Special Relativity: Traveling to Regulus

[iknowsigularity]

Homework Statement

A visit to Regulus is on my bucket list. However, it is 77 light years away. Assuming I will live only another 40 years, can I make it to Regulus? How fast would I have to travel (at constant speed) to get there in 40 years?

Homework Equations

t1 = t0 (1/(sqrt(1-(u^2/c^2)))
L1 = L0 (sqrt(1-(u^2/c^2))

The Attempt at a Solution

I attempted to sub t0 out for L0/u in the time dilation equation but couldn't get the equation to solve for u?
I also attempted to logic my way to v1 = d1/t1 but just got confused.
It just seems like no matter what i do i have two variables to solve for or no way to get u out.
(btw by t1 and L1 i mean prime)
Thanks for any help.

 

[Orodruin]

It just seems like no matter what i do i have two variables to solve for or no way to get u out.

What two variables? Your only unknown is the speed.

Also, please show your work in terms of your resulting expressions.

 

[iknowsigularity]

What two variables? Your only unknown is the speed.

Also, please show your work in terms of your resulting expressions.

Assuming i sub t0 for L0/u then the only variable unknown is u but I am left with t1 = L0/u (1/(sqrt(1- (u^2/c^2))) and i don't know how to solve this for u. (Sorry I am on mobile so i can't add this to post)

 

[Orodruin]

From there it should be a simple question of arithmetic. To start you off: How can you get rid of the square root?

 

[iknowsigularity]

From there it should be a simple question of arithmetic. To start you off: How can you get rid of the square root?

You square both sides correct? How do you deal with the u^2/c^2 though?

 

[Orodruin]

Please write down what you have. You should have an equation for u^2. Just solve it.

Also, make sure that you use the correct quantities to define the speed. You want distance/time in the Earth frame.

 

[iknowsigularity]

Please write down what you have. You should have an equation for u^2. Just solve it.

Also, make sure that you use the correct quantities to define the speed. You want distance/time in the Earth frame.

So i solved for u^2 and got = L0/(t1^2(1-(1/c^2) but doing unit analysis gives me s^2 over m so this can't be correct :/

 

[Orodruin]

First of all: Check the dimensions of all your intermediate steps. A good place to start is wherever you got 1 - 1/c^2 as you cannot subtract something with dimensions T^2/L^2 from something dimensionless.

Second: Please state explicitly what you identify L0 and t1 as.

 

[iknowsigularity]

First of all: Check the dimensions of all your intermediate steps. A good place to start is wherever you got 1 - 1/c^2 as you cannot subtract something with dimensions T^2/L^2 from something dimensionless.

Second: Please state explicitly what you identify L0 and t1 as.

Looking again i see that i did it wrong the problem I am having is dealing with the 1-u^2/c^2 i don't know how to separate the u out. I was defining t1 as 40 as its the number of years he would need to reach into be alive and L0 i defined as 77 light years.

 

[iknowsigularity]

First of all: Check the dimensions of all your intermediate steps. A good place to start is wherever you got 1 - 1/c^2 as you cannot subtract something with dimensions T^2/L^2 from something dimensionless.

Second: Please state explicitly what you identify L0 and t1 as.

would the proper equation be u1 ^2 = L0^2/t1^2 * (1/(1-u0^2/c^2)) where u0 is just simply 77ly / 40 years? edit: nvm that can't be it either

 

[Orodruin]

If you want to learn relativity, or indeed any advanced physics topic, you will need to stop for some time to really think about what you are doing and how it fits into the theory rather than "plug and chug" with a few formulas. You need to understand where the formulas come from in order to use them properly rather than guessing.

 

[iknowsigularity]

If you want to learn relativity, or indeed any advanced physics topic, you will need to stop for some time to really think about what you are doing and how it fits into the theory rather than "plug and chug" with a few formulas. You need to understand where the formulas come from in order to use them properly rather than guessing.

I'm not trying to guess i am taking the course and i understand where the time dilation equation and length contraction equation come from, maybe it is as simple as saying 77lys / 40 years produces a speed over c so therefore it is not possible but i thought that maybe time dilation would have an affect on this result. All i was looking for was clarification not a speech on how i should take the subject seriously. Thanks for your help though. I have been thinking on this question for 2 days now.

