# Oxbridge Physics Prelims Revision

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LennonMcCartney
yeah that's what i mean with the dashes the wrong way round. will try and sort it out in a bit. this hayfever is killing me.

well thats exactly the thing. I have done that exact working you've presented over and over and over, and I can never quite understand why l' = l/gamma. It's got to be to do with the Time T not being the same (which RichE is on about) but the thing is RichE basically is using the equations to arrive T, rather than making it clear why the T's are different and why the T is as it is (maybe he is but it isnt clear to me!)
Willa
well thats exactly the thing. I have done that exact working you've presented over and over and over, and I can never quite understand why l' = l/gamma. It's got to be to do with the Time T not being the same (which RichE is on about) but the thing is RichE basically is using the equations to arrive T, rather than making it clear why the T's are different and why the T is as it is (maybe he is but it isnt clear to me!)

I can't make it any plainer than I did in post 671

The events the O' observer sees as the ends of the rods aren't simultaneous in the O frame.

In fact if O' sees one end as the origin O=O' at time t'=0, the event associated with the other end (where x = L and t =T) at that same time t'=0 is at some x'=L' and I give you the equation that first calculates T and then you can convert (T,L) into (0,L') to get the answer
RichE

Consider the origin O' at which t'=0. One end of the rod is observed as being there at t'=0 by O'. Simultaneously he sees (0,L') the other end of the rod - but this isn't a t=0 event - this is when

g(t-vL/c^2) = 0 (*)
g(L-vt) = L'

So

t = vL/c^2

and

L' = g (L -vt) = g(L - v^2 L/c^2) = g L (1-v^2/c^2) = L/g

(*) is the equation I use to calculate T (from the O frame) - which corresponds to a t'=0 event

Once I have x=L and t = T then I can work out x'
~Raphael~
Oh stop deluding yourself. Oxford's ALWAYS awful, full-stop.

How you've changed
RichE
How you've changed

The realisation was brutal but I am better for it
~Raphael~
The realisation was brutal but I am better for it

Bold from the boy who's only seen pictures of Cambridge

Nah, you'll love it
RichE
Bold from the boy who's only seen pictures of Cambridge

It's not my fault
~Raphael~
It's not my fault

Well you were saying you were 99.9% probable of applying there, even before you missed the Open Day. Many of the colleges are beautiful, I admit.
RichE
Well you were saying you were 99.9% probable of applying there, even before you missed the Open Day. Many of the colleges are beautiful, I admit.

~Raphael~

They'll still be there in the summer
RichE
(*) is the equation I use to calculate T (from the O frame) - which corresponds to a t'=0 event

Once I have x=L and t = T then I can work out x'

problem is that is working from the equation, rather than working from trying to understand the situation. I am struggling to understand why simultaneity or time comes into any of it when it's not an event we're measuring it's just a distance, time shouldnt be involved in distances (if they were talking about change in x...it's like how in galilean transformation Delta x is delta x in both frames...time isnt a factor!!)
Willa
problem is that is working from the equation, rather than working from trying to understand the situation. I am struggling to understand why simultaneity or time comes into any of it when it's not an event we're measuring it's just a distance, time shouldnt be involved in distances (if they were talking about change in x...it's like how in galilean transformation Delta x is delta x in both frames...time isnt a factor!!)

Surely it's reasonable that an observer will measure distances between events s/he perceives as simultaneous - and then measures distance in the usual way - so in that sense time isn't involved

In galilean transformations there is no interaction between space and time co-ordinates - but that is a crucial difference with relativity and the idea of space-time
arghhhhh only 141hr and 27minutes until the CP4 paper! eek!!! I have a feeling it's going to be bloody awful since all the others have been! <groans>

Nice of you to say hi today LM - you definitely were who I thought you were! F1 Fanatic, still not sure about you...did you get the bus back with the others?

Anyways, I probably should sleep now...nite xxx
Hoofbeat
arghhhhh only 141hr and 27minutes until the CP4 paper! eek!!! I have a feeling it's going to be bloody awful since all the others have been! <groans>

Nice of you to say hi today LM - you definitely were who I thought you were! F1 Fanatic, still not sure about you...did you get the bus back with the others?

