# Trying to Understand the Lorentz Transformation

#### kvnsmnsn

Taking a look at "http://www.space.com/30026-earth-twin-kepler-452b-exoplanet-discovery.html" I observe that planet Kepler-452b (judged to be somewhat Earth-like) is 1400 light years from Earth. If a spaceship leaves Earth at a fifth of the speed of light, traveling toward Kepler-452b, from Earth's perspective it will take 7000 years to get to its destination. Or, perhaps better stated, if the spaceship sends a signal from Kepler-452b the moment it arrives there, and the signal travels at the speed of light, Earth observers will receive the signal 8400 years after the spaceship left Earth. But using the formula for the Lorentz factor on "https://en.wikipedia.org/wiki/Lorentz_transformation" I get gamma = 1 / sqrt( 1 - v^2/c^2) which would be gamma = 1.02062, which would mean to astronauts on the spaceship it would appear that the trip to Kepler-452b took 6858.57 years, right?

On the other hand, if the spaceship traveled at .3c instead of the .2c of the previous example, from Earth's frame of reference the trip would take 4666.7 light years (by the same reckoning as above, a signal sent out from the spaceship upon arriving would reach the Earth 6066.7 years after the spaceship left). But now the Lorentz factor would have gamma be 1.04828, so the duration of the trip from the frame of reference of the astronauts on the ship would be 4451.72 years, correct?

I'm just trying to get a feel for how the Lorentz Transformation affects how an astronaut perceived time as her/his spaceship traveled at speeds close to the speed of light.

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#### Ibix

All looks right to me.

#### kvnsmnsn

The thing that confused me about the "https://en.wikipedia.org/wiki/Lorentz_transformation" website was the section labeled "Physical formulation of Lorentz boosts", subsection labeled "Coordinate transformation", the box labeled "Lorentz boost (x direction)" that has four lines: "t' = gamma * (t - vx/c^2) / x' = gamma * (x - vt) / y' = y / z' = z" where gamma is defined as I stated, "1 / sqrt( 1 - v^2 / c^2)". What does that first line, defining t', mean? What is the variable x doing there? Doesn't that denote relative position? What does relative position have to do with how much time has passed? If anyone can clear this up for me I'd really appreciate it.

#### Ibix

You should probably look up "relativity of simultaneity". In short, there is a simple process to synchronise two clocks that are not right next to each other. If two people who are in motion relative to one another each carry out this synchronisation procedure to synchronise a long line of clocks at rest with respect to them, they will both say that the other's clocks are not correctly synchronised - being either ahead or behind depending on where they are (not to mention gaining or losing due to time dilation). That is where the x comes into the t transform.

Length contraction and time dilation get all the air time, but you can't make relativity consistent without the relativity of simultaneity.

#### kvnsmnsn

So the x value in the mentioned box is the distance on the x axis between the two references that are moving away from each other at velocity v?

#### Mister T

Gold Member
What does relative position have to do with how much time has passed?
Suppose two clocks, synchronized and separated by a distance $x$ in their rest frame, move relative to some observer with speed $v$ along a line joining them. To that observer, the clock in the rear will be ahead of the clock in front by an amount equal to $\frac{vx}{c^2}$.

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#### kvnsmnsn

By synchronized, do you mean that for the first reference point when the time $t$ equals the time $t'$ for the second reference point, the position for the first reference point also equals the position for the second reference point? Would it be erroneous to say that when $t$ equals zero then $t'$ also equals zero, and the distance between the two reference points is also zero? I'm just trying to understand what synchronized means.

#### Mister T

Gold Member
By synchronized, do you mean that for the first reference point when the time $t$ equals the time $t'$ for the second reference point, the position for the first reference point also equals the position for the second reference point?
We call them events, not points. Usually a reference event is an event where $t$ and $t'$ are both set to zero, and also $x$ and $x'$ are set to zero. This is synchronization in the sense that frames $S$ and $S'$ each have a single clock and when they share the same location they are set to the same time. This is the "easy" type of synchronization. Difficulty arises when we need to synchronize two clocks in different locations, and that is the type of synchronization I was discussing in Post #6.