 

[Orodruin]

I'm not trying to guess i am taking the course and i understand where the time dilation equation and length contraction equation come from, maybe it is as simple as saying 77lys / 40 years produces a speed over c so therefore it is not possible but i thought that maybe time dilation would have an affect on this result.

Yes, time dilation does have an effect. When I say that you need to learn where time-dilation and length contraction comes from I am saying so because I have several years of experience in teaching special relativity at Master level, going quite deep into the subject, and one of the more common misconceptions among students who have taken introductory SR in a modern physics course or similar is that they think they understand time-dilation and length contraction, but in reality they have completely missed the prerequisites and basic understanding of what they mean. Your problems in this thread seem to indicate the same problem.

It would also be easier to help you if you wrote out all your steps along with your logical argumentation instead of just referring to your end result.

 

[iknowsigularity]

Yes, time dilation does have an effect. When I say that you need to learn where time-dilation and length contraction comes from I am saying so because I have several years of experience in teaching special relativity at Master level, going quite deep into the subject, and one of the more common misconceptions among students who have taken introductory SR in a modern physics course or similar is that they think they understand time-dilation and length contraction, but in reality they have completely missed the prerequisites and basic understanding of what they mean. Your problems in this thread seem to indicate the same problem.

It would also be easier to help you if you wrote out all your steps along with your logical argumentation instead of just referring to your end result.

My apologies this question has given me a bad mood. Steps:

Assuming we are in ship's inertial frame as the ship goes toward regulus, the time we see for the ship to make the trip should be equal to L1 (the contracted distance between the ship and regulus) / u (the velocity of the ship) aka Distance/Time.

so we should be able to make this relationship L1/u = t0 since this is the length the person on the ship would see, so his clock would correlate.

so substituting L1 = L0 * Sqrt(1-(u^2/c^2)) gives us L0 *Sqrt(1-(u^2/c^2)) / u = t0 where L0 = 77 lys.

so since the person will only live for 40 years we would want to set t0 = to 40 and possibly solve for u?

I've rethought it. how is this logic?

 

[Orodruin]

Assuming we are in ship's inertial frame as the ship goes toward regulus, the time we see for the ship to make the trip should be equal to L1 (the contracted distance between the ship and regulus) / u (the velocity of the ship) aka Distance/Time.

Ok, I will stop you right here from the beginning. It may seem pedantic but it is an important point. The velocity of the ship is 0 in its rest frame. In the ship's rest frame, it is Regulus (and the Earth) that has a velocity. Due to symmetry, the speed of Regulus in the ship's rest frame is the same as the speed of the ship in Regulus' rest frame. It will help you avoid confusion if you always take some extra time in the beginning to think about which velocities you are dealing with and in what frames these velocities refer to.

so we should be able to make this relationship L1/u = t0 since this is the length the person on the ship would see, so his clock would correlate.

Assuming that t0 is the proper time on the ship (please write this out explicitly), yes. It takes Regulus L1/u to travel to the ship in the ship's rest frame.

A major issue is that you have used t0 to refer to the ship's proper time and L0 to refer to the proper distance between Earth and Regulus. This is also something that will easily confuse you as you are using the same subscript (0) for quantities that refer to different frames.

so since the person will only live for 40 years we would want to set t0 = to 40 and possibly solve for u?

I've rethought it. how is this logic?

Yes, what do you get when you solve for u?

 

[iknowsigularity]

Ok, I will stop you right here from the beginning. It may seem pedantic but it is an important point. The velocity of the ship is 0 in its rest frame. In the ship's rest frame, it is Regulus (and the Earth) that has a velocity. Due to symmetry, the speed of Regulus in the ship's rest frame is the same as the speed of the ship in Regulus' rest frame. It will help you avoid confusion if you always take some extra time in the beginning to think about which velocities you are dealing with and in what frames these velocities refer to.
Assuming that t0 is the proper time on the ship (please write this out explicitly), yes. It takes Regulus L1/u to travel to the ship in the ship's rest frame.

A major issue is that you have used t0 to refer to the ship's proper time and L0 to refer to the proper distance between Earth and Regulus. This is also something that will easily confuse you as you are using the same subscript (0) for quantities that refer to different frames.Yes, what do you get when you solve for u?

