Relativity Revision Thread

I'm not sure that anyone watching this thread will know the answers- it seems relatively empty. :getmecoat:

Reply 2

Essex_B

4a)i Yep I put that, it's a logical argument so don't see why it's wrong.

ii) http://http://en.wikipedia.org/wiki/Special_relativity

Look at the force section, it gives a full derivation of it - quite useful.

It also explains iii) as well.

5a)

ii) This is just substitution, you'll do it eventually if you just play around with it.

iii) I dont think you normally see it, the eV can also be a unit of mass. its more common in particle physics, where mass and energy afre interchnaged, to use eV/c^2, or simply eV with c set to 1, where cs just a unit of mass.

I just quickly glanced at this. Going to do some linear algebra which I don't like...

Reply 3

01perryd

for 5bii you say it is just a simple substitution but I have tried various different things for hours and still havent got it right. Any further hints?

Reply 4

Essex_B

01perryd

for 5bii you say it is just a simple substitution but I have tried various different things for hours and still havent got it right. Any further hints?

Yep, it's not too bad.

Try

E = E_1 + E_2

Where E_1 and E_2 are the relativistic energies of the particles.

So, when you look at the energies

E^2 = (p_0)^2.c^2 + M_(0A)^2.c^4

But

p_0 = 0

So

E^2 = M_(0A)^2.c^4

And

[br]E^2 = (E_1 + E_2)^2 [br]=(E_1)^2 + 2(E_1).(E_2) + (E_2)^2 [br]=2(E_1)^2 + 2(E_1)(E_2) + (E_2)^2 - (E_1)^2 = E^2[br]

Don't read on if you want to try it yourself from here I dont know how to use spoilers.

From the last equation you can get

(E_1)x2(E_1+E_2) + (E_2)^2 - (E_1)^2 = E^2

So

E_1 = \frac {(E^2 + (E_1)^2 - (E_2)^2)}{2(E_1 + E_2)} [br][br]= \frac {c^2[(M_{0A})^2.c^2 - (m_{01})^2.c^2 + (m_{02})^2.c^2]}{2c^2.(M_{0A})} [br][br]

You can see that I havent included a momentum bit for the Equation for m_1 and m_2, this is because they have the same momentum so they cancel out.

And then from there you cancel the c squareds and you have your answer!

Reply 5

01perryd

Thank you so much for that. I can never see things like replacing E1^2 with 2E1^-E1^2.

Reply 6

Essex_B

No worries Dave, val doesnt teach you that in tutorials... :P

It's easier alot of the time if you just work backwards with the show that questions, because that question didn't require any real understanding just really crap algebra.

Reply 7

Is there an easy way to get from relativistic momentum to relativistic mass without solving for the velocity in order to find

\gamma

? Just that that way seems a bit of effort for the number of marks (can't remember which question).

Reply 8

OP

Essex_B

Yep, it's not too bad.

Try

E = E_1 + E_2

...
and you have your answer!

Wow that's a pain to fiddle with, I should of attempted working backwards. I guess if you look at what it wants and turn it into the E's as you did it makes it easier to see what to try to do.

.matt

Is there an easy way to get from relativistic momentum to relativistic mass without solving for the velocity in order to find \gamma? Just that that way seems a bit of effort for the number of marks (can't remember which question).

Have you got the question? If you have only a value for relativistic momentum then you can't do anything to find the rest mass. If by relativist mass you mean

m = \gamma m_0

then you have to solve for v in terms of m0 I'd think and then divide by it. But it depends on the question a lot, usually if the marks are like that it's because there's a trick somewhere.

