Special Relativity

1. Sep 10, 2007

Ray

Can anyone explain exactly how the diagram in Schutz 'Invariance of the interval' actually shows the rod appears the same length in the two different frames and how in any case that shows the direction of the relative velocity between two frames is irrelevant? I can see that the worldline of the clock is parallel to the t' axis, but why that particular clock?

2. Sep 10, 2007

pmb_phy

In general the length of the rod is different in different frames. It is the spacetime interval which remains the same. To get a better grasp on this you need to first define the events that you are considering before you can make sense of the spacetime interval. Consider a rod at rest in frame S. Set a fire cracker off at each end at the same time as observed in this frame. The magnitude of the spacetime interval regarding these two events will have the same magnitude of the rod's proper length, i.e. that length that is intrinsic to the rod, more commonly known as the rest length, i.e. the length of the rod as measured in a frame of reference in which the rod is at rest.

Pete

3. Sep 10, 2007

bernhard.rothenstein

Schutz diagram

4. Sep 11, 2007

Ray

Reply to Pmb. Many thanks, but Schutz's diagram and explanation are independent of rest length (after all one frame is moving and both lengths are still equal). It's his diagrammatic proof that the lengths are equal that I cannot understand. Any ideas?
Ray

5. Sep 11, 2007

pmb_phy

Please post the page number and figure number on Schutz's text that you're referring to.

Pete

6. Sep 11, 2007

robphy

7. Sep 11, 2007

JesseM

I think Ray is talking about the bit about the rod and the clock on p. 12.

8. Sep 11, 2007

Ray

Schutz 1985 Section 1.6 pp12-13 diagrams 1.6 and 1.7

9. Sep 11, 2007

Ray

10. Sep 12, 2007

pmb_phy

Okay. I have the pages and diagrams. Can you tell me where on these pages that Schutz indicates that "actually shows the rod appears the same length in the two different frames"?

In the meantime I'll see if I can find it and help explain what he's speaking of. This is interesting enough to make a whole new web page on my web site. Thanks for asking this question! :)

Best regards

Pete

11. Sep 12, 2007

Ray

Pete, Many thanks for your reply. Bit sloppy on my part, Schutz actually states 'the result is:' (length of rod in O') squared = figh(v)(length of rod in O) squared.' But then on page 14 he states 'Notice that from the first part of the proof we can also conclude now that the length of a rod..... is the same in either frame.' On second reading I think his word 'now' means 'together with the second part of the proof'. Sorry about that should have been more careful. Nevertheless I still don't understand the diagrammatic construction nor his justification that figh(v) =figh(mod v).
Best regards Ray

12. Sep 13, 2007

pmb_phy

If you are referring to Eq. (1.5) then it is not saying what you think it is. "delta s" is not the length of the rod. If is the spacetime interval. What led you to believe that this was speaking about the length of the rod??

Pete

13. Sep 14, 2007

Ray

The refence is to the un-numbered result appearing part way down page 13 just before eq 1.6. The text concludes: delta s squared = length of rod squared in each frame. But I do not understand how the diagram is constructed nor how it proves the un-numbered result nor Schutz's use of the principle of relativity to prove figh(v) =figh(mod v). You're right when you say it's interesting, but it's also very difficult.

14. Sep 14, 2007

pmb_phy

Okay. I see what's going on now. The inertial frame O' is moving relative to the inertial frame O in the +x direction. The book tells you to consider a rod which is oriented perpendicular to the velocity of v of O' relative to O. Such a rod is illustrated in this link
http://www.geocities.com/physics_world/sr/lorentz_contraction.htm
at the bottom of the web page in the section labeled Measure Length Perpendicular to Motion. As you can see from that section a rod which is moving perpendicular to its motion will have an invariant length and that is what the equation above Eq. (1.6). The web page and the book arrive at the same result using two different methods regarding the invariance of the length of the rod.

Pete

Last edited: Sep 14, 2007
15. Sep 15, 2007

Ray

Yes, you're right the web page arrives at a result in section 1.6 of the book by a different method, but that's precisely the problem. I do not understand how the diagram in Schutz is constructed nor how it proves the un-numbered result nor Schutz's use of the principle of relativity to prove figh(v) =figh(mod v). Any ideas?

Regards

Ray

16. Sep 15, 2007

pmb_phy

Okay. We'll start from scratch. First off its not "figh(v)" its phi(v). It took me a while to recognize your spelling of it so let's use phi(v) (i.e. $\Phi(v)$) so that those who are following won't be as confused as I was. Okay?

Let's start from the beginning. We are trying to completely comprehend Section 1.6 in Schutz's text A first course in general relativity. This section is covered from page 10 to 15. This section shows that the spacetime interval is a Lorentz invariant.

Schutz starts out by describing two events on the world line of the same light beam. He calls these events E and P. Inframe O the events themselves are labeled (t, x, y, z). The difference between two events is labeled $(\Delta t, \Delta x, \Delta y, \Delta z)$. The two events Eand P satisfy the relation (Choosing units so that c = 1)

$$-(\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 = 0$$

Since the speed of light is a Lorentz invariant we must also have

$$-(\Delta t')^2 + (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 = 0$$

Where events in O' are labeled (t', x', y', z'). This motivates Schutz to define the spacetime interval, $\Delta s^2$ (which is the square of $\Delta s$) as

$$\Delta s^2 = -(\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 = 0$$

It is readily seen that for the spacetime interval between the two events E and P] is $\Delta s^2$ = 0. This is true wheher you choose coordinates in O or in O'. In O' this reads $\Delta s'^2$ = 0. Therefore this is a geometric relationship between the events, i.e. one that is not dependant on coordinates.

The steps from Eq. (1.2) to 1.5 are difficult. I recommend that you follow this with pen and paper and convince yourself that these steps are correct. If you have a problem in doing so then come back and let me know and I'll help you through it. However it appears that you've already done this so on we go. What we end up with is Eq. (1.5)

(1.5) $$\Delta s'^2 = \Phi(v)\Delta s^2$$

The next step is to show that $\Phi(v)[/tex] = 1. Now consider the rod which is lying on the y axis. This is shown in Fig. (1.6). This diagram is supposed to show that a rod is at rest in O and perpendicular to the x-axis. The length of the rod will equal the square root of the spacetime interval between the events A and B as shown shown in Fig (1.6). How do we know this? Consider the difference in time between those events as measured in O. The diagram shows that they are simultaneous events as observed in O. Therefore the spacetime interval for these two events, given that $$\Delta t = \Delta x = \Delta z$$= 0 $$\Delta s^2 = \Delta y^2$$ [itex]\Delta y$ is the proper length of the rod. Why? Because in O the rod is at rest and since $\Delta t$ = 0 the magnitude of the spacetime interval is the magnitude of the length of the rod which is at rest in frame O. Thus the term "rest length" is used for the intrinsic length of the rod. I prefer to use the term proper length. So if you see that term in the future you'll know what it means. Now take a look at the clock's worldline as shown in Fig. (1.7). The clock starts at the midpoint between events A and B. If you carefully study this diagram you'll see that the clock is at rest in O' and is thus moving with respect to O. The worldline is parallel to the xt-plane. However the distance spatial distance between events A and B are equal and this means that if at these events a flash of light is emitted from each end of the rod the flashes will reach the clock at the same time which means $\Delta t'$ = 0 as well. If we substitute $\Delta t'$ = $\Delta x$ = $\Delta z$ = 0 into the spacetime interval for $\Delta s'^2$ we get the following relationship

$$\Delta y'^2 = \Phi(v)\Delta y^2$$

which is exactly what the equation above Eq. (1.6 on page 13 reads). Now switch the direction that O' is moving and you will obtain

$$\Delta y'^2 = \Phi(-v)(\Delta y)^2$$

This is interpreted to mean that $\Phi(v) = \Phi(|v|)$. Follow the next paragraph to where it shows that $\Phi(v)$ = +-1. This results in

$$\Delta s'^2 = \Delta s^2$$

This means that the spacetime interval is a Lorentz invariant. This can also be shown rather easily by taking the expression for $\Delta s^2$ and substitute the variables for system O' by using the Lorentz transformation. This will yield the same result.

Let me know if you need more help.

Best wishes

Pete

Last edited: Sep 15, 2007
17. Sep 16, 2007

Ray

Pete, Many thanks for your reply. Your rendition is about a million times clearer than the book's. Among the points you made two points in particular made it all fit together for me: the symmetry between the rod and world line, and the reversal of the direction of the relative motion. Neither of these two points appears explicitly as far I can see in Schutz. Apologies for my rubbish presentation.
Once again very many thanks

Ray

18. Sep 16, 2007

pmb_phy

Its an honor to help.

Pete

19. Sep 16, 2007

bernhard.rothenstein

Sr

Are you happy with the way in which the Lorentz transformation is derived in that textbook?

20. Sep 17, 2007

pmb_phy

I don't recall it. I try to stay away from those deivations. Its one of those things in physics that I hate the most. Sorry.

Pete