I'm having trouble interpreting the Lorentz transformations

There are two frames which I will call O and O'. Suppose I'm someone in O. Are the lorentz transformations a means by which to calculate coordinates on O' in a way that only frames like me would calculate? Or is it a calculation of what is actually happening in O'. Or is this last statement meaningless in relativity?

I think that Lorentz transformations are specific to each frame (this means there would be different conclusions) and shouldn't be viewed as "what's actually happening in O' ". I think the main point of relativity is that there are multiple ways of seeing things, and if different observers (say O and S) calculate different space-time coordinates of the same frame (namely O') this isn't a contradiction, it's just relativity.

Is this interpretation correct?

So when I'm given problems in SR such as "calculate the time that has elapsed on the shuttle travelling at 0.8c with respect to you" then this means "calculate the time that has elapsed on the shuttle ACCORDING TO YOUR FRAME".

Thanks.

It is worth reminding yourself what "a frame of reference" means: an agreement on how to measure when and where some event happened. For the purposes of special relativity we generally only talk about inertial frames which have some constant velocity (less that $c$) with respect to one another.

If you have the coordinates $(x,y,z,t)$ you know how far over, up, and forward you had to go to get from the origin to the location of that event and how long after an arbitrary starting time that event occurred.

Now, you apply the transformation and get out another set of coordinates $(x',y',z',t')$. Those tell you how to find the same even starting from a different origin.

The two set of coordinates refer to the same event.1 That is important.

When you start talking about length contraction or time dilation you are now talking about the distance or duration between two events; call these $\Delta L = \sqrt{(\Delta x)^2+ (\Delta y)^2+ (\Delta z)^2}$ and $\Delta t$ and comparing those values to the distance or duration between the same two events as measured using a different set of measurement conventions ($\Delta L'$ and $\Delta t'$).

Your job, is to figure out what the events that mark the beginning and the end of the interval are and then construct the right delta's in both frames.

Pro tip

A useful fact is that measurements made in every single frame will measure2 $$s^2 = (c\Delta t)^2 - (\Delta L)^2 \tag{interval}$$ to have the same value.3 That gives $$(c\Delta t)^ 2- (\Delta L)^2 = (c\Delta t')^2 - (\Delta L')^2 \,. \tag{*}$$

If you chose a pair of events for which $\Delta L = 0$ in the unprimed frame (i.e. that occur in the same spot, then you must have $(\Delta t')^2 > (\Delta t)^2$ because $(\Delta L')^2 > 0$ (assuming that the frame are not co-moving). This is a way to reassure yourself that you are applying the factor of $\gamma$ in the correct sense.

To use the same kind of reasoning for length contraction take $\Delta t' = 0$ (the length in the primed frame is measured at a single moment in that frame), then re-arrange (*) to get $(\Delta L)^2 = (\Delta L')^2 + (\Delta t)^2$ (with $\Delta t \ne 0$ because of the relativity of simultaneity) leading to $(\Delta L)^2 > (\Delta L')^2$.

1 Because a single event is given two different names this kind of transformation is sometimes referred to as an alias transformation.

2 Or possibly $s^2 = (\Delta L)^2 - (c\Delta t)^2$ depending on the sign convention preferred by the source you are using.

3 Checking this for yourself by direct application of the Lorentz transformation is algebraically tedious, but straight forward. Recognizing intuitively that it is a necessary consequence of the postulated constancy of the speed of light may be a sign that you are starting to get the hang of relativity.

There are two frames which I will call O and O'. Suppose I'm someone in O.

By that, I suppose, you mean I'm someone at rest in $O$.

Are the lorentz transformations a means by which to calculate coordinates on O' in a way that only frames like me would calculate?

The Lorentz transformations transform the coordinates (of events) in $O$ to the coordinates (of events) in $O'$ and vica versa.

The events (and the associated interval of any pair of events) are independent of coordinates and so, are invariant under a Lorentz transformation.

So when I'm given problems in SR such as "calculate the time that has elapsed on the shuttle travelling at 0.8c with respect to you" then this means "calculate the time that has elapsed on the shuttle ACCORDING TO YOUR FRAME".

No, this isn't correct.

The time elapsed on the shuttle is invariant. That is to say, all observers agree on the elapsed time according to a clock on the shuttle.

However, relatively moving observers disagree on the amount of coordinate time elapsed in their frame of reference.

So, if Alice is moving with respect to you, Bob, then you and Alice will agree that, e.g., 1 second has elapsed on the shuttle clock between two events but you will disagree on the coordinate time elapsed in your respective frames of reference.

• The reason I say stuff like what was put in CapsLock, is because I find no other way to explain to myself how time dilation can be reciprocal, and maybe other apparent paradoxes I am not aware of right now… How can I think about reciprocal time dilation in terms of Lorentz transformations not being relative as well? Thanks. – DLV Feb 22 '15 at 3:49
• It surely means "according to your frame" since both frames measure time dilation with respect to the other, so one clearly has to say "the time on S' according to S". Thanks. – DLV Feb 22 '15 at 17:06
• @David, the elapsed time on the shuttle is a proper time so there is no ACCORDING TO YOUR FRAME qualifier; the proper time is an invariant (is the same in all inertial reference frames). – Alfred Centauri Feb 22 '15 at 17:27
• I wouldn't like to spam you with questions here. Could you maybe tell me a good place to read about this? I've read the chapter in my book by Serway, but I wonder if you've seen a good exposition elsewhere? Thanks. – DLV Feb 22 '15 at 18:13