what is the difference between speed(units of c) and the Reciprocal in special relativity? Link below for reference

[can't_think_of_a_name 0]

https://en.wikipedia.org/wiki/Lorentz_factor
 

Just scroll a tiny bit to find what I am referring to.

Edited October 25, 2020 by can't_think_of_a_name

[swansont 7443]

Could you be more specific? I don't see a use of the reciprocal for speed. I see a reciprocal for gamma mentioned, which they call alpha. Is that what you're referring to? The only mention of "reciprocal" is in this context.

[Endy0816 452]

Going based off the table in the numerical values section; one is v/c and the other the multiplicative inverse or reciprocal of the lorentz factor.

v is always less than c, so for v/c you end up with: 0 ≤ β < 1

Now for the reciprocal of the lorentz factor you're doing the equivalent of finding the length of one side of a square with an area equal to the shaded section below.

Sqrt(1^2 - β^2)

Edited October 25, 2020 by Endy0816

[can't_think_of_a_name 0]

6 hours ago, swansont said:

Could you be more specific? I don't see a use of the reciprocal for speed. I see a reciprocal for gamma mentioned, which they call alpha. Is that what you're referring to? The only mention of "reciprocal" is in this context.

I start with gamma.
The part that confuses me is that I want velocity = , from gamma . I get v =  c/gamma.(I could have made a math mistake.)  It never says I need B = v/c. I know B isn't the correct symbol I am just using it here.

[swansont 7443]

48 minutes ago, can't_think_of_a_name said:

I start with gamma.
The part that confuses me is that I want velocity = , from gamma . I get v =  c/gamma.(I could have made a math mistake.)  It never says I need B = v/c. I know B isn't the correct symbol I am just using it here.

Invert gamma (which is alpha) and square it. Rearrange 

v^2/c^2 = 1 - alpha^2

Take a square root

[can't_think_of_a_name 0]

I know I got v = gamma/c. Like stated earlier why does  b= v/c give a different answer. I was solving a problem problem 3 A. I used V = c/gamma but they want b = c/v. This confuses me why does one works and  not the other? I guess I really didn't explain this well. In my course they never explain the difference. If the link doesn't work I will post the question.  

From here https://d3c33hcgiwev3.cloudfront.net/_d37cb29a797de375eb7866e695098a79_Wk7_problemsetsolutions.pdf?Expires=1603756800&Signature=SUmoIwTy2VAIs1CfSGO~F7C3BD7lIJKyMQJuedeO4RUqiCeo9HTLJk50r~oKjI6pAFoaSG5p-Pu3FRHuPiNizcGdd6EdCj4Eer1tU4BqbIBsdzW0WjLXxR8E~-5gZx3LDteO6L4ruR80eOQi1EzUdBj8Z9lnXJ6kQuj1fHpJGik_&Key-Pair-Id=APKAJLTNE6QMUY6HBC5A

[can't_think_of_a_name 0]

I meant v = c/ gamma

[joigus 356]

6 hours ago, can't_think_of_a_name said:

I meant v = c/ gamma

No. It's,

\[\gamma^{2}\left(1-\beta^{2}\right)=1\]

Gamma is a number always bigger than one. Beta is a number always less than one (in absolute value.) The absolute value of beta determines gamma.

[swansont 7443]

9 hours ago, can't_think_of_a_name said:

I meant v = c/ gamma

What speed does this apply to? I don't recognize this as a valid equation.

[can't_think_of_a_name 0]

I apologize for asking basic question I guess when I was originally in school either I didn't pay attention or wasn't taught properly. I assume the former but I can't be certain. I appreciate the help on basic questions.

why is this wrong?

 

Gamma = 1 / √ (1) - v^2 / c^2 =

1 / Gamma = √ 1 - v^2 c^2/ c^2 =

c / Gamma = √ 1 - v^2 +v^2 =

c + v / gamma = √ 1 =

switch the c + v to v + c =

v + c / gamma = √ 1=

v + c / gamma = √ 1 =

v + c gamma / gamma = √ 1 (gamma^2) =

v + c -c = √ gamma^2 =

v = √ gamma^2 - c

or

1 / Gamma - c^2 = √ 1 - v^2 / c^2 - c^2 =

1 / Gamma - c^2 = √ 1 - v^2 =

1 + v / Gamma - c = √ 1 - v^2 + v^2 =

1 +v / Gamma - c = √ 1 =switch the 1+ v to v + 1 =

v + 1 / gamma - c = √ 1=

v + 1 gamma / gamma - c = √ 1 (gamma) =

(v + 1) -c / - c = √ gamma -c^2 =

- (v+1) = √ gamma -c^2 =

-v -1 = gamma -c^2 =

-v = √ gamma^2 - c^2 + 1^2 =

- v / - = √ gamma^2 - c^2 + 1^2/ - =

v = √ gamma^2 - c^2 + 1^2 / - =

v = √ -gamma^2 + c^2 - 1^2

 

[Janus 1093]

46 minutes ago, can't_think_of_a_name said:

I apologize for asking basic question I guess when I was originally in school either I didn't pay attention or wasn't taught properly. I assume the former but I can't be certain. I appreciate the help on basic questions.

why is this wrong?

