Spaceship, Earth and proton problem in special relativity

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

My solution to part (a) is $u'_x = 0.3c$, however, I'm confused by (b) and (c). By definition in (b) and (c), the proper velocity of the proton is the speed of the proton in the proton-frame (frame where proton is at rest). So would the answer for (b) and (c) not both be zero?

Does someone please know what I'm missing here?

Thanks!

 

[Ibix]

I'm afraid that your answer to (a) is obviously wrong. The ship is doing 0.8c towards Earth and fires a proton towards Earth. The proton's speed in the Earth frame must be greater than the ship's.

"Proper velocity" (sometimes called "celerity") is a strange beast. Your supposition that it's zero is reasonable; unfortunately proper velocity doesn't follow the usual meaning of "proper". The velocity (what the question calls "ordinary velocity") is the distance travelled by the object in some frame divided by the time taken in that same frame. But "proper velocity" is the distance moved by the object in some frame divided by the proper time of the object.

Hint: the proper time of the object is a factor of ##\gamma# less than the coordinate time.

 

[member 731016]

I'm afraid that your answer to (a) is obviously wrong. The ship is doing 0.8c towards Earth and fires a proton towards Earth. The proton's speed in the Earth frame must be greater than the ship's.

"Proper velocity" (sometimes called "celerity") is a strange beast. Your supposition that it's zero is reasonable; unfortunately proper velocity doesn't follow the usual meaning of "proper". The velocity (what the question calls "ordinary velocity") is the distance travelled by the object in some frame divided by the time taken in that same frame. But "proper velocity" is the distance moved by the object in some frame divided by the proper time of the object.

Hint: the proper time of the object is a factor of ##\gamma# less than the coordinate time.

Thank you for your reply @Ibix!

Sorry how can the protons speed be greater than the ships?

Using the formula for velocity addition $u'_x = \frac{u_x - v}{1 - \frac{v}{c^2}}u_x = \frac{0.8c - 0.6c}{1 - \frac{0.6c \times 0.8c}{c^2}} = 0.38c$

There is another formula I think $u'_x = \frac{u_x + v}{1 + \frac{v}{c^2}}u_x$ but the one that they taught me was the $u'_x = \frac{u_x - v}{1 - \frac{v}{c^2}}u_x$ I think.

Also do you please know whether there is a formula for proper velocity? I have never seen it written down anywhere.

Thanks!

 

[member 731016]

I'm afraid that your answer to (a) is obviously wrong. The ship is doing 0.8c towards Earth and fires a proton towards Earth. The proton's speed in the Earth frame must be greater than the ship's.

"Proper velocity" (sometimes called "celerity") is a strange beast. Your supposition that it's zero is reasonable; unfortunately proper velocity doesn't follow the usual meaning of "proper". The velocity (what the question calls "ordinary velocity") is the distance travelled by the object in some frame divided by the time taken in that same frame. But "proper velocity" is the distance moved by the object in some frame divided by the proper time of the object.

Hint: the proper time of the object is a factor of ##\gamma# less than the coordinate time.

Also sorry, I'm confused how to find the proper velocity since we are given no distance or time to use in a equation of the form speed = distance/time. Maybe you please know whether this may be a purely symbolic problem?

Thanks!

 

[Ibix]

Sorry how can the protons speed be greater than the ships?

How can it not be? If the ship fires the proton towards Earth, which one is going to reach Earth sooner, the ship or the proton? So which one is going faster relative to the Earth?

Using the formula for velocity addition $u'_x = \frac{u_x - v}{1 - \frac{v}{c^2}}u_x = \frac{0.8c - 0.6c}{1 - \frac{0.6c \times 0.8c}{c^2}} = 0.38c$

What is the speed of the Earth in the ship's rest frame? What is its velocity? What is the speed of the proton in the ship's rest frame? What is its velocity? Does that formula use speed or velocity?

There is another formula I think $u'_x = \frac{u_x + v}{1 + \frac{v}{c^2}}u_x$ but the one that they taught me was the $u'_x = \frac{u_x - v}{1 - \frac{v}{c^2}}u_x$ I think.

You can use either formula as long as you keep track of what signs you are using. The first one is probably the better one.

Also do you please know whether there is a formula for proper velocity? I have never seen it written down anywhere.

If my velocity is $v$ and I travel for time $\Delta t$, how far do I travel? How much proper time elapsed for me? Can you combine those two answers to get my proper velocity?

 

[member 731016]

How can it not be? If the ship fires the proton towards Earth, which one is going to reach Earth sooner, the ship or the proton? So which one is going faster relative to the Earth?

What is the speed of the Earth in the ship's rest frame? What is its velocity? What is the speed of the proton in the ship's rest frame? What is its velocity? Does that formula use speed or velocity?

You can use either formula as long as you keep track of what signs you are using. The first one is probably the better one.

If my velocity is $v$ and I travel for time $\Delta t$, how far do I travel? How much proper time elapsed for me? Can you combine those two answers to get my proper velocity?

Thank you for your reply @Ibix!

Proton should be moving faster than spaceship so will reach earth sooner.

Formula uses velocity and it is confusing that both give different answer's. Speed of earth in ships rest frame is 0.8c. Velocity of earth in ships rest frame is -0.8c.

