Relativity: Cosmic ray proton traveling to Earth

[Mawl]

Homework Statement

Q: One of the proton energies detected in cosmic rays coming to Earth from space has a kinetic energy of 3.0×10²⁰ eV. Determine:
Measure the time needed for proton to travels all the way to the earth if it's measured relative to the earth's and the proton's frame of reference itself.

Relevant Equations

Given:
Ek= 3.0×10²⁰ eV
∆x= 9.8×10⁴ ly
v=0.99c

Now, the teacher had already given me the key answers that is:
1. ∆t relative to the earth's=9800 years
2. ∆t relative to the proton's=9.7 s
3. For the (a) answer, v= 0.99c

For the first answer, and if my understanding were truly correct, I can just input the given ∆x from the question and divide it by 1c (I rounded up the given v):
∆t relative to the earth's=∆x/v
=(9.8×10⁴ cy)/1c
=98000 years
Hence giving me the answer, 98000 years.

For the second answer though, I have literally zero idea about am I supposed to find the ∆t relative to the proton's. Am I supposed to change the units? If so, how? And how can the result be 9.7 s?

I've been looking for the answer, but I just couldn't wrapped it around my head. It feels like I'm missing a knowledge to solve this problem, but I have no clue of what's that supposed to be.

 

[Orodruin]

Are you familiar with time dilation?

 

[Mawl]

Are you familiar with time dilation?

I understand some gist of it. But the units this question gave makes me lost from the track. Based on the equation:
∆t'=∆tƔ
I thought I could've just changed 98000 years into seconds and then just plug in the rest of it to find the answer. But this was blantantly wrong since the numbers were too big.
Even though I know it was so obviously wrong, I just can't figured it out where exactly my mistakes are...

 

[PeroK]

I understand some gist of it. But the units this question gave makes me lost from the track. Based on the equation:
∆t'=∆tƔ
I thought I could've just changed 98000 years into seconds and then just plug in the rest of it to find the answer. But this was blantantly wrong since the numbers were too big.
Even though I know it was so obviously wrong, I just can't figured it out where exactly my mistakes are...

The Earth and proton rest frames have a high relative speed - must be very close to $c$. You could calculate this speed and apply time dilation and/or length contraction to transform the data you have in the Earth frame to the proton frame.

That said, it would be nice if you could get the answer more directly from knowing the energy of the proton. If we take the proton mass to be $938 MeV$ and its KE to be $3 \times 10^{20}eV$ in the Earth frame, what does that tell us?

 

[PeroK]

PS it is quite tricky to get an accurate value for the speed of the proton - because it is so close to $c$. I would recommend avoiding this calculation and using the energy method. And, even if you calculate $v$, what precisely are you going to do with it?

I hope this makes some sort of sense. Just say if it doesn't!

 

[Mawl]

PS it is quite tricky to get an accurate value for the speed of the proton - because it is so close to $c$. I would recommend avoiding this calculation and using the energy method. And, even if you calculate $v$, what precisely are you going to do with it?

I hope this makes some sort of sense. Just say if it doesn't!

I'm going to put the v inside the Ɣ equation, where Ɣ=1/(√(1-v²/c²)),
because the results are really close to 1c, I'm going to round it up into 0.99c and then put the results into the time dilation or length contraction which going to gives me... a really big number.
This is probably the exact part that gets me really confused. What am I supposed to do next to make the results into a really small amounts, which is 9.7 s. The result values were even less than 10 s.
Should I change the units first? I'm pretty sure it won't affects the calculations at all. My teacher haven't really showed us how to solve these kind of problems and I'm afraid that my current knowledges couldn't really comprehend on how to solve it.. sorry.

 

[PeroK]

I'm going to put the v inside the Ɣ equation, where Ɣ=1/(√(1-v²/c²)),
because the results are really close to 1c, I'm going to round it up into 0.99c and then put the results into the time dilation or length contraction which going to gives me... a really big number.

That's not accurate enough. $v$ is much closer to $c$ than that. In any case, do you need $v$ to calculate $\gamma$? Or, is there another way to find $\gamma$?