 

Gamma = 1 / √ (1) - v^2 / c^2 =

1 / Gamma = √ 1 - v^2 c^2/ c^2 =

c / Gamma = √ 1 - v^2 +v^2 =

c + v / gamma = √ 1 =

switch the c + v to v + c =

v + c / gamma = √ 1=

v + c / gamma = √ 1 =

v + c gamma / gamma = √ 1 (gamma^2) =

v + c -c = √ gamma^2 =

v = √ gamma^2 - c

or

1 / Gamma - c^2 = √ 1 - v^2 / c^2 - c^2 =

1 / Gamma - c^2 = √ 1 - v^2 =

1 + v / Gamma - c = √ 1 - v^2 + v^2 =

1 +v / Gamma - c = √ 1 =switch the 1+ v to v + 1 =

v + 1 / gamma - c = √ 1=

v + 1 gamma / gamma - c = √ 1 (gamma) =

(v + 1) -c / - c = √ gamma -c^2 =

- (v+1) = √ gamma -c^2 =

-v -1 = gamma -c^2 =

-v = √ gamma^2 - c^2 + 1^2 =

- v / - = √ gamma^2 - c^2 + 1^2/ - =

v = √ gamma^2 - c^2 + 1^2 / - =

v = √ -gamma^2 + c^2 - 1^2

 

To start off.

1 / Gamma = √ 1 - v^2 c^2/ c^2 =

c / Gamma = √ 1 - v^2 +v^2

You have to multiply all the factors under the radical by c^2 if you want to move c to the other side of the equation

This leaves

c/Gamma =√ (c^2 - v^2)

And since c^2  ≠  1+v^2 you can't get to where you got.

And at the end, your answer is not a multiple of c, so that right there should have been a tip-off that you did something wrong along the way.

To solve for v from

1 / Gamma = √ 1 - v^2/c^2

You first square both sides:

1/Gamma^2 = 1- v^2/c^2 (you square both the 1 and gamma, but since 1^2 = 1...)

v^2/c^2  = 1-1/gamma^2

v^2 = c^2(1-1/gamma^2)

take the square root of both sides:

v= c√(1-1/gamma^2)

Thus if v = 0.6c

Then

Gamma = 1/√(1- 0.6c^2/c^2) =  1.25

and

v = c√(1-1/1.25^2) = 0.6c

[can't_think_of_a_name 0]

1 / Gamma = √ (1 - v^2) c^2/ c^2  is the same as

1 / Gamma = √ (c^2 - v^2) c^2/ c^2 

 

shouldn't this give  c / Gamma = √ 1 - v^2   if I do it my way. Or is that just not how algebra works?

Edited October 30, 2020 by can't_think_of_a_name

[Janus 1093]

1-v^2/c^2 is not the same as (1-v^2)/c^2 

So for example, again using v= 0.6c

1- (0.6c)^2/c^2 = 1- 0.6^2 = 0.64

but

(1-(0.6c)^2)/c^2 =  1/c^2- 0.6^2  = 1/c^2 - 0.36

1-v^2/c^2 = (1-v^2)/c^2

is like saying

(1-1/2) = (1-1)/2

but solving the left side gives 0.5 and solving the right side gives 0

Edited October 30, 2020 by Janus

[can't_think_of_a_name 0]

Hopefully the last question about basic math.  Does √1 = 1^-1. Also I can go √1^2 = 1. Are there any other ways to remove a square root?

[can't_think_of_a_name 0]

I also have one more question if I have c+t =g. If I want to move g to the left of the = sign. I don't care how pointless this is. I go c+t-g =  0?

[joigus 356]

8 hours ago, can't_think_of_a_name said:

Hopefully the last question about basic math.  Does √1 = 1^-1. Also I can go √1^2 = 1. Are there any other ways to remove a square root?

Number 1 is a bit misleading, because \( \sqrt{1} = 1 \). Generally there are two basic ways of undoing a square root. One is squaring a root; e.g.,

\[\sqrt{a}=2\]

which gives,

\[a=4\]

and the other is the one you suggest --rooting a square--, but with that one you must be careful:

\[\sqrt{a^{2}}=4\]

which gives,

\[a=\pm4\]

Another possible way to get square roots out of the way is to remember that sums times differences give differences of squares. As in,

\[\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=\sqrt{a}^{2}-\sqrt{b}^{2}=a-b\]

You can prove quite amazing identities with this:

\[\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\frac{2a}{a-b}\]

\[\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}=-\frac{2b}{a-b}\]

There's almost no end to fun with square roots!

3 hours ago, can't_think_of_a_name said:

I also have one more question if I have c+t =g. If I want to move g to the left of the = sign. I don't care how pointless this is. I go c+t-g =  0?

Exactly.