Can't Tell If you are shrinking?

[TheScienceOrca]

(I've tried to guess what you're asking and have clarified accordingly. You really have to get in the habit of never stating a velocity or speed without also considering what it is relative to)

If I have guessed correctly what you're asking, Alice will see the ship moving at a speed of .9c relative to her, and will see the bullet moving at a speed of .994c relative to her.

I got this result from the relativistic velocity addition formula: $w=\frac{u+v}{1+uv}$ (and measuring distance in light-seconds and time in seconds so that $c=1$ and don't need to clutter things up with factors of $c$ and $c^2$)

Thanks, now Let's just imagine a ship traveling at .9c relative to Alice. On that ship a bullet is shot in the same direction of motion at .9c. Let's say the bullet ship stop exactly 1 second relative to Alice.

Would the distance from the bullet to Alice be .994 LS or 1.8 LS If we stop all of the objects 1 second relative to the objects then the bullet would be .994 LS or 1.8 LS away from alice?

 

[pbuk]

(I've tried to guess what you're asking and have clarified accordingly. You really have to get in the habit of never stating a velocity or speed without also considering what it is relative to)

If I have guessed correctly what you're asking, Alice will see the ship moving at a speed of .9c relative to her, and will see the bullet moving at a speed of .994c relative to her.

I got this result from the relativistic velocity addition formula: $w=\frac{u+v}{1+uv}$ (and measuring distance in light-seconds and time in seconds so that $c=1$ and don't need to clutter things up with factors of $c$ and $c^2$)

Alice will see the ship moving at a speed of .9c relative to her, and will see the bullet moving at a speed of .994c relative to her. The ship will appear shortened along the line of flight, and the bullet will appear shortened even more. If Alice can see a clock on the ship it will appear to move slowly, and the spinning bullet will appear to spin even more slowly. However Alice won't actually see the ship at all because any light from the ship will be shifted into the near infrared by the Doppler effect (approximately 4x wavelength distortion), and any light from the bullet even more so (approx 20x wavelength distortion).

Bob who is on the ship will see...

You can't possibly get your head around all this stuff working it out from thinking about the fact that the speed of light in a vacuum is invariant - even Einstein didn't do that, he started with the maths which showed him that some weird things would be observed and his non-mathematical thought experiments to make sense of it came later.

Instead you need to find a book or online resource or tutorial or something (can't anyone recommend one?), learn this stuff and then if you need to come back and ask questions.

 

[Nugatory]

Thanks, now Let's just imagine a ship traveling at .9c relative to Alice. On that ship a bullet is shot in the same direction of motion at .9c. Let's say the bullet ship stop exactly 1 second relative to Alice.

Would the distance from the bullet to Alice be .994 LS or 1.8 LS


If we stop all of the objects 1 second relative to the objects then the bullet would be .994 LS or 1.8 LS away from alice?

In one second of Alice's time the bullet moving at .9c relative to Alice will move a distance of .9 light-seconds according to Alice and therefore be .9 light-seconds away according to Alice.

You on the planet will see the bullet approaching at .994c relative to you, and Alice approaching at .9c relative to you, so the bullet is only gaining on Alice by .094c - for every second of your time that passes after the gun is fired, the bullet will be .094 light-seconds closer to you than Alice is.

Thanks to time dilation, Alice's .436 seconds is one second of your time and her one second is 2.3 seconds of your time. When Alice says that the bullet is .9 light-seconds away from her one second after the gun is fired, you will say that the bullet is $2.3\times{.094}=.22$ light-seconds away from her 2.3 seconds after the gun is fired.

 

[TheScienceOrca]

In one second of Alice's time the bullet moving at .9c relative to Alice will move a distance of .9 light-seconds according to Alice and therefore be .9 light-seconds away according to Alice.

You on the planet will see the bullet approaching at .994c relative to you, and Alice approaching at .9c relative to you, so the bullet is only gaining on Alice by .094c - for every second of your time that passes after the gun is fired, the bullet will be .094 light-seconds closer to you than Alice is.

Thanks to time dilation, Alice's .436 seconds is one second of your time and her one second is 2.3 seconds of your time. When Alice says that the bullet is .9 light-seconds away from her one second after the gun is fired, you will say that the bullet is $2.3\times{.094}=.22$ light-seconds away from her 2.3 seconds after the gun is fired.

I think I am misscommunicated I am not in this scenario, I will try to restate it.