 

[Mawl]

Ah I see, I can use
E=KE+Eo
Which means

ymoc²=KE+moc²
y=KE+moc²/moc²
Then I can input the numbers:
y=48.06+(1.673×10^-27)(3×10^8)²/(1.673×10^-27)(3×10^8)²
y≈3.19187×10^11

And then plug it into length contraction's:
L=yLo
L=(3.19187×10^11)(9.8×10^4 ly)
L≈3.1280326×10^16 ly

...am I in the right track?

 

[PeroK]

Ah I see, I can use
E=KE+Eo
Which means

ymoc²=KE+moc²
y=KE+moc²/moc²
Then I can input the numbers:
y=48.06+(1.673×10^-27)(3×10^8)²/(1.673×10^-27)(3×10^8)²
y≈3.19187×10^11

Yes.

And then plug it into length contraction's:
L=yLo
L=(3.19187×10^11)(9.8×10^4 ly)
L≈3.1280326×10^16 ly

...am I in the right track?

That looks like length extension.

The idea is that you treat the proton like a moving clock, which runs slower by a factor of $\gamma$ between the two events. Note that it's important that the two events are on the proton's worldline: i.e. on its trajectory through spacetime.

 

[Mawl]

Yes.

That looks like length extension.

The idea is that you treat the proton like a moving clock, which runs slower by a factor of $\gamma$ between the two events. Note that it's important that the two events are on the proton's worldline: i.e. on its trajectory through spacetime.

Whoa, I was too overwhelmed and didn't even realize it's actually extended, haha.

Okay, so based on that idea, it means by definition that the more faster things are moving by y, the more stretched it'll be relative to its own point of view compared to Earth's. Correct?

Oh wait, but how to convert light-years into seconds only? I mean, the numbers were too big, I just don't really see how it can be 9.7 seconds.

 

[PeroK]

Whoa, I was too overwhelmed and didn't even realize it's actually extended, haha.

Okay, so based on that idea, it means by definition that the more faster things are moving by y, the more stretched it'll be relative to its own point of view compared to Earth's. Correct?

No. The $\Delta x$ you were given was actually a length in the Earth's frame. I assume it was the constant distance to the source of the proton. That $\Delta x$ is contracted by a factor of $\gamma$ in the proton frame. You could calculate $\Delta x'$, the distance between the source and Earth in the proton frame. And, in that frame, the Earth is moving at close to the speed of light. That gives you the time in the proton's frame for the Earth to move to it.

Oh wait, but how to convert light-years into seconds only? I mean, the numbers were too big, I just don't really see how it can be 9.7 seconds.

Alternatively, you could take the time between the events in the Earth's frame (98,000 years) and that time is less by a factor of $\gamma$ in the proton's frame.

In either case, you have to convert years to seconds.

 

[Mawl]

No. The $\Delta x$ you were given was actually a length in the Earth's frame. I assume it was the constant distance to the source of the proton. That $\Delta x$ is contracted by a factor of $\gamma$ in the proton frame. You could calculate $\Delta x'$, the distance between the source and Earth in the proton frame. And, in that frame, the Earth is moving at close to the speed of light. That gives you the time in the proton's frame for the Earth to move to it.

Alternatively, you could take the time between the events in the Earth's frame (98,000 years) and that time is less by a factor of $\gamma$ in the proton's frame.

In either case, you have to convert years to seconds.

AHH I GET ITT
So basically, the real reason why I couldn't get the correct answer was because I had always forgotten to change the ∆t to seconds. I've always had thought that I could just change the units at the end of the result and left the ∆t in years, which ended up screwing all of my calculations ahaha.

So,
∆t=98000(365.25)(24)(3600)
∆t≈3.09×10¹² s

∆t'=∆ty
∆t'=(3.09×10¹²)/(3.19×10¹¹)
∆t'≈9.7 s

Thank you so much sir! I might've probably gonna stuck in this sloppy mistake for the rest of the week if it wasn't because of your enlightenment.

 

[Orodruin]

It doesn’t really matter when you convert your units. You should get the same result regardless.