity in the particle frame?

 

[member 731016]

See #3. What is the lab's velocity in the particle frame?

In particle frame, particle is at rest, lab is moving. In lab frame, lab is at rest, particle is moving.

In particle frame, particle is at rest, lab is moving at $-bc$ ($bc$ to the left).

In lab frame, lab is at rest, particle is moving at $bc$ to the right.

Is that please what you were getting at?

Thanks!

 

[Ibix]

In particle frame, particle is at rest, lab is moving at $-bc$ ($bc$ to the left).

Yes. So how far does the lab move according to the particke frame?

 

[member 731016]

Yes. So how far does the lab move according to the particke frame?

Thank you for yor reply @Ibix! I'll think about it overnight and get back to you in tomorrow!

 

[member 731016]

Yes. So how far does the lab move according to the particke frame?

THank you for your reply @Ibix! $v\Delta t = d$

$-bca = d$.

Is that please correct?

Thanks!

 

[Ibix]

THank you for your reply @Ibix! $v\Delta t = d$

$-bca = d$.

Is that please correct?

Thanks!

Looks right to me.