Jump to content

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

Recommended Posts

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.

Link to comment
Share on other sites

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)

1570445349_2020-10-2514_24_47.thumb.jpg.c37cd663f201c516d2f7b4b886782ed3.jpg

Edited by Endy0816
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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

Link to comment
Share on other sites

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

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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

 

Link to comment
Share on other sites

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

Link to comment
Share on other sites

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 by Janus
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.