Speed and velocity of proton in ships rest frame are 0.6c and 0.6c.

We want to find velocity of proton in earth-frame. I think the formula for $u'_x$ with the positive sign in the correct formula to use, so I think I have been using wrong formula. Should be $u'_x = \frac{u_x + v}{1 + \frac{v}{c^2}}u_x = \frac{0.8c + 0.6c}{1 + \frac{0.6c \times 0.8c }{c^2}}$

Now for finding proper velocity formula,

$v \Delta t_0 = d$ where proper time $t_0 = \frac{\Delta t}{\gamma}$

Solve for $v$, then $v = \frac{\gamma d}{\Delta t}$

Are these statements please correct?

Thanks!

 

[Ibix]

Formula uses velocity and it is confusing that both give different answer's.

Unfortunately, there's no law that says you must always use the same sign convention, and you have to get used to the fact that formulae can be based on subtly different conventions. That's one reason why it's a good idea to think about whether your answer is reasonable or not, not just grab a formula and plug numbers in.

We want to find velocity of proton in earth-frame. I think the formula for $u'_x$ with the positive sign in the correct formula to use,

Either formula will work as long as you understand their sign conventions. The one with the plus signs treats positive $u_x$ and $v$ as meaning the objects are moving towards each other. The one with the negative signs treats $u_x$and $v$ as in the same direction if they have the same sign. So you could also have used$$\begin{eqnarray*}
u'_x&=&\frac{u_x-v}{1-u_xv/c^2}\\
&=&\frac{0.6-(-0.8)}{1-0.6\times(-0.8)}
\end{eqnarray*}$$and you would get the same answer. I find this form easier to use since you usually refer to velocities with the same sign in the same direction, but you can use whichever works for you.

Are these statements please correct?

Assuming you are using $v$ to mean proper velocity there, yes. I'd suggest something like $v_p$ for proper velocity and $v$ for ordinary velocity, and then your last formula becomes $v_p=\gamma v$.

 

[member 731016]

Unfortunately, there's no law that says you must always use the same sign convention, and you have to get used to the fact that formulae can be based on subtly different conventions. That's one reason why it's a good idea to think about whether your answer is reasonable or not, not just grab a formula and plug numbers in.

Either formula will work as long as you understand their sign conventions. The one with the plus signs treats positive $u_x$ and $v$ as meaning the objects are moving towards each other. The one with the negative signs treats $u_x$and $v$ as in the same direction if they have the same sign. So you could also have used$$\begin{eqnarray*}
u'_x&=&\frac{u_x-v}{1-u_xv/c^2}\\
&=&\frac{0.6-(-0.8)}{1-0.6\times(-0.8)}
\end{eqnarray*}$$and you would get the same answer. I find this form easier to use since you usually refer to velocities with the same sign in the same direction, but you can use whichever works for you.

Assuming you are using $v$ to mean proper velocity there, yes. I'd suggest something like $v_p$ for proper velocity and $v$ for ordinary velocity, and then your last formula becomes $v_p=\gamma v$.

Thank you for your reply @Ibix!

That is a good idea to use that notation. I wil doing those problems and I should have the solution in no more than 14 hours.

Thanks!

 

[member 731016]

Sorry, I am getting something strange using the formula @Ibix . I would please appreciate you guidence.

For (b) I get $v_p = \gamma v = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{0.95c}{\sqrt{1 - \frac{(0.95c)^2}{c^2}}} = 3.04c$ which is non-physical. This is using $0.95c = u'_x = v$ from (a).

Do you please know whether the Lorentz factor is meant to be the other way (dividing v_p)?

Thanks for any help!

 

[jbriggs444]

For (b) I get $v_p = \gamma v = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{0.95c}{\sqrt{1 - \frac{(0.95c)^2}{c^2}}} = 3.04c$ which is non-physical.

"proper velocity" has no speed limit at $c$. It is not a velocity as measured in a single inertial frame.

Consider, for example, a passenger speeding 4 light years to Alpha Centauri at a ludicrous fraction of the speed of light so that the trip takes mere seconds of his elapsed proper time. The "proper velocity" or celerity will be four light years divided by seconds. It will vastly exceed the speed of light.

I see that you have rounded the intermediate result (0.945945945... c) to 0.95 c before using it in further calculations. You used the intermediate result with its two significant digits to produce a final result with claimed three digit precision. By my calculations, the correct result is slightly different.

 

[Ibix]

As @jbriggs444 says, proper velocity can have any value, and it tends to infinity as the velocity approaches $c$.

I'm not actually sure what the point of proper velocity is. I don't recall seeing it used outside of introductory exercises such as this one. $\gamma v$ will occasionally turn up in algebra, but that's all. It's also the three momentum per unit mass, but how often do you neeeld that...?

In short, don't worry about it unless you need it for an exam.

 

[member 731016]

Thank you for your replies @jbriggs444 and @Ibix!

That is so strange to think about that proper velocity can have values greater than c. The answer to (c) is more intuitive since

$v = \gamma v = \frac{1}{\sqrt{1 - \frac{(0.6c)^2}{c^2}}} \times 0.6c = 0.75 c$, is that please correct?

Thanks!

 

[Ibix]

Yes.

Proper velocity isn't a velocity in the ordinary sense of the word. As I say, unless you need it for an exam don't worry about it