Relativistic motion and length contraction

[the farmer]

TL;DR Summary: An electron with kinetic energy of 50 MeV, such as an electron produced in a linear accelerator, can be shown to have a speed parameter of 0.999949c. A beam of such electrons moves along the axis of an evacuated tube that is 10 m long, measured in the reference frame S fixed in the lab. Imagine a second frame S', attached to an electron in the beam and moving with it.

(a) How long would this tube seem to be to an observer riding on the electron?

KE = gamma × mc^2 - mc^2
From the I make gamma the subject of the formula and get a solution which is 1.21 ×10^13
Then calculate the length contraction which is L = proper length ÷ gamma factor. The ni got an answer of 8.19 × 10^-13. I am stuck with this question because I did not use the speed = 0.999949c which I was given in the question.

 

[Orodruin]

The speed is superfluous if you know the mass of the electron and the kinetic energy. However, the problem does not provide the mass of the electron and therefore probably expects you to compute gamma from the speed alone.

 

[Ibix]

That's a fine approach if you know the electron mass (I haven't checked your numbers). You could calculate the gamma factor from the velocity as a check.

 

[Orodruin]

I haven't checked your numbers

OP should check their numbers though ... The electron rest mass is around 0.5 MeV/c^2 so gamma is certainly not 10^13

 

[the farmer]

That's a fine approach if you know the electron mass (I haven't checked your numbers). You could calculate the gamma factor from the velocity as a check.

What if I use 9.11 x 10^31

 

[Orodruin]

What if I use 9.11 x 10^31

9.11 x 10^31 what? Please use units. They are essential in physics.

 

[the farmer]

9.11 x 10^31 what? Please use units. They are essential in physics.

9.11× 10^-31 kg

 

[Orodruin]

9.11× 10^-31 kg

That is the same as 0.5 MeV/c^2 so you should get the same result. (Assuming that you do all of the unit conversions correctly.)

 

[Orodruin]

To add to that: Now would be a good moment to provide your full calculation, not only the results, but also what you put into all of your formulas.

 

[the farmer]

What do you meant by speed is superfluous

 

[the farmer]

Fine a

That's a fine approach if you know the electron mass (I haven't checked your numbers). You could calculate the gamma factor from the velocity as a check.

Fine approach but I didn't use the velocity that I was given in the question somehow I am in correct..

 

[PeroK]

What do you meant by speed is superfluous

In relativistic kinematics, energy and momentum are usually more useful quantities than speed. If you know the mass and energy of a particle, then you don't need to calculate its speed to do the calculations.

You can do the same in classical mechanics, in fact. It's just more often taught using speed and momentum, rather than energy and momentum.

 

[Orodruin]

somehow I am in correct..

You are incorrect because you did the math wrong. Hence me asking you to actually write out what you did:

To add to that: Now would be a good moment to provide your full calculation, not only the results, but also what you put into all of your formulas.

 

[the farmer]

In relativistic kinematics, energy and momentum are usually more useful quantities than speed. If you know the mass and energy of a particle, then you don't need to calculate its speed to do the calculations.

You can do the same in classical mechanics, in fact. It's just more often taught using speed and momentum, rather than energy and momentum.

So in short, you mean that I must ignore the given speed in my question?

 

[Orodruin]

So in short, you mean that I must ignore the given speed in my question?

No. You can use either the provided kinetic energy or the provided speed. They both correspond to the same gamma factor because the speed was computed given the correct electron mass.

 

[the farmer]

I will redo it now

 

[Orodruin]

KE = gamma × mc^2 - mc^2
From the I make gamma the subject of the formula and get a solution which is 1.21 ×10^13

For what it is worth: The first line is correct. The second line is not. Unless you provide more details on your calculation in between, it is impossible to say exactly where the error occurs.

 

[PeroK]

So in short, you mean that I must ignore the given speed in my question?

You can use either, but computing a gamma factor from a speed so close to $c$ is a bit messy. The gamma factor comes out more easily from the energy. In fact, that's why $MeV$ are useful units for high-energy particles and why mass is given in $MeV/c^2$. The electron mass is approx $0.51 MeV/c^2$. Note that often the $c^2$ is dropped. The gamma factor is simply the energy in $MeV$ divided by the mass in $MeV/c^2$.

