Is it okay if I summarize special relativity in my own words and examples and someone reviews it to make sure I didn't make any mistakes?

[can't_think_of_a_name]

In my first post I just want to confirm length contraction in special relativity at a later date I will post my entire summary of special relativity. It is a work in progress.

I travel to a planet from earth it takes 5 years.
Bob is located on earth. Alice is moving at v = .986c on a spaceship to the other planet.
If I want to know the length for Bob in Bob's frame  d= vt = (.986c)(5 years) = 4.93 LY
If I want to know the length for Alice in Bob's frame. L = L'/ gamma =  L' = (L) (gamma) = L' = (4.93)(5 years) = 29.58 LY
 

If I want to switch velocity I use Einstein''s velocity addition formula and I get -.986c for planet earth and the other planet.
My main question is why time  for Alice in Alice's frame  = 5 years.
 If I am not mistaken.
Then I Calculate the length for Alice in Alice's frame I get d = vt
Then I calculate d = vt  = (-.986c)(5 years) =  -4.93 LY
Then I calculate  the length for Bob in Alice's frame .  L' = (L) (gamma = -29.58 LYLY.

So to summarize I have 2 frames Alice and Bob and 2 measurements in each frame a moving and stationary frame that I need to calculate.

If I made any mistakes please correct me if I didn't then tell me?

 

 

[swansont]

1 hour ago, can't_think_of_a_name said:

In my first post I just want to confirm length contraction in special relativity at a later date I will post my entire summary of special relativity. It is a work in progress.

I travel to a planet from earth it takes 5 years.
Bob is located on earth. Alice is moving at v = .986c on a spaceship to the other planet.
If I want to know the length for Bob in Bob's frame  d= vt = (.986c)(5 years) = 4.93 LY
If I want to know the length for Alice in Bob's frame. L = L'/ gamma =  L' = (L) (gamma) = L' = (4.93)(5 years) = 29.58 LY
 

What do you mean by “length for Alice in Bob's frame”? You have Alice’s frame, and Bob’s frame.

 

[can't_think_of_a_name]

Maybe I am just really confused but I have Bob's frame and Alice's frame. In Bob's frame I have Bob who is stationary and Alice who is moving. In Alice's frame I have Alice who is stationary and Bob who is moving in the opposite direction. Length contraction deals with either Bob's frame or Alice frame but it doesn't switch from Alice's stationary frame to Bob's stationary frame.

To answer your question exactly "Alice in Bob's frame" I just mean Alice is moving and Bob is stationary.

[Janus]

2 minutes ago, can't_think_of_a_name said:

 Length contraction deals with either Bob's frame or Alice frame but it doesn't switch from Alice's stationary frame to Bob's stationary frame.
 

If you by this you mean that Bob can measure Alice as being length contracted, but Alice can't measure Bob as being length contracted, then you're wrong.

Or let's put it this way:  Assume there is a planet 4.93 ly from Bob as measured by Bob, to which Alice is traveling. Then, as measured by Alice, this planet and Bob are only ~0.822 ly apart.

Bob and this planet is traveling at  0.986c relative to Alice, and thus Alice measures 0.822 ly/0.986 = ~0.834 yr between Bob passing and the planet passing.

 

[swansont]

21 minutes ago, can't_think_of_a_name said:

Maybe I am just really confused but I have Bob's frame and Alice's frame. In Bob's frame I have Bob who is stationary and Alice who is moving. In Alice's frame I have Alice who is stationary and Bob who is moving in the opposite direction. Length contraction deals with either Bob's frame or Alice frame but it doesn't switch from Alice's stationary frame to Bob's stationary frame.

To answer your question exactly "Alice in Bob's frame" I just mean Alice is moving and Bob is stationary.

So it’s Alice’s frame. Does it make sense that length contraction gives you a distance much larger than in Bob’s frame?

[can't_think_of_a_name]

 No. If I am not mistaken there length needs to be the same to maintain symmetry for stationary Bob and stationary Alice. So how do I switch frames? Just to confirm L_stationary = L_moving / gamma. That doesn't switch frames? I assume the problem is the - sign.

Edited November 16, 2020 by can't_think_of_a_name

[swansont]

9 hours ago, can't_think_of_a_name said:

 No. If I am not mistaken there length needs to be the same to maintain symmetry for stationary Bob and stationary Alice. So how do I switch frames? Just to confirm L_stationary = L_moving / gamma. That doesn't switch frames? I assume the problem is the - sign.

No, the lengths would not be the same. In Alice's frame, the planets are moving, so the length between them would be contracted. If moving lengths are contracted, what does that say about the equation?

[can't_think_of_a_name]

I Calculate the length for Alice in Alice's frame I get d = vt
Then I calculate d = vt  = (-.986c)(5 years) =  -4.93 LY. Alice sees the planets moving so I go L=L'/gamma. -4.93/5 = -0.986 LY

The part that was confusing me is the difference between d = vt and L=L'/y. I thought d = vt and L=L'/y are the same. How are they different?

Thanks

gamma should be 6 but ignore that just go with 5.

[can't_think_of_a_name]

I mean d = L.