Imagine looking at an x and y grid.

Lets say Alice is floating in the middle of the universe at 0,0

a ship at 5,0 is moving up the y-axis at .9c parallel to alice (x will always stay 5).

Their z is equal.

If this rocket also shoots a bullet in the same direction (up the y-axis at .9c relative to itself).

You stated the bullet would be moving .994c relative to alice and the rocket still .9c.


If the ship fires its bullet right when it's at (5,0).

In 1 second relative to Alice, according to your statements the bullet would be .994c LS away and the rocket .9c LS away. Which means the bullet is only .094c away from the rocket even though the bullet is traveling .9c relative to the rocket.

This is because as 1 second has passed for Alice, but not a full second for the bullet right?




1 second relative to the rocket what would the coordinates be from 5,0 start point?

1 second relative to the bullet what would the coordinates be form 5,0 start point?

I am slowly putting this together, I just don't understand if light travels at the speed of light shouldn't it be instant because time is traveling infinitely slow around it?

 

[PeterDonis]

if light travels at the speed of light shouldn't it be instant because time is traveling infinitely slow around it?

No. Light travels at the speed of light in any frame; that is, it travels at the speed of light relative to any observer (since an observer must have nonzero rest mass and so can't move the way light does). The speed of light is not instantaneous.

 

[phinds]

TheScienceOrca, since you have persistently had some difficulty in understanding what you are being told, let me expand slightly on what Peter has said.

Let's take an observer at, say, the origin of his own grid system. In his grid system, that is, his frame of reference, he looks out and sees numerous spaceships traveling at all different directions relative to him and at all different speeds relative to him. They are all also traveling at different speeds relative to each other and they are all different distances from our observer.

Now, our observer sets of a "light bomb" that shoots light off in all directions at once.

The light will reach all of the above mentioned spaceships at different times.

Our observer AND ALL OF THE SPACESHIPS will see the light traveling at c when it reaches them.

It will be blue-shifted for some of them and red-shifted for some of them but that is irrelevant to speed of the beam of light that reaches them. EVERYONE sees the light as traveling at c.

 

[PeterDonis]

Lets say Alice is floating in the middle of the universe at 0,0

a ship at 5,0 is moving up the y-axis at .9c parallel to alice (x will always stay 5).

The x dimension is superfluous here, and it will make the math easier if we leave it out and only deal with the $t$ and $y$ coordinates. That's what I'll do below; everything I do will be valid regardless of what $x$ coordinates (or $z$ coordinates, for that matter) we assign to Alice and the ship and the bullet, as long as they are all constant.

If this rocket also shoots a bullet in the same direction (up the y-axis at .9c relative to itself).

You stated the bullet would be moving .994c relative to alice and the rocket still .9c.

Yes. (More accurately, 0.9945c, which is the accuracy I'll use in calculations below.)

In 1 second relative to Alice, according to your statements the bullet would be .994c LS away and the rocket .9c LS away.

Yes.

Which means the bullet is only .094c away from the rocket

Relative to Alice; *not* relative to the rocket. Distances get transformed when you change frames, just like velocities do. Also there is relativity of simultaneity to consider. See below.

even though the bullet is traveling .9c relative to the rocket.

But not relative to Alice. You have to be very careful not to switch frames in mid-stream, so to speak, which is what you did in the sentence I just quoted (in two parts so you can see exactly where you switched--in between the two parts I quoted).

This is because as 1 second has passed for Alice, but not a full second for the bullet right?

No, it's more than that. Let's work out the coordinates that you asked for.

We have the following events, given with their coordinates in Alice's frame:

Event A: Alice and the ship start out co-located at (0, 0), and the ship fires the bullet at the same instant. The ship moves at 0.9c relative to Alice, and the bullet moves at 0.9945c relative to Alice.

Event B: The ship is located at (1, 0.9) after 1 second relative to Alice. (Note that we're using coordinates in which time is in seconds and distance is in light-seconds.)

Event C: The bullet is located at (1, 0.9945) after 1 second relative to Alice.

