# Length Contraction and what not

## Recommended Posts

hello

if im going at a reltavistic speed, .9c, and im travelling towards point A, does pt A become a shorter distance for me due to length contraction. i see that it cant because the distance isnt moving, but im unsure about what this person would see outside. .e.g length contraction of pt A which lets say is a sun. if the sun would contract doesnt this make the distance even longer. thanks

• Replies 70
• Created

#### Popular Days

Yes.

yes to wat?

##### Share on other sites

Yes to "if im going at a reltavistic speed, .9c, and im travelling towards point A, does pt A become a shorter distance for me due to length contraction."

##### Share on other sites

You would be wrong. How do you reconcile the time dilation observed by an observer in a different frame if the length contraction does not take place?

##### Share on other sites

If a spaceship travels at v=.9c toward a star 5.4 LY away, it wll reach it in

6 years. (5.4/.9)

##### Share on other sites

If a spaceship travels at v=.9c toward a star 5.4 LY away' date=' it wll reach it in

6 years. (5.4/.9)[/quote']

Yes, but less than 3 in "Spaceship Standard Time".

Of course, it will be less than 2.7 LY in "Spaceship Standard Distance".

##### Share on other sites

but if the sun's length contracts does that make the distance longer away from the spaceship.

##### Share on other sites

but if the sun's length contracts does that make the distance longer away from the spaceship.

You can't do the contractions separately - it's not like the length contracts, and then the sun's diameter changes, moving it away.

Pick your distance, e.g. to the center of the sun, and it will shorten by a factor of gamma. So will the size of the sun, in the direction of motion.

o right thanks

##### Share on other sites

• 1 month later...
Yes' date=' but less than 3 in "Spaceship Standard Time".

Of course, it will be less than 2.7 LY in "Spaceship Standard Distance".[/quote']

J.C.McSewell,

I understand the dilation and contraction aspects of the question. Starting from Meir Achuz' statement,
Originally Posted by Meir Achuz

If a spaceship travels at v=.9c toward a star 5.4 LY away, it wll reach it in

6 years. (5.4/.9)
.

The moving observer's clock willread 2.7 ly when the ship arrives. I assume you mean that the clock reading will show "2.7ly"l, which is a substantially shorter period of time than the 6 ly on the stationary frame clock. This seems like the dilated time of the space ship imposes a problem in symmetry. If a fleet of space ships move from A to B at 100 different velocities there will be 100 different clock rate readings for all the time of flight measurements between two fixed points in space which is a well defined inertial frame, correct?

In a classical world the difference in times for the travel is explained simply by the difference in absolute velocity of the moving frames, absolute velocity wrt a v = 0 , or near v = 0 frame of reference.

Then how can anyone say that the physics on each of the 100 inertial frames will give the same result when it is theoretically clear that clocks do not behave the same on each frame, that is the clocks do not tick at the same rate in all inertial frames of reference?

Perhaps performing acceleration experiments on different frames is one thing , but when the frame itself produces contradiction of expected confirmation, what is this supposed to mean?

Geistkiesel

##### Share on other sites

Then how can anyone say that the physics on each of the 100 inertial frames will give the same result when it is theoretically clear that clocks do not behave the same on each frame' date=' that is the clocks do not tick at the same rate in all inertial frames of reference?

[/quote']

Where did anyone say that they would give the same result?

##### Share on other sites

J.C.McSewell' date='

I understand the dilation and contraction aspects of the question. Starting from Meir Achuz' statement,

.

The moving observer's clock willread [b']2.7 ly
[/b]
when the ship arrives. I assume you mean that the
"2.7ly"
l, which is a substantially shorter period of time than the 6 ly on the stationary frame clock.

Geistkiesel

2.7 LY is a distance. Clocks do not read distance.

I meant that the distance will be less than 2.7 LY in the spaceships frame of reference.

##### Share on other sites

Where did anyone say that they would give the same result?

If the physics inall inertial frames is the same, identical, equivalent, then a tick rate in frame A is the same as the tick rate in frame B. This is what equivalence of physical law in inertial frames means to me.