 

[Orodruin]

The gamma factor is simply the energy in MeV divided by the mass in MeV/c2.

Note however that the OP is given kinetic energy.

Pet peeve: units should generally not be typeset in math font in typical LaTeX. For example, the mass unit would be ${\rm MeV}/c^2$ and be preceeded by a space to separate from the numerical value.

 

[the farmer]

Alright, you said the formula in the first line is correct right? Step 2 I make my gamma the subject of the formula let say gamma is = y, y= KE + mc^2 / mc^2 = 8×10^-12 J (9.11×10^-31 kg ) ×( 3×10^8m/s)^3 ÷ 9.11×10^-31×(3×19^8)^2 then my final answer for gamma is 98.57
From there i have to find l= L0/ y , L0 is proper length which is 10m, then L = 10m/ 98.57 = 0.10 m,
What I am confused of is that I didn't use speed of 0.999949c which I was given. Whic method should use if I want to use it for accuracy.

 

[Orodruin]

Whic method should use if I want to use it for accuracy.

Either. Given your input data has only one or at most two significant digits, the result will not be more accurate in either case.

 

[the farmer]

Okay that I know, my question is what method should use if I want to use both kinetic and speed that I was give to find that length in a)

 

[the farmer]

Either. Given your input data has only one or at most two significant digits, the result will not be more accurate in either case.

Okay that I know, my question is what method should use if I want to use both kinetic and speed that I was give to find that length in a)

 

[Ibix]

Okay that I know, my question is what method should use if I want to use both kinetic and speed that I was give to find that length in a)

You don't need both. The speed alone is enough; the kinetic energy and rest mass is enough. As Orodruin noted, they're probably expecting you to use the speed because they didn't give you the mass, but either eay works.

I presume there's a part (b) and (c) etcetera where you will need the KE.

 

[the farmer]

No there is no other questions, but which formula do need to use so that I can use the speed only without using kinetic energy. I feel like is this but I am not sure L = L0/y. Y i gamma factor then it will be L = 10m/ square root of 1- (0.999949)^2 where by c^2/c^2 will cancel each other then the solution is 990.16016776 that is my length for a

 

[Ibix]

990.16016776 what? And is that a sensible number of decimal places?

 

[the farmer]

990 m in 3 significant figures

 

[Ibix]

Right - agreed.

For future reference, note that we have $\LaTeX$ here, so you can write things like $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$. You may need to refresh the page to get the renderer to kick in. If you're not familiar with $\LaTeX$ there's an introductory guide linked just below the reply box.

 

[the farmer]

Thanks so much? L =10 m ÷ 99.01601676
= 0.10099376117m
0.101m in 3 significant figures
Is this the right answer In a)

 

[Orodruin]

FWIW Anything more than two significant digits is wildly overstating accuracy here.

 

[the farmer]

But my lecturer told us to write in three significant figures

 

[the farmer]

FWIW Anything more than two significant digits is wildly overstating accuracy here.

But let ne write my answers in 2 significant it will be 0.10 m that's a length . Is my answers correct in a) ?

 

[the farmer]

I know the question is asking me to find a length, L = Lo/ y. Y is my gamma factor. In the question I am given kinetic energy. So to find my gamma I need to make it the subject of formula. Y= mc^2 + KE / mc^2
Y = 1 + (50×10^6) × ( 1.6×10^-19) J ÷ 9.11×10^-31 kg ( 3× 10^8m/s)^2
Y = 98. 57287474
L = 10m/98.5787474
=0.10 in to 2 decimal place.
I tried to calculate my length again but this time I am using the given speed and used this formula l= Lo/Y ,Y is gamma factor then L = 10m × (1 -(0.999949)^2)^1/2 ,c^2/c^2 canceled each other then my final answer is 0.10 m in 2 decimal place. I concluded that both these two methods are giving me same answer. Am I on the right truck or there is other method that I can use to solve my a).

