Trying to understand the Lorentz transformation.

[Hellabyte]

Ok so I am attempting to get a "feel" of the Lorentz equations. For a observer O' moving with velocity v respect to a observer O along the x direction the transformed variables are x and t.

$$x' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x - vt)$$
$$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(t - \frac{vx}{c^2})$$

So I think I am starting to understand x'. When the observer O looks at the point x, where an object or event would be, The distance from O' to that object or event is contracted. Let's say that the observer O looks out and sees that O' is 5 meters from an object at x. This length will have to be scaled by gamma in the O' frame because O is actually seeing a a contracted length that appears to be 5 meters. the uncontracted length that o' observes will be 5m*gamma which is what the formula for x' says.

However time is giving me more difficulties. It appears that the formula is saying that the time interval (t') that has elapsed for O' is longer by a factor of gamma than the interval that O sees has elapsed (t) on his watch.
This seems opposite to how I usually think of time dilation which is that time for a moving observer will pass more slowly. When a spaceship flies by, the clock on board appears to only make a half turn while ours on Earth makes a full turn. To me, the Lorentz transformation seems to be saying that more time will elapse on the spaceship than for us, or that our clock will only make a half turn while theirs make a full turn. What in my thinking is wrong?

Thanks for the assistance.

 

[JesseM]

So I think I am starting to understand x'. When the observer O looks at the point x, where an object or event would be, The distance from O' to that object or event is contracted. Let's say that the observer O looks out and sees that O' is 5 meters from an object at x. This length will have to be scaled by gamma in the O' frame because O is actually seeing a a contracted length that appears to be 5 meters. the uncontracted length that o' observes will be 5m*gamma which is what the formula for x' says.

I'm not clear on what you mean above, but length contraction only applies when you measure the distance between two ends of an object simultaneously in the frame where the object is moving (since for a moving object, you're interested in its length at a single instant), and then compare with the distance between two ends of the object in its rest frame (in this frame it's not important if you pick simultaneous events since the ends stay at the same positions). So, if the object is at rest in O and moving in O', and you pick two simultaneous events on either end which are 5 meters apart in O', like (x'=0, t'=0) and (x'=5, t'=0) then when you transform these events back into O, the difference in x coordinates between them should be gamma*5.

However time is giving me more difficulties. It appears that the formula is saying that the time interval (t') that has elapsed for O' is longer by a factor of gamma than the interval that O sees has elapsed (t) on his watch.
This seems opposite to how I usually think of time dilation which is that time for a moving observer will pass more slowly.

The usual time dilation equation can only be used if the clock is at rest in one of the two frames and you're considering two events on the worldline of the clock, then if the time between them in the clock's rest frame is t_R and the time between them in the frame where the clock is moving is t_M, the time dilation equation says the t_M will be greater since the clock is ticking more slowly in the frame where the clock is moving, by the equation $$t_M = \frac{t_R}{\sqrt{1 - v^2/c^2}}$$

You'll find that this works in terms of the Lorentz transformation, for example if you say the clock is at rest in O at position x=0, then two events on the worldline of that clock might be (x=0, t=0) and (x=0, t=t_R), in which case the first event will have (x'=0, t'=0) in O' while the second will have (x'=gamma*(0-v*t_R), t'=gamma*(t_R - v*0)). So, the time between these events is gamma*t_R.

 

[ghwellsjr]

However time is giving me more difficulties. It appears that the formula is saying that the time interval (t') that has elapsed for O' is longer by a factor of gamma than the interval that O sees has elapsed (t) on his watch.
This seems opposite to how I usually think of time dilation which is that time for a moving observer will pass more slowly.

Is your problem that time dilation means time is getting stretched out for a moving observer but you thought that time passes more slowly for a moving observer?

Yes, when your clock runs slow, it means the ticks occur further apart, correct? So when time is getting longer, the clock is running slower. They mean the same thing.

 

[grav-universe]

In this case, t and t' do not mean quite the same thing that they do with time dilation, although it is similar and time dilation still factors in. With time dilation only, t would be the time that O measures upon the clock of O while t' would be the time that O measures upon the clock of O' for the same event, whereas t' = sqrt[1 - (v/c)^2] t. However, in this case, the same terms are used to describe what each observer reads upon their own clocks for the event instead, t for what O reads upon the clock of O, which is the same, then, but t' is now the time that O' reads upon the clock of O' rather than what O reads upon the clock of O'. The same goes for the distances x and x' measured with each of the observer's own rulers.

The time dilation O observes of the clock of O' would only work out the same as what O' observes of the clock of O' if the event occurs in the same place as O'. Then whatever clock reading O observes upon the clock of O' will be t' since the event and the clock reading of O' coincide in the same place so all observers must agree that the clock reading of O' and the event are simultaneous to O'. Then we have

t' = (t - v x / c^2) / sqrt[1 - (v/c)^2],

where x = v t according to O if the event occurs in the same place as O', so

t' = (t - v^2 t / c^2) / sqrt[1 - (v/c)^2]

= t (1 - v^2 / c^2) / sqrt[1 - (v/c)^2]

= sqrt[1 - (v/c)^2] t

 

[jason12345]

Time dilation and length contraction are the LTs applied to special cases:

1. For two events at the same time in one frame, you get the Lorentz contraction relationship between the space difference in the two frames.

2. For two events at the same location in one frame, you get a time dilation relationship between the time difference in the two frames.

Notice also that two events that are simultaneous in one frame aren't in the other, which is extremely important in resolving so called contraditions skeptics look for in SR.