Anyways, I probably should sleep now...nite xxx

Good luck for tomorrow, Chloe - and everyone else
Hoofbeat
arghhhhh only 141hr and 27minutes until the CP4 paper! eek!!! I have a feeling it's going to be bloody awful since all the others have been! <groans>

Nice of you to say hi today LM - you definitely were who I thought you were! F1 Fanatic, still not sure about you...did you get the bus back with the others?

Anyways, I probably should sleep now...nite xxx

Yup it was awful. 5 out of 10 of us here at Keble think we failed it so... Thankfully Im not among them and Im pretty sure I got a pass. But not much else. Today Im expecting an evil paper, why should they change now?

As for bus no I didnt, otherwise I would definitely have said hello. Did you see 2 keble physicists walking back? 1 on a bike and one walking next to him. Cos I was the guy not on the bike. I'll definitely say hi today if I see you. Bad paper or not. You should be let out before me anyway if its in reverse alphabetical order so I may well see you at the bus-stop.

Best of luck for everyone with exams today

Another few hours and it will all be over - save the pointless short option.

Edit: in fact theres a picture of me in my profile. If you look you'll definitely know who I am.
F1 fanatic
Yup it was awful. 5 out of 10 of us here at Keble think we failed it so... Thankfully Im not among them and Im pretty sure I got a pass. But not much else.

I seem to be the only person who thought it was ok - hopefully not a sign that I completely misunderstood it
F1 fanatic
Edit: in fact theres a picture of me in my profile. If you look you'll definitely know who I am.

Ahhh...yes I do know who you are! And I do know your name, so I can say hello! hehe

Good Luck guys with today's paper - just think, it's sort of the end for us as we don't have to pass the short option! Anyways, I have a quick question:

====
How many different values are there for the cube root of i? Find their values
====
Surely there are 3 values. they're not by any chance exp(i3pi/2), exp(i15pi/2) and exp(i27pi/2)....I'm fairly certain I'm wrong!!!

Anyways, off to the library!!!
Bezza
I seem to be the only person who thought it was ok - hopefully not a sign that I completely misunderstood it

Well I thought it was a hard paper but I think I nailed section A; section B was what I found really hard. I think I probably passed it, but I still think it was hard and another unusual paper, and I probably didn't do as well on it (in fact I'm certain!!!) as I have done on other past papers.
Willa
I am struggling to understand why simultaneity or time comes into any of it when it's not an event we're measuring it's just a distance, time shouldnt be involved in distances

well if you're measuring somethings length by taking the difference between the positions of it's 2 ends, then if it's moving relative to you, you obviously need to measure where the 2 ends are at the same time - if i measured where a car's back end was now, then where it's front end is 10s later, the difference between the 2 isn't going to give me the car's length because it's moving. If I was in the car though and measured where it's front and back were relative to me, it wouldn't matter when I measured them because they'd always be in the same position. It's the same in galilean transformations because x' = x - vt so $\Delta x' = \Delta x - v\Delta t$ so you've got to measure at the same time in S for the lengths to be the same in both frames.
Hoofbeat
Ahhh...yes I do know who you are! And I do know your name, so I can say hello! hehe

Good Luck guys with today's paper - just think, it's sort of the end for us as we don't have to pass the short option! Anyways, I have a quick question:

====
How many different values are there for the cube root of i? Find their values
====
Surely there are 3 values. they're not by any chance exp(i3pi/2), exp(i15pi/2) and exp(i27pi/2)....I'm fairly certain I'm wrong!!!

Anyways, off to the library!!!

Well I got e^(ipi/6), e^(i5pi/6), e^(i3pi/2)

Which may be the same as yours. I just said z^3 = i = e^(ipi/2+2mpi)
z=e^(ipi/6+2mpi/3)
then m= 0, 1, 2

but I dont know if its any more right than yours.

And my name is Stu btw. I prefer it to F1F or whatever on here anyway. As I say Ill definitely try to say hi today, provided I see you.

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