 

[Orodruin]

As has already been said, the velocity given has quite obviously been computed by the problem author to correspond to the given kinetic energy. This is done by computing the gamma factor to find what velocity it corresonds to. Obviously that velocity is going to give the correct gamma factor - it was chosen that way by the author.

 

[the farmer]

As has already been said, the velocity given has quite obviously been computed by the problem author to correspond to the given kinetic energy. This is done by computing the gamma factor to find what velocity it corresonds to. Obviously that velocity is going to give the correct gamma factor - it was chosen that way by the author.

Still don't understand what you meant

 

[Orodruin]

Still don't understand what you meant

That is not a constructive response. A constroctive response would be pointing out which part you do not understand.

 

[the farmer]

I don't understand the way you responded. My question was:
Since the question is asking me to find a length, L = Lo/ y. Y is my gamma factor. In the question I am given kinetic energy. So to find my gamma I need to make it the subject of formula. Y= mc^2 + KE / mc^2
Y = 1 + (50×10^6) × ( 1.6×10^-19) J ÷ 9.11×10^-31 kg ( 3× 10^8m/s)^2
Y = 98. 57287474
L = 10m/98.5787474
=0.10 in to 2 decimal place.
I tried to calculate my length again but this time I am using the given speed and used this formula l= Lo/Y ,Y is gamma factor then L = 10m × (1 -(0.999949)^2)^1/2 ,c^2/c^2 canceled each other then my final answer is 0.10 m in 2 decimal place. I concluded that both these two methods are giving me same answer. Am I on the right truck or there is other method that I can use to solve my a)?

 

[Ibix]

Am I on the right truck or there is other method that I can use to solve my a)?

Those are the two methods available. That they give the same answer suggests you've done it right.

 

[jbriggs444]

I don't understand the way you responded. My question was:
Since the question is asking me to find a length, L = Lo/ y. Y is my gamma factor. In the question I am given kinetic energy. So to find my gamma I need to make it the subject of formula. Y= mc^2 + KE / mc^2

As it stands, that formula is not correct. Remember PEMDAS. The formula above parses as: $\gamma = mc^2 + \frac{\text{KE}}{mc^2}$. That is obviously dimensionally incorrect. Gamma is a pure number. You've given a formula for an energy plus a pure number.

Presumably you meant to write $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$ That formula would be correct. Total energy is rest energy plus kinetic energy. Gamma is total energy divided by rest energy.

Let us look at your math.

Y = 1 + (50×10^6) × ( 1.6×10^-19) J ÷ 9.11×10^-31 kg ( 3× 10^8m/s)^2

I had trouble making sense of this. Oh, I think I get it now. You've done a simplification behind the scenes.

You set out to evaluate: $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$
You actually evaluated $\gamma = \frac{mc^2}{mc^2} + \frac{\text{KE}}{mc^2} = 1 + \frac{\text{KE}}{mc^2}$

You convert the $50 \text{ MeV}$ to Joules for the numerator of the right hand side.

You multiply the rest mass of the electron, $9.11\times 10^{-31}\text{ kg}$ by $c^2$ to get Joules for the denominator.

That is a valid calculation for gamma.

Y = 98. 57287474
L = 10m/98.5787474
=0.10 in to 2 decimal place.

Seems sound.

So you have computed $\gamma$ based on the energy ratio that you were given (50 MeV from the problem and $9.11 \times 10^{-31} \text{ kg}$ from other references). Personally I would have looked up the rest mass of the electron in Mev (about 0.5 MeV) and skipped both conversions to joules. Then you can practically read off gamma without even using a calculator.

I tried to calculate my length again but this time I am using the given speed and used this formula l= Lo/Y ,Y is gamma factor then L = 10m × (1 -(0.999949)^2)^1/2 ,c^2/c^2 canceled each other then my final answer is 0.10 m in 2 decimal place. I concluded that both these two methods are giving me same answer. Am I on the right truck or there is other method that I can use to solve my a)?

Yes inceed. Since you are given that $\frac{v}{c} = .999949$ then $\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-.999949^2}$.