Geistkiesel

##### Share on other sites

If the physics inall inertial frames is the same, identical, equivalent, then a tick rate in frame A is the same as the tick rate in frame B. This is what equivalence of physical law in inertial frames means to me.

That's precisely backwards. If the physics is the same in all inertial frames, then it is necessarily implied that space and time are relative.

##### Share on other sites

That's precisely backwards. If the physics is the same in all inertial frames, then it is necessarily implied that space and time are relative.

How so?

I don't think so.

Regards

##### Share on other sites

If the physics inall inertial frames is the same' date=' identical, equivalent, then a tick rate in frame A is the same as the tick rate in frame B. This is what equivalence of physical law in inertial frames means to me.

Geistkiesel[/indent']

And the laws of physics say that time depends on your frame of reference, so that the tick rates are not the same in all frames.
##### Share on other sites

And the laws of physics say that time depends on your frame of reference, so that the tick rates are not the same in all frames.

I have performed an extremely thorough analysis of the special theory of relativity, and the inescapable analytical conclusion is that both the Lorentz contraction formula, and the time dilation formula are false in all inertial reference frames.

I even showed you how to prove it in some other thread.

Don't be mad at the messenger.

Kind regards Dr. Swanson

##### Share on other sites

How so?

I don't think so.

The deductions that lead from the assumptions of experimental indistinguishibility of inertial frames to the Lorentz transformations have been well known for over 100 years now.

Don't be mad at the messenger.

##### Share on other sites

I have performed an extremely thorough analysis of the special theory of relativity' date=' and the inescapable analytical conclusion is that [b']both the Lorentz contraction formula, and the time dilation formula [/b] are false in all inertial reference frames.

I even showed you how to prove it in some other thread.

Don't be mad at the messenger.

Kind regards Dr. Swanson

Don't sell yourself short Johnny. You also proved reality was wrong as well in that thread because it also contradicted your "Galilean transformations are always correct" assumptions.

##### Share on other sites

2.7 LY is a distance. Clocks do not read distance.

I meant that the distance will be less than 2.7 LY in the spaceships frame of reference.

OOPs' date=' got off track. The ship's clock will read "6 years" is this correct?

Geistkiesel[/indent']
##### Share on other sites

That's precisely backwards. If the physics is the same in all inertial frames, then it is necessarily implied that space and time are relative.

If the clocks are ticking at different rates in different frames of reference, then the physics is not the same in all inertial reference frames and necessarily, using your rules space and time aren't relative they are absolute.

I will restate the problem. Reference frame A moving at .9999c wrt reference frame Ve, the embankment and B moving at 10^-6c wrt Ve.The A and B frames are moving toward each other. Assume at time t = 0 the three inertial reference frames have their respective clocks set to zero at the instant the A and B frames are separated by 90 lh (light hours) measured by observers in the stationary frame or reference, Ve. The instant the clocks pass by each other the clocks are stopped and their total number of one second tick counts are printed on paper in all three frames All three clocks started simultaneously at t = 0 and all three frames stopped simultaneously

The Ve, A and B frames transmit a continuous signal of consecutive one second clock ticks, measured within their own frames of reference, to the other frames.

As I understand SRT both the A and B frames may consider themselves at rest with respect to the other frame. ISs this a correct assumption?

My questions:

What clock rate will each frame record in their respectrive frames of the other frame's received tick rate:
1. A sees B rates faster or slower, the same?

2. A sees Ve clock rates faster or slower, the same?

3. B sees A tick rate faster or slower, the same?

4. B sees the Ve clock rates faster or slower, the same?

5. Ve sees the A rate faster than B, the B rate faster than A, the same.

6. compare all three (at least the relative) rates to the Ve frame.

7. What does a comparison of the A and B clock data show wrt the measured rate of the Ve clock rate?

Absolute values are OK if you prefer.

The only rates I am concerned with are the one second constinuous pulse the the three frames radiate.

Geistkiesel

##### Share on other sites

I even showed you how to prove it in some another thread

Where in some other thread did you prove this Johnny5? : That time dilation and Lorentz contraction are false.

Geistkiesel

##### Share on other sites

Johnny is right. The velocity 4-vector is not a physical velocity. Check out Lorentz invariant.

## Create an account

Register a new account