In this case we simplify things a bit by placing the two clocks in the same reference frame. That is, the two clocks are at rest relative to each other. And in that reference frame they are separated by a distance $x$. The easy way to do this is to have a light signal leave one of the clocks at time $t=0$ and have the other clock read $x/c$ when that signal arrives.

#### Ibix

Synchronised clocks are clocks that read the same time. If the clock right in font of me is reading 12:00:00 and a clock one light second away is reading 11:59:59 and a clock one light minute away is reading 11:59:00, and they are all ticking at the same rate then these are all synchronised - they show the same time when I correct for the lightspeed delay. If the clocks are at rest with respect to me then I can set this up - in fact I can fill space with a grid of clocks that are at rest with respect to me and behave like this. Furthermore, I can move to any clock and stop next to it, and all of the clocks will have the same behaviour. This is exactly the same as you synchronising your watch with the station clock so you don't miss the trains. Formally you should allow for the fact that the clock is further away from your eye than your watch, but a couple of nanoseconds either way don't matter for that application.

Now an observer moving at constant speed with respect to me and my clocks comes along, bringing her own grid of clocks that are (for her) synchronised in the same sense as above. I look at her clocks, correct for the light speed delay, and what will I find?

The first thing is that the clocks don't tick at the same rate - they tick slower. That's time dilation. The second thing is that her clocks don't all show the same time as each other even after I correct for the light speed delay - they are not synchronised. If she's moving in the x direction, all the clocks in any plane perpendicular to the x direction agree with each other, but no two planes of clocks are in sync.

In short, the x term in the time transformation tells you how out of sync the travelling observer's clocks are with respect to each other, from my point of view. Literally, it's how much time you'll have to add to the reading on my clock that is positioned at x to get the reading on her clock that is passing mine at time t.

#### kvnsmnsn

Then in my example, the reference event would be the launch of the mentioned space ship from planet Earth, bound for Kepler-452b. The value $t$ is the time as seen from Earth, while $t'$ is the time seen as astronauts on the space ship, and at the launch of the space ship they are both set to zero; similarly $x$ (the position of the launch pad) and $x'$ (the position of the space ship) are also set to zero. For simplicity I'll assume that the space ship travels at $.3c$ from the instant that it launches until the instant that it arrives at Kepler-452b. So that would be the value of $v$, namely $.3c$. And at the moment the space ship arrives $x$ still has the value zero, but $x'$ has the value $1400$ (using light years as units). So the formula becomes $t'$ equals $γ (t - \frac{vx}{c^2})$ which equals $γ (4666.7 - \frac{.3(1400)}{1})$. Here I'm also using years as my units of time, so the speed of light becomes 1 (that is, one light year per year). Now $γ$ is defined as $\frac{1}{√(1 - \frac{v^2}{c^]2})}$ which would be $\frac{1}{√(1 - \frac{.3^]2}{1^2})} = \frac{1}{√(1 - .09)} = \frac{1}{.953939} = 1.04828$ so the value of $t'$ becomes $1.04828 (4666.7 - 420)$ which equals $4451.71627$. Which is pretty close to the result I got in the original post to this thread. Anybody care to check my math to make sure I got all that right?

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#### kvnsmnsn

That last post came out really garbled. Can anybody tell me what I did wrong with my mathematical expression? I tried to say t prime equals gamma times (t minus the fraction of vx over c squared), but it looks like the third pound pound is not working.

#### Ibix

You've tried to get superscripts using BBCODE SUP tags. That's what's broken it. To get $c^2$ write c^2. Braces are needed for more than one character - so to get $c^{1/2}$ write c^{1/2}

#### kvnsmnsn

Ibix, thanks for showing me what was breaking my math equations. I went back and I think I fixed it.

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