As you say, this time you used velocity as the starting point instead of kinetic energy.

Yes, both methods are giving you the same answer. Which is what @Orodruin was trying to point out. The person who set the question purposely chose the kinetic energy (50 MeV) and the velocity (0.999949 c) so that the two figures were consistent.

 

[the farmer]

As it stands, that formula is not correct. Remember PEMDAS. The formula above parses as: $\gamma = mc^2 + \frac{\text{KE}}{mc^2}$. That is obviously dimensionally incorrect. Gamma is a pure number. You've given a formula for an energy plus a pure number.

Presumably you meant to write $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$ That formula would be correct. Total energy is rest energy plus kinetic energy. Gamma is total energy divided by rest energy.

Let us look at your math.


I had trouble making sense of this. Oh, I think I get it now. You've done a simplification behind the scenes.

You set out to evaluate: $\gamma = \frac{mc^2 + \text{KE}}{mc^2}$
You actually evaluated $\gamma = \frac{mc^2}{mc^2} + \frac{\text{KE}}{mc^2} = 1 + \frac{\text{KE}}{mc^2}$

You convert the $50 \text{ MeV}$ to Joules for the numerator of the right hand side.

You multiply the rest mass of the electron, $9.11\times 10^{-31}\text{ kg}$ by $c^2$ to get Joules for the denominator.

That is a valid calculation for gamma.

Seems sound.

So you have computed $\gamma$ based on the energy ratio that you were given (50 MeV from the problem and $9.11 \times 10^{-31} \text{ kg}$ from other references). Personally I would have looked up the rest mass of the electron in Mev (about 0.5 MeV) and skipped both conversions to joules. Then you can practically read off gamma without even using a calculator.

Yes inceed. Since you are given that $\frac{v}{c} = .999949$ then $\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-.999949^2}$.

As you say, this time you used velocity as the starting point instead of kinetic energy.

Yes, both methods are giving you the same answer. Which is what @Orodruin was trying to point out. The person who set the question purposely chose the kinetic energy (50 MeV) and the velocity (0.999949 c) so that the two figures were consistent.

Wow thanks,so meaning I answered the question correctly right?

 

[Herman Trivilino]

Wow thanks,so meaning I answered the question correctly right?

One of the skills needed for success in the course is the ability to judge the validity of your answers for yourself. Focus on sense-making rather than answer-making.

 

[the farmer]

One of the skills needed for success in the course is the ability to judge the validity of your answers for yourself. Focus on sense-making rather than answer-making.

But,I need that to discuss with my colleagues, I want to ensure that is correct

 

[Herman Trivilino]

But,I need that to discuss with my colleagues, I want to ensure that is correct

I know. That's my point. You need to learn how to do that. Otherwise you have to take it on someone else's authority and you never learn. You won't achieve success that way.

 

[the farmer]

I know. That's my point. You need to learn how to do that. Otherwise you have to take it on someone else's authority and you never learn. You won't achieve success that way.

You a right? It correct or i need to redo it again?

 

[PeroK]

Still don't understand what you meant

In classical physics, we have four quantities associated with a particle: mass, kinetic energy, magnitude of momemtum and speed. If you know any two of those, you can calculate the others.

E.g. if you have mass ($m$) and speed ($v$), then you have:

$KE = \frac 1 2 mv^2$ (i.e. kinetic energy is determined by mass and speed), and $p = mv$ (likewise, magnitude of momentum is determined by mass and speed).

And, if you know KE and magnitude of momentum, then:

$v = \frac{2KE}{p} = \frac{mv^2}{mv}$ (i.e. speed is determined by KE and magnitude of momentum)

In SR, you often add the rest energy to the kinetic energy to give energy instead and you have the gamma factor. Technically, we now have six possible quantities. But, again, you only need to know two to determine all six.

I suspect your professor assumes you know this or can work this out for yourself. In any case, I strongly suggest you work through the permutations and check that if you know, say, the magnitude of momentum and the mass, then you can calculate the total energy, kinetic energy, gamma factor and speed.