Now what we want are the coordinates of events B and C relative to the ship (note that event A has the same coordinates relative to the ship, since it's the origin of both frames). This is easily obtained via the Lorentz transformation; if $t, y$ are the coordinates relative to Alice, and $t', y'$ are the coordinates relative to the ship, then we have:

$$
t' = \gamma \left( t - \frac{v y}{c^2} \right)
$$
$$
y' = \gamma \left( y - v t \right)
$$

where $\gamma = 1 / \sqrt{1 - v^2 / c^2}$. For $v = 0.9c$, we have $\gamma = 2.294$, and this gives the following event coordinates relative to the ship:

Event B: (0.4359, 0) (note that we expect $y' = 0$ here because the ship is at rest at $y' = 0$ in its own frame)

Event C: (0.2408, 0.2168)

Note carefully several things:

(1) In the ship's frame, event B happens *less* than 1 second after event A. This is an example of time dilation: only 0.4359 seconds elapse on the ship between two events that are 1 second apart for Alice.

(2) In the ship's frame, event C happens *before* event B (whereas in Alice's frame, they happen at the same time). This is an example of relativity of simultaneity: events that are simultaneous in one frame are not simultaneous in another frame. But it also means that, if we want to know how far away the bullet is from the ship at event B, in the ship's frame, event C is the *wrong* event to look at. Instead, we need to look at:

Event D: (0.4359, 0.3923) is the event where the bullet is, in the ship's frame, when the ship is at event B (note that the time of this event, in the ship's frame, is the same as the time of event B). The bullet is 0.3923 light seconds away from the ship, in the ship's frame, when 0.4359 seconds have elapsed, because the bullet is moving at 0.9c relative to the ship. (We can verify this, by the way, by taking the ratio of $y'$ to $t'$ for event C; $0.2168 / 0.2408 = 0.9$, as expected.)

And just for completeness, we can transform the coordinates of event D back to Alice's frame, simply by inverting the sign of $v$ in the equations above; this gives

Event D: (1.81, 1.8) in Alice's frame. Notice that the $y$ coordinate here is 1.8 = 0.9 + 0.9; this is not a coincidence. It has to be that way because of how all the math combines: the way velocities add, and the way coordinates transform. It might be instructive for you to work out, from the various equations already given, how this comes about.

 

[Simon Bridge]

Recap:

A ship moving at 0.9c relative to Alice fires a bullet in the direction of it's motion at 0.9c relative to the ship.
(The bit in italics is what got left out of the original statement and makes all the difference.)

After one second by Alice's clock, Alice acquires a "snapshot" of the situation.
There is no need to stop the ship and bullet - Alice just wants to look at the the two at an instant of time.

After 1s, Alice finds that the ship has traveled a distance of 0.9 light-seconds from before, and the bullet has traveled a distance of 0.994 light-seconds. The distance between the ship and the bullet, as measured by Alice, is 0.094 light-seconds.
Now: Alice could naively deduce that the bullet is traveling 0.094c with respect to the ship but it is more accurate to say that the separation between the ship and the bullet increases at the rate of 0.094c.

The point of this example was to discuss how things could travel FTL with respect to each other without violating special relativity.
This is a common but somewhat misleading way to describe the effect. To see what I mean, we need to look at another situation:

This time, two ships pass each other, and Alice, at t=0, but traveling in opposite directions at 0.9c (wrt Alice).
At t=1s, the first ship has gone 0.9 light-seconds in one direction while the second ship has gone 0.9 light-seconds in the opposite direction.
This means that, in one second, the ships are 1.8 light-seconds apart.
Now: Alice could naively deduce that the second ship is traveling 1.8c with respect to the 1st ship but it is more accurate to say that the separation between the ships increases at the rate of 1.8c.

All this sort of thing is why we have to be careful with our descriptions when relativity is involved.

A similar issue comes up in post #1, where there was an unspoken assumption that absolute velocity can be determined... I'm not sure that idea has quite been shaken but, even so, some future soul may google here...

There is no experiment you can do to determine your constant velocity - but it is not because time dilation and length contraction conspire to prevent it. Rather it is the other way around: time dilation and length contraction are the consequence of there being no experiment to determine you own constant velocity. This means that the concept of velocity is meaningless without also specifying what it is relative to. This is an idea that can take some students quite a long time to wrap their heads around.

It is similar to how an object will have a different height depending on how far away the ruler it - this effect is called "perspective". We define the "proper height" to be that measured by a ruler that is right next to the object ... and we just go around calling it "height". Special relativity shows us there is also a perspective-like effect at different speeds. To get the proper height of an object, we now need to say the rule has to be stationary with respect to the object as well as being right next to it.

Relativity and perspective are both consequences of the laws of geometry that Nature happens to use, it's just that we are not used to needing to use all of them in one go and our everyday language is too vague to cope.