Jordan Abbott

@Blue oh I see I’ll have a crack at it I guess

Just confusing as we’ve never explicitly done an expansion of cosh x before, or ever been told that we’d need to know standard Taylor expansions

I’ll check through the paper to see if they give us that

That’s weird then

Alright thanks

I just took the $e^{-\lambda}$ out of the sum

Where $P_i$ is the Poisson distribution, $P_i=\frac{\lambda^i}{i!}e^{-\lambda}$

@Blue Question asks to show that $$\sum_{i=0}^{\infty}P_{2i}=\frac{1}{2}(1+e^{-2\lambda})$$

Also, hi.

The question asks about the sum so proving it conversely probably wouldn't do

Is there any way to know that $$\sum_{i=0}^{\infty}\frac{x^{2i}}{(2i)!}=\cosh x$$ holds unless you just know that it holds? I can't think of a way to work it out.

It gives a weird answer to my question from the other day

Hmmm okay I’ll take a look at that q

coincident origins at t=0

Can you assume all reference frames have

Glad my uni's creative smh

Okay that's exactly the same question

Ahh couldn't find something like that - thanks I'll take a look

The only thing I could think was something to do with space-time interval as that's invariant between frames

I don't quite get how to approach it?

Basically it says that a group of people want to hold a party for a length such that their 'celebrations' are simultaneous with the impact of a comet on Jupiter in all inertial frames.

It's about one I asked earlier today can't get my head around it

sure sure sure...

Anyone help me with a question?

but is it really just that?

My thoughts is that it has to go on forever as some inertial frames move at $c$

This q. says that a group of people want to hold a party for a length such that their 'celebrations' are simultaneous with the impact of a comet on Jupiter in all inertial frames.

And after expanding horribly the part that isn't in the $(h(t))^{-1/2})$ (i.e the only part that can equal $0$) comes out to be equal to $r\cdot v$.

Which is just repeatedly using product/chain rule

Yeah that was what I did

I just needed to not be lazy - classic

It was actually quite easy

Oh I did it

I like this.

(Definitely neither true or quotable or said by anyone, ever)

@Avantgarde In Manchester - anything is possible

@Blue Actually I didn't look at the second answer on that post which gives a better explanation I think - might just use that but I'll still have a go with it algebraically.

I'm gonna try and crack on with it tonight - thanks.

Although maybe that's not plausible

@Blue I did see that but I kind of wanted a more algebraic approach

I mean they're measured to be moving from the perspective of one stationary inertial frame of reference

Trust I'm no where near to the point of being able to lorentz transform this

At least i got that part correct

I'm trying to prove that for two objects at their point of closest approach $\vec r_{B/A}\cdot\vec v_{B/A}=0$ but it's not working out

And the same holds for $r_{A/B}$, where $r$ is displacement right?

This might be a really dumb question but if $\vec v_A(t)=f_i(t)\hat i+f_j(t)\hat j$ and $\vec v_B(t)=g_i(t)\hat i+g_j(t)\hat j$ where $v_x$ is the velocity of $x$, then is the relative velocity of $B$ w.r.t $A$ just given by $\vec v_{B/A} = (g_i(t)-f_i(t))\hat i+(g_j(t)-f_j(t))\hat j\ ?$

Nice one that’ll be handy thanks

I’ll check it out in a second currently in a lecture

@SirCumference does it cover Lorentz transformations etc.?

@JohnRennie Okay thanks :).