Lorentz Transforms - Understanding Time Dilation & Velocity Slope

[calis]

I am new here

didnt find the reply to this before

I hope I get lorentz transformations right

he said, that not only we have to shift placement axis, but also shift time axis

I know that the slope for velocity and time dalation must be same, but what formula to use to prove that the slopes are the same. (time delation formula doesn't work)

 

[calis]

I am talking about the slopes for the axes for moving body in coordinate system impied by lorentz transformations

en.wikipedia.org /wiki/Lorentz_transformation

I get not equal slopes using time delation formula

 

[JesseM]

You mean, the slopes for the time and space axes of the second frame, as seen in the first frame? The slopes should not be equal, but inverse, like 2 and 1/2, when expressed in units where c=1 (like x=light-seconds and t=seconds). And what do you mean by "time dilation formula"? How would you calculate slope from the time dilation formula?

 

[calis]

I understand how to draw velocity axis as seen in first frame, but about time axis I jut know it must be inverse... OK but why is it inverse

 

[JesseM]

I understand how to draw velocity axis as seen in first frame, but about time axis I jut know it must be inverse... OK but why is it inverse

What do you mean by "velocity axis" and "time axis"? Is the velocity axis the x'-axis of the second frame as seen in the first frame, while the time axis is the t'-axis of the second frame? The t'-axis is very simple, it behaves just like an object at rest in that frame (since the t'-axis is a line of constant x' coordinate, and an object at rest in that frame will have a constant x' coordinate)...if the second frame is moving at 0.6c relative to the first, then you'd just draw a line that moves sideways 0.6 light-seconds for every 1 second you go upward, so the slope is just 0.6. The x'-axis is the inverse because it goes sideways 1 light-second for every 0.6 seconds you go upward. You can see this by picking two points on the x'-axis, like (x'=0, t'=0) and (x'=10, t'=0) and then using the Lorentz transform to see where they'd be when drawn in the first frame's (x, t) coordinate system:

x = gamma*(x' + vt')
t = gamma*(t' + vx'/c^2)

(x'=0, t'=0) would be mapped to (x=0, t=0), i.e. the origin, while (x'=10, t'=0) would be mapped to:

x = gamma*(10)
t = gamma*(0.6*10) = gamma*(6)

Since gamma is the same in both cases (it's equal to 1/sqrt(1 - v^2/c^2)), we can see the point will be less far from the origin on the t-axis than it is on the x-axis, by a factor of 0.6. So, the x'-axis drawn in the first coordinate system must be a straight line through the origin which goes up 0.6 seconds for every 1 light-second you move sideways.

 

[jtbell]

To put it a more mathematical way:

For simplicity, use units in which c = 1 so the Lorentz transformation between inertial reference frames S and S' is

$$x^{\prime} = \gamma (x - vt)$$

$$t^{\prime} = \gamma (t - vx)$$

In S, the x-axis is the set of points that have t = 0. To find the corresponding set of points in S':

$$x^{\prime} = \gamma (x - v \cdot 0)$$

$$t^{\prime} = \gamma (0 - vx)$$

Combining these two equations gives

$$t^{\prime} = -vx^{\prime}$$

which is a straight line through the origin in S', with slope -v.

Returning to S, the t-axis is the set of points that have x = 0. Proceeding similarly to the above, the corresponding set of points in S' is

$$t^{\prime} = -x^{\prime} / v$$

which is a straight line through the origin in S', with slope -1/v. The two slopes are reciprocals.