Solving for u:

L0 *Sqrt(1-(u^2/c^2)) / u = t0

First square both sides giving us L0^2 / u^2 * (1-(u^2/c^2)) = t0^2

next we distribute the L0^2 / u^2 so we get

L0^2 / u^2 - L0^2/c^2 = t0^2 (u^2 gets canceled out in the second term)

so solving for u we get

u^2 = L0^2 / (t0^2 + (L0^2/c^2)) (skipping some smaller steps so there might be a mistake here)

take the square root

u = sqrt( L0^2 / (t0^2 + (L0^2/c^2)) )

correct?

also for future reference since the ships intial position is Earth can you assume the proper length between the ship and regulus is = to the proper length between Earth and regulus?

 

[Orodruin]

u = sqrt( L0^2 / (t0^2 + (L0^2/c^2)) )

Looks fine. I would tidy it up a bit:
$$
u = \frac{c}{\sqrt{1 + \frac{c^2 t_0^2}{L_0}}}.
$$

There is also a much faster way to do this problem using the space-time interval $\Delta s^2 = c^2 \Delta \tau^2 = c^2 \Delta t^2 - \Delta x^2$, where $\Delta \tau$ is the proper time elapsed for an object following the straight world line with coordinate differences $\Delta t$ and $\Delta x$.

can you assume the proper length between the ship and regulus is = to the proper length between Earth and regulus?

There is no such thing as a proper length between the ship and Regulus. Proper length only makes sense for distances between objects at mutual rest. Trying to go down that road is essentially guaranteed to lead you astray.

 

[iknowsigularity]

Looks fine. I would tidy it up a bit:
$$
u = \frac{c}{\sqrt{1 + \frac{c^2 t_0^2}{L_0}}}.
$$

There is also a much faster way to do this problem using the space-time interval $\Delta s^2 = c^2 \Delta \tau^2 = c^2 \Delta t^2 - \Delta x^2$, where $\Delta \tau$ is the proper time elapsed for an object following the straight world line with coordinate differences $\Delta t$ and $\Delta x$.There is no such thing as a proper length between the ship and Regulus. Proper length only makes sense for distances between objects at mutual rest. Trying to go down that road is essentially guaranteed to lead you astray.

Thanks you so much! Sorry for the bad manners. Last question i would have is whether in this case you need to convert the units from lys to meters and years to seconds? i would assume so.

 

[Orodruin]

Thanks you so much! Sorry for the bad manners. Last question i would have is whether in this case you need to convert the units from lys to meters and years to seconds? i would assume so.

If you want to obtain a numerical value that makes sense, you need to take care to use the same units everywhere. However, the only thing that matters is that you are consistent. In your given units, $c = 1$ ly/year and so it is actually much easier not to convert to meters and seconds.

Edit: Of course, you can still do the conversion. If you are worried I suggest that you do it both ways and verify that you get exactly the same result.

 

[iknowsigularity]

If you want to obtain a numerical value that makes sense, you need to take care to use the same units everywhere. However, the only thing that matters is that you are consistent. In your given units, $c = 1$ ly/year and so it is actually much easier not to convert to meters and seconds.

Edit: Of course, you can still do the conversion. If you are worried I suggest that you do it both ways and verify that you get exactly the same result.

So i drop c to 1 and plug in 40 and 77 for the values. i got 0.21 would this be in percent of the speed of light or is it wrong?

 

[Orodruin]

So i drop c to 1

While it is common that physicists work in units where $c = 1$ (unitless), you are not. In your given units $c = 1$ ly/year.

and plug in 40 and 77 for the values.

The units are important. You are plugging in 1 ly/year, 40 years, and 77 ly, not dimensionless numbers. The units will cancel inside the square root of my expression and leave $c$ divided by the value of the square root.

i got 0.21 would this be in percent of the speed of light or is it wrong?

I suspect you did a numerical mistake. The numerical value should be 0.89.

If you keep the units properly, you will get 0.89 ly/year. Of course, this is indeed 0.89c, since c = 1 ly/year.

 

[iknowsigularity]

While it is common that physicists work in units where $c = 1$ (unitless), you are not. In your given units $c = 1$ ly/year.The units are important. You are plugging in 1 ly/year, 40 years, and 77 ly, not dimensionless numbers. The units will cancel inside the square root of my expression and leave $c$ divided by the value of the square root.I suspect you did a numerical mistake. The numerical value should be 0.89.

If you keep the units properly, you will get 0.89 ly/year. Of course, this is indeed 0.89c, since c = 1 ly/year.

I redid the numbers and got 0.89c. :D thank you so much so excited to finally understand this. Thanks for sticking with me.