Reply 9

benwellsday

6b) I need this checked because I'm not really sure if it's right...I'll just post what I've got and see if anyone disagrees
i) 900 years
ii)

\frac{2\sqrt{2}}{3}c

iii) 150 light years

Same with 6c), I get

\frac{13}{14}c

I get:
i) 900 years
ii)

t = \gamma t_0 \rightarrow \gamma = \frac{t}{t_0} = \frac{900}{30} = 30

so

v = c\sqrt{1-\frac{1}{30^2}} = 0.999c

(3sf)
iii)

l = \frac{l_0}{\gamma} = \frac{450}{30} = 15

light years

For 6)c), putting

v_x' = 0.8c

and

u = -0.5c

and subbing into the formula gives

v_x = 0.5c

...but I really have no idea what I'm doing! :p:

Reply 10

OP

.matt

I get:
i) 900 years
ii)

t = \gamma t_0 \rightarrow \gamma = \frac{t}{t_0} = \frac{900}{30} = 30

so

v = c\sqrt{1-\frac{1}{30^2}} = 0.999c

(3sf)
iii)

l = \frac{l_0}{\gamma} = \frac{450}{30} = 15

light years

For 6)c), putting

v_x' = 0.8c

and

u = -0.5c

and subbing into the formula gives

v_x = 0.5c

...but I really have no idea what I'm doing! :p:

You're right for (ii) and (iii), I crossed out one too many 0s when scribbling. I never liked the crossing out method with fractions...still, that's no excuse, I are an retard.

For 6)c) I think u = 0.5c, so you get

v_x = \frac{v' _x + u}{1 + \frac{uv' _x}{c^2}} = \frac{1.3}{1.4} c

.
I keep getting confused between whos the observer and whos at rest...that better clear up by tommorow...

Reply 11

2007 Paper (dodgyness may follow):

Spoiler

benwellsday

For 6)c) I think u = 0.5c, so you get

v_x = \frac{v' _x + u}{1 + \frac{uv' _x}{c^2}} = \frac{1.3}{1.4} c

.
I keep getting confused between whos the observer and whos at rest...that better clear up by tommorow...

Yeah, I just figured for 4 marks it would need at least some thought, as the mark allocations seem a lot more reasonable in Relativity than Mechanics. But yeah, I really need to revise/learn this stuff! :|

Reply 12

rowzee

Are we really expected to know the full derivation of the relativistic force as on the wiki page? It's not exactly straightforward, besides in the lectures we didn't even evaluate the velocity and gamma derivatives. It could be that the syllabus has changed since 05/06 and we shouldn't be faced with a question like 4.ii.

Reply 13

OP

rowzee

Are we really expected to know the full derivation of the relativistic force as on the wiki page? It's not exactly straightforward, besides in the lectures we didn't even evaluate the velocity and gamma derivatives. It could be that the syllabus has changed since 05/06 and we shouldn't be faced with a question like 4.ii.

Probably not the full derivation that takes into account both components, but that was 4) (iii), which it only asked you to comment on. The derivation for force along the line of movement isn't too bad,

F = \frac{dp}{dt} = m_0(\frac{dv}{dt} \gamma + v \frac{d\gamma}{dt})

It's rearranging it that's a bit of a pain (and differentiating gamma is way too easy to make mistakes on).

Reply 14

rowzee

Also, what kind of calculators may we use? I'm pretty sure that according to regulations my TI-84 is out of the question, but that's the only one I have.

Reply 15

Next question from 2007 paper (and much more dodgyness):

Spoiler

Reply 16

OP

.matt

2007 Paper (dodgyness may follow):

Spoiler

Yeah, I just figured for 4 marks it would need at least some thought, as the mark allocations seem a lot more reasonable in Relativity than Mechanics. But yeah, I really need to revise/learn this stuff! :|

I agree with everything from this except the last part I got 0.234c, which is probably just a typo.

Reply 17

OP

.matt

Next question from 2007 paper (and much more dodgyness):

Spoiler

Also agree with these. I can't work out (ii) and (iii) if the first bit, the figures are craaazzzyyy. Do you think we have to convert to kg, because I don't know how and generally masses are given in MeV / c^2 or GeV / c^2.

Reply 18

benwellsday

Also agree with these. I can't work out (ii) and (iii) if the first bit, the figures are craaazzzyyy. Do you think we have to convert to kg, because I don't know how and generally masses are given in MeV / c^2 or GeV / c^2.

I don't think we do, I just have a habit of retreating to SI units after Physics A Level...

(ii) would be 40.0 GeV/c and (iii) would be 40 GeV/c^2 if using those units I think.

(if you're interested, to convert from eV to J just multiply by e

\approx

1.6*10^-19)

Reply 19

OP