Mathematica Lorentz Transformations Help Mathematica

[woodysooner]

Ok I need some help. I have.

t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}

and i rearrange to get:

v=\sqrt{c^2(1-\frac{t_0^2}{t^2})}

Ok I am setting C=1
tnot = 518400
t = 4.7304 *10^17

as you can imagine my answer i know will be .999999999999 with a lot of 9'sV.

The problem is I can't get anything anywhere to solve this exactly. Everything rounds to 1, even mathematica, somebody help me with this because after i solved this i wanted to graph this function and examine it closer but if i can't even get an exact answer i don't think it's going to be easy to graph.

 

[woodysooner]

and this does have to do with SR not just math so mentors please don't warn me. i thought if i posted it in math no one would know what lorentz was. so sorry if this is bad post.

 

[woodysooner]

and if you don't understand why i need exact solution it's cause an answer of .9999c might correspond to 200 years where .999999999999999999 to 2 billion so i need exacts. sorry to bother yall.

thanx
Woody

 

[pmb_phy]

v=\sqrt{c^2(1-\frac{t_0^2}{t^2})}

Ok I am setting C=1
tnot = 518400
t = 4.7304 *10^17

That relation can be re-written using the following approximation

(1 + d )^n \approx 1+d

The ratio v/c becomes

v/c = 0.9999999999999999999999994

That's 29 0's followed by a 4

Pete

 

[woodysooner]

sweetness, but how did you do that, can you show me a derivation if its not so long you don't want to.?

 

[woodysooner]

what is the d for? and where did the v and t's go?

 

[pmb_phy]

what is the d for? and where did the v and t's go?

The first thing I did was to factor out the c and divide through by c since it sems that you're looking for the ratio v/c right? d is any number such that |d| << 1 and n is any number You are now left with a relation of the form

(1 - r^2)^(1/2)

where r = t₀/t. So if I want to use that relation I'd have to set

d = -r^2

n = 1/2

As we say in the world of math and physics plug and chug, which means to place in the actual values and calculate.

Pete

 

[woodysooner]

so you just said 1 + (-r^2)^1/2 and said r = 518400. 4.730 *10^17

 

[woodysooner]

Pete you know way more than me so I am not saying this is bad solution, but that still doesn't help me because I get 1- tiny tiny number and stinking mathmatica after .0000001 goes to 0 that was why i posted this thread, do you see what I am saying, btw, how are you getting the answer like you calculator or what, or maybe I am misunderstanding you still.

Thanx

 

[woodysooner]

pmbphy

your message box is full lol it won't let me text ya

ok my calc and mathmatica are rounding it's drivin me batty how are you inputing it into. I know i keep asking you the same question, but are you putting it in as d or actual numbers. cause i get sqrt. of 1 - something tiny so it just cause to one takes root and answer is one but we both know that its like .99999999999999999999

 

[pmb_phy]

Pete you know way more than me so I am not saying this is bad solution, but that still doesn't help me because I get 1- tiny tiny number and stinking mathmatica after .0000001 goes to 0 that was why i posted this thread, do you see what I am saying, btw, how are you getting the answer like you calculator or what, or maybe I am misunderstanding you still.

Thanx

The approximation is

(1-r^2)^2 ~ 1 - (1/2)r^2

(1/2)r^2 ~= 6x10^-29

Notice:

1- (1/2)r^2 = 1 - 6x10^-1 = 0.4
1- (1/2)r^2 = 1 - 6x10^-2 = 0.94
1- (1/2)r^2 = 1 - 6x10^-3 = 0.994
1- (1/2)r^2 = 1 - 6x10^-4 = 0.9994
1- (1/2)r^2 = 1 - 6x10^-5 = 0.99994
...
1- (1/2)r^2 = 1 - 6x10^-29 = 0.99999999999999999999999999994

I think I missed one of the 9's at first. There should be 28 9s followed by a 4

Pete

 

[woodysooner]

I just keep on don't I,

The approximation is (1-r^2)^2 ~ 1 - (1/2)r^2

where does this come from. not that it's wrong

 

[pmb_phy]

I just keep on don't I,



where does this come from. not that it's wrong

Its rooted in calculus and power series expansion. It's a truncated Maclurin series expansion of a function. Any function which has all of its derivatives well defined has a series expansion. So all one does is tak f(x) = (1 + x)^n and expand it in powers of x. When |x| << 1 only the first power of x need be retained. Then, as in this problem, you can let x = -(t<sub>0/t)²

Pete</sub>

 

[woodysooner]

that is what i was thinkin but not sure. thanx

i'm going to work with this but i still think my calc is not going to handle it, butyou've done more than your job.

Thanx alot

 

[woodysooner]

do you square the t's afterwards or before.

 

[pmb_phy]

do you square the t's afterwards or before.

Iit doesn' matter.

Pete

 

[DW]

That relation can be re-written using the following approximation

(1 + d )^n \approx 1+d

The ratio v/c becomes

v/c = 0.9999999999999999999999994

That's 29 0's followed by a 4

Pete

Actually,
(1 + d )^n \approx 1+nd and only for |d|<<1.

 

[DW]

The approximation is

(1-r^2)^2 ~ 1 - (1/2)r^2

(1/2)r^2 ~= 6x10^-29

Notice:

1- (1/2)r^2 = 1 - 6x10^-1 = 0.4
1- (1/2)r^2 = 1 - 6x10^-2 = 0.94
1- (1/2)r^2 = 1 - 6x10^-3 = 0.994
1- (1/2)r^2 = 1 - 6x10^-4 = 0.9994
1- (1/2)r^2 = 1 - 6x10^-5 = 0.99994
...
1- (1/2)r^2 = 1 - 6x10^-29 = 0.99999999999999999999999999994

I think I missed one of the 9's at first. There should be 28 9s followed by a 4

Pete

Actually,
(1-r^2)^(1/2) ~ 1 - (1/2)r^2

 

[woodysooner]

I keep getting x to be 6.00488*10^-25
and my answer to have 24 0's where's the discrpancy, doing it just like you.

i'm off by factor of 4 o's but have the same numbers other than o's

 

[pmb_phy]

I keep getting x to be 6.00488*10^-25
and my answer to have 24 0's where's the discrpancy, doing it just like you.

i'm off by factor of 4 o's but have the same numbers other than o's

What did you use as x? I used x = (1/2)(t<sub>0/t)².

Pete</sub>

 

[woodysooner]

same as you, its odd we are getting somehting diff. my prof looked at it today and thanks to all your help he was like Whoa good job, but that's cause I'm a youngin.

 

[pmb_phy]

same as you, its odd we are getting somehting diff. my prof looked at it today and thanks to all your help he was like Whoa good job, but that's cause I'm a youngin.

What did you get for

x = -\frac{1}{2}(t_0/t)^2

If it was different than mine then please do the calculation step by step.

Pete

 

[Nenad]

the equation is derived from binomial expansion.

 

[woodysooner]

x = -\frac{1}{2}(t_0/t)^2

-6.00488 * 10^-25

same as you but you have four more 0's.

 

[Zanket]

I haven’t used Mathematica, but I’m surprised it doesn’t give you the precision you need. I would think a math package would offer that. I found this quote on the web:

Numerical computation in Mathematica is conducted to arbitrary precision controlled by the user. The user selects the precision he requires in the result and Mathematica automatically selects the precision needed in intermediate results to give that precision in the end result.

For more info google for: mathematica precision.

I also found info on plotting high precision numbers with this search: "High precision numbers and Plot" mathematica

 

[woodysooner]

it is highly precise but not with things that approach 0 after.0000001 it becomes 0 it sux, asked my prof about it he said it is just the way it is he hates it to, but with numbers not going to 0 your fine andits highly accurate and precise.