[Janus]

7 hours ago, can't_think_of_a_name said:

I Calculate the length for Alice in Alice's frame I get d = vt
Then I calculate d = vt  = (-.986c)(5 years) =  -4.93 LY. Alice sees the planets moving so I go L=L'/gamma. -4.93/5 = -0.986 LY

The part that was confusing me is the difference between d = vt and L=L'/y. I thought d = vt and L=L'/y are the same. How are they different?

Thanks

gamma should be 6 but ignore that just go with 5.

In d = vt, all three are measured by Bob.  v = 0.986c and t = 5 y, so d = 4.93 LY

Alice would measure 'd = vt',  for her v = 0.986c and t' equals 0.833 y, so d' ( the distance between planets) is ~0.822 ly

To use length contraction,   Alice would use d/y = d'  or 4.93/y = 0.822 LY, which is the same answer.

 

[can't_think_of_a_name]

I uploaded a picture just to make this clear for me. I think I got the correct answer.

But I have 1 question. L =  4.93 in Bob's frame. In Alice frame L' = 4.93. Therefore I apply Length contraction. L = L' / gamma.  This makes L = .82.

Why doesn't L' in Bob's frame become L = 29.58 in Alice's frame?
 

I know moving objects are smaller from a stationary observer.

But isn't stationary object bigger for a moving observer?
I apologize if this repetitive.

 

Ignore this picture below.

 

Edited November 18, 2020 by can't_think_of_a_name

[swansont]

L isn't moving relative to Bob.

But why are you expecting the value to be 29.58? That's taking L and multiplying by gamma. There's no reason for this to happen. 

Bob can ask "What does Alice see?" and transform his measurements into her frame. That maps L to L', and t to t'. Alice can do the converse and map L' to L, and t' to t.

What you are doing is taking what Bob measures and doing an inverse of a transform, which doesn't get him into Alice's frame.

 

IOW, there is no "L' in Bob's frame" or "L in Alice's frame" That's mixing frames

L' is in Alice's frame. L is in Bob's frame. 

[Janus]

11 hours ago, can't_think_of_a_name said:

 

I know moving objects are smaller from a stationary observer.

But isn't stationary object bigger for a moving observer?
I apologize if this repetitive.

You are assuming that there is such a thing as absolute motion, and that there is a way to tell who is really "moving" and who isn't.  This is not the case.  Motion is only relative.  Thus, you can only say that something is moving relative to some chosen reference frame frame.  In this case, there are two reference frames to choose from:*

1. The one Bob and the Planet are at rest with respect to

2. The One Alice is at rest with respect to.

If you are working from the Bob/planet frame, then Alice is moving and this frame measures Alice as length contracted.

If you are working from Alice's frame, then it is Bob and the Planet that are moving, and it is they and the distance between them that is measured as being length contracted.**

 

 

* There are an infinite number of reference frames we could use, it is just that these two are the most convenient to work from in this scenario.

** in addition, Bob would measure Alice's clock as ticking slower than his own, while Alice would measure clocks on Earth and the Planet as ticking slow compared to hers.

[can't_think_of_a_name]

So what your saying is the Lorentz transform doesn't change from a stationary frame to another stationary frame?

It goes from stationary object to moving object,

but the moving object doesn't become stationary it stays moving. This always works from moving object to stationary object.

Does the Galilean transform still work in special relativity in changing  from Bob's stationary frame to Alice stationary frame ? What is now confusing me is why the speeds in Bob's frame doesn't go from .983c to Alice frame -.983c.  Is this where I use Einstein velocity addition formula?

 

Edited November 19, 2020 by can't_think_of_a_name

[can't_think_of_a_name]

Ignore the post right ahead of this.

So what your saying is the Lorentz transform doesn't change form a stationary frame to another stationary frame?

It seems like you have to start with L = L'/gamma in whatever reference frame you are in. Why can't you start with L' = L gamma?

Does the Galilean transform still work in special relativity in changing from Bob's stationary frame to Alice stationary frame? What is now confusing me is why the speeds in Bob's frame doesn't go from .983c to Alice frame -.0983. is this where I use Einstein velocity addition formula

 

[Janus]

8 hours ago, can't_think_of_a_name said:

Ignore the post right ahead of this.

So what your saying is the Lorentz transform doesn't change form a stationary frame to another stationary frame?

It seems like you have to start with L = L'/gamma in whatever reference frame you are in. Why can't you start with L' = L gamma?

Does the Galilean transform still work in special relativity in changing from Bob's stationary frame to Alice stationary frame? What is now confusing me is why the speeds in Bob's frame doesn't go from .983c to Alice frame -.0983. is this where I use Einstein velocity addition formula

 

"Speed" doesn't have negative or positive values, as it is just a measure of magnitude.  "velocity" is magnitude and direction, and thus can take on negative and positive values.

So, if you assume that Bob measures a velocity for Alice of +0.983c , then using the same velocity convention, Alice would measure a velocity of -0.983c for Bob. However, the +/- in these velocities have no bearing on gamma factor each measures effecting the other.  This is because gamma contains v^2/c^2.    and v^2 is v x v.    so if v= +0.983c, you have (+0.983c) x (+0.983c)/c^2 =   0.966...* , and if v = -0.983, you have ( -0.983c) x (-0.983c)/c^2 = 0.996... , since a negative times a negative yields a positive.  You get the same answer if v is positive or negative.

 

* (+0.983c) x (+0.983c)/c^2 =  (+0.983)x(+0.983) c^2/c^2.   The c^2 cancels out.