# I'm having trouble understanding Lorentz transforms

1. Feb 15, 2015

### davidbenari

Do lorentz transformations calculate what is really happening in the primed reference frame? Or do they calculate what the non-primed reference would think the primed reference's coordinates should be?

I might be doing things wrong algebraically but it seems that:

If I calculate from O things in O' knowing certain things about O. Then I calculate from O' using what I got in the previous calculation, and I don't get back what I know about O!

Also, In problems dealing with particle decay, or something of the kind, they usually tell me the mean lifetime of the particle. My interpretation is this: the particle's lifetime has been measured at rest with respect to my frame to be X. If that particle is travelling near C with respect to myself, I can assume that its meanlife time in ITS frame is the one I measured when it was at rest with respect to me.

Is this right?

Also: With respect to both frames measuring time dilation of the other, does the same happen with length contraction? Are all contradictions of this kind solved by the typical "you have to consider accelerations to solve this apparent paradox" ?

I have other questions, but I don't really have them well thought out. I hope I wasn't unclear here.

Thanks.

2. Feb 15, 2015

Staff Emeritus
That's what it sounds like.

3. Feb 15, 2015

### phinds

Well, think about it this way: you, right now as you are reading this, are MASSIVELY time dilated according to an accelerated particle at CERN, you are mildly time dilated according to Haley's Comet, and you are not time dilated at all according to your desk. Are you feeling all of these at the same time? Are you feeling ANY of these?

4. Feb 15, 2015

### davidbenari

phinds: I guess something that bothers me is this: I'm massively time dilated according to an accelerated particle. He says the clock in my computer measures (10:30 AM) (I don't think details are important, right now), but right now I see 9:32 PM. So if Lorentz transformations do indeed transform coordinates in one frame to another one, then theyre doing a horrible job! Theyre saying something false!

5. Feb 15, 2015

### davidbenari

Vanadium: So are you implying Lorentz transformations should be able to go both ways? Working O -> O', will yield the same data as O' -> O, or something like that? I think (I'm probably wrong) thats in contradiction to what phinds says.

Last edited: Feb 15, 2015
6. Feb 15, 2015

### davidbenari

phinds: I think I get your idea. Are you saying that these transformations say what someone in O thinks someone in O' measures? And that the whole point of relativity is that all these point of views are equally valid?

7. Feb 15, 2015

### davidbenari

So is relativity saying that what I'm doing right now as I type this message isn't any more valid than what the particle at CERN thinks I'm doing?

8. Feb 15, 2015

### phinds

This is a meaningless statement given that you have not said what time it was when you and the particle synchronized your clocks while you were in the room together. Had you done so, the Lorentz transforms would be giving you the right answers.

9. Feb 15, 2015

### phinds

Yes, it says that no Frame of Reference is any more valid than any other. You, after all, think that the particle is massively time dilated but it doesn't feel a thing and wonders what the hell you are talking about

10. Feb 15, 2015

Staff Emeritus
If changing your velocity by v takes you from O to O-prime, changing it by -v will take you from O-prime to O. If you're not getting that to happen, you are making an algebra or arithmetic error somewhere, and it probably doesn't make sense to go on until you have that straightened out.

11. Feb 15, 2015

### davidbenari

I don't get how Lorentz transforms giving "the right answers" is compatible with all "frames are equally valid (and can say different things)". Namely, the way I see it right now is that if O works his way to know what's happening in O', and O' works his way to know what's going on in O (using the data I got when I did O->O') then I don't have to get stuff which is "consistent" with a "common-sense" view. And so Lorentz transforms don't give "the right answers" If by "right answers" you meant "consistency".

Thanks.

12. Feb 15, 2015

### phinds

13. Feb 15, 2015

### davidbenari

Ok, I'm working through what I've done before. But I have a question. If I have three observers then what you guys say can't be true (I think, most probably wrong). You guys say that O->O' will be consistent with O'->O

But if I have three observers and apply transformations to each combinations then wouldn't I arrive at some contradiction?

14. Feb 15, 2015

### phinds

No, you would have to do transforms between each set of observer's FoR's and all would be well. In other words, call your observers A, B, and C. Now A and B see each other symmetrically, as do A and C and C and B, but when you are looking at any of those symmetries the 3rd observer is not involved.

15. Feb 15, 2015

### davidbenari

I think I've found one way in which O->O' isn't the same as O'->O.

Say O' moves in the x direction at speed v with respect to O. O' measures the ends of a rod at rest in O simultaneously and gets the following equation $\Delta x = \gamma \Delta x'$ which is rewritten as $\Delta x' = \frac{\Delta x}{\gamma}$. If he travels at .99c and x2=5 and x1=3 then $\Delta x'= 0.3 meters$

Suppose we want to go now in the O->O' direction and want to see what O' measures as the ends of the rod. We get the equation $\Delta x'= \gamma \Delta x$ which is in contradiction with the above paragraph. You might say I'm wrong in that I've assumed the measurement done by x' as observed by x was simultaneous. I agree that its wrong, but if it is wrong then t1'≠t2', and there you would have that Lorentz transforms don't work both ways.

Whats wrong here?

16. Feb 15, 2015

### davidbenari

if you think I've written that horribly please tell me so I can correct it.

17. Feb 16, 2015

### Staff: Mentor

You're leaving out the $t$ coordinate. You can't just transform one coordinate; you have to transform all of them. This is because of relativity of simultaneity; the distance $\Delta x$ between the two ends of the rod represents a spacetime interval in the unprimed frame with $\Delta t = 0$; but in the primed frame, $\Delta t'$ for the same interval is not zero. So if you only transform the $x$ coordinate but not the $t$ coordinate, you are leaving out crucial information.

18. Feb 16, 2015

### davidbenari

Hey but O can measure both ends of the rod simultaneously. And O' can measure both ends of the rod simultaneously. One gets equation $\Delta x' =\gamma \Delta x$ the other gets $\Delta x =\gamma \Delta x'$ which are contradictory. I'm thinking that the person making the measurement is on the RHS of each equation and his terms for time vanish. What am I missing? How should it look like?

19. Feb 16, 2015

### Staff: Mentor

Yes, but those measurements refer to two different spacetime intervals. You are getting nonsensical answers because you are not taking that into account. See below.

No, they aren't; they're just referring to different spacetime intervals.

The first interval, referring to the measurement made by O, is between events which have coordinates, in the O frame, of $(0, 0)$ and $(x, 0)$ (where we assume that the measurement takes place at time $t = 0$ in the O frame).

The second interval, referring to the measurement made by O', is between events which have coordinates, in the O' frame, of $(0, 0)$ and $(x', 0)$ (where we assume that the measurement takes place at time $t' = 0$ in the O' frame).

If you Lorentz transform the endpoint events of either interval into the other frame, you will see that they are different pairs of events. (To make it simple, I have made one endpoint the origin in both cases, which transforms to itself; so the only thing you need to check is that the event $(x, 0)$ in the O frame transforms to something that is not $(x', 0)$ in the O' frame. That is simple to do, and doing it should help you understand what is going on.)

20. Feb 16, 2015

### davidbenari

Wow. I'm not sure I get all of this. Say I'm holding a pencil horizontally in the O frame. A rocket flies past me in the O' frame and travels in the x direction. Why is the valid transformation for the length of the pencil $\Delta x = \gamma \Delta x'$ and not the other way around? Namely, the other way would say length contraction happens for me and not for him. Or is this where your discussion on spacetime intervals takes place?

Is it normal that my class sees special relativity without seeing Minkowski diagrams at all? I think nobody in my class really understands SR. I definitely do not!

21. Feb 16, 2015

### pervect

Staff Emeritus
Why not draw yourself a Minkowskii diagram of a pencil then? Start in frame O.

Suppose the eraser of the pencil is at the origin , i.e at t=0 and x=0. Draw a space-time or Minkowskii diagram by choosing a dozen or so points on the worldines of the eraser and another dozen or so the tip of the pencil.

I'll give you some coordinates that should work out nicely. Assume the pencil is 1 light-second long in its rest frame, and use units of seconds for time t and light-seconds for distances, x.

Eraser: (t=-.5, x=0), (t=-.4, x=0) ...(t=0,x=0)...(t=.4, x=0), (t=.5, x=0)
Point: (t=-.5, x=1), (t=-.4, x=1) ... (t=0, x=1) ... (t=.4, x=1), (t=.5, x=1)

Now let v=(3/5)*c, .6 times lightspeed.

Use the Lorentz transform on each of the couple of dozen points, and make a graph. Using a velocity of v = .3 c, the gamma factor will be 1 / sqrt( 1 - .36) = 1/sqrt(.64) = 1 / .8 = 5 /4 = 1.25

The Lorentz transform will then be:

t' = gamma*(t - v*x/c^2) = gamma*(t-v*x), because we've chosen to measure distances in units of light seconds and times in units of seconds, which makes c numerically equal to one in that system of units, which makes our calculations a lot easier.

x' = gamma(x - v*t)

Now I'm not going to do all two dozen points for you, but if you want to figure out what's going on, you'll need to chug through them with your graphing calculator (or whatever you use), and plot them.

I'll do one point: (t= .5, x=0)

then t' = gamma (t - v x), but we've chosen v = (3/5) which implies gamma = 5/4 as previously mentioned.

so t' = (5/4) * (0.5 - (3/5)*0) = .625
and
x' = (5/4)*(0 - (3/5)*.5) = -.375

Repeat for the other two dozen points, graph the resulting pair of lines, and graphically find the length of the pencil.

Then look at how you would algebraically solve for the length if you wish.

ps - I won't guarantee I didn't make any typos, but the basic scheme should be OK.

22. Feb 16, 2015

### Ibix

I certainly didn't see them until I started reading here. I really recommend them, though. Once you see how the diagrams for two frames relate, you have a visual model for what's going on and most student-level relativity problems get a lot easier.

Maybe that's why they don't get taught...

23. Feb 16, 2015

### Fredrik

Staff Emeritus
Unfortunately yes. If you want to understand it, you should study spacetime diagrams. It's a small effort that makes everything else easier. My standard recommendation is to read the first two chapters of "A first course in general relativity" by Schutz.

Consider the diagram below. Let's call the observer who uses the unprimed coordinates O, and the observer who uses the primed coordinates O'. The t' axis and the line that's parallel to it represent the motion of the endpoints of a rod. What O considers the length of the rod at time 0 is the difference between the position coordinate of the right endpoint at time 0, minus the position coordinate of the left endpoint at time 0. So what he considers the length of the rod is a property of the blue line. Similarly, what O' considers the length of the rod is a property of the red line.

24. Feb 16, 2015

### harrylin

This is quite simple, if you look carefully at how S and S' arrived at their respective equations and if you don't mix up the labels (which I sometimes do myself, I hope that this time I'm not mixing them up!). Note that I follow the convention of using S for "system of reference" and O for "origin". "The following also answers your last question in #20 (and while it can be helpful, for the derivation you don't need Minkowski diagrams!):

- S' measures the ends x'=a' and x'=b' of a rod at rest in S simultaneously according to S'.
Try for yourself: set as condition that t'a' = t'b'. For t'a' = t'b', $\Delta x' = \frac{\Delta x}{\gamma}$

- S measures the ends x=a and x=b of a rod at rest in S' simultaneously according to S.
Try for yourself: set as condition that ta = tb. For ta = tb, $\Delta x'= \gamma \Delta x$

No contradiction here. As a matter of fact, S and S' agree on these relationships.

25. Feb 16, 2015

### davidbenari

But equation with t'a=t'b yields something of the form x'<x. While equation with ta=tb yields something of the form x<x'. Why